The Relativistic Force-Norm and the Proof of E=mc2

[velo city]

What is the proof of Einsteins famous equation E = mc² ?

 

[dipstik]

There is a derivation:
https://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[Matterwave]

I believe this ancillary paper is a more direct proof of what you want: https://www.fourmilab.ch/etexts/einstein/E_mc2/www/

Einstein does use results from his previous paper though, so you might want to read both.

 

[russ_watters]

In science, "proof" generally refers to evidence. Nuclear energy and particle accelerators are two good pieces of evidence.

 

[bcrowell]

There are many different possible ways of proving this. For a critique of some of the technical issues in Einstein's original derivation, see Ohanian, "Einstein's E = mc^2 mistakes," http://arxiv.org/abs/0805.1400 . In my SR book http://www.lightandmatter.com/sr/ I give a recap of the Einstein argument in section 4.2, and resolutions of the issues Ohanian complains about in sections 4.4 and 9.2. It's hard to give a really rigorous treatment without using the stress-energy tensor, which hadn't been invented when Einstein wrote his 1905 proof.

 

[stevendaryl]

There is a heuristic derivation of the relativistic expressions for energy and momentum (not a rigorous proof) that goes something like this:

1. Assume that the energy of a particle is proportional to the mass and is a function of the magnitude of the velocity: $E = g(v) m$ (By "mass", I mean rest mass).
2. Assume that the momentum of a particle is proportional to the mass and is in the same direction as the velocity: $\vec{p} = f(v) m \vec{v}$. I'm also assuming that the unknown function $f(v)$ depends only on the magnitude of the velocity, and not it's direction.
3. Assume that the expression for momentum approaches the Newtonian result, $\vec{p} = m \vec{v}$ for low speeds. That implies that $f(0) = 1$
4. Assume that the expression for kinetic energy (total energy minus rest energy) approaches the Newtonian result: $m g(v) - m g(0) \Rightarrow 1/2 m v^2$
5. Assume that in a collision of particles, the energy and momentum are conserved.

With all those assumptions, we can prove that
$f(v) = \gamma = 1/\sqrt{1-v^2/c^2}$
$g(v) = \gamma m c^2$

To see this, consider an experiment where we collide two identical pieces of clay, each of mass $m$, so that they stick to form a single, larger chunk of clay of mass $M$. We look at the collision in three different rest frames:

1. Frame 1: Pick a frame in which one mass is traveling at velocity $-v$ in the y-direction, and the other at velocity $+v$ in the y-direction. Neither has any velocity in the x-direction.
2. Frame 2: Pick another frame whose spatial origin is moving at velocity $u$ in the x-direction, relative to the first frame. Assume that $u \ll v$.
3. Pick a third frame whose spatial origin is moving at velocity $u$ in the y direction. Again, choose $u \ll v$.

In Frame 1: By symmetry, afterward the composite mass $M$ is stationary. Conservation of energy tells us that $2 g(v) m = g(0) M$.

In Frame 2: The total momentum in the x-direction before the collision is approximately (ignoring higher-order terms in $u$) $-2 f(v) m u$. The total momentum afterward is approximately (using the Newtonian approximation, since $u$ is small): $-M u = -2 g(v) m u/g(0)$. Conservation of momentum in the x-direction gives: $f(v) = g(v)/g(0)$.

In Frame 3: One mass is moving at velocity $v_1 = \dfrac{v-u}{1-uv/c^2} \approx v - u (1-v^2/c^2)$ in the y-direction. (In the approximation, I did a Taylor expansion, and only kept the lowest order terms in $u$). The other mass is moving at velocity $v_2 = -\dfrac{v+u}{1+uv/c^2} \approx -v - u (1-v^2/c^2)$. After the collision, mass $M$ is moving with velocity $-u$ in the y-direction.

