Why c2 (speed of light squared)?

[Kaushik96]

In relativistic physics momentum isn't defined as $p=mv$ but rather as $p= \gamma m v$ where $\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}$.

You are right ! Maybe i have to search some more about this derivation .

 

[DiracPool]

You can't do something like that because In E=mc^2 relation 'm' stands for the mass of light !

That's about as wrong as it gets in special relativity. E=mc^2 is actually the abbreviated form of the full equation which is E=mc^2+pc. The abbreviated version is used in cases that deal with energies that specifically EXCLUDE light photons, that is, rest mass. Situations dealing with light quanta will typically exclude the mc^2 term and be in the form E=pc.

And in this equation v=E/Mc , 'M' stands for the mass of the cylinder ! Even if you put 'c' in the place of 'v' it won't become E=mc^2

Of course it will, this is simple algebra.

 

[Kaushik96]

Of course it will, this is simple algebra.

You actually mistook the whole derivation . v=E/Mc not E/mc ! 'M' and 'm' are different ... Go through the derivation once more . This derivation is not circular .

 

[WannabeNewton]

The Lagrangian L = -mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}}, p = \frac{\partial L}{\partial v} = \frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}, E = p\cdot v - L = \frac{mv^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} + mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}} = \frac{mc^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} so in the rest frame we have that E = mc^{2}. The speed of light squared quantity is just a unit conversion. If you work in natural units then you won't even see it explicitly.

 

[Kaushik96]

The Lagrangian L = -mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}}, p = \frac{\partial L}{\partial v} = \frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}, E = p\cdot v - L = \frac{mv^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} + mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}} = \frac{mc^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} so in the rest frame we have that E = mc^{2}. The speed of light squared quantity is just a unit conversion. If you work in natural units then you won't even see it explicitly.

Agreed . But c^2 is not just an unit conversion . It is also known "the specific energy" .

And I don't know what is "ρ" in the equation . Could you explain?

 

[Vorde]

P is momentum: which is standard notation.

 

[Kaushik96]

P is momentum: which is standard notation.

Sorry bro ! I mistook it as "rho(ρ)"