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A Einstein Hilbert action integral

  1. Jun 24, 2016 #1
    hi, when I see the einstein hilbert action I really started to be curious about that equation $$S=\int{\sqrt{g}d^4xR}$$. How is this action derived?? Is there a any proof using action integral involving the lagrangian density ??? If there is not a derivation from lagrangian action, What is the real proof of it ?? I am looking forward to your responses. Thanks in advance....
     
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  3. Jun 24, 2016 #2

    Paul Colby

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    Not much to say (because I'm not really very knowledgable on the subject) but, the action given is an invariant background independent functional which depends on the metric and it's derivatives. With this and the fact that variation of this action leads to the field equations should be enough to be a suitable action.
     
  4. Jun 24, 2016 #3
    Thanks Paul colby, your response is beneficial for me but I need also some mathematical demonstration to convince myself....
     
  5. Jun 24, 2016 #4
    Is there a really good proof of einstein hilbert equation ??? I mean I really tried to dig something up, but no proper proof exist...
     
  6. Jun 25, 2016 #5

    vanhees71

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    The Ricci scalar ##R## is the only scalar that's a function of the pseudo-metric components up to their 2nd derivatives, where the 2nd derivatives appear only linearly with coefficients that depend only on the pseudo-metric components and not any of their derivatives, which implies that it's the only scalar to be built out of the pseudo-metric components leading to an action with a Lagrangian that is a function of the pseudo-metric components and their 1st derivatives. You find the latter form by integrating by parts the terms containing the 2nd-order derivatives linearly, but that's not necessary since you find of course the correct Einstein field equations also from the form with ##R## as the Lagrangian.

    Of course, this argument is not complete, because you can as well add a constant to the action too. That's the famous cosmological constant. So the Einstein-Hilbert action with cosmological constant is the only generally covariant action of the desired type, and in this sense the Einstein field equations are somewhat unique.
     
  7. Jun 25, 2016 #6

    haushofer

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    What do you mean by "proof"?
     
  8. Jun 25, 2016 #7
    I consider there are some required mathematical steps to reach that equation, and I try to find them. If you have, could you explain to me ???
     
  9. Jun 25, 2016 #8
    I found this post that I believe answers your question.
     
  10. Jun 25, 2016 #9

    haushofer

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    There is no proof, only principes which guide you in constructing an action, hence my question. Vanhees gave you these principles.
     
  11. Jun 25, 2016 #10
    I have to admit that it is very remarkable derivation, but in the attachment I can not understand how the lagrangian takes such a weird form in terms of metric tensor ???? I could not capture the logic...
     

    Attached Files:

  12. Jun 25, 2016 #11

    Paul Colby

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    That may take some investment on your part. There are quite a collection of GR books out there.
     
  13. Jun 25, 2016 #12

    haushofer

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    You are asking quite standard textbook material, so maybe you should consult a good textbook on GR. From there you can ask questions.
     
  14. Jun 25, 2016 #13
    By the way, the attachment at post 10 is totally about the euler lagrange formula including second derivative of a variable (metric tensor in attachment). I consider it does not have to involve GR, just about lagrangian equation. But, I have to admit that I have never seen the second derivative concept of euler lagrangian formula. Also when it involves second derivative concept, there is a (-) sign. In short, Could you provide me with the second derivative concept of euler lagrangian formula or derivation of it ??? I just know the proof of at most first derivative concept....
     
  15. Jun 25, 2016 #14

    robphy

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  16. Jun 25, 2016 #15

    Paul Colby

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    doesn't usually happen but conceptually it's identical to the usual case. No GR was evoked in the equation you posted which was an application of variational calculus. You might read up on that.
     
  17. Jun 25, 2016 #16

    haushofer

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    See e.g. Ortin's Gravity and Strings, or d'Inverno. You consider a lagrangian up to second order derivatives, vary it and use partial integration. This calculation can be found in the textbooks and explains the alternation of the signs. Try it also for yourself.
     
  18. Jun 25, 2016 #17

    robphy

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  19. Jun 26, 2016 #18

    haushofer

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    I suggest the following exercise. Consider a Lagrangian density (!) ##\mathcal{L}## which depends on one field ##\phi(x)## and its first AND second derivatives. The action is then written as

    ##
    S = \int d^n x \mathcal{L} (\phi, \partial_a \phi, \partial_a \partial_b \phi)
    ##

    The variation of the action is then

    ##
    \delta S = \int d^n x \delta \mathcal{L}
    ##

    The variation ##\delta \mathcal{L}## is

    ##
    \delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi} \delta\phi + \frac{\partial \mathcal{L}}{\partial (\partial_a \phi) } \delta (\partial_a \phi) + \frac{\partial \mathcal{L}}{\partial (\partial_a \partial_b \phi)} \delta (\partial_a\partial_b \phi)
    ##

    Now you should write this, using partial integration, as

    ##
    \delta \mathcal{L} = [EOM] \delta \phi + \partial_a B^a
    ##

    where [EOM] are all the terms which multiply ## \delta \phi ## and hence will give the equations of motion (EOM), whereas ## B^a## is a boundary term which will vanish upon plugging into the action and using Stokes' theorem and your boundary conditions. You can use the fact that the variation and the partial derivatives commute, i.e.

    ##
    \delta (\partial_a \phi) = \partial_a (\delta\phi)
    ##

    (why?)
     
  20. Jun 28, 2016 #19
    Initially Thanks for all your valuable Responses, I have a very little question by the way: We admit that lagrangian is a scalar function. But sometimes I think why it is scalar. Lagrangian is comprised of time, first order derivative of SOMETHİNG, SOMETHİNG . If We talk about velocity, SOMETHİNG must be a position vector. Therefore if lagrangian function has a vector variable how we can strictly say that lagrangian is a scalar function? ?? May it not be a scalar????
     
    Last edited: Jun 28, 2016
  21. Jun 28, 2016 #20

    Paul Colby

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    A non-scalar Lagrangian would result in frame dependent equations of motion. Since things are not frame dependent at a level we can measure, the resulting model wouldn't fit reality as we measure it.
     
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