Register to reply

Newtonian force as a covariant or contravariant quantity

Share this thread:
TrickyDicky
#91
Jan30-13, 11:58 AM
P: 3,001
Quote Quote by micromass View Post
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
#92
Jan30-13, 12:26 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,423
Quote Quote by TrickyDicky View Post
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
#93
Jan30-13, 12:43 PM
P: 3,001
Quote Quote by robphy View Post
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.

Quote Quote by robphy View Post
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
#94
Jan30-13, 12:49 PM
Sci Advisor
P: 1,945
Quote Quote by micromass View Post
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
#95
Jan30-13, 12:53 PM
P: 3,001
Quote Quote by WannabeNewton View Post
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.
Quote Quote by WannabeNewton View Post
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
#96
Jan30-13, 12:58 PM
P: 3,001
Quote Quote by stevendaryl View Post
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
#97
Jan30-13, 01:11 PM
Sci Advisor
P: 1,945
Quote Quote by TrickyDicky View Post
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
#98
Jan30-13, 01:31 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,584
Quote Quote by robphy View Post
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
#99
Jan30-13, 02:00 PM
Mentor
micromass's Avatar
P: 18,036
Quote Quote by stevendaryl View Post
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
#100
Jan30-13, 05:42 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,584
Quote Quote by William Burke
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, [itex]p_x=\dot{x}[/itex], [itex]p_y=\dot{y}[/itex]. The force of constraint is [itex]F_a=dp_a/dt[/itex]. Let [itex]w^a[/itex] be a vector parallel to the wall. Since we do happen to have a metric in this example,we can say that [itex]F_aw^a=0[/itex]; 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 [itex]\theta[/itex] and the lower arm raised at an angle [itex]\phi[/itex] (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 [itex]\theta \le \phi[/itex]. The conjugate momenta (which are actually angular momenta) turn out to be [itex]p_\theta=\dot{\theta}+\cos(\phi-\theta)\dot{\phi}[/itex] and a similar expression for [itex]p_\phi[/itex]. The force of constraint is [itex]F_a=dp_a/dt[/itex]. The surface of constraint can be represented by a vector [itex]w^a[/itex], which is on a diagonal line in the [itex](\phi,\theta)[/itex] plane. Since there is no metric, it doesn't make sense to say that [itex]F_a[/itex] is perpendicular to [itex]w^a[/itex].

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
#101
Jan30-13, 06:04 PM
Sci Advisor
P: 8,370
Wouldn't one use a metric in writing cosine of an angle?
bcrowell
#102
Jan30-13, 07:01 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,584
Quote Quote by atyy View Post
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 [itex](\phi,\theta)[/itex].
DrGreg
#103
Jan30-13, 07:23 PM
Sci Advisor
PF Gold
DrGreg's Avatar
P: 1,843
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
[tex]
ds^2 = g^2 z^2 \, dt^2 - dx^2 - dy^2 -dz^2
[/tex](c=1) the contravariant 4-momentum of a particle at rest is
[tex]
P^\alpha = \begin{bmatrix}
\frac{m}{gz} \\
0 \\
0 \\
0
\end{bmatrix}
[/tex]whereas the covariant 4-momentum is
[tex]
P_\alpha = \begin{bmatrix}
mgz &&
0 &&
0 &&
0
\end{bmatrix}
[/tex]Of course [itex]mgz[/itex] is the correct formula for "gravitational" potential energy in Rindler coordinates, whereas [itex]m/(gz)[/itex] has no significance that I know of.

Another non-relativistic example I can think of is cylindrical polar coordinates in Euclidean 3-space
[tex]
ds^2 = dr^2 + r^2 \, d\theta^2 + dz^2
[/tex]For an arbitrary particle
[tex]
x^i = \left( r(t), \theta(t), z(t) \right)
[/tex]we have a contravariant 3-momentum
[tex]
p^i = \begin{bmatrix}
m \dot{r} \\
m \dot{\theta} \\
m \dot{z}
\end{bmatrix}
[/tex]whereas the covariant 3-momentum is
[tex]
p_i = \begin{bmatrix}
m \dot{r} &&
m r^2 \dot{\theta} &&
m \dot{z}
\end{bmatrix}
[/tex]Here again we see that the covariant component [itex]m r^2 \dot{\theta}[/itex] is the conserved angular momentum, whereas the contravariant component [itex]m \dot{\theta}[/itex] 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
#104
Jan30-13, 07:37 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,423
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
#105
Jan30-13, 09:44 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,584
Quote Quote by DrGreg View Post
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 [itex]p'_a=p_a[/itex] (final=initial), but [itex]p'^a \ne p^a[/itex], because the particle can be in different places at the initial and final times.

Quote Quote by DrGreg View Post
we have a contravariant 3-momentum
[tex]
p^i = \begin{bmatrix}
m \dot{r} \\
m \dot{\theta} \\
m \dot{z}
\end{bmatrix}
[/tex]whereas the covariant 3-momentum is
[tex]
p_i = \begin{bmatrix}
m \dot{r} &&
m r^2 \dot{\theta} &&
m \dot{z}
\end{bmatrix}
[/tex]Here again we see that the covariant component [itex]m r^2 \dot{\theta}[/itex] is the conserved angular momentum, whereas the contravariant component [itex]m \dot{\theta}[/itex] is not conserved.
I guess the Killing vectors [itex]\partial/\partial \theta[/itex] and [itex]\partial/\partial z[/itex] 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
#106
Jan30-13, 11:54 PM
Sci Advisor
P: 8,370
Quote Quote by bcrowell View Post
When I say there's no metric, I mean that there's no metric on the two-dimensional space of [itex](\phi,\theta)[/itex].
I see. And regarding the definition of parallel - is it defined by a connection?
WannabeNewton
#107
Jan31-13, 12:02 AM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,423
Quote Quote by bcrowell View Post
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
#108
Jan31-13, 06:38 AM
Sci Advisor
P: 1,945
Quote Quote by WannabeNewton View Post
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 [itex]X^\mu[/itex] is a Killing vector field if for all vectors [itex]Y[/itex] and [itex]Z[/itex],

[itex]g(\nabla_Y (X),Z) + g(Y , \nabla_Z (X)) = 0[/itex]

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

[itex]\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0[/itex]

which is an equation on covector fields [itex]X_\mu[/itex]. 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 [itex]X_\mu[/itex].


Register to reply

Related Discussions
When is something covariant or contravariant? Special & General Relativity 7
Covariant vs contravariant Differential Geometry 7
Covariant vs contravariant Differential Geometry 62
Covariant vs. contravariant Special & General Relativity 6
Covariant and contravariant ? General Physics 15