Linear Gravity Field - Lagrangian

In summary: The terms involving ##h_{,\nu}## and ##\partial_\nu h## would be separate. I think that way you could maybe see more clearly where the terms of the form ##\partial_\mu \partial_\nu h## would come from. I don't know if taking the derivatives term-by-term will help you or not. I would think that ##h_{,\nu}## terms are the only way to go, really, if ##h## is the only variable that appears explicitly in the Lagrangian. I think you would have to rewrite the Euler-Lagrange equation as a ##\partial_\mu \partial_\nu## derivative in order to get all the terms.
  • #1
ProfDawgstein
80
1
Another Exercise... Ohanian Exercise 6

Book said:
Exercise 6 Show that ##\mathcal{L}_{(0)}## leads to the field equation (3.38) with ##T^{\mu\nu} = 0## (Hint: When working out the partial derivatives of ##\mathcal{L}_{(0)}## with respect to ##h^{\alpha\beta}_{\ \ \ ,\mu}## you can treat ##h_{\alpha\beta}## as independent of ##h_{\beta\alpha}##; that is, you need not worry about the symmetry of the tensor field. The terms in the Lagrangian (A.34) are already arranged in such a way that they directly lead to a symmetric field equation.)

(A.34) ##\mathcal{L}_{(0)} = \frac{1}{4} \left( h_{\mu\nu ,\alpha} h^{\mu\nu ,\alpha} - 2 h_{\mu\nu ,\alpha} h^{\alpha\mu ,\nu} + 2 h_{\mu\nu}^{\ \ \ \ ,\mu} h^{,\nu} - h_{,\nu} h^{,\nu} \right)##

(3.38) ##\partial_\lambda \partial^\lambda h^{\mu\nu} + \partial^\mu \partial^\nu h - (\partial_\lambda \partial^\nu h^{\mu\lambda} + \partial_\lambda \partial^\mu h^{\nu\lambda}) - \eta^{\mu\nu} \partial_\lambda \partial^\lambda h + \eta^{\mu\nu} \partial_\lambda \partial_\sigma h^{\lambda\sigma} = -k T^{\mu\nu} = 0##

The Euler-Lagrange Equation given:

##\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})} - \frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0##

##\frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0##, because there is no explicit ##h^{\alpha\beta}## dependence

Now, since we have ##h## here, which is the trace of ##h^{\alpha\beta}##

##h = \eta^{\alpha\beta} h_{\alpha\beta} = \eta_{\alpha\beta} h^{\alpha\beta}##

Expanding all ##h## in the Lagrangian using

##\partial^\nu h = \eta_{\alpha\beta} \partial^\nu h^{\alpha\beta}##

##\partial_\nu h = \eta_{\alpha\beta} \partial_\nu h^{\alpha\beta}##

results in the 'new' Lagrangian

##\mathcal{L}_{(0)} = \frac{1}{4} \left( (\partial_\alpha h_{\mu\nu}) (\partial^\alpha h^{\mu\nu}) - 2 (\partial_\alpha h_{\mu\nu}) (\partial^\nu h^{\alpha\mu}) + 2 (\partial^\mu h_{\mu\nu}) (\partial^\nu \eta_{\alpha\beta} h^{\alpha\beta}) - (\partial_\nu \eta_{\alpha\beta} h^{\alpha\beta}) (\partial^\nu \eta_{\rho\sigma} h^{\rho\sigma}) \right)##

Now after doing lots of product rules and triple ##\delta##'s I can't get 2nd order derivatives...

[Doing: ##\frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})}##]

Replacing the ##\partial_a h^{bc}## with ##\delta \delta \delta##

I always have something like ##\eta \partial_{\alpha} h^{ab}##,

but never ##\eta \partial \partial h^{ab}##.

And somehow I still end up making all terms look like ##\partial_{\mu} h^{\alpha\beta}## for the Euler-Lagrange derivatives.

If we use ##\eta_{\mu\nu}## twice on ##h^{\alpha\beta}## we shouldn't change the tensor

##h^{\alpha\beta} = h_{\alpha\beta}## | (-1)(-1) = +1 and

for the partials I just use ##\partial_\alpha = \eta_{\alpha\mu} \partial^\mu## ?

I am getting like 80 indices in the Lagrangian, doing all the details... (a,b,sigma,tau,alpha,beta,mu,nu,rho,lambda,epsilon) [lol]

Also never had any 'symmetry problems', as he says in the HINT.

