Solving Geodesic Equations with Euler-Lagrange and Noether's Theorem

[binbagsss]

Homework Statement

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?)

Many thanks in advance

 

[PeroK]

The problem states that the light ray is moving in the x direction, so initially $\dot{x} \ne 0$.

 

[binbagsss]

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]

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?

 

[binbagsss]

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]

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.

 

[binbagsss]

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?