kinetic energy

A thought experiment for the relativity skeptics

esiegel — Wed, 09 Jan 2013 17:22:39 +0000

A thought experiment for the relativity skeptics

"Great spirits have always encountered violent opposition from mediocre minds." -Albert Einstein

It may be hard to believe, seeing as how it's been our leading theory of gravity for nearly a century now, but Einstein's General Relativity is possibly the most frequently challenged scientific idea of all-time. Of course, it's emerged victorious from each and every one of those challenges, making a slew of unintuitive predictions that have been spectacularly confirmed each time they've been tested.

Image credit: Miloslav Druckmuller (Brno U. of Tech.), Peter Aniol, and Vojtech Rusin.

This includes the bending of distant starlight by the predicted amount by gravitationally strong objects such as the Sun, as is only visible during a total solar eclipse (as shown above), and goes well-beyond to gravitational lensing (of both the strong and weak varieties) of extremely distant galaxies by intervening sources of mass.

Weak Lensing (L) by Mike Hudson of http://mhvm.uwaterloo.ca/, Strong Lensing (R) by NASA, ESA, K. Sharon and E. Ofek.

There are plenty of other tests that have succeeded, ranging from test of the gravitational time delay of light to the decay of binary pulsar systems to the Lense-Thirring effect (relativistic Frame-Dragging) and more.

But perhaps the test you're most familiar with comes every single time you use your GPS device.

Image credit: RC Davison of BrightHub.

Without relativity, the errors in a GPS signal, even if you calibrated it daily, would accumulate to give you an incorrect position of 10 kilometers after just 24 hours! In order to compensate and make our GPS devices work properly, we need an understanding of two things:

Special relativistic time dilation, or the fact that objects moving more quickly experience the passage of time differently, and
General relativistic gravitational redshift, or the fact that light red-or-blueshifts its frequency dependent on the relative gravitational field of the observer and the emitter.

Image credit: wikipedia users Vlad2i and mapos.

Gravitational redshift is a little bit counterintuitive to most: a photon (or light-wave) that has to climb out of a gravitational field loses energy and becomes longer-wavelength, or lower in energy, while one that plummets into a gravitational field gains energy and becomes shorter-wavelength, or higher energy.

But it's not going to be counterintuitive to you any longer, because you're going to follow one of the simplest thought experiments (that I first heard described in a talk by Mark Trodden) that tells you exactly why gravitational redshifts-and-blueshifts must be real.

Image credit: MHSA Physical Science Review.

If you begin at rest, high up in a gravitational field, you have plenty of gravitational potential energy, but no kinetic energy. And if you let that object free-fall, it gains energy, meaning that it's more energetic at the bottom.

In other words, an object at rest that's in a shallower gravitational potential well has the same amount of energy as an object with some kinetic energy that's deeper in a gravitational potential well. The three objects below -- effects exaggerated for clarity -- all have the same amount of total energy, and that's just classical, basic mechanics.

Image credit: Ray Shapp / Mike Luciuk, modifications by me.

But now, instead of a single particle, imagine we had two particles: an electron (which is a form of matter) and a positron (an anti-electron, a form of antimatter). When electrons and positrons collide at rest, they produce two photons that are exactly equal in energy to the rest mass (via E=mc²) of the electron/positron.

Image credit: NASA's Imagine the Universe, by Goddard Space Flight Center.

Consider these three facts, now:

If I had an electron/positron annihilate from rest high up in a gravitational field, they would make two photons of a very specific energy at the point where they annihilated, high in that field.
If I had an electron/positron annihilate from low down in a gravitational field, they would make two photons of that same energy at the point where they annihilated, low down in that field.
If I had an electron/positron at rest high up in a gravitational field and I released them, letting them fall, they would gain energy, turning that gravitational potential energy into kinetic energy as they fell.

Agreed? So tell me, when that electron/positron pair that I released reaches that lower point in that gravitational field, where they have kinetic energy, and now they find each other and annihilate, what is the energy of the photons that they produce?

Image credit: Ray Shapp / Mike Luciuk, modifications by me.

It's going to be greater than in case two, because you have the rest-mass energy of the electron-positron, plus the gravitational potential energy that must be turned into kinetic energy of the photons, which means we have bluer (higher-frequency, higher-energy) light the deeper we are in the gravitational field!

It also means that if we produce light of a certain energy deep in the gravitational field, it's going to gravitationally redshift -- and lose energy -- as it climbs out of that field!

Image credit: Pearson Prentice Hall, Inc.

If light didn't change its frequency in a gravitational field, it'd be possible to build a perpetual motion machine simply by having electrons/positrons annihilate deep in a gravitational field, building a mirror to reflect those photons upwards and out of the gravitational potential well, re-form them into electrons and positrons again (which you could do if and only if they didn't lose energy as they climbed out of the gravitational field), and then let them fall back to Earth, gaining kinetic energy which you used to turn your turbine/produce power.

And you know what I think of schemes like that.

It may seem obvious now, but the idea that light would gain or lose energy and change frequency/wavelength as it reached different points in a gravitational potential well was brand new with the arrival of general relativity, and was resisted by a great many scientists. Thankfully, experiments -- both of the thought-variety and the physical kind -- always reveal the nature of physical reality, and relativity had it right.

So the next time someone questions relativity, ask them about the gravitational redshift; there's no wriggling your way out of what science tells us is true about the natural world!

esiegel Wed, 01/09/2013 - 12:22

Tags

gravity

Physics

Relativity

annihilation

anti-matter

antimatter

blueshift

conservation of energy

You misunderstand relativity, and thus should not defend it, because it needs to be defended correctly or it shoots you in the foot. The argument that you give holds just as well for gravity that is not general relativistic! Especially, you should never ever write stuff like "slow clock", because it misses the whole point of relativity. It is the clock process that is compared and defines time. You need a synchronization proceedure to compare clock states at different space-time points, which are four (!) different points at least if you want to compare two clocks. There simply are no slow or fast clocks (except for broken ones). The "flow of time", i.e. the rate at which time changes, is always precisely a meaningless dt/dt=1, even at the event horizon of a black hole.

OK, matter becomes more energetic because it is being accelerated by gravity. Light on the other hand, isn't accelerated, its wavelength is decreased without decreasing its velocity. To do that, there has to be something funny going on with time, surely?

Changing the photon into a pair of particles makes it much easier to understand - I've never understood light. Or gravity for that matter. Combine the two and you have what looks to me like black magic.

@Vince: You can take the view that light _is_ accelerated when falling into a gravity well: specifically, it gains ("kinetic") energy (and momentum). This can be quantified when you talk in terms of photons: the momentum is related to the wavelength (technically, it is proportional to the wavenumber or 1/wavelength), and the energy is proportional to frequency. Increasing frequency (blue shift) means increasing energy. The velocity doesn't change because photons are massless, and therefore always travel at 'c' in vacuum.

Now, this description/interpretation brings in quantum mechanics, while the underlying general relativistic mathematics doesn't require that. It makes use of the differing behaviour of clocks as a function of gravitational potential to derive the frequency shift.

One of the cool things about physics is that the facts don't change depending on your interpretation. In this case, both methods will give you the same final result.

haha Ethan that does make sense as much as it could to my non physic trained mind... I wonder if we could ever have a convo without me constantly asking you to explain!!
Loving your info & so glad I found someone that explains so simply & easily...Thanks!

"One of the cool things about physics is that the facts don’t change depending on your interpretation. In this case, both methods will give you the same final result."

I would rather put this as if the facts change depending on your interpretation, then there is a way to tell which interpretation is correct.

E.g. Mercury's perihelion precession.

For Newtonian motion, it is an anomaly and needs explaining. For GR, it is expected.

The facts (perihelion) change from anomaly to concordant, depending on interpretation.

"it’d be possible to build a perpetual motion machine simply by having electrons/positrons annihilate deep in a gravitational field, building a mirror to reflect those photons upwards and out of the gravitational potential well,"

If zero energy loss were managed you'd be able to do it on the earth's gravitational field (or, better, the Moon's).

It's just the power production would be lower for the same investing of "motor".

In the Falling Steel Ball diagram, you say it "gains energy" but it has exactly the same energy at the top and the bottom: 600 J. It is just in a different form.

Sascha
I go to you site sometimes; because you do seem to understand phyiscs well. But I do not go to your site every day or even every month because you get tied up in philosophical/physical knots that may be important to you but to little or nothing to educate me about physics.

I go to ethan's site Starts with a Bang almost every day; because on his site I learn.

Now for my criticism of you Sascha. You are a professional physicist, you know that Ethan's site is a teaching site. And yet every time you come here and post a comment; it seems that you come to criticize Ethan's understanding of some aspect of physics.

I grant you might have a better explanation about some aspect of physics than Ethan. But I will not grant that you Sascha ever give a better explanation of physics than Ethan when you make a comment as a guest on Ethan's site.

What you do Sascha is give Ethan a stick in the eye. A stick in the eye is very seldom useful or necessary criticism. Among civil people( and Ethan demostrates and brings a very high degree of civility to discussion on his site),; among civil people a deeper explanation, a helpful additional insight, a link to an explanation that further clarifies such and such detail of physical importance is the civil response.

Sascha Vongehr be civil; bring more to the table than a stick in the eye!

Michael Kelsey is a professional physicists, he always brings deeper insight and clarification to the table; never a stick in the eye, never even a harsh word. Always his words are civil and focused upon more clearly explaning the science.

Wow is a professional astronomer. He indeed is harsh about pseudoscience and nonsense. But there is a line he does not cross. He mocks but never the stick in the eye.

Look above read Michael Kelsey's explanation. Nice.

Look above at Wow's explanation and professional slightly different opinion and Michael Kelsey and also Ethan. Nice.

Where is your informative professional explanation and insight Sascha? Ethan is a professional. and no professional knows and understands or can explain everything perfectly.

So make your point, add your insight to Ethan's explanation without the stick in the eye.

I must admit, I learned from Ethan's post. Yes his explanation was good, clarified my thinking. And yes your piddling stick in the eye carried a smidgeon of a science idea (though unexplained). so I did some google to better understand the use of special and general relativity in GPS.

Sascha, you are a smart knowledgeable philosophical physicist PhD with a wide range of understanding. So bring some science to the discusion; and leave the stick in the eye for use on your own web site.

Oh look at Sascha's current post on his own site.
SUICIDE ALERT!!!

The most current post on Sascha Vongehr' site alpha meme is
New-Year Resolution: Should I Kill Myself In 2013
http://www.science20.com/alpha_meme/newyear_resolution_should_i_kill_my…
Some of Sascha's comments in response to other commenters in this post of Sascha's is:
Sascha says, "If I commit a rational suicide, all that understand me will be fine with it. In fact, I will make the less people cry the sooner I finish... I will kill myself very slowly so to say in order that we may all enjoy the pain. ;-)... last time I checked, prozac was basically for free in the UK."

So Sascha drop the stick in the eye pointed at others.
And drop the stick in the eye pointed at yourself.

Destructiveness and Self-destructiveness are two sides of the same coin.

You need to talk with someone professionally. Not just a friend; but also a psychiatrist.

Sascha, your philosophy is bullshit if you think it is leading you to suicide.

SUICIDE ALERT!!!
All talk of suicide must be taken seriously!!!
If you know Sascha personally contact him immediately.

Sascha, I will write more to you on your blog.

I have tried to post on Sascha's blog; three times now.

Each time unsuccessfully. I get the message "Service unavailable"

I am concerned about Sascha!

"There simply are no slow or fast clocks (except for broken ones). "

It's patently obvious that the description "fast" and "slow" is with respect to one another, not compared to themselves where indeed there can be no such thing as "fast" or "slow".

I really can't stand this kind of lame attempted pedantry. "Gee, there are two ways in which to interpret this statement -- one which is patently false, and one which is clearly correct. Even though I go through this exercise automatically a thousand times each day, correctly interpreting other people's sentences in the context which makes sense, I will now suddenly do the opposite and call that being smart."

Well, it isn't.

@ Daniel Clements:
You mean like he said literally in the next sentence? He excluded the word "kinetic" before the word "energy" in one sentence. I'll grant that at least this nitpick is technically correct.

The familiar graphical 'illustrations' of gravity are getting a bit long in the tooth, especially in this age of computer-generated 3-D plots. Not only that, they're wildly inaccurate; what's reallly happening is that (roughly speaking), an object in free fall is actually moving normal to the surface you see depicted at the top. Far away from where the Earth dents it, this is nearly perpendicular to the two spatial dimensions, i.e., the objects worldline has almost no spatial component. Closer in, normality to the surface gives you a much more noticeable spatial component, and further, this component always points towards the Earth.

And that is gravity, modulo the +++- signature of space-time, the coupling to stress-energy rather than mass, etc.

Now, why can't someone come up with an easy graphic to illustrate this?

I thought that GR effect on GPS clock is due to it being in a weaker gravitational field than the clock on earth. It's a time effect but opposite the one of SR.

The blue/red shift might be interesting in signal processing. But the clocks have no relation to blueshift.. no?

The blueshift and the clock change are tied together as intimately as frequency and energy of the photon.

