B Measuring Light's Brightness & Power Flux Relative to Observer's Speed

[alan123hk]

TL;DR

The relationship between the movement speed of the observer's measuring instrument and the measured brightness and/or power flux density.

Assuming that the observer moves along the direction of the light, does the speed of the observer's measuring instrument relative to the light source (which may be close to the speed of light) affect the brightness and/or power flux density measured by the observer?

I'm not sure about this, thanks for helping.

 

[vanhees71]

Just consider a plane wave and Lorentz boost it from the restframe of the source to the rest frame of the observer. You'll get a plane wave again with changed $\omega$ and $\vec{k}$, from which you get the Poynting vector the observer measures.

 

[Ibix]

And qualitatively, imagine you 10m behind me coming towards me at 10m/s. At the end of one second you'll be next to me, so you'll have received all the light that I have, plus the bit that had passed me but not reached you at the start of the second. It's not a particularly significant effect unless you are doing a large fraction of light speed.

 

[alan123hk]

Just consider a plane wave and Lorentz boost it from the restframe of the source to the rest frame of the observer. You'll get a plane wave again with changed ω and k→, from which you get the Poynting vector the observer measures.

Thank you for your concise, direct and valuable reply.

 

[alan123hk]

For the sake of simplicity, only the momentum of light is considered. Applying the Lorentz Transformation of the Momentum-Energy Four Vectors of Light (https://hepweb.ucsd.edu/ph110b/110b_notes/node54.html).

If the movement direction of the observer is opposite to the direction of the light

$$ E'_x = \gamma \left( E_x - \beta c \vec P_x \right) = \gamma \left( E_x + \beta E_x \right) = \gamma E_x \left( 1+ \beta \right) $$ $$ \text {Therefore,}~~~ E_x'=E_x \left( \frac {1+\frac{v}{c} }{\sqrt{1-\frac {v^2}{c^2}}} \right)~~~~\Rightarrow~~~~\frac {\text {Power Flux Density'}} {\text {Power Flux Density}} =
\frac {1+\frac{v}{c}} {\sqrt{1-\frac {v^2}{c^2}}} $$
I believe that even if the above calculation is not very complete and strict, there should be no serious errors.

From the above mathematical formula, although the observer needs infinite energy to accelerate his spacecraft to close to the speed of light, it seems that he can receive infinite energy from a distant light source at the same time. For example, a huge distant galaxy. The question is whether it can operate like wind power, that is, convert the received light energy into kinetic energy and let it move in the direction opposite to the direction of the received light.

Edit: - I admit that the above idea is very impossible to realize, because when the spacecraft is close to the speed of light, the infinite light pressure received by the spacecraft will destroy it, and those high-energy photons will also completely kill the lives on the spacecraft .