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I am a bit confused on the way Landau derives the Lagrangian of the free particle in SR (L. Landau, E. Lifshitz - The Classical Theory of Fields) and his conclusions about the equivalence between mass and energy.

He claims that there exists an integral that assumes its minimum value on the actual trajectory of the particle. Since the actual trajectory in space-time must be the same in every reference frame, this integral must be:

$ S = \alpha \int \mathrm{d}s $, where the integral is taken between two fixed points in space-time.

$\alpha$ is just a constant that can be found comparing this Lagrangian in the limit $c \to \infty $ to the classical one. It is found $\alpha = mc^2 $.

One can then express $\mathrm{d}s$ in an inertial reference frame. Collecting $\mathrm{d}t$ we get:

$$ S = - \int mc^2 \sqrt{1-\frac{v^2}{c^2}} \mathrm{d}t $$

Therefore, we conclude that, in an inertial reference frame, the Lagrangian is just:

$$L=- mc^2 \sqrt{1-\frac{v^2}{c^2}}$$

We can then derive the energy of a free particle with the formula we borrow from Classical Mechanics:

$$E=\sum_{i} \dot{q}_i \frac{\partial L}{\partial \dot{q}_i}-L$$

and we get:

$$E=\frac{mc^2}{\sqrt{1-v^2/c^2}}$$

He then claims that in SR this energy is NOT defined up to a constant, and therefore we can conclude that a mass at rest has an energy of $mc^2$. I do not understand why. After all, I can always add a constant $C$ to the Lagrangian. This would not change the equations of motion in this reference frame (because it is a total derivative of the function $Ct$). It wouldn't even change the equations of motion in any reference frame. This is because changing reference frame means putting $t=f(\textbf{x}', t')$, therefore $\mathrm{d}t=\mathrm{d}f=\frac{\mathrm{d}f}{\mathrm{d}t'}\mathrm{d}t' $. In the action integral, this would become

$$S' = \int \left( - mc^2 \mathrm{d}s + \frac{\mathrm{d}f}{\mathrm{d}t'}\mathrm{d}t' \right) $$

That does not change the equations of motion because $\frac{\mathrm{d}f}{\mathrm{d}t'} $ is a total derivative in time. Also, this term would change the energy in the non primed reference frame, making the energy:

$$E=\frac{mc^2}{\sqrt{1-v^2/c^2}}-C$$

which would prove that the energy is indeed defined up to a constant. What am I missing?

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Here is one argument:

  1. OP has already argued that the energy $E$ is of the form $$ E~=~ m_0 \gamma c^2+C, $$ where $C$ is a constant.

  2. In SR, the $4$-momentum $p^{\mu}=(E/c,{\bf p})$ transforms as $4$-vector under Lorentz transformations. In particular, the length-square of the $4$-vector should be an invariant: $$ {\rm const.}~=~\left(\frac{E}{c}\right)^2-{\bf p}^2~=~\left(\frac{m_0 \gamma c^2+C}{c}\right)^2 - (m_0 \gamma{\bf v})^2.$$ It is straightforward to see that this is only possible if the constant $C=0$ is zero.

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  • $\begingroup$ Basically I am postulating that the four-momentum must be a four-vector, right? $\endgroup$
    – Masterme
    Oct 26 '20 at 13:23
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    $\begingroup$ @Masterme: Right. $\endgroup$
    – Qmechanic
    Oct 26 '20 at 13:24
  • $\begingroup$ Can it be argued without invoking the 4-vector nature of $p^\mu$ ? I wanted to follow Landau's discussion as close as possible $\endgroup$
    – ohneVal
    Oct 26 '20 at 13:25
  • $\begingroup$ He is fully correct to use this Lorenz invariance argument. Look, forget 4-vector, let’s look at action only. If I add C to energy E, then I need to add it to L as well. But then the new action is S= \int mc^2 ds + C \int dt. The last term is obviously not Lorenz invariant, what should not be (otherwise your physics is different in every frame) $\endgroup$
    – Alex
    Oct 27 '20 at 7:56
  • $\begingroup$ I am not saying it is wrong by any means. I was just checking the book of Landau and he doesn't mention or impose any Lorentz symmetry at that point, however he dismisses the possible constants. So perhaps Landau saw another argument... $\endgroup$
    – ohneVal
    Oct 27 '20 at 10:12
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A way to think about this is the following. Consider you have not one but two particles. For which you can follow the same derivation that was made in order to set the proportionality constant $\alpha$ for each. As we now, it will be related to the mass of each particle (take the case they are different). Now you can see that no matter what constant you add you wont be able to cancel all the constant terms. So the issue remains, there is a piece which when compared to any reference you take, it doesn't go away.


There are other cases where there is more controversy or discussion. If you were to try adding a constant in GR, you will see that the factor $\sqrt{-\det g}$ actually has an impact on the e.o.m.'s.

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  • $\begingroup$ Shouldn't it be $E(v)-E(0)=\frac{mc^2}{\sqrt{1-v^2/c^2}}-C-mc^2+C=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2$, which goes to $0$ as $v \to 0$? $\endgroup$
    – Masterme
    Oct 26 '20 at 13:11
  • $\begingroup$ What I took as ground state was $p=0$ in the Legendre transform if you want, that is why I tried speaking about ground state and not $v=0$ as reference. And just at the end compare the result in the $v\rightarrow 0$ limit. I corrected the text a bit. $\endgroup$
    – ohneVal
    Oct 26 '20 at 13:12
  • $\begingroup$ Okay, but then $ E(0) = \textbf{p} \cdot \textbf{v} - L = - L = \frac{mc^2}{\sqrt{1-v^2/c^2}} - C$, and since $\textbf{p}=\textbf{0} $ implies $v=0$ we still get $E(0) = mc^2 - C$. If I misunderstood could you please elaborate a bit more? Thanks a lot. $\endgroup$
    – Masterme
    Oct 26 '20 at 13:20
  • $\begingroup$ Oh I see yes I made the mistake there, so I believe the argument cannot not be done so simply. I will leave argument 2 only. $\endgroup$
    – ohneVal
    Oct 26 '20 at 13:24
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I repeat my comment here with the citation of LL: You're not free to add constant to the energy while it'll break the Lorenz invariance. If $E\to E+C$ then also $L\to L+C$, then $mc \int ds \to mc \int ds + C \int dt$. It is not relativistic invariant anymore. I repeat what is written in LL: $\int ds$ in the only possible relativistic invariant expression.

The citation of the LL which basically answers your question

P.S. All credits to Qmechanic, he pointed out in his answer the necessity of the relativistic invariance.

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  • $\begingroup$ Well, the Lagrangian wouldn't be the equations of motion would $\endgroup$
    – Masterme
    Oct 27 '20 at 19:08

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