The photon isn't being shifted in frequency from its point of view. What's happening is that time is slowing down for it. Therefore what you think is 1 second, it thinks is 2 seconds and therefore puts twice as many beats as it would in 1 second.

So you see a shift to double the frequency.

A shift to blue.

"The photon isn’t being shifted in frequency from its point of view. What’s happening is that time is slowing down for it."

ahhh.. didn't look at it from that perspective. thanx.

Its still inaccurate, but it's close enough to get you thinking correctly.

Like many things, the only solid explanation of the situation is the maths.

It's better than Calvin's Dad's "The trees are sneezing", though.

Well my comment on Sascha's site finally got posted. So his site is still up.
I emailed and left a phone message to the Science 2.0 site where Sascha's blog is. No response yet.

When I hear if Sascha is OK or not: I will let you know.

Either way, I am so pissed at Sascha.
If his post is meant to be serious intellectual philosophy then publish it in a journal; not on a blog where his philosophical bullshit might influence some depressed teenager.

I know depression first hand; Sascha's post is irresponsible.

Many years ago, y shrink said to me, "If you kill yourself; I will kill you."

Sorry for being off topic.

I am not a special or general relativity skeptic. The evidence to me is pretty strong. It'd be nice to have a quantum gravity; but that's not being a skeptic.

But I do have questions that I can't answer with words or mathematics. Maybe someone can help clarify for me.

I read recently, "since (delta lambda) is finite in every ionertial frame pa free electron cannot emit or absorb a photon" This was on pg 85 of Special Relativity for Physicists by G. Stephenson and C.W. Kilmister, 1958

Question 1)
Can somebody explain this to me. I assumed a free electron (i.e. one in an interial frame) could emit and absorb photons and gravitons.

Question 1b)
But if it can't; then I don't understand how it it could emit or absorb gravitons (i.e. the hypothetical graviton). Yes, yes, I understand that gravity is the curvature of space. But there still is a gap in my thinking; if I can't more correctly think of a spin 2 graviton being absorbed or emitted.

Now my second question is more complicated.
1) I assume that every photon in interstellar space is in an inertial reference frame (Is this correct?)
2) But from any particular photon A's point of view; any length measure (on a geodesic parallel to its motion) is zero due to special relativity length contraction
----a) since the photon A from its own inertial reference frame has a velocity of zero
----b) but for photon A's point of view all other phtonic and baryonic matter must be moving at the speed of light c relative to photon A (maybe this isn't exactly correct)
BUT NOW MY QUESTION: So it appears to me that (from photon A's point) of view there is no time measure and there is no length measure. But the photon A must sense gravity somehow. So my question, what is the fundamental thing that a photon A is feeling? I know the easy answer is that photon A sense curvature. But my question is does photon A in some sense sense Energy as in = (m^2c^4 + p^2c^2)^1/2. As in the summation of such energies of various objects other photons or galaxies.

OK those are my questions. I've been searching the internet and textbooks to better understand the idea of a photon in a kind of intergalactic inertial frame of reference. But I can't find anything.

So. I am not looking for any non standard answer to my questions 1 and 2. I am looking for the standard special and general relativistic answer, insight or just clearer way to think upon these two questions.

thanks for any insight. On my question 2 in particular; I have gotten myself to the point of thinking that I understand and then back to square one where I realize I am totaly confused.

Any clarity will be appreciated. Thanks.

There is an implicit distinction in relativistic physics between moving very close to the speed c, (particles in particle accelerators), and moving *at* the speed c (propagating electromagnetic energy).

Moving at speed c is disjoint from moving close to speed c, like the graph of the relation y=1/x is discontinuous for x=0

For any speed that is not c there is an inertial frame co-moving at that speed. For propagating electromagnetic energy there is *no co-moving frame*.

Another example: the wave-character of propagating electromagnetic energy. As we know, there is a periodicity, and with different pathlengths you get phase differences, hence interference effects.

Now, if you assume that 'for a photon time stands still' then the periodicity in electromagnetic waves cannot be a property of the photons. But it must be, how else can you get the interference effects. So that's a self-contradiction. The assumption 'for a photon time stands still' leads to self-contradiction.

The point is that you're sitting near the photon and the reference frame has a different metric than you.

Like I said, it's not accurate, but it gives you the right idea.

Oh, sorry, thought you were improving the explanation I'd given.

Seems like the world isn't solely for my benefit :-)

Cleon
Thanks and yes.
But here is my conceptual problem.

Thus is photon A (from its own point of view) in an inertial frame or not in an inertial frame?

I'm not trying to be argumentative; I'm trying to understand.

So I reason that photon A must be in an inertial frame from photon A's point of view; or else it is not in an inertial frame. But if it is not in an inertial frame then it is experiencing some kind of inherent acceleration (i.e. some kind of persistent redshift or blueshift). So this is a problem.

Now I assume that my thinking is wrong. Because among other things, I can't make it clear to myself.

So any help in, helping me to understand more clearly will be appreciate.

But yes, yes, I have reread what you have said and you make sense.

So I must still be misunderstanding something. You see an electron or a galaxy or any object has wave properties; but we do talk about an inertial system of Einstein or an electron or a photon in a falling elevator.

So though yes I see the periodicity problem that you bring up; it doesn't seem to be a problem for an electron.

I can accept that my question is entirely non sequitur, impossible, contradictory, whatever; I'd just like to understand a bit more why?

From a technical point of view, Cleon has nailed it: one cannot "ride" the photon in relativity. There is a reason it is called a null vector in relativity: from its point of view, photons traverse an arbitrarily large distance instantaneously, as lengths all contract down to zero and time dilates infinitely.

So no, OKThen, you cannot create an inertial frame for a photon due to the fact that it -- being light -- moves at the speed of light.

But I get the analogy that Wow is trying to make and the conclusion is this: rather than speed, the property of the photon that changes is its wavelength as you change the curvature of the spacetime it passes through. Something (sorry, Sascha) that newtonian physics makes no prediction for.

I don't quite understand this exercise because it seems sort of tautological to me. Not that it's not correct — it's self-consistent and empirically correct.

But the implication, if I understand this correctly as a pedantic rhetorical exercise, is that you can start from a Newtonian perspective and this thought experiment will strongly imply GR. But it won't, because the particle physics portion has relativity implicitly already built into it — this isn't made clear.

On the other hand, this is a valuable exercise for anyone who already has some understanding because it elegantly demonstrates a simple, expected relativistic result. But that's not how this is presented — it's presented as a means of convincing relativity skeptics. But they won't understand the particle physics or, if they do, they'll have some crankish alternative view.

When I read the GPS bit, I also had a bit of trouble because of the weird formulation. I would have said something like, "GPS is designed to include relativity and so, if relativity weren't true, GPS would accumulate huge errors every day."

I think that GPS is very handy as a means of demonstrating to the general public that relativity has been proven correct in a way that they encounter on a daily basis — they hear about all the counter-intuitive relativistic weirdness and they think that maybe it's just scientists bullshitting and it's all theory irrelevant to anything real. GPS counters this.

But only mildly.

Because the relatively very uninformed/uneducated public will, nevertheless, simply find relativity absurd. They believe in absolute time and space, full stop. You're not going to change their minds until you accelerate one of them personally to a significant fraction of c and have them experience time dilation.

And the cranks? You'll never convince them, either. I spent a number or years reading sci.relativity. I have an interest in cranks. The people that actually work hard at refuting relativity or, say, Cantor's Diagonal Method, will just keep on keeping on. Their whole process is to continually refine entirely alternate self-consistent worldviews — they can't be defeated, just ignored. They cause some damage, I suppose. But that's really quite insignificant compared to the much more pervasive willful ignorance about huge swaths of basic science that allows a huge number of people to think that any scientifically established idea that they find absurd is, really and truly, most likely absurd.

Honestly, I think that with regard to SR, at least, Michelson–Morley is well within the range of popular, lay comprehension. We're working against commonsensical intuition, here; and Michelson–Morley proves that this particular intuition is simply wrong. That's the enabling first step.

Most contemporary scientists and educators are opposed to much, or any, historical contextual pedagogy in science. But I think this is a mistake because very often (though not always, of course) the historical progression is one of moving away from common intuition to things that seem, in comparison, to be weird and difficult — it can be very helpful to move students over that hump in the same ways that past scientists did because often (though not always) the historical progression matches the most natural progression of dawning comprehension in individual people. Conversely, as I think this post slightly demonstrates, ahistorical presentations from those who are very aware and versed in the contemporary comprehension tend to unwittingly be a bit tautological and/or misidentify what the uneducated need to understand in order to have that moment of epiphany.

Yes, yes Ethan.
And it does seem that there is no such thing as an inertial frame for a photon. Though I don't really understand why.

Also, I don't know what you mean by "null vector of photon"; I can't seem to google anything that I can understand on it.

"And it does seem that there is no such thing as an inertial frame for a photon. Though I don’t really understand why."

That's implicit from a whole bunch of different perspectives. Start with the fact that the photon has no mass. And the invariance of c. When thinking about "inertial frames" you're already in a context that invalidates the idea of a photon having an inertial frame.

I'm having trouble following your arguments because they seem very odd to me — knowledgeable in some sense, very uninformed in others. It's strange and perhaps I'm the one who is underinformed or I'm just misreading you.

There was never a proper debate as to whether Einstein's relativity genuinely replaced Newtonian physics. The effect you refer to is usually deemed a test of the equivalence principle and that is implicit in Newtonian physics anyway, so no comparison in the test. Newtonian physics is itself “a” relativity theory because incorporates Galilean relativity. So to call many skeptics of Einstein's relativity as “relativity skeptics” is a misrepresentation because they can accept “a” relativity theory it is just may not be Einstein's relativity they accept. As far as the starlight bending circa 1919 it means under Newtonian physics that light would have mass, and if it bends more than expected then mere extra effect to be added; nothing prevents it to be still worked upon from the Newtonian physics way of dealing with things by forces in Euclidean space, gravity operating through mass by inverse square law etc., anything extra as I said just add as an extra effect. What Einstein has is a long series of changes, he starts from special relativity, changes for light he decides it must have zero rest mass, so when he gets light bending under gravity he can't then give Newton's answer that it would mean it had mass, so changes gravity to being space-time curvature. As far as the maths is concerned whatever gets interpreted in Einstein's way can just be interpreted back to how Newtonian physics would interpret it. The starting point of Einstein seems to be assume that light speed (in vacuum) free of influences (such as gravity) is a constant, from that it compels him to make the changes as to how Newtonian physics deals with things. But treating it all as mere convention as per Poincare who worked on relativity theory before Einstein, then its just a maths trick. A maths trick added to Newtonian physics is no real change in Newtonian physics. Set clocks by Einstein synchronization process is just meaning - are hiding what would otherwise being variable lightspeed as time dilation and related effects. So all it all its just: manipulating the maths and referring to the maths terms in a different way. What Newtonians would refer to as forces in Euclidean space just gets changed to the Einsteinian language of spacetime curvature.

" a null vector is a non zero vector that is orthogonal to itself"

Hmm, a spacelike vector is positive, a timelike vector is negative, but a null vector seems to be neither but not zero. "It is particularly important to understand that a null vector is not a zero vector"

"If the 'length squared' of a vector computed with this pseudo-metric is positive, the vector is called space-like; if negative the vector is called time-like; if zero, the vector is called null or light-like."

"Two nonzero vectors __ and ___ are said to be orthogonal if their inner product __ is zero. Null vectors are thus self orthogonal."

OK searching google books online finds stuff for null vectors and relativity.

OK, I'm beginning to understand; and I need some persistent studying.

I'm thinking maybe A First Course in General Relativity, 2nd edition. by Bernard F Schutz.

Robert M. Wald's General Relativity
and Weinberg's Gravitation and Cosmology
I've tried, they are too difficult, at least to start with.

Any other suggestions besides Schultz?
thanks.

@Ethan: Not so fast. You argue via energy (not geometry), and the relation between energy and wavelength is QM, so you could do the same for non-relativistic gravity. That special relativity kind of includes a lot of QM already is an important point that I also try to put across (with no support yet from you), however, that argument is not the one you are trying to make here.
@OKThen: You are the only one here who needs a shrink. If my stuff is beyond many punters reading comprehension, so be it, but I will address issues straight and not treat my readers like little children. Suicide rates are skyrocketing because those contemplating it are talked to like little children. I take them seriously and will not bow to the idiocy of dismissing their perspective as irrational, not given a background of exploiting, warfaring, mass imprisoning society telling us to stay alive in order to procreate it.
@CB: Look at the comments and realize that no, people do not all get that there are no slow clocks. And no, they are not slow relative to each other, not as long s a proper comparison proceedure is not given (see twin paradox! - it is not certain which clock is slower). Time has no further time to allow it to do anything, so it does not flow at all, not fast, not slow.

"and the relation between energy and wavelength is QM,"

No, the relation between energy and wavelength is Classical.

Maxwell.

Sascha
I'm glad you haven't suicided.

I will comment on your irresponsible, dillusional, self indulgent, dangerous and irrational thinking on your blog.

