Why do particles move along longest proper time trajectories

[MeJennifer]

If you define "staying at the same spatial location" as standing on the surface of the Earth, yes. ...

Well, it was not my definition.

I was merely pointing out that proving or demonstrating things in GR with coordinate dependent examples is generally not a good idea.

 

[stevendaryl]

Well, it was not my definition.

Well, you weren't using my definition, either. I clearly said that I was talking about locally Minkowskian coordinate systems.

I was merely pointing out that proving or demonstrating things in GR with coordinate dependent examples is generally not a good idea.

The whole point about covariance is that you can use whatever coordinate system is convenient to compute invariant quantities. You get the same answer, regardless of the choice of coordinates.

 

[PAllen]

There's only one straight line between any two nearby points, and light in vacuum always travels in a straight line (any other path would imply that the light changes its speed through space, but we know that that speed is c and it doesn't change), so there's only one possible path light can follow. Thus, the zero proper-time path that light follows is extremal in a trivial sense - it is the only path so it is both the longest and the shortest path.

You can trivially have null paths that are not geodesics, so the statement that light travels on a null geodesic is substantive statement about physics. Just consider circular motion at c, for example. It is a null path but not a null geodesic. In this case, using an action principle, you are picking out a saddle point. Using a parallel transport definition, you are picking a path that is everywhere locally as straight as possible.

 

[Orodruin]

Well, it was not my definition.

I was merely pointing out that proving or demonstrating things in GR with coordinate dependent examples is generally not a good idea.

Which is why stevendaryl explicitly stated that he was providing a special case in order to help intuition, he was never claiming that this was a completely general proof. In Minkowski space, what he said was that you can find an inertial frame where it is the case that the events have the same spatial coordinates. There is absolutely nothing wrong with that and I think you are reading too much into what he was saying. Your argument certainly does not come across as simply pointing out the GR case, but rather argumentative.

 

[stevendaryl]

You can trivially have null paths that are not geodesics, so the statement that light travels on a null geodesic is substantive statement about physics. Just consider circular motion at c, for example. It is a null path but not a null geodesic. In this case, using an action principle, you are picking out a saddle point. Using a parallel transport definition, you are picking a path that is everywhere locally as straight as possible.

But if you pick two nearby events $e_1$ and $e_2$, then if the separation is null, then there is only one path connecting the events that is everywhere null. Any null path passing through $e_1$ that is not a geodesic will fail to pass through $e_2$. I think that's right.

 

[PAllen]

Hi vanhees71,

I do not agree with "That's the action principle". The action principle only says you take an extreme (min or max) of the action. In optics this normally leads to the fastest path and often shortest path a photon/wave can propagate. Here it is said that the paths are those with the longest proper time. That is counter-intuitive, at least to me. I've figured out by now that proper time really is a bit special. I was wondering if there nevertheless is a easy way to understand why the extreme action always leads to the longest proper time...

do you have any thoughts on this?

Bas

PS: Yes, I mean when no external forces are applied.

Well, it is the nature of the metric. Instead of Riemannian space, with metric signature (+, +, +, ..) you have pseudo-Riemannian spactime with signature (+, -, -, -) or (-, +, +, +) depending on your convention. This is exactly what makes an extremal for timelike path a local maximum rather than a minimum.

 

[PAllen]

But if you pick two nearby events $e_1$ and $e_2$, then if the separation is null, then there is only one path connecting the events that is everywhere null. Any null path passing through $e_1$ that is not a geodesic will fail to pass through $e_2$. I think that's right.

But this is rather tautological. Null separation means connected by a null geodesic, and for nearby points, geodesic is unique. I was simply emphasizing that the statement that light travels on null geodesics has just as much physical content as that inertial motion follows timelike geodesics.

 

[write4u]

I am sorry, but I must be missing something here.

As I understand it photons travel a maximal speed and arrive at the target in minimal time allowed by spacetime.
Example: a photon inside the sun may spend hundreds of years inside the sun , being bounced around until it escapes the interior and then it takes only 8 minutes to reach earth.
OTOH, a slug travels at minimal speed and uses maximal time to arrive at the same target.

