It is said that things like the rest mass and internal energy are measured the same by any inertial observers regardless of their relative speeds, however, I cannot get my head around the following problem:

Assume that a point laser diode moves at a speed very close to that of light. Two similar photons with the same frequency of $\nu_0$ are emitted in two opposite directions simultaneously from the diode. The observer in the diode's rest frame asserts that the rest mass of the diode would be reduced by $\Delta E_0/c^2=2h\nu_0/c^2$. It is anticipated that the observer, who moves at $v$ relative to the diode, measures the same reduction in the rest mass of the diode; however, the frequency of the blue-shifted photon does not exactly compensate for that of the red-shifted one as far as there is a Doppler effect for each photon to be considered by the moving observer:

$$\Delta E=h\nu_0 \sqrt{\frac{c+v}{c-v}}+h\nu_0 \sqrt{\frac{c-v}{c+v}}=2\gamma h\nu_0 \not=2h\nu_0$$

This means that the change in the rest mass of the diode ($\Delta E/c^2=2\gamma h\nu_0/c^2$) is not necessarily an invariant from the viewpoint of inertial observers. Where is the problem? Does this violate the law of energy conservation?

  • $\begingroup$ Very fast moving slightly warm object emits x-rays with very high power, and loses kinetic energy at very high rate. The rate at which the rest mass of the object is reduced is highly time dilated. $\endgroup$
    – stuffu
    Oct 29, 2020 at 9:53

2 Answers 2


If the initial mass is $M$, then the moving observer sees a total energy of:

$$p^0 = \gamma Mc^2 $$

After the photon emission occurs, the rest mass becomes:

$$ m = M - 2\frac{h\nu}{c^2} $$

so that the total energy of the diode is now:

$$\gamma mc^2 = \gamma (M- 2\frac{h\nu}{c^2})c^2 = Mc^2 -2\gamma\hbar\nu$$

and as you have shown, the photon energy in the moving frame is

$$ E_{\gamma} = 2\gamma\hbar\nu $$

so that the total energy of the diode + photon system has not changed:

$$ {p^0}' = mc^2 + E_{\gamma} = Mc^2 $$

  • $\begingroup$ Hi, It is possible that you have posted the right answer, however, in line 3 of the equations, it seems that you have missed a $\gamma$ to be multiplied by $M$ on the right-hand side of the equation. On the other hand, your results do not seem to satisfy the final equation. $\endgroup$ Oct 29, 2020 at 8:15

The internal energy of a moving body is either: The internal energy of the body according to an observer moving with the body.

or: The internal energy of the body according to an observer moving with the body, divided by the time dilation factor

The first one is the more conventional one.

I like the latter one, because it is consistent with the idea that time and energy are related.

Let's say in a lab a fully charged battery is swung around in a centrifuge at speed close to c. Now if we drain all the energy out of the battery, we only receive electric energy E/gamma in the lab frame. Where E is the number of Volt-Ampere-hours that is printed on the side of the battery.

I let the reader figure out how energy is conserved in the above example.

(The battery should be drained using wires running from the battery to the center of the centrifuge)


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