Newtonian force as a covariant or contravariant quantity

[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.

 

[WannabeNewton]

In GR you deal with levi civita connections but the right way to think of killing vector fields is as vector fields that result in vanishing lie derivative of the metric tensor along their flow i.e their flows are local isometries on the riemanniam manifold. The killing equation with the convectors is just a computational consequence of that.

 

[stevendaryl]

In GR you deal with levi civita connections but the right way to think of killing vector fields is as vector fields that result in vanishing lie derivative of the metric tensor along their flow i.e their flows are local isometries on the riemanniam manifold. The killing equation with the convectors is just a computational consequence of that.

The easiest way for me to picture a Killing vector field is to have a coordinate system in which the components of the metric tensor are independent of one or more coordinate. In that case, the vector field is just the basis vector for that coordinate. What I'm not sure about is whether this is perfectly general---that is, if there is a Killing field, can we always cook up a coordinate system (at least for a region) in which the metric is independent of one of the coordinates?

 

[martinbn]

The easiest way for me to picture a Killing vector field is to have a coordinate system in which the components of the metric tensor are independent of one or more coordinate. In that case, the vector field is just the basis vector for that coordinate. What I'm not sure about is whether this is perfectly general---that is, if there is a Killing field, can we always cook up a coordinate system (at least for a region) in which the metric is independent of one of the coordinates?

Just take the integral curves of the Killing field for one of the coordinates.

 

[WannabeNewton]

If you have access, the appendix on killing fields and flows in Carroll's GR text gives a very accessible and excellent treatment of these things.

 

[bcrowell]

If you have access, the appendix on killing fields and flows in Carroll's GR text gives a very accessible and excellent treatment of these things.

Access shouldn't be an issue, unless there's something in the printed book that's not in the online version: http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html

 

[PAllen]

Access shouldn't be an issue, unless there's something in the printed book that's not in the online version: http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html

There is a lot in the printed book that is not in the online version.

See:

http://preposterousuniverse.com/spacetimeandgeometry/
and
http://preposterousuniverse.com/grnotes/

discussion of the differences. The Appendices and several chapters are only in the book.

 

[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?

Browsing Wiki, there's a comment on an induced metric http://en.wikipedia.org/wiki/Newtonian_dynamics#Embedding_and_the_induced_Riemannian_metric : "Geometrically, the vector-function (7) implements an embedding of the comfiguration space M of the constrained Newtonian dynamical system into the 3N-dimensional flat configuration space of the unconstrained Newtonian dynamical system (3). Due to this embedding the Euclidean structure of the ambient space induces the Riemannian metric onto the manifold M." Aren't (\phi,\theta) the q of Wiki?

Google also came up with http://www.mast.queensu.ca/~andrew/teaching/math439/pdf/439notes.pdf which seems useful. Force is defined as a covector on p100: "Thus a force ... is a covector in the cotangent space ... Just why a force should naturally be regarded as a covector rather than a tangent vector is a question best answered by saying, 'Because that's the way it works out.'". Example 2.6.1 (p109) which is continued as Example 2.6.13 (p118) gives an example of a constrained system. It does mention a Riemannian metric.

Maybe one difference is that in general the Riemannian metric is curved. In a flat space, the Riemannian metric can be used to talk about "position vectors" being perpendicular after we have assigned an origin and cartesian coordinates. However, in general, a Riemannian metric acts on velocity vectors, and there is no such thing as a "position vector", just position coordinates.

 

[TrickyDicky]

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


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.

There are higher-index contravariant derivatives too, what you call natural is just pure convention.

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.

So much for what looks natural.

 

[TrickyDicky]

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

I never said otherwise, I was clearly referring to the Killing field vectors in the context of Riemannian manifolds, (like in the Wikipedia page BTW) as was clear from my #115.

 

[bcrowell]

Browsing Wiki, there's a comment on an induced metric http://en.wikipedia.org/wiki/Newtonian_dynamics#Embedding_and_the_induced_Riemannian_metric : "Geometrically, the vector-function (7) implements an embedding of the comfiguration space M of the constrained Newtonian dynamical system into the 3N-dimensional flat configuration space of the unconstrained Newtonian dynamical system (3). Due to this embedding the Euclidean structure of the ambient space induces the Riemannian metric onto the manifold M." Aren't (\phi,\theta) the q of Wiki?

