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I am having trouble understanding how the following statement (taken from some old notes) is true:

For a 2 dimensional space such that $$ds^2=\frac{1}{u^2}(-du^2+dv^2)$$ the timelike geodesics are given by $$u^2=v^2+av+b$$ where $a,b$ are constants.

When I see "geodesics" I jump to the Euler-Lagrange equations. They give me $$\frac{d}{d\lambda}(-2\frac{\dot u}{u^2})=(-\dot u^2+\dot v^2)(-\frac{2}{u^3})\\ \implies \frac{\ddot u}{u^2}-2\frac{\dot u^2}{u^3}=\frac{1}{u^3}(-\dot u^2+\dot v^2)\\ \implies u\ddot u-\dot u^2-\dot v^2=0$$ and $$\frac{d}{d\lambda}(2\frac{\dot v}{u^2})=0\\ \implies \dot v=cu^2$$ where $c$ is some constant.

Timelike implies $$\dot x^a\dot x_a=-1$$ where I have adopted the $(-+++)$ signature.

I can't for the life of me see how the statement results from these. Would someone mind explaining? Thanks.

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Why did you choose the Euler-Lagrange equations instead, of, say, the geodesic equation for general relativity? –  Muphrid May 13 '13 at 17:14
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@Muphrid: Because the geodesic equation for GR are derived from the E-L equations -- so I a basically using the same thing. Only, the E-L equations are more convenient, imo, because I don't have to compute the Christoffel symbols first and then... –  val May 13 '13 at 17:18

2 Answers 2

up vote 5 down vote accepted

I prefer to use Killing vectors and conservation laws to solve stuff like this, so let's analyze the problem using Killing vectors, and see if the results agree with your Euler-Lagrange equations.

Notice that the metric is invariant under translations of $v$. The associated killing vector is $\partial_v$ which in turn gives the following conserved quantity: $$ c_v = \dot x\cdot \partial_v = \frac{\dot v}{u^2} $$ This agrees precisely with your second Euler-Lagrange equation; so far so good. The timelike condition $\dot x\cdot\dot x = -1$ can be written in components as $$ \frac{-\dot u^2 + \dot v^2}{u^2} = -1 $$ Using the above conservation equation to eliminate $\dot v$ then gives a first order differential equation for $\dot u$ $$ \frac{-\dot u^2+c_v^2u^4}{u^2}=-1 $$ which simplifies to $$ \dot u^2 = c_v^2u^4 + u^2 $$ This is a first order, separable differential equation that can be solved by separation of variables and integration. Once you solve this for $u$, you can plug the solution back into the conservation equation $c_v = \dot v/u^2$ and solve this equation by integration as well. This yields the general solution to the system of differential equations, and then you can relate $u$ and $v$ in the way stated in the quote.

Warning; there may be simpler ways of showing what you want to show.

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Thank you so very much, Josh!! :D –  val May 13 '13 at 20:20
    
@val Sure thing! –  joshphysics May 13 '13 at 22:23

Method 1: Implicit differentiation without explicitly solving ODEs:

$$\frac{d(u^2)}{dv}~=~\frac{1}{\dot{v}} \frac{d(u^2)}{dt}~=~\frac{2u\dot{u}}{\dot{v}}~=~\frac{2\dot{u}}{cu} $$

$$\Downarrow $$

$$\frac{d^2(u^2)}{dv^2}~=~\frac{1}{\dot{v}} \frac{d}{dt}\left(\frac{2\dot{u}}{cu} \right)~=~\frac{2}{\dot{v}} \frac{\ddot{u}u-\dot{u}^2}{cu^2}=\frac{2}{\dot{v}} \frac{\dot{v}^2}{cu^2} ~=~2,$$

which in turn implies OP's sought-for equation. Above we have only used the two Euler-Lagrange equations $\ddot{u}u=\dot{u}^2+\dot{v}^2$ and $\dot{v}=cu^2$.

Method 2: Explicitly solving ODEs:

Rescale the variables as

$$ U~:=~cu, \qquad V~:=~cv. $$

Then the two equations $\dot u^2 = c^2u^4 + u^2$ and $\dot{v}=cu^2$ become

$$ \dot{U}^2 ~=~U^4+U^2, \qquad \dot{V}~=~U^2 ,$$

with full solution

$$U(t)~=~\pm {\rm csch}(t-t_0), \qquad V(t)~=~\coth(t-t_0)+V_0. $$

OP's sought-for equation now follows from

$$(V-V_0)^2 ~=~ U^2 +1.$$

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