GR geodesics

Homework Equations

There are 5 equations we can use.
We have the fact that Lagrangian is a constant for an affinely parameterised geodesic- 0 in this case for a light ray : ##L=0##
And then the Euler-Lagrange equation for each of the 4 variables.

The Attempt at a Solution

The worked solution proceeds by letting ## \dot{y}=0## , where a dot denotes the derivative wrt the affine parameter ##s##.
It uses ##L=0## and then eliminates ##\dot{t}## and ##\dot{x}## via the constants of motion found from Noether's theorem/Euler Lagrange equations.

My question:

How do we know that we have the freedom to set one of the variables ## \dot{y}=0##. (I see by symmetry we could have equally chose ## \dot{x}=0##)..?

(I have worked through without setting ## \dot{y}=0## and then by the symmetry of x and y, using Noether's theorem/Euler - Lagrange we redefine another constant = C_1 + C_2 where C_1 and C_2 are the different constants associated with x/y respectively, however, from just looking at the question, before I start my working out, how do I know I have this freedom?) Thanks

Why do we not have the freedom to set both ## \dot{y}=0## and ## \dot{x}=0##?

(I see that we have 5 equations and 4 variables, does this have something to do with why have freedom to set one of ##\dot{x}## / ##\dot{y}## equal to zero?)

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PeroK
Homework Helper
Gold Member
The problem states that the light ray is moving in the x direction, so initially ##\dot{x} \ne 0##.

The problem states that the light ray is moving in the x direction, so initially ##\dot{x} \ne 0##.
Sorry replying on my phone so let x' represent the derivative as a pose to a dot..

At t=0 do we not have ##y' , z' = 0 ## and ## x' \neq 0 ## and so therefore for the constant of motion associated with ##y## it is ##0## at t=0 and so must be 0 for all of t ?

PeroK
Homework Helper
Gold Member
Sorry replying on my phone so let x' represent the derivative as a pose to a dot..

At t=0 do we not have ##y' , z' = 0 ## and ## x' \neq 0 ## and so therefore for the constant of motion associated with ##y## it is ##0## at t=0 and so must be 0 for all of t ?
Since only ##dy^2## appears in the metric, what would cause a change in the +y direction rather than the -y direction, or vice versa?

Since only ##dy^2## appears in the metric, what would cause a change in the +y direction rather than the -y direction, or vice versa?
i'm not sure? so post 3 is wrong?
the metric components would have to have y dependence?

PeroK
Homework Helper
Gold Member
i'm not sure? so post 3 is wrong?
the metric components would have to have y dependence?
What happens if you apply Euler-Lagrange? You should get something that looks like ##f(z)\dot{y} = C##. If ##\dot{y} = 0## initially, then clearly the solution is that it remains ##0##.

So, this should fall out of the equations easily enough. But, you could also ask what it is physically about this metric that would cause motion in the y-direction if there is none initially? You could flip your y-coordinates and get the same metric. Then, if the solution to the equations is motion in the +y direction, then that implies two different physical solutions depending on which direction is +y.

Whereas, you can't flip your x-coordinate without changing the initial ##\dot{x}## as well.

In other words, overall the problem is symmetrical in ##y## but not in ##x## due to the non-zero initial velocity; and not in ##z## because it's in the metric.

What happens if you apply Euler-Lagrange? You should get something that looks like ##f(z)\dot{y} = C##. If ##\dot{y} = 0## initially, then clearly the solution is that it remains ##0##.
is this not what i said?