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Newtonian limit of covariant derivative of stress-energy tensor(schutz ch7)

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    For a perfect fluid verify that the spacial components of [itex]T^{\mu \nu};_{\nu} = 0[/itex] in the newtonian limit reduce to

    [tex](\rho v^{i}),_{t} + (\rho v^{i} v^{j}),_{j} + P,_{i} + \rho \phi ,_{i}[/tex]

    2. Relevant equations

    [tex]ds^{2} = -(1+2 \phi )dt^{2} + (1-2 \phi) (dx^{2} + dy^{2} + dz^{2})[/tex]

    Christoffel Symbols
    [tex] \Gamma^{t}_{t \alpha} = \phi ,_{\alpha}[/tex]
    [tex] \Gamma^{t}_{ij} = - \phi ,_{t} \delta _{ij}[/tex]
    [tex] \Gamma^{i}_{tt} = \phi ,_{i}[/tex]
    [tex] \Gamma^{i}_{tj} = - \phi ,_{t} \delta _{ij}[/tex]
    [tex] \Gamma^{i}_{jk} = \phi ,_{i} \delta _{jk} - \phi ,_{k} \delta _{ij} - \phi ,_{j} \delta _{ik}[/tex]

    Stress-energy tensor for a perfect fluid
    [tex] T^{\alpha \beta} = (\rho + P)U^{\alpha}U^{\beta} + Pg^{\alpha \beta}[/tex]

    In newtonian limit,

    [tex]T^{tt} = \rho[/tex]
    [tex]T^{ti} = \rho v^{i}[/tex]
    [tex]T^{ij} = \rho v^{i}v^{j} + P\delta ^{ij}[/tex]

    3. The attempt at a solution

    Spacial components are

    [tex]T^{\i \nu};_{\nu} [/tex]
    [tex] = T^{it};_{t} + T^{ij};_{j}[/tex]
    [tex] = T^{it},_{t} + T^{\alpha t} \Gamma ^{i}_{\alpha t} +T^{i \alpha}\Gamma ^{t}_{\alpha t} + T^{ij},_{j} + T^{\alpha j} \Gamma ^{i}_{\alpha j} + T^{i \alpha} \Gamma ^{j}_{\alpha j}[/tex]
    [tex] = T^{it},_{t} + T^{tt} \Gamma ^{i}_{tt} + T^{jt} \Gamma ^{i}_{jt} + T^{it}\Gamma ^{t}_{tt} + T^{ij}\Gamma ^{t}_{jt} + T^{ij},_{j} + T^{tj} \Gamma ^{i}_{tj} + T^{kj} \Gamma ^{i}_{kj} + T^{it} \Gamma ^{j}_{tj} + T^{ik} \Gamma ^{j}_{kj}[/tex]

    Then substituting in the above values and simplifying I get

    [tex](\rho v^{i}),_{t} + (\rho v^{i} v^{j}),_{j} + P,_{i} + \rho \phi ,_{i} + P\phi,_{i} + \rho v^{k}v^{j} \delta _{kj} \phi ,_{i} - 4 \rho v^{i} \phi ,_{t} - 2 \rho v^{i} v^{j} \phi ,_{j}[/tex]

    and as [itex]T^{\i \nu};_{\nu} = 0[/itex], this expression is also = 0.

    As you can see the first 4 terms are what I'm looking for, but I somehow have a load of terms on the end which should vanish but I just don't know how.....

    (if it's helpful I'll post my full working with the simplification etc)
    Last edited: Feb 12, 2009
  2. jcsd
  3. Feb 12, 2009 #2
    I haven't looked in details yet, but your extra terms seem to be contain stuff that would become very small in the Newtonian limit, like pressure and speeds. For example you can take [tex] \rho \phi ,_{i} [/tex] in common b/w third and fourth terms and then neglect the pressure compared to density. Through switching dummy indices you can add the fifth and the final term. This sum is again small compared to the third term : [tex] \ v^{i} v^{j} << 1 [/tex]

    By the way, have you forgotten the [tex] \ g^{\mu \nu} [/tex] terms while evaluating the Christoffel symbols?
  4. Feb 13, 2009 #3
    do you mean that I can write [itex]\rho \phi ,_{i} + P\phi,_{i}=\phi,_{i}(\rho + P)[/itex] and then [itex](\rho + P)\approx \rho[/itex]?