Moreover, Einstein (in his example of the man in the box) proved that relatively it makes no difference if a photon appears to move in a curve or a straight line. It has no effect on the time it takes to travel the same distance.

Am I caught in a semantic misunderstanding?

 

[stevendaryl]

But this is rather tautological. Null separation means connected by a null geodesic, and for nearby points, geodesic is unique.

I think you misunderstood the claim that was being made: Given nearby null-separated events $e_1$ and $e_2$ there is exactly one null path connecting them. That's a stronger claim than saying that there is exactly one null geodesic connecting them.

That's a different situation from timelike separations. There are many timelike paths connecting two timelike-separated events, but there is only one null path connecting null-separated events.

 

[PAllen]

I think you misunderstood the claim that was being made: Given nearby null-separated events $e_1$ and $e_2$ there is exactly one null path connecting them. That's a stronger claim than saying that there is exactly one null geodesic connecting them.

That's a different situation from timelike separations. There are many timelike paths connecting two timelike-separated events, but there is only one null path connecting null-separated events.

Ok, yes I agree with this. In fact, it is timelike separated events that can have non-geodesic null paths between them.

 

[pervect]

Hi,

I am working my way thought Hartle's Gravity. In Section 5.4 he states that "The straight lines along which free particles move in spacetime are paths of longest proper time" and proceeds to proof that "in flat space time the proper time is a curve of extremal proper time".

Can someone explain why it actually are the paths with longest proper time? Light travels along path with zero proper time. Is that a longest path (normal paths being negative)?

Thanks
Bas

A slightly more precise statement is one that says the curve of a body in "natural motion", i.e. force-free motion, extremizes proper time. Note that this is what Hartle said, formally. (The excrutiatingly correct statement, according to the referece I just read, is that the curve of natural motion is a curve of stationary action).

http://www.eftaylor.com/leastaction.html has some helpful articles on the action principle, specifically "When action is not least" addresses the issue of why we say the action is stationary rather than minimal.

As discussed in "When action is not least", the idea that action is "least" rather similar to the idea that a spatial geodesic minimizes length. This is always true on a plane, it's true only in a sufficiently small local region on a curved spatial geometry such as the surface of the sphere. I'll leave it at that rather than try to explain further, and refer the interested reader to the article.

As far as your particular question goes, I'd say that the short and simple version is that proper time is maximized only if a) one restricts oneself to a set of time-like curves, and b) consider only a "sufficiently small" region of space-time. The problem in your case is a). Things do break down when you try to apply the principle of least action to light-like curves. The solution to this is not to do it - instead, apply the correct principle of stationary action.

 

[MeJennifer]

Moreover, Einstein (in his example of the man in the box) proved that relatively it makes no difference if a photon appears to move in a curve or a straight line. It has no effect on the time it takes to travel the same distance.

Time taken for which observer?

For instance consider the following thought experiment.

In a Schwarzschild solution stationary observers A_r1 and B_r2 (r2 > r1 > r-event horizon) send each other light signals and clock the time it takes light to go from A to B and back to A and from B to A and back to B.

Would you think light takes exactly the same round trip time for both observers?

 

[Gerry99]

Hi,

I am working my way thought Hartle's Gravity. In Section 5.4 he states that "The straight lines along which free particles move in spacetime are paths of longest proper time" and proceeds to proof that "in flat space time the proper time is a curve of extremal proper time".

Can someone explain why it actually are the paths with longest proper time? Light travels along path with zero proper time. Is that a longest path (normal paths being negative)?

Thanks
Bas

Only need to prove the proper time between p0 and p2 is larger than p0 to p1 + p1 to p2.
We can make p0 to p1: {t1, x1} = τ1{cosh a1, sinh a1}, p1 to p2: {t2, x2}=τ2{cosh a2, sinh a2}, so p0 to p2 is {t3, x3}={τ1cosh a1 + τ2 cosh a2, τ1 sinh a1 + τ2 sinh a2}
only need to prove |S₁| + |S₂| < |S₃|