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accdd

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accdd

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- #2

Dale

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It is a path that goes straight from one place to another.

- #3

martinbn

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He is asking about a physical meaning. This seems more like a geometric meaning.It is a path that goes straight from one place to another.

- #4

Dale

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Geometry is an important part of physics. I would even say it is central

- #5

martinbn

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Well, technically geometry is not part of physics. Of course it is important for physics, but it is not part of it. That aside, I think that when people ask about physical meaning, they don't mean geometry.Geometry is an important part of physics. I would even say it is central

- #6

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If one exists, a spacelike geodesic passing through two events is the longest (usually) distance between those two events. In curved spacetime there can be multiple geodesics through two events and each one is an extremum locally, meaning that all nearby paths are longer or all nearby paths are shorter.

- #7

accdd

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Dale answered my question.

For me, the geometric point of view counts as physical meaning.

- #8

martinbn

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I'd argue that spacelike geodesics do not have physical meaning.

- #9

PeterDonis

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I would argue that they do. If you don't want to accept @Dale's statement as it stands, consider:I'd argue that spacelike geodesics do not have physical meaning.

Arc length along a timelike geodesic represents the proper time of an object in free fall.

Arc length along a spacelike geodesic represents the proper length of an object in free fall.

- #10

Dale

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I completely disagree with this. Any time you have an equation with an ##x## or a ##t## or something similar then geometry is part of that equation. In physics there are many equations that have geometry built into the equation, so geometry is indeed part of physics. I haven't gone through to count, but I would suspect that in physics there are more equations with geometry than without. So geometry is most definitely part of physics, as much as mass or energy or any other quantity that we put into our equations.Well, technically geometry is not part of physics. Of course it is important for physics, but it is not part of it. That aside, I think that when people ask about physical meaning, they don't mean geometry.

There are substantially fewer equations that use spacelike geodesics, so I wouldn't argue strongly against this, but I don't share that opinion. The phrase "physical meaning" is the slippery part here.I'd argue that spacelike geodesics do not have physical meaning.

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- #11

martinbn

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That I understand and it makes perfect sense to me. It has physical meanin (whatever that may mean) a clock with that worldline would be measuring the arc length.I would argue that they do. If you don't want to accept @Dale's statement as it stands, consider:

Arc length along a timelike geodesic represents the proper time of an object in free fall.

This I don't understand this withoud some further specification. It seems that it has implicit underlining assumptions and conventions. My guess would be the following. An extended object is represented by its worldsheet in spacetime. I am guessing that you need a spacelike foliation and the intersection with each leaf and the worldsheet would give the length. There is a lot of freedom in the choice of a folitation without the necessity of it having any physical meaning. I suppose that you have in mind a special choice that reflects the being at rest for the extended object and hence giving proper length, but I am not sure what that might be.Arc length along a spacelike geodesic represents the proper length of an object in free fall.

But I still cannot see what the possible phycial meaning could be. Any atempt to measure any sort of length would fail (it seems to me) unless you have a static or stationary spacetime and/or have some convention in mind. It is very similar to claiming that there is physical meaning to some implicit choice of simultaneity convention in special relativity.

- #12

martinbn

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May be I misunderstood what you meant by geometry is part of physics. I thiught you meant it in the same way as in electrodynamics is part of physics, that the subject geometry is a subfield of physics. In any case I agree that a lot og geometry is used in physics. But I still disagree that geometric meaning answers the question about physical meaning. Well, the OP clarified that it was fine, so that's that.I completely disagree with this. Any time you have an equation with an ##x## or a ##t## or something similar then geometry is part of that equation. In physics there are many equations that have geometry built into the equation, so geometry is indeed part of physics. I haven't gone through to count, but I would suspect that in physics there are more equations with geometry than without. So geometry is most definitely part of physics, as much as mass or energy or any other quantity that we put into our equations.

I agree that physical meaning is unclear and can lead us in any of a number of directions.There are substantially fewer equations that use spacelike geodesics, so I wouldn't argue strongly against this, but I don't share that opinion. The phrase "physical meaning" is the slippery part here.

- #13

Dale

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Oh, I can see how my statement would have made you think I intended to say that. My apologies. I don’t mean as an academic discipline that geometry is a sub discipline within physics. Unfortunately I don’t know a better way to say what I intended.May be I misunderstood what you meant by geometry is part of physics. I thiught you meant it in the same way as in electrodynamics is part of physics, that the subject geometry is a subfield of physics.