Don't waste our time HERE with your "intellectual" blather.

At least for the velocity dependent time dilation, this was observed 60 years ago after the invention of the synchrotron when fast muons were seen to have longer lifetimes then slow muons.

@SLC:
But that's an SR-effect, isn't it?
Here we're talking about a GR-effect: the change of photon's energy induced by acceleration (or the easy-to-handle mathematical equivalent of bending spacetime, as you may prefer) due to objects with rest mass.

Question for the professionals on here (and a bit OT):

OKThen made reference to electrons emitting gravitons. If they did, would that not violate conservation of angular momentum? Electrons are spin 1/2 particles, and can only exist in a spin +1/2 or -1/2 state. Therefore, during an interaction, the maximum change in spin angular momentum an electron can undergo is 1. A gravition is a spin 2 particle. Therefore, an electron emitting a graviton would result in an overall increase in angular momentum.

Ordinary matter is made up of conglomerations of spin 1/2 particles. The same argument applies to neutrons and protons as well. By this argument, no ordinary matter could emit gravitons.

Obviously, I'm missing something here. I am sure that professional physicists have considered this and that there's an answer. Could someone please fill me in? Thanks.

muon experiment does not show time dilation independent of a belief in the maths manipulated to suit SR. Given SR maths then speeds are relativistic speeds and to account for muons existing further that they are expected to get as per that belief, then time dilation is introduced to try to account for it. Going by Newtonian physics the muons have gone faster than expected without the relativistic speed addition restriction. Its just all playing in the maths manipulation.

@SeanT:
Spin 2 is just a symmetry-description needed to fulfill the postulate that even contrarily charged particles always gravitationally attract.
Spin is not an angular momentum of the kind to be conserved: [Wiki:
However, in quantum physics, there is another type of angular momentum, called spin angular momentum, represented by the spin operator S. Almost all elementary particles have spin. Spin is often depicted as a particle literally spinning around an axis, but this is a misleading and inaccurate picture: Spin is an intrinsic property of a particle, fundamentally different from orbital angular momentum.]

Schwar_A,

I understand that in QM "spin" is not a literal spin. However, I thought that QM spin did have an associated angular momentum, and that this is subject to conservation.

For instance, an electron bound in an atom can make transitions from higher energy states to lower energy ones thereby emitting a photon. The change in orbital quantum number, l, must be + or -1 for this transition to occur. I was always taught in my p. chem class that this was because of angular momentum conservation, namely the fact that the photon is a spin 1 particle, so the electron's angular momentum must change accordingly. Is this only the case for a bound state, and not applicable generally? Is that what I'm missing?

AFAIK, angular momentum is NOT one of the conservation laws that has ever been observed to be broken, so I am assuming that it must still be conserved for interactions involving gravitons. Even if spin angular momentum and orbital angular momentum are separately conserved, the original question still stands: how can ordinary matter emit a graviton since normal fermions can only change their spin angular momentum by +/-1 whereas the gravition has a spin angular momentum of 2?

I think it's really cool that photons don't experience time (so to speak). So there it is boiling off some whatnot in the Hubble Deep Field, and wham there it is in my eye. I am in direct contact with the most distant stars, as if it had reached out and touched me!
... but timeless photons make the concept of "frequency" a little problematic. Oh well, none of this stuff means what it looks like it means.
.... the most frequently challenged scientific idea?? Sorry, that would be Darwinian Evolution.

Schwar_A,

I read you link a bit more closely, and it seems that my example above is the real answer to my question. Total angular momentum is conserved, but spin and orbital angular momenta are not separately conserved. So I guess, the short answer to my original question is that the electron can emit a gravition only if it experiences as corresponding decrease in its orbital angular momentum. Thanks for the help.

the comment: I think it’s really cool that photons don’t experience time (so to speak).

which is nonsense, because given the thought experiment setup of a beam of light being observed at c from an inrtial frame "A " then by the Principle of relativity there should be an inertial frame "B" for which the light is stationary and observes instead "A" travelling at speed c. It could not observe a speed if time had ceased.

Ethan, is it possible to post images here? I thought about it for a bit, then decided I might as well do a graphic of my own. Here's my own first-drafty attempt. The green lines are lines of simultaneity and the blue lines are normal to them at every point. To a first approximation, objects in free fall move along these normal lines and this is what is commonly known as 'gravity'.

" Spin is often depicted as a particle literally spinning around an axis, but this is a misleading and inaccurate picture:"

Indeed for an electron to spin with the momentum we give it and the size we KNOW it has, it would require that the edge of the electron be moving faster than light.

Therefore it cant be macroscopic spin.

Re SCHWAR_A

Read carefully, I specifically said velocity dependent time dilation which = special relativity.

Re Wow

Electrons, Protons, and Neutrons have intrinsic angular momentum = 1/2 in natural units. Photons have intrinsic angular momentum = 1 and gravitons have intrinsic angular momentum = 2 in natural units. The concept of a physical object spinning makes no sense in QM as "particles" are represented by fields.

Re anti-Einstein

Mr. anti, your comment is totally nonsensical. The reason why a synchrotron is needed is because of the mass of the muons is a function of velocity. If Newtonian physics held, the mass would be independent of the velocity and there would be no need of a synchrotron, which amounts to a delay in applying the electric field that accelerates the muons. The time delays required are exactly as predicted by special relativity.

By the way, if special relativity is wrong, then please explain how quantum electrodynamics is able to predict the anomalous magnetic moment of the electron to an agreement with experimental observations to 10 significant digits.

@ Wow: Yes, the relation could have been curvature (geometry) and the Maxwell field if Ethan would have argued that, which is what I explicitly wrote (read my comment again)! What he writes is basically that energy is preserved in a certain theory. However, it would be also preserved in a gravity theory that has a different red shift, and the bringing in of the creation of particles serves here onlu to confuse via their quantized energy requirements! You cannot argue red shift via energy conservation between two different, equally self-consistent theories.

"However, it would be also preserved in a gravity theory that has a different red shift"

What does that mean?

The Earth has a different redshift to an incoming photon than the Sun does.

But that's pretty empty of any new meaning.

So what new meaning are you trying to introduce with your "different red shift"?

And if you knew that your statement

“and the relation between energy and wavelength is QM,”

was incorrect, why the hell did you say it?

"Electrons, Protons, and Neutrons have intrinsic angular momentum = 1/2 in natural units."

They aren't. however, real spin angular momentum. To do so, the edge of the electron would be going faster than light.

Coment: By the way, if special relativity is wrong, then please explain how quantum electrodynamics is able to predict the anomalous magnetic moment of the electron to an agreement with experimental observations to 10 significant digits.

Calling SR wrong can be problematic, in a sense it is. What I actually referred to was the maths trick of convention – Set clocks by Einstein synchronization process is just meaning – are hiding what would otherwise being variable lightspeed as time dilation and related effects. So all it all its just: manipulating the maths and referring to the maths terms in a different way. What Newtonians would refer to as forces in Euclidean space just gets changed to the Einsteinian language of spacetime curvature. – So the answer is the maths has been bodged.

comment: They aren’t. however, real spin angular momentum. To do so, the edge of the electron would be going faster than light.

hence why SR is just bodged maths, when it fits things its because its bodged, abd when things don't fit then it has to be bodged some more.

Given two inertial frames "A" and "B" if A observes B at non-zero speed v then we would expect by Newtonian physics that B observes from its frame that A has speed v.

But when v =c we descend into madness of SR and as per earlier comment of someone saying time stops

Einstein built on nonsensical thought experiments and then the maths had to be bodged to fit them, and then real experiments have to be bodged to fit that bodged maths.

"hence why SR is just bodged maths"

Nope, it's because there is energy in spinning momentum and terms were used that were ANALOGUES of what was considered to be happening at the time the terms were coined.

I'm afraid the only bodging being done here is by you.

Re anti-Einstein

Excuse me, what experiments were "bodged"? The Michelson/Morley experiment was performed in 1889, when Einstein was 10 years old. The result was totally incompatible with Newtonian mechanics and Galilean relativity. The difference in measurements of the earths velocity taken 6 months apart is 37 miles/second. If Galilean relativity was correct, the difference in the observed speed of light of 37 miles/second was eminently observable by Michelson's equipment which was capable of measuring a difference in light speed of 4 miles/second. Michelson got a null result, within experimental error. Thus, the results of the experiment didn't agree with the theory. Einstein's 1905 papers, which posited that the speed of light was independent of the speed of the observer agreed with the result of the Michelson experiment. That's all one can expect from a theoretical prediction, agreement with experiment. There is nothing "bodged" in the calculation, it falls out naturally from the speed invariance assumption of light (e.g. Lorentz invariance).

Furthermore, another fact which Prof. Siegal didn't mention was the discrepancy in the precession rate of the elliptical orbit of the planet Mercury. When the effects of the interplanetary effects on the motion of Mercury were calculated, using perturbation theory and the Hamiltonian formulation of classical mechanics, the result differed from observations by 43 seconds of arc/century. these calculations and observations were done before Einstein was born. This was a mystery until General Relativity was proposed in 1915, calculations from which explained the discrepancy. The notion that the observations were "bodged", whatever the hell that means or that the calculations were "bodged" is preposterous. There are no adjustable parameters in the calculation, which is the only way they could be fiddled to fit the data.

By the way, General Relativity also predicts the existence of black holes, which have been unambiguously observed. The notion of black holes makes no sense in Newtonian Mechanics.

“hence why SR is just bodged maths”
Wow: Nope, it’s because there is energy in spinning momentum

Read the comment properly it was referring to the speed of light bit being bodged

SLC: Excuse me, what experiments were “bodged”?

The maths of SR is bodged and the experiments then bodged to fit with that maths.

SLC: The Michelson/Morley experiment was performed in 1889, when Einstein was 10 years old. The result was totally incompatible with Newtonian mechanics and Galilean relativity.
Thats just a typical false claim. The experiment is consistent with the classical principle of relativity as per Newtonian physics. When we look at the maths of SR it is just based on the bodge of lightspeed constancy, which leads to a long series of bodges involving time dilation etc. If we don't do that then we can still deal with it by the maths of Newtonian physics.
SLC: the discrepancy in the precession rate of the elliptical orbit of the planet Mercury.
Can be dealt with in the context of maths of Newtonian physics, as I have earlier stated is context of forces in Euclidean space with extra effects added when required (I.e the adjustment parameters you talk about).

SLC: By the way, General Relativity also predicts the existence of black holes
The concept that became called blackholes were dealt with in context of Newtonian physics long before, Mitchell considered what would be the effect of light trying to escape from a very massive star.

Re anti

Mr. anti is totally irrational. How could the experiments be "bodged" to agree with special relativity when they were performed 16 years before Einstein published his papers on special relativity in 1905.

Mr. anti is ignorant, as well as irrational. There are no parameters to be adjusted in the calculation of the discrepancy in the precession rate of the orbit of the planet Mercury. All the constants are computed independently (e.g. the universal gravitational constant, the mass of the sun, the masses of the other planets, etc.). The result is that there is a 43 seconds of arc which Newtonian mechanics cannot account for in the observed precession rate of the orbit of the planet Mercury. This problem was was known for at least 50 years before Einstein's 1915 paper and was not solved until Schwartzchild's 1920 paper which utilized General Relativity.

Furthermore, Newtonian mechanics cannot predict the existence of a black hole because the velocity of light, c, is not a constant. Thus, the Schwartzchild radius is meaningless because it relies on the value of c being constant.

I must say that I find Mr. anti to be remarkably ignorant for a grown man.

Re anti-Einstein

By the way, there is nothing in General Relativity that requires that a star must be massive to collapse into a black hole. If the Sun, which is not a massive star, were to collapse inside its event horizon (radius ~ 4 miles), it would become a black hole. In fact, there is nothing in General Relativity preventing a golf ball from becoming a black hole if it were to collapse inside its event horizon.

SLC: How could the experiments be “bodged” to agree with special relativity when they were performed 16 years before Einstein published his papers on special relativity in 1905.

Its being bodged to fit with the maths that came in 1905.

SLC: There are no parameters to be adjusted in the calculation of the discrepancy in the precession rate of the orbit of the planet Mercury.
That is false. Oblateness of the sun is one factor that can be introduced to a calculation based on idealisation of perfect symmetric sun.

SLC Furthermore, Newtonian mechanics cannot predict the existence of a black hole because the velocity of light, c, is not a constant.
Mitchell dealt with black holes in context of Newtonian physics. And as pointed out constancy lightspeed bit is a bodge in the maths.
SLC:I must say that I find Mr. anti to be remarkably ignorant for a grown man.
You make it a mistake it is you that is ignorant. Sinec you are unable to cope with the facts you are reduced to acting rude and stupid.

SLC: How could the experiments be “bodged” to agree with special relativity when they were performed 16 years before Einstein published his papers on special relativity in 1905.

