Recent content by Wotso

Thanks, I think I got it. Love you all =D

But I'm not there yet. the last step is also a bit too much for me. help?

I got it now =D I was looking totally in the wrong direction. \frac{d}{dv}\,\frac{1}{\sqrt{z}} = \left( \frac{d}{dz} \, \frac{1}{\sqrt{z}} \right)\frac{dz}{dv} z=1-v2/c2 Y=1/sqrt(z) first the dm/dt becomes dm/dv * dv/dt dm/dv becomes m0 *dY/dz *dz/dv and then first part. dz-0,5/dz...

Well, I think I'm almost there... I just can't figure out what to make of the dv (and looking ahead, I probably won't be able to do the next step either, but that depends on the solution =/ ) Thanks alot.

well, it does explain where that dv/dt comes from.. But I still have problems, Now I am thinking like this: ( which might be completely wrong) dm/dv multiplied with v = what is on the left of the + sign without the dv/dt and dm = m - m0 and dv must be something like v -v0 which...

First of all, I'm not sure where to post this. Now, I am reading some lectures about relativity, I came quite far and understand the concept. The steps in the following I just can't follow. "Kinetic Energy and Mass for Particles of Arbitrary Speed For a particle of rest mass m0...