# Homework Help: Having trouble with Einstein's Derivation of E=mc^2

1. Dec 27, 2009

### jwdink

Also posted this in the SR and GR forum, but wasn't sure which was more appropriate.

Hi, I'm a non-physics student trying to use primarily Einstein's original 1905 essay to write a paper on the philosophical questions that E=mc^2 brings up. However, I've hit a snag which I can't seem to get past.

1. The problem statement, all variables and given/known data

Einstein essentially uses what he's proven about the Doppler shift to show that a moving body emits a more intense light flash than does a stationary one. Since the internal energy of the body remains the same, he asserts that the increased energy of light must have come from the kinetic energy of the body-- and since KE=1/2mv^2, and v didn't change, therefore m changed.

That is,

1) E(bef)=IE1 E(aft)=IE2

2) E'(bef)=IE1+KE1 E'(aft)=IE2+KE2

Subtracting 1) from 2) yields:

E'(bef-aft)-E(bef-aft)=ΔKE

My question: how can we be sure that the internal energy of the body does not change when the body moves? I suppose we could say that, "by definition, 'internal energy' means 'that energy which does not change in motion.'" But then we'd be questioning the equation for KE energy above-- we'd be saying that kinetic energy is 1/2mv^2 PLUS some other energy which doesn't necessarily relate to the mass, but increases with increased velocity. We could, I suppose, assert that the equation above is an approximation for very low velocities, but this runs us into trouble. Why? Well, because the Doppler shift only occurs at very high velocities! So while KE=1/mv^2 could be said to be an approximation of kinetic energy for low velocities, so could A'/A=1 be an approximation for the doppler shift of intensity at low velocity. If we want to say that KE only deviates from the classical equation by a negligible amount, we have to say that the light's energy only deviates a negligible amount as well.

2. Relevant equations

Attached.

3. The attempt at a solution

Attached.

Thanks!!

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2. Dec 28, 2009

### Andrew Mason

You have an interesting perspective on relativity.

Einstein's 1905 paper on inertia follows from his previous 1905 paper on special relativity. So he uses the Lorentz transformation for energy as measured in two different frames of reference and compares them.

It may be helpful if you approach the problem from the point of view of Newton's laws of motion and the conservation of momentum. E=mc^2 follows directly from the fact that light carries momentum p = E/c = $h\nu/c$. See: https://www.physicsforums.com/showpost.php?p=470117&postcount=11

AM

3. Dec 28, 2009

### jwdink

I should have said "original essays". I'm aware of his previous essay-- indeed, I wrote half my paper on it. That's why I don't want to use a different derivation of e=mc^2: all my work explaining the Maxwell equations, his reformulation of the Maxwell equations, how this leads to doppler shift, why this means that light will look more intense, etc. is for THIS original derivation.

4. Dec 28, 2009

### Andrew Mason

You are using "intensity" here when you probably should be using "frequency". Intensity of light is usually used to denote the number of photons/unit time. As the moving observer approaches the stationary source, the frequency of the light is doppler shifted upward, not the intensity (at least as the term "intensity" is normally used in relation to light).

I will have a closer look at your paper. I am not sure I understand your problem yet.

AM

5. Dec 28, 2009

### jwdink

If you look at the original Einstein paper "On the Electrodynamics of Moving Bodies," both the frequency AND the intensity get shifted upwards in the doppler shift. I think it has something to do with length contraction, ultimately. Or something. See http://www.fourmilab.ch/etexts/einstein/specrel/www/#SECTION22

"It follows from these results that to an observer approaching a source of light with the velocity c, this source of light must appear of infinite intensity."

I should mention that I think I figured out the problem. See the other thread I made on this topic: https://www.physicsforums.com/showthread.php?p=2508638#post2508638

EDIT: In which case, the draft you're reading has an old formulation, which I've redone in a much more satisfactory manner. See next post.

I'd love any extra input you can give, though.

6. Dec 28, 2009

### jwdink

Here's the reformulation:

7. Dec 29, 2009

### Andrew Mason

I don't think Einstein was saying this. Besides, whether the moving body emits a more 'intense' flash depends on whether the source is moving toward or away from the observer. If it is moving away, the light is red-shifted, meaning that to the moving observer the light has less energy/unit time. The point that Einstein makes is simply that there is a difference in the kinetic energy of the emitting system as measured by the moving observer. Since speed has not changed, the only explanation is that the mass of the emitting system has decreased by an amount $\Delta m = L/c^2$. This has nothing really to do with doppler shift.

