# A question about geodesics and notion of parallelism

• I
• Antarres
Antarres
I think I have a slight misconception maybe, but I was wondering about this question.

Usually when we say that the vectors are parallel, we say that it means that there's an equation ##k = \alpha l##, for the vectors ##k## and ##l## and some scalar ##\alpha##. In the context of differential geometry, the notion of a scalar means a scalar function of coordinates.

But if we take the geodesic equation ##u^\mu\nabla_\mu u^\nu = 0##, then if we consider a parallel vector ##\alpha u##, we have
$$\alpha u^\mu\nabla_\mu(\alpha u^\nu) = \alpha (u^\mu\partial_\mu\alpha)u^\nu + \alpha^2 u^\mu\nabla_\mu u^\nu = 0$$

So in general this is not equal to zero, although we can argue that it's proportional to ##u^\mu##, we have:
$$u^\mu\nabla_\mu u^\nu = f(x)u^\nu$$

But the geodesic equation in the general form ##u^\mu\nabla_\mu u^\nu = k(\lambda)u^\nu## is only preserved under affine transformations of the parameter, and so this new equation isn't the geodesic equation? So that means that congruences parallel to geodesics are not geodesics themselves? I kinda feel that this is trivial, but I had some conception in my head that this shouldn't be the case, that parallel vectors to the geodesic vector should also be geodesic, so I figured it's a misconception, but I wanted to check if I missed something in this conclusion.

vanhees71
Rather than consider the space-time case, I'd suggest considering geodesics on a sphere as a warm-up exercise.

Then geodesics on this sphere are great circles.

If we consider a pair of geodesics at the "equator" of the sphere, both heading north, loosely speaking we can consider them to be parallel at the equator. It might be worthwhile to be more formal, though. The notion of "parallel curve" is a bit fuzzy, while the notion of "parallel transport" is something you'll find in a textbook and is much clearer. More below on this below, but I'm going to skip this issue for now.

Now, consider these two geodesics, as the move towards the north pole - "move" in the sense that the affine parameter(s) that determine a point on the geodesic change. Eventually they intersect. And when they intersect, they are no longer parallel, they intersect at an angle.

So, in the curved space of the sphere, we see that geodesics that are initially parallel do not remain parallel. This process is related to something called geodesic deviation.

If we interpret our affine parameter that characterizes our geodesics as some sort of time (note that we are not doing general relativity here, we are considering 2d spatial metric with a ++ signature), we can talk about the relative acceleration of our two initially parallel geodesics via the geodesic deviation equation. Here we'd find the relative acceleration between geodesics is proportional to the velocity of each geodesic, and the initial distance between them, the proportionality constant being the Riemann curvature tensor.

In tensor notation we might write:

$$A^a = \frac {D^2 x^a}{d\tau^2} = -R^a{}_{bcd} \, u^b x^c u^d$$

here A is the relative acceleration , u are velocities (one for each geodesic), and x is the separation between them. The left hand side gives a sort of acceleration between two parallel geodesics.

Most GR texts should discuss this equation, which is probably familiar anyway, in a GR context. It might take a bit of context-switching to go back to our non-GR example, though.

The separation "vector" x is not frequently discussed in depth, to my mind it only makes sense if the separation is infinitesimal however.

Now, to my mind, one obvious question comes up about your original question. Make one geodesic that's a "reference" geodesic, a great circle that goes from the equator to the north pole. What curve do we (or perhaps you) consider to be "parallel" to this reference curve?

In our two dimensional example, for a sufficiently close curve, we might say that the parallel curve keeps a constant distance (called x perviously) from our reference curve. But I think this is an inadequate definition, that only works for our 2d case and only for nearby curves. I'm not sure there is a general notion. Some thought into exactly and precisely what you mean by a "parallel curve" would be fruitful, I think.

For example, if we go to three dimensions (still a +++ metric) is a helix that stays a constant distance away from the reference curve what we mean by "parallel"?