I think what they're saying is that you can associate a metric with it of the form ds^2=d\theta^2+d\phi^2. But that isn't a metric that has any physical meaning in the example I made up in #100 of the arm holding a weight. E.g., it doesn't tell you how far the hand moved, or any other physically interesting piece of information. An equally plausible metric would be ds^2=3d\theta^2+d\phi^2, or virtually anything else you cared to make up.

 

[DrGreg]

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.

The convention we have of using the symbol \partial_t in this context is, I think, rather confusing. Here the symbol is being used, by convention, to represent a vector, not a co-vector, nor a component of a co-vector, viz.
<br /> \partial_t = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}<br />(expressed in a (t, x^1, x^2, x^3) coordinate system). It's unfortunate the symbol has a "downstairs" suffix, which suggests covariance instead of contravariance. To avoid further confusion, I will denote this vector by \textbf{e}, and its components by e^\alpha, for the remainder of this post.

(The symbol \partial_t also has a different meaning as the t-component of \partial_\alpha.)

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

In the notation above combined with post #103, the "covariant energy" is P_\alpha e^\alpha, so if \textbf{e} ( = \partial_t) is a Killing vector...

 

[atyy]

I think what they're saying is that you can associate a metric with it of the form ds^2=d\theta^2+d\phi^2. But that isn't a metric that has any physical meaning in the example I made up in #100 of the arm holding a weight. E.g., it doesn't tell you how far the hand moved, or any other physically interesting piece of information. An equally plausible metric would be ds^2=3d\theta^2+d\phi^2, or virtually anything else you cared to make up.

The notes from Lewis I linked to in #127 seem to define "simple" Lagrangian dynamics with a metric. I think it might be better to say that in a general curved space, a metric acts on velocity vectors. In flat space the metric can be carried over to "position vectors", but in a curved space there are only "position coordinates". The "simple" includes constrained systems, Example 2.6.13 (p118).

 

[Ben Niehoff]

Since there seems to be some confusion on Killing vectors...

A Killing vector is a vector that preserves the metric under Lie dragging, i.e.

\mathcal{L}_K \, g = 0
In this case, K is a vector, specifically, and not a covector. This equation is actually defined even in the absence of a connection; Lie dragging does not depend on it.

Now, obviously we have a metric, so we can use it to "lower the index" of K to make it a covector. In this case, the above equation boils down to

\nabla_\mu K_\nu + \nabla_\nu K_\mu = 0
where \nabla is the Levi-Civita connection (this is important). Because this equation specifically depends on choosing the Levi-Civita connection, it is really an equation that involves the metric and its derivatives, just as the first equation does.

If you choose some other connection \hat{\nabla} (or if you have no metric, so there is no natural choice of connection), then you can still write the equation

\hat{\nabla}_\mu K_\nu + \hat{\nabla}_\nu K_\mu = 0
but it is no longer Killing's equation! It is just some equation satisfied by some covector K.

Killing's equation is specifically "The Lie derivative along K of the metric is zero", in whatever form you choose to write it.

P.S. To see how

g(\nabla_X K, Y) + g(X, \nabla_Y K) = 0
gives the usual form of Killing's equation, simply put in X = \partial_\mu and Y = \partial_\nu.

 

[bcrowell]

The convention we have of using the symbol \partial_t in this context is, I think, rather confusing. Here the symbol is being used, by convention, to represent a vector, not a co-vector, nor a component of a co-vector, viz.
<br /> \partial_t = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}<br />(expressed in a (t, x^1, x^2, x^3) coordinate system). It's unfortunate the symbol has a "downstairs" suffix, which suggests covariance instead of contravariance. To avoid further confusion, I will denote this vector by \textbf{e}, and its components by e^\alpha, for the remainder of this post.

(The symbol \partial_t also has a different meaning as the t-component of \partial_\alpha.)