Since we are assuming that momentum has the form $p = f(v) m \vec{v}$, if we change frames so that the velocity changes slightly by $\delta v$, the momentum will change by an amoung $\delta p = m (\frac{d}{dv}(v f(v))) \delta v$ (ignoring higher-order terms in $\delta v$)

So the first particle will have momentum in the y-direction: $f(v) m v - m u (\frac{d}{dv}(v f(v))) (1-v^2/c^2)$

The second particle will have momentum in the y-direction: $-f(v) m v - m u (\frac{d}{dv}(v f(v))) (1-v^2/c^2)$

The total momentum before the collision in the y-direction is:

$-2mu (\frac{d}{dv}(v f(v))) (1-v^2/c^2)$

The momentum after the collision in the y-direction is:
$- Mu = -2mg(v)/g(0) u = -2m f(v) u$
(where I used the results from the other two frames).

So conservation of momentum in this frame gives:

$-2mu (\frac{d}{dv}(v f(v))) (1-v^2/c^2) = -2m f(v) u$

So we have a first-order differential equation for $f(v)$:

$(\frac{d}{dv}(v f(v))) (1-v^2/c^2) - f(v) = 0$

The unique solution satisfying $f(0) = 1$ is

$f(v) = 1/\sqrt{1-v^2/c^2}$

Now going back to $g(v)$. We showed that $g(v)/g(0) = f(v) = 1/\sqrt{1-v^2/c^2}$. Expanding for low-velocities gives: $g(v) = g(0) (1 + 1/2 v^2/c^2 + ...)$. So the kinetic energy at low velocities is $m (g(v) - g(0)) = m g(0) 1/2 v^2/c^2 + ...$. For this to agree with the Newtonian result, $g(0) = c^2$

So we have the two results:
$\vec{p} = 1/\sqrt{1-v^2/c^2} m \vec{v} = \gamma m \vec{v}$
$E = g(v) m = m g(0) f(v) = m c^2 1/\sqrt{1-v^2/c^2} = \gamma m c^2$

 

[bcrowell]

There is a heuristic derivation of the relativistic expressions for energy and momentum (not a rigorous proof) that goes something like this[...]

Since this is only for point particles, I think it's subject to the same criticisms that Ohanian makes of Einstein's 1905 derivation of E=mc².

 

[stevendaryl]

Since this is only for point particles, I think it's subject to the same criticisms that Ohanian makes of Einstein's 1905 derivation of E=mc².

Yeah, as I said, it's a heuristic derivation, not a proof.

 

[DrStupid]

[*]Assume that the energy of a particle is proportional to the mass and is a function of the magnitude of the velocity: $E = g(v) m$ (By "mass", I mean rest mass).

Is there a justification for this assumption? Why not $E = E_0 + g(v) m$?

 

[stevendaryl]

Is there a justification for this assumption? Why not $E = E_0 + g(v) m$?

Right off the top of my head, I don't remember an argument for why rest energy should be proportional to mass.

 

[PeterDonis]

Right off the top of my head, I don't remember an argument for why rest energy should be proportional to mass.

If energy and momentum form a 4-vector, then rest energy has to be proportional (equal, if you choose units properly) to the invariant length of the 4-vector, i.e., to invariant (rest) mass.

 

[PeterDonis]

Is there a justification for this assumption? Why not $E = E_0 + g(v) m$?

$E_0$ is just $m$ (in units where $c = 1$, which you can always choose), so this equation reduces to stevendaryl's equation (you just redefine $g(v)$ to include the constant $1$).

 

[DrStupid]

I don't remember an argument for why rest energy should be proportional to mass.

Me too.

Here is a derivation of the relativistic momentum without any assumptions apart from the basics of classical mechanics and Lorentz transformation (and without energy at all):

https://www.physicsforums.com/showpost.php?p=4737048&postcount=4

Starting with the result of this derivation I get

$$dE = F \cdot ds = v \cdot dp = \frac{{m \cdot v \cdot dv}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} ^3 }}$$

The integration results in

$$E - E_0 = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} - m \cdot c^2$$

With your assumption #1 you simply set the constant of integration to zero resulting in Eo=m·c². That turns your derivation into a circular argumentation.

 

[Fredrik]

...without any assumptions apart from the basics of classical mechanics and Lorentz transformation (and without energy at all):

This is a bit of an overstatement. You're using that $F=\frac{dp}{dt}$ and that $p=\gamma m v$. These are non-trivial assumptions. They can be considered definitions of the left-hand sides, but you can also take $E^2=p^2c^2+m^2c^4$ to be the definition of E, and then you don't have to do any calculations at all. This definition needs no justification other than the fact that the theory makes very accurate predictions (as already mentioned by Russ).

