Newtonian force as a covariant or contravariant quantity

[TrickyDicky]

You need a connection in order to make sense of this, as wbn mentioned.

Exactly, and I like to think of this affine connection that is needed in order to differentiate vector fields in differentiable manifolds as a structure that takes advantage of the common property of all manifolds, being locally like an affine Euclidean space.
So it is clear that some extra structure is needed, it just seems natural to me that this connection needed to parallel transport is the canonical Riemannian connection in the vast majority of classical physical problems, be it in configuration space (configuration manifold), Euclidean space, Lorentzian manifolds, etc.
My point was that I don't understand Burke's prejudice against metric tensors and what is it "not geometric" about them that might bother anyone. I agree with him that it should be more enphasized when the Euclidean metric is being implicitly used and when it is not.

 

[WannabeNewton]

Exactly, and I like to think of this affine connection that is needed in order to differentiate vector fields in differentiable manifolds as a structure that takes advantage of the common property of all manifolds, being locally like an affine Euclidean space.
So it is clear that some extra structure is needed, it just seems natural to me that this connection needed to parallel transport is the canonical Riemannian connection in the vast majority of classical physical problems, be it in configuration space (configuration manifold), Euclidean space, Lorentzian manifolds, etc.
My point was that I don't understand Burke's prejudice against metric tensors and what is it "not geometric" about them that might bother anyone. I agree with him that it should be more enphasized when the Euclidean metric is being implicitly used and when it is not.

How is it goin Tricky Dicky. Just to clarify, a connection is not the ONLY way to differentiate vector fields on a smooth manifold. Take, for example, the lie derivative. Both offer different geometric notions of "parallelism". I think his issue is just that in coordinate - based computations, there is so much blurring of what quantities should naturally be taken as one - forms and what as vectors because people tend to use the metric tensor to go from one to the other readily.

 

[TrickyDicky]

Projective geometry (that of perspective and vanishing points, etc...) has no metric (akin to the Euclidean metric).

Yes, that is a good example of not needing the concepts of angle and distance in geometry generalizations. A geometry only based in those concepts would be very poor. But my point was rather that there wasn't anything wrong with metrics if one wants to stress the geometrical side of something.

With regard to the volume of parallelepipeds, I think the correct statement is that whether you choose the Euclidean, Minkowski, or Galilean metric, you get the same value for the volume [possibly, up to signs]. So, it really doesn't depend on the metric... but instead on something common to those metrics (something describing the affine structure). Does the "determinant" rely on the choice of metric among these three space[time]s... or just the parallelepiped's vectors themselves?

Of course it doesn't. But I always think of the cross product when talking about the determinant, is there anything more Euclidean than that?

 

[stevendaryl]

It is very dangerous to pretend that a topic is much easier than it actually is.

I'm not sure that I agree. Progress in physics really was only possible because physicists oversimplified in a way that was good enough for the problems that they were interested in. I think it's important to have a feel for what the limitations are of a particular approach, and sometimes going forward means going backwards and redoing what you've already done, but in a more careful way.

Anyway, I was aiming my comments at the level of someone who is familiar with vectors in the context of Euclidean space, but doesn't realize the full implications of lacking a metric. I think going into the full complexities of differential geometry is the wrong level for the discussion.

 

[TrickyDicky]

How is it goin Tricky Dicky. Just to clarify, a connection is not the ONLY way to differentiate vector fields on a smooth manifold. Take, for example, the lie derivative. Both offer different geometric notions of "parallelism".

Yes, the only difference is that the Lie derivative admits torsion. I was jst thinking that most classical physics situations don't need to include torsion. An exception is the non-mainstream Einstein-Cartan theory that relates torsion with QM spin, but QM spin is not a classical concept.

I think his issue is just that in coordinate - based computations, there is so much blurring of what quantities should naturally be taken as one - forms and what as vectors because people tend to use the metric tensor to go from one to the other readily.

Yeah, agreed. But again, sure, you don't need metrics to talk about differential forms and vectors, but I still don't know if I buy the notion that classical physics quantities must be taken naturally as one or the other independently of the geometry of the physical problem at hand.

