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Derivative of metric with respect to metric

  1. Apr 27, 2015 #1
    I'm hoping someone can clarify for me, I have seen the following used:

    [itex]\frac{\partial}{\partial g^{ab}}\left( g^{cd} \right) = \frac{1}{2} \left( \delta_a^c \delta_b^d + \delta_b^c \delta_a^d\right)[/itex]

    I understand the two half terms are used to account for the symmetry of the metric tensor.

    However, it appears to only half account for terms where [itex]a \ne b[/itex].

    To demonstrate what I mean by example, consider a 2D metric where [itex](a,b,c,d = 1,2)[/itex]

    [itex]\frac{\partial}{\partial g^{11}}\left( g^{11} \right) = \frac{1}{2} \left( \delta_1^1 \delta_1^1 + \delta_1^1 \delta_1^1\right) = 1[/itex]

    but

    [itex]\frac{\partial}{\partial g^{12}}\left( g^{12} \right) = \frac{1}{2} \left( \delta_1^1 \delta_2^2 + \delta_2^1 \delta_1^2\right) = \frac{1}{2}[/itex]

    I expected (possibly incorrectly) the latter would also be 1.

    Is the contribution of [itex]g^{cd}[/itex] evenly distributed between [itex]\frac{\partial}{\partial g^{ab}}[/itex] and [itex]\frac{\partial}{\partial g^{ba}}[/itex], and thus for [itex]a=b[/itex] the value is double the others?
     
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  3. Apr 27, 2015 #2

    ShayanJ

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    ## \frac{\partial x}{\partial x}=1 ## is a logical result of the definition of derivative regardless of what is x.
    Your formula is wrong. The correct formula is:
    ## {\Large\frac{\partial g^{ab}}{\partial g^{cd}}=}(1-\delta^a_c \delta^b_d)\delta^a_d\delta^b_c+\delta^a_c\delta^b_d##
     
    Last edited: Apr 27, 2015
  4. Apr 27, 2015 #3

    stevendaryl

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    I think that the author is using a kludge in order to get a simple expression for the change in a quantity when the metric changes.

    Suppose you have some metric-dependent quantity [itex]Q[/itex]. Let me use 2-D spacetime for simplicity, and let's assume it is linear in the metric components:

    [itex]Q = A_{00} g^{00} + A_{01} g^{01} + A_{10} g^{10} + A_{11} g^{11}[/itex]

    We would like the change in [itex]Q[/itex] due to varying the metric to behave like this:

    [itex]\delta Q = \sum_{\mu \nu} \frac{\partial Q}{\partial g_{\mu \nu}} \delta g_{\mu \nu}[/itex]

    So let's calculate the partial derivatives, without the factor of 1/2:

    • [itex]\frac{\partial Q}{\partial g_{00}} = A_{00}[/itex]
    • [itex]\frac{\partial Q}{\partial g_{01}} = A_{01} + A_{10}[/itex]
    • [itex]\frac{\partial Q}{\partial g_{10}} = A_{10} + A_{01}[/itex]
    • [itex]\frac{\partial Q}{\partial g_{11}} = A_{11}[/itex]
    Then the above expression for [itex]\delta Q[/itex] gives us:

    [itex]\delta Q = A_{00} \delta g_{00} + (A_{01} + A_{10}) \delta g_{01} + (A_{01} + A_{10}) \delta g_{10} + A_{11} \delta g_{11}[/itex]

    But looking at the definition of [itex]Q[/itex], it's clear that the answer should be:

    [itex]\delta Q = A_{00} \delta g_{00} + A_{01} \delta g_{01} + A_{10} \delta g_{10} + A_{11} \delta g_{11}[/itex]

    So without the 1/2, you get the wrong answer. Now, the actual problem is that the general formula

    [itex]\delta F = \sum_j \frac{\partial F}{\partial x^j} \delta x^j[/itex]

    is really only applicable if the variations [itex]\delta x^j[/itex] are independent. In the case of the metric tensor, they aren't independent, since [itex]\delta g^{\mu \nu} = \delta g^{\nu \mu}[/itex]. I'm not sure what is the mathematically elegant way to handle this situation, but sticking in the 1/2 is an ad-hoc approach that might give the right answer in (all? most?) cases.
     
