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In Landau & Lifshitz The Classical Theory Of Fields there's a statement:

$$\mathscr E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\tag{9.4}$$ This very important formula shows, in particular, that in relativistic mechanics the energy of a free particle does not go to zero for $v=0$, but rather takes on a finite value $$\mathscr E=mc^2.\tag{9.5}$$ This quantity is called the rest energy of the particle.

But this is the direct result of choosing Lagrangian as $$L=-mc^2\sqrt{1-\frac{v^2}{c^2}}.$$

We could as easily subtract $-mc^2$ from the Lagrangian to have $L^\prime=L+mc^2$ and have nothing changed in equations of motion, but now having the energy $$E^\prime=mc^2\left(\frac1{\sqrt{1-\frac{v^2}{c^2}}}-1\right),$$ which would give rest energy $E^\prime_0=0$.

So my question is what is the physical significance of using $(9.4)$ and corresponding Lagrangian to define rest energy? Why is such a seemingly special-case result makes the authors say "in relativistic mechanics the energy of a free particle does not go to zero" as if it's a general result?

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  • $\begingroup$ Are you sure that a Lagrangian shift leads to the energy shift? What about generalized momentum and other things involved? $\endgroup$ Commented Jan 28, 2014 at 20:38
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    $\begingroup$ @VladimirKalitvianski momentum being a partial derivative of Lagrangian $\frac{\partial L}{\partial \vec v}$ remains the same. Then, as $E=\vec p\vec v-L$, this leads to shift of energy. $\endgroup$
    – Ruslan
    Commented Jan 29, 2014 at 4:53
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    $\begingroup$ It is Hamiltonian, not energy. $\endgroup$ Commented Jan 29, 2014 at 8:08
  • $\begingroup$ @VladimirKalitvianski It is the way L&L derive energy (unnumbered equation just above $(9.4)$), without saying anything about Hamiltonian. $\endgroup$
    – Ruslan
    Commented Jan 29, 2014 at 8:29

2 Answers 2

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So my question is what is the physical significance of using (9.4) and corresponding Lagrangian to define rest energy?

If the energy is defined as in 9.4, then this energy together with the three components of the momentum transform by Lorentz transformation in the same way as spacetime displacement four-vector, which is nice and preferable. If the energy was defined only as the kinetic energy, this could not be regarded as a component of such four-vector.

There are other reasons why the rest energy $mc^2$ is included, for example Einstein wrote many papers about the fact that change of energy of system in its rest frame leads to corresponding change in the rest mass according to relation $\Delta \mathscr{E} = \Delta mc^2$. This also suggests the above definition of energy.

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  • $\begingroup$ OK, so this could be explained later in text when four-vector of momentum is introduced. But why is it stated as a general result from a special-case equation? Do I not understand something here? $\endgroup$
    – Ruslan
    Commented Jan 29, 2014 at 5:24
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    $\begingroup$ No, Landau and Lifshitz are just being too quick. An explanation of why it is good to define energy so that for $v=0$ it gives $\mathscr{E} = mc^2$ is more involved than what they do around 9.4; Einstein wrote papers about it. The considerations behind this cannot be skipped in one line derivation from some Lagrangian. $\endgroup$ Commented Jan 29, 2014 at 9:33
  • $\begingroup$ In fact, I seem to have now understood what my mistake was. I supposed that I may subtract a constant from Lagrangian. But this way I make the action non-invariant with respect to Lorentz rotations (which was not obvious to me), and thus original demand for isotropic action is not satisfied. So, L&L are right making this general statement, because the formula for energy is indeed unique. I'm going to make an answer for my own question. $\endgroup$
    – Ruslan
    Commented Jan 29, 2014 at 9:45
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When deriving Lagrangian, L&L demanded that action must be isotropic.

If we select Lagrangian as $$L=-mc^2\left(\sqrt{1-\frac{v^2}{c^2}}-1\right),$$

we automatically lose isotropy of infinitesimal action $L\text{d}t$. To see this, let's make a Lorentz transform on this form:

$$L^\prime\text{d}t^\prime=-mc^2\left(\sqrt{1-\frac{\left(\mathscr L_x(\vec v)\right)^2}{c^2}}-1\right)\text{d}\mathscr L_t(t)=-mc^2\Bigg(\sqrt{1-\frac{{v^\prime} ^2}{c^2}}-\underbrace{\frac{1+\frac{\Delta v}{c^2}v_x}{\sqrt{1-\frac{\Delta v^2}{c^2}}}}_{anisotropic}\Bigg)\text{d}t^\prime.$$

So, this means that the Lagrangian given by L&L in $(8.2)$ is unique, and thus energy given by $(9.4)$ is also unique, so there's always a rest energy, which can't be simply subtracted compatibly with Einstein relativity principle.

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    $\begingroup$ Invariance of action does lead to 9.4, but there is no good physical reason why action should be invariant. In Newtonian mechanics, action based on the Lagrangian function $L_N = \frac{1}{2}mv^2$ is not invariant with respect to change of inertial frame (there is no such function that would give correct equations of motion). Yet $L_{N}$ gives correct equations of motion. There is no reason to require there is $unique$, "the" Lagrangian and that it has to give Lorentz-invariant action. One can add/subtract any function of coordinates and time, and $mc^2$ qualifies. $\endgroup$ Commented Jan 29, 2014 at 10:15
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    $\begingroup$ Consequently, even if one can derive 9.4 from the invariance of action, there is no way to know it is the energy. One could call it "invariant action based value of the Hamiltonian" to express what really happened. The word "energy" has different connotations and requirements, and rather than invariance of action, $these$ were decisive for acceptance of the idea that energy of moving body is $mc^2/\sqrt{1-v^2/c^2}$. $\endgroup$ Commented Jan 29, 2014 at 10:20
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    $\begingroup$ Even in the Newtonian case, the Lagrangian function may get such non-isotropic term dependent on the component of velocity of the particle. This term is a derivative of particle's coordinates and has no consequence on the equations of motion. Similarly in the relativistic case. $\endgroup$ Commented Jan 29, 2014 at 10:57
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    $\begingroup$ Indeed, isotropy of the Lagrangian leads to convenient and the simplest Lagrangian. But it is used only because it is convenient, not because it is some fundamental principle. Invariance of Lagrangian with respect to some operation is of no consequence if it has no impact on the equations of motion. We may use $L' = \frac{1}{2}mv_x^2 + Cv_x$ if we wish with equal results. Landau and Lifshitz attempt to reconstruct mechanics based on some "obvious" invariance requirements, but it seems all this approach accomplishes is reconstruction of the laws of mechanics in a more dogmatic way. $\endgroup$ Commented Jan 29, 2014 at 11:45
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    $\begingroup$ In $Fields$, the action principle and invariance is important. In physics, these are important too, but rather as a consequence of equations of motion. We have no direct experimental experience with action and symmetries of Lagrangian - the equations of motion are much more important. $\endgroup$ Commented Jan 29, 2014 at 12:20

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