 

[TrickyDicky]

I think going into the full complexities of differential geometry is the wrong level for the discussion.

This looks a bit condescending with the people in this forum. (Not with poor ignorant me but with many others very capable of handling that level).

 

[stevendaryl]

This looks a bit condescending with the people in this forum. (Not with poor ignorant me but with many others very capable of handling that level).

I don't mean to be condescending, it's just that in an informal setting, as opposed to writing a textbook, it seems to me that pitching at the right level is more important than being perfectly general. Yes, there's the danger that someone will take something more literally than it was intended, and could be led astray, but I think that people generally can recognize the difference between a textbook style formal definition and an intuitive argument, and should know that the latter is only meant to steer someone in the right direction, rather than to be the definitive last word on the subject.

 

[bcrowell]

Here are some quotes from Burke's Applied Differential Geometry that support my statement above

The quotes are great, robphy -- thanks for going to the trouble of posting them. I found the one about forces of constraint to be particularly helpful.

 

[micromass]

I don't mean to be condescending, it's just that in an informal setting, as opposed to writing a textbook, it seems to me that pitching at the right level is more important than being perfectly general. Yes, there's the danger that someone will take something more literally than it was intended, and could be led astray, but I think that people generally can recognize the difference between a textbook style formal definition and an intuitive argument, and should know that the latter is only meant to steer someone in the right direction, rather than to be the definitive last word on the subject.

If you don't want to be completely literal and precise, then that's perfectly ok. But you should say that you're being imprecise. Nobody benefits from people getting misconceptions.

 

[bcrowell]

A further payoff for having a clear geometric picture of force comes in situations involving constraints. A slippery constraint is one having no frictional forces. One usually says that the constraint force is perpendicular to the constraint surface. Now if you are truly learning to think covariantly with respect to linear transformations, you will see that this is a nonsense statement. Perpendicularity is a meaningless concept in any configuration space that does not accidentally happen to have a metric. The correct geometric view of a constraint is sketched in Figure 18.4. As a 1-form, the constraint force is parallel to the constraint surface, and parallelism is a properly covariant notion. Again, the non-covariant language comes from excessive attention to the peculiar features of particle mechanics in Euclidean space, features which do not generalize.

I wanted to see if I could work out an example of this that was as simple as possible. I think my example makes sense, but maybe others here could tell me if this makes sense and help me smooth out the stuff I'm unsure of.

I actually made two examples. Example #1 is a billiard ball of unit mass in two dimensions, constrained by a diagonal wall to have y<x. The Lagrangian formalism just leads to the expected Newtonian expressions, p_x=\dot{x}, p_y=\dot{y}. The force of constraint is F_a=dp_a/dt. Let w^a be a vector parallel to the wall. Since we do happen to have a metric in this example,we can say that F_aw^a=0; most people would say that the force was perpendicular to the wall.

Example #2 is a human arm with a heavy mass gripped in the hand. The upper arm is raised at an angle \theta and the lower arm raised at an angle \phi (both measured relative to the vertical). The arm's weight is negligible compared to the unit mass of the gripped weight, and both the upper and lower arm have unit length. Because of the construction of the elbow joint, we have a constraint \theta \le \phi. The conjugate momenta (which are actually angular momenta) turn out to be p_\theta=\dot{\theta}+\cos(\phi-\theta)\dot{\phi} and a similar expression for p_\phi. The force of constraint is F_a=dp_a/dt. The surface of constraint can be represented by a vector w^a, which is on a diagonal line in the (\phi,\theta) plane. Since there is no metric, it doesn't make sense to say that F_a is perpendicular to w^a.

Geometrically, Burke has a nice representation of a 1-form as a pair of parallel lines, with one of the two lines marked with an arrowhead (see http://www.scribd.com/doc/37538938/Burke-DivGradCurl ). In this representation, the parallel lines representing the force of constraint are clearly parallel to the surface of constraint.

But I'm not clear on how to notate this idea that in both examples, the force 1-form is parallel to the surface. Do you represent the surface as, say, a 1-form created by taking the (infinite) gradient of a step function across the wall?

Apart from my notational confusion, does the rest of this seem right?

