# General relativity, geodesic question

1. Jun 25, 2017

### binbagsss

1. The problem statement, all variables and given/known data

question attached

2. Relevant equations

3. The attempt at a solution

Attempt :

Check if $V^{\alpha}\nabla_{\alpha}V^u=0$

Since Minkowski space, connection tensors/christoffel symbols are zero so this reduces to:

$V^{\alpha}\partial_{\alpha}V^u=0$

where $\partial_{\alpha}=\frac{\partial}{\partial x^{\alpha}}$

Where $V^{\alpha}=\frac{\partial x^{\alpha}}{\Lambda}$ is the tangent vector

$\frac{\partial x^{\alpha}}{\Lambda} \partial_{\alpha}V^u = \frac{\partial x^{0}}{\Lambda} \partial_{0}V^u + \frac{\partial x^{1}}{\Lambda} \partial_{1}V^u +\frac{\partial x^{2}}{\Lambda} \partial_{2}V^u +\frac{\partial x^{3}}{\Lambda} \partial_{3}V^u = 4 \frac{\partial V^u}{\Lambda}$ (using the chain rule)

which is 4 separate equations for $u=0,1,2,3,4$, obviously these are not zero.

Have I done something wrong and should I have expected a summation ?

Or would I argue that these can not simultaneously all be zero for the same $\Lambda$ since cosh and sinh are independent?

Many thanks.

2. Jun 26, 2017

### binbagsss

Apologies typo edit:

Where $V^{\alpha}=\frac{\partial x^{\alpha}}{ \partial \Lambda}$ is the tangent vector

$\frac{\partial x^{\alpha}}{\partial \Lambda} \partial_{\alpha}V^u = \frac{\partial x^{0}}{\partial\Lambda} \partial_{0}V^u + \frac{\partial x^{1}}{\partial\Lambda} \partial_{1}V^u +\frac{\partial x^{2}}{\partial\Lambda} \partial_{2}V^u +\frac{\partial x^{3}}{\partial\Lambda} \partial_{3}V^u = 4 \frac{\partial V^u}{\partial\Lambda}$ (using the chain rule)

3. Jun 28, 2017

### strangerep

Which part of the question are you trying to answer in your attempt in post #1? (If you're trying to answer part (c), that depends on the answer to part (a) which you haven't shown.)

Also, which textbook are you working from?

[And btw, I notice in several of your HW posts you make no attempt to write down any "relevant equations". That's not helping.]

4. Jun 29, 2017

### binbagsss

I have not listed relevant equations since they are either included in the question or attempt.

I am attempting part c)

For part a) you simply differentiate w.r.t $\Lambda$ and then compute $g_{uv}V^vV^u$ and check it's sign. $< 0$ here being time-like from the chosen metric convention.

Therefore I would identify a failure to make a connection between these two parts of the question as a mistake in the attempt, rather than purposely not showing to attempt question a, because I thought this wasn't as issue to me. Pretty much then, my approach to the question is wrong, which is all needed to be said.

5. Jun 29, 2017

### strangerep

Are you sure? What are your components $V^\mu$ explicitly?

For part (c), can't you simply evaluate $dV^\mu/d\lambda$ (since the connection is 0 in Minkowski spacetime) ?

And you didn't tell me which textbook the question is from. Or is it from private lecture notes?

6. Jul 1, 2017

### binbagsss

My second post was exactly this? To evaluate $dV^\mu/d\lambda=0$. My question was this is 4 seperate equations? So to check whether it's a geodesic or not you check whether all four equations are consistent in the $\lambda$ which solves them?

For part a)
$V^u=( sinh \Lambda, \frac{1}{\sqrt 3} coshh \Lambda, \frac{1}{\sqrt{2}}cosh \Lambda, \frac{1}{\sqrt{6}}cosh \Lambda)$

So the check for part c) returns the components of the original vector as four seperate equations. My question was concerning solving them anyway. THIS: So to check whether it's a geodesic or not you check whether all four equations are consistent in the $\lambda$ which solves them?

7. Jul 1, 2017

### strangerep

I think you're overcomplicating this. To be a geodesic you'd need $dV^\mu/d\lambda=0$ for all $\lambda$.

8. Jul 2, 2017

### binbagsss

but $\mu$ runs over $\mu=0,1,2,3$ so need to solve, there are four equations to check, what am i over-complicated, don't we need to solve the following system of equations? To which the answer will be inconsistency due to the non-linear dependence of sinh and cosh?

$\frac{dV^{0}}{d\Lambda}=0$
$\frac{dV^{1}}{d\Lambda}=0$
$\frac{dV^{2}}{d\Lambda}=0$
$\frac{dV^{3}}{d\Lambda}=0$

9. Jul 3, 2017

### strangerep

I get the feeling you're not reading my reply hints carefully enough. I'll try one more time:
Note the "for all $\lambda$" part.

10. Jul 4, 2017

### binbagsss

oh okay.
but there are four equations?
so this would have only happened if $V^{u}$ components were constants indepedent of $\Lambda$?

11. Jul 4, 2017

### strangerep

So? It's a single vector (which has 4 components).

If by "this", you mean "the curve would be a geodesic only if..." then yes -- because you're working with Minkowski spacetime in this question.

12. Jul 5, 2017

### binbagsss

As in this is the expected result since we are in flat space-time?
I don't think I'm picking up what you are saying properly, could you explain a tad more?
greatly appreciated, thanks

13. Jul 5, 2017

### strangerep

In Minkowski spacetime, with metric $(-1,+1,+1,+1)$, the Christoffel connection coefficients are all 0. So the equation for a geodesic curve (in terms of an arbitrary parameter $\lambda$) is just $$0 ~=~ \frac{d^2 x^\mu(\lambda)}{d\lambda^2} ~\equiv~ \frac{dV^\mu(\lambda)}{d\lambda} ~.$$ So if this equation is not satisfied (for any component), then the curve is not a geodesic.