 

[DrStupid]

$E_0$ is just $m$

Remember the topic! Eo=m·c² has to be proofed. Thus you can't take it as a given.

 

[DrStupid]

You're using that $F=\frac{dp}{dt}$ and that $p=\gamma m v$. These are non-trivial assumptions.

$F=\frac{dp}{dt}$ is Newton's second law and $p=\gamma m v$ the result of the derivation.

 

[Fredrik]

$F=\frac{dp}{dt}$ is Newton's second law

Right, but so is $F=m\frac{d^2x}{dt^2}$, and this formula doesn't lead to the desired result. If you want the calculation to be considered a proof, or just a fairly solid argument, you have to explain why you're using that specific statement of Newton's 2nd. If it's just because it leads to the result we want, you might as well just write down the result we want and call it a definition.

and $p=\gamma m v$ the result of the derivation.

Can you justify the last equality in the first line without this?

 

[DrStupid]

Right, but so is $F=m\frac{d^2x}{dt^2}$

No. That results from the second law in classical mechanics but it is not identical with the second law. In SR you get another result.

you have to explain why you're using that specific statement of Newton's 2nd

The use of the original statement of Newton's second law does not need to be explained. It is the special case you mentioned above that would need a justification.

If it's just because it leads to the result we want, you might as well just write down the result we want and call it a definition.

Stick to facts!

 

[Matterwave]

If the argument is regarding whether Newton's second law states $F=ma$ or $F=\frac{dp}{dt}$, the answer is actually that his language is explicitly given as that of the second formula, although he explicitly assumes that the momentum in the second formula is given by $p=mv$.

Reading the Principia you find that Newton stated his second law thus (taken from the translation by Andrew Motte):

"The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed".

Here "motion" is roughly Newton's word for "momentum". He also only ever used the word "proportional" and not equal to, so there could be some as yet defined constant. So his language technically says $\frac{dp}{dt}\propto F$. However, earlier in the text, Newton defined "motion" (momentum) thus:

"The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly".

He states that if you double the quantity of matter (his word for mass) you double the "motion", and if you double the velocity you also double the motion. THIS explicitly assumes $p \propto mv$ and NOT $p=\gamma mv$ (obviously he would have no idea about the second case as that arose 300 years after him).

So, although his language technically might agree with DrStupid, I think the spirit of his language agrees with Frederik.

 

[Fredrik]

No. That results from the second law in classical mechanics but it is not identical with the second law. In SR you get another result.

The use of the original statement of Newton's second law does not need to be explained. It is the special case you mentioned above that would need a justification.

Is the stuff you feed into the calculation (of the work required to accelerate a massive point particle in SR) motivated by analogy with non-relativistic classical mechanics, or defined in the framework of SR? If it's the former, you certainly do have to explain why you favor dp/dt over ma. If it's the latter, then you have to explain why you consider your definitions (of momentum, work, etc) more fundamental than the end result, which can also be taken as a definition.

 

[DrStupid]

He states that if you double the quantity of matter (his word for mass) you double the "motion", and if you double the velocity you also double the motion. THIS explicitly assumes $p \propto mv$ and NOT $p=\gamma mv$ (obviously he would have no idea about the second case as that arose 300 years after him).

I can't follow your argumentation. Could you please explain why his statement excludes $p=\gamma mv$?

 

[DrStupid]

Is the stuff you feed into the calculation (of the work required to accelerate a massive point particle in SR) motivated by analogy with non-relativistic classical mechanics, or defined in the framework of SR?

Did you read the link above?

 

[Matterwave]

I can't follow your argumentation. Could you please explain why his statement excludes $p=\gamma mv$?

He says explicitly that if you double the mass you double the motion. If you double the velocity you double the motion. If you double both mass and velocity you quadruple the motion. This is not compatible with $p=\gamma mv$ because $\gamma$ depends on $v$ and his statements will no longer be true. If you double v, you certainly more than double $\gamma mv$ (assuming v<.5c).

