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Covariant four vector

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Show that [tex]\frac{\partial\phi}{\partial x^{\mu}}[/tex] is a covariant four vector .


    2. Relevant equations
    All covariant four vector transformations .


    3. The attempt at a solution
    I really didn't understand what question implies . How can this vector be showed as being a covariant vector ?
    If you could give me a hint at least about what is really asked .

    Thanks .
     
    Last edited: Dec 12, 2007
  2. jcsd
  3. Dec 12, 2007 #2

    nrqed

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    You must write the corresponding expression expressed in a different frame, viz

    [tex]\frac{\partial\phi'(x')}{\partial x'^{\mu}}[/tex]
    and then relate all the quantities to the unprimed coordinate system.
     
  4. Dec 12, 2007 #3
    How can i write it in a different frame since I don't know how to transform the derivate from one frame to another ?
     
  5. Dec 12, 2007 #4

    George Jones

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    I would call the

    [tex]\frac{\partial \phi}{\partial x^\mu}[/tex]

    components of a covector field with respect to a particular basis, but physicists often think terms of components (or abstract index notation).

    In any case, switch to a primed coordinate system and use the multivariable chain rule to express

    [tex]\frac{\partial \phi}{\partial x'^\nu}[/tex]

    in terms of

    [tex]\frac{\partial \phi}{\partial x^\mu}.[/tex]

    [edit]Was typing while nrqed's post came in. His notation is better than mine. Why?[/edit]
     
    Last edited: Dec 12, 2007
  6. Dec 12, 2007 #5
    I don't know how to switch from one inertial frame to another on differential form .
    I have Lorentz transformation of (x0,x1,x2,x3) linear terms but I really have no clue how to obtain derivate form of that .
     
  7. Dec 12, 2007 #6

    George Jones

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    Are yuo working only with inertial coordinate systems in special relativity, and do you have something like (I might have the staggering wrong)

    [tex]x^\nu = \Lambda^\nu{}_\mu x'^\mu?[/tex]

    If so, differentiate this with respect to [itex]x'^\alpha[/itex] to get something that you can use in conjunction with the chain rule.
     
  8. Dec 12, 2007 #7
    Should superscript alpha be nu ? Or is alpha for another indice ?
     
  9. Dec 12, 2007 #8

    George Jones

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    No to the first; yes to the second.

    If you differentiate with respect to a nu index, nu will appear twice in each side of the expression, and there will be possible problems with the summation convention. If you differentiate with respect to mu, the mu will appear three times.

    What does

    [tex]\frac{\partial x'^\mu}{\partial x'^\alpha}[/tex]

    equal?
     
  10. Dec 12, 2007 #9
    What about the lamdba term ? It contains some x0 to x3 term that should be derivated .

    Actually , this is the key problem that confuses my mind from the beginning .
     
  11. Dec 12, 2007 #10

    George Jones

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    Again, I'm not sure of the context in which we're working. If talking about Lorentz transformations between inertial coordinate systems, Lambda is a matrix of constants (i.e., the constants that multiply the x coordinates or x' coordinates). It has no x dependence
     
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