# Conservation of energy in SR: Is internal energy measured the same from the viewpoint of all inertial observers?

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?

• 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. Commented Oct 29, 2020 at 9:53

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:

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

• 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. Commented 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)