No, I'm pretty sure you're wrong about this. The operator \partial_a=\partial/\partial x^a is written with a lower index precisely because it's a covector and transforms like one. The notation isn't misleading. The notation shows exactly how the thing transforms. For instance, you can convince yourself of this by writing down a rescaling of the coordinate and applying the tensor transformation law.

 

[micromass]

No, I'm pretty sure you're wrong about this. The operator \partial_a=\partial/\partial x^a is written with a lower index precisely because it's a covector and transforms like one. The notation isn't misleading. The notation shows exactly how the thing transforms. For instance, you can convince yourself of this by writing down a rescaling of the coordinate and applying the tensor transformation law.

The notation \frac{\partial}{\partial x^i}\vert_p represents a tangent vector,and not a 1-form. A 1-form would look like dx^i\vert_p. See "Introduction to Smooth Manifolds" by Lee.

 

[WannabeNewton]

No, I'm pretty sure you're wrong about this. The operator \partial_a=\partial/\partial x^a is written with a lower index precisely because it's a covector and transforms like one. The notation isn't misleading. The notation shows exactly how the thing transforms. For instance, you can convince yourself of this by writing down a rescaling of the coordinate and applying the tensor transformation law.

Hello again misturr :). While they do transform in that way, at a given point p\in M and a coordinate chart (U,(x^{i})) the n derivations \left \{ \partial _{1}|_{p},...,\partial _{n}|_{p} \right \} actually form a basis for T_{p}(M) so they actually lie in the tangent space at that point.

 

[Ben Niehoff]

A Killing vector is a vector that preserves the metric under Lie dragging, i.e.

\mathcal{L}_K \, g = 0
In this case, K is a vector, specifically, and not a covector. This equation is actually defined even in the absence of a connection; Lie dragging does not depend on it.

I wanted to expand on this, because it might seem mysterious, given that the equation usually called "Killing's equation" has nablas in it.

Remember what Killing's equation actually means. The solution K is a vector field that generates an isometry of the manifold. So for example, on a sphere, the vector K points along lines of constant lattitude, and describes the velocity at each point on the sphere if the sphere as a whole were to be rigidly rotated about its axis. Notice that the flow of K along itself does NOT follow parallel transport (especially near the poles).

Of course, in order for K to solve a meaningful differential equation, there must be some way to relate nearby tangent spaces to each other. It is the rigid motion itself that links tangent spaces via the pullback. The connection is never needed.

The point is that ANY vector field generates some 1-parameter flow, called Lie dragging. The flow lines are the integral curves of that vector field. So, using pullbacks along this 1-parameter flow, you can relate different tangent spaces and compare vectors at different points. If that 1-parameter flow is a rigid motion, then you have an isometry (which essentially means, the manifold can be rigidly moved into itself). The conditions for a flow being a rigid motion are precisely Killing's equation*.

* Note: "Rigidity" is really the wrong concept here, as rigidity is an extrinsic property. Consider the rigidity of a triangular lattice in R^2, and now put the same triangular lattice into R^3; it becomes quite floppy. The correct concept is "metric-preserving", which is the intrinsic property that most closely corresponds to rigidity.

 

[Ben Niehoff]

Hello again misturr :). While they do transform in that way, at a given point p\in M and a coordinate chart (U,(x^{i})) the n derivations \left \{ \partial _{1}|_{p},...,\partial _{n}|_{p} \right \} actually form a basis for T_{p}(M) so they actually lie in the tangent space at that point.

It's important to remember that the indices on \partial_a, \partial_b, etc. are indices that indicate "which vector", not "which component". So to indicate contravariant vectors, the indices must be down in order for X = X^a \partial_a to make sense.

 

[DrGreg]

Edit: in the time it took me to write this, I see several others have already made the same point, but I might as well post this anyway.

No, I'm pretty sure you're wrong about this. The operator \partial_a=\partial/\partial x^a is written with a lower index precisely because it's a covector and transforms like one. The notation isn't misleading. The notation shows exactly how the thing transforms. For instance, you can convince yourself of this by writing down a rescaling of the coordinate and applying the tensor transformation law.

