MathB.in
New
Demo
Tutorial
About
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. \[{m}_{rel} = \frac{h \cdot f}{c^2} \]
ERROR: JavaScript must be enabled to render input!
Fri, 29 Nov 2019 23:01 GMT