Einstein Hilbert action integral

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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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  • #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.
 
  • #3
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Thanks Paul colby, your response is beneficial for me but I need also some mathematical demonstration to convince myself....
 
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Is there a really good proof of einstein hilbert equation ??? I mean I really tried to dig something up, but no proper proof exist...
 
  • #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.
 
  • #6
haushofer
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Is there a really good proof of einstein hilbert equation ??? I mean I really tried to dig something up, but no proper proof exist...
What do you mean by "proof"?
 
  • #7
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What do you mean by "proof"?
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 ???
 
  • #8
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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....
I found this post that I believe answers your question.
 
  • #9
haushofer
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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 ???
There is no proof, only principes which guide you in constructing an action, hence my question. Vanhees gave you these principles.
 
  • #10
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I found this post that I believe answers your question.
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...
 

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  • #11
Paul Colby
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I could not capture the logic...
That may take some investment on your part. There are quite a collection of GR books out there.
 
  • #12
haushofer
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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...
You are asking quite standard textbook material, so maybe you should consult a good textbook on GR. From there you can ask questions.
 
  • #13
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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
Paul Colby
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But, I have to admit that I have never seen the second derivative concept of euler lagrangian formula.
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.
 
  • #16
haushofer
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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....
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.
 
  • #17
robphy
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  • #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?)
 
  • #19
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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????
 
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  • #20
Paul Colby
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May it not be a scalar
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.
 
  • #21
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hi, I would like to express that when I look at the shared derivation ( called "this") in post 8, I can not understand the equation in my attachment. Could you explain this equation??? Why are the metric tensors with partial derivatives equal in different coordinates???
 

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  • #23
Ibix
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(((if you barely understand equation, you can look at my attachment.... ))))
\bar g_{ab,cd} A^{abcd} = g_{ab,cd} A^{abcd}
Just add two hashes before and after for inline maths like ##x=1## or two dollars for maths on its own line like $$F=ma $$ Quote my post to see what I've done if you want. Your equation is
$$\bar g_{ab,cd} A^{abcd} = g_{ab,cd} A^{abcd}$$
 
  • #24
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hi, I would like to express that when I look at the shared derivation ( called "this") in post 8 and fourth equation of "this", I can not understand the equation below. Could you explain this equation??? Why does the capital A remain same in the equation??? ( I think it should not remain same, because it consists of metric tensors, and if we change the coordinates, metrics change and hence A will change )

$$\bar g_{ab,cd} A^{abcd} = g_{ab,cd} A^{abcd}$$
 
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  • #25
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hi, I have another question: Initially I would like to express that when I again look at the shared derivation ( in the link https://www.physicsforums.com/threads/einstein-hilbert-action-integral.876680/ and called "this" in post 8), I can not understand the equation below. Could you explain this equation??? How do we get this equation??? Could you provide me with a derivation??? Thanks in advance...

$$x^a=\bar x^a + \frac 1 6 \eta^{ae} C_{ebcd} \bar x^b \bar x^c \bar x^d$$
 

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