Hints to make things even simpler are always appreciated :)

I really don't want to do things like 'contravariant ##\leftrightarrow## covariant'.
(Staying consistent with the Langrangian derivative)

I really hope I didn't forget any product or chain rules...

Thank you in advance.
 
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  • #2
ProfDawgstein said:
Now after doing lots of product rules and triple ##\delta##'s I can't get 2nd order derivatives...

[Doing: ##\frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})}##]

Replacing the ##\partial_a h^{bc}## with ##\delta \delta \delta##

I always have something like ##\eta \partial_{\alpha} h^{ab}##,

but never ##\eta \partial \partial h^{ab}##.

The 2nd derivatives come from the partial derivative in the E-L equation

$$ \underbrace{\frac{\partial}{\partial x^\mu}} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})} - \frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0.$$

I'm not sure if that is your main problem. I haven't gone through the algebra of this particular computation recently.
 
  • #3
fzero said:
The 2nd derivatives come from the partial derivative in the E-L equation

$$ \underbrace{\frac{\partial}{\partial x^\mu}} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})} - \frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0.$$

I'm not sure if that is your main problem. I haven't gone through the algebra of this particular computation recently.

Wow, the obvious again...
I have seen way too many indices today, thanks alot!
Kind of stressed to finish this, not my own book...

I will do the whole thing again tomorrow, and will post my results here.

Any tips/hints to make the calculation easier?
I think my starting point should be correct.

##\mathcal{L}_{(0)} = \frac{1}{4} \left( (\partial_\alpha h_{\mu\nu}) (\partial^\alpha h^{\mu\nu}) - 2 (\partial_\alpha h_{\mu\nu}) (\partial^\nu h^{\alpha\mu}) + 2 (\partial^\mu h_{\mu\nu}) (\partial^\nu \eta_{\alpha\beta} h^{\alpha\beta}) - (\partial_\nu \eta_{\alpha\beta} h^{\alpha\beta}) (\partial^\nu \eta_{\rho\sigma} h^{\rho\sigma}) \right)##

Could have pulled the ##\eta_{\alpha\beta}## out, though...

I would love to know a few more tricks for this type of calculations :P

Also, does it help / is it a good idea
-to lower / raise the indices to the same level? (but I don't want to insert too many ##\eta##'s)
-rewrite the euler equation to have 2 derivatives (1 with covariant and 1 with contravariant ##\partial##'s) to reduce the ##\eta##'s
-what about the ##h_{,\nu}## terms? any way to make that easier / better looking?

I want everything to be consistent, but not too complicated.

[I want to improve my 'style']
 
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  • #4
ProfDawgstein said:
I would love to know a few more tricks for this type of calculations :P

Also, does it help / is it a good idea
-to lower / raise the indices to the same level? (but I don't want to insert too many ##\eta##'s)
-rewrite the euler equation to have 2 derivatives (1 with covariant and 1 with contravariant ##\partial##'s) to reduce the ##\eta##'s
-what about the ##h_{,\nu}## terms? any way to make that easier / better looking?

I want everything to be consistent, but not too complicated.

[I want to improve my 'style']

You could certainly take the derivatives term-by-term. Perhaps one convenient way of doing the calculation, that would simplify taking derivatives of the traces, would be to write the Lagrangian in the form

$$ \mathcal{L}_{(0)} = C^{\alpha \mu \nu \beta \rho \sigma} \partial_\alpha h_{\mu\nu} \partial_\beta h_{\rho\sigma},$$

where ##C^{\alpha \mu \nu \beta \rho \sigma}## can be written as a sum of terms involving ##\eta##s. Because of the symmetry, you really only take one derivative here and then contract ##C^{\alpha \mu \nu \beta \rho \sigma}## and your ##\delta##s into a single factor of ##\partial_\mu \partial_\beta h_{\rho\sigma}##.
 
  • #5
Okay, after 1 hour of 'index porn' I got this result...

##\partial_\mu \partial^\mu h - \partial_\mu \partial_\nu h^{\mu\nu} = 0##

1) I treated ##h^{\mu\nu}## as symmetric
2) I contracted out the ##\eta##'s
3) Divided by the common factors (0.5 in mine, 4 in 3.38)

If I do the same for (3.38) I get exactly the same as mine.

Does anybody know what this equation means?
The first part looks like a wave equation...

It took me about 3 pages with 20 lines each.

