# Why do particles move along longest proper time trajectories

• I
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
Gold Member
2021 Award
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}##.

• Dale
Nugatory
Mentor
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
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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 ;-)).

• 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.
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.

• Nugatory
stevendaryl
Staff Emeritus
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.

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?

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
Staff Emeritus
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
Gold Member
2021 Award
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.

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.

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
Mentor
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
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
Staff Emeritus
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 $\tau = (t_2 - t_1)$

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
Staff Emeritus
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.

• Ibix
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
Staff Emeritus
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).

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.

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PeterDonis
Mentor
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.

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
Mentor
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
Staff Emeritus
Homework Helper
Gold Member
2021 Award
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.

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!

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stevendaryl
Staff Emeritus
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
Mentor
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.)