What Is the Shortest Path on a Sphere If Not a Line of Latitude?

[DeDunc]

Simple question about geodesics.

I have a question which I guess will be easy to answer for anyone who is familiar with the geometry involved in GR. Firstly, I have a numbered list which shows my (current) understanding of geodesics. If there is any wrong with my understanding please let me know by referring to the point(s) which is incorrect. After this I have posed my question. If you could explain the answer that would be great.

My understanding of geodesics.

1. Straight lines and great circles also represent the shortest path between two points.

2. A geodesic is the analogue of straight lines and great circles in a general Riemannian space.

3. If a pathway satisfies geodesic equation then it is said to be a geodesic.

4. If we imagine the surface of sphere (2-d) and draw a line of latitude defined by the following: theta=pi/4 and 0<phi<2pi. The parameterised pathway does not satisfy the geodesic equation.

5. This means that the line of latitude described above is not a geodesic.

My Question

If the line of latitude above is not a geodesic. Then it is not the shortest path between the two points defined by (theta=pi/2 , phi=0 & theta=pi/2 , phi= 2pi). These two points coincide. If it is not the shortest path between the two points described. Then what is? From a purely intuitive standpoint there does not seen to be a shorter path which links these two points.

Thanks

 

[bcrowell]

Hi, DeDunc,

Welcome to Physics Forums!

1. Straight lines and great circles also represent the shortest path between two points.

In spherical or elliptic geometry, a segment of a great circle may be the shortest *or* longest path between two points.

2. A geodesic is the analogue of straight lines and great circles in a general Riemannian space.

When doing this generalization, you need to generalize the shortest/longest thing. In fact, a geodesic need only be a curve of extremal length compared to all other curves that can be obtained from it by infinitesimal deformations.

4. If we imagine the surface of sphere (2-d) and draw a line of latitude defined by the following: theta=pi/4 and 0<phi<2pi.[...]

If the line of latitude above is not a geodesic. Then it is not the shortest path between the two points defined by (theta=pi/2 , phi=0 & theta=pi/2 , phi= 2pi). These two points coincide. If it is not the shortest path between the two points described. Then what is? From a purely intuitive standpoint there does not seen to be a shorter path which links these two points.

Why did you switch from theta=pi/4 in the original example to theta=pi/2 later?

Anyway, there is clearly no way to distinguish geodesic from non-geodesic curves by only examining one point on the curve. More generally, just because some segment of a curve is geodesic, that doesn't mean that parts lying outside that segment are geodesic as well. If you like, you can define a point to be a degenerate geodesic.

-Ben

 

[DeDunc]

Thanks Ben for the reply. Very useful. I made a mistake with theta. It should be pi/4 all the way, well spotted.
Another way to rephrase the question would be :

If the line of latitude ( on the 2-d Sphere) I described is not a geodesic. How can I intuitively understand the geodesic that goes through the any point on this line and returns to the same point?

 

[bcrowell]

If the line of latitude ( on the 2-d Sphere) I described is not a geodesic. How can I intuitively understand the geodesic that goes through the any point on this line and returns to the same point?

The geodesic is a geodesic regardless of whether it has *end*-points. A geodesic is a curve such that no matter which two points P and Q you pick on it, the length of the segment from P to Q is a max or min with respect to infinitesimal changes.

 

[phyzguy]

DeDunc,

Perhaps another way to look at it is to ask what is the shortest distance between the point theta=pi/4, phi=0 and theta=pi/4, phi=pi. Following the latitude line is not the shortest path. A great circle which goes significantly north of theta=pi/4 (in fact, directly over the north pole) will be a lot shorter. Try getting a globe and stretching a string over it and you will see.

Does this help?

 

[pervect]

DeDunc,

Following the latitude line is not the shortest path. A great circle which goes significantly north of theta=pi/4 (in fact, directly over the north pole) will be a lot shorter. Try getting a globe and stretching a string over it and you will see.

Does this help?

