Varying the action of a point particle in curved spacetime

In summary, the homework asks for the equation of motion for a particle following a given trajectory, with arbitraryparameters. After varying the action, one must use the Euler-Lagrange equation to derive the geodesic equation. Roughly, this is a derivative with respect to ##x^\mu## and ##\dot x^\mu##.
  • #1
Verdict
117
0

Homework Statement


What I have to do is vary the action
[tex] S =-mc \int ds = -mc \int \sqrt{-g_{\mu\nu}(x(t))\frac{dx^\mu(t)}{dt} \frac{dx^\nu(t)}{dt}} \ \ dt [/tex]
to find the equation of motion for [tex]x^\mu(t)[/tex]


The Attempt at a Solution



Now, to begin with, I have to admit that I am having quite some trouble with this course. I have done some exercises varying the action before, but they were a lot easier. What I 'learned' from that was that I'll probably have to linearize using a taylor expansion, and in the end I'll have to use partial integration.

But my main problem is, how do I start? I want to take S(x+dx) - S(x), but how do I do that here?

I apologize if this is not what one would call attempting to solve the question. I'm just pretty much lost when it comes to this. Pointing me towards a textbook in which this is explained is just as awesome (or even more) as giving me a helping hand as to start solving this problem.

(note: the actual question is much longer than this, but if I can't solve this part, I'll get nowhere)
 
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  • #2
Verdict said:

Homework Statement


What I have to do is vary the action
[tex] S =-mc \int ds = -mc \int \sqrt{-g_{\mu\nu}(x(t))\frac{dx^\mu(t)}{dt} \frac{dx^\nu(t)}{dt}} \ \ dt [/tex]
to find the equation of motion for [tex]x^\mu(t)[/tex]


The Attempt at a Solution



Now, to begin with, I have to admit that I am having quite some trouble with this course. I have done some exercises varying the action before, but they were a lot easier. What I 'learned' from that was that I'll probably have to linearize using a taylor expansion, and in the end I'll have to use partial integration.

But my main problem is, how do I start? I want to take S(x+dx) - S(x), but how do I do that here?

I apologize if this is not what one would call attempting to solve the question. I'm just pretty much lost when it comes to this. Pointing me towards a textbook in which this is explained is just as awesome (or even more) as giving me a helping hand as to start solving this problem.

(note: the actual question is much longer than this, but if I can't solve this part, I'll get nowhere)

What you have to use is the Euler-Lagrange equations. And what you are doing here is deriving the geodesic equation from the action S. That should be in any number of books.
 
  • #3
Well, I checked with the teacher actually, and he really wants us to actually vary the action, were not allowed to use the Euler-Lagrange equations. But I will try and look up a source that does this for the geodesic equation, thank you :)
 
  • #4
Hm, one thing I forgot to add is that t is an arbitrary parameter here, so after the variation I can replace it with tau, proper time, and clear most of the things out, just to be left with a c. But sadly, I still don't see how to get there. (Does this make sense?)
 
  • #5
Verdict said:
Hm, one thing I forgot to add is that t is an arbitrary parameter here, so after the variation I can replace it with tau, proper time, and clear most of the things out, just to be left with a c. But sadly, I still don't see how to get there. (Does this make sense?)

It's hard to say until you show more of your work.
 
  • #6
I understand, I just don't know where to begin really.
Looking at my syllabus, it seems as if the variation is effectively just a derivative. So would it be right to say that
[tex]\delta S[/tex] is just 1 divided by two times the square root, multiplied by three terms (from the chain rule)? I apologize for such rough work, I am really not comfortable with these calculations as of now.
 
  • #7
Verdict said:
I understand, I just don't know where to begin really.
Looking at my syllabus, it seems as if the variation is effectively just a derivative. So would it be right to say that
[tex]\delta S[/tex] is just 1 divided by two times the square root, multiplied by three terms (from the chain rule)? I apologize for such rough work, I am really not comfortable with these calculations as of now.

Roughly, sure. You are going to have derivatives with respect to ##x^\mu## and ##\dot x^\mu## and yes, variation is effectively a derivative. It's explicitly spelled out in the Euler-Lagrange equation. They are the same thing as a 'variation'. Look up some examples of using that.
 
  • #8
Well, the exercise is really deriving those equations, I guess.
What I've done so far can be written as
[tex] \delta S = -mc \int \frac{-(\frac{g_{\mu\nu}(x(t))}{dx^ \lambda }\delta x^ \lambda \frac{dx^\mu(t)}{dt} \frac{dx^\nu(t)}{dt})-(g_{\mu\nu}(x(t))\frac{d\delta x^\mu(t)}{dt} \frac{dx^\nu(t)}{dt})-(g_{\mu\nu}(x(t))\frac{dx^\mu(t)}{dt} \frac{d \delta x^\nu(t)}{dt})}{2\sqrt{-g_{\mu\nu}(x(t))\frac{dx^\mu(t)}{dt} \frac{dx^\nu(t)}{dt}}} \ \ dt [/tex]

which is a bit neater with xdot notation and such, but I'm not that good at latex yet, so I apologize.

But from here on, I don't know how to continue. I can combine two of the terms up top if the metric is symmetric (which I think it is?), but then..

(I did this literally applying the chain rule, I hope that's ok)
 
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  • #9
Ah, I got it. I can use a trick to make the square root term into a c, replacing t by proper time. Then up top, I just need to combine the two terms at the end, partially integrate and relabel a bit from there. I do indeed end up with the equations of motion for curved space. Awesome :) Thanks for taking the time to look at this, next time I'll try and post more work. (Where I got stuck was that you could just take [tex]\delta S[/tex] to be a derivative of some sorts, instead of having to do the taylor expansions)
 

FAQ: Varying the action of a point particle in curved spacetime

What is a point particle?

A point particle is a theoretical object that has no size or shape, and is treated as having all of its mass concentrated at a single point.

What is curved spacetime?

Curved spacetime is a concept in Einstein's theory of general relativity, where the presence of mass and energy causes the fabric of space and time to bend or curve.

How does a point particle behave in curved spacetime?

A point particle follows the shortest possible path in curved spacetime, known as a geodesic. This path is determined by the curvature of spacetime and the mass of the point particle.

How can we vary the action of a point particle in curved spacetime?

The action of a point particle in curved spacetime can be varied by changing its mass, its velocity, or the curvature of spacetime itself. This can be done through external forces or by changing the distribution of mass and energy in the surrounding spacetime.

What are the implications of varying the action of a point particle in curved spacetime?

Varying the action of a point particle in curved spacetime can affect its trajectory and the overall shape of spacetime. It can also impact the behavior of other particles and objects in the surrounding spacetime, leading to phenomena such as gravitational lensing and the bending of light.

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