What you say is true of the second meaning I listed. I agree that
<br /> \begin{bmatrix}<br /> \frac{\partial}{\partial t} &amp;&amp; <br /> \frac{\partial}{\partial x} &amp;&amp; <br /> \frac{\partial}{\partial y} &amp;&amp; <br /> \frac{\partial}{\partial z}<br /> \end{bmatrix}<br />transforms as a co-vector, where the symbol \partial / \partial t denote one component of a covector.

But the symbol is not being used in that sense. It's being used (in an abuse of notation denoting an isomorphism to another space) to denote a vector, not a component.

It's like writing<br /> \textbf{V} = V^\alpha \textbf{e}_\alpha = V^0 \textbf{e}_0 + V^1 \textbf{e}_1 + V^2 \textbf{e}_2 + V^3 \textbf{e}_3<br />where \textbf{e}_0, \textbf{e}_1, ... are four different vectors, not covectors, nor 4 components of one covector.

 

[bcrowell]

Edit: in the time it took me to write this, I see several others have already made the same point, but I might as well post this anyway.

Actually, I think yours was the only post that actually cleared up the issue, so I'm glad you posted it anyway!

What you say is true of the second meaning I listed. I agree that
<br /> \begin{bmatrix}<br /> \frac{\partial}{\partial t} &amp;&amp; <br /> \frac{\partial}{\partial x} &amp;&amp; <br /> \frac{\partial}{\partial y} &amp;&amp; <br /> \frac{\partial}{\partial z}<br /> \end{bmatrix}<br />transforms as a co-vector, where the symbol \partial / \partial t denote one component of a covector.

But the symbol is not being used in that sense. It's being used (in an abuse of notation denoting an isomorphism to another space) to denote a vector, not a component.

It's like writing<br /> \textbf{V} = V^\alpha \textbf{e}_\alpha = V^0 \textbf{e}_0 + V^1 \textbf{e}_1 + V^2 \textbf{e}_2 + V^3 \textbf{e}_3<br />where \textbf{e}_0, \textbf{e}_1, ... are four different vectors, not covectors, nor 4 components of one covector.

IMO this hits the nail on the head and clears up something that I'd found confusing and that IMO #135 and #136 failed to deal with by resorting to appeals to authority. I think the key words here are "abuse of notation."

Let me see if I can lay out my present understanding and see if others agree with me.

We want a bunch of different things:

(1) We want to be able to use upper and lower indices to notate the real physical differences between two different types of vectors, and as physicists we define these types by their transformation properties.
(2) We want the transformation properties of an object to be apparent from the way we write its indices.
(3) We want to have grammatical rules that make it apparent when we're writing nonsense, e.g., we want to be able to recognize that there's something wrong in an equation like u_a=v_a w_{abc}.
(4) We want a notation that's compact and expressive, so we'd like to use Einstein summation notation and avoid writing sigmas.
(5) We want a notation that is widely accepted and understood by other physicists.

Appeals to authority are useful only with respect to #5. Since we don't all have the same printed books on our bookshelves, it's useful to have an online reference that's accessible to all of us, so let's use the online version of Carroll: http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll2.html

an arbitrary one-form is expanded into components as \omega=\omega_\mu dx^\mu [...] We will usually write the components \omega_{\mu} when we speak about a one-form \omega.

This violates #2, because the l.h.s. is written without indices, so we identify it as a scalar, and yet it's not a scalar, it's a covector. It satisfies #4 by being written in compact Einstein summation notation. It obeys #2 by writing the covector as \omega_{\mu}, but violates it by writing the basis covector as dx^\mu.

Carroll (and by proxy physicists in general) is trying to simultaneously satisfy incompatible desires 1-5. A clear way of seeing this is that he wants to write \omega_{\mu} as a synonym for \omega. Now suppose that (in old-fashioned non-abstract index notation, indicated by the Greek indices), we have \omega_0=1 for \mu=1 and all the other components zero. Then we have \omega=dx^\mu. But \omega_{\mu} is a synonym for \omega, so can substitute in and make it \omega_{\mu}=dx^\mu. Oops, this violates #3.

I think what's really happening here is that in a misguided attempt to satisfy 3 and 4, we write expressions like \omega_\mu dx^\mu. This reads like the scalar product of a vector and a covector, which isn't what it is. Maybe a better notation would be something like \omega_{(.)}=\sum_\mu\omega(\mu) \partial_\mu. It violates 4 and 5, but it obeys 1-3.

