# Energy of light emission disparity from moving and stationary frames

I am going over Einstein's original paper. In it, he derives $$\frac{E'}{E} = \sqrt{ \frac{1-v/c}{1+v/c}}$$ for the observed energy of light that is being emitted from a spherical body as measured by moving and stationary observers $$K'$$ and $$K$$ where $$K'$$ moves along the common $$x$$ axis at speed $$v$$ relative to $$K$$. So the body appears to move relative to $$K'$$ with $$-v$$.

1. My first confusion is how there is a sign dependence on $$v$$. $$E' < E$$ if $$v > 0$$ but $$E' > E$$ if $$v < 0$$. I'll remind myself that the sign of $$v$$ is determined by how $$K'$$ moves relative to $$K$$, and not how the body moves in $$K'.$$ Why should the energy of light as observed by $$K'$$ depend on whether the body is moving towards or away from the $$K'$$ observer?

2. Can the energy disparity be explained in terms of the Doppler effect? For a light wave we have $$E' = h f'$$, and the waves at the front of the moving body will be compressed and therefore higher frequency (blueshift) while the waves in its wake will be stretched out and therefore of lower frequency (redshift). But perhaps the averaged frequency of the waves will be identical to that measured by observer $$K$$, where the body appears stationary?

3. Does length contraction explain the energy disparity? If so, length contraction is independent of the sign of $$v$$, so I am confused again. Whether the body moves towards or away from $$K'$$, it will appear compressed in the $$x$$ direction relative to what $$K$$ sees. Similarly, will the emitted light waves also appear compressed, so that we have Doppler on top of length-contracted light waves? That would serve to increase $$f'$$.

There are some symmetries under $$v \mapsto -v$$ that I am aiming to consider. Length contraction and time dilation are unaffected by this replacement, but the energy difference $$E'-E$$, and redshift or blueshift is.

I would prefer not to think of spherical waves, but of plane waves, since it is simpler. However, this may be necessary as a sphere to $$K$$ is an ellipsoid to $$K'$$, so that may factor into the explanation somehow.

Thank you.

(1) plus/minus in $$K'$$ is a blue/red shift vs the energy in $$K$$

(2) Yes, it's the Doppler shift. The geometric average of approaching and receding Doppler shifts is $$1$$.

(3) How you want to explain the Doppler shift is up to you:

(3.1) You can Lorentz transform with 4-wave vector: $$k_{\mu} = (\omega/c, \vec k) = (k, 0, 0, k)$$, where I used a common $$z$$-axis.

(3.2) You can time dilating the frequency $$\omega = ck$$ or length contracting the wavelength $$\lambda = 2\pi/k$$ does not depend on the sign of $$v$$, so it's clearly problematic. You have to consider the simultaneity of starting and ending points of what you're contracting.

(3.2) Phase, $$\phi = -k_{\mu}x^{\mu} = kx - \omega t$$ is a Lorentz invariant, which may help. Since Lorentz transformations mix $$x$$ and $$t$$, the distance/time between nodes is frame dependent, but the world lines of the nodes are not.

Obviously 3.1 is easiest, but you should try them all.

• In (2) you say the energy disparity is due to the Doppler shift but the geometric average of approaching and receding shifts is 1. If the observer in $K$ sees no Doppler shift, they will observe a certain energy $E = hf$. For $K'$ if the light-emitting object is moving away from them, they will see $E' = hf'$, where $f' > f$ (blue-shift) on the side of the body which is in its direction of travel, and $f' < f$ (red-shift) on the side of the body that is adjacent to its wake. But to me, the average $f'$ is equal to $f$, so $E' = E$, i.e., the Doppler shift doesn't explain the energy disparity. Aug 18, 2023 at 0:01
• The relativistic Doppler shift is $f_{\pm} = ((1\pm v)/(1\mp v))^{\frac 1 2}$ for $\pm$ blue (red) shifted. So: $f_+f_=1$. This is all in $K'$, $K$ is stationary wrt to the source, so $f=1$.
– JEB
Aug 19, 2023 at 4:13
• Thank you, I understand your answer after following the derivation here: en.wikipedia.org/wiki/… Basically, a moving observer will see different wavelengths and periods which is also true in Galilean relativity but the differences in the differences are... different for Einsteinian relativity. Aug 21, 2023 at 18:48
• I'm glad it helped. That the Doppler shift formula is symmetric and doesn't reference an absolute source/receiver velocity in the medium pure SR. See "Bondi's $k$-calculus" for a unique derivation of SR based on it.
– JEB
Aug 22, 2023 at 16:00