Hi to everyone. I want to ask something, if my rest mass is 81 kg and if i move with the speed of light, can i still use the formula for energymomentum relation to calculate my energy?
''r'' is gamma
E= (г.m.v).c+m.c2
E= (г.m.c2)+m.c2 ???

Hi to everyone. I want to ask something, if my rest mass is 81 kg and if i move with the speed of light, can i still use the formula for energymomentum relation to calculate my energy?
''r'' is gamma
E= (г.m.v).c+m.c2
E= (г.m.c2)+m.c2 ???
Last edited by Sanev; 05102018 at 06:33 PM.
So your question is: "If I violate the laws of physics, will the laws of physics still apply?"
I'm beginner so i'm still learning. Sorry for that.
My question is if i'm moving at the speed of light what is my energy?
And also why me moving hypothetically with the speed of light is violation?
Hint:
What is the correct expression for energy? Your original equation was not relativistically correct (I see you have edited it without comment; that is not good). Why not? What is this parameter "r" you have now added?
As you proceed in selfstudy, it is critically important to understand explicitly in which domains given formulas are valid.
''r'' is gamma because the momentum is relativistic far as i know.
If r = gamma, and gamma= 1/sqrt(1v^2/c^2), then what happens to gamma if v=c? It reduces to 1/0. But 1/0 is undefined.
Anything with a nonzero rest mass can never achieve a velocity of c.
The sum of the kinetic energy and rest mass equivalent energy for such an object for a given velocity v is E= mc^2/sqrt(1v^2/c^2) or rmc^2, using r for gamma.
The KE part is equal to mc^2(r1) and the rest mass equivalent energy is mc^2.
As v approaches c, E approaches infinity. v can be as close to c as you want, but it can never equal c, because this brings in the division by zero problem again.
So you added gamma after your initial post, but you didn't consider what that means. As Janus pointed out, when you actually use gamma, you are led to the mathematical absurdity I hinted at in my first posts.
Summary: An object with mass cannot travel at c.
Hopefully you now see why your question in its original, unedited form, where you used the Newtonian mv as the momentum, and then merely substituted c for v to arrive at E = 2mc^2 or some such thing, violates the laws of physics.
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