# Thread: Could you help to integrate dE = (m+dm)*v*dv + c^2*dm

1. ## Could you help to integrate dE = (m+dm)*v*dv + c^2*dm

* Could you help to integrate
dE = (m+dm)*v*dv + (c^2)*dm

I would like to compare the result with relativistic kinetic energy
which as you know is calculated by equation

E = mc^2( 1/(1-v^2/c^2)^(1/2) -1 )
Kinetic energy - Wikipedia, the free encyclopedia

Thank you very much for any possible help.

2. ## Re: Could you help to integrate dE = (m+dm)*v*dv + c^2*dm

I am sorry, likely my initial question was wrong formulated.
Maybe this question would be more correct and easier:
How to prove
dm = m*v*dv / (c^2 - v^2)
is equivalent to
m = m0 / (1-(v^2 / c^2))^(1/2)
?

It is easy to write a program to calculate
dm = m*v*dv / (c^2 - v^2)
numerically by expression
m += m*v*dv / (c^2 - v^2)
inside a loop.
But I need analytical derivation.

Thank you

3. ## Re: Could you help to integrate dE = (m+dm)*v*dv + c^2*dm

It's a fairly straight-forward differential equation. I'm not sure I agree with your results. Here's WolframAlpha's solution (this DE is a first-order separable ODE, with decently straight-forward integrals).

4. ## Re: Could you help to integrate dE = (m+dm)*v*dv + c^2*dm

Originally Posted by Ackbeet
It's a fairly straight-forward differential equation. I'm not sure I agree with your results. Here's WolframAlpha's solution (this DE is a first-order separable ODE, with decently straight-forward integrals).
Thank you very much. It looks it works well.
Now I can write better equation for kinetic energy:

dE = ( m*c^2*v / (c^2 – v^2) ) dv = ( (m0/(1-v^2/c^2))*c^2*v / (c^2 – v^2) ) dv

I think this is equal to relativistic kinetic energy
E = m0*c^2( 1/(1-v^2/c^2)^(1/2) -1 )
I was mentioned by my first post.

By using WolframAlpha I have got :
E = m0*c^2 / (1-v^2 / c^2)^(1/2) + constant.

http://www.wolframalpha.com/input/?i=int+m_0*c^2*v%2F%28%281-v^2%2Fc^2%29^%281%2F2%29*%28c^2+-+v^2%29%29+dv

Maybe you have some hint how to prove this constant equals to
-m0*c^2 ?

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Thank you, I have solved it by myself. You may close the tread.