Euler lagrange equation and Einstein lagrangian

In summary, the conversation revolves around the derivation of the Euler-Lagrange equation in D'inverno chapter 11 using the Hilbert-Einstein Lagrangian. The conversation includes equations and calculations, with one person asking for help and the other providing guidance and corrections. The final result confirms the equation (11.26) of D'inverno, with the reminder to remember the symmetry in the indices.
  • #1
shadi_s10
89
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Dear everyone

can anyone help me with the euler lagrange equation which is stated in d'inverno chapter 11?
in equation (11.26) it is said that when we use the hilbert-einstein lagrangian we can have:

∂L/(∂g_(ab,cd) )=(g^(-1/2) )[(1/2)(g^ac g^bd+g^ad g^bc )-g^ab g^cd ]

haw can we derive this out of (11.24)?


I tried to derive ∂L/(∂g_(ab,cd) ) out of (11.24), but I really don't know how to work with (∂g_(ab,cd) ?

please help me!
 
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  • #2
shadi_s10 said:
Dear everyone

can anyone help me with the euler lagrange equation which is stated in d'inverno chapter 11?
in equation (11.26) it is said that when we use the hilbert-einstein lagrangian we can have:

∂L/(∂g_(ab,cd) )=(g^(-1/2) )[(1/2)(g^ac g^bd+g^ad g^bc )-g^ab g^cd ]

haw can we derive this out of (11.24)?


I tried to derive ∂L/(∂g_(ab,cd) ) out of (11.24), but I really don't know how to work with (∂g_(ab,cd) ?

please help me!

So let's think about it!

In the equation (11.24), you have to forget about the terms of the last two lines because they are all made of the first derivatives of the metric tensor or the products of 'em. Keeping this in mind, make use of the relation between the determinant of the metric tensor and the christoffel symbols that is a simple equation involving first derivatives of gab. Remember that?!

AB
 
  • #3
Thanks for helping me again!

mmmmmmmmm
ok
let me see
here is what I do to get the equation:

(∂L_G)/(∂g_(ab,cd) )=∂/(∂g_(ab,cd) ) {g^cd [1/2 g^ef {g_(cf,de)+g_(df,ce)-g_(cd,fe) }-1/2 g^ef {g_(cf,ed)+g_(ef,cd)-g_(ce,fd) } ] }

now could you tell me what do you write for a term like this:
∂/(∂g_(ab,cd) )(g^cd g^ef g_(cf,de))

maybe I do not know how to work with this... :confused:
 
  • #4
shadi_s10 said:
Thanks for helping me again!

mmmmmmmmm
ok
let me see
here is what I do to get the equation:

(∂L_G)/(∂g_(ab,cd) )=∂/(∂g_(ab,cd) ) {g^cd [1/2 g^ef {g_(cf,de)+g_(df,ce)-g_(cd,fe) }-1/2 g^ef {g_(cf,ed)+g_(ef,cd)-g_(ce,fd) } ] }

A nice work up to here! So you've taken in what exactly I said about keeping those terms including a single metric tensor differentiated twice w.r.t the coordinates.

now could you tell me what do you write for a term like this:
∂/(∂g_(ab,cd) )(g^cd g^ef g_(cf,de))

maybe I do not know how to work with this... :confused:

I give you this sample and wish to come to me with the final result and calculation procedure:

http://img513.imageshack.us/img513/907/calf.jpg

AB
 
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  • #5
Thanks again for helping. :smile:


I did the exact thing but there is a minus sign which ruins everything!
here is what I have:

(∂L_G)/(∂g_(ab,cd) )

=1/2(g^ac g^db+g^ca g^db-g^ab g^dc-g^ad g^cb-g^cd g^ab+g^ad g^bc)

=1/2 √(-g)(2g^ac g^db-2g^ab g^dc)

which clearly is not (11.26)

a problem is that -g^ad g^cb and +g^ad g^bc vanish together! (the third term and the last one)
and another one is that g^ac g^db+g^ca g^db=2g^ac g^db

am I wrong?
 
  • #6
shadi_s10 said:
Thanks again for helping. :smile:


I did the exact thing but there is a minus sign which ruins everything!
here is what I have:

(∂L_G)/(∂g_(ab,cd) )

=1/2(g^ac g^db+g^ca g^db-g^ab g^dc-g^ad g^cb-g^cd g^ab+g^ad g^bc)

=1/2 √(-g)(2g^ac g^db-2g^ab g^dc)

which clearly is not (11.26)

a problem is that -g^ad g^cb and +g^ad g^bc vanish together! (the third term and the last one)
and another one is that g^ac g^db+g^ca g^db=2g^ac g^db

am I wrong?

I want you to check your calculation again. Your result has a major error.

AB
 
  • #7
I'm confused! This is exactly what I get from the deltas!
would you please tke a look at this:

(∂L_G)/(∂g_(ab,cd) )

=1/2 g^cd g^ef [(δ_c^a δ_f^b δ_d^c δ_e^d+δ_d^a δ_f^b δ_c^c δ_e^d-δ_c^a δ_d^b δ_f^c δ_e^d )-(δ_c^a δ_f^b δ_e^c δ_d^d+δ_e^a δ_f^b δ_c^c δ_d^d-δ_c^a δ_e^b δ_f^c δ_d^d )]

=1/2(g^ac g^db+g^ca g^db-g^ab g^dc-g^ad g^cb-g^cd g^ab+g^ad g^bc)

and this is 1/2 √(-g)(2g^ac g^db-2g^ab g^dc)
= √(-g)(g^ac g^db-g^ab g^dc)

:rolleyes:
 
  • #8
shadi_s10 said:
I'm confused! This is exactly what I get from the deltas!
would you please tke a look at this:

(∂L_G)/(∂g_(ab,cd) )

=1/2 g^cd g^ef [(δ_c^a δ_f^b δ_d^c δ_e^d+δ_d^a δ_f^b δ_c^c δ_e^d-δ_c^a δ_d^b δ_f^c δ_e^d )-(δ_c^a δ_f^b δ_e^c δ_d^d+δ_e^a δ_f^b δ_c^c δ_d^d-δ_c^a δ_e^b δ_f^c δ_d^d )]

=1/2(g^ac g^db+g^ca g^db-g^ab g^dc-g^ad g^cb-g^cd g^ab+g^ad g^bc)

and this is 1/2 √(-g)(2g^ac g^db-2g^ab g^dc)
= √(-g)(g^ac g^db-g^ab g^dc)

:rolleyes:

Everything is okay with your calculation now! The only thing to be remembered is that (∂L_G)/∂g_(ab,cd) is symmetric in the indices c and d. So you need to just put

[tex]g^{ac} g^{db}[/tex] = [tex]1/2(g^{ac} g^{db}+g^{ad} g^{cb})[/tex]

to get the equation (11.26) of D'inverno!

AB
 
  • #9
oh my god!
that's it!
thank you very very much!
(I don't know why, but I usually forget the symmetry! :biggrin:)

Thanks!
 
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