Solving for x(t) in the Equation $\frac{d^2 \xi^\alpha}{d \tau^2}=0$

[latentcorpse]

I found a question in one of my past papers that I was unsure about:

Show by explicit solution x(t) that the equation
$\frac{d^2 \xi^\alpha}{d \tau^2}=0$
where $\xi^\alpha=(ct, \vec{x})$ and $\tau$ is proper time, represents a straight world line for a massive particle.

My initial thoughts were just to integrate twice with respect to tau but it's worth 4 marks and so i think it must be a bit more involved than that. Besides if we integrated wrt tau then our answer would be in terms of tau and I think (but not 100%) that we are looking to imply that $\xi^i(t)=A_it+B_i$.

Also, what does $\xi^\alpha=(ct, \vec{x})$ actually mean?

thanks.

 

[Phrak]

It is a very involved problem. This is a 'shortest distance problem' (but in this case maximal).
By the calculus of variations, integral xi is extremal between two spacetime events for a freely falling body. You've been given the solution which you might recognize as an unaccelerated particle in Minkowski space.

You might begin with a spacetime path parametric in lambda. However, you may be able to assume that lambda=tau, the proper time, right off and avoid some pitfalls. I don't recall the details.

 

[latentcorpse]

It is a very involved problem. This is a 'shortest distance problem' (but in this case maximal).
By the calculus of variations, integral xi is extremal between two spacetime events for a freely falling body. You've been given the solution which might be recognizable for the case of an unaccelerated particle in Minkowski space.

You might begin with a spacetime path parametric in lambda. However, you may be able to assume that lambda=tau, the proper time, right off and avoid some pitfalls. I don't recall the details.

thanks for the reply. but i still don't really understand where i start/ where I'm going with this

 

[Phrak]

OK. First, it's late for me, and I wouldn't mind if someone else helped you out, and I'm afraid I could get in over my head. It's been a long time since I had this sort of problem.

I assume you are in flat Minkowski space rather than curved spacetime because I don't see any connection terms. Is that correct?

 

[latentcorpse]

OK. First, it's late for me, and I wouldn't mind if someone else helped you out, and I'm afraid I could get in over my head. It's been a long time since I had this sort of problem.

I assume you are in flat Minkowski space rather than curved spacetime because I don't see any connection terms. Is that correct?

i guess. so. it's part a) of the question and we haven't been given any other information yet so I think that's a fair assumption.

 

[Phrak]

This is my last post tonight, but is this a current homework problem, or one you are going back to and revisiting?

Perhaps you can restate the problem. I'm still confused. Is this a 'shortest distance' problem?

 

[latentcorpse]

i have an exam on GR in a couple of weeks and I'm just doing some past papers. i copied out the question exactly so i don't really know what else i can say...sorry.

 

[Phrak]

I'm afraid this is a matter of the blind leading the blind, but I'll see what I can learn.

Xi is a 4D vector--a spacetime displacement vector. d\xi/d\tau is the change in displacement over the proper time of the particle. You are given the geodesic of xi in Minkowski space. I think you are to show that x_i is also a straight line in 3D space. You might look up parallel transport of a displacement vector. The question may be asking you to prove that d/dt(dx_idt)=0.

 

[Ben Niehoff]

The OP's first hunch is correct: You are given the differential equation, all you need to do is integrate it. In fact, you don't even need to integrate it all the way. Once you have

$$\frac{d\xi^\mu}{d\tau} = \mathrm{const}$$

then this should obviously be a straight line, because its velocity vector is constant.

 

[stevenb]

I have a question which maybe is really just a question of terminology.

The title of the thread is "General Relativity", so the given equation is not the geodesic equation of GR. If a particle is obeying that equation with the presence of gravity, then there must be another external force that keeps the massive particle on a "straight worldline" line in the observer's reference frame.

