I Using the derivative of a tangent vector to define a geodesic

I hope I'm asking this in the right place!! I'm making my way through the tensors chapter of the Riley et al Math Methods book, and am being tripped up on their discussion of geodesics at the very end of the chapter. In deriving the equation for a geodesic, they basically look at the absolute derivative along a curve parameterized by its arc length and ask that the derivative of the tangent to the curve be zero.

$$\frac {d \vec t} {ds} = \vec 0,$$ where ##\vec t = \frac {d\vec r}{ds}## and ##\vec r(s)## is the position vector parameterized by arc length.

Then they just write out the derivative ##\frac {d\vec t}{ds} = t^{i}{}_{;k} \frac {du^{k}}{ds} \vec e^{}{}_{i} = \vec 0##. I'm totally new to all of this so I'm not sure how standard their notation is, but ##t^{i}{}_{;k}## is the covariant derivative. This all comes from differentiating ##\vec t = t^{i}\vec e^{}{}_{i}## with respect to arc length. Some chain rule gets you to the final expression.

Writing out the covariant derivative, the equation is $$(\frac{dt^{i}}{ds} + \Gamma^{i}{}_{jk}t^{j} \frac{du^{k}}{ds})\vec e^{}{}_{i} = 0.$$

Up to that point I follow, but then to simplify they assert that ##t^{j} = \frac{du^{j}}{ds}##, which substituting for ##t^{i}## and ##t^{j}## gives their final equation for a geodesic: $$\frac{d^{2}u^{i}}{ds^{2}} + \Gamma^{i}{}_{jk}\frac{du^{j}}{ds}\frac{du^{k}}{ds}=0.$$

My problem is specifically on the use of ##t^{j} = \frac{du^{j}}{ds}##. Is this always true, even in general curvilinear coordinates? Why doesn't the fact that basis vectors aren't constant throughout the space complicate that derivative so that you can't always just take the derivative of the components and ignore the basis vectors? Why don't we have to take an absolute derivative with respect to arc length of ##\vec r(s)## to get the components ##t^{i}## of ##\vec t##?

When I treat the position vector ##\vec r(s)## as I would any other vector, I get $$\vec t = \frac{d\vec r}{ds} =\frac{d}{ds}(u^{i}\vec e^{}{}_{i})= \frac{du^{i}}{ds}\vec e^{}{}_{i} + u^{i}\frac{d\vec e^{}{}_{i}}{ds} = \frac{du^{i}}{ds} + u^{m}\Gamma^{i}{}_{mk}\frac{du^{k}}{ds}.$$

Why does that last term in the last expression go to zero? I know I must be overlooking something (probably obvious) to do with the fact that I'm differentiating a position vector with respect to arc length. Any help would be much appreciated, and thanks a ton in advance. Please let me know if I can clarify or expand on anything!
 

Orodruin

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Is this always true, even in general curvilinear coordinates?
Yes. It is true by definition of the tangent vector basis, which is that
$$
\vec E_a = \frac{\partial \vec x}{\partial \xi^a}
$$
(using ##\xi^a## for coordinates and ##\vec x## for the position vector, I do not like the notation using ##u^i##). From this follows that
$$
\vec t = \frac{d\vec x}{ds} = \frac{d \xi^a}{d s} \frac{\partial \vec x}{\partial \xi^a} = \frac{d\xi^a}{ds} \vec E_a
$$
by the chain rule.

Note that the position vector is not given by ##\vec x = \xi^a \vec E_a## in general curvilinear coordinates.
 
Thanks for the reply!!

Note that the position vector is not given by ##\vec x = \xi^a \vec E_a## in general curvilinear coordinates.
Ah ok, so I was overlooking the fact that the basis I'm working in is defined by tangent vectors to the coordinate curves (I think?). Can I think of the tangent ##\frac {d\vec x}{ds}## geometrically as being a linear combination of the ##\vec E^{}{}_{a}## coordinate tangent vectors, weighted by how much each coordinate ##\xi^{a}## is changing with respect to ##s##?

Also, a major problem in my reasoning was my attempt to represent the position vector as ##\vec x = \xi^a \vec E_a##. Would you mind elaborating on why this isn't the case generally? Is it only the case in an orthogonal coordinate system? How can you write down a position vector in general coordinates?

Thanks again for your help!
 

Orodruin

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the basis I'm working in is defined by tangent vectors to the coordinate curves (I think?)
Yes, this is correct.

