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.
\[{m}_{rel} = \frac{h \cdot f}{c^2} \]
Wow used count growth in #QOTO is still going up, I wonder where this surge is from. We had 74 new users 24 hours ago and of the last 24 houra its up to 99 new users per day.
Up up up we go.
Welcome new users!
@gaffen i don't want to prove anything because i am here to know something other than celebs or politicians.
Hello everyon,
Is there anything like verified people or verified profile on an open source platform.
#lefttwitter