# How can the gradient of a scalar field be covarient?

1. Feb 1, 2013

### dEdt

According to Carroll, $\nabla \phi$ is covariant under rotations. This really confuses me. For example, how could equations like $\vec{F}=-\nabla V$ be rotationally covariant if force is a contravariant vector?

I know this is strictly speaking more of a mathy question, but I still figured this was the best place to ask this question.

2. Feb 1, 2013

### WannabeNewton

Hi! There was a very, very recent thread on this: https://www.physicsforums.com/showthread.php?t=666861. I have no idea what you mean by covariant under rotations so if you could explain that I would be much obliged thanks. $\omega = df$ is actually a one - form because if we take $(dx^{i})$ to be the dual basis for $T^{*}_{p}(\mathbb{R}^{n})$ then we can write $df(p) = \partial _{i}f(p)dx^{i}$ so that for all u,v in $T_{p}(\mathbb{R}^{n})$ holds
$$\begin{eqnarray*} (df(p))(av + bu) & = & \partial _{i}f(p)dx^{i}(av + bu)\\ & = & \partial _{i}f(p)dx^{i}(av) + \partial _{i}f(p)dx^{i}(bu)\\ & = & a\partial _{i}f(p)dx^{i}(v) + b\partial _{i}f(p)dx^{i}(u)\\ & = & av^{j}\partial _{i}f(p)dx^{i}(\partial _{j}) + bu^{j}\partial _{i}f(p)dx^{i}(\partial _{j})\\ & = & av^{i}\partial _{i}f(p) + bu^{i}\partial _{i}f(p)\\ & = & a(df(p))(v) + b(df(p))(u) \end{eqnarray*}$$ so the differential of f at p is actually a linear functional on the tangent space at p so it is actually a one - form.

3. Feb 1, 2013

### atyy

Maybe he meant the old fashioned 3D vector calculus gradient which requires a metric to be defined, and he meant "covariant" analogous to Newton's equations written in component form in an inertial frame are "covariant" under Galilean transforms?

Maybe the OP can give a link so we can see Carroll's text?

4. Feb 2, 2013

### WannabeNewton

Yes that is probably what is being said here; covariance in the sense of general covariance if I understood you correctly, right? I searched my copy of the book and couldn't find the phrase itself as stated by the OP anywhere so indeed if the OP could give what page / section he/she is talking about that would be splendid. G'day to you atyy =D.

5. Feb 2, 2013

### atyy

Yes, that's what I was thinking.

6. Feb 2, 2013

### stevendaryl

Staff Emeritus
I've read some authors who talk about thingies with indices as being "covariant" or "contravariant" under coordinate changes. I can never remember which is which, but when changing from one set of coordinates $x^i$ to another set $x'^j$, one of them works like:

$(V^j)' = \dfrac{\partial x'^j}{\partial x^i} V^i$

and the other works like:

$(V_j)' = \dfrac{\partial x^i}{\partial x'_j} V_i$

Components of velocity vectors and accelerations work like the first, while components of gradients of scalars works like the second.

Last edited: Feb 2, 2013
7. Feb 2, 2013

### stevendaryl

Staff Emeritus
Of course, the convention for using raised or lowered indices automatically insures that you're going to use the correct transformation, because the wrong kind is syntactically invalid (a summation over indices requires one upper index and one lower index).

8. Feb 2, 2013

"Co is Lo"

9. Feb 2, 2013

### WannabeNewton

Oh if that is what he means, and it might be if the OP could clarify, then the statement "covariant under rotations" seems very specific considering a one - form must transform in that way under any diffeomorphism.

10. Feb 2, 2013

### vanhees71

It's easy to remember, if you just keep in mind that the coordinates $q^{\mu}$ have upper indices. Then the one-forms $\mathrm{d} q^{\mu}$ are the dual basis to the holonomous basis $\partial_{\mu}$ induced by the coordinates.

You can read these one-forms in a intuitive way as infinitesimal changes of the coordinates. Then switching to new coordinates $q'^{\mu}$ with a local diffeomorphism gives
$$\mathrm{d} q'^{\mu}=\frac{\partial q'^{\mu}}{\partial q^{\nu}} \mathrm{d} q^{\nu}.$$
This defines the contravariant transformation law from one dual basis to the other, i.e., you have objects with an upper index.

