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A General wave equation

  1. Jun 14, 2016 #1
    hi, I know the wave equation in terms of minkowski metric, and we use ordinary derivative in that equation. I also know this wave equation in terms of general metric form using covariant derivative, but I do not know the derivation of it. Could you spell out the steps of derivation?? How do we derive the general metric form of wave equation???
     
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  3. Jun 14, 2016 #2

    Orodruin

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    Your question is unclear. What do you want to derive the wave equation for? The wave equation itself is just the divergence of the covariant derivative. It is not really much to derive.
     
  4. Jun 15, 2016 #3
    Ok, I try to make it more clear. In this video (, at 35.13)

    You can see a wave equation and it's minkowski metric interpretation at 35.13. Also at 52.00, we have the general metric interpretation of wave equation(leonard suskind said so) with covariant derivatives. So, I wonder: how does this metric transformation yield a covariant derivatives??????? ( by the way suskind tells something but I can not fully understand)......Could you spell this situation out??????
     
  5. Jun 15, 2016 #4
    So many views but I can not received response.... I hope my question is explicit....
     
  6. Jun 15, 2016 #5

    stevendaryl

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    I watched most of the video, but I don't understand what part you are having trouble with. He's trying to figure out two things about a wave field:

    1. How does gravity (metric + Christoffel symbols) affect a wave.
    2. How does a wave affect gravity?
    The second part is about how to write the stress-energy tensor [itex]T_{\mu \nu}[/itex] for a scalar field [itex]\phi[/itex]. It doesn't have much to do with the covariant derivative.

    The first part, though, is about the covariant derivative. The wave equation in terms of the covariant derivative is just:

    [itex]g^{\mu \nu} \nabla_\mu \nabla_\nu \phi = 0[/itex]

    where [itex]\nabla_\mu[/itex] is the covariant derivative.

    So what is your question about this?
     
  7. Jun 15, 2016 #6
    Initially I understand the wave equation with minkowski metric, but my struggle is to derive wave equation in terms of metric tensor and covariant derivative.so How can I derive it?? How can I reach it??
     
  8. Jun 16, 2016 #7

    Orodruin

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    So we come back to the original problem. That it is unclear what you mean. The wave equation itself is not something you derive, it is just a name you give to a certain equation. The wave equation in curved space-time is its natural generalisation.

    Whether or not a physical quantity obeys the wave equation is a diffrent question and the derivation might in general depend on the quantity in question.
     
  9. Jun 16, 2016 #8

    stevendaryl

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    I'm not sure what you mean by deriving it. Deriving it starting from what?

    As Susskind says in the video, the wave equation looks the same, whether you have curved spacetime or curvilinear coordinates in flat spacetime. So if you start with the flat spacetime, Cartesian coordinates form for the wave equation, you can derive the form in arbitrary curvilinear coordinates by just doing some calculus:

    [itex]\eta^{a b} \partial_a \partial_b \phi = 0[/itex] in Cartesian coordinates. Now let's switch to curvilinear coordinates. I'll use Greek letters for the curvilinear indices, and Roman letters for the Caresian indices.

    The relationship between the two kinds of derivatives is: [itex]\partial_b \phi = L^\mu_b \partial_\mu \phi[/itex] where [itex]L^\mu_b = \frac{\partial x^\mu}{\partial x^b}[/itex]. When you take a second derivative, you end up derivatives of [itex]L^\mu_b[/itex] in addition to derivatives of [itex]\phi[/itex]. The derivatives of [itex]L^\mu_b[/itex] become part of the Christoffel symbols in the curvilinear coordinates.
     
  10. Jun 16, 2016 #9
    Stevendarly thank you I got your last answer, you understand my question truly, by the way I can not reach covariant form of wave equation using the notion that you shared, because I can not compile all the christoffel symbols stuff so, Would you mind explaining your last response in more detail using little bit more mathematical demonstration??????
     
  11. Jun 16, 2016 #10

    stevendaryl

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    Okay. In Cartesian coordinates, there is no difference between partial derivatives and covariant derivatives. So if [itex]V^b[/itex] is a vector field, then
    [itex]\nabla_a V^b = \partial_a V^b[/itex]

    Now, switch to a curvilinear coordinate system. Let [itex]L^\mu_a = \partial_a x^\mu[/itex] and letting [itex]\tilde{L}^b_\nu = \partial_\nu x^b[/itex]. [itex]L[/itex] and [itex]\tilde{L}[/itex] are inverses, in the sense that:

    [itex]L^\mu_a \tilde{L}^b_\mu = \delta^b_a[/itex]
    [itex]L^\mu_a \tilde{L}^a_\nu = \delta^\mu_\nu[/itex]

    Then since [itex]\nabla_a V^b[/itex] is a tensor, we can rewrite it in terms of the curvilinear coordinates:

    [itex]\nabla_a V^b = L^\mu_a \tilde{L}^b_\nu \nabla_\mu V^\nu[/itex]

    We can also write:

    [itex]\partial_a V^b = L^\mu_a \partial_\mu (\tilde{L}^b_\nu V^\nu)[/itex]
    [itex] = L^\mu_a (\partial_\mu \tilde{L}^b_\nu) V^\nu + L^\mu_a \tilde{L}^b_\nu \partial_\mu V^\nu[/itex]

    Here's where I think we have to do some unmotivated manipulation. We want to pull out a factor of [itex]\tilde{L}^b_\nu[/itex]. So we can rewrite:

    [itex](\partial_\mu \tilde{L}^b_\nu) V^\nu [/itex]
    [itex]= (\partial_\mu \tilde{L}^b_\lambda) V^\lambda [/itex]
    [itex]= \delta^b_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda[/itex]
    [itex]= \tilde{L}^b_\nu L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda[/itex]

    So we can write:
    [itex]\partial_a V^b = L^\mu_a \tilde{L}^b_\nu [ L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu][/itex]

    Putting the facts together, we have:
    [itex]L^\mu_a \tilde{L}^b_\nu \nabla_\mu V^\nu = L^\mu_a \tilde{L}^b_\nu [ L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu][/itex]

    which implies that
    [itex]\nabla_\mu V^\nu = L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu[/itex]

    Comparing that with the usual expression for [itex]\nabla_\mu V^\nu[/itex] tells us that:

    [itex]\Gamma^\nu_{\mu \lambda} = L^\nu_c (\partial_\mu \tilde{L}^c_\lambda)[/itex]
     
  12. Jun 17, 2016 #11
    stevendarly thank you for your last return, by the way I tried to merge the information you had given in "post 8" with the information you had given in "post 10", but still I can not entirely switch the wave equation in cartesian coordinates to curvilinear coordinates or covariant form of wave equation. I need your a little bit more support. Could you help me a bit more to obtain wave equation in curvilinear coordinates or covariant form of wave equation ?????
     
  13. Jun 17, 2016 #12

    stevendaryl

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    It's just [itex]g^{\mu \nu} \nabla_\mu \nabla_\nu \phi = 0[/itex]. Expanding in terms of Christoffel symbols, it's:

    [itex]g^{\mu \nu} (\partial_\mu \partial_\nu \phi + \Gamma^\lambda_{\mu \nu} \partial_\lambda \phi) = 0[/itex]

    [edit] I got a sign wrong. The covariant derivative of a covector has a minus sign:

    [itex]g^{\mu \nu} (\partial_\mu \partial_\nu \phi - \Gamma^\lambda_{\mu \nu} \partial_\lambda \phi) = 0[/itex]
     
    Last edited: Jun 18, 2016
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