  5. Apr 27, 2015 #4

    ShayanJ

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    We don't have to sum over all the components. Also I think we can say that for a symmetric metric, any quantity depending on the metric components, should only contain one of the two symmetric components. It means we can say ## \frac{\partial F}{\partial g_{\mu\nu}}=\frac{\partial F}{\partial g_{\nu\mu}}## and the sum you mentioned is only on the independent components.
     
  6. Apr 27, 2015 #5

    stevendaryl

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    Right, you could take the point of view that anything that is a function of the metric is really only a function of 10 independent quantities:
    4 of the form [itex]g_{\mu \mu}[/itex], and 6 of the form [itex]g_{\mu \nu}[/itex] with the restriction that [itex]\nu >\mu[/itex]. But that's kind of messy, because it involves arbitrarily declaring [itex]g_{01}[/itex] as preferred over [itex]g_{10}[/itex]

    I guess another approach might be to let all 16 components of the metric tensor be independent, and then at the end, only choose variations for which [itex]g_{\mu \nu} = g_{\nu \mu}[/itex]
     
  7. Apr 27, 2015 #6

    stevendaryl

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    So rather than viewing the symmetry of the metric as true by definition, you would view it as a symmetry of initial conditions (which would be preserved by the equations of motion).
     
  8. Apr 27, 2015 #7

    ShayanJ

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    Yeah, that's nice. But then we should deal with cosmologists asking:"Why metric was symmetric at the beginning?":rolleyes:
     
  9. Apr 29, 2015 #8

    ShayanJ

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    There is something wrong with this. When we want to vary Einstein-Hilbert action, we consider a variation of the metric which is symmetric and that gaurantees that the varried metric is still symmetric. But if we accept what you say, then there is no reason to consider symmetric variations and there is no gaurantee that the varried metric remains symmetric.
     
  10. Apr 29, 2015 #9
    If I understand correctly, the original formula I had seen:

    [itex]\frac{\partial}{\partial g^{ab}} ( g^{cd} ) = \frac{1}{2} ( \delta_a^c \delta_b^d + \delta_b^c \delta_a^d) [/itex]

    splits the contribution of the [itex] a \ne b [/itex] derivatives in half to account for Einstein notation summing over them twice.

    The correct formula requires an additional term:

    [itex]\frac{\partial}{\partial g^{ab}} ( g^{cd} ) = ( \delta_a^c \delta_b^d + \delta_b^c \delta_a^d - \delta_a^c \delta_b^d \delta_b^c \delta_a^d) [/itex]

    and could be used if the summation was explicitly only for the 10 independent quantities of the metric.

    Thank you both for your responses, I appreciate it! :biggrin:
     
  11. Apr 29, 2015 #10

    stevendaryl

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    I wasn't making any kind of claim, I was simply thinking out loud about a possible approach to handling the symmetry. Maybe Lagrange multipliers to enforce symmetry?
     
  12. Apr 29, 2015 #11

    ShayanJ

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    Yeah I know, and I just participated in your thoughts. No problem! And yeah, that's a good idea.
     
  13. Apr 30, 2015 #12

    samalkhaiat

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    Yes, when you differentiate any tensor [itex]\mathbf{T}[/itex] with respect to itself, you get the identity tensor of the tensor product space: [tex]\frac{\delta \mathbf{T}}{\delta \mathbf{T}} = \mathbf{I} \otimes \mathbf{I} .[/tex] So, for symmetric tensor you have [tex]\frac{\delta \mathbf{g}}{\delta \mathbf{g}} = \frac{1}{2} \left( \mathbf{I} \otimes \mathbf{I} \oplus \mathbf{I} \otimes \mathbf{I} \right) .[/tex] In terms of components, the above equations become [tex]\frac{\delta T^{a b}}{\delta T^{c d}} = \delta^{a}_{c} \delta^{b}_{d} , \ \ \ \ \ \ (1)[/tex] [tex]\frac{\delta g^{ab}}{\delta g^{cd}} = \frac{1}{2} ( \delta^{a}_{c} \delta^{b}_{d} + \delta^{a}_{d} \delta^{b}_{c} ) . \ \ \ (2)[/tex] Okay, now you try to write the following equations in components: [tex]\frac{\delta \mathbf{T}}{\delta \mathbf{T}^{-1}} = - \mathbf{T} \otimes \mathbf{T} ,[/tex] [tex]\frac{\delta \mathbf{g}}{\delta \mathbf{g}^{-1}} = - \frac{1}{2} \left( \mathbf{g} \otimes \mathbf{g} \oplus \mathbf{g} \otimes \mathbf{g} \right) .[/tex]

