# Confusion with lower/raising indices

1. Jun 6, 2011

### vaibhavtewari

will the following actions be appropriate

$$g^{\mu\sigma}\frac{dA_{\sigma}}{d\tau}=\frac{dA^{\mu}}{d\tau}$$

Please let me know, I will be glad if you can explain too.

thanks

2. Jun 6, 2011

### bcrowell

Staff Emeritus
Yes, this looks right to me (except that you forgot the backslash on the final mu, so it looks wrong in the browser).

Please tell us what aspect of this you don't understand.

3. Jun 6, 2011

### henry_m

We have that $A^\mu=g^{\mu\nu}A_\nu$, essentially by definition. But as soon as you try to take a derivative, you will get a term involving the derivative of the metric from the Leibniz rule.

4. Jun 6, 2011

### Matterwave

Unless your g is independent of tau then you can't really write it that way. You have to write the g inside the derivative.

5. Jun 6, 2011

### vaibhavtewari

Well I am getting two answers...which one should I trust ?

To me the derivative looks fine(just some intuition)...I believe we have similar things for partial derivative...instead derivative with respect to proper time. I want to clarify all these things once for all

6. Jun 6, 2011

### bcrowell

Staff Emeritus
I could be wrong, but I think this is incorrect. It seems to me that $dA_\sigma$ is a four-vector, and $d\tau$ is a Lorentz scalar, and therefore when you divide you get a patriotic and God-fearing four-vector. Since $dA_\sigma/d\tau$ is a four-vector, you can raise an index on it just as you would with any other four-vector.

7. Jun 6, 2011

### vaibhavtewari

Your explanation seems quite promising to me, along the same lines I wanted to ask, is this also true....

$$g^{\mu\sigma}\partial_{\rho} A_{\sigma}=\partial_{\rho}A^{\mu}$$

8. Jun 6, 2011

### WannabeNewton

Yes that is perfectly fine. Its easier to work with raising and lowering if you used commas and colons for partial and covariant derivatives respectively.

EDIT: Actually I'm not 100% positive thats right. I was mixing up raising and lowering in linearized theory as opposed to full GR. I'm sure you can only do that if you are working with covariant derivatives because you can always pull the metric out of a covariant derivative (covariant derivative of the metric tensor is zero so the leibnez rule permits that). When working with partial derivatives you can only do that with the minkowski tensor (which is what one does in flat background linearized theory). Correct me if I am wrong, sorry.

Last edited: Jun 6, 2011
9. Jun 6, 2011

### Sam Gralla

Trust the one who gives an explanation that convinces you! (That is, trust yourself ;p.) Remember that raising an index is just a definition. Try plugging in that definition into your proposed equation and see if the equation is true.

10. Jun 6, 2011

### bcrowell

Staff Emeritus
One thing to point out is that since we're differentiating with respect to a scalar rather than a coordinate, the distinction between a partial derivative and a covariant derivative doesn't apply.

A similar example is raising an index on a velocity four-vector. It's a four-vector, so you raise its index in the usual way. But the velocity four-vector can also be written as the derivative of position with respect to proper time.

11. Jun 6, 2011

### WannabeNewton

But in his second example he is talking about partial derivatives with respect to the coordinates.

12. Jun 6, 2011

### Ben Niehoff

This is wrong. It depends on the value of $dg^{\mu\nu}/d\tau$, by applying the Leibniz rule, as mentioned earlier. You raise indices by multiplying $g^{\mu\nu}$ outside any partial derivatives. And if you are in curved space, you should be using covariant derivatives anyway, in which case it wouldn't matter.

Furthermore, if $A_\sigma$ is a 4-vector, then $dA_\sigma$ is a vector-valued 1-form. Similarly, if $\tau$ is a Lorentz scalar, then $d\tau$ is an ordinary 1-form. And the derivative operator

$$\frac{d}{d\tau} = \frac{dx^\mu}{d\tau} \frac{\partial}{\partial x^\mu}$$

is a vector field.

13. Jun 6, 2011

### Ben Niehoff

No, the distinction always applies.

Given some scalar $\varphi$, the partial derivative is the unique derivative for which we can write the chain rule,

$$\frac{d}{d\varphi} = \frac{dx^\mu}{d\varphi} \frac{\partial}{\partial x^\mu}$$

and hence we always have

$$\frac{d}{d\varphi} (g^{\mu\nu} A_\nu) = g^{\mu\nu} \frac{dA_\nu}{d\varphi} + \frac{dg^{\mu\nu}}{d\varphi} A_\nu$$

You may have been thinking of

$$\frac{D}{d\varphi} \equiv \frac{dx^\mu}{d\varphi} \nabla_\mu$$

which is a special shorthand.

14. Jun 6, 2011

### Sam Gralla

Coordinates are scalars. But you can't differentiate with respect to one of them until you have all of them (so you know what to hold fixed while differentiating). So the derivative with respect to a scalar is not defined. In this case tau is almost certainly defined as a parameter along a worldline, although the poster didn't specify. (This may be what you meant by "scalar", but it's not standard.) In that case you can differentiate the components in the orindary sense. The four-velocity is then defined by $$u^\mu = d x^\mu / d\tau$$, and you're of course free to lower, $$u_\mu = g_{\mu \nu} u^\nu$$. But does it does not then follow that $$u_\mu = d x_\mu / d\tau$$. (Here I've lowered the index on the coordinate in the usual way, even though coordinates shouldn't really be thought of as four-vectors.)

However, my guess is that the poster didn't really mean ordinary derivative but actually meant the $$u^\mu \nabla_\mu$$ worldline derivative that comes up in applications, sometimes written as $$D/d\tau$$. In that case his formula is correct because the metric commutes with the covariant derivative.

15. Jun 6, 2011

### bcrowell

Staff Emeritus
Hmm...by "scalar" I simply meant something that is invariant under boosts. By that definition, a coordinate isn't a scalar, right?

16. Jun 6, 2011

### Ben Niehoff

Coordinates are functions on a manifold, and hence are scalar fields.

17. Jun 6, 2011

### Matterwave

Well, it is certainly true that:

$$A^\mu=g^{\mu\nu}A_\nu$$

And so, taking derivatives on both sides:

$$\frac{dA^\mu}{d\tau}=\frac{d}{d\tau}(g^{\mu\nu}A_{\nu})$$

So, unless the metric is independent of tau, I don't see how you can simply pull it out or else you are saying that the derivative is not Leibniz.

18. Jun 6, 2011

### bcrowell

Staff Emeritus
Do you mean "scalar" the same way I'm using it, as something that's invariant under a boost?

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 was assuming from the OP's notation that $\tau$ referred to the proper time measured along a certain world-line. Therefore $d\tau$ is a scalar, in the sense of being invariant under boosts.

I agree that if $\tau$ is just some arbitrary number defined as a function on a certain world-line, then the index manipulation in #1 is invalid. The way I would put it is that then, $dA_\sigma/d\tau$ is not a four-vector; when you divide a four-vector by something that isn't a Lorentz scalar, you don't get a four-vector.

Last edited: Jun 6, 2011
19. Jun 6, 2011

### bcrowell

Staff Emeritus
I'm not familiar with the notation $D/d\tau$, but your description of it sounds the same as what I thought the OP was describing.

20. Jun 6, 2011

### Ben Niehoff

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 $\varphi(x)$ transforms as

$$\varphi(x) \rightarrow \varphi(x(y))$$

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

$$A^\mu(x) \rightarrow \frac{\partial y^\mu}{\partial x^\nu} A^\nu(x(y))$$

where you see there is an additional change of basis.

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 $x^\mu$ 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.