# I Why do particles move along longest proper time trajectories

1. Jun 22, 2016

### Staff: Mentor

No, I'm not. Do the math.

Which is the path of maximal proper time between two fixed events, for a small enough region of spacetime. In the case under discussion, the two events are the ball being thrown and the ball being caught. The ball travels on a geodesic between these two events. The person standing on Earth does not. The region of spacetime containing the two events and both worldlines between them is small enough that we can use the "geodesic has maximal proper time" rule. So the ball's proper time is longer between the two events. (Or we can do the math explicitly, as I've asked you to do.)

2. Jun 22, 2016

### Orodruin

Staff Emeritus
Yes, and this geodesic locally maximises the proper time. This is true in Minkowski space as well as in any other Lorentzian manifold just as a geodesic locally minimises the path in a Riemannian manifold.

3. Jun 22, 2016

### MeJennifer

Actually I meant to say increase proper time, sorry for that!

My point stands though!

Maximizing proper time is in this example not staying at "the same spatial location" as was used in the example:

Tossing a ball (or easier a clock) and catching it demonstrates the opposite!

Last edited: Jun 22, 2016
4. Jun 22, 2016

### stevendaryl

Staff Emeritus
No, it doesn't. Near the Earth, the inertial frames are freefall frames, not frames that are fixed relative to the Earth. If you toss a ball up into air, then there is a local inertial frame in which the tossed ball is stationary, and in that frame, the ball in your hand is accelerating upward.

5. Jun 22, 2016

### Staff: Mentor

Yes, that makes a big difference.

If you define "staying at the same spatial location" as standing on the surface of the Earth, yes. But you can construct coordinates in which the thrown ball is at rest and stays at the same spatial location, and the person standing on the Earth does not. (The simplest way would be to construct Fermi normal coordinates centered on the ball's worldline.)

6. Jun 22, 2016

### MeJennifer

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.

7. Jun 22, 2016

### stevendaryl

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

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.

8. Jun 22, 2016

### PAllen

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.

9. Jun 22, 2016

### Orodruin

Staff Emeritus
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.

10. Jun 22, 2016

### stevendaryl

Staff Emeritus
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.

11. Jun 22, 2016

### PAllen

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.

Last edited: Jun 22, 2016
12. Jun 22, 2016

### PAllen

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.

13. Jun 22, 2016

### 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?

14. Jun 22, 2016

### stevendaryl

Staff Emeritus
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.

15. Jun 22, 2016

### PAllen

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

16. Jun 22, 2016

### pervect

Staff Emeritus
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.

17. Jun 22, 2016

### MeJennifer

Time taken for which observer?

For instance consider the following thought experiment.

In a Schwarzschild solution stationary observers Ar1 and Br2 (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?

18. Nov 5, 2017

### Gerry99

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|