Its being bodged to fit with the maths that came in 1905.

add: because when we look at the maths we find that for the experiment where we had light going at c+v in one direction and c-v in the other direction etc with c and v non-zero, in the maths pre-1905, we find that from 1905 that is bodged to being c with what would be variable lightspeed then made an effect on time; a bodge in the maths 1905 which an experiment pre1905 is made to fit

"Read the comment properly it was referring to the speed of light bit being bodged"

Oh, I get it, you're bollocking just like chelle did about how the meter is defined as something to do with the speed of light, right?

Anti-science bullshit and completely uninteresting.

wow: Anti-science bullshit and completely uninteresting.

which sums up your position.

Lets assume gravitons being permanently emitted equally into all directions by an object with positive rest mass to cause this object's gravity field.

To keep the object's original energy the emitted energy of all gravitons has to completely return. If it does not, the object should start moving towards the less returning energy direction.

This is my speculation - but how is the graviton's energy flow and balance actually handled theoretically?

They are not really emitted like light emitted from an incandescent source.

They are virtual and only when something is nearby and exchanges a similar *on with the other object that the *on has an effect.

So the electric field (Photon) swaps with another electrically charged particle (another Photon) and that swap causes the force of attraction or repulsion.

That is the time when the emitted energy "has to return".

anti try something a little more grown up than "I know you are, but what am I?", hmm?

wow: anti try something a little more grown up than

that would be beyond your ability to comprehend, so I have to talk down to you

Re anti

The value of c was measured by Michelson some 10 years before the Michelson/Morley experiments were performed, using a light source that was at rest relative to the interferometer. He won a Nobel Prize for the measurement. In Newtonian terms the apparent result of the M/M experiment was that v = 0, i.e. the earth is at rest. Since this is nonsensical, the only other possible explanation was that c was a constant, independent of the speed of the observer.

As for the oblateness of the sun, the interior would have to be spinning at least 100 times as fast as the atmosphere to generate a quadrupole moment sufficient to explain away the 43 seconds of arc. Calculations by Dicke showed that a rotation speed of 10 times that of the atmosphere yielded a quadrupole moment only sufficient to account for some 3.5 seconds of the 43 seconds observed.

Re wow

I'm beginning to think that Mr. anti is a Poe. Nobody could be that stupid.

SLC: The value of c was measured by Michelson some 10 years before the Michelson/Morley experiments were performed, using a light source that was at rest relative to the interferometer.

I have no problem with that, and it shows the principle of relativity, and that is consistent with Newtonian physics. Adding the constancy of lightspeed is just an unacceptable bodge.

SLC: As for the oblateness of the sun, the interior would have to be spinning at least 100 times as fast as the atmosphere to generate a quadrupole moment sufficient to explain away the 43 seconds of arc.

You wanted an example of one of the “other” factors involved, and I gave it. Of course more can be taken into consideration. GR gives 43 secs arc take away 3.5 for oblateness and its not such a good match. Add more factors and it becomes less so.

So you and your buddy wow stop acting like idiots and pay attention to what is being said, rather than make stuff up.

SLC: The value of c was measured by Michelson some 10 years before the Michelson/Morley experiments were performed, using a light source that was at rest relative to the interferometer.
I have no problem with that, and it shows the principle of relativity, and that is consistent with Newtonian physics. Adding the constancy of lightspeed is just an unacceptable bodge.
Add on because its like what your buddy wow said

wow : bollocking just like chelle did about how the meter is defined as something to do with the speed of light, right

I.e bodging or bollocking (as wow calls it) the maths as used by the Michelson-Morley experiment.

Anti Einstein

Yes you are anti-science.

You neither understand nor want to understand science and scientific evidence.

Trying to reason with you Anti Einstein (i.e. anti science) is a waste of every one's time; because of the breath and depth of inconsistency, incomprehesion and intellectual dishonesty in your anti-science (i.e. anti Einstein's) position.

So Anti Einstein please go to Ethan's comment policy web page
http://scienceblogs.com/startswithabang/2012/09/23/weekend-diversion-yo…

Read it carefully and understand it.
Because if you continue with your nonsense; you will be asked to confine your nonsense to that web page alone.

Relativity-deniers are deep cranks. I compared them to anti-Cantor cranks for good reason. Their commonsense intuition is deeply insulted — they *know* these things are wrong. Like all such, they will be impervious to evidence or reason. There is no benefit to engaging with them

Ellis: Relativity-deniers are deep cranks. I compared them to anti-Cantor cranks for good reason. Their commonsense intuition is deeply insulted — they *know* these things are wrong. Like all such, they will be impervious to evidence or reason. There is no benefit to engaging with them

calling someone with good intuition a crank is just being insulting, and good intuition means having a good understanding of logic, and no evidence disproves logic.

OKThen: Read it carefully and understand it.

You should try re-reading it because its you being insulting.

Keith, are you surprised that your prediction would turn out to be proven so quickly?

@Wow:
Thank you.

"someting is nearby" seems to me very relative due to the huge reach of gravitation.

One Photon would accelerate exactly once - to have a permanent acceleration you need a "stream" of photons.
Something very similar seems to exist related to gravity.

Graviton as a kind of instantaneously swapped virtual particle seems to contradict speed of gravity c...

"“someting is nearby” seems to me very relative due to the huge reach of gravitation."

Being massless, the energy of the graviton exchanged that is "allowed" to exist long enough to do that is very small. But it can be arbitrarily small. And so the force falls off with distance to infinity.

This is how we figure EM and gravity to be carried by massless particles. If they had mass, there would be a maximum range because the field would have to borrow at least the rest mass energy of the exchange particle.

Light speed is not Constant (to observer) !!

All that we receive with our eyes are the facts of the past (unchangeable). Wavelength of incident light is coming from the past. On incident light, a formula c = λ f stands up. And λ is unchangeable (by our motion). Terms f and c change.

Sorry, I can't receive E-mail. I don't have PC.

http://www.geocities.co.jp/Technopolis/2561/eng.html

Light speed is not Constant (to observer) !!

Yes it is!

All that we receive with our eyes are the facts of the past (unchangeable).

our eyes can't see infra-red, but we still know the speed of that! And the fact of the past is that the speed of light is constant!

Terms f and c change.

No, only f changes, c is a constant!

Graviton as a kind of instantaneously swapped virtual particle seems to contradict speed of gravity c…

Meh, while I'm here, in case you're still around:

Nothing is being instantaneously swapped. The virtual particle only exists once it has swapped with the other object another virtual particle. This is why there's no need to have as you put it:

To keep the object’s original energy the emitted energy of all gravitons has to completely return.

it doesn't exist unless it gets involved in an exchange, and then its existence causes the change in momentum required to create the view of a force acting between the two participants.

This is for the model that has particle exchange as the mediator of force.

Quantum fields are another model and have a different explanation within that model.

No, only f changes, c is a constant!

Then, on incident light, c = λ f doesn't stand up ? I show a one more picture below.

In outer space, from the right and the left, plane waves of star lights are coming vertically. λ of coming light cannot be varied by observer's motin. f and c vary.

No, c is a constant. Wavelength changes because frequency changes but not velocity.

Then, how about λ , when there is no observer ?
It's my last post.

Even when there's no observer, wavelength changes and frequency changes and c remains constant.

Also nothing was in your last post. It was meaningless. You didn't give an illustration and then just flat out claimed that wavelength could not change, why not explained, then frequency changes, why again not explained, and therefore c changed, except without explaining why frequency changed but not wavelength, you are running the very definition of "begging the question".

In outer space, a mirror is reflecting a star ray. On incident ray and reflected ray each, a formula c =λf will stand up (seen from the mirror). In this two formulae, term f is the same always. So, when λ is not the same (usually, it's not the same), c must not be the same.

Nope, f can be different.

c=λf

If f changes and λ, then c can be the same.

Since f does change when λ changes, this is just fine.

We'll assume that a frequency f (of two rays) is measured at the point one meter away from the mirror. Where will the remainder of the different value go?

If we assume that the frequency of the two rays is measured one meter away from the mirror,there is no remainder of the different value.

They're two different photons. There is no "remainder" to find.

Why did you not ask yourself "If the speed of light is measured at one meter from the mirror, where will the remainder of the different value go?"?

It's just as valid a question. Yet not one you asked.Ever. Even of yourself.

Allow me to write once again, please (it is my last post).

If f (frequency) of two rays (at two points) is different, number of waves that exists within 1 meter will increase or decrease endlessly. Plainly, it is impossible.

Plainly it is not. You can do an actual experiment to show it's possible.

What IS shown impossible is light going slower than the speed of light in that medium.

Even you know it's possible.

If wavelength changes, then number of waves that exists within 1 meter will increase or decrease endlessly.

Einstein's accelerating chest thought experiment is flawed.
Its foundational to General Relativity.

One chest at rest in Earth's gravity field. One chest pulled by a genie in a region of the Universe where there is no gravity. Both have an acceleration equal to the gravitational pull of Earth. Never mind the fact that this place is imagined and this experiment cannot be and has not been done. This invalidates the free fall part. The free fall part is also invalid because falling is accelerated and not a state of a constant velocity, so he has to make the accelerated state a constant one with sleight of hand.

In any case the chest at rest relative to the Earth's surface is not the same as the chest being pulled by a genie in an imagined zero g environment.

Here's why: gravity on Earth is at 9.8 m/s2
An object dropped will start falling with zero velocity and will see that velocity increase as it falls. Each and every object dropped from a height X would take the same amount of time to hit the floor below.

In the imagined Zero G environment, the genies pulls the chest with a constant acceleration that mimics gravity. 9.8 m/s2
Each object dropped from the same height X will not fall at the same rate. Each subsequent object will fall as if the force of gravity were increasing. IE the acceleration means specifically that the ship is moving progressively faster with each instant of time.

For example:
9 balls dropped in succession by someone standing in the chest on Earth means each ball falls at 32 feet per second, each one taking say half a second to drop about 4 feet.
Each ball takes the same amount of time to hit the floor.

9 balls dropped in succession by someone standing in the chest being pulled by a genie in a zero-g environment at an accelerated rate means each ball falls faster than the one dropped prior.

It starts at say 32 feet per second and then increases exponentially. Each subsequent ball falls to the floor with greater velocity, covering that distance in less and less time. Unlike the experiment done on Earth. On Earth the rate of drop 'resets' with each ball.

Equivalence is wrong

"Einstein’s accelerating chest thought experiment is flawed."

No it isn't.

"Its foundational to General Relativity."

Not that either.

"9 balls dropped in succession by someone standing in the chest being pulled by a genie in a zero-g environment at an accelerated rate means each ball falls faster than the one dropped prior."

No it doesn't. It starts off at zero and is left behind by the hand accelerating away at the rate of 9.81m/s^2.

Whereas the earth gravity one, the ball starts off at zero and is drawn away from the hand accelerating away at the rate of 9.81m/s^2

Each of the nine do that. Not one changes. Both scenarios stay the same.

You are wrong.

Lightspeed is not Constant (to observer) !!

Imagine spherical waves of light (or light sphere) that are sent from two sources in relative motion. Except the emission theory, what explanation is possible ?

Sorry, I can’t receive E-mail. I don’t have PC.
http://www.geocities.co.jp/Technopolis/2561/eng.html

Interesting concept of a realativistic-redshift.

Would no the best "proof" for this be made for the frequency of sunlight vs. other sized stars vs. man-made fusion.

Surely quantum mechanics can calculate the theoretical wavelength of fusion reaction.

"Lightspeed is not Constant (to observer) !!"

No, the speed of light IS a constant (to the observer)!!!!

FOUR exclamation marks, therefore I'm twice as right!!

"Imagine spherical waves of light (or light sphere) that are sent from two sources in relative motion. "

I did.

What's the problem?

Lightspeed is not Constant (to observer) !!

To a swinging stick, speed of light varies. Like sound waves, water waves. Follows Galilean transformation.

Sorry, I can’t receive E-mail. I don’t have PC.

Lightspeed IS constant to observer!

Sorry, I can't explain your stupidity, it won't listen, because you don't have a brain.

nakayama,

Perhaps you've never been exposed to the theory of special relativity. Let me try an explanation. The special theory of relativity is really nothing more than the statement that the laws of physics are the same for any observer in an inertial reference frame (that is, for all observers whose motion is not accelerated). It was well understood prior to Einstein that the laws of mechanics were the same for all observers in an inertial reference frame, but Einstein extended this to include the laws of electromagnetism that had been discovered by Maxwell near the end of the 19th century.

Maxwell's laws unified the previously separate fields of electricity and magnetism. One of the key findings of Maxwell was that electric and magnetic fields could oscillate and by doing so produce a travelling wave. His equations worked out that the speed of this wave was independent of anything other than two constants that were previously known from electrical and magnetic measurements. That is, the speed of these waves was a constant value. From previous measurements of light, it was known that the speed predicted for Maxwell's waves was precisely the speed at which light travelled, and thus we came to understand that light was an example of these waves.

