Time derivative of relativistic momentum

1. Sep 16, 2008

diewlasing

How does one take the time derivative of ϒmv ?

I tried treating gamma and mv as separate functions but it just gets messy and ultimately wrong.

2. Sep 17, 2008

gabbagabbahey

Well, the product rule simply gives:
$$\dot {\vec{p}} = \dot {\gamma} m \vec{v} + \gamma \dot {m} \vec{v} + \gamma m \dot {\vec{v}}$$

For systems that conserve mass, $$\dot m =0$$. While,
$$\dot {\gamma} = \frac{d}{dt} \left( 1- \frac{v^2}{c^2} \right) ^{-1/2} = \left( \frac{-1}{2} \right) \left( \frac{2v \dot {v} }{c^2} \right) \left( 1- \frac{v^2}{c^2} \right) ^{-3/2} = - \gamma ^3 \left( \frac{v \dot {v}}{c^2} \right)$$

And so,

$$\dot {\vec{p}} = \gamma m \dot {\vec{v}} -\gamma ^3 m \left( \frac{v \dot {v}}{c^2} \right) \vec{v} =\gamma m {\vec{a}} -\gamma ^3 m \left( \frac{v a}{c^2} \right) \vec{v}$$

3. Sep 18, 2008

snoopies622

Why is $$\left( \frac{2v \dot {v} }{c^2} \right)$$ not $$\left( \frac{-2v \dot {v} }{c^2} \right)$$? What happened to the minus sign?

4. Sep 19, 2008

gabbagabbahey

OOps,. my bad. Yes I accidentally dropped a negative sign. The correct answer should be
$$\dot {\vec{p}} = \gamma m \dot {\vec{v}} +\gamma ^3 m \left( \frac{v \dot {v}}{c^2} \right) \vec{v} =\gamma m {\vec{a}} +\gamma ^3 m \left( \frac{v a}{c^2} \right) \vec{v}$$

5. Sep 20, 2008

diewlasing

Ah yes, thank you

6. Oct 31, 2011

sgwayne

i can't get (−2*v*v/c^2). what rule do we use on this? can anybody show a more detail step on this part?

Last edited: Oct 31, 2011
7. Oct 31, 2011

MasterNewbie

Chain rule. Since v is a function of t, the time derivative of 1-v^2/c^2 is -2v(dv/dt)/c^2.

I posted a similar question a year after this one if you want a second angle.

Last edited: Oct 31, 2011
8. Nov 1, 2011

sgwayne

thanks master newbie, i was confused on the c^2 actually. so we just leave it as it is bcoz we want the derivative of v not c. correct me if i'm wrong.

9. Nov 1, 2011

MasterNewbie

It is left there because c (and c^2) is a constant.