 

[DrStupid]

If you double the velocity you double the motion.

I read that as "If you double the velocity [and keep quantity of matter constant] you double the motion."

Without this assumption it won't even be correct in classical mechanics. In SR you will need an open system (or two different systems) to meet this condition.

Edit: I added the possibility of two different systems because the explanation could also mean: If you have a system A with quantity of matter q and velocity v and a system B with the same quantity of matter but the velocity 2·v, than system b will have twice the motion of system A.

 

[Matterwave]

I read that as "If you double the velocity [and keep quantity of matter constant] you double the motion."

Without this assumption it won't even be correct in classical mechanics. In SR you will need an open system to meet this condition.

? Yes, keep m constant, $\gamma m v$ still more than doubles when you double v...I don't see how you can conclude $\gamma m v$ is a valid possible expression of the "motion" as defined by Newton.

 

[stevendaryl]

He says explicitly that if you double the mass you double the motion. If you double the velocity you double the motion. If you double both mass and velocity you quadruple the motion. This is not compatible with $p=\gamma mv$ because $\gamma$ depends on $v$ and his statements will no longer be true. If you double v, you certainly more than double $\gamma mv$ (assuming v<.5c).

I don't quite understand the point of this argument. Obviously, Newton's laws of physics don't imply SR. What I was looking for with my derivation, and Dr. Stupid presumably was looking for in his derivation, was an alternate theory of energy, momentum and force that reduces to the Newtonian case in the low-velocity limit and which is invariant under Lorentz transformations. The distinction between $\dfrac{dp}{dt} = F$ and $m \dfrac{dv}{dt} = F$ doesn't matter in the non-relativistic limit.

In my derivation, I didn't explicitly mention forces at all, but instead just assumed that in collisions there were two quantities that were conserved: (1)A scalar-valued quantity, energy, and (2)a vector-valued quantity, momentum.

 

[DrStupid]

? Yes, keep m constant, $\gamma m v$ still more than doubles when you double v

Remember: Motion is defined the product of quantity of matter and velocity. That means in SR quantity of matter is not m but $\gamma m$.

 

[Matterwave]

Remember: Motion is defined the product of quantity of matter and velocity. That means in SR quantity of matter is not m but $\gamma m$.

This is a stretch...and is certainly NOT in the spirit of Newton. Saying things such as "quantity of matter" is $\gamma m$ is certainly NOT something Newton would agree with in his time. If you want to say it NOW, you are ASSUMING special relativity is right. Even in special relativity, there is very limited utility of defining the relativistic mass, as this approach only works for forces which are co-linear with the object's motion.

But this is beyond the point.

My point is that, as formulated by Newton, the second law should read, given modern notation, $F=\frac{dp}{dt}$ but it MUST be appended with the explicit assumption that $p=mv$.

If you want to explore what Newton meant by "m", then you need to read his definition I in his Principia. A problem arises here; however, because mass was mysterious to Newton himself. Newton basically defined mass as density times volume, which just begs the question of defining the density.

But given Newton's definition, there is no way to define quantity of matter as $\gamma m$, both the density and volume should be measured in the rest frame.

 

[stevendaryl]

Remember: Motion is defined the product of quantity of matter and velocity. That means in SR quantity of matter is not m but $\gamma m$.

That's a possible interpretation. The first derivation of E = mc² that I saw used relativistic mass (that is, it assumed that $\vec{p} = m(v)\vec{v}$, and derived the form of $m(v)$). I didn't like that derivation, since I tended to think of "relativistic mass" as an old-fashioned concept that was no longer used. However, I have seen some respectable physicist (on the sci.physics.research newsgroup---I can't remember his name) argue that there are good mathematical reasons to have a relativistic mass. I don't remember the details, but there was some group-theoretic reason showing that both SR and galilean relativity were special cases of a more general kind of relativity.

 

[DrStupid]

What I was looking for with my derivation, and Dr. Stupid presumably was looking for in his derivation, was an alternate theory of energy, momentum and force that reduces to the Newtonian case in the low-velocity limit and which is invariant under Lorentz transformations.

My derivation also works for relativistic velocities. E=m·c² doesn't make much sense in the non-relativistic case.