## Why c2 (speed of light squared)?

 It's not circular, it comes from conservation of momentum and the known (through observation) fact that radiation transfers momentum when it strikes a surface.
Hmmm, OK, can you give me link where I can see this equation, v=E/(Mc), used somewhere by someone with a straight face who derived it independently from Einstein's derivation, and for some other purpose? I mean, all we have to do is set the velocity here to c and we have E=mc^2. I think Einstein had conservation of momentum and radiation transfer in mind when he derived it too. What I'm taking issue with is that it seems the DrPhysics is using E=mc^2 to derive E=mc^2. I don't see what novel insight he is bequeathing upon us here. Have I missed something?
 Initial momentum or the momentum with which the light hits the cylinder will be equal to TOTAL ENERGY OF THE LIGHT / SPEED OF LIGHT (E/c). Final momentum or the momentum with which the cylinder moves after colliding with light is just THE MASS OF THE CYLINDER * VELOCITY WITH WHICH IT MOVES (Mv) According to Momentum conservation principle Initial momentum will be equal to Final momentum .So E/c = Mv (OR) v=E/Mc
 Recognitions: Gold Member That's a nice point. You are missing a factor of gamma on the right side though - so it doesn't quite work. As a novice's derivation it works well though.

 Quote by Vorde You are missing a factor of gamma on the right side though
I cant quite understand . What do you mean by this ?
 Recognitions: Gold Member 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}}##. So your derivation doesn't quite work. It's been a while since I've had to derive E=mc^2 (I had to do it for an extra credit problem on a test) so I'm not sure whether or not the derivation on that site can be simply expanded to be correct or not, but in it's presented form it does not work. As I said though I approve of it as a teaching method - it's quite simple - as long as readers are aware that a more thorough proof is required.

 Quote by DiracPool all we have to do is set the velocity here to c and we have E=mc^2. I think Einstein had conservation of momentum and radiation transfer in mind when he derived it too. What I'm taking issue with is that it seems the DrPhysics is using E=mc^2 to derive E=mc^2. I don't see what novel insight he is bequeathing upon us here. Have I missed something?
You cant do something like that because In E=mc^2 relation 'm' stands for the mass of light !!! 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 wont become E=mc^2
 Landau's derivation of the equations of relativistic mechanics in his book Classical Field Theory might help you better understand relativity.

Recognitions:
Gold Member
 Quote by Kaushik96 In E=mc^2 relation 'm' stands for the mass of light !!!
You are fundamentally wrong about this, and you won't get anywhere with relativity unless you do research (wikipedia is a great place to start) and realize that the mass can be anything.

 Quote by Vorde You are fundamentally wrong about this, and you won't get anywhere with relativity unless you do research (wikipedia is a great place to start) and realize that the mass can be anything.
In the derivation from the external site 'm' refers to the mass of the light . Thats what i meant

 Quote by Vorde 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 cant 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 wont become E=mc^2
Of course it will, this is simple algebra.

 Quote by DiracPool 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 .
 Recognitions: Gold Member Science Advisor 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.

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

And I dont know what is "ρ" in the equation . Could you explain?
 Recognitions: Gold Member P is momentum: which is standard notation.

 Quote by Vorde P is momentum: which is standard notation.
Sorry bro !! I mistook it as "rho(ρ)"