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Four gradient operator, covariance/contravariance

  1. Dec 11, 2008 #1
    I'm doing a selfstudy on relativistic electrodynamics and stumbled over a problem (which i find rather important) i can't solve. It's concerning problem 12.55 in Griffiths introduction to electrodynamics.
    One needs show that the four gradient:
    \frac{\partial}{\partial x ^\mu} functions as a covariant fourvector, using a scalar function (call it \phi) and the chain rule.

    Any help would be much appreciated!
  2. jcsd
  3. Dec 11, 2008 #2
    What have you tried so far? What are the differences between covariant and contravariant vectors?

  4. Dec 12, 2008 #3
    The differential of the scalar function is a four-vector, because the differentials are over time, x, y and z, but in contravariant notation. Try understanding how the covariant position compares to the contravariant position (it's pretty simple). Then use the chain-rule to rewrite the contravariant four-vector into a four-vector with a covariant form, to show it's contravariant.
    Last edited: Dec 12, 2008
  5. Dec 12, 2008 #4

    Ben Niehoff

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    The connection is fairly obvious if you write the transformation matrix as

    [tex]\frac{\partial \bar x^i}{\partial x^j}[/tex]

    and its inverse as

    [tex]\frac{\partial x^i}{\partial \bar x^j}[/tex]

    These are just Jacobian matrices.
  6. Dec 12, 2008 #5
    Thanks for your replies. Now, here is what i've tried:

    I consider the Lorentz transformation:

    \bar{x}^{\mu} = \Lambda ^{\mu} _{\nu} x^{\nu}.
    Differentiating i obtain:
    \Lambda ^{\mu} _{\nu} = \frac{\partial \bar{x} ^{\mu}}{\partial x^{\nu}}.
    Acting upon the transformation with the inverse transformation i obtain:
    {\Lambda ^{\mu} _{\nu}}^{-1} \bar{x} ^{\mu} = x ^{\nu}.
    Differentiating this i obtain:
    {\Lambda ^{\mu} _{\nu} }^{-1} = \frac{\partial x^{\nu}}{\partial \bar{x}^{\mu}}.
    The four-gradient of the scalar function \phi is:
    \frac{\partial \phi}{\partial x^{\mu}}.

    In a boosted frame this must be:
    \frac {\partial \bar{\phi}}{ \partial \bar{x} ^{\mu}} = \frac{\partial \phi}{\partial \bar{x} ^{\mu}}
    , since \phi is a scalar function.

    Using the chain rule i obtain for the four-gradient in the boosted frame:
    \frac{\partial \phi}{\partial \bar{x} ^{\mu}} = \frac{\partial \phi}{\partial x^{\nu}} \frac{\partial x ^{\nu}}{\partial \bar{x} ^{\mu}}.

    And inserting the inverse transformation i get:
    \frac{\partial \bar{\phi}}{\partial \bar{x}^{\mu}} = {\Lambda ^{\mu} _{\nu} }^{-1} \frac{\partial \phi}{\partial x ^{\nu}}.

    This tells how the four-gradient of the scalar function transform under lorentz transformation. But im not sure how this tells us that the four-gradient transforms like a covariant four vector. I guess i'm not confident with the whole covariant/contravariant concept.
    As far as i know a contravariant four vector transforms the same way as:
    x^ \mu = (ct, x, y, z)
    And a covariant four vector transforms the same way as:
    x_ \mu = (ct,-x,-y,-z)
    (For the metric i use..).
  7. Dec 12, 2008 #6
    You're overdoing it. I suggest you look at Ben's reply, which gives you the chain term you get from applying the chain rule as a result of changing the variables (this terms will turn out to be the metric for co- to contravariant and back (since it is its own inverse)). Also, the question only asks whether taking the contravariant gradient of a scalar function results in a contravariant 4-vector. It doesn't ask you anything about transforming from one inertia system to another.
  8. Dec 12, 2008 #7

    Ben Niehoff

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    It seems your course is covering this only in the context of Lorentz transformations, rather than the wider context of general coordinate transformations, which is unfortunate. The wider context is applicable in many areas, including ordinary Euclidean 3-space (when changing between spherical & Cartesian systems, for example).

    A general coordinate transformation of a vector (or vector field) can always be represented by a Jacobian matrix. Lorentz transformations are a special case.

    "Covariance" and "contravariance" are meant to hold in the case of general coordinate transformations; not just Lorentz transformations.
  9. Dec 13, 2008 #8
    "Also, the question only asks whether taking the contravariant gradient of a scalar function results in a contravariant 4-vector. It doesn't ask you anything about transforming from one inertia system to another."

    Hmm.. I think i'm getting confused. The contravariant gradient is:
    \frac{\partial \phi}{\partial x ^{\mu}} = ( \frac{ \partial \phi}{ c\partial t } ; \frac{ \partial \phi}{ \partial x }; \frac{ \partial \phi}{ \partial y }; \frac{ \partial \phi}{ \partial z }).
    This does not look like the covariant form: x_{\mu} = (ct, -x, -y, -z).
    So it must be a matter of how it transforms.
    As far as i know what matters is that the inner product of a contravariant and a covariant four vector is invariant under Lorentz transformations. It so happens that the inner product of \frac{\partial \phi}{\partial x^{\mu}} and a contravariant fourvector is a Lorentz invariant, which means that frac{\partial \phi}{\partial x^{\mu}} must transform like a covariant, although its explicit form looks rather contravariant.
    Am i completely wrong?
  10. Dec 13, 2008 #9
    Oh i think i found my error. What Ben wrote must be transformation between covariants and contravariants? i thought it was Lorentz transformations. Well, if i'm right here is what i get. In my notation: x^{\mu} = contravariant x_{\mu} = covariant g_{\mu \nu} = metric.

    Connection between covariant and contravariant:
    x_{\nu} = g_{\nu \mu} x ^{\mu}

    g_{\nu \mu} = \frac{\partial x _{\nu}}{\partial x^{\mu}}

    For the four gradient, using the chain rule i get:
    \frac{\partial \phi}{\partial x^{\mu}} = \frac{\partial \phi}{\partial x_{\nu}} \frac{\partial x_{\nu}}{\partial x^{\mu}}


    \frac{\partial \phi}{\partial x^{\mu}} = g_{\nu \mu} \frac{\partial \phi}{\partial x_{\nu}}.

    From this i see that the connection between \frac{\partial \phi}{\partial x^{\mu}} and \frac{\partial \phi}{\partial x_{\nu}}, is the same as between x^{\mu} and x_{\mu}.
    Therefore \frac{\partial \phi}{\partial x_{\nu}} must be a covariant fourvector, although it appers contravariant when written out explicitly. And \frac{\partial \phi}{\partial x_{\nu}} must be a contravariant fourvector.

    Is this the right way to do it?
  11. Dec 13, 2008 #10
    Yep, that's the correct way.
  12. Dec 13, 2008 #11
    Ok. Thanks alot for your help and inspiration.
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