- #1

mcas

- 24

- 5

- Homework Statement
- Check invariance under rotation group in spacetime of a relativistic Newton equation of a charged particle in e-m field with Lorentz force.

- Relevant Equations
- ##\frac{dp^\mu}{ds}=\frac{e}{c}F^{\mu \nu} u_\nu##

I started by inserting ##ds=\sqrt{dx'^{\mu} dx'_{\mu}}## and ##p'^{\mu}=mc \frac{dx'^{\mu}}{ds}##.

So we have:

$$\frac{dp'^{\mu}}{ds}=mc \frac{d}{dx'^{\mu}} \frac{d}{dx'_{\mu}} (x'^{\mu})$$

Now I know that

##dx'^{\mu}=C_\beta \ ^\mu dx^\beta##

and

##dx'_{\mu}=C^\gamma \ _\mu dx_\gamma##

where ##C## is the transformation and ##C_\beta \ ^\mu C^\gamma \ _\mu = \delta^\gamma _\beta##.

Taking this, we have

$$\frac{dp'^{\mu}}{ds}=mc \frac{d}{C_\beta \ ^\mu dx^\beta C^\gamma \ _\mu dx_\gamma} (x'^{\mu})=mc \frac{d}{dx^\beta dx_\beta} (x'^{\mu})=mc \frac{d}{ds} (x'^{\mu})$$

And now if I were to write ##x'^{\mu}=C_\delta \ ^\mu dx^\delta##, this equation wouldn't be an invariant but I think it should be.

So we have:

$$\frac{dp'^{\mu}}{ds}=mc \frac{d}{dx'^{\mu}} \frac{d}{dx'_{\mu}} (x'^{\mu})$$

Now I know that

##dx'^{\mu}=C_\beta \ ^\mu dx^\beta##

and

##dx'_{\mu}=C^\gamma \ _\mu dx_\gamma##

where ##C## is the transformation and ##C_\beta \ ^\mu C^\gamma \ _\mu = \delta^\gamma _\beta##.

Taking this, we have

$$\frac{dp'^{\mu}}{ds}=mc \frac{d}{C_\beta \ ^\mu dx^\beta C^\gamma \ _\mu dx_\gamma} (x'^{\mu})=mc \frac{d}{dx^\beta dx_\beta} (x'^{\mu})=mc \frac{d}{ds} (x'^{\mu})$$

And now if I were to write ##x'^{\mu}=C_\delta \ ^\mu dx^\delta##, this equation wouldn't be an invariant but I think it should be.