Einstein notes that the difference in energy of the system (from before emission compared to after emission) in the stationary frame is L whereas in the moving frame it is $\gamma L$. However, he notes, the difference in the energy of the system (before emission) as measured in the two frames is just the kinetic energy of the system as measured in the moving frame (kinetic energy being 0 in the stationary frame). Similarly, after emission, the difference in energy of the system is the kinetic energy of the system measured in the moving frame. The difference in energy ($\gamma L - L$) is the difference in the before and after kinetic energies of the system as measured in the moving system. Since v has not changed, this difference represents a difference in the system's mass.

Einstein assumes conservation of energy. He makes no statement about the measure of the internal energy of the system is or what it consists of. He just says that the difference between the system's energy as measured in the stationary and moving frames is the kinetic energy of the system in the moving frame. The doppler shift has nothing to do with kinetic energy of the system.

AM

8. Dec 29, 2009

### jwdink

How are you talking about "doppler shift"? I was using it to refer to an altered frequency OR altered intensity, since they both happen. I'm confused about your contention.

Okay, say before emission the difference (in the body's energy) between the stationary and moving system is equal to KE(bef), while after emission the difference between systems equals KE(aft). This is Einstein's original argument. He then says that 1/2mv^2 is an approximation of KE. Since the velocity didn't change (the body emitted light from both ends) then the only thing that could've changed is the mass. That's his argument.

My contention can be looked at in two ways:

1) KE might not equal to 1/2mv^2, that is, a moving body might have internal sources of energy which increase at high velocities, and therefore could explain the extra energy of the light, as measured by one of the observers. Conservation of energy means that the extra light energy had to come from somewhere-- but under this explanation, it doesn't have to come from the mass, it comes from the extra energy the body possesses.
2) You could object that KE is defined as 1/2mv^2, but then I could just ask: how do you know that the difference between the energy content of a stationary body and that of a moving body is exactly the classical KE? (Exactly 1/2mv^2?) What if this is just a classical approximation? But if its an approximation, then the extra energy of emitted light could have come from the "negligible" variables that the classical approximation leaves out (such as some augmented internal energy, see first phrasing).

By saying that the difference between the moving and the stationary system body energy is equal to classical kinetic energy, he is tacitly making an assumption about the internal energy of the system: namely, that it stayed the same, so that the light's augmented intensity could not have come from it, but instead must have came from kinetic energy by way of the mass. But to assume that "the difference between the system's energy as measured in the stationary and moving frames is the kinetic energy of the system in the moving frame", AND to assume that KE=1/2mv^2-- these assumptions are necessary for his argument to work, but not necessarily justified.

EDIT:
In fact:

This is what I said:
This is what you said:
Which is pretty much the same thing. What are you objecting to?

Last edited: Dec 29, 2009
9. Dec 29, 2009

### Andrew Mason

All I am pointing out is that Einstein does not mention it. You seem to be attributing a significance to the doppler shift that Einstein was not.

At the end of his paper, Einstein says that the difference in the kinetic energy of the system, as measured by the moving observer, is $L(\gamma -1)$. He then says that this change in kinetic energy ($\frac{1}{2}L\frac{v^2}{c^2})$ can only be accounted for by a loss of mass, $\Delta m = L/c^2$

Internal energy sources are immaterial. The system obviously has the energy required to emit the light. Einstein does not say what this energy is. He doesn't care what it is. All his is interested in is the change in that total energy. What he says about it is that the total energy content changes only by the amount of energy that is released as light. He notes that kinetic energy is the difference in total energy as measured in the moving frame and the total energy as measured in the rest frame. (By definition that is kinetic energy). In the moving observer's frame of reference the system has kinetic energy equal to H-E which is the difference between the total energy as measured in the moving frame (H) and the total energy as measured in the rest frame (E).

I don't see where Einstein talks about this "extra" light energy. What "extra" light energy are you referring to?

