What Does the Derivative of the Lorentz Gamma Function Indicate?

[Orion1]

What is the signifigance of the first derivative of the Lorentz transformation gamma function with respect to dv?

What type of system does this derivative represent?

\gamma'(v) = \frac{d}{dv} \left( \frac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} \right) = \frac{v}{c^2 \left[ 1 - \left( \frac{v}{c} \right)^2 \right]^\frac{3}{2}}

\boxed{\gamma'(v) = \frac{v}{c^2 \left[ 1 - \left( \frac{v}{c} \right)^2 \right]^\frac{3}{2}}}

 

[Mortimer]

I don't see any usefulness for this particular derivative. Why do you ask?

\frac{d(1/\gamma)}{dv}
might be meaningful. It represents the change of the time-velocity cd\tau/dt}=c/\gamma (see e.g. Brian Greene's "The elegant universe") as a function of the change of the spatial velocity v. The function is goniometric.

 

[Orion1]

Reletive Relation...


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

\boxed{\gamma'(v)^{-1} = - \frac{v}{c^2 \sqrt{1 - \left( \frac{v}{c} \right)^2 }}}

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

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

Are these equation solutions correct?