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Paul Colby

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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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What do you mean by "proof"?

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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 ???What do you mean by "proof"?

- #8

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I found this post that I believe answers your question.

- #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.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 ???

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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...I found this post that I believe answers your question.

- #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.I could not capture the logic...

- #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.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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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.But, I have to admit that I have never seen the second derivative concept of euler lagrangian formula.

- #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.

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https://books.google.com/books?id=W...120&dq=schouten+lagrange+derivative&source=bl

for details:

https://books.google.com/books?id=r...agrange+derivative&source=bl&pg=111#v=onepage

- #18

haushofer

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##

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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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.May it not be a scalar

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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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https://www.physicsforums.com/help/latexhelp/

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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(((if you barely understand equation, you can look at my attachment.... ))))

\bar g_{ab,cd} A^{abcd} = g_{ab,cd} A^{abcd}

$$\bar g_{ab,cd} A^{abcd} = g_{ab,cd} A^{abcd}$$

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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 )

$$\bar g_{ab,cd} A^{abcd} = g_{ab,cd} A^{abcd}$$

$$\bar g_{ab,cd} A^{abcd} = g_{ab,cd} A^{abcd}$$

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- #25

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

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