All Einstein points out is that the moving observer measures a difference in total energy that is different (by $\Delta m = L/c^2$) than the energy measured by the stationary observer. The moving observer interprets this as a decrease in kinetic energy resulting from a loss of mass (since v is unchanged). The stationary observer does not measure a similar change in kinetic energy because it has 0 kinetic energy both before and after emitting the light.
No. Einstein assumes it comes from the energy the body possesses. He does not assume it comes from its mass. But he shows that, regardless of the source of energy, the moving observer will measure the system as having a change in mass (by measuring a change in kinetic energy of .5Lv^2/c^2 while v remains unchanged) as a result of emitting the light energy.

It is.

What Einstein is saying is that the total energy of the system (including its internal and kinetic energies) changes only by the energy of the light that is emitted. To the stationary system, this is L. To the moving system, it is $\gamma L$. He is also saying that the difference in total system energy between the moving and stationary observers is by definition the kinetic energy of the system as measured by the moving observer. He simply notes that there is a difference in the kinetic energy of the system as measured by the moving observer (ie. a difference in the kinetic energy before and the kinetic energy after the system emits the light).

AM

10. Dec 29, 2009

### jwdink

...which would, in itself, tell him NOTHING unless he thought that kinetic energy depends ONLY on mass and velocity. In that case, he could say that, since the velocity stayed the same, the decrease had to be a decrease in mass. But neither he, nor you, have shown why one ought to believe that kinetic energy ONLY depends on mass and velocity.

What? Then where does he get this equation?:

This is from the previous paper on Electrodynamics, where he talks about how the Doppler shift will affect frequency AND intensity of light:

I think you have an understanding of the paper that's not commensurable with mine. Perhaps you'd better understand my position if you read this:

http://www.mathpages.com/home/kmath600/kmath600.htm

Which interprets the paper in the same way as I do.

Last edited: Dec 29, 2009
11. Dec 29, 2009

### Andrew Mason

As a result of how it is defined, kinetic energy derives only from mass and speed.

Kinetic energy is defined as the amount of work that a body moving at speed v can perform by virtue of its motion. Such a body can apply a force = ma = mdv/dt over a distance s (or from speed = v to speed = 0) where:

$$W = \int_0^S \vec{F}\cdot d\vec{s} = \int_0^S -Fds = -\int_0^S F\frac{ds}{dt}dt = -\int_0^S m\frac{dv}{dt}\frac{ds}{dt}dt = -\int_0^S mdv(\frac{ds}{dt}) = -\int_v^0 mdv(v) = \frac{1}{2}mv^2$$

This is the Lorentz transformation for radiant energy. Since it includes light emitted in both directions, the doppler shifts are in both directions. It is greater by the factor $\gamma$ regardless of the direction in which the light moves. Indeed, it could be emitted vertically so that there is no doppler shift at all ($cos\phi = 0$). The energy of the vertically emitted radiation would still be greater when measured by the moving observer by the factor $\gamma$. For that reason, I don't see how doppler shift enters into the analysis here. The end result is that, with no doppler shift, the total radiation energy as measured in the moving observer's frame is greater (by the factor $\gamma$) than the radiation energy as measured in the moving observer's frame.

AM

12. Dec 29, 2009

### Andrew Mason

This paper provides an interesting perspective. For the reasons I have given above, I do not agree that E=mc^2 depends on the doppler shift. I don't think this paper suggests it does. Doppler shift is a directional phenomenon. What Einstein is relying on is the upward shift in frequency (energy) of light due to the relative motion regardless of the direction of the radiation. This is a shift (by the factor $\gamma$ that is independent of direction so it is not a doppler shift.

Ultimately, E=mc^2 is the result of the fact that electromagnetic radiation can cause matter to experience a force. This can be deduced without resort to relativity or to Lorentz transformations. It can be deduced from the simple assumption that momentum must be conserved AND from the fact that light takes a finite amount of time to move through space so the radiation reaction force of the emitter and that of the receiving mass do not occur at the same time. This necessarily results in a shift of the centre of mass between the emitting and receiving bodies. In order to conserve momentum, there has to be a transfer of mass between the emitting and receiving bodies (see the link in my first post in this thread).