What I did:
1) make all terms look like ##\partial_\mu h^{\alpha\beta}##
2) re-arrange -> ##\eta##'s in front
3) do the lagrange derivative
4) do the ##\delta## contractions
5) do the ##\eta## contractions
6) contract out the remaining ##\eta##'s
7) group terms
8) treat ##h^{\alpha\beta}## as symmetric
9) done

I don't know if it was allowed to treat ##h^{\alpha\beta}## as symmetric, if not the whole thing is a bit longer...

Without contracting out the ##\eta##'s and not treating ##h^{\alpha\beta}## as symmetric I get this (divided by 1/2)

##\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \eta_{\alpha\beta} \partial_\mu \partial_\nu h^{\nu \mu} + \partial_\alpha \partial_\beta h - \eta_{\alpha\beta} \partial_\mu \partial^\mu h = 0##

-----------------------------------------------------------------------------------

EDIT:

4cd9a6370e09dd01be98a59e8e80e232.png


It seems like I just derived ##R = 0##.
 
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  • #6
ProfDawgstein said:
Without contracting out the ##\eta##'s and not treating ##h^{\alpha\beta}## as symmetric I get this (divided by 1/2)

##\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \eta_{\alpha\beta} \partial_\mu \partial_\nu h^{\nu \mu} + \partial_\alpha \partial_\beta h - \eta_{\alpha\beta} \partial_\mu \partial^\mu h = 0##

This agrees with 3.38, so is the result that you wanted to obtain. The first equation seems to be a result of some incorrect manipulations. In particular, you seem to be setting

$$\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \partial_\alpha \partial_\beta h = 0$$

by symmetry, which is incorrect. Perhaps you can give a few more details on what makes you think these terms cancel amongst themselves.

4cd9a6370e09dd01be98a59e8e80e232.png


It seems like I just derived ##R = 0##.

That might be correct, but the Einstein field equation is

$$ R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = k T_{\mu\nu},$$

so you end up with the extra terms in 3.38.

For some sense of how this becomes a wave equation, you might try to read through the discussion of the de Donder gauge here. It might also be discussed in another section of O'Hanian.
 
  • #7
fzero said:
This agrees with 3.38, so is the result that you wanted to obtain. The first equation seems to be a result of some incorrect manipulations. In particular, you seem to be setting

$$\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \partial_\alpha \partial_\beta h = 0$$

by symmetry, which is incorrect. Perhaps you can give a few more details on what makes you think these terms cancel amongst themselves.



That might be correct, but the Einstein field equation is

$$ R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = k T_{\mu\nu},$$

so you end up with the extra terms in 3.38.

For some sense of how this becomes a wave equation, you might try to read through the discussion of the de Donder gauge here. It might also be discussed in another section of O'Hanian.

In this exercise ##T_{\mu\nu} = 0##.

##\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \eta_{\alpha\beta} \partial_\mu \partial_\nu h^{\nu \mu} + \partial_\alpha \partial_\beta h - \eta_{\alpha\beta} \partial_\mu \partial^\mu h = 0##

I contracted this one with ##\eta^{\alpha\beta}## and then after renaming some indices I get

##- \partial_\mu \partial^\mu h - \frac{1}{2} \partial_\mu \partial_\nu h^{\mu\nu} + \frac{3}{2} \partial_\mu \partial_\nu h^{\nu\mu} = 0##

The order of the ##\partial##'s shouldn't matter, so (##h_{\mu\nu} = h_{\nu\mu}##)

##\partial_\mu \partial^\mu h - \partial_\mu \partial_\nu h^{\mu\nu} = 0##
 
  • #8
ProfDawgstein said:
In this exercise ##T_{\mu\nu} = 0##.

##\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \eta_{\alpha\beta} \partial_\mu \partial_\nu h^{\nu \mu} + \partial_\alpha \partial_\beta h - \eta_{\alpha\beta} \partial_\mu \partial^\mu h = 0##

I contracted this one with ##\eta^{\alpha\beta}## and then after renaming some indices I get

##- \partial_\mu \partial^\mu h - \frac{1}{2} \partial_\mu \partial_\nu h^{\mu\nu} + \frac{3}{2} \partial_\mu \partial_\nu h^{\nu\mu} = 0##

The order of the ##\partial##'s shouldn't matter, so (##h_{\mu\nu} = h_{\nu\mu}##)

##\partial_\mu \partial^\mu h - \partial_\mu \partial_\nu h^{\mu\nu} = 0##


OK, that's fine, but you have to remember that when you take the trace of an equation, the result does not contain as much information as the original equation. 3.38 is a set of 10 independent equations (accounting for symmetry of indices, but not diffeomorphism invariance), while the trace is a single linear combination of these. It turns out that one can use diffeomorphism invariance to relate the system of equations to a single wave equation for 2 polarizations of the gravity wave, but the details are a bit more subtle (as you can see in that link) than just taking a trace.
 
