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How can the gradient of a scalar field be covarient?

  1. Feb 1, 2013 #1
    According to Carroll, [itex]\nabla \phi[/itex] is covariant under rotations. This really confuses me. For example, how could equations like [itex]\vec{F}=-\nabla V[/itex] 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.
     
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  3. Feb 1, 2013 #2

    WannabeNewton

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    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. [itex]\omega = df[/itex] is actually a one - form because if we take [itex](dx^{i})[/itex] to be the dual basis for [itex]T^{*}_{p}(\mathbb{R}^{n})[/itex] then we can write [itex]df(p) = \partial _{i}f(p)dx^{i}[/itex] so that for all u,v in [itex]T_{p}(\mathbb{R}^{n})[/itex] holds
    [tex]\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*}[/tex] 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.
     
  4. Feb 1, 2013 #3

    atyy

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    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?
     
  5. Feb 2, 2013 #4

    WannabeNewton

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    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.
     
  6. Feb 2, 2013 #5

    atyy

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    Yes, that's what I was thinking.
     
  7. Feb 2, 2013 #6

    stevendaryl

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    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 [itex]x^i[/itex] to another set [itex]x'^j[/itex], one of them works like:

    [itex](V^j)' = \dfrac{\partial x'^j}{\partial x^i} V^i[/itex]

    and the other works like:

    [itex](V_j)' = \dfrac{\partial x^i}{\partial x'_j} V_i[/itex]

    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
  8. Feb 2, 2013 #7

    stevendaryl

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    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).
     
  9. Feb 2, 2013 #8

    Nugatory

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    "Co is Lo"
     
  10. Feb 2, 2013 #9

    WannabeNewton

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    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.
     
  11. Feb 2, 2013 #10

    vanhees71

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    It's easy to remember, if you just keep in mind that the coordinates [itex]q^{\mu}[/itex] have upper indices. Then the one-forms [itex]\mathrm{d} q^{\mu}[/itex] are the dual basis to the holonomous basis [itex]\partial_{\mu}[/itex] 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 [itex]q'^{\mu}[/itex] with a local diffeomorphism gives
    [tex]\mathrm{d} q'^{\mu}=\frac{\partial q'^{\mu}}{\partial q^{\nu}} \mathrm{d} q^{\nu}.[/tex]
    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 [itex]\Phi[/itex]. By definition, it's transformation property is
    [tex]\Phi'(q')=\Phi(q).[/tex]
    Then
    [tex]A_{\mu}=\frac{\partial}{\partial q^{\mu}} \Phi=\partial_{\mu} \Phi[/tex]
    obviously give covariant vector-field components. They transform covariantly (i.e., contragrediently wrt. to the co-basis vectors):
    [tex]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).[/tex]
    In other words the holonomous basis vectors transform like
    [tex]\partial_{\mu}'=\frac{\partial q^{\nu}}{\partial q'^{\mu}} \partial_{\nu},[/tex]
    i.e., contragrediently to the co-basis vectors, i.e., you have covariant objects with a lower index.
     
  12. Feb 2, 2013 #11

    Bill_K

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    Newton had a clear advantage - he'd never heard of one-forms and diffeomorphisms.
     
  13. Feb 2, 2013 #12

    stevendaryl

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    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.
     
  14. Feb 2, 2013 #13

    WannabeNewton

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    Or did he? Dun dun duuuuuun...:tongue2:
     
  15. Feb 2, 2013 #14

    dx

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    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.
     
  16. Feb 2, 2013 #15
    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 [itex]\vec {F} [/itex] in one Cartesian coordinate system (CCS) is equal to the representation of [itex] -\nabla V[/itex] 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
  17. Feb 2, 2013 #16

    WannabeNewton

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    If [itex]X = \triangledown f[/itex] 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 [itex]X[/itex]. Try this simple counter - example: let [itex]f:\mathbb{R}^{2} \rightarrow \mathbb{R}, (x,y) \rightarrow x^{2}[/itex] and [itex]X = \triangledown f = 2x\partial _{x}[/itex] and change to polar coordinates. This is why we instead use [itex]df(p) = \partial _{i}f(p)dx^{i}[/itex] because this is coordinate independent.
     
  18. Feb 2, 2013 #17
    That's kinda my point. There doesn't seem to be any reason why [itex]\vec {F}=-\nabla V[/itex] should hold in all coordinate systems, but it does...
     
  19. Feb 2, 2013 #18

    WannabeNewton

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    Who said it does? Try the example I gave you; you will see it doesn't.
     
  20. Feb 2, 2013 #19

    atyy

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    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
  21. Feb 6, 2013 #20
    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.
     
  22. Feb 6, 2013 #21

    WannabeNewton

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    What I posted is an exercise straight out of Lee's Smooth Manifolds book. Try it, it shouldn't take too long. I can refer you to the pages the discussion takes place in the text if you can get access to it so that this thread doesn't get derailed.
     
  23. Feb 7, 2013 #22

    dx

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    The gradient of a function is simply the local linear behavior of the function and has nothing to do with any particular coordinate system, and therefore must be an object which is coordinate independent. This is not the case when one thinks of it as ∇f. One can illustrate this using a simple scaling of coordinates in one dimension. Let X and Y be two coordinate functions on the line, with X = 2Y, and let f be a function on the line with f = X = 2Y

    Calculating the vector gradient in the X system, we have

    ∇f = (df/dX)eX = eX = (1/2)eY

    If we calculate it in the Y system we get

    ∇f = (df/dY)eY = = 2eY = 4eX

    These are different vectors. So which one is the gradient? The representation is not even covariant with respect to scaling of the coordinate.

    On the other hand, let's calculate the 1-form df in X:

    df = (df/dX)dX = dX = 2dY​

    Calculating in Y:

    df = (df/dY)dY = 2dY = dX​

    They are the same, as they should be. The gradient is a 1-form, not a vector.


    It is not covariant under linear transformations, as the example above shows. When one says something is covariant under some group G, one means that it is covariant under all transformations belonging to the group. A rotation is a linear transformation, but a linear transformation is usually not a rotation.
     
    Last edited: Feb 7, 2013
  24. Feb 7, 2013 #23
    See below.

    True but in your example you used a non-uniform scaling (you scaled differently one coordinate wrt the other) wich is a non-linear transformation, only uniform scaling is linear. See http://en.wikipedia.org/wiki/Scaling_(geometry)
    Can you give me a true example of a linear transformation that changes the components of a covector wrt a vector in Euclidean space?
     
  25. Feb 7, 2013 #24
    Huh? So you are saying that T(x,y)=(2x,y) is not a linear transformation?? You seem to have a very weird definition of linear transformations...
     
  26. Feb 7, 2013 #25
    That is a linear transformation of course, but I believe that is not what dx is presenting, I think he is giving two functions with two arguments, and the transformation is linear in each argument separately but not for both. At least if it is truely a non-uniform scaling.
    EDIT:
    Admittedly I might also have misunderstood the Wikipedia article about scaling.
     
    Last edited: Feb 7, 2013
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