 

[atyy]

Wouldn't one use a metric in writing cosine of an angle?

 

[bcrowell]

Wouldn't one use a metric in writing cosine of an angle?

When I say there's no metric, I mean that there's no metric on the two-dimensional space of (\phi,\theta).

 

[DrGreg]

While searching for information relative to this thread, I stumbled across this on Google Books (Richard A Mould, Basic Relativity, p.258). It claims that, under some circumstances, "covariant energy" (time component of covariant 4-momentum) is globally conserved, but "contravariant energy" (time component of contravariant 4-momentum) is conserved only locally but not globally. Unfortunately the next page is missing from the preview, so it's not clear what the "some circumstances" actually are, or how you prove the claim.

Does anyone know what is the correct statement and its proof?

Well, in the case of Rindler coordinates
<br /> ds^2 = g^2 z^2 \, dt^2 - dx^2 - dy^2 -dz^2<br />(c=1) the contravariant 4-momentum of a particle at rest is
<br /> P^\alpha = \begin{bmatrix}<br /> \frac{m}{gz} \\<br /> 0 \\<br /> 0 \\<br /> 0<br /> \end{bmatrix}<br />whereas the covariant 4-momentum is
<br /> P_\alpha = \begin{bmatrix}<br /> mgz &amp;&amp;<br /> 0 &amp;&amp;<br /> 0 &amp;&amp;<br /> 0<br /> \end{bmatrix}<br />Of course mgz is the correct formula for "gravitational" potential energy in Rindler coordinates, whereas m/(gz) has no significance that I know of.

Another non-relativistic example I can think of is cylindrical polar coordinates in Euclidean 3-space
<br /> ds^2 = dr^2 + r^2 \, d\theta^2 + dz^2<br />For an arbitrary particle
<br /> x^i = \left( r(t), \theta(t), z(t) \right)<br />we have a contravariant 3-momentum
<br /> p^i = \begin{bmatrix}<br /> m \dot{r} \\<br /> m \dot{\theta} \\<br /> m \dot{z} <br /> \end{bmatrix}<br />whereas the covariant 3-momentum is
<br /> p_i = \begin{bmatrix}<br /> m \dot{r} &amp;&amp;<br /> m r^2 \dot{\theta} &amp;&amp;<br /> m \dot{z}<br /> \end{bmatrix}<br />Here again we see that the covariant component m r^2 \dot{\theta} is the conserved angular momentum, whereas the contravariant component m \dot{\theta} is not conserved.

So all of the above seems to be more evidence to suggest that momentum and its time-derivative, force, are naturally covariant rather than contravariant.

 

[WannabeNewton]

Is the statement that the 0 component of contravariant 4 - momentum is NEVER globally conserved and can only ever be locally conserved whereas the 0 component of covariant 4 - momentum CAN be globally conserved under the appropriate conditions?

 

[bcrowell]

While searching for information relative to this thread, I stumbled across this on Google Books (Richard A Mould, Basic Relativity, p.258). It claims that, under some circumstances, "covariant energy" (time component of covariant 4-momentum) is globally conserved, but "contravariant energy" (time component of contravariant 4-momentum) is conserved only locally but not globally. Unfortunately the next page is missing from the preview, so it's not clear what the "some circumstances" actually are, or how you prove the claim.

Does anyone know what is the correct statement and its proof?

I see. Since the metric can depend on position, we can have p&#039;_a=p_a (final=initial), but p&#039;^a \ne p^a, because the particle can be in different places at the initial and final times.

we have a contravariant 3-momentum
<br /> p^i = \begin{bmatrix}<br /> m \dot{r} \\<br /> m \dot{\theta} \\<br /> m \dot{z} <br /> \end{bmatrix}<br />whereas the covariant 3-momentum is
<br /> p_i = \begin{bmatrix}<br /> m \dot{r} &amp;&amp;<br /> m r^2 \dot{\theta} &amp;&amp;<br /> m \dot{z}<br /> \end{bmatrix}<br />Here again we see that the covariant component m r^2 \dot{\theta} is the conserved angular momentum, whereas the contravariant component m \dot{\theta} is not conserved.