Nugatory
pervect said:
The notion of "parallel curve" is a bit fuzzy
No, it isn't. The standard geometric definition works just fine: two geodesics are parallel if they both intersect a third geodesic at the same angle. (More precisely, they both intersect a third geodesic in such a way that alternate interior angles are equal.)

pervect said:
in the curved space of the sphere, we see that geodesics that are initially parallel do not remain parallel. This process is related to something called geodesic deviation.
Yes, and this is not a "fuzzy" concept; it is precise. The only thing that changes in spacetime, i.e., in a Lorentzian manifold, is that we now have to consider timelike and null geodesics as well as spacelike ones and have a precise definition of "angle of intersection of geodesics" for all possible cases. Most GR textbooks don't get to that level of explicit rigor, they just assume it, but a mathematical treatise on differential geometry will.

vanhees71
pervect said:
The separation "vector" x is not frequently discussed in depth, to my mind it only makes sense if the separation is infinitesimal however.
In a curved manifold, vectors are not spacetime objects, they are objects in the tangent space at a point. So are tensors such as the Riemann tensor that involve objects like the separation vector. The usual tangent space definition of the separation vector is that it is orthogonal to the "reference" geodesic at the given point, and points in the direction of the "nearby" geodesic. The reason for restricting the "nearby" geodesic to be nearby (I'm not sure it has to be infinitesimally close but it has to be close) is to ensure that the separation vector is unique, in the sense that it generates a unique geodesic that intersects the "nearby" geodesic at a specified point. Any vector in the tangent space generates a unique geodesic, but if the "nearby" geodesic is far enough away from the "reference" geodesic, the point of intersection might be a conjugate point, meaning a point that is connected to the "reference" geodesic by multiple geodesics. (The antipodal point on a 2-sphere from a given point, e.g., the South Pole from the North Pole, is an example of a conjugate point.)

vanhees71
pervect said:
if we go to three dimensions (still a +++ metric) is a helix that stays a constant distance away from the reference curve what we mean by "parallel"?
The question is not well-defined because a helix is not a geodesic, and the definition of "parallel" only applies to geodesics. Trying to come up with a definition that extends well to non-geodesic curves is indeed "fuzzy" (and is possibly what you were thinking of in your earlier comment that I referred to before).

vanhees71
pervect said:
Make one geodesic that's a "reference" geodesic, a great circle that goes from the equator to the north pole. What curve do we (or perhaps you) consider to be "parallel" to this reference curve?
Another great circle from the equator to the North Pole that intersects the equator at a slightly different point. These two geodesics will be parallel at the equator. Of course they do not stay parallel, since they intersect at the North Pole. But that is just a consequence of the manifold being curved; parallel geodesics do not stay parallel. So "parallel" has to be defined with respect to a specific third geodesic that they both intersect (in this case the equator).

cianfa72 and vanhees71
Antarres said:
Usually when we say that the vectors are parallel
Are you getting this definition from somewhere? A reference would be helpful. And you might want to compare with the definition of "parallel" that I gave in post #3. There are a lot of traps lurking in these parts for the unwary.

vanhees71
PeterDonis said:
The usual tangent space definition of the separation vector is that it is orthogonal to the "reference" geodesic at the given point, and points in the direction of the "nearby" geodesic.
Probably nitpicking, but I've usually seen the separation vector defined by the condition that its Lie bracket with the reference geodesic's tangent vector is zero.

Nugatory and vanhees71
strangerep said:
I've usually seen the separation vector defined by the condition that its Lie bracket with the reference geodesic's tangent vector is zero
Hm. That's more general than the definition I gave, since it only requires that the inner product of the separation vector and the tangent vector is constant, which of course it will be by parallel transport (the Lie bracket being zero is the same as parallel transport). It doesn't require that the inner product is zero.

MTW's definition uses the affine parameter; basically it says that if you have a family of geodesics, each with affine parameter ##\lambda## and each one labeled by a label parameter ##n##, the the separation vector at a given geodesic with label ##n## at affine parameter ##\lambda## points towards the point on a neighboring geodesic with slightly different ##n## that has the same affine parameter. That could be true without the separation vector being orthogonal to the tangent vector. However, later on MTW talk about initially parallel geodesics not staying parallel as the definitive sign of curvature, and their description of "initially parallel" does have the separation vector (which is basically the tangent vector to the third geodesic that both initially parallel geodesics intersect) being orthogonal.

So it might be that the literature is not quite definite on this point.

vanhees71
PeterDonis said:
The question is not well-defined because a helix is not a geodesic, and the definition of "parallel" only applies to geodesics. Trying to come up with a definition that extends well to non-geodesic curves is indeed "fuzzy" (and is possibly what you were thinking of in your earlier comment that I referred to before).