    You say that I can combine these terms by relabelling indices
    [tex]\rho v^{k}v^{j} \delta _{kj} \phi ,_{i} - 2 \rho v^{i} v^{j} \phi ,_{j}[/tex]
    But I'm not sure that I can as the first term here is summed on [itex]\delta[/itex] and the second on [itex]\phi[/itex] how do I combine them?
    I do see now though that [itex]v^{k}v^{j}[/itex] is zero in the limit so that gets rid of both terms anyway, Thanks.

    Do you mean in the calculation of the stress-energy tensor?
    I have included this term taking into account that [itex]\phi P\approx 0[/itex] in the limit.
    I didn't use the approximation [itex](\rho + P)\approx \rho[/itex].

    So using
    [tex] T^{\alpha \beta} = (\rho + P)U^{\alpha}U^{\beta} + Pg^{\alpha \beta}[/tex]

    what I did is:
    [tex]T^{tt} = (\rho + P)U^{t}U^{t} + Pg^{tt}[/tex]
    [tex]= (\rho + P)(1X1) + P(-1+2\phi)[/tex]
    [tex]= (\rho + P) - P +2P\phi[/tex]
    [tex]\approx (\rho + P) - P [/tex]
    [tex]=\rho [/tex]

    and similar for the other terms, So I think I'm ok there.

    You have helped me a lot... Now I have only one more term left to get rid of: [itex]- 4 \rho v^{i} \phi ,_{t}[/itex] which doesn't look like it's going to disappear in the limit... Hmmm I'll check my working carefully to see if I made a little mistake somewhere.

    Thanks loads for your help so far :)
    Last edited: Feb 13, 2009
  5. Feb 13, 2009 #4
    Oops, my mistake.
  6. Feb 13, 2009 #5
    I can't see a mistake in my working......
    The offending term comes from the 7th and 9th terms in
    [tex]T^{\i \nu};_{\nu} = T^{it},_{t} + T^{tt} \Gamma ^{i}_{tt} + T^{jt} \Gamma ^{i}_{jt} + T^{it}\Gamma ^{t}_{tt} + T^{ij}\Gamma ^{t}_{jt} + T^{ij},_{j} + T^{tj} \Gamma ^{i}_{tj} + T^{kj} \Gamma ^{i}_{kj} + T^{it} \Gamma ^{j}_{tj} + T^{ik} \Gamma ^{j}_{kj}[/tex]

    These two:
    [tex]T^{tj} \Gamma ^{i}_{tj} + T^{it} \Gamma ^{j}_{tj}[/tex]

    [tex]T^{tj} = \rho v^{j}[/tex]
    [tex]T^{it} = \rho v^{i}[/tex]
    [tex] \Gamma^{i}_{tj} = - \phi ,_{t} \delta _{ij}[/tex]
    This last christoffel symbol also implies that
    [tex] \Gamma^{j}_{tj} = - \phi ,_{t} - \phi ,_{t} - \phi ,_{t} = -3\phi ,_{t}[/tex]

    [tex]T^{tj} \Gamma ^{i}_{tj} + T^{it} \Gamma ^{j}_{tj} = \rho v^{j} (- \phi ,_{t}) \delta _{ij} + \rho v^{i}(-3\phi ,_{t}) = \rho v^{i} (- \phi ,_{t}) + \rho v^{i}(-3\phi ,_{t}) = - 4 \rho v^{i} \phi ,_{t}[/tex]

    Which is my rogue term....
    How do I get rid of it?
  7. Feb 16, 2009 #6
    Did you take care of the [tex] T^{it}\Gamma ^{t}_{tt} [/tex]? This should give another time derivative of phi term...
    Last edited: Feb 16, 2009
  8. Feb 16, 2009 #7

    Yes it cancels with the term before it!

    [tex]T^{jt} \Gamma ^{i}_{jt} + T^{it}\Gamma ^{t}_{tt}[/tex]
    [tex]= -\phi ,_{t} \delta _{ij} \rho v^{j} + \rho v^{i} \phi ,_{t}[/tex]
    [tex]= -\rho v^{i} \phi ,_{t} + \rho v^{i} \phi ,_{t}[/tex]
    [tex] = 0 [/tex]

  9. Feb 16, 2009 #8
    I remembered that the metric was derived with some assumption about the field being slowly varying, in which case time derivative of phi is small compared to space derivatives. Check equation 8.46 of Schutz for instance.
  10. Feb 17, 2009 #9
    I'm afraid that I haven't read that far yet and it don't understand what that equation means...

    It should be possible to do this using the material up to and including chapter 7 shouldn't it?

