I Why do particles move along longest proper time trajectories

[Bas73]

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

 

[vanhees71]

That's the action principle. For a free particle in special relativity the only Lorentz invariant action is given by the Lagrangian
$$L=-mc^2 \sqrt{1-\dot{\vec{x}}^2/c^2}.$$
This you can write in terms of an arbitrary world-line parameter as
$$A=\int \mathrm{d} t L=-m c^2 \int \mathrm{d} \lambda \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}},$$
where here the dot means the derivative wrt. $\lambda$.

Now from the principle of equivalence this translates into the generally covariant action
$$A=-m c^2 \int \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{nu}}.$$
The Euler-Lagrange equations leads to the equations for the geodesics in the pseudo-Riemannian manifold with pseudo-metric components $g_{\mu \nu}$.

 

[Nugatory]

You need an additional qualifier: it's not "a particle follows a path of maximum proper time", it is "a particle not subject to any external forces follows a path of maximum proper time". Because the paths of extremal proper time are straight lines this statement is the curved spacetime equivalent of Newton's first law - the particle is moving along a "straight" line (actually a geodesic, because that's the curved spacetime equivalent of a straight line) because there's no force acting on it to push it off that path.

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.

 

[vanhees71]

One should stress that it's a geodesic in spacetime, not in space! E.g. the Kepler orbit of the Earth around the sun is also described by a geodesic in spacetime, but for sure it's not a geodesic in space for an (asymptotic inertial) observer. It's a nice excercise to calculate this as the motion of a body in the Schwarzschild spacetime (outside the Schwarzschild radius of course ;-)).

 

[MeJennifer]

You need an additional qualifier: it's not "a particle follows a path of maximum proper time", it is "a particle not subject to any external forces follows a path of maximum proper time". Because the paths of extremal proper time are straight lines this statement is the curved spacetime equivalent of Newton's first law - the particle is moving along a "straight" line (actually a geodesic, because that's the curved spacetime equivalent of a straight line) because there's no force acting on it to push it off that path.

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.

The question is how can both be true?

What you are saying is that and a massive particle and a massless travel on a geodesic in spacetime and that (for nearby points) there is only one path.

But a massless and a massive particle leaving the same event will never ever meet in a future event. Thus they must follow different paths.

So that's kind of a dilemma? Right?

Or perhaps a massive particle does not travel on a geodesic at all?
Because perhaps it constantly couples and modulates the background?

 

[Ibix]

If the particle isn't massless then it can't travel between the same start and end events as a photon, not in flat spacetime. So I don't see anything contradictory in what Nugatory wrote.

 

[stevendaryl]

The question is how can both be true?

What you are saying is that and a massive particle and a massless travel on a geodesic in spacetime and that (for nearby points) there is only one path.

But a massless and a massive particle leaving the same event will never ever meet in a future event. Thus they must follow different paths.

I think what he meant is that there is exactly one geodesic path connecting two (nearby) events. There are infinitely many geodesics that go through a single event.

 

[MeJennifer]

If the particle isn't massless then it can't travel between the same start and end events as a photon, not in flat spacetime. So I don't see anything contradictory in what Nugatory wrote.

Between event A and B (which are nearby) how many paths are there that are geodesics?

 

[MeJennifer]

I think what he meant is that there is exactly one geodesic path connecting two (nearby) events. There are infinitely many geodesics that go through a single event.

That is of course true.

 

[stevendaryl]

Between event A and B (which are nearby) how many paths are there that are geodesics?

Exactly one. If you use locally Minkowskian coordinates, then the geodesic can be parametrized by:

x^\mu(s) = A^\mu + s (B^\mu - A^\mu)

 

[vanhees71]

Of curse in this case you have
$$u \cdot u = (B-A) \cdot (B-A),$$
and the geodesic cannot be both time and lightlike (null), but that's self-evident anyway :-).

 

[Ibix]

Between event A and B (which are nearby) how many paths are there that are geodesics?

One, as you just said to Steven.

 

[Bas73]

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.

 

[Bas73]

Hi Nugatory,

You are correct, I indeed meant a particle that is not subject to any external forces. My question however was, to put it in your own words: why are paths of extremal proper time always straight lines/the straightest lines possible?

Bas

 

[Nugatory]

If the particle isn't massless then it can't travel between the same start and end events as a photon, not in flat spacetime. So I don't see anything contradictory in what Nugatory wrote.

Exactly - thx.

 

[Ibix]

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.

Carroll points out that you can always approximate a path in spacetime by a series of null paths (on a Minkowski diagram, approximate a curve by a connected series of 45-degree lines). By shortening the null segments you can get arbitrarily close to the curve - thus there is always a path of length zero arbitrarily near any other path. Thus the extremised path length must be a maximised one.

 

[stevendaryl]

I was wondering if there nevertheless is a easy way to understand why the extreme action always leads to the longest proper time...