AM

Last edited: Dec 29, 2009
13. Dec 29, 2009

### jwdink

But then we're back to the other way of formulating my objection:

"2) You could object that KE is defined as 1/2mv^2, but then I could just ask: how do you know that the difference between the energy content of a stationary body and that of a moving body is exactly the classical KE? (Exactly 1/2mv^2?)"

Which still stands. If KE=1/2mv^2, then I don't see why it's the case that "the difference in total system energy between the moving and stationary observers is by definition the kinetic energy [1/2mv^2] of the system as measured by the moving observer." Why does the difference in total energy between two bodies depend ONLY on our Newtonian approximations of how much work the moving body could do (KE)?

Say γ is the difference in what a moving and stationary observer measure as a body's energy content. It is certainly conceivable that γ could depend on more than just the mass of the body. Say we have an object with mass M=10 grams, and an observer moving with v=10m/s measures its energy content: he will say it can potentially do a certain amount of work, so its KE=500gm/s (sorry if those units are wrong). Now, imagine another body, with the same ostensible properties, but it also has some weird type of internal energy which increases with v. If IE is its internal energy when observed by a stationary observer, and IE' is its internal energy when observerd by a moving observer, then it's perfectly plausible that its energy content can be given the equation IE/IE'=1/(1+v/c). Then γ ("the difference in total system energy between the moving and stationary observers") would not by definition be 1/2mv^2, but instead would be IE'-IE+1/2mv^2. The amount of useful work it could do might still be 500, but the difference in observation between our two observers (γ) would be 500+(IE'-IE)

My contention is essentially that of the linked mathpages guy:

Except I think his response is insufficient.

Last edited by a moderator: Apr 24, 2017
14. Dec 29, 2009

### Andrew Mason

It doesn't depend on Newtonian approximations. The difference in total energy must be kinetic if the principle of relativity applies.

There are no changes to the inertial frames of reference here (the stationary system emitting the light and the observer moving at velocity v). The only change in energy is the change caused by the emission of radiation energy. There can be no other change in the total energy of the emitting system in either frame of reference. Otherwise the laws of physics would depend on a body's motion, which is prohibited by the relativity principle.

I have to stop you here. The principle of relativity, (which Einstein is using here since his E=mc^2 paper is an addendum to his earlier relativity paper) says that all inertial frames of reference are equivalent. There is no such thing as absolute motion. So, by that very principle, there cannot be any change to the laws of physics within each system that results from uniform relative motion. The difference in energy measurements between the two systems is, therefore, kinetic energy (a difference resulting from the relative motion of the two frames of reference) and cannot be internal energy.

Such a result would violate the principle of relativity. - see comment above. This touches an interesting and difficult area - the application of relativity to thermodynamics.

AM

15. Dec 29, 2009

### jwdink

Okay, now we're getting somewhere. (Pretty sure I'll soon figure out how and why I was horribly wrong, or I'll soon convince you.)

My hypothesis doesn't state that the internal energy of the body increases due to absolute motion, it states that a moving observer's measurements of the body's internal energy are greater than a stationary observer's measurements of the body's internal energy. Sorry for not being clearer about that. If there's nothing egregiously misguided about that assertion, then re-read what I said and see if that's a plausible contention.

...that wasn't my argument. My argument was that one of the frames of reference sees more emitted energy [than the other one], but this could be plausibly explained by saying that that frame of reference ALSO sees more internal energy (instead of that frame of reference extrapolating that the extra emitted energy came from the mass).

Last edited: Dec 29, 2009
16. Dec 29, 2009

### jwdink

Forgot to respond to this:

Well, Einstein adds a contrivance to take the directionality out, namely, his light source emits in opposite directions. But then he just plugs in the the equation for doppler intensity shift from the electrodynamics paper. This all seems odd if he's just working with the Lorentz transformation for radiant energy. I agree that either works, I just think that historically, this is how it was come up with. I could be wrong-- do you know the original paper in which he derived the Lorentz transformations for radiant energy?

17. Dec 29, 2009

### Andrew Mason

Previously you stated: "Now, imagine another body, with the same ostensible properties, but it also has some weird type of internal energy which increases with v." Now you are asserting that it doesn't really increase. It is just that the observer moving at constant relative speed v measures it to be greater than does the stationary observer.