 

[atyy]

I don't agree. One has to distinguish between a representation of an object in particular coordinates, and the object itself. The typical physics index gymnastics notation refer to representations (components) of an object.

It is true that you could say that a tuple of of basis covariant vectors transforms as the components of a single contravariant vector. But this is not standard in physics, where we refer to transformation of the components. (See the Warning on p42 of http://books.google.com/books?id=DUnjs6nEn8wC&source=gbs_navlinks_s.)

 

[Ben Niehoff]

Carroll (and by proxy physicists in general) is trying to simultaneously satisfy incompatible desires 1-5. A clear way of seeing this is that he wants to write \omega_{\mu} as a synonym for \omega. Now suppose that (in old-fashioned non-abstract index notation, indicated by the Greek indices), we have \omega_0=1 for \mu=1 and all the other components zero. Then we have \omega=dx^\mu.

No, you can't write that. But you can write \omega = dx^1, which is perfectly sensible.

But \omega_{\mu} is a synonym for \omega, so can substitute in and make it \omega_{\mu}=dx^\mu.

That is also nonsense. At best you can write \omega_\mu = \delta_\mu{}^1.

There is nothing wrong or abusive about writing \omega = \omega_\mu \, dx^\mu. The summation rule is being followed ("one upper and one lower index are summed"), and the indices are in places that indicate their purpose:

1. Lower indices on real-number-type quantities indicate which component of a covariant object.

2. Upper indices on 1-form-type quantities indicate which 1-form. Such indices are written upper in order to be consistent with the summation rule.

If you were to consistently stick with index notation (either abstract or not), you would never write something like \omega = \omega_\mu \, dx^\mu. You would just talk about "the 1-form \omega_\mu".

At any rate, the constant arguing about what exactly indices "mean" is one of the reasons I hate this notation. Such discussions don't lead anywhere useful. It's even worse when you try to use orthonormal frames and you get into "flat space indices" vs. "curved space indices".

 

[WannabeNewton]

Just like we think of vectors at a point as a linear combination of derivations, we can write the dual vectors at a point as a linear combination of the dual basis to the derivations. We write V = V^{i}\partial _{i}, \omega = \omega _{i}dx^{i} so that \omega(V) = \omega _{i}dx^{i}(V^{j}\partial _{j}) = \omega_{i}V^{j}dx^{i}(\partial _{j}) = \omega_{i}V^{j}\delta ^{i}_{j} = \omega _{i}V^{i}. The initial two expressions are not meant to be inner products of some form. The indices on the derivations and their dual represent the elements of the basis and dual basis respectively, not components.

 

[micromass]

Appeals to authority are useful only with respect to #5. Since we don't all have the same printed books on our bookshelves, it's useful to have an online reference that's accessible to all of us, so let's use the online version of Carroll: http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll2.html

Well, your version of Carroll says the same thing as we have been saying:

We are now going to claim that the partial derivative operators \{\partial_{\mu}\} at p form a basis for the tangent space T_p. (It follows immediately that T_p is n-dimensional, since that is the number of basis vectors.)

and

Therefore the gradients \{dx^\mu\} are an appropriate set of basis one-forms; an arbitrary one-form is expanded into components as \omega = \omega_{\mu} dx^\mu.

 

[dx]

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?

The constraint surface can be represented by an equation of the form f(Xⁱ) = 0. The constraint surface in your second example is

f(φ, θ) = θ - φ = 0

A constraint force which is parallel to the constraint surface can be written as a multiple of df, since df is parallel to the constraint surface.

F = λdf = λ(dθ - dφ)

 

[stevendaryl]

No, I'm pretty sure you're wrong about this. The operator \partial_a=\partial/\partial x^a is written with a lower index precisely because it's a covector and transforms like one. The notation isn't misleading. The notation shows exactly how the thing transforms. For instance, you can convince yourself of this by writing down a rescaling of the coordinate and applying the tensor transformation law.

Yeah, these conventions are confusing, but they make some sense if you think about it.