    Maybe I should leave it for now and come back to this question later after I have read some more.... I don't like doing that though.... But I suppose sometimes it is the only way....
  11. Feb 18, 2009 #10
    I've been looking in the dark depths of the internet and found this statement:

    for a perfect fluid,
    [tex]T^{\mu \nu};_{\nu} = 0 [/tex]
    [tex]\Rightarrow P,_\mu + ((\rho + P) U_\mu U^\nu),_\nu = -(\rho+P)(\Gamma^{\lambda}_{\nu \lambda} U^\nu U_\mu + \Gamma_{\nu \lambda \mu}U^\nu U^\lambda)[/tex]

    If I look at the spacial components and take to the newtonian limit, I get exactly the correct result!
    But I just cant prove the above.... I always get some extra terms that just won't cancel.

    I get this:

    [tex]T^{\mu \nu};_{\nu} = 0[/tex]

    [tex]\Rightarrow T^{\mu \nu} ,_\nu + T^{\mu \sigma} \Gamma^\nu _{\sigma \nu} + T^{\sigma \nu} \Gamma^\mu _{\sigma \nu} = 0[/tex]

    [tex]\Rightarrow ((P + \rho)U^\mu U^\nu), _\nu + (Pg^{\mu \nu}),_\nu = -(P+ \rho)(U^\mu U^\sigma \Gamma^\nu _{\sigma \nu} + U^\sigma U^\nu \Gamma^\mu _{\sigma \nu}) - P(g^{\mu \sigma} \Gamma^\nu _{\sigma \nu} + g^{\sigma \nu} \Gamma^\mu _{\sigma \nu})[/tex]

    now if I contract everything with [itex]g_{\alpha \mu}[/itex] I nearly get it apart from the extra [itex]g_{\alpha \mu}[/itex] and the raised [itex]\mu[/itex] index in the first term,
    the [itex]Pg_{\alpha \mu} g^{\mu\nu}, _\nu}[/itex] from the second term and the whole of the second term on the right hand side.

    I'm obviously doing something wrong... but what??
  12. Feb 18, 2009 #11


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    It may be easier to try and derive this form the [tex]{{T_\mu}^\nu}_{;\nu}=0[/tex] version. So,

    [tex]{{T_\mu}^\nu}_{;\nu}=\left[(\rho + P) U_\mu U^\nu+P{\delta_\mu}^\nu \right]_{;\nu}=((\rho + P) U_\mu U^\nu)_{;\nu}+P_{,\mu}{\delta_\mu}^\nu[/tex]

    Then, expand the first covariant derivative in terms of Christoffel symbols. You then shouldn't get as many terms (since we have used the fact that the covariant derivative of the metric tensor is zero, in the second equality sign.
  13. Feb 18, 2009 #12
    Ahh yes...That makes it quite easy....

    but it should work the other way too right?

    if I use the fact that the covariant derivative of g is zero

    [tex]{T^{\mu \nu}}_{;\nu}=\left[(\rho + P) U^\mu U^\nu+Pg^{\mu \nu} \right]_{;\nu}=((\rho + P) U^\mu U^\nu)_{;\nu}+P_{,\mu} g^{\mu \nu} = 0[/tex]

    if I expand and then contract with [itex]g_{\alpha \mu}[/itex] (as opposed to the other way around)

    [tex]\Rightarrow P,_\alpha + g_{\alpha \mu}((\rho + P) U^\mu U^\nu),_\nu = -(\rho+P)(\Gamma^{\lambda}_{\nu \lambda} U^\nu U_\alpha + \Gamma_{\nu \lambda \alpha}U^\nu U^\lambda)[/tex]

    Very nearly there, but that second term just doesn't seem to be right...
  14. Feb 19, 2009 #13
    Actually, I take that back... I still cant get it!

    [tex]{{T_\mu}^\nu}_{;\nu}=\left[(\rho + P) U_\mu U^\nu+P{\delta_\mu}^\nu \right]_{;\nu}=((\rho + P) U_\mu U^\nu)_{;\nu}+P_{,\nu}{\delta_\mu}^\nu[/tex]

    [tex]=((\rho + P) U_\mu U^\nu)_{,\nu}+(\rho + P)(U_\mu U^\sigma \Gamma^\nu _{\sigma \nu} - U_\sigma U^\nu \Gamma^\sigma _{\mu \nu}) + P_{,\mu}[/tex]