All I can say is that Einstein allowed for that. He simply said let the total energy (internal + kinetic) of the system as measured by the moving observer be H0 before emission and H1 after (E0 and E1 as measured in the stationary system frame). He then said that the difference between the two frames is kinetic energy: K0 = H0-E0 (before emission) and K1=H1-E1 (after emission). The difference in kinetic energy due to emission of light is:

$$\Delta K = K1-K0 = (H1-E1) - (H0-E0) = (H1-H0) - (E1-E0) = L(\gamma -1)$$

Einstein never says what H1 and H0 are. All he says is the difference in the before and after total energy of the system as measured in the moving frame was H1-H0. He says that this must equal the energy of the radiation as measured in the moving frame (which we can determine from the Lorentz transformation): $H1-H0 = \gamma L$.

So if the internal energy of the system as measured by the moving observer was something 'weird' it doesn't matter. The only assumption is that the whatever it is, it changes only by the energy that was released.

Again, Einstein allowed for that. He was not concerned with how much more energy the moving observer sees. He was only concerned with the change in that energy. He applied conservation of energy and said that the difference was equal to the energy of the radiation emitted by the system. Since v doesn't change, the measurement of H1 differs from H0 only by the measured energy of the released radiation - and we know what that is.

AM

18. Dec 30, 2009

### jwdink

Not sure what you mean here; according to relativity, 1 and 2 are equivalent statements. My measurements in a stationary coordinate system of a moving body (with velocity v) are symmetrical with my measurements in a moving system (with velocity -v) of a stationary body. If an observer moving at v measures a stationary body to have more energy than it did while stationary, then a stationary observer will measure that moving body as having more energy than it did when it was stationary. In other words, it looks like it has more energy while its moving: it has some weird type of internal energy which increases with v.

But THAT is what I'm objecting to! If you assume their energy difference (before and after) is equal to 1/2mv2, then you are assuming H0 and H1 won't have some augmented internal energy. So, by stating that K0 = H0-E0 and K1=H1-E1, Einstein is tacitly asserting something about the internal energy (namely, that different observers don't measure it differently), even if he's not explicitly making any statements about it. So Einstein implicitly does NOT allow for the situation I described above, conceivable as it is.

Last edited: Dec 30, 2009
19. Dec 30, 2009

### Andrew Mason

I don't want to belabour the point. You had stated that the stationary emitting system had some kind of "weird" internal energy that increased with v. I inferred from that that it was an energy which most other systems did not have. In any event, it does not matter. Einstein allowed for it.

No. This is where I think you are going astray. H0 and H1 can have any kind of augmented internal energy. If it depends on v, then there is no change in that augmented energy since v does not change. All Einstein said was that the difference between H0 and H1 is the energy of the radiation that is emitted, as measured by the moving observer. Why would the actual value of H0 and H1 matter so long as we know what the difference is?

Actually, he is allowing for the situation you described above, as I explained in my previous post.

Let's say that H0 = IE + f(v) where IE is the internal energy measured in the rest frame of the emitter and f(v) is some arbitrary function of velocity. Then

$$H_0 = IE + f(v)$$
$$H_1 = IE + f(v) - \gamma L$$

$$H_0-H_1 = \gamma L$$

AM

Last edited: Dec 30, 2009
20. Dec 30, 2009

### jwdink

It seems like you are suggesting i'm objecting to gamma. But I'm not. H0-H1=energy emitted, and that's perfectly fine.

That gives him these steps, and ONLY these steps:

That is very clearly NOT what I'm objecting to. I'm objecting to these later steps:

Which, do NOT make the internal energy irrelevant, but-- for the presupposition that the difference in energy content to be equal to 1/2mv2 to work--depend on it being the same whether regarded from E or H. So you're right that your f(v) is irrelevant when talking about energy before and after emission in a given system, but its NOT irrelevant when asserting the difference between two systems at a given moment is classical KE. I'm not objecting to H0-H1=gamma, I'm objecting to K0 = H0-E0 and K1=H1-E1.

-----

This can be shown in great detail. IF we presuppose the difference in energy between the two systems is KE, then the following steps are valid:

E0TE=E0IE+E0KE (sta syst: total energybef = internal +kinetic)
E1TE=E1IE+E1KE (sta syst: total energyaft = internal +kinetic)

H0TE=H0IE+H0KE (mov syst: total energybef = internal +kinetic)
H1TE=H1IE+H1KE(mov syst: total energyaft = internal +kinetic)

Now we subtract these two rows.

