- #1
Fisherlam
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- Homework Statement
- For the two-dimensional metric ##ds^2 = [dx^2 + c^2dt^2] /(\alpha t^{-2})##, with ##\alpha## being a constant of appropriate dimensions, show that $$\frac{dx/dt}{\sqrt{1-(dx/dt)^2}}$$ is constant and hence, or otherwise, find all timelike geodesic curves.
- Relevant Equations
- $$L=g_{ab}\dot{x}^a\dot{x}^b $$ $$\frac{\partial L}{\partial x}=\frac{\partial }{\partial u}\left(\frac{\partial L}{\partial \dot{x}}\right) $$
Using EL equation, $$L=\left(\frac{t^2}{\alpha}\dot{x}^2-\frac{c^2t^2}{\alpha}\dot{t}^2\right)^{0.5} \Longrightarrow \mathrm{constant} =\left(\dot{x}^2 -c^2 \dot{t}^2\right)^{-0.5} \left(\frac{t^2}{\alpha}\right)^{0.5} \dot{x}$$.
Get another equation from the metric: $$ds^2=-\frac{c^2t^2}\alpha dt^2+\frac{t^2}\alpha dx^2=c^2d\tau^2\quad\Longrightarrow\quad-\frac{c^2t^2}\alpha t^2+\frac{t^2}\alpha\dot{x}^2=c^2\quad\Longrightarrow\quad\frac{t^2}\alpha=\frac{c^2}{\dot{x}^2-c^2\dot{t}^2}$$
Substitution and set ##c=1##: $$\mathrm{constant}=\left(\dot{x}^2-c^2\dot{t}^2\right)^{-0.5}\left(\frac{t^2}\alpha\right)^{0.5}\dot{x}=\frac{c\dot{x}}{\dot{x}^2-c^2\dot{t}^2}=\frac{\dot{x}}{\dot{x}^2-\dot{t}^2}=\cdots?$$
I think I am close but clearly missing something...
Get another equation from the metric: $$ds^2=-\frac{c^2t^2}\alpha dt^2+\frac{t^2}\alpha dx^2=c^2d\tau^2\quad\Longrightarrow\quad-\frac{c^2t^2}\alpha t^2+\frac{t^2}\alpha\dot{x}^2=c^2\quad\Longrightarrow\quad\frac{t^2}\alpha=\frac{c^2}{\dot{x}^2-c^2\dot{t}^2}$$
Substitution and set ##c=1##: $$\mathrm{constant}=\left(\dot{x}^2-c^2\dot{t}^2\right)^{-0.5}\left(\frac{t^2}\alpha\right)^{0.5}\dot{x}=\frac{c\dot{x}}{\dot{x}^2-c^2\dot{t}^2}=\frac{\dot{x}}{\dot{x}^2-\dot{t}^2}=\cdots?$$
I think I am close but clearly missing something...
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