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For the sake of simplicity, let's imagine that the entire universe is empty except for a single lump of (classical) matter with mass $m$. In its center of momentum frame, it is clear that the total energy is simply $E_\text{CoM}=mc^2$. However, in a frame moving relative to it with speed $v$, we have $E_\text{moving}=\sqrt{m^2c^4+p^2c^2}> mc^2$ for all nonzero $v$.

Are we to infer that the energy content of a system can be different relative to the observer? Does this not violate the first law of thermodynamics?

ACuriousMind
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Caio Belussi
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3 Answers3

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Yes, the observed energy content differs between different observers:

Energy is manifestly no Lorentz invariant quantity, as it is the zeroth component of the momentum four vector, and hence differs between different inertial frames. Thus, rather trivially, different frames will observe different energy contents for the same system.

This does not mean that conservation of energy is violated, since conservation of energy simply states that the time derivative of energy is zero - and it is, within every frame. Only when you change frames, the energy you observe changes, and again, in every momentarily comoving frame in that process, energy is indeed conserved.

ACuriousMind
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This is really just a footnote to ACuriousMind's answer, since I can't improve on his discussion of energy.

In relativity, both special and general, there is an invariant quantity analogous to energy called the stress-energy tensor. This does not depend on the observer, that is all observers in all frames will measure the same stress-energy tensor (though its representation will vary for different observers).

John Rennie
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Newtonian Mechanics: A particle of mass $m$ is moving with constant speed $v$ in a given inertial frame A. The energy of the particle is:

$$E_A=E^{kin}_A=\frac{1}{2}m v^2 $$

Energy of the same particle in a frame B moving with speed $v$ relative to the first frame:

$$E_B=E^{kin}_B=0 $$

Rexcirus
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