- #1
sspitz
- 21
- 0
A point mass has a force on it in its rest frame (F). Now go to a frame moving in the +x direction (F'). EM book claims the forces can be related like this:
[tex]
f'_{x'}=f_{x}\\f'_{y'}=\frac{f_{y}}{\gamma}\\f'_{z'}=\frac{f_{z}}{\gamma}
[/tex]
I would like to be able to see this with four vectors, but am having trouble. Four vectors have arrows. Tau is proper time.
[tex]
\vec{f}=\frac{dp}{d\tau}=\Gamma(\frac{dp_{x}}{dt},\frac{dp_{y}}{dt},\frac{dp_{z}}{dx},\frac{dE}{dt} \frac{1}{c})=(f_{x},f_{y},f_{z},\frac{dE}{dt} \frac{1}{c})\\
\vec{f'}=\frac{dp'}{d\tau}=\gamma(f'_{x'},f'_{y'},f'_{z'},\frac{dE}{dt'} \frac{1}{c})
[/tex]
In this case, big gamma=1 because there is no velocity in the rest frame at the time of interest.
Transform
[tex]
\vec{f'}=(\gamma f_{x} - \gamma \beta \frac{dE}{dt} \frac{1}{c},f_{y},f_{z},...)
[/tex]
I can see for y and z, but not for x.
[tex]
f'_{x'}=f_{x}\\f'_{y'}=\frac{f_{y}}{\gamma}\\f'_{z'}=\frac{f_{z}}{\gamma}
[/tex]
I would like to be able to see this with four vectors, but am having trouble. Four vectors have arrows. Tau is proper time.
[tex]
\vec{f}=\frac{dp}{d\tau}=\Gamma(\frac{dp_{x}}{dt},\frac{dp_{y}}{dt},\frac{dp_{z}}{dx},\frac{dE}{dt} \frac{1}{c})=(f_{x},f_{y},f_{z},\frac{dE}{dt} \frac{1}{c})\\
\vec{f'}=\frac{dp'}{d\tau}=\gamma(f'_{x'},f'_{y'},f'_{z'},\frac{dE}{dt'} \frac{1}{c})
[/tex]
In this case, big gamma=1 because there is no velocity in the rest frame at the time of interest.
Transform
[tex]
\vec{f'}=(\gamma f_{x} - \gamma \beta \frac{dE}{dt} \frac{1}{c},f_{y},f_{z},...)
[/tex]
I can see for y and z, but not for x.