I mean that there is a geometrical relationship between physical objects. Geometry is therefore part of the description of how physical objects interact. This is reflected in the equations we use to describe the objects and their interactions which almost universally include geometrical variables.

- #14

cianfa72

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That's my viewpoint too. Unlike the timelike geodesic case, there is no way to attach any "intrinsic/independent" physical meaning to a spacelike geodesic until you pick some convention for a spacelike foliation of the extended object worldtube (as you said it is alike the simultaneity convention).An extended object is represented by its worldsheet in spacetime. I am guessing that you need a spacelike foliation and the intersection with each leaf and the worldsheet would give the length. There is a lot of freedom in the choice of a folitation without the necessity of it having any physical meaning. I suppose that you have in mind a special choice that reflects the being at rest for the extended object and hence giving proper length, but I am not sure what that might be.

But I still cannot see what the possible physical meaning could be. Any attempt to measure any sort of length would fail (it seems to me) unless you have a static or stationary spacetime and/or have some convention in mind. It is very similar to claiming that there is physical meaning to some implicit choice of simultaneity convention in special relativity.

We had a thread on this topic some time ago...

- #15

PeterDonis

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In general, yes, but if you have a specific worldline that is a geodesic, then there is a unique foliation defined on at least some "world tube" surrounding that worldline: just take the spacelike 3-surface that is orthogonal to the worldline at each event. That "stack" of 3-surfaces, within the "world tube" where none of them cross (eventually some will in a curved spacetime, but there will be some finite "world tube" surrounding the worldline where they do not), is the unique foliation. It is unique because of the orthogonality condition.There is a lot of freedom in the choice of a folitation

Technically you can do this for a non-geodesic worldline as well, but because of the path curvature of the worldline, the surfaces of the foliation might cross sooner.

- #16

cianfa72

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From what we said in other threads on this topic, if the worldtube geodesic congruence has not zero vorticity then the spacelike hypersurfaces othogonal to the congruence's worldlines at each event will not 'join' together (i.e. Frobenius's theorem condition will not be met hence the worldlines' orthogonal spacelike 3D distribution at each event will not be integrable).just take the spacelike 3-surface that is orthogonal to the worldline at each event. That "stack" of 3-surfaces, within the "world tube" where none of them cross (eventually some will in a curved spacetime, but there will be some finite "world tube" surrounding the worldline where they do not), is the unique foliation. It is unique because of the orthogonality condition.

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- #17

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This is not true for saddle points (such as the long part of the great circle passing through two points on a sphere).there can be multiple geodesics through two events and each one is an extremum locally, meaning that all nearby paths are longer or all nearby paths are shorter.

- #18

PeterDonis

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If extended out far enough, yes. But even in this case, there will still be a finite "world tube" around the worldline in which the foliation I described can be used without the issue you describe arising. As the vorticity increases, the finite size of this "world tube" will in general decrease.if the worldtube geodesic congruence has not zero vorticity then the spacelike hypersurfaces othogonal to the congruence's worldlines at each event will not 'join' together

- #19

cianfa72

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Ah ok, basically you are saying that if we pick a worldline and consider the neighboring worldlines in the congruence in a small region of spacetime around it, then the orthogonal 3D spacelike distribution at each event along them meetsBut even in this case, there will still be a finite "world tube" around the worldline in which the foliation I described can be used without the issue you describe arising. As the vorticity increases, the finite size of this "world tube" will in general decrease.

- #20

PeterDonis

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No. The congruence still has nonzero vorticity and still does not meet the Frobenius conditions. But the non-integrability due to those conditions not being met is not a local property. (In general, anything involving integrals is not a local property; the whole point of integrals is to evaluate properties that involve extension over some region, not just properties at a point.)basically you are saying that if we pick a worldline and consider the neighboring worldlines in the congruence in a small region of spacetime around it, then the orthogonal 3D spacelike distribution at each event along them meetslocallythe Frobenius's conditions.

- #21

cianfa72

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So basically what you are saying amounts to the following: take the set of orthogonal 3D spacelike subspaces in tangent spaces at each event along the given worldline and for each of them evaluate the exponential map. Such maps will define in a finite region of spacetime around the given worldline the spacelike foliation you were talking about.But the non-integrability due to those conditions not being met is not a local property. (In general, anything involving integrals is not a local property; the whole point of integrals is to evaluate properties that involve extension over some region, not just properties at a point.)