Well, in flat spacetime it is pretty obvious:

Two events e_1 and e_2 can have three types of separation:

1. Timelike, meaning that it is possible for a slower-than-light particle to move from e_1 to e_2.
2. Lightlike, meaning that it is possible for a light signal to move from e_1 to e_2
3. Spacelike, meaning that neither of the above is possible.

If e_1 and e_2 are timelike separated, then that means that there is an inertial frame in which e_1 and e_2 take place at the same spatial location, at different times. In that case, the proper time in going from e_1 to e_2 is given by:

\tau = \int_{t_1}^{t_2} \sqrt{1-\frac{v^2}{c^2}} dt

It's clear that \tau \leq (t_2 - t_1). The largest that \tau can be is when v=0, which gives \tau = (t_2 - t_1)

 

[MeJennifer]

Two events e_1 and e_2 can have three types of separation:

1. Timelike, meaning that it is possible for a slower-than-light particle to move from e_1 to e_2.
2. Lightlike, meaning that it is possible for a light signal to move from e_1 to e_2
3. Spacelike, meaning that neither of the above is possible.

If e_1 and e_2 are timelike separated, then that means that there is an inertial frame in which e_1 and e_2 take place at the same spatial location, at different times. In that case, the proper time in going from e_1 to e_2 is given by:

\tau = \int_{t_1}^{t_2} \sqrt{1-\frac{v^2}{c^2}} dt

It's clear that \tau \leq (t_2 - t_1). The largest that \tau can be is when v=0, which gives \tau = (t_2 - t_1)

You might want to qualify what I highlighted in bold as it is a coordinate dependent quality.
There is no absolute sense of a "same spatial location" in relativity.

 

[stevendaryl]

You might want to qualify what I highlighted in bold as it is a coordinate dependent quality.
There is no absolute sense of a "same spatial location" in relativity.

I said "there is an inertial frame in which the two events take place at the same spatial location". Yes, it's frame-dependent.

The logic is this:

1. Pick a coordinate system such that e_1 and e_2 have the same location, but different times.
2. In that coordinate system, convince yourself that the extremal value for \tau is the largest, not the smallest value.
3. If the conclusion is true in one coordinate system, then it is true in every coordinate system, since the proper time integral is invariant.

 

[MeJennifer]

I said "there is an inertial frame in which the two events take place at the same spatial location". Yes, it's frame-dependent.

The logic is this:

1. Pick a coordinate system such that e_1 and e_2 have the same location, but different times.
2. In that coordinate system, convince yourself that the extremal value for \tau is the largest, not the smallest value.

...

Well actually that is true only in SR but not generally in GR.
Think for instance about a ball thrown up while standing on Earth!

 

[Ibix]

Well actually that is true only in SR but not generally in GR.
Think for instance about a ball thrown up while standing on Earth!

...as @stevendaryl said in #17.

 

[stevendaryl]

Well actually that is true only in SR but not generally in GR.
Think for instance about a ball thrown up while standing on Earth!

It works well enough for curved spacetime if you take a small enough region of spacetime. Divide spacetime up into little tiny regions. Then within each region, you can come up with an approximately Minkowskian coordinate system, and use the definition of geodesic as the path with the largest proper time. A geodesic path is a path such that in each region it passes through, it is a geodesic in the sense of flat spacetime (in the limit as the sizes of the regions goes to zero).

 

[MeJennifer]

It works well enough for curved spacetime if you take a small enough region of spacetime. Divide spacetime up into little tiny regions. Then within each region, you can come up with an approximately Minkowskian coordinate system, and use the definition of geodesic as the path with the largest proper time. A geodesic path is a path such that in each region it passes through, it is a geodesic in the sense of flat spacetime (in the limit as the sizes of the regions goes to zero).

It seems to me you are missing the point.

You are trying to prove something using coordinate dependent properties (same place in space), I just gave you an example where that would not be valid.

Throwing up a ball and catching it while standing on Earth will reduce the total proper time for the ball as compared to leaving it in your hands.

Edited to add: please read increase instead of reduce the total proper time

 

[PeterDonis]

Throwing up a ball and catching it while standing on Earth will reduce the total proper time for the ball as compared to leaving it in your hands.

No, it won't. It will increase the ball's proper time. Please check your math.

 

[MeJennifer]

No, it won't. It will increase the ball's proper time. Please check your math.

You are wrong!

When you throw the ball it will travel on a geodesic!

 

[PeterDonis]

You are wrong!

No, I'm not. Do the math.

When you throw the ball it will travel on a geodesic!

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.)

 

[Orodruin]

When you throw the ball it will travel on a geodesic!

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.

 

[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!

 

[stevendaryl]

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!

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.

 

[PeterDonis]

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

Yes, that makes a big difference.

Maximizing proper time is in this example not staying at "the same spatial location"

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.)

 

[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|