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Why c2 (speed of light squared)?

by neoweb
Tags: light, speed, squared
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Kaushik96
#55
Jan8-13, 07:09 PM
P: 18
Quote Quote by Vorde View Post
You are missing a factor of gamma on the right side though
I cant quite understand . What do you mean by this ?
Vorde
#56
Jan8-13, 07:20 PM
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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.
Kaushik96
#57
Jan8-13, 07:39 PM
P: 18
Quote Quote by DiracPool View Post
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
kostas230
#58
Jan8-13, 08:08 PM
P: 77
Landau's derivation of the equations of relativistic mechanics in his book Classical Field Theory might help you better understand relativity.
Vorde
#59
Jan8-13, 08:11 PM
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Quote Quote by Kaushik96 View Post
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.
Kaushik96
#60
Jan8-13, 08:35 PM
P: 18
Quote Quote by Vorde View Post
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
Kaushik96
#61
Jan8-13, 08:42 PM
P: 18
Quote Quote by Vorde View Post
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
#62
Jan8-13, 11:54 PM
P: 534
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.
Kaushik96
#63
Jan9-13, 01:48 AM
P: 18
Quote Quote by DiracPool View Post
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
#64
Jan9-13, 02:20 AM
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The Lagrangian [itex]L = -mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}}[/itex], [itex]p = \frac{\partial L}{\partial v} = \frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/itex], [itex]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}}}}}[/itex] so in the rest frame we have that [itex]E = mc^{2}[/itex]. 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
#65
Jan9-13, 02:36 AM
P: 18
Quote Quote by WannabeNewton View Post
The Lagrangian [itex]L = -mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}}[/itex], [itex]p = \frac{\partial L}{\partial v} = \frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/itex], [itex]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}}}}}[/itex] so in the rest frame we have that [itex]E = mc^{2}[/itex]. 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?
Vorde
#66
Jan9-13, 03:44 PM
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P: 784
P is momentum: which is standard notation.
Kaushik96
#67
Jan10-13, 10:02 AM
P: 18
Quote Quote by Vorde View Post
P is momentum: which is standard notation.
Sorry bro !! I mistook it as "rho(ρ)"


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