- #1
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Homework Statement
I have derived the metric in for 2D Rindler space in a previous problem and it is explicitly given again here:
[tex] ds^2 = dx^2 - (dx^0)^2 = dw^2 - (1+gw/c^2)^2(dw^0)^2 [/tex],
where [itex] (x^0, x)[/itex] are Minkowski coordinates in an intertial system I and [itex] (w^0,w) [/itex] the Rindler coordinates of a system of reference R with constant acceleration g relative to I.
In the question I have furthermore derived the solution for a particle starting at w=w~ and with velocity zero for w^0=0:
[tex] w(w^0) = \frac{c^2}{g}\left( (1+g\tilde w/c^2)\frac{1}{\cosh(gw^0/c^2)} -1 \right) [/tex]
2. Question
Calculate the velocity [tex] v = c \frac{dw}{dw^0}[/tex]. Find the maximum velocity of the particle and compare with the speed of light (calculate that too). What happens for [itex] w^0 \rightarrow \infty [/itex]?
The Attempt at a Solution
I have solved all the questions and I am sure they are correct; only the interpretation is lacking:
[tex] v_{\mathrm{particle}} = -c(1+g\tilde w/c^2)\frac{\sinh(gw^0/c^2)}{\cosh^2(gw^0/c^2)}[/tex]
[tex] v_{\mathrm{particle, max}} = \pm c/2(1+g\tilde w/c^2)[/tex]
[tex] v_{\mathrm{light}} = \pm c(1+g\tilde w/c^2)\frac{1}{\cosh(gw^0/c^2)}[/tex]
[tex] v_{\mathrm{light, max}} = \pm c(1+g\tilde w/c^2)[/tex]
Now for the interpretation:
-First of all: I presume the minus sign in the equation for the speed of the particle just reflects the fact that for w^0<0 it travels in one direction and for [itex] w^0>0 [/itex] in another direction?
-I find it weird that the speed of light is not a fixed c. I know that the for light [itex] ds^2 \equiv 0[/itex], but I still don't find this answer rather comforting. Can anyone elaborate on this?
-In the limit [itex] w^0 \rightarrow \infty [/itex] the speed of the particle and of the light go to zero. I completely do not understand what is happening here.
-What does it mean that the maximum speed of light is two times larger than the maximum particle velocity?