Breaking the conservation law with relativistic doppler effect?
Imagine we have a red photon of $780\text{ nm}$.
Using this from wikipedia: $${\displaystyle E={\frac {hc}{\lambda }}}$$ $${\displaystyle E(eV)={\frac {1.2398}{\mathrm {\lambda } ({\mu }m)}}}$$ $${\displaystyle E(eV)={\frac {1.2398}{\mathrm {0.780 } ({\mu }m)}}}$$ $${\displaystyle E(eV)=1.58}$$ So a red photon of $780\text{ nm}$ has $1.58\text{ eV}$ of energy.
So if we launch the photon to a solar panel (100% efficiency) it would give us $1.58\text{ eV}$. But...
Now we push the solar panel and it's travelling at $v = 0.99\text{c}$.
Using the relativistic doppler effect the photon can "change" of color. (Idea from xkcd).
I can't found how to use the Relativistic Doppler Effect with wavelength instead of frequency so: $$f = \frac{c}{\lambda}$$ $$f = \frac{299,792,458\text{ m/s}}{780\text{ nm}}$$ $$f = 384,349.30\text{ hertz} = 384.34\text{ KHz}$$ Using the formula: $${\displaystyle f_{o}=f_{s}{\sqrt {\frac {1-v/c}{1+v/c}}}}$$ Where here $v = -296,794,533.42$. $${\displaystyle f_{o}=384,349.30{\sqrt {\frac {1-\frac{-296,794,533.42}{299,792,458}}{1+\frac{-296,794,533.42}{299,792,458}}}}}$$ $${\displaystyle f_{o}= 384,349.30{\sqrt {\frac {1-(-0.99)}{1+(-0.99)}}}}$$ $${\displaystyle f_{o}= 384,349.30{\sqrt {\frac {1.99}{0.01}}}}$$ $${\displaystyle f_{o}= 384,349.30{\sqrt {199}}}$$ $${\displaystyle f_{o}= 384,349.30 \times 14.10}$$ $${\displaystyle f_{o}= 5,419,325.13} = 5.41\text{ MHz}$$ Or using all the decimals: ${\displaystyle f_{o}= 5,421,914.17 = 5,42 \text{ MHz}}$
Now converting frequency in wavelengh: $$f = \frac{c}{\lambda}$$ $$\lambda = \frac{c}{f}$$ $$\lambda = \frac{299,792,458\text{ m/s}}{5,421,914.17}$$ $$\lambda = \frac{299,792,458\text{ m/s}}{5,421,914.17}$$ $$\lambda = 55.29 \text{ nm}$$
Now the photon is EUV (Extreme ultraviolet).
And using: $${\displaystyle E(eV)={\frac {1.2398}{\mathrm {\lambda } ({\mu }m)}}}$$ $${\displaystyle E(eV)={\frac {1.2398}{\mathrm {0.05529 } ({\mu }m)}}}$$ $${\displaystyle E(eV)={\frac {1.2398}{\mathrm {0.05529 } ({\mu }m)}}}$$ $${\displaystyle E(eV)=22.43}$$
So if we throw the photon to a super-fast solar panel it would give us $22.43 \text{ eV}$? Can I produce infinite energy with a red laser and solar panel travelling really fast?
How can this be possible? The solar panel would get more energy?
Is possible break conservation law with this? Obviously it isn't possible, but why? My theory is that when the photon collide to the panel it slow down the solar panel by the same amount of energy produced, am I right?: $${\displaystyle P_{\text{absorb}}={\frac {\langle S\rangle }{c}}={\frac {E_{f}}{c}}}$$ Where $E_{f} = 22.43 \times <PHONE_NUMBER> \times 10^{-19})$. 565