We can calcuate this - it's convenient to use

$$ds^2 = r^2 d \theta^2 + r^2 sin^2 \theta d\phi^2$$

The distance going around the latitude line

$$\int_{\phi=0}^{\phi=2 \pi} r \sin \, \pi/4 \, d\phi = r \sin \, \pi/4 \int_{\phi=0}^{\phi=2 \pi} d\phi = 2 \pi r \sin \pi/4= 1.414\, \pi\, r$$

The distance going up to the north pole and back down - you can circle the pole with zero distance while you're at the pole

$$2 \, \int_{\theta=\pi/4}^{\theta=\pi/2} r d\theta = 2 \, r \, \int_{\theta=\pi/4}^{\theta=\pi/2} d\theta = (\pi/2) \, r = .5\, \pi \, r$$

 

[DeDunc]

Thanks pervect, phyzguy and Brian. Your contributions have been very helpful to me.

Pervect your comment highlights what it probably the misunderstanding underlying my question.

I perfectly accept your calculation of your distance going around the latitude line is given by 1.414pi*r. In particular, the limits of the integral (phi=0 and phi=2pi) coincide.

However the second calculation going though the North Pole raises some questions for me. To ensure that the limits of integral coincide, shouldn’t theta go from pi/4 to -7pi/4 (going through the north pole and returning to the same point) or pi/4 to 9pi/4 ( going through the south pole and returning to the same point) .

As I see it, the limits of second calculation do not represent the same start and end points as the limits of the first calculation.

Where am I going wrong?

 

[Dale]

In spherical or elliptic geometry, a segment of a great circle may be the shortest *or* longest path between two points.

Is this correct? I think that the "long" segment of a great circle is still a local minimum. E.g. if you added a "sine wave wiggle" of infinitesimal amplitude to the great circle then it would surely be slightly longer.

 

[bcrowell]

Is this correct? I think that the "long" segment of a great circle is still a local minimum. E.g. if you added a "sine wave wiggle" of infinitesimal amplitude to the great circle then it would surely be slightly longer.

Yep, you're right and I'm wrong. Clearly there can't be a globally longest path :-)

 

[George Jones]

I think that the "long" segment of a great circle is still a local minimum.

It is not a local minimum, either. Consider what happens to an elastic band that is fixed at its ends and stretched along the long segment of a great circle. It starts sliding back towards the short short segment of the great circle while decreasing continuously in length.

 

[bcrowell]

Hmm...it's actually pretty tough to come up with a formulation in terms of path length that works. In a Riemannian signature, I'm guessing that the following works: A geodesic is a curve such that for any two points P and Q on the curve that are sufficiently close together, the length of the segment from P to Q is a max or min with respect to infinitesimal changes in the curve. George and Dale, do you think this is right, and if so, how does one generalize it to a non-Riemannian signature?

 

[Dale]

It is not a local minimum, either.

Interesting. Then it must be some sort of a local "saddle point" since it is stationary.

It starts sliding back towards the short short segment of the great circle while decreasing continuously in length.

Are you sure about that? I always thought it would have to get slightly longer first before it could start decreasing in length. I am not sure how to parameterize this.

 

[George Jones]

Interesting. Then it must be some sort of a local "saddle point" since it is stationary.

Yes.

I always thought it would have to get slightly longer first before it could start decreasing in length. I am not sure how to parameterize this.

No. The point antipodal to the starting point of the long segment is a conjugate point to the starting point. Any geodesic that goes past a conjugate point is not a local minimum.

 

[sheaf]

Hmm...it's actually pretty tough to come up with a formulation in terms of path length that works. In a Riemannian signature, I'm guessing that the following works: A geodesic is a curve such that for any two points P and Q on the curve that are sufficiently close together, the length of the segment from P to Q is a max or min with respect to infinitesimal changes in the curve. George and Dale, do you think this is right, and if so, how does one generalize it to a non-Riemannian signature?

Isn't the only possible statement from the length-variational definition of geodesic simply that the length will be an extremum ? i.e. there's no indication of whether it's a maximum or a minimum.

 

[HallsofIvy]

Thanks Ben for the reply. Very useful. I made a mistake with theta. It should be pi/4 all the way, well spotted.
Another way to rephrase the question would be :

If the line of latitude ( on the 2-d Sphere) I described is not a geodesic. How can I intuitively understand the geodesic that goes through the any point on this line and returns to the same point?

