- #1
Sissy
- 6
- 0
Hello
I have huge problems with the following exercise. Please give me some hints. No complete Solutions but a little bit help.
Find the differential equations of the paths of test particles in the space-time of which the metric ist
[tex]\mathrm{d}s^2 = e^{2kx} \left[- \left( \mathrm{d}x^2 + \mathrm{d}y^2 +\mathrm{d}z^2 \right) + \mathrm{d}t^2 \right][/tex],
where [tex]k[/tex] is a constant. If
[tex] v^2 = \left( \dfrac{\mathrm{d} x }{\mathrm{d} t } \right)^2 + \left( \dfrac{\mathrm{d}y }{\mathrm{d}t } \right)^2 + \left( \dfrac{ \mathrm{d} z }{\mathrm{d} t } \right)^2 [/tex]
and if [tex]v=V[/tex] when [tex]x=0[/tex], show that
[tex] 1-v^2 = \left( 1-V^2 \right) e^{2kx} [/tex].
Now I have no idea how to start. I do not want a solution. I will calculate it on my own but I need some assistance.
greetings
I have huge problems with the following exercise. Please give me some hints. No complete Solutions but a little bit help.
Find the differential equations of the paths of test particles in the space-time of which the metric ist
[tex]\mathrm{d}s^2 = e^{2kx} \left[- \left( \mathrm{d}x^2 + \mathrm{d}y^2 +\mathrm{d}z^2 \right) + \mathrm{d}t^2 \right][/tex],
where [tex]k[/tex] is a constant. If
[tex] v^2 = \left( \dfrac{\mathrm{d} x }{\mathrm{d} t } \right)^2 + \left( \dfrac{\mathrm{d}y }{\mathrm{d}t } \right)^2 + \left( \dfrac{ \mathrm{d} z }{\mathrm{d} t } \right)^2 [/tex]
and if [tex]v=V[/tex] when [tex]x=0[/tex], show that
[tex] 1-v^2 = \left( 1-V^2 \right) e^{2kx} [/tex].
Now I have no idea how to start. I do not want a solution. I will calculate it on my own but I need some assistance.
greetings