Now consider a scalar field $\Phi$. By definition, it's transformation property is
$$\Phi'(q')=\Phi(q).$$
Then
$$A_{\mu}=\frac{\partial}{\partial q^{\mu}} \Phi=\partial_{\mu} \Phi$$
obviously give covariant vector-field components. They transform covariantly (i.e., contragrediently wrt. to the co-basis vectors):
$$A'_{\mu}=\partial_{\mu}' \Phi'=\frac{\partial}{\partial q'^{\mu}} \Phi'(q')=\frac{\partial q^{\nu}}{\partial q'^{\mu}} \frac{\partial}{\partial q^{\nu}} \Phi(q).$$
In other words the holonomous basis vectors transform like
$$\partial_{\mu}'=\frac{\partial q^{\nu}}{\partial q'^{\mu}} \partial_{\nu},$$
i.e., contragrediently to the co-basis vectors, i.e., you have covariant objects with a lower index.

11. Feb 2, 2013

### Bill_K

Newton had a clear advantage - he'd never heard of one-forms and diffeomorphisms.

12. Feb 2, 2013

### stevendaryl

Staff Emeritus
Okay, that's a good way to remember it.

Someone explained the co- and contra- to me in terms of category theory, but I'm not sure it clarified anything to me.

13. Feb 2, 2013

### WannabeNewton

Or did he? Dun dun duuuuuun...:tongue2:

14. Feb 2, 2013

### dx

As long as one thinks about the gradient as a vector ∇φ instead of the 1-form dφ, one is restricted to Euclidean space.

One can say that the representation of the gradient as a vector ∇φ is covariant under rotations and translations, but not under linear transformations or diffeomorphisms.

Note that the use of the word covariant here is different from its use in the word 'covariant vector.' Here we are talking about the covariance of a representation.

15. Feb 2, 2013

### dEdt

To clarify some ambiguities:
By "rotationally covariant" I meant something analogous to generally covariant. Given the context of the question this was probably a stupid term to use, and I should have used something like rotationally invariant. My point, though, is that if the representation of $\vec {F}$ in one Cartesian coordinate system (CCS) is equal to the representation of $-\nabla V$ in the same CCS, then the representation of the two things cannot be equal in another CCS because they transform differently under a change of coordinate system.

At least, this was what I thought until about two minutes ago when I actually computed how the components of the gradient of a scalar change under a change of coordinate system. I found that they change in exactly the same way as the components of force do. So now I'm wondering what the difference between covariance and contravariance really is. Is seems like they're the same thing.

I think that dx may have addressed some of these questions, but I didn't really understand what he wrote.

Last edited: Feb 2, 2013
16. Feb 2, 2013

### WannabeNewton

If $X = \triangledown f$ in some coordinate system and you define a change of coordinates then the relation doesn't have to hold once you transform the components of $X$. Try this simple counter - example: let $f:\mathbb{R}^{2} \rightarrow \mathbb{R}, (x,y) \rightarrow x^{2}$ and $X = \triangledown f = 2x\partial _{x}$ and change to polar coordinates. This is why we instead use $df(p) = \partial _{i}f(p)dx^{i}$ because this is coordinate independent.

17. Feb 2, 2013

### dEdt

That's kinda my point. There doesn't seem to be any reason why $\vec {F}=-\nabla V$ should hold in all coordinate systems, but it does...

18. Feb 2, 2013

### WannabeNewton

Who said it does? Try the example I gave you; you will see it doesn't.

19. Feb 2, 2013

### atyy

As dx said, the traditional vector calculus gradient is a vector - not a covector.

The gradient that is a covector is a different object from the traditional vector calculus gradient. For clarity dx wrote gradient with df for the former and ∇f for the latter. One can get the latter from the former in a space which has a Euclidean metric.

Last edited: Feb 2, 2013
20. Feb 6, 2013

### TrickyDicky

The reason one can think of the gradient as a vector in Euclidean space, and in the case one uses orthogonal coordinates is that in this particular case due to the Euclidean metric form, vector and covectors have the same components and are therefore indistinguishable.
Well rotations are linear transformations so it seems to be covariant under linear transformations. And yes the representation is not diffeomorphism invariant due to what I mentioned about the requirement to use orthogonal coordinates.

Polar coordinates are orthogonal so it should be covariant under such transformation.

As commented above in euclidean space and as long as orthogonal coordinates are used the distinction is superfluous.