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Confusion with lower/raising indices

by vaibhavtewari
Tags: confusion, indices, lower or raising
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bcrowell
#19
Jun6-11, 10:51 PM
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Quote Quote by Sam Gralla View Post
However, my guess is that the poster didn't really mean ordinary derivative but actually meant the [tex]u^\mu \nabla_\mu[/tex] worldline derivative that comes up in applications, sometimes written as [tex] D/d\tau[/tex]. In that case his formula is correct because the metric commutes with the covariant derivative.
I'm not familiar with the notation [itex] D/d\tau[/itex], but your description of it sounds the same as what I thought the OP was describing.
Ben Niehoff
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Jun6-11, 10:51 PM
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Quote Quote by bcrowell View Post
Do you mean "scalar" the same way I'm using it, as something that's invariant under a boost?
No, that's why I said "scalar field". A scalar field is invariant under local Lorentz transformations, including boosts. That is, under a general coordinate transformation, a scalar field [itex]\varphi(x)[/itex] transforms as

[tex]\varphi(x) \rightarrow \varphi(x(y))[/tex]

where y are the new coordinates. By contrast, a vector field [itex]A^\mu(x)[/itex] would transform as

[tex]A^\mu(x) \rightarrow \frac{\partial y^\mu}{\partial x^\nu} A^\nu(x(y))[/tex]

where you see there is an additional change of basis.

Let's forget about GR for a second and just talk about SR. We have Minkowski coordinates (t,x,y,z). Under a boost, t changes, so it's not a scalar in the sense I'm talking about, right?
I think this will only confuse the issue. Minkowski space is special, because it can be identified with its own tangent space at the origin. As a result, points in Minkowski space can be identified with vectors, and under linear transformations, the coordinates [itex]x^\mu[/itex] behave somewhat like a 4-vector.

But note that linear coordinate transformations are subsumed under the first equation above. And if you do a non-linear coordinate change (such as changing from Cartesian to spherical coordinates), you will find that only the transformation law for a scalar field makes sense.
bcrowell
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Jun6-11, 11:06 PM
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Quote Quote by Ben Niehoff View Post
I think this will only confuse the issue. Minkowski space is special, because it can be identified with its own tangent space at the origin. As a result, points in Minkowski space can be identified with vectors, and under linear transformations, the coordinates [itex]x^\mu[/itex] behave somewhat like a 4-vector.
OK, instead let's talk about some other example, say the four-force, which is defined as the derivative of a particle's four-momentum with respect to its proper time, [itex]F^\mu=dp^\mu/d\tau[/itex]. Are you saying that I can't do [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex]? I'm pretty sure I can. If I can't do that, then the four-force isn't actually a four-vector.

I think the whole issue here revolves around different interpretations of the OP's notation [itex]\tau[/itex]. As far as I can tell, we're all in agreement that if [itex]d/d\tau[/itex] just means some kind of differentiation with respect to an arbitrary real parameter (which is a function of God-knows-what), then the manipulation in #1 is invalid. On the other hand, if it represents differentiation with respect to proper time along a world-line, then I'm pretty sure that the manipulation is valid.
Matterwave
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Jun6-11, 11:35 PM
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Quote Quote by bcrowell View Post
OK, instead let's talk about some other example, say the four-force, which is defined as the derivative of a particle's four-momentum with respect to its proper time, [itex]F^\mu=dp^\mu/d\tau[/itex]. Are you saying that I can't do [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex]? I'm pretty sure I can. If I can't do that, then the four-force isn't actually a four-vector.

I think the whole issue here revolves around different interpretations of the OP's notation [itex]\tau[/itex]. As far as I can tell, we're all in agreement that if [itex]d/d\tau[/itex] just means some kind of differentiation with respect to an arbitrary real parameter (which is a function of God-knows-what), then the manipulation in #1 is invalid. On the other hand, if it represents differentiation with respect to proper time along a world-line, then I'm pretty sure that the manipulation is valid.
I have a related question about this. Would it then be true that:

[tex]F_\mu=\frac{dp_\mu}{d\tau}[/tex]

?