  • #9
fzero said:
OK, that's fine, but you have to remember that when you take the trace of an equation, the result does not contain as much information as the original equation. 3.38 is a set of 10 independent equations (accounting for symmetry of indices, but not diffeomorphism invariance), while the trace is a single linear combination of these. It turns out that one can use diffeomorphism invariance to relate the system of equations to a single wave equation for 2 polarizations of the gravity wave, but the details are a bit more subtle (as you can see in that link) than just taking a trace.

Oh yes, I know.
I just wanted to see what I get.
Taking the trace / contracting with ##\eta## made it easier to compare both equations.

If you have ##h_{ab}## or ##h^{ab}## shouldn't matter, right?
It just has to be on the same levels as in the other equation.

At the moment I am just happy that I finished this one :D

--------------------------------------------------------------------

Could I also have done something like this?

##\frac{\partial (\partial_a h^{bc})}{\partial (\partial_\mu h^{\alpha\beta})} = \delta^{\mu}_{a} \delta^{b}_{\alpha} \delta^{c}_{\beta}##

##\frac{\partial (\partial^a h_{bc})}{\partial (\partial^\mu h_{\alpha\beta})} = \delta^{a}_{\mu} \delta^{\alpha}_{b} \delta^{\beta}_{c}##

I didn't because it's not consistent with the Euler-Lagrange derivative, so I made all terms to look like

##\partial_\mu h^{\alpha\beta}##

Is there a nice way to do the whole thing notationally correct but simpler?

-> less ##\eta## contractions

Having this

##\mathcal{L}_{(0)} = \frac{1}{4} \left( (\partial_\alpha h_{\mu\nu}) (\partial^\alpha h^{\mu\nu}) - 2 (\partial_\alpha h_{\mu\nu}) (\partial^\nu h^{\alpha\mu}) + 2 (\partial^\mu h_{\mu\nu}) (\partial^\nu \eta_{\alpha\beta} h^{\alpha\beta}) - (\partial_\nu \eta_{\alpha\beta} h^{\alpha\beta}) (\partial^\nu \eta_{\rho\sigma} h^{\rho\sigma}) \right)##

and doing this

##\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})} - \frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0##.

I need to insert like 10 ##\eta##'s to make that consistent with the EL derivative...

Could somebody show how to do this thing nicely, but keeping the whole thing together?

Thanks
 

FAQ: Linear Gravity Field - Lagrangian

What is a Linear Gravity Field?

A linear gravity field is a mathematical model that describes the gravitational forces acting on objects in a specific region. It assumes that the gravitational force varies linearly with distance from a central point, such as the center of a planet or other massive object.

What is the Lagrangian in a Linear Gravity Field?

The Lagrangian in a linear gravity field is a function that describes the total energy of a system, taking into account both kinetic and potential energy. It is used in the Lagrangian formalism, a mathematical framework for understanding the dynamics of a system, and is essential in solving problems related to the motion of objects in a linear gravity field.

How is the Lagrangian derived in a Linear Gravity Field?

The Lagrangian in a linear gravity field is derived using the principle of least action, which states that the path taken by a system between two points is the one that minimizes the total action (a measure of the system's energy). This principle is applied to the equations of motion in the linear gravity field to derive the Lagrangian.

What are the advantages of using the Lagrangian in a Linear Gravity Field?

Using the Lagrangian in a linear gravity field has several advantages. It allows for a more elegant and concise description of the system's dynamics, and it is also more generalizable to different types of coordinate systems. Additionally, the Lagrangian formalism allows for the use of powerful tools such as variational calculus to solve complex problems.

What are some real-world applications of Linear Gravity Field and Lagrangian?

Linear gravity fields and the Lagrangian formalism have numerous applications in physics and engineering. They are essential in understanding the motion of planets, satellites, and other celestial bodies in our solar system. They are also used in the design of spacecraft trajectories and in the study of celestial mechanics. In engineering, the Lagrangian is used in the design of control systems for airplanes, robots, and other complex systems.

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