# Lorentz transformations

1. Jun 21, 2005

### 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?

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

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

2. Jun 22, 2005

### 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.

Last edited: Jun 22, 2005
3. Jun 23, 2005

### 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?