What I mean is, is the lowered index on the F so simple as just lowering the index on the p?
TrickyDicky
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Jun7-11, 05:16 AM
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Quote Quote by bcrowell View Post
I think the whole issue here revolves around different interpretations of the OP's notation [itex]\tau[/itex]. As far as I can tell, we're all in agreement that if [itex]d/d\tau[/itex] just means some kind of differentiation with respect to an arbitrary real parameter (which is a function of God-knows-what), then the manipulation in #1 is invalid. On the other hand, if it represents differentiation with respect to proper time along a world-line, then I'm pretty sure that the manipulation is valid.
Since the OP never specified he was referring to Minkowski space or to SR, (but he might as well be referring to it, it would be interesting to know) the correct thing to do is to suppose he is talking about the general case. In this sense I agree with Ben Niehoff that you are confusing the issue by introducing an interpretation valid only for flat spacetime.

It is remarkable how often this misuse of SR notions confuses differential geometry issues.
Sam Gralla
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Jun7-11, 08:36 AM
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Quote Quote by bcrowell
Let's forget about GR for a second and just talk about SR. We have Minkowski coordinates (t,x,y,z). Under a boost, t changes, so it's not a scalar in the sense I'm talking about, right?
No, t changes correctly under a boost, as does any coordinate under any coordinate transformation. Scalars are invariant in the sense that when you evaluate one at the new coordinate value the value agrees with what you had at the *old* coordinate value. This happens automatically for any coordinate or any function of the coordinations. Since any function on the manifold can be written as a function of the coordinates, your argument would mean that functions aren't scalars. (Of course, they are.)

This point is often confused because there are four coordinates, and hence they come with an index. But it's really just a collection of scalars. I'm not too dogmatic about the "geometric" versus "coordinate-based" definitions of tensor fields, but in this case I think the geometric approach makes things much clearer.

Quote Quote by bcrowell View Post
OK, instead let's talk about some other example, say the four-force, which is defined as the derivative of a particle's four-momentum with respect to its proper time, [itex]F^\mu=dp^\mu/d\tau[/itex]. Are you saying that I can't do [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex]? I'm pretty sure I can. If I can't do that, then the four-force isn't actually a four-vector.
You're free to define [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex] after defining [itex]F^\mu=dp^\mu/d\tau[/itex]. But it does not then follow that [itex]F_\mu=dp_\mu/d\tau[/itex], where [itex]p_\sigma=g_{\sigma\mu}p^\mu[/itex], if "d/dtau" means ordinary differentiation, sometimes written [itex]u^\mu \partial_\mu[/itex]. If "d/dtau" means the covariant worldline derivative, usually written as [itex]u^\mu \nabla_\mu[/itex] or D/dtau, then it does follow.

EDIT: In reading quickly I didn't notice you're thinking of F as the four-force and p as the four-momentum. In this case your expression should indeed be using [itex]F^\nu = u^\mu \nabla_\mu p^\nu = D p^\mu / d\tau[/itex], not the ordinary derivative. This is the standard definition of four-force, so that F^\mu = 0 is the geodesic equation.
bcrowell
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Jun7-11, 09:35 AM
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Quote Quote by Sam Gralla View Post
You're free to define [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex] after defining [itex]F^\mu=dp^\mu/d\tau[/itex]. But it does not then follow that [itex]F_\mu=dp_\mu/d\tau[/itex], where [itex]p_\sigma=g_{\sigma\mu}p^\mu[/itex], if "d/dtau" means ordinary differentiation, sometimes written [itex]u^\mu \partial_\mu[/itex]. If "d/dtau" means the covariant worldline derivative, usually written as [itex]u^\mu \nabla_\mu[/itex] or D/dtau, then it does follow.
Cool, thanks for the explanation!


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