Time derivative of relativistic momentum help

[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.

 

[gabbagabbahey]

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.

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}

 

[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?

 

[gabbagabbahey]

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?

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}

 

[diewlasing]

Ah yes, thank you

 

[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?

 

[MasterNewbie]

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?

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

https://www.physicsforums.com/showthread.php?t=343032
I posted a similar question a year after this one if you want a second angle.

 

[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.

 

[MasterNewbie]

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