Mostly here I'm trying to figure out what the Original Poster (OP) meant by "parallel curves". Parallel transport is well defined, but it's less clear what "parallel curves" might mean in a curved geometry. The sphere is a useful test bed as one of the simplest and most familiar curved geometries, so some simple examples of what "parallel curves" on a sphere might be seems like it'd be a good way to understand the OP's question and also to prompt him to think about the topic.

Another couple of example in the same spirit for our simple test case, the sphere. Are lines of longitude (great circles that intersect at the poles) "parallel"? How about lines of lattitude (which are not great circles except at the equator, and thus in general not geodesics) - are they "parallel curves"?

I do recall now that I actually have read a more mathematical sophisticated discussion of the "separation vector" in Wald, but mostly forgotten it, favoring less mathematical ideas inspired mainly by MTW's "Gravitation". The Lie bracket approach mentioned by Strangerep is new to me, but the commutation of the separation vector with the "one parameter family of geodesics" which generated the separation vectors was mentioned in Wald's treatment, but more as an afterthought.

pervect said:
Parallel transport is well defined, but it's less clear what "parallel curves" might mean in a curved geometry.
That's why I gave the standard definition of "parallel curves" that is used in geometry: to clear up that very question.

The limitation in curved geometries is not that it's "less clear" what "parallel curves" means, but that, as already noted, initially parallel curves don't stay parallel. So it's not enough to just say that two curves are parallel, as it would be in a flat geometry. You have to specify exactly where the two curves are parallel (which normally means specifying a particular third geodesic that both curves intersect at the same angle). Also, as already noted, the definition only works for geodesics; for non-geodesic curves, even in flat geometries, there is no well-defined notion of "parallel" curves in general--although one can say that two non-geodesic curves that both cross a given third geodesic are parallel at that geodesic if the geodesic curves that have the same tangent vectors at the crossing points are parallel.

pervect said:
The sphere is a useful test bed as one of the simplest and most familiar curved geometries, so some simple examples of what "parallel curves" on a sphere might be seems like it'd be a good way to understand the OP's question and also to prompt him to think about the topic.
Yes, that's why I gave a simple example for a 2-sphere in post #6. That example illustrates all of the issues I have mentioned for geodesic curves.

pervect said:
Are lines of longitude (great circles that intersect at the poles) "parallel"?
At the equator, yes. Elsewhere, no. This is the question I already answered in post #6.

pervect said:
How about lines of lattitude (which are not great circles except at the equator, and thus in general not geodesics) - are they "parallel curves"?
Since there is no well-defined notion of "parallel curves" for non-geodesics, this question is not answerable as it is stated.

However, we can restate the question slightly to make it answerable: if we pick a given line of longitude (which, as you note, is a geodesic), are lines of latitude (other than the equator) parallel at that line of longitude? The answer to that question is yes, since all lines of latitude intersect a given line of longitude at the same angle.

Since the answer above is the same for any line of longitude, one might wonder whether that in itself does not make lines of latitude parallel curves everywhere. However, as already noted, different lines of longitude are not parallel except at the equator. So lines of latitude have to curve to intersect different lines of longitude at the same angle. (This is another way of saying that lines of latitude--other than the equator--are not geodesics.) This is similar to concentric circles on the Euclidean plane: they all intersect any straight line through their common center at the same angle, and they have to curve in order to do so.

Normally we do not refer to concentric circles on a plane as being parallel, so we would not refer to lines of latitude on a 2-sphere as being parallel either.

PeterDonis said:
Since there is no well-defined notion of "parallel curves" for non-geodesics, [...]
I'm not so sure about that.

Consider the treatment of geodesic deviation in Carroll, section 3.10. He starts off by considering a congruence of geodesics -- but suppose they're not geodesics. His definition of "deviation vector" ##S^\mu## and "relative velocity of geodesics" seem not to depend on the congruence curves being geodesic, so these terms still seem sensible for non-geodesics. E.g., the "relative velocity" of neighbouring curves can still be defined as $$V^\mu ~=~ T^\rho \nabla_\rho S^\mu ~.$$ Moreover, the requirement ##\,[S,T] = 0\,## does not depend on the curves being geodesics, afaict. That only comes in later when he massages the equation for the relative acceleration ##A^\mu## into an expression involving the Riemann tensor -- that manipulation uses the geodesic condition ##\,T^\rho \nabla_\rho T^\mu = 0\,##.