Now, Einstein's postulate was that the laws of physics, including Maxwell's laws, were the same for all inertial observers. That would then include the notion that light has the same velocity for all unaccelerated observers. Based on common sense, this seems preposterous. If you are holding a flashlight, and I am moving away from you at 90% of the speed of light, how can I measure the same light speed from your flashlight as you do? Well, Einstein's genius was that he didn't let this bother him; he just worked out what the consequences would be assuming that his postulate was true. It turns out that in my example above, my measurement of time and space would differ from yours. If you were wearing a watch in my example, I would observe that your watch ran slower than what you said it did. I would also measure distances to be smaller than you would. Because of these differences in distance and time measurements, my measured value for the speed of light from your flashlight would match up perfectly with yours.

Of course, this doesn't do much toward having it all make sense. Common sense, however, is a poor guide. What we need to do is some experimental work to see if time dilation (as the change in time measurement is known) and/or length contraction is a real phenomenon. Well, experimentally we find that time dilation is real. We can observe, for instance, unstable particles in cosmic rays that should not have survived to reach the surface of the earth because they decay too rapidly to survive. These particles are very energetic, however, which means they move close to light speed. The time that we measure for them to reach the surface of the earth is longer than the time that an observer moving with the particle would measure. Therefore, the particles survive. If that's not convincing, we've actually used an atomic clock being flown around the earth on an airplane to confirm time dilation. No experimental result has ever been found that contradicts relativity. Therefore, we conclude that light really does have a constant velocity for inertial observers.

What if everyone jumped?

rallain — Thu, 26 Aug 2010 13:48:47 +0000

What if everyone jumped?

Might as well jump. Jump. Go ahead, jump. - Van Halen

Suppose everyone in the world got together and jumped. Would the Earth move? Yes. Would it be noticeable? Time for a calculation. Note: I am almost certain that I have done this before, but I can't find where.

Starting assumptions.

7 billion people.
Average weight: 50 kg (you know, kids and stuff)
Average vertical jump (center of mass): 0.3 meters - and I think that is generous.
Mass of the Earth: 6 x 10²⁴ kg
Gravitational field near the surface of the Earth is constant with a magnitude of 9.8 N/kg
Ignore the interaction with the Sun and Moon

Basic physics

Suppose I take the Earth and the people as my system. In this case, there are essentially no external forces on the system (see assumptions above). There will be two conserved quantities - momentum and energy. Here, the term conserved means that that quantity does not change. I can write:

What does the "1" and "2" mean? These could be any two times. For this situation, let me say that time 1 is right after the people jump (and still moving up) and time 2 is when the people are at their highest point.

Energy is also conserved. If I take the people plus the Earth as the system, then I can have both kinetic energy (K) and gravitational potential energy (U_g). Using the 1 to represent the people just jumping and 2 to represent them at their highest point, then:

About gravitational potential. First, it is the potential energy of the system, not of each object. Second, in this approximate linear form (mgh), the change is what really matters. This means that I can set the potential at point 1 as 0 Joules. Also, the mass of the Earth does matter in this potential - that is where the 9.8 N/kg comes from.

The calculation

A couple of important things to start with. At position (and time) number 1, the Earth and the people are moving but there is zero gravitational potential energy. At position 2, the Earth and the people are 0.3 meters apart and not moving (at the highest point). Finally, momentum is a vector - but this is a one-dimensional problem. I am going to let the y-direction be in the direction the people jump.

This gives a momentum conservation equation of:

Now, I can use the energy equation to get an expression for the initial velocity of the people:

Just a quick check with reality. If you want to jump a height h, you would need a speed of:

This is what you get if you assume the velocity of the Earth is super small from above. Ok, I am going to put these two equations (momentum and energy) together. This looks bad, but it really isn't too bad. The problem is the velocity of the people from the work-energy method still has the velocity of the Earth. Avert your eyes if you are algebra-allergic.

Not finished quite yet - I need to now solve for the velocity of the Earth.

See, that wasn't too bad. You can open your eyes now. Now for the numbers. If I use the values form above, I get a recoil speed of the Earth as:

Maybe you don't like my starting values. But you know what? It doesn't really matter - the mass of the Earth is so huge that it is going to be pretty darn difficult to get a detectable speed. Also, there is the whole issue of getting everyone at the same place at the same time and getting them to jump at the same time.

I seem to recall the last time I did this calculation (that I can't find) that I also estimated how many people you could get in one spot of the Earth.

rallain Thu, 08/26/2010 - 09:48

Tags

conservation of momentum

analysis

Turn or go straight? Quick!

rallain — Thu, 05 Aug 2010 14:07:24 +0000

Turn or go straight? Quick!

This is a classic problem. You are in a car heading straight towards a wall. Should you try to stop or should you try to turn to avoid the wall? Bonus question: what if the wall is not really wide so you don't have to turn 90 degrees?

Assumption: Let me assume that I can use the normal model of friction - that the maximum static friction force is proportional to the normal force. Also, I will assume that the frictional coefficient for stopping is the same as for turning.

Stopping

I am going to start with the case of trying to stop. Suppose the car is moving towards the wall at a speed v₀ and an initial distance s away from the wall. Diagram time:

This is a 1-d problem. So, let me consider the forces in the direction of motion. There is only one force - friction. Now - you might be tempted to use one of the kinematic equations. Well, I guess that is just fine. The following equation is appropriate here.

Really though, I would think - hey distance. That means use the work-energy equation. It gives you the same thing though - essentially. Since I already started with this kinematic equation, let me proceed. In the direction of motion, I get:

Putting this into the above kinematic equation (with the change in x-distance as just s). Oh, note that I am using the maximum static frictional force. I am assuming that this will be the shortest distance you could stop. Also, I am assuming that I the car stops without skidding.

There you have it. That is how far the car would need to stop. Quick check - does it have the right units? Yes.

Turning

Now, how far away could the car be and turn to miss the wall? Really, the question should be: if moving at a speed v_o, what is the smallest radius turn the car could make?

For an object moving in a circle, the following is true:

Here is my review of acceleration of an object moving in a circle. Key point: I said I could have used work-energy for the stopping part. I could NOT have used work energy for this turning part (well, I could use it but it wouldn't give me anything useful). There are two reasons why the work-energy principle won't do you any good. First, the speed of the car doesn't change during this motion. This means that there is no change in kinetic energy. Second, the frictional force is perpendicular to the direction of motion so that it does no work (we can discuss work done by static friction later).

Back to the turning calculation. I know an expression for the frictional force and I want the radius of the circle to be s. This gives:

And there you have it. If a car is traveling at a certain speed, it can stop in half the distance that it would take to turn.

I kind of like this result. Long ago, I took a driving class. You know, to learn how to drive. One think stuck in my mind. While driving, something came out in the road in front of me (I can't remember what it was). I reacted by swerving just a little into the next lane. The driving instructor used that annoying brake on the passenger side (that he would sometimes use just to show he was in control - I was going to stop, but he didn't give me a chance). Anyway, he said "always stay in your lane". He probably said that because he was so wise in physics even though he did smell funny.

Oh, it is probably a good idea to stay in your lane not only for physics reasons but also because you don't want to hit the car next to you (unless you are playing Grand Theft Auto - then that is encouraged).

Another question

I wonder if you could stop in even shorter distance? Is stopping the best way? Is there some combination of stopping and turning that could work?

Let me try the following. What if the car brakes for the first half and then turns for the second half. Would it hit the wall? First, how fast would it be going after braking for s/2 distance? The acceleration would be the same as before:

Using the same expression for the stopping distance from above, I get:

And this makes sense. If the car is stopping just half the distance, then it should have half the kinetic energy (which is proportional to v²). Ok, so if that is the new speed, what radius of a circle would it be able to move in? Again, using the expression from above:

Using this with half the distance - the total distance it would take to stop would be:

This is still greater than the stopping distance for just braking (which is s). But, did I prove that just stopping is the shortest distance? No. Maybe I just convinced myself to stop for now.

Bonus

Here is a short bonus. Let me show that the work-energy principle is the same as that kinematic equation I was using. So, a car is stopping with just friction. The work done on the car by friction (and I can do this if I consider the car to be a point particle):

The work-energy principle says this will be the same as the change in kinetic energy of the car. If the car starts at a speed of v₀ and stops at rest then:

See. Same thing.

Homework

How wide would the wall have to be so that it wouldn't matter if you brake or turn? Either way you would miss?

rallain Thu, 08/05/2010 - 10:07

Tags

One of the things I remember being drilled into me during driver training was brake and avoid.

There's an interesting dynamics aspect to this, in that braking results in a transfer of the car's weight forwards, which gives you more traction to make the turn. In the absence of ABS, slamming on the brakes will result in your wheels locking up and thus a complete inability to steer (not to mention, an increased stopping distance).

What about hard left, then brake? After all, in America, that would get the driver's side away from the wall, so that if your stopping distance is too much, you at least avoid a head on collision, and provided you have no passenger, there shouldn't be serious injury. Because if s> the distance between the car and the wall, you really need to think in terms of minimizing damage.

For that matter, if you hit the wall at an angle instead of head-on, you'll slide along it instead of stopping instantly. This should reduce the deceleration forces acting on you, decreasing your odds of injury. So it seems to me that combined steering and braking is the way to go.

Don't forget the crumple zone you have if you hit front on

In Sweden you learn to break and swerve. Reason being, your normal obstacle (a moose, a human) is not all that wide, and the road is pretty likely to be empty apart from yourself. If there's lots of traffic, it's unlikely that there'd be a static obstacle on it after all.

There's a secondary consideration too. If you just swerve and fail, then you're likely to hit the object at close to your original speed. And as speed is by far the most important determinant of injury and death, that's a Bad Thing.

On the other hand, obstacles are very rarely really wide. Anything remotely likely to show up on a street is pretty narrow - a moose, a car or something like it. So once you're slowed down a bit chances are still pretty good that you'll be able to avoid it. And if you don't, you'll still hit at much lower speed than if you just swerved and failed.

@Samantha,

Turn and then brake? Why didn't I think about that? I did brake then turn. Hmmm.. I will have to do a follow up post.

if you have a choice of a head on collision with, say an oncoming car, or swerving a bit so it isn't head on, choose the swerve option.

your momentum can be broken up into components parallel and perpendicular to the other objects motion. by avoiding headon, you can decrease the parallel component by the cosine of the angle, and thus decrease the magnitude of your change in momentum, the acceleration and thus the damage.

If your wall is infinitely wide (as stated in the original problem), you can show that braking is always the best option if you want to avoid a collision with the greatest car-wall separation (assuming the car is a point).

Just decompose your velocity vector into the component normal to the wall and a component along the wall.

If you turn (and avoid the wall) or do some combination of braking and turning:
1) the component normal must decelerate to zero at the point of closest approach
and
2) the component along the wall will increase to something nonzero.

If you just brake, you still have to do #1, but you avoid any forces required to achieve #2. Thus braking requires less total force.

So, for the infinite wall (and a point-mass car) braking along a straight line is superior to any braking/turning combination.

For the finite wall (or moose, as discussed above) it would probably require some algebra.

I agree with break and trun. With that, how fast would your speed have to be to flip your vehicle during your turn (depending on weight, center of gravity, width of tires, etc.)?

A lane change can be done in significantly less distance than either of the two extant options, but that will not work in the classic homework version of this problem where the road is completely blocked but there is a side street open if you turn. This works because you don't turn 90 degrees. Not even close. However, see my final observation below.

Minor detail in your solution: The standard problem assumes the same coefficient of friction for braking and turning, but this is only an approximation. Most tires have more grip under braking than when turning, which favors the braking solution even more.

What Anonymous Coward says is a result of what is known as the "friction circle" in racing. You trade sideways acceleration for fore-aft acceleration, which is why you brake and then turn ... leaving out all sorts of fine details. (Tires are complicated. Kinetic friction is a bit more than static friction as long as the slip is a small fraction of the velocity, then falls off rapidly once the tire is not rolling at all. This applies to both forward motion and the "slip angle" when cornering and is significantly different, depending on tire design, when cornering. The "friction circle" is usually an ellipse with the coefficient bigger for braking as noted above.)

The coefficient of friction when rolling, or (even better) rolling and slipping slightly, is bigger than when the tire is sliding sideways. This means that tossing the car so it slides sideways is a really bad solution and will result is significantly larger values for the variable "s" in your solution.

All of this ignores vehicle dynamics and traffic. If you are in the right car and are fully aware of the space around you, a lane change is the best way to avoid a problem in the road ahead. If you are not fully aware of the space around your vehicle, changing lanes can make things worse, and even skilled drivers are not always paying full attention to the road. Hence the driving instructor advice to stay in your lane. Worse, you might be in a van or SUV where sudden turning movements, particularly lane changes, with under-inflated tires can cause the vehicle to roll. Even passenger cars can get in trouble this way if one tire is off the pavement, which is why you should have been taught to slow down in a straight line before pulling the car back on the pavement from an unpaved shoulder.

@comment 1- You can still steer when braking, independent of whether or no the car has ABS. It's called the Emergency Brake, and in most vehicles it is attached only to the rear wheels.

I see I've been beaten to the Brake and Turn. I have done just that in a car- E-brake and a turn so that I do a complete 180. That would seem the best choice- and the distance needed to stop is far less than braking alone.