In differential geometry, a vector V is defined via its effects on scalar fields. Specifically, if we choose a particular coordinate basis, we can write:

V(\Phi) = V^\mu \partial_\mu \Phi

with the implicit summation convention. Now, if V happens to be a basis vector e_\sigma, then that means that V_\sigma = 1, and all the other components are equal to 0. So we have, in this case:

e_\sigma(\Phi) = \partial_\sigma \Phi

Since this is true for any \Phi, it's an operator equation:

e_\sigma = \partial_\sigma

 

[stevendaryl]

The notation \frac{\partial}{\partial x^i}\vert_p represents a tangent vector,and not a 1-form. A 1-form would look like dx^i\vert_p. See "Introduction to Smooth Manifolds" by Lee.

Yeah, this stuff is pretty confusing. Even though \frac{\partial}{\partial x^\mu} is a vector, the prototypical covector is \nabla \Phi for some scalar field \Phi, which has components \frac{\partial}{\partial x^\mu} \Phi

To complete the confusion, even though dx^\mu is a 1-form, the prototypical vector is a tangent vector to a parametrized path P(\lambda), which has components \dfrac{dx^\mu}{d\lambda}.

 

[TrickyDicky]

Yeah, this stuff is pretty confusing. Even though \frac{\partial}{\partial x^\mu} is a vector, the prototypical covector is \nabla \Phi for some scalar field \Phi, which has components \frac{\partial}{\partial x^\mu} \Phi

To complete the confusion, even though dx^\mu is a 1-form, the prototypical vector is a tangent vector to a parametrized path P(\lambda), which has components \dfrac{dx^\mu}{d\lambda}.

It is confusing, but even more if one doesn't distinguish the geometrical object from its component representation. dx^\mu is a 1-form, but transforms contravariantly, however if it had components other than unity they would transform covariantly.

So I'm still waiting for someone to explain why the rate of change of momentum is not naturally a tangent vector.

 

[stevendaryl]

It is confusing, but even more if one doesn't distinguish the geometrical object from its component representation. dx^\mu is a 1-form, but transforms contravariantly, however if it had components other than unity they would transform covariantly.

So I'm still waiting for someone to explain why the rate of change of momentum is not naturally a tangent vector.

If momentum is defined to be \vec{P} = m \frac{\vec{dU}}{dt}, then it is naturally a vector. If it is defined via a Lagrangian L by P_\mu = \frac{\partial}{\partial U^\mu}L, then it is naturally a covector.

Another way to see that momentum should be considered contravariant might be:

In curvilinear coordinates, the equations of motion for a free particle is

\dfrac{d}{dt} U_\mu = 0

not

\dfrac{d}{dt} U^\mu = 0

 

[bcrowell]

I think the basic cause of the ugly inconsistency in Einstein notation is not that there's anything wrong with Einstein notation but that it's being mixed inappropriately with Sylvester's notation. Sylvester was the one who originally introduced terms like "contravariant" and "covariant." The pure Sylvester notation is described in this WP article: http://en.wikipedia.org/wiki/Contravariant_vector Unfortunately Sylvester's notation and terminology are much more cumbersome than Einstein notation, and it doesn't correspond to the way physicists customarily describe the objects themselves as having transformation properties, rather than the objects staying invariant while their representations transform.

In Einstein notation, a sum over up-down indices represents a tensor contraction that reduces the rank of an expression by 2. In Sylvester notation, a vector is expressed in terms of a basis as a sum over up-down indices, which is not a tensor product. Mixing Einstein and Sylvester notation produces these silly-looking things where it looks like we're expressing a covector as a sum of vectors or a vector as a sum of covectors.

The abstract index form of Einstein notation is carefully designed so that you *can't* inadvertently refer to a coordinate system; anything written in abstract index notation is automatically coordinate-independent. This means that in abstract index notation, you can't express a vector as a sum over basis vectors. You don't want to and you don't need to.

There is also an issue because in Einstein notation we refer to vectors as covariant and contravariant, whereas in Sylvester notation a given vector or 1-form is written in terms of components and basis vectors that are covariant and contravariant. Neither is right or wrong, but mixing them leads to stuff that looks like nonsense.