    No, here you need to be careful with the functional dependence. For any fixed value of [itex]c[/itex] and [itex]d[/itex], you have the following function [tex]g^{c d} = \bar{g}^{c d} ( g^{1 1} , g^{(1 2)} , g^{22} ) \equiv g^{c d} \left( g^{11} , g^{12} ( g^{(12)} ) , g^{21}( g^{(12)} ) , g^{22} \right) ,[/tex] where [tex]g^{(12)} = \frac{1}{2} ( g^{12} + g^{21} ) = g^{12} = g^{21} .[/tex] So, [tex]\frac{\partial \bar{g}^{cd}}{\partial g^{(12)}} = \frac{\partial g^{cd}}{\partial g^{12}} \frac{\partial g^{12}}{\partial g^{(12)}} + \frac{\partial g^{cd}}{\partial g^{21}} \frac{\partial g^{21}}{\partial g^{(12)}} ,[/tex] or [tex]\frac{\partial \bar{g}^{cd}}{\partial g^{(12)}} = \frac{\partial g^{cd}}{\partial g^{12}} + \frac{\partial g^{cd}}{\partial g^{21}} = \delta^{c}_{1} \delta^{d}_{2} + \delta^{c}_{2} \delta^{d}_{1} .[/tex]

    In practise the metric tensor comes always contracted with other tensors, so you never need to use Eq(2), because the (symmetric) metric kills the anti-symmetric part of the contraction. So, it is always safe to use Eq(1) for the metric. To see that, let’s consider two examples and try to calculate the functional derivative with respect to [itex]g^{n m}[/itex]

    The first is the simple trace equation [tex]S = A^{a}_{a} = A_{ab}g^{ab} .[/tex] Since [itex]g^{ab} = g^{ba}[/itex], the anti-symmetric part of [itex]A_{ab}[/itex] will get killed by the metric. So, we may take [itex]A_{ab}[/itex] to be symmetric, i.e. [itex]A_{ab} = A_{(ab)}[/itex]. Now [tex]\frac{\delta S}{\delta g^{nm}} = A_{(ab)} \frac{\delta g^{ab}}{\delta g^{nm}} = A_{(ab)} \delta^{a}_{n} \delta^{b}_{m} = A_{(nm)} .[/tex] Notice that Eq(1) and Eq(2) give you the same answer. The second example which is a bit more complicated (but important), is the square of rank-2 tensor [tex]S = F^{2} = F_{ab} F^{ab} = F_{ac} F_{bd} g^{ab} g^{cd} .[/tex] I will use Eq(1) to calculate the derivative

    [tex]\frac{\delta S}{\delta g^{nm}} = F_{ac} F_{bd} \left( g^{ab} \frac{\delta g^{cd}}{\delta g^{nm}} + \frac{\delta g^{ab}}{\delta g^{nm}} g^{cd} \right) .[/tex] Thus, by Eq(1), [tex]\frac{\delta S}{\delta g^{nm}} = F_{ac} F_{bd} ( g^{ab} \delta^{c}_{n} \delta^{d}_{m} + g^{cd} \delta^{a}_{n} \delta^{b}_{m} ) .[/tex] Finally, by changing the dummy indices in the second term, we get [tex]\frac{\delta S}{\delta g^{nm}} = g^{ab} ( F_{an} F_{bm} + F_{na} F_{mb} ) .[/tex] This is the final result for arbitrary rank-2 tensor. Furthermore, for symmetric or anti-symmetric [itex]F_{ab}[/itex], the above expression simplifies to [tex]\frac{\delta S}{\delta g^{nm}} = 2 g^{ab} F_{an} F_{bm} .[/tex] Now you repeat the calculation using Eq(2) instead.

    Sam
     
  14. Apr 30, 2015 #13

    samalkhaiat

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    When x is a rank-n tensor, the “1” is a rank-2n identity tensor. See my previous post.
    There is nothing correct about that "thing" which I can't even call it a equation. Where did you get this piece of garbage from?
     
  15. May 1, 2015 #14

    ShayanJ

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    Oh...sorry. I wasn't thinking in terms of tensors at all. I treated the metric components only as a bunch of functions and wrote a formula that gives only the correct numerical value for those bunch of functions! It was a mixture of several mistakes I suppose. I apologize anyway.
     
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