@11 - Unless you're a trained stunt driver, I wouldn't advise yanking the hand brake and then turning, because you're going to end up locking up the rear wheels and then going into an uncontrolled skid/slide.

But sure, it will look really cool.

@11 - The only advantage to doing a 180 is that you hit the wall going backwards. Your stopping distance will be greater because the coefficient of friction is less than with threshold braking or modern ABS while going straight.

The disadvantage might be that you kill the kids in the backseat because cars are not designed to crash that way.

Sure would make a great Mythbusters episode, however.

Collisions: Kinetic energy or momentum?

rallain — Thu, 10 Jun 2010 08:57:03 +0000

Collisions: Kinetic energy or momentum?

In the last episode of MythBusters, they wanted to see if a tornado could make some glass cut off a person's head. The first attempt was just to take some glass and through it at a simulated human neck. Clearly, this wasn't quite the same as a tornado.

So, here was their plan. If they want to simulate glass moving at 300 mph, they could get a bigger piece of glass and put it on a truck moving at 80 mph. The result would give a piece of glass with the same kinetic energy as a smaller piece moving at 300 mph. Their calculations look to be correct. However, the question is: would this make the same type of collision?

Let me just write an example. Suppose I want to simulate a 2 kg piece of glass moving at 100 m/s. This would have a kinetic energy of:

Now, what if I want an object with the same kinetic energy, but just moving at 1/4^th the speed of the original object, but with a larger mass?

If you want it to go 1/4^th as fast, it would have to be 16 times more massive. Now, here is the possible problem. What about momentum? Here is the momentum of these two objects (well, the magnitude of the momentum)

Not the same momentum. Now, here is the real question. Does it matter? What matters in a collision, the energy, the momentum, or both? I am not really sure of the answer in the case of a decapitation. I am thinking only the energy matters (but I am ok with the possibility that I am in correct). Why would I say this? In this particular situation, the MythBusters have the fake neck attached to some holder. During the collision, there momentum will not be conserved because there is an external force (from the ground) on the fake-neck. So maybe it doesn't even matter that the momentums are not the same.

Now, if this were a collision between two free objects I think the momentum would be important.

rallain Thu, 06/10/2010 - 04:57

Tags

"Does it matter? What matters in a collision, the energy, the momentum, or both"

'Rule of thumb' is: momentum for penetration, energy for damages.

It's easy to illustrate: just look at bullets.

@Alex,

I would think that penetration would depend on energy. If you assume a constant force acting on the bullet (or object) while it is interacting with the material, then you could think about work as W = Fd = change in kinetic energy.

But, if that is the rule of thumb then it must be based on experience.

I don't think you can take such a shortcut, you'd have to work out the collision in at least some detail. In any case even with a tornado wind at 300mph, it is unlikely a solid object picked up by it would be traveling that fast. There should be a rough time constant during which an object in a fluid will come to have the same velocity of the fluid. I would bet that time is longer than the time for the tornado (vector) wind velocity to remain constant.

The answer is almost certainly 'neither' or perhaps 'some combination of both'. The penetration of a chunk of glass will be driven by both the work required to cut through the neck (linear in energy) and the energy wasted by fluid dynamics interactions with tissue (generally second order in velocity and thus actually scaling with sectional density, not energy or momentum).

"But, if that is the rule of thumb then it must be based on experience."

Yes. In essence, this rule of thumb means that among the projectiles of the same energy a projectile with the largest momentum generally has best penetration power.

And among the projectiles with the same momentum, a projectile with the largest energy generally inflicts more damage.

A scaling complication is that the materials are not linear. Tendons and other biological material respond differently in different time frames. They might fail by tensile tearing at low speed, and by transverse cutting at high speed.

The geometry of the glass edge is also a factor which might not be linearly scalable. A thin, sharp edge will blunt (melt?) during a high-speed cut. A thick edge will retain its geometry at higher speeds.

"...and through it at a simulated human neck"

Through it? Is that an attempt to combine "throw it" and "through the neck"?! Hey it could catch on....

excellent work, ive just finished my kinetic energy paper but ill update with the maths thanks.

Ask a ScienceBlogger: Evaporating Water

rallain — Wed, 09 Jun 2010 14:31:03 +0000

Ask a ScienceBlogger: Evaporating Water

Here is another question from Ask a ScienceBlogger. Reader Uday Panta asks:

How does water evaporate in the seas? Doesn't water evaporate at 100 C?

There were some very good responses in the comments where the question was, but I am going to answer it with some more details.

Small Particle Model

This is where we need to start - the small particle model of liquids and gases. This model treats the liquid as being made up of a lot of particles (well, obviously). If there is a gas (or liquid) at a certain temperature, then there are particles moving around at different speeds. Often it is said that temperature is a measure of the average kinetic energy of the particles in a gas. This isn't too bad of a definition, but the point is that some particles are moving fast and some are slow. They are not all going at the same speed.

Check out this great applet from the PhET simulators.

This is a snap shot of the Gas Properties simulator. The cool thing is the histogram it displays showing the distribution of both the particle energies and particle speeds.

Liquids

Here is another simulator from PhET. This one shows something in different states of matter. I am just going to focus on the liquid and the gas phases for now. I am going to show a movie of this running. In this video, you will see some water in liquid form evaporating. (Go play with the simulator online too - it is pretty cool.)

So, the liquid water has some water molecules in it. The water is at a certain temperature, but some of the liquid water molecules have more energy that others. That means that some of these molecules have enough energy to break free from the other water molecules and fly free. You can see this in the above animation.

That is really the answer to the question from Uday. The water does not have to be at 100 C for some of the molecules to evaporate.

But won't the water keep evaporating? Well, this depends. First, if you keep adding energy to the water then yes it will. Watch the video again. Follow one water molecule that evaporates. Eventually, it will collide with the liquid water and become 'trapped' again. So, the water will reach an equilibrium between the water molecules that become gas the the gas molecules that become water.

It would be cool if the PhET app could show a graph of the number of gas particles vs. time and the number of liquid particles vs. time.

rallain Wed, 06/09/2010 - 10:31

Tags

This would be a fine time to bring up the concept of vapor pressure. That is the partial pressure at which evaporation and condensation are in equilibrium. (Relative humidity measures how much water vapor is actually present compared with this maximum.) We say that water boils at 100 C because that is the temperature at which its vapor pressure equals the average atmospheric pressure at sea level. If normal atmospheric pressure were lower, water would boil at a lower temperature--which it does if you are in, say, Denver (1600 m above sea level) or La Paz (4000 m above sea level). Conversely, if you put it in a pressure cooker it boils at a higher temperature.

Vapor pressure also exists in the solid phase. You probably don't see this very often near the Gulf Coast, but it's a commonplace here in New England: the temperature remains below freezing for several days in a row, yet the amount of snow on the ground decreases through evaporation. If you were in a place where the atmospheric pressure were lower than the vapor pressure at the freezing point, you will not see a liquid phase: the substance will go directly from solid to gas (this is called sublimation). Carbon dioxide ("dry ice") does this because its vapor pressure at its putative melting point is significantly higher than our normal atmospheric pressure. In an environment with sufficiently high pressure, you can find liquid CO2.

The most important concept about temperature that I learned in my Physics 1 class is that temperature is an average measure of the energy of all particles involved. As soon as I got this, I was able to figure out why water can evaporate even though it's not 100C. Macro scale can sometimes only be explained by looking at the micro.

Good suggestion for the PhET sim - I've sent it on to the developers. Don't forget you can always send suggestions for PhET sims to phethelp@colorado.edu.

I thought you might want to see this video I did in -30C in Yellowknife, NWT to compliment your theory.

I have a son who is doing a science project about different states of matter and this simulator is the perfect thing for him right now. Anyways great post as usual ;)

These simulators are very interesting! While I tried to take your advice and watch the video, follow one molecule and watch it evaporate, I couldn't see it. I didn't see any molecules actually disappear. Regardless, this explanation makes so much sense to me now. Does this mean that each water molecule is technically a different temperature? Does the energy of that molecule directly reflect the temperature of that particular molecule?

-W.M.

How hot would the space jumper get?

rallain — Wed, 19 May 2010 12:56:30 +0000

How hot would the space jumper get?

A new video from the Red Bull Stratos Jump guys came out. Here it is:

This reminds me of an unanswered question about the Stratos jump that I didn't address on my last post on this topic. Commenter Long Drop asked about how much Felix would heat up as he falls from 120,000 feet. This is a great question. The first, off the bat answer is that he won't heat up too much. Why do I say this? Well, when Joe Kittinger jumped from over 100,000 feet and didn't melt. Still, this is a great thing to calculate.

How do you calculate something like this? I will look at this in terms of energy. For simplicity, I will consider the jump from 120,000 feet down to about 30,000 feet. After that, Felix will pretty much be a normal sky diver. Here is the plot of speed vs. height from my previous post.

Just a note, the green line is the speed of sound, the red line is his speed if he jumped from 100,000 feet and the blue is from 120,000 feet. Think about this fall. If there were no air resistance, he would be going much faster and would have much more kinetic energy. So, without air resistance I could use the work-energy principle. If the Earth and the jumper are in the system, then:

But with air resistance, the jumper will not actually be going that fast with that much kinetic energy. So the missing energy had to go into an increase in thermal energy. This increase in thermal goes both into heating up the air and the jumper. But, how much goes into the air and how much into the jumper? I am just going to make a basic assumption that half of the energy goes into the air and half into the jumper. Simple, right? Now I just need to re run my numerical calculation and get the difference between the no air kinetic energy and with air kinetic energy. Here is a plot of kinetic energy vs. height (both with and without air resistance).

From this, I get the values:

That seems like a lot, even if only half of that went to the jumper. Instead of calculating the change in temperature, let me think about this in terms of power. That can give me the change in thermal energy, but how long did it take? From the numerical calculation, falling to 30,000 feet takes about 150 seconds. This would give an average power (so I could compare to an electric heater) of:

Still not very good. I just can't imagine having a 70,000 watt heater hooked up to you for even 2 minutes. Maybe the time is so short, it doesn't matter. Here is an idea. What if I do the same thing for a normal sky diver? Let me assume a sky diver jumps from 10,000 feet to 3,000 feet falling at 120 mph (constant the whole way for simplicity). This is a little bit simpler. I can calculate the change in gravitational potential energy for the fall and compare it to the kinetic energy of a dude going 120 mph.

Assuming the fall is at 120 mph, this would take about 40 seconds. The power for this case would be about 19,000 watts. Ok. I guess the stratos jump isn't too bad. Yes, it is more - but not way out of this range. So, maybe the jumper will get a little hotter - but he does have a space suit on.

One - one more thing

I asked the Red Bull Space jump guys for acceleration data when Felix actually does the jump, but I never officially heard back from them. Red Bull, if you read this - please?

rallain Wed, 05/19/2010 - 08:56

Tags

Google tells me that a joule is 0.239 calories, and one calorie will heat one gram of liquid water by 1 K. So if half of that energy goes to the jumper--lets call it 10^7 J for a 100 kg man. That's 2.39*10^6 calories going into 10^5 g of jumper, raising his average temperature from 37 C to 61 C if he's not wearing a space suit. (In reality, the leading edge of his body could get quite a bit hotter.) So he'd better be wearing that space suit, or else he will be toast.

How about a "space dive" from an orbiting vehicle? Say from the ISS.

At a mere 40 km, is the air really thin enough for a human body to hit Mach One? I would expect the wave drag to put a pretty serious limit on velocity at that point.

Secondly, I suspect that your 50/50 distribution of energy is pretty far off. First, for the near-sonic portions of the fall the compression heating of the air would be significant. Secondly, although the energy of impact between air molecules and the body would be distributed more or less evenly, there is also heat transfer from the body to the air -- and that air is cold. What's the wind chill of -40 C air at 300 m/s?

@D.C. Sessions,

I agree that maybe the 50/50 thing is a bit off. Look at a normal sky diver, they don't really get hot. It was just my first guess.

Re: Orbital Jump(#2)

Jumping from orbit is a little more complicated. You are moving at least 15,000 mph parallel to the ground. You will need a small rocket to change your orbit enough to intersect the atmosphere. When you hit the atmosphere, you will burn up unless you have a heat shield or tiles(i.e. shuttle) to convert all that velocity to heat that doesn't kill you.

An alternative would be a bigger rocket to dump all your orbital velocity, discard the rocket, and then fall with zero velocity relative to the ground. Would a small 3rd stage solid rocket give enough delta-V to kill your orbital velocity? Add in the control system also, unless manual control would be part of the challenge :-)

Tedd

It was just my first guess.

I would expect it to be a very good one from first principles. The problem is that there are other heat transfer processes which seem likely to dominate.

The good news is that it shouldn't be all that hard to calculate the loss rates. Fun assignment for students.

here is an article i just stumbled across today:

American Journal of Physics

"High-altitude free fall revised"

Jan Benackaa
Faculty of Natural Sciences, Constantine the Philosopher University, Tr. A Hlinku 1, SK-94974 Nitra,Slovakia
Received 24 September 2009; accepted 4 January 2010

@rob,

Thanks for the article - I am reading it now. How do I miss these things?

here is another one you might have missed twisted one 151 weblog. it has to do with heating of a skydiver. the coincidences abound today!

http://twistedone151.wordpress.com/2010/05/14/physics-friday-119/

May I ask what software you use to create the equation images?