So, if such a deviation vector remains constant along the first curve, is that sufficient to reasonably consider the adjacent curve as "parallel" to the first curve?

Ibix and vanhees71
strangerep said:
I'm not so sure about that.
Would you consider concentric circles on a plane to be parallel curves? (Note that the standard Euclidean geometry definition of "parallel" requires the lines to be straight lines.) Do they meet Carroll's definition if we extend it to non-geodesics?

Same question for lines of latitude other than the equator on a 2-sphere.

PeterDonis said:
Would you consider concentric circles on a plane to be parallel curves?
Yes, at least in the context considered here: Parallel curve

renormalize said:
Yes, at least in the context considered here: Parallel curve
Ok, but then the concept of "parallel" becomes useless when assessing whether the manifold is curved or not. Which might be ok for the OP of this thread, but it shows why this definition is not generally used in GR.

strangerep said:
if such a deviation vector remains constant along the first curve, is that sufficient to reasonably consider the adjacent curve as "parallel" to the first curve?
Before even considering this for the case I asked about, concentric circles in a plane, you should first check to see whether the "if" condition is even met. In a plane, the "remains constant" condition for vectors is simple: they keep pointing in the same direction and they maintain the same length. But the "separation vector" between two concentric circles obviously does not meet that condition, since its direction changes as you go around the circles. (You are welcome to check this using the formulas you gave.) So the "if" condition above is not met for concentric circles. (A somewhat more complex computation, since you can't use the nice property of Cartesian coordinates in a plane that the connection coefficients are all zero, will show you that the condition is not met for latitude lines on a 2-sphere either.)

Note that this is not the condition that is used in the WIkipedia article on "parallel curves" that you linked to: the condition there is "constant distance between the curves". I.e., the length of the separation vector has to be constant, but its direction does not. This is a weaker condition than the Carroll condition you referred to earlier.

PeterDonis said:
Before even considering this for the case I asked about, concentric circles in a plane, you should first check to see whether the "if" condition is even met. In a plane, the "remains constant" condition for vectors is simple: they keep pointing in the same direction and they maintain the same length. But the "separation vector" between two concentric circles obviously does not meet that condition, since its direction changes as you go around the circles.
When I said "if such a deviation vector remains constant along the first curve", I meant under parallel transport along the first curve. (Apologies for my sloppiness.)

One way to interpret this is by preservation of the angle between S and T under such parallel transport, but that presumes a metric. Another way to interpret it is in terms of preservation of the Lie bracket under such parallel transport, and that's trivial, since the Lie bracket is a vector, and zero in this case. But whichever way,... yes, we do still seem to need a metric to establish a length of S, since, e.g., S and kS give the same vanishing Lie bracket with T.

strangerep said:
When I said "if such a deviation vector remains constant along the first curve", I meant under parallel transport along the first curve. (Apologies for my sloppiness.)
Then the deviation vector for concentric circles on a flat plane or for latitude lines on a 2-sphere is not constant by this criterion. That's what I meant by saying that those vectors change direction.

strangerep said:
I meant under parallel transport along the first curve
For non-geodesic curves, I assume you intended to say Fermi-Walker transport, since that's the appropriate generalization of parallel transport for non-geodesic curves. (That doesn't change my answer in post #18, but I wanted to clarify.)

renormalize said:
Yes, at least in the context considered here: Parallel curve

We can see from our 2d spherical example that curves parallel to a reference geodesic in a curved space are not necessarily themselves geodesics using the above definition of parallel curves. Without loss of generality we can rotate the 2-sphere so the reference geodesic is the equator. Then the parallel curves to the reference geodesic (the equator) will be curves of constant latitude. Because curves of constant latitude are not great circles, they won't be geodesics.

This isn't surprising. In GR, we would say that space-time curvature causes tidal forces, and these tidal forces will cause geodesics to accelerate relative to one another via the geodesic deviation equation, making them not satisfy the above definition of "parallel curves" which requires a constant separation.

In our 2-sphere example, the Riemann has only one component. In more complex examples it could be possible to have a pair of geodesics that did not accelerate to each other if positive and negative terms in the geodesic equation canceled each other out for a net zero geodesic deviation for some particular pair of geodesics. Absent such a fortuitous cancellation, though, geodesics in a curved space-time will (usually) accelerate towards or away from each other. This implies that they won't be curves with a constant separation.

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