@zeccano
NOTE: for those who cant read the below equations (they aren't rendered) you can follow this link to see: http://mathb.in/38381
Here you go a full derivation from the math equations you started with to the one I provided you several times...
axioms (given by you, agreed by physicist):
\[E = {m}_{rel} \cdot {C}^{2}\]
\[{E}^{2} = {p}^{2} \cdot {C}^{2} + {{m}_{rest}}^{2} \cdot {C}^{4}\]
De Broglie's Equation for the momentum of a photon:
\[p = \frac{h}{\lambda}\]
Formula to convert wavelength \(\lambda\) into a frequency:
\[\lambda = \frac{C}{f}\]
Simplify given axioms:
\[{({m}_{rel} \cdot {C}^{2})}^{2} = {p}^{2} \cdot {C}^{2} + {0}^{2} \cdot {C}^{4}\]
\[{{m}_{rel}}^{2} \cdot {C}^{4} = {p}^{2} \cdot {C}^{2}\]
\[{{m}_{rel}}^{2} = \frac{{p}^{2} \cdot {C}^{2}}{{C}^{4}}\]
\[{{m}_{rel}}^{2} = \frac{{p}^{2}}{{C}^{2}}\]
\[{m}_{rel} = \sqrt{\frac{{p}^{2}}{{C}^{2}}}\]
then use de broglie's equation for the momentum...
\[{m}_{rel} = \sqrt{\frac{{(\frac{h}{\lambda})}^{2}}{{C}^{2}}}\]
\[{m}_{rel} = \sqrt{\frac{\frac{{h}^{2}}{{\lambda}^{2}}}{{C}^{2}}}\]
\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {\lambda}^{2}}}\]
Now convert wavelength to frequency...
\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {(\frac{C}{f})}^{2}}}\]
\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot \frac{C^2}{f^2}}}\]
\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{\frac{C^4}{f^2}}}\]
\[{m}_{rel} = \sqrt{{h}^{2} \cdot \frac{f^2}{C^4}}\]
\[{m}_{rel} = \sqrt{\frac{{h}^{2} \cdot f^2}{C^4}}\]
and bam you arrive at the equation for the relativistic mass of a photon.
\[{m}_{rel} = \frac{h \cdot f}{c^2} \]
#physics #relativity #einstein #math #maths #mathematics
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