Can I think of the tangent ##\frac {d\vec x}{ds}## geometrically as being a linear combination of the ##\vec E^{}{}_{a}## coordinate tangent vectors, weighted by how much each coordinate ##\xi^{a}## is changing with respect to ##s##?
Yes. In fact, it is in essence how (tangent) vectors are defined on more general spaces (manifolds).

Also, a major problem in my reasoning was my attempt to represent the position vector as ##\vec x = \xi^a \vec E_a##. Would you mind elaborating on why this isn't the case generally?
As you have touched upon, the tangent vector basis changes from point to point. The position vector at each point is the translation vector from the origin to the point in question. If you have a general coordinate system, it is not at all certain that it is constructed in such a way that the coordinates of the point are the components of the position vector in the tangent vector basis, in fact, it will not be true in most coordinate systems.

A particular example of where this is not the case is polar coordinates in ##\mathbb R^2## where ##\vec x = r \vec E_r##, i.e., there is no component in the ##\vec E_\phi## direction. This is due to the direction from the origin to a point is in the ##\vec E_r## direction of that point.

Is it only the case in an orthogonal coordinate system? How can you write down a position vector in general coordinates?
The coordinate systems where it is true are the affine coordinate systems, which is slightly more general than orthogonal coordinate systems. These are the coordinate systems where the basis vectors do not change from point to point, i.e., all the coordinate lines are straight lines.

In a general manifold (such as on the two-dimensional sphere) there is no position vector and tangent vectors are defined in a different way. In affine spaces (such as Euclidean space), you can find the position vector in curvilinear coordinates by expressing your Cartesian basis in the curvilinear basis and expressing the Cartesian coordinates in terms of the curvilinear coordinates. For example, for polar coordinates in ##\mathbb R^2##, you would have
$$
x = r\cos(\phi), \quad y = r \sin(\phi).
$$
This leads to
$$
\vec E_r = \cos(\phi) \vec e_1 + \sin(\phi) \vec e_2, \quad
\vec E_\phi = r[-\sin(\phi) \vec e_1 + \cos(\phi) \vec e_2].
$$
The position vector in Cartesian coordinates is ##\vec x = x\vec e_1 + y \vec e_2## and so
$$
\vec x = r\cos(\phi) \vec e_1 + r \sin(\phi) \vec e_2 = r[\cos(\phi)\vec e_1 + \sin(\phi)\vec e_2] = r \vec E_r.
$$

I am a bit surprised if Riley et al do not discuss these issues at all.
 
This is very helpful, thank you!

A particular example of where this is not the case is polar coordinates in ##\mathbb R^2## where ##\vec x = r \vec E_r##, i.e., there is no component in the ##\vec E_\phi## direction. This is due to the direction from the origin to a point is in the ##\vec E_r## direction of that point.
I thought of polar coordinates shortly after posting my question. (And felt dumb for not having thought about them immediately!!) A clear case of when the coordinates at a point aren't the components of a position vector.

The coordinate systems where it is true are the affine coordinate systems, which is slightly more general than orthogonal coordinate systems. These are the coordinate systems where the basis vectors do not change from point to point, i.e., all the coordinate lines are straight lines.
This is really helpful, thank you.

In affine spaces (such as Euclidean space), you can find the position vector in curvilinear coordinates by expressing your Cartesian basis in the curvilinear basis and expressing the Cartesian coordinates in terms of the curvilinear coordinates.
So we can have a coordinate system that is not affine, but defined on an affine space, and this is the situation in which we are able to write down a position vector by beginning from Cartesian coordinates and then converting to the curvilinear coordinates? But we can't do this in more general spaces, correct? I think I ought to devote some time to a differential geometry text; the Riley et al chapter specifically treats tensors in 3D Euclidean space.

One other thing that's been bothering me: in a general curvilinear coordinate system in Euclidean space, if the position vector is a displacement from the origin - which involves two points: the origin and the point ##P## of which we want to know the position - do we always describe the position vector using the basis vectors at ##P##? In polar coordinates this distinction seems not to matter because every point along the line connecting the origin to our point has the same ##\vec E^{}{}_{r}##. I can pick a direction ##\vec E{}{}_{r}## at the origin, move along it by ##r##, and end up at ##P## described by ##r\vec E{}{}_{r}##. I can obviously follow a similar procedure in Cartesian. But I can imagine systems in which this is not the case.