I guess the Killing vectors \partial/\partial \theta and \partial/\partial z lead to the 2 conserved components of the lower-index momentum.

Here's another thing I have to try to wrap my head around: I think Killing vectors are naturally lower-index vectors, and even though you do have a metric, it doesn't make sense to try to raise their indices. The reason is that the Killing vector is a field, not just a vector defined at a point, and the metric is varying...Does that make sense!?

 

[atyy]

When I say there's no metric, I mean that there's no metric on the two-dimensional space of (\phi,\theta).

I see. And regarding the definition of parallel - is it defined by a connection?

 

[WannabeNewton]

Here's another thing I have to try to wrap my head around: I think Killing vectors are naturally lower-index vectors, and even though you do have a metric, it doesn't make sense to try to raise their indices. The reason is that the Killing vector is a field, not just a vector defined at a point, and the metric is varying...Does that make sense!?

I'm not sure I understand this. A killing vector field is defined as a vector field such that the lie derivative of the metric tensor along this vector field is zero. We can talk of killing vector fields in terms of their flows and so on. The lie derivative measures the rate of change of a tensor field along the flow of a vector field so it makes sense to take a killing vector field as, at each point, naturally an element of the tangent space. Of course you can use the metric to raise and lower indices as usual but the way a killing field is defined it uses the notion of a vector field as per the lie derivative and the notion of a vector field as a derivation. I don't see immediately why they would instead be naturally co -vector fields a priori. Note that we raise and lower indices of vector fields all the time as it is a point wise operation done at each point in space - time. This isn't an issue.

 

[stevendaryl]

I'm not sure I understand this. A killing vector field is defined as a vector field such that the lie derivative of the metric tensor along this vector field is zero.

In the Wikipedia article on Killing vectors, it is said that X^\mu is a Killing vector field if for all vectors Y and Z,

g(\nabla_Y (X),Z) + g(Y , \nabla_Z (X)) = 0

which is an equation on vector fields X^\mu. However, it also says that in "local coordinates", this is equivalent to

\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0

which is an equation on covector fields X_\mu. I don't quite understand this, because it seems that the latter equation doesn't even involve the metric (except indirectly, through the covariant derivative). It would seem to me that the latter could be defined even for a manifold without a metric, with just a connection, as a definition of a Killing co-vector field X_\mu.

 

[TrickyDicky]

In the Wikipedia article on Killing vectors, it is said that X^\mu is a Killing vector field if for all vectors Y and Z,

g(\nabla_Y (X),Z) + g(Y , \nabla_Z (X)) = 0

which is an equation on vector fields X^\mu. However, it also says that in "local coordinates", this is equivalent to

\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0

which is an equation on covector fields X_\mu. I don't quite understand this, because it seems that the latter equation doesn't even involve the metric (except indirectly, through the covariant derivative). It would seem to me that the latter could be defined even for a manifold without a metric, with just a connection, as a definition of a Killing co-vector field X_\mu.

Huh?, Killing (co)vectors are those that preserve the metric tensor, how can you define them if there is no metric?

 

[WannabeNewton]

In the Wikipedia article on Killing vectors, it is said that X^\mu is a Killing vector field if for all vectors Y and Z,

g(\nabla_Y (X),Z) + g(Y , \nabla_Z (X)) = 0

which is an equation on vector fields X^\mu. However, it also says that in "local coordinates", this is equivalent to

\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0

which is an equation on covector fields X_\mu.

Yeah you can express them that way in local coordinates as you prolly already know because when you compute the first expression in coordinates the metric tensor ends up lowering the indices so I don't know if that would count as making the co - vector expression any more natural.

 

[TrickyDicky]

Here's another thing I have to try to wrap my head around: I think Killing vectors are naturally lower-index vectors, and even though you do have a metric, it doesn't make sense to try to raise their indices. The reason is that the Killing vector is a field, not just a vector defined at a point, and the metric is varying...Does that make sense!?

I don't understand this. What varying metric are you referring to? why doesn't make sense to raise or lower indices for fields if there is a metric?, fields assign a vector or covector to each point in the manifold.
Why the insistence on what is natural for a certain field when examples keep popping up that contradict any abstract "naturalness" in the presence of a metric of the covariance or contravariance of (co)vector fields?

