Chris @chrism@qoto.org

If you are a lay person and dont understand physics but want to in a fun easy way without math, check this out. Feynman was a wonderful mind and great at explanations:

https://youtu.be/P1ww1IXRfTA

@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

@CCoinTradingIdeas