- #1
jk4
[SOLVED] Force using relativistic momentum
I have an example problem in a textbook I'm reading:
"Find the acceleration of a particle of mass m and velocity v when it is acted upon by the constant force F, where F is parallel to v.
then it proceeds to show the solution:
F = \frac{d}{dt}(\gamma mv) = m\frac{d}{dt}(\frac{v}{\sqrt{1-v^{2}/c^{2}}})
I get all that so far. The next step is where it loses me:
= m[\frac{1}{\sqrt{1-v^{2}/c^{2}}} + \frac{v^{2}/c^{2}}{(1-v^{2}/c^{2})^{3/2}}] \frac{dv}{dt}
so I don't know how they got there... and then the next step confuses me also. They go from above to here:
= \frac{ma}{(1-v^{2}/c^{2})^{3/2}}
and then of course there a few more steps after that one, but I can get those, I'm just confused about those two steps. Please help clarify it for me, thank you.
I have an example problem in a textbook I'm reading:
"Find the acceleration of a particle of mass m and velocity v when it is acted upon by the constant force F, where F is parallel to v.
then it proceeds to show the solution:
F = \frac{d}{dt}(\gamma mv) = m\frac{d}{dt}(\frac{v}{\sqrt{1-v^{2}/c^{2}}})
I get all that so far. The next step is where it loses me:
= m[\frac{1}{\sqrt{1-v^{2}/c^{2}}} + \frac{v^{2}/c^{2}}{(1-v^{2}/c^{2})^{3/2}}] \frac{dv}{dt}
so I don't know how they got there... and then the next step confuses me also. They go from above to here:
= \frac{ma}{(1-v^{2}/c^{2})^{3/2}}
and then of course there a few more steps after that one, but I can get those, I'm just confused about those two steps. Please help clarify it for me, thank you.