1) IF KE is 0 in the stationary system, then E0KE=0 and E1KE=0.

2) IF E1IE=H1IE, and IF H0IE=E0IE

THEN and ONLY THEN is this true:

H1TE-E1TE=H1KE (Total energy of moving system(bef) - total energy of stationary system(bef) = 1/2m(b)v2)
H0TE-E0TE=E0KE (Total energy of moving system(aft) - total energy of stationary system(aft) = 1/2m(a)v2

i.e., the difference in energy of the two systems is kinetic-- based on step 2). Then we could say that the difference in these two kinetic energies is equal to the change in energy upon emission

The argument doesn't work if step 2) isn't true. Equations above become:

E0TE=E0IE(sta)+E0KE (sta syst: total energybef = internal(sta) +kinetic)
E1TE=E1IE(sta)+E1KE (sta syst: total energyaft = internal(sta) +kinetic)

H0TE=H0IE(mov)+H0KE (mov syst: total energybef = internal(mov) +kinetic)
H1TE=H1IE(mov)+H1KE(mov syst: total energyaft = internal(mov) +kinetic)

EnKE still is 0, but now, when we try and subtract, we get:

H1TE-E1TE=H1KE + (H1IE(mov)-E1IE(sta)) (TE of moving system(bef) - TE of stationary system(bef) = 1/2mv2+[diff in internal energy(bef) between two systems])
H0TE-E0TE=E0KE + (H0IE(mov)-E0IE(sta)) (TE of moving system(aft) - TE of stationary system(aft) = 1/2mv2+[diff in internal energy(aft) between two systems])

In this case, the kinetic energy might've changed upon emission, OR the difference in internal energy might've decreased.

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21. Dec 30, 2009

### Andrew Mason

I don't think you can say that:

$$H_0^{TE}=H_0^{IE} + H_0^{KE}$$

Einstein did not distinguish different forms of energy. He simply observed that the kinetic energy of the stationary system as measured by the moving observer was:

$$H_0^{KE} = H_0^{TE} - E_0^{TE}$$

You are trying to argue that $H_0^{KE} = H_0^{TE} - H_0^{IE}$ which, as you observe, would only be true if $E_0^{TE} = E_0^{IE}= H_0^{IE}$.

But the problem is that $H_0^{IE}$ is a measure by the moving observer of the internal energy of the stationary system. Any difference between $H_0^{IE}$ and $E_0^{IE}$ would appear to be kinetic energy - energy by virtue of the relative motion between the stationary and moving observer.

AM

22. Dec 30, 2009

### jwdink

That's true by definition. The total energy of a body (in a given perspective) is all of its internal energy plus all of its kinetic energy. What other energy would it have?

See, I'm pretty sure you're thinking about this backwards. You assert here that $H_0^{KE} = H_0^{TE} - E_0^{TE}[/tex] is obviously true, while [itex]H_0^{KE} = H_0^{TE} - H_0^{IE}$ isn't necessarily true. It seems to be clear that it's the other way around: I have no reason to believe that the difference in the body's total energy content from system to system will be its kinetic energy (where KE=1/2mv^2, mind you), while its inconceivable that the body's total energy content in a given system could be anything but its kinetic plus internal energy. How are you thinking of it?

Now you're playing fast and loose with the term "Kinetic Energy." You can say EITHER a) KE is by definition 1/2mv2, OR you can say b) KE is by definition energy by virtue of the relative motion between the stationary and moving observer. But you can't assert BOTH by definition. So prove to me ONE of these statements, and then I will be willing to accept the OTHER one by definition.