    [tex]=((\rho + P) U_\mu U^\nu)_{,\nu}+(\rho + P)(U_\mu U^\sigma \Gamma^\nu _{\sigma \nu} - g_{\alpha \sigma} U^\alpha U^\nu \Gamma^\sigma _{\mu \nu}) + P_{,\mu}[/tex]

    [tex]=((\rho + P) U_\mu U^\nu)_{,\nu}+(\rho + P)(U_\mu U^\sigma \Gamma^\nu _{\sigma \nu} - U^\alpha U^\nu \Gamma_{\mu \nu \alpha}) + P_{,\mu}[/tex]

    Whereas I should get

    [tex] ((\rho + P) U_\mu U^\nu),_\nu + (\rho+P)(U_\mu U^\sigma \Gamma^{\nu}_{\sigma \nu} + U^\alpha U^\nu \Gamma_{\alpha \nu \mu}) + P,_\mu [/tex]

    and [tex]\Gamma_{\alpha \nu \mu} \neq -\Gamma_{\mu \nu \alpha}[/tex]

  15. Feb 19, 2009 #14


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    This looks right to me. Where did you get this identity from? Perhaps there's a typo there or something.
  16. Feb 19, 2009 #15
    Unfortunately I have lost the link to the place I found that identity as my history gets reset every night.

    But I think you may be right that there is a mistake in the text as the newtonian limit of this gives the correct solution to the original problem (although there are some covariant indices where I should have contravariant)... I'll have to have a closer look over the weekend or next week.....

    Thank you both, you have really been a lot of help.
  17. Feb 23, 2009 #16
    I have now calculated this problem in about 5 different ways and always get the same result:
    [tex]T^{\mu \nu};_{\nu} = 0[/tex]
    [tex]\Rightarrow (\rho v^{i}),_{t} + (\rho v^{i} v^{j}),_{j} + P,_{i} + \rho \phi ,_{i} - 4 \rho v^{i} \phi ,_{t} = 0[/tex]

    So I'm pretty convinced that this is correct and as xboy says, the [itex] - 4 \rho v^{i} \phi ,_{t}[/itex] term must tend to zero in the newtonian limit. I suppose that I'll find out why as I read more about this.

    I'm also sure that the equation I found on the internet is wrong!
    [tex] ((\rho + P) U_\mu U^\nu),_\nu + (\rho+P)(U_\mu U^\sigma \Gamma^{\nu}_{\sigma \nu} + U^\alpha U^\nu \Gamma_{\alpha \nu \mu}) + P,_\mu = 0[/tex]

    It should be
    [tex]((\rho + P) U_\mu U^\nu)_{,\nu}+(\rho + P)(U_\mu U^\sigma \Gamma^\nu _{\sigma \nu} - U^\alpha U^\nu \Gamma_{\mu \nu \alpha}) + P_{,\mu} = 0[/tex]

    as the first here gives me a negative [itex]\rho \phi ,_{i}[/itex] in the newtonian limit, which is clearly wrong. It also fails for a static fluid whereas the second equation doesnt, so i'm doubly convinced.

    Now I'll just have to keep reading :)

    Thanks again!
  18. Mar 6, 2009 #17
    I have come across another question with a similar outcome (schutz P.193 Q9b)

    The answer should be
    [tex]-\phi_{,ij} \xi^j[/tex]

    but I get
    [tex]-\phi_{,ij} \xi^j - \phi_{,tt}\xi^j[/tex]

    My original question and this one make me think that maybe in the newtonian limit [itex]\phi[/itex] is independent of time, which sort of makes sense to me as Newton probably didnt concern himself with a gravity field that changes over time... But what do I know?

    Is this right?
    Last edited: Mar 6, 2009
  19. Mar 9, 2009 #18
    More thoughts on this:

    If [itex]\phi[/itex] is a funtion of t, then [itex]g_{\alpha \beta}[/itex] is a function of t and so [itex]p_t =energy[/itex] is not conserved.
    do we need energy conserved in newtonian limit and so [itex]\phi[/itex] independent of t?
  20. Mar 18, 2009 #19
    I've found the answer in Schutz. When deriving the metric he says
    This means for a stress-energy tensor independent of t, the t partial derivatives have v to 1st order. so v multiplied by a t derivative is second order and so very small and a second t derivative is also of second order and so very small.

    This sorts out the problem

    but why is [itex]\partial / \partial t [/itex] is of the same order as [itex]v \partial / \partial x [/itex]?

    I can see that [itex]d / dt = v^\alpha \partial / \partial x^\alpha [/itex]
    Is it because of this?
    Last edited: Mar 18, 2009
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