Lorentz Transformation: Finding d(gamma)/dt

[PlutoniumBoy]

How do we find d(gamma)/dt?

 

[HallsofIvy]

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

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

[tex]= \frac{v}{c^2\sqrt{1- \frac{v^2}{c^2}}}= \frac{v}{\sqrt{c^2- v^2}}[/tex]

 

[CompuChip]

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

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

 

[facenian]

the factor [tex]\frac{dv}{dt}[/tex]
isn't missing in the equation?

[tex]\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)[/tex]

 

[Fredrik]

the factor [tex]\frac{dv}{dt}[/tex]
isn't missing in the equation?

[tex]\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)[/tex]

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