Last edited: Dec 30, 2009
23. Dec 31, 2009

### Andrew Mason

Essentially, if I understand you correctly, you are taking issue with this part of Einstein's analysis and, in particular, the additive constant C being the same before and after emission of the light:
All Einstein is saying here is (using your nomenclature):

$$H_0^{IE} = E_0^{IE} + C$$ and

$$H_0^{TE} = E_0^{IE} + C + H_0^{KE}$$

He could have used f(v) instead of C:

$$H_0^{TE} = E_0^{IE} + f(v) + H_0^{KE}$$ and

$$H_1^{TE} = E_1^{IE} + f(v) + H_1^{KE}$$

Since $E_0^{IE} = E_0^{TE}$ you have Einstein's result:

1. $$H_0^{TE} - E_0^{TE} = f(v) + H_0^{KE}$$

2. $$H_1^{TE} - E_1^{TE} = f(v) + H_1^{KE}$$

The result is (subtracting 2. from 1.):

$$(H_0^{TE} - E_0^{TE}) - (H_1^{TE} - E_1^{TE})= H_0^{KE} - H_1^{KE}$$

The LS is $L(\gamma -1)$ and the RS is $\Delta KE$ and you end up with L = mc^2.

AM

Last edited: Dec 31, 2009
24. Dec 31, 2009

### jwdink

Einstein's C is just a constant having to do with arbitrary zero point in units, if I understood it correctly. You can't use it here in place of f(v), because C stays constant before and after emission by definition. But whether f(v)'s value is variable depending on energy content, or whether m is variable depending on energy content-- this is what's under debate. You can't make f(v) by definition the same before and after emission. The point of this contention is that, instead of mass diminishing by L/c2, f(v)'s value does.

It might be better to think of f(v) as a sort of coefficient of the internal energy. Rewriting what you wrote:

$$H_0^{TE} = f(v)E_0^{IE} + H_0^{KE}$$ and

$$H_1^{TE} = f(v)E_1^{IE} + H_1^{KE}$$

Since $E_0^{IE} = E_0^{TE}$ and $E_1^{IE} = E_1^{TE}$, you have:

1. $$H_0^{TE} = f(v)E_0^{TE} + H_0^{KE}$$ and

2. $$H_1^{TE} = f(v)E_1^{TE} + H_1^{KE}$$

Which is much less amenable to proving that kinetic energy had to change, since, if its the case that $H_0^{TE} = f(v)E_0^{TE}$ and $H_1^{TE} = f(v)E_1^{TE}$, then there need be no change in kinetic energy at all. In other words, let's try and show what the difference between the total energy of the moving and stationary systems is equal to, as before. It's no longer simply kinetic energy. We subtract $E_0^{TE}$ and $E_1^{TE}$ from both sides of 1 and 2, respectively.

3. $$H_0^{TE}-E_0^{TE} = f(v)E_0^{TE}-E_0^{TE} + H_0^{KE}$$

4. $$H_1^{TE}-E_1^{TE} = f(v)E_1^{TE}-E_1^{TE} + H_1^{KE}$$

3.' $$H_0^{TE}-E_0^{TE} = (f(v)-1)E_0^{TE} + H_0^{KE}$$ [Simplified.]

4.' $$H_1^{TE}-E_1^{TE} = (f(v)-1)E_1^{TE} + H_1^{KE}$$ [Simplified.]

Now let's postulate that the kinetic energy didn't change, so that $H_0^{KE}=H_1^{KE}$. We subtract 3' and 4' from each other, much like before.

$$(H_1^{TE}-E_1^{TE})-(H_0^{TE}-E_0^{TE})=(f(v)-1)E_1^{TE}-(f(v)-1)E_0^{TE}$$

OR:

$$(H_1^{TE}-E_1^{TE})-(H_0^{TE}-E_0^{TE})=(f(v)E_1^{TE}-E_1^{TE})-(f(v)E_0^{TE}-E_0^{TE})$$

In plain english:

The body's energy content decreased upon emission MORE in the moving system than in the stationary. We'll call this value L/c2.

The body's internal energy was GREATER in the moving system than in the stationary. We'll call this value IE0 before emission, and IE1 after.

The above equation states that IE0-IE1=L/c2

So you see, if you allow for the internal energy to augment upon movement, it will not just cancel out, but can actually be the thing that decreases instead of mass. Since there's a little extra IE due to movement, it can account for the little extra energy light has due to movement.

Last edited: Dec 31, 2009
25. Jan 1, 2010

### Andrew Mason

I will study your post a little more. You seem to be saying that Einstein was wrong in asserting that the additive constant C was not the same after emission as before. However, your last sentence (quoted above) seems to violate the relativity principle. We know what the effect of movement has on the measurement of the energy of light. Light cannot have a different or 'a little extra' energy because of movement.

AM