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.