Given a single point there exist an infinite number of such geodesics, just as there exist an infinite number of straight lines through a given point in the plane. Draw the line through the given point and the center of the sphere. Imagine a plane rotating around that line. It continually intersects the sphere in "great circle" geodesics through that given point as it rotates.

If you are given two points on the sphere, whether on the same line of latitude or not, those two points, together with the center of the sphere, determine a unique plane. That plane intersects the sphere in the "great circle" geodesic through those two points.

 

[Sourabh N]

Thanks pervect, phyzguy and Brian. Your contributions have been very helpful to me.

Pervect your comment highlights what it probably the misunderstanding underlying my question.

I perfectly accept your calculation of your distance going around the latitude line is given by 1.414pi*r. In particular, the limits of the integral (phi=0 and phi=2pi) coincide.

However the second calculation going though the North Pole raises some questions for me. To ensure that the limits of integral coincide, shouldn’t theta go from pi/4 to -7pi/4 (going through the north pole and returning to the same point) or pi/4 to 9pi/4 ( going through the south pole and returning to the same point) .

As I see it, the limits of second calculation do not represent the same start and end points as the limits of the first calculation.

Where am I going wrong?

I *think* what pervect is doing is, move on the geodesic and come back. Mathematically, I *think* it looks like going from Theta = Pi/4, Phi = 0 to Theta = Pi/4, Phi = Pi via north pole and then back, again via north pole. This ensures we stay on the geodesic and take the shorter route (the longer route is not return, keep moving and return via south pole). It'll be Pi*r instead of 0.5*Pi*r what pervect got, which still, fits the bill.

 

[DeDunc]

At the risk of hi-jacking this thread can I ask the question in a different way. I fully expect that the question reveals a misunderstanding of geodesics so would be happy if someone could correct my thinking.

Question below:

If we imagine the surface of sphere (2-d) and draw a line of latitude defined by the following: theta=pi/4 and 0<phi<2pi. ( like salami slicing through a ball) . A line which starts and ends on the same point on this curve is not a geodesic.

What is the description of the curve which starts and ends on the same point on the line of lattitude but minimizes the distance travelled?

 

[Sourabh N]

The http://en.wikipedia.org/wiki/Great_circle" . It (obviously?) doesn't lie on the latitude, unless you're on the equator, where the great circle is the latitude.

 

[DeDunc]

Oh great. Thanks HallofIvy & Sourabh. Both posted very useful replies just before I sent my last reply.

HallsofIVy- I understand how the plane going through the centre intersecting the curve defines the great circle. But I couldn't see how this defines a shorter path

I think Sourabh has clarifed it for me. The curve goes through the north pole, but rather than continuing past the south pole and returning to the same point. You simply reverse direction ( once you reach the line of latitude) and go back through the north pole to return to the same point.

Do I have that correct now?

 

[Sourabh N]

That's what I said. It'll be better if a third person approves.

 

[A.T.]

Consider what happens to an elastic band that is fixed at its ends and stretched along the long segment of a great circle. It starts sliding back towards the short short segment of the great circle while decreasing continuously in length.

If it is placed accurately along the long segment it will not slide. Isn't the long segment a global maximum, among the connections which are locally a minimum along their entire length?

 

[korkscrew]

Simple question about geodesics.
My Question

If the line of latitude above is not a geodesic. Then it is not the shortest path between the two points defined by (theta=pi/2 , phi=0 & theta=pi/2 , phi= 2pi). These two points coincide. If it is not the shortest path between the two points described. Then what is? From a purely intuitive standpoint there does not seen to be a shorter path which links these two points.

Short answer:
If you consider one point on a sphere and wish to draw a geodesic on the sphere including this point, this line would start at the point, go around the fattest part of the sphere, and then connect up with the starting point again.

A bit more detail:
Imagine you make two dots on a grapefruit with a Sharpie. If you then draw a line connecting them in the most direct way (no tricks here), you've drawn a section of a geodesic. If you extend that line without curving its path, you'll arrive at the first point again. This is a geodesic. Looking at this great circle around your grapefruit, you see you have two ways to connect the dots: the short way and a long way. Both these paths satisfy the geodesic equation.

Does this help?