I guess my concern is that in polar coordinates (for example), ##r\vec E{}{}_{r}## means "this far from the origin, in this direction." But if we allow the directions of our basis vectors not to be constant along the line connecting the origin to the point, it's not clear to me how we can define a displacement. When we say "this far from the origin, in this direction" (or this far in this direction, and this far in this direction, etc), do we imagine plucking our basis vectors from ##P##, setting them down at the origin (though the basis vectors at the origin may well be different) and then translating to ##P## using its own basis vectors? Apologies if that doesn't make sense; I can try to explain further if it doesn't. Maybe this worry is a bit pedantic!!

If I think of just starting in Cartesian and converting to the new coordinates, it's clear that the relevant basis vectors will be the ones at ##P##, so I guess my concern is about how to understand that vector as describing a translation from the origin.

I am a bit surprised if Riley et al do not discuss these issues at all.
They certainly discuss coordinate transformations of the position vector - the fault lies with me there!! Thank you very much for helping me to become unconfused. As for more general discussions of tensors, Riley et al suggest going elsewhere.
 

Orodruin

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So we can have a coordinate system that is not affine, but defined on an affine space, and this is the situation in which we are able to write down a position vector by beginning from Cartesian coordinates and then converting to the curvilinear coordinates?
Technically, you can only have a Cartesian coordinate system on Euclidean space (it requires an inner product). In a more general affine space you can start with the position vector in any affine coordinate system and relate your new coordinates to that system.

But we can't do this in more general spaces, correct? I think I ought to devote some time to a differential geometry text; the Riley et al chapter specifically treats tensors in 3D Euclidean space.
Correct. There is no position vector in more general spaces (your typical manifold). I think it is a good idea to start with Euclidean space, it gives a certain amount of intuition for the concepts before moving on to general manifolds. In fact, it is also the way I do it in an early chapter of my book before introducing calculus on manifolds in the penultimate chapter.

One other thing that's been bothering me: in a general curvilinear coordinate system in Euclidean space, if the position vector is a displacement from the origin - which involves two points: the origin and the point P of which we want to know the position - do we always describe the position vector using the basis vectors at P?
Typically yes. It is natural to describe any vector field in terms of the basis at the point where the field is being considered. In the case of the position vector field, it is a vector field taking the value of the displacement vector necessary to get to the point P from the origin and it is natural to describe that using the basis at the point P. In a general manifold, it would not make any sense at all to describe a vector at a point P using the basis at a different point Q because the bases at different points span different tangent spaces. Thus, using the basis at the given point is the concept that generalises with the least amount of headache.

Note that the only important thing in the case of the affine space and curvilinear coordinates are the point P and the origin. It is not a question of following a line outwards from the origin to the point, but has more to do with the bijection between the tangent space and the affine space itself that exists in an affine space through the translation map.

do we imagine plucking our basis vectors from PPP, setting them down at the origin (though the basis vectors at the origin may well be different) and then translating to PPP using its own basis vectors?
Yes. In an affine space, there is nothing ambiguous about this procedure because the affine space is flat and tangent spaces at different points have a unique correspondence to each other. You would have a problem doing this in a curved space, which is one of the reasons there is no position vector in curved spaces.

Another way of thinking about it (made equivalent by the properties of the translation map) is "the origin is displaced by ##-\vec x## relative to the point where I am".
 
In fact, it is also the way I do it in an early chapter of my book before introducing calculus on manifolds in the penultimate chapter.
Is this the book pictured in your avatar?

It is natural to describe any vector field in terms of the basis at the point where the field is being considered. In the case of the position vector field, it is a vector field taking the value of the displacement vector necessary to get to the point P from the origin and it is natural to describe that using the basis at the point P. In a general manifold, it would not make any sense at all to describe a vector at a point P using the basis at a different point Q because the bases at different points span different tangent spaces. Thus, using the basis at the given point is the concept that generalises with the least amount of headache.
This phrasing is particularly helpful, thank you!

Yes. In an affine space, there is nothing ambiguous about this procedure because the affine space is flat and tangent spaces at different points have a unique correspondence to each other. You would have a problem doing this in a curved space, which is one of the reasons there is no position vector in curved spaces.
This makes a lot of sense. Thanks again for all of your help, it's really made things much clearer for me!!
 

Orodruin

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Is this the book pictured in your avatar?
Yes. You can read more here.

Thanks again for all of your help, it's really made things much clearer for me!!
Happy to be of help. The subject of calculus on manifolds can be daunting and difficult to grasp, but it is certainly rewarding once you start developing some intuition for the concepts.
 

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