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- #22

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Fair enough. Is a great circle arc really a saddle point? I'd have thought that by symmetry nearby paths on one side of the arc had the same lengths as ones on the other side of it. Or is there some sense of 'variation' that I'm missing?This is not true for saddle points (such as the long part of the great circle passing through two points on a sphere).

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They do. But a functional space is infinite dimensional. You cannot restrict yourself to a single shape of the perturbation.Fair enough. Is a great circle arc really a saddle point? I'd have thought that by symmetry nearby paths on one side of the arc had the same lengths as ones on the other side of it. Or is there some sense of 'variation' that I'm missing?

Easy way to see that long way around is a saddle point: Use coordinates such that great circle is the equator. Expand perturbation in Fourier series. If there is less than ##\pi R## in distance between the points, all modes will lead to a longer path. If there is more, then the first mode will lead to a shorter path and the others to longer. Hence, a saddle point.

- #24

watcher1

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- #25

DrGreg

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We are discussing geodesics between events in 4D spacetime, not geodesics between points in 3D space. In spacetime light follows a null geodesic.

Whether or not light follows a geodesic in space depends on how you choose to extract a "space" from spacetime.

- #26

PeterDonis

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Yes, that's it.take the set of orthogonal 3D spacelike subspaces in tangent spaces at each event along the given worldline and for each of them evaluate the exponential map. Such maps will define in a finite region of spacetime around the given worldline the spacelike foliation you were talking about.

- #27

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To make it a bit more hands-on, I managed to dig out this post from my first months on PF:They do. But a functional space is infinite dimensional. You cannot restrict yourself to a single shape of the perturbation.

Easy way to see that long way around is a saddle point: Use coordinates such that great circle is the equator. Expand perturbation in Fourier series. If there is less than ##\pi R## in distance between the points, all modes will lead to a longer path. If there is more, then the first mode will lead to a shorter path and the others to longer. Hence, a saddle point.

Post in thread 'Understand the major arc connecting two points on a sphere'

https://www.physicsforums.com/threa...ng-two-points-on-a-sphere.767025/post-4829841

- #28

cianfa72

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You mean take two points on the great circle on the sphere (e.g. on the equator) and evaluate the distance between them along the great circle via the short way and the long way around. In the general case (i.e. not antipodal points) the two will be different.Easy way to see that long way around is a saddle point: Use coordinates such that great circle is the equator. Expand perturbation in Fourier series. If there is less than ##\pi R## in distance between the points, all modes will lead to a longer path.

Now for the long (geodesic) path the evaluated distance will be greater than ##\pi R##. For this case your following argument applies (i.e. the long path around is actually a saddle point).

If there is more, then the first mode will lead to a shorter path and the others to longer. Hence, a saddle point.

- #29

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A straight answer.It is a path that goes straight from one place to another.

- #30

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Yes, the long way around is a saddle point. The short way is the global minimum.You mean take two points on the great circle on the sphere (e.g. on the equator) and evaluate the distance between them along the great circle via the short way and the long way around. In the general case (i.e. not antipodal points) the two will be different.

Now for the long (geodesic) path the evaluated distance will be greater than ##\pi R##. For this case your following argument applies (i.e. the long path around is actually a saddle point).

- #31

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AA straight answer.

- #32

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How would you build the optimal road from one city to another? I would build a straight road, that is a spacelike geodesic.I'd argue that spacelike geodesics do not have physical meaning.

- #33

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Well, it's a worldsheet with several spacelike geodesics lying in it. The intersection of that sheet with a member of some foliation of spacetime (i.e. given some definition of "now") may be a spacelike geodesic.How would you build the optimal road from one city to another? I would build a straight road, that is a spacelike geodesic.

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- #34

cianfa72

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I believe the only physical requirement for aWell, it's a worldsheet with several spacelike geodesics lying in it. The intersection of that sheet with a member of some foliation of spacetime {i.e. given some definition of "now") may be a spacelike geodesic.

- #35

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Yes. That's why I said its intersection with the road's worldsheetI believe the only physical requirement for a spacelike foliation to be effective is that the events on it cannot be connected by a light rays or by the paths of massive objects.

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