@Swulf,

I use LaTeXit - http://chachatelier.fr/programmation/latexit_en.php

It is a small latex equation editor for mac. I then take screen shots of the equations. I know there are probably better ways, but this is quick for me. Also, you can drag vector graphic equations into Keynote for pictures.

MythBusters' energy explanation

rallain — Thu, 06 May 2010 16:32:57 +0000

MythBusters' energy explanation

I already mentioned the MythBusters' crashing two cars episode where they correctly doubled the speed of a pendulum type object. Overall, this was a very visual (although expensive) demo. There was one part that left a sour taste in my mouth - the final explanation from the narrator. First, they showed this.

And then they had an explanation that went something very similar to to this (after restating what the sign above said)

"Although the two-car crash doubles the speed, the energy the crash is transferred to twice the mass resulting in a crash that looks like just one car hitting a wall at 50 mph."

Here is the graphic that went with that.

I had to re-listen to this narration a couple of times because something seemed not quite right. First, let me comment on the last diagram. Really, maybe it would have been better to leave this off. It doesn't really add any useful explanation other than to point out that the kinetic energy is dependent on the square of the velocity. And what about the narrative? I think what bothered me is that they said the two-car crash doubles the speed, but what they probably should have said is "the two car crash doubles the kinetic energy and this energy is spread out over 2 cars." Using their statement, you would say "oh, double speed means 4 times as much energy."

Here is the real question: what concept are you trying to get across? Newton's third law? The idea of kinetic energy? Conservation of momentum? I would just pick one and stick with it. Otherwise, you are kinda implying that "action and reaction" have something to do with kinetic energy. Oh, have I ever mentioned how much I hate "action and reaction" explanation of Newton's 3rd law? Action? Reaction? How about this for Newton's third law:

Forces are an interaction between two objects. Forces come in pairs. Or, if you must: For every force there is an equal and opposite force.

Oh sure - you can make the action reaction thing work, but it can also cause problems. Either way, I would suggest sticking with the energy explanation. Here, I want to help. I am going to give an energy explanation that the MythBusters could use and a force explanation for the two-car collision.

Two-car crash, energy explanation

Why are two cars crashing into each other not the same as one car going into a wall at twice the speed?

Explanation: In terms of energy, the energy of motion is called kinetic energy. Kinetic energy depends on the square of the velocity. This means a car moving at twice the speed with have 4 times the kinetic energy.

So 10 + 10 is not the same as 40.

Two-car crash, force explanation

This is a little more complex, but I will try to make it simple. First, two key points:

Forces are an interaction between two objects. Object 1 pushes on object 2 the same as object 2 pushes on object 1 (same interaction).
A force on an object changes the object's momentum where momentum is mass times velocity.

Suppose a car crashes into a wall with a velocity v. While it is interacting with the wall, the wall exerts a force (F) on the car and the car exert a force F on the wall.

Where the force the wall exerts on the car and the car on the wall have the same magnitude. Now, what if I replace the wall with another identical car traveling at the same speed?

Since the initial momentums are the same and the forces are same, the effects are the same on the two cars. So, two cars are the same as one car into a wall. If I now double the speed, I will have different initial momentum, so it will not be the same.

Other stuff

I actually forgot that I have talked about the MythBusters colliding two cars before. I wrote this post the first time they did this myth. That is a little more detail than I have here. Also, a similar thing came up when the MythBusters tried to pull two phone books apart. Here is my discussion of forces in that situation.

One more thing. I would like to emphasize how awesome this demonstration was. You hear people discussing things just like this all the time, but no one actually does it. For many physicists, the actual experiment doesn't mean much. However, to many people this experiment is important. They just need to fix their final explanation (call me next time and I will be glad to help).

rallain Thu, 05/06/2010 - 12:32

Tags

I can't wrap my brain around how Newton's 3rd law logically proves Jamie's hypothesis wrong.

Doubling the mass (action) results in what type of reaction? (Sorry for using those terms.)

@C. Felix,

I am not sure I proved he was wrong, maybe instead I just showed that two cars colliding is the same as 1 into a wall at the same speed. If that works, then I guess it follows that double the speed would not be the same.

I started to think about this in terms of reference frames and this really makes me believe that energy is the wrong way to explain this problem to the average person.

If you look at the proper reference frame of the orange car, because the event is happening to it. The person in the car would see a yellow car with E=40 hitting them or a wall with E=40 hitting them.

Am I thinking about the reference frame incorrectly?

Got this idea from Al Guenther: to teach newton's third, cut a plastic pear in half and put a force vector in each half. HA! Forces come in pears. Thought you might like that if you hadn't seen it before.

@Matt,

Great idea. How about this modification. Make a plastic arrow for a vector. Take a real pear and embed this plastic arrow inside. Then when you cut it open in class you can say vectors come in pears.

Ummm... Isn't the velocity in the kinetic energy equation suppose to the relative velocity between two objects, not relative to a fixed point in space or the point of the collision. Isn't the relative velocity between two cars traveling at 50 mph the same as one car travelling toward a wall at 100 mph.

Me thinks a few mistakes were made.

@Evan,

Kinetic energy is not independent of reference frame. If you move to the reference frame of one of the cars, you would see a different total KE before the collision. What you would agree on is the changes in energy during the collision.

My statement was that the v in kinetic energy equation is based on relative velocity. Relative velocity is just that, Relative. Using the earth as a universal reference is a very common mistake. Each object's kinetic energy relative to the earth is irrelevant. We only care about the kinetic energy relative to the objects in the experiment. We're forced to choose one of the objects as a frame of reference. With the wall the choice is obvious. We have to use the wall as a frame of reference. With two cars travelling towards each other we have to pick one of the cars as our frame of reference and consider that car stationary. Thus, the relative velocity is the same between the two experiments. If the mass of each car is the same and the relative velocity is the same, then the kinetic energy is the same. It's like saying 1/2m(100)^2 = 1/2m(100)^2.

Basically, I'm saying your explanation using kinetic energy is wrong because you calculated kinetic energy relative to the earth and not relative the objects in the experiment.

I'm also saying that a car hitting the wall at 100 mph hit that wall with the same kinetic energy as two cars hitting head on at 50 mph. Maybe not relative to the earth, but relative to one of the cars.

Because kinetic energy is dependent on frame of reference, it's absolutely critical that you choose a valid frame of reference.

But nobody has addressed deformation, which Adam and Jamie used *as their measurement* of impact (along with G-forces i.e. deceleration). Of course the clay between the weights on two moving hammers squished only as much as the clay in one hammer did... there was twice as much total clay to absorb the impact. Likewise, when one moving car hits the wall, the wall shows no deformation - only the car is squished. When two cars hit with the same relative velocity as in car vs wall, *both* cars get deformed - so each individual car deforms less.

@Pteryxx

I agree 100%. The difference in damage had more to do with elasticity then it did with force. Elasticity was not constant as the wall was far more elastic then the cars. The amount of force due to relative kinetic energy was the same for the 2 cars traveling at 50 and the one car traveling at 100. The difference was the one car traveling at 100 was the only object in the collision that could absorb energy through deformation. Thus it was damaged far more. The energy had to go somewhere.

I agree with Rhett that the action-is-minus-reaction is a particularly awkward way to formulate that law of motion.

Whatever Newton's reasons were to formulate the third law in that form, a far superior form is present in the Principia. In the Principia, the three laws are followed by corrollaries, and the fourth corrolary states the third law in dynamical form:

"The common center of mass of two or more objects does not alter its state of motion or rest by the actions of the bodies among themselves."

In retrospect we would have been in a far better situation if Newton would have asserted the third law in that dynamical form. (In fact many textbooks do present that dynamical form as the third law.)

I can't resist fantasizing what things would be like if that dynamical form would be the canonical form. Then that whacky "action-is-minus-reaction" thing would not be there to confuse the guys who write the Mythbuster narrator's script.

Going back once more to the dynamical form:
"The common center of mass of two or more objects does not alter its state of motion or rest by the actions of the bodies among themselves."

This form emphasizes the mutualness, and it is uncommitted as to what force is. It's a statement about motion, not a statement about force. For better education that "action-is-minus-reaction" thing ought to be phased out, and replaced with the dynamical form.

You write: "The two-car crash doubles the kinetic energy and this energy is spread out over 2 cars." I think this is the clearest explanation for why 2 cars hitting each other @ 50 mph might do the same amount of damage (to Car #1) as would Car #1 hitting a stationary wall @ 50 mph.

For me, the tricky part is convincing myself why, in the wall case, we don't need to worry about some of that initial KE getting transferred to the wall instead of going into damaging Car #1. One model that seems to work is if you think of the wall as a spring with spring constant k >> F_max/L, where F_max is the max force w/ which the car pushes back while being crushed, and where L is a characteristic length scale in the process of crushing the car (F_max*L ~ energy that goes into damaging Car #1 during crash).

Then, in the two-car collision, where each car has initial speed v, all the initial KE is shared equally between Car #1 and Car #2 during the crash:

2 * 1/2 m v^2 = 2 * (Energy that goes into damaging Car #1),

=> Energy that goes into damaging Car #1 = 1/2 m v^2.

But in the Car#1-hitting-wall collision, where Car #1 again has initial speed v, the initial KE is shared unequally between Car #1 and wall during the crash:

1/2 m v^2 = E_wall + (Energy that goes into damaging Car #1)

Now, IF the wall acts like an ideal spring w/ k >> F_max/L, and if we can assume the wall/spring ends up compressed by a (tiny) distance x, which is how far it had to compress to push back on the car with the required crushing force F_max, we have

E_wall = 1/2 k x^2
= 1/2 k (F_max/k)^2
= F_max^2 / (2k)
= F_max*L / (2 kappa)
= O(1/kappa)

where kappa = kL/F_max >> 1 is the dimensionless large parameter in our approximation. And recall that, by definition,

Energy that goes into damaging Car#1 ~ F_max*L
= O(1);

that is, in the limit kappa->infinity, E_wall is negligible in comparison with the energy that goes into damaging Car #1. Going back to our energy-transfer eqn for car-hitting-wall, and taking the 1st term on the RHS to zero, we thus have

Energy that goes into damaging Car #1 ~= 1/2 m v^2,

which is the same amount of energy that went into damaging Car #1 in the two-car collision when *both* cars had initial speed v.

But is it accurate to treat the wall like a large-k spring in this manner? I don't know!

Re the comments about reference frames: Remember that the COM reference frames of the cars are NONinertial. (As the cars get crushed, they decelerate.) So the physics can't be expected to work properly in those frames. From his own COM frame, Car #1 thinks there's twice as much energy to go around than there actually is (1/2 m*(2v)^2 = 2 mv^2 instead of 2* 1/2 mv^2 = mv^2), and he'll therefore be confused about why there isn't more damage to the two vehicles.

@Carol,

I am not ignoring your comments, I am just taking time to think of a response. Sorry for the delay.

1. One almost always has to make simplifying assumptions when "modeling".

2. One must "model" in order to keep the math reasonable.

3. Sometimes the real world answer is lost in the over simplifications of the models.

4. In this case, I think the key is to ignore the crumple factor, focus on the choosing the inertial reference frame for each case and let the kinetic energy tell the tale.

They should have made a wall traveling at 50mph and a car traveling at 50mph hit each other.

The energy argument is wrong, by the way. The momentum one is correct. The problem with the energy argument is that you have all assumed that there is such a thing as "absolute kinetic energy". Unfortunatley, that does not exist. Only changes in energy are meaningful - this is 1st year physics. I'll give you an example. Say you are in a train moving at speed v (not accelerating) and the windows are blacked out. You would say that you have zero kinetic energy. Now, and observer on the ground says that you have 1/2 m v^2 kinetic energy. Who is right? It's meaningless since the notion of absolute KE is just as meaningless as absolute potential energy (PE). If you considered changes in energy, then things would be more meaningful.

Finally, this is an in-elastic collision, so what you can apply is conservation of momentum. Mechanical energy is not conserved in this collision.

Summary: Newton's third law, and the cons. of momentum explain this result. That is all you need. This whole issue of different frames of reference (cars frame, ground frame) is a red herring.

I have a queston if there were to eople in the wouds how would last longer the skiny one or the fat one ??????

MythBusters and double the speed

rallain — Thu, 06 May 2010 10:51:47 +0000

MythBusters and double the speed

In the last episode of MythBusters, Adam and Jamie wanted to test something that Jamie had said earlier:

"Two cars crashing head on at 50 mph is the same as one car crashing into a wall at 100 mph"

Jamie was wrong, but that is not what I am going to talk about. Instead, I am going to talk about Adam's small scale test of this situation. Really, it was a nice set up. Basically, he wanted to collide something into a wall at one speed and then double that speed. Then he was going to collide two things together at the lower speed. He had a cool way of measuring the collision. He put a piece of clay between two masses. When the object collided with something, the clay would get smashed and he could measure how smashed it got. Here is a simple diagram of his apparatus.