 

[stevendaryl]

Huh?, Killing (co)vectors are those that preserve the metric tensor, how can you define them if there is no metric?

I'm just saying that the equation for a Killing co-vector doesn't mention the metric tensor, and so the equation would make sense even in the absence of a metric, it would seem.

 

[bcrowell]

Trying to clear up my foggy thinking re Killing vectors...

The Killing equation \nabla_a \xi_b+\nabla_b \xi_a=0 does use the metric, because the covariant derivative is defined in terms of the metric. Likewise if you define Killing vectors in terms of preserving distances, you're still appealing to a metric. On a bare manifold without a connection, there is no way to define a Killing vector. For instance, if I take the spacetime in and around the planet earth, it has Killing vectors such as rotation and time translation. If I take away the metrical information, then all I have is a topological space that's topologically isomorphic to R^3, and there's no way to say what its Killing vectors are.

You can raise the indices like this \nabla^a \xi^b+\nabla^b \xi^a=0. Note how the structure of the equation forces you to raise the indices on the derivatives as well as the Killing vector. If we're thinking of derivatives as being "naturally" lower-index quantities, then this does provide some motivation for saying that Killing vectos are naturally lower-index. But the fact that this is a *covariant* derivative means that you must have a metric defined, and therefore there's nothing to stop you from raising its index. This is different from the case of a plain old partial derivative.

People normally express Killing vectors using partial derivatives as a basis, e.g., \partial_t for a metric that doesn't change over time. So if there is a "natural" way to write them, it would definitely have to be lower-index.

On the other hand, it would also seem extremely natural to me to express the same symmetry as a translation of the time coordinate, x^t \rightarrow x^t+dt.

When you have a Killing vector \xi_a, test particles follow trajectories that conserve \xi_a v^a. This seems odd to me because we normally think of momentum as being what's conserved, not velocity. We can multiply by m to get a momentum, but then it would be an upper-index momentum, which is not the natural way to express momentum.

I would like to connect this to DrGreg's #103, which wasn't explicitly written in terms of a discussion of Killing vectors.

 

[stevendaryl]

Trying to clear up my foggy thinking re Killing vectors...

The Killing equation \nabla_a \xi_b+\nabla_b \xi_a=0 does use the metric, because the covariant derivative is defined in terms of the metric.

But covariant derivative only requires a connection, not a metric. A metric is sufficient for a connection, but not necessary.

 

[TrickyDicky]

But covariant derivative only requires a connection, not a metric. A metric is sufficient for a connection, but not necessary.

The covariant dervative used with Killing fields is not a general connection but the Levi-Civita unique metric connection.

 

[TrickyDicky]

I'm just saying that the equation for a Killing co-vector doesn't mention the metric tensor, and so the equation would make sense even in the absence of a metric, it would seem.

The equation doesn't make sense without a metric, the metric is implicit in the Christoffel coefficients of the covariant derivatives of the equation, you need the metric to obtain them.

 

[stevendaryl]

The equation doesn't make sense without a metric, the metric is implicit in the Christoffel coefficients of the covariant derivatives of the equation, you need the metric to obtain them.

No, connections are more basic than metrics. It's possible to have a connection even when you don't have a metric.

 

[bcrowell]

But covariant derivative only requires a connection, not a metric. A metric is sufficient for a connection, but not necessary.

OK, but the examples we've come up with so far are all either ones with metrics or ones with no connection at all. Can you come up with a physically well motivated example where there's a connection but no metric, and tie it into this topic?

 

[stevendaryl]

OK, but the examples we've come up with so far are all either ones with metrics or ones with no connection at all. Can you come up with a physically well motivated example where there's a connection but no metric, and tie it into this topic?

Newton-Cartan theory of gravity has a connection, but no metric (or I should say, it doesn't have a full metric that allows the raising and lowering of indices).

 

[martinbn]

stevendaryl, the useful property of Killing fields is that their flows are local isometries. You can make any definition you'd like, but what would it be good for?

The wikipedia article has a comment at the end of a generalization to a manifold with no metric, but a group action to substitute for the lack of isometries.