Why Does Integrating Relativistic Kinetic Energy Lead to Different Results?

[omiros]

I am doing the integral DK = ∫pdv = mv/(√1-v²/c²)dv , in the same way as they do the ∫Fds = ∫mvdv if you separate the integrals. Where is my mistake and istead of (γ-1)mc² I get -mc²/γ . I know that I am mistaken, I did see some equations but I don't get the 'theoretical' part.

 

[PAllen]

I am doing the integral DK = ∫pdv = mv/(√1-v²/c²)dv , in the same way as they do the ∫Fds = ∫mvdv if you separate the integrals. Where is my mistake and istead of (γ-1)mc² I get -mc²/γ . I know that I am mistaken, I did see some equations but I don't get the 'theoretical' part.

I don't think this approach is valid. The equivalence of ∫Fds to ∫pdv is based on dp/dt being m(dv/dt) which is false in SR because dp/dt needs to include terms from dγ/dt. So your starting point: that you can use ∫pdv as long as you use relativistic momentum is false.

 

[Bill_K]

The integral you should be doing is ∫Fds = ∫(dp/dt)ds = ∫v dp. Nonrelativistically, ∫v dp = ∫p dv since they are both equal to ∫mv dv, but relativistically they are different.

Start from the beginning. The energy-momentum 4-vector is p = (γmc, γmv). Its proper time derivative is

dp/dτ = (mc dγ/dτ, m d(γv)/dτ)

Since its magnitude is constant, p·dp/dτ = 0. Written out, this is

0 = γm²c² dγ/dτ - γm²v d(γv)/dτ
divide by γm: mc² dγ/dτ = mv d(γv)/dτ. Integrate:

∫mc² dγ/dτ dτ = ∫mv d(γv)/dτ dτ
mc² γ + const = ∫v d(mγv)

Choosing a constant of integration, this identity expresses the fact that. even relativistically, the kinetic energy is equal to ∫v dp