Where would you have to put the first object so that the second position is twice as fast?

How fast as a function of angle?

Since these objects are not moving in straight lines, they actually don't have constant accelerations. I know the distance they travel though, so that hints that I should use the work-energy principle. In short, this says that if I look at the mass plus the Earth as my system, there will be no change in total energy. (Gravity won't do any work because it is an internal force to my system). This means:

Going from Position 1 to the bottom (which I will call position 0), I can write this change in energy equation. For simplicity, I will call the bottom of the motion the zero gravitational potential energy. This gives:

Notice that the mass cancels - that is a good thing. Also, in order to double the velocity at the bottom, y₂ would have to be 4 times higher than y₁. Or, if you want to work backwards. Suppose you put the one mass at the horizontal position where position 2 is. What angle would position 1 be at? Here is the expression for the velocity from position 2:

Now, how high would y₁ be so that v_0-1 is half of that value?

If height is determined by an angle measured from the vertical axis, then (and the radius of the circle is R):

Ok, I admit. I first looked at the video and thought Adam had put the position 1 at 45 degrees (which wouldn't even be half the height). I was wrong. Here is a screen shot of the set up.

See the mark that says "49"? That must be the angle as measured from the top. Hats off to you, MythBusters.

rallain Thu, 05/06/2010 - 06:51

Tags

So what is it you're trying to say exactly?

rijkswaanvijand, Mythbusters frequently gets criticism from various people for not getting the physics right, either through misuse of terms (force, velocity, and acceleration, mostly) or through simple mistakes like they avoided above. Most people would have simply put the 1x height at 45 degrees instead of the correct one as calculated. It looks like Rhett expected them to as well, and was pleasantly surprised when they didn't.

I'm just glad to see this because I noticed the "49" while watching, but didn't understand why it was done that way. Now I know.

Rhett, what took you so long? That episode aired 17 hours ago!! (haha). I was thinking exactly of the 45 degree angle, and your impending dissection of the physics, as I watched it last night.

It makes me wonder why they don't bother to have a physics consultant on the show's production team -- and if Grant is really an electrical engineering grad, then shouldn't he be catching some of this too?! (or, as a civil engineer myself, I should realize that the EE's didn't really spend too much time focusing on kinetics and kinematics!).

Ok, I'm going to ask a stupid question and really display my ignorance. But what the heck.

When Adam did the clay test with 2 objects at speed X, he put one piece of clay in each object. Each clay piece showed the same compression as 1 clay object hitting a wall at X speed. That made sense to me even thought I didn't know the formulas.

But I yelled out to my wife that he did the test wrong. If he had put 1 piece of clay between the two objects, wouldn't that have compressed as much as the clay in the 2X speed test? That piece of clay would have absorbed an X mile an hour crash on each side. Wouldn't that be the same as one 2X crash?

That's how I've always imagined this thought experiment.

If I'm standing still in the middle of an intersection, wouldn't getting hit by one car traveling 50 miles an hour be equivalent to getting sandwiched between 2 cars going 25? Assuming that I'm super-glued in the spot and can't move from the impact.

Or am I just ignorant? Hopefully I'm not stupid as well.

@Bob,

I was going to talk about the clay, but I decided not to include it. In short, you are right in that you need to think about energy. If the clay exerted a constant force while it was being compressed, then the compression amount would be proportional to the amount of work the clay did on the mass.

@Bob "If I'm standing still in the middle of an intersection, wouldn't getting hit by one car traveling 50 miles an hour be equivalent to getting sandwiched between 2 cars going 25? Assuming that I'm super-glued in the spot and can't move from the impact."

Superglued to the spot might be more like being sandwiched against a bridge abutment. The trick in thinking about how much you'd get squished is to think about the energy you'd have to dissipate. The energy in 2 25mph cars hitting you, one on either side of you, would only be 1/2 of the kinetic energy (E=.5mv^2) you'd have to absorb from single 50mph car squishing you against wall.

I have to admit I guessed the outcome wrong too. I had figured double the energy, energy goes by speed squared, same as 70 miles an hour into the wall. But the doubled energy is being distributed into two cars, so it doesn't make a difference.

I'm confused. Regarding:

"Two cars crashing head on at 50 mph is the same as one car crashing into a wall at 100 mph"

Why is that wrong?

I mean, sure, it'll be wrong if you stick a compressible object of non-negligible mass at 0 mph between the colliding objects (maybe like a lump of clay?). But that hardly seems to be what's implied by the original statement.

If there's nothing but two identical cars and you're measuring the damage done to the car, shouldn't the two situations be the same? (Assuming an incompressible wall, neglecting friction with the ground, yadda yadda yadda.)

Whoops. My bad.

Somehow I parsed the original statement as "Two cars crashing head on at 50 mph each (in opposite directions) is the same as one car crashing into an incompressible wall at 50 mph."

Nevermind.

Dude, what is up with the tiny type?

@Art,

Oops - I missed a close tag for my subscript - I fixed.

Thanks for pointing that out.

"and if Grant is really an electrical engineering grad, then shouldn't he be catching some of this too?!"

Adam/Jaime work at M5 - Jaime's original work shop. Kari/Tory/Grant work down the street at M7 (another building) which would make it difficult to know the minutiae of what Jaime/Adam are doing. Plus, film schedules seem to be tight enough that he doesn't have time to worry about them either.

BUT WAIT!!!

I watched the episode and did the same analysis with the 49 vs 41 degrees. But I also did a screenshot of the two-pendulum (head-on) collision and realized that Adam measured 49 degrees from the bottom, not the top. (I simply held the corner of a piece of paper up to the vertex, and the combined angle was greater than 90 degrees, so each must have been greater than 41, i.e. 49.

So, despite getting everything calculated right, he measured the angle from the wrong side and lifted the pendulum to 49 degrees from the vertical instead of only 41 degrees.

Oops.

so "Two cars crashing head on at 50 mph is the same as one car crashing into a wall at 100 mph" is a correct statement or not?

Some People Running

rallain — Mon, 03 May 2010 13:40:27 +0000

Some People Running

This video seems like it is getting popular, but maybe that is because it is so awesome.

Maybe it is just me, but I find this video very visually satisfying. I love the way they compare the different runners.

Anyway, there is some physics here. Commenter Ben sent me the link to this video (thanks!) with some great questions. Which of the runners has a greater kinetic energy? What about the power? These aren't too difficult to answer, but the first thing is to get the data. There are several options (including just using a stop watch). But no, that is not good enough for me. Instead, I used my favorite free, multi-platform video analysis tool Tracker Video.

It turns out that there were a couple of problems analyzing this video. First, it was tedious. Second, there is some perspective error as the camera pans along the motion. To fix this, I broke the video into three segments with different coordinate systems and then kind of "stitched" them together. The video above shows several people running, I only plotted the "normal" guy, Jacoby Ford, and Terrance Cody. Here is what I get using LoggerPro.

I went ahead and fit a linear function for these three runners for the portion that they were at a constantish speed. If you can't read the image, I have:

Normal dude: 7.3 m/s (16.3 mph)
Jacoby Ford: 11.5 m/s (25.7 mph) (damn that is fast)
Terrance Cody: 8.4 m/s (18.8 mph)

To calculate the kinetic energy, I will need the masses. Let me use 75 kg for the normal dude. Cody has a wikipedia page that lists his mass at 161 kg. Jacoby's wikipedia page gives a mass of 84 kg. So, here is a plot of kinetic energy as a function of time.

Note: since I used Logger Pro to take the derivative of the position to get the speed, I also used the "smoothAve" function to smooth the data out. Looking at the graph, it is interesting that Jacboy and Cody have similar kinetic energies. I am not sure why (or really if) Jacoby is slowing down. When you look at the position time graph for Jacoby, it doesn't look like he is slowing down that much. Either way, these two athletes have some serious energy compared to the normal guy.

What about power? Power is difficult with running people. The problem is that there are two things going on. First, there is some air resistance. Second, when you are running at a constant speed, it is not the same as moving a particle at a constant speed. You have to keep accelerating your legs in order to maintain this motion. So, when running and speeding up, you have to do three things: increase your kinetic energy, have a greater rate at which you change the kinetic energy of your legs, and last fight against air resistance.

I am going to have to come back to running in a future post because it is quite interesting. However, for now I will calculate the power just due to change in kinetic energy. In a short time interval, think of power as:

So, this would be the slope of the above KE plot. I know I already "smoothed" it once, but this will give an idea of the power of these people.

First, negative power. That is not negative power. That means the rate of energy going into KE is going down. There is still energy going into moving the legs and air resistance. Second, are these ok for power of a human? Well, it is not 56,000 watts (ESPN Sport Science) - and that is a good, right? This wikipedia page lists the sprinting output of a cycler at about 2000 watts. So, I don't see anything crazy in this graph.

rallain Mon, 05/03/2010 - 09:40

Tags

Nice, that is some well put together analysis. And I agree the video is just great, for reasons I can't explain.

In terms of power, there are a couple of ways to look at it. Many years ago I wondered why I could ride a bicycle at 10 m.p.h. for hours on end but running at 10 m.p.h for more than a very short time would be nearly impossible (for me). After all, bike + me weighs more and presents at least as much area to the wind as me alone. But clearly, it's the fact that I'm continuously lifting myself up and then coming back down when I run. So if you look at power required to accomplish it (i.e., food energy burned*efficiency of turning that into muscular activity) running at a given speed uses much more power. Whereas if you look at energy used to do "useful" work (i.e., moving forward), not so much.

@Rob,

I think you hit it - that is the key. So, it couldn't just be air resistance or change in KE of the whole body (or bike). It must be the change in motion of the legs and the change in center of mass of the body (you don't move up and down while riding a bike).

This is a topic I will certainly be coming back to.

I think the biggest factors in bike vs. run are a) that if you stop expending energy while biking, you decelerate slowly due to bearing/ground friction and air resistance, whereas if you stop expending energy while running you stop; b) the limit on the speed of legs when running vs. the ability to match optimum speed of the legs to the ground speed through the gearing and drive wheel.

What could you do with 54,000 watts?

rallain — Tue, 23 Mar 2010 22:43:18 +0000

What could you do with 54,000 watts?

I already looked at ESPN's Sport Science episode where they calculate that Marshawn Lynch produces 54,000 watts when pulling some tires. Yes, that is way too high. However, what would happen if some was actually that powerful? What could that person do? How fast could they run 100 meters? That is what I am going to calculate.

First, I am going to assume that Marshawn has a mass of about 100 kg. Also, let me say that he can produce 54,000 watts no matter what his speed.

Take a short time interval. During this time, Marshawn will increase his speed from say v₁ to v₂ this would be a change in energy of:

And this would relate to the power by:

So, if I know this small time interval and the velocity he started at (at the beginning of the interval) then I can find the final velocity:

If the time interval is short, then the velocity is essentially constant (for very short time intervals) so that I can use the average velocity to write:

You see where I am going don't you? This is all set up for a numerical calculation. Here it is - I made it as simple as I could:

I changed my mind. Instead of using the average velocity to find the new position, I just used the velocity. Trust me, it is ok. Here - you can check. One good way of checking your calculations is to make the time interval (dt in this case) smaller and see if you get the same result.

So, what do I get. Here is a plot of the speed as a function of time:

There you go - 100 meter dash in under 3 seconds. Take that Usain Bolt. Note that Usain not only has a cool name (Bolt) but has the world record at 9.58 seconds. Another note - I just noticed that lists the wind speed for these records. Boom. That is another blog post.

Not only would 54,000 watts give you a 100 meter time under 3 seconds, you would be going over 50 m/s. Yes, that is like 120 mph.

How about another check. What if I put in a more reasonable power of 2000 watts? Here is what I get:

Seems better, doesn't it? Still a world-record time, but I did not take into account air resistance and I assumed the power would be constant. Oh, also that would give a speed of 40 mph - so that isn't quite right.

rallain Tue, 03/23/2010 - 18:43

Tags

The caption of the second plot should be Speed of 2000 Watts dude, no?

@anon,

Thanks for catching that - I fixed the title.

"What could you do with 54,000 watts?"

Produce one research paper on AGW?

Hmmm... On a quick glance there would seem to be a problem. 54000 watts is about 72 horsepower. Most cars have more than this and do not accelerate to 120 m.p.h. in the space of 100 yards.

Being at work and all, I haven't had time to think through whether there's an error in your calculations, your assumptions or my comparison.

@Rob,

Cars don't have a mass of 100 kg.

*smacks forehead* duh...

The ability of human muscle to produce power is very velocity dependent (obviously, otherwise everyone would ride 1-speed bikes).

Which is why the ability to apply 2 kW continuously might get you a world record time DESPITE the ability of sprinters to produce well over 2 kW of power. I think in a previous comment I mentioned that Usain Bolt's split times demonstrate a peak mechanical power production over 4 kW in his early acceleration. But he doesn't have gearing on his legs, so biomechanical inefficiencies at high velocities (and wind resistance) prevent him from reaching your times.

I still agree 54 kW is baloney.