Lorentz Transformation: Finding d(gamma)/dt

[PlutoniumBoy]

How do we find d(gamma)/dt?

 

[HallsofIvy]

gamma is \sqrt{1- v^2/c^2}?

Differentiating that is fairly basic "Calculus I".
\gamma= \left(1- \frac{v^2}{c^2}\right)^{\frac{1}{2}}
so
\frac{d\gamma}{dt}= \frac{1}{2}\left(1- \frac{v^2}{c^2}\right)^{-\frac{1}{2}}\left(2\frac{v}{c^2}\right)

= \frac{v}{c^2\sqrt{1- \frac{v^2}{c^2}}}= \frac{v}{\sqrt{c^2- v^2}}

 

[CompuChip]

HallsOfIvy, I think you differentiated with respect to v, instead of t.
For the t derivative, use the chain rule
\frac{d\gamma}{dt} = \frac{d\gamma}{dv} \frac{dv}{dt}

And I didn't check, but you might have missed a minus sign (it's -v^2 giving -2v isn't it?)

 

[Meir Achuz]

Except \gamma=\frac{1}{\sqrt{1-v^2/c^2}}
which I'm sure you knew before you answered the question.

 

[facenian]

the factor \frac{dv}{dt}
isn't missing in the equation?

\frac{d\gamma}{dt}= \frac{1}{2}\left(1- \frac{v^2}{c^2}\right)^{-\frac{1}{2}}\left(2\frac{v}{c^2}\right)

 

[Fredrik]

the factor \frac{dv}{dt}
isn't missing in the equation?

\frac{d\gamma}{dt}= \frac{1}{2}\left(1- \frac{v^2}{c^2}\right)^{-\frac{1}{2}}\left(2\frac{v}{c^2}\right)

It is, along with a minus sign. See CompuChip's post.