Is this truly a "straight worldline" in general? That is, with gravity, is it correct to say that the particle is moving on a "straight worldline"? Clearly it is straight in SR no matter what inertial reference frame is chosen. However, I thought that, in GR, the geodesic was the "straightest possible path" and it is clear that this path is not a geodesic. Are these conflicting statements;

1. The worldline is straight
2. The particle is not on a geodesic

or, do they just say different things and don't contradict or conflict at all?

 

[Phrak]

Is this truly a "straight worldline" in general?

That's correct.

$$\frac{d^2 \xi^\alpha}{d \tau^2}=0$$

isn't a geodesic in general relativity, in general, but rather this one is:

$${\frac{d^2\xi^\alpha}{d\tau^2} + \Gamma^{\alpha}_{\mu\nu} \frac{\xi^\mu}{d\tau} \frac{\xi^\nu}{d\tau} = 0$$

 

[Phrak]

The OP's first hunch is correct: You are given the differential equation, all you need to do is integrate it. In fact, you don't even need to integrate it all the way. Once you have

$$\frac{d\xi^\mu}{d\tau} = \mathrm{const}$$

then this should obviously be a straight line, because its velocity vector is constant.

I don't see it. I know that $$\partial_t \xi^i$$ is constant for constant velocity, but I don't see it in your equation.

 

[stevenb]

That's correct.

$$\frac{d^2 \xi^\alpha}{d \tau^2}=0$$

isn't a geodesic in general relativity, in general, but rather this one is:

$${\frac{d^2\xi^\alpha}{d\tau^2} + \Gamma^{\alpha}_{\mu\nu} \frac{\xi^\mu}{d\tau} \frac{\xi^\nu}{d\tau} = 0$$

Ok, that's what I expected. The first equation is basically the definition of a straight line for an observer.

The main point of confusion for me was that almost all books on GR use the phrase "geodesics are the straightest possible path". I always found this phrasing confusing, but I guess they are using a different definition of "straight" when they say this. An observer may not see a geodesic path as being straight, and a geodesic will not match the first definition in GR.

 

[latentcorpse]

ok. we usually use coordinates $\xi^\alpha$ for a particle moving in a locally inertial frame if that helps?

 

[nickjer]

You were right in your first post latentcorpse since this is flat spacetime. So a geodesic would be a straight line as you solved for by integrating twice (also linear momentum is conserved and hence why you get the first derivative to be constant). But if this were curved spacetime then you would have the more complicated equation that Phrak listed 4 posts up.

 

[stevenb]

You were right in your first post latentcorpse since this is flat spacetime. ...

Now I'm really confused. Perhaps the OP can clarify if this problem assumes flat Minkowski spacetime. The title of the tread is "General Relativity", hence that is why I asked my first question above.

 

[Phrak]

Now I'm really confused. Perhaps the OP can clarify if this problem assumes flat Minkowski spacetime. The title of the tread is "General Relativity", hence that is why I asked my first question above.

In a course on General Relativity the mathematical machinery is established that is capable of making some sense of intrinsically curved manifolds. It makes a great deal so sense to first apply the machinery to the simpler problems of constant metric that will be generalized later, wouldn't you agree?

 

[stevenb]

In a course on General Relativity the mathematical machinery is established that is capable of making some sense of intrinsically curved manifolds. It makes a great deal so sense to first apply the machinery to the simpler problems of constant metric that will be generalized later, wouldn't you agree?

Yes, I definitely agree. I just was asking for clarification. I take it that is the explanation then. I normally expect the assumption to be stated. The OP also said that the coordinate system was locally inertial which also seems to imply curved space with the use of Riemann normal coordinates.

Anyway, if the assumption is SR, then my questions don't make sense and should be withdrawn with an apology for missing the boat.

 

[Phrak]

Yes, I definitely agree. I just was asking for clarification. I take it that is the explanation then. I normally expect the assumption to be stated. The OP also said that the coordinate system was locally inertial which also seems to imply curved space with the use of Riemann normal coordinates.

Anyway, if the assumption is SR, then my questions don't make sense and should be withdrawn with an apology for missing the boat.

Hey, no worries. I think you may have inadvertently read 'locally inertial' into the question--which puts us in the same boat.