How Do You Approach the Variational Problem in GR from Misner's Exercise 7.1?

  • Thread starter Thread starter Karl G.
  • Start date Start date
  • Tags Tags
    Gr
Karl G.
Messages
40
Reaction score
0

Homework Statement


From Gravitation by Misner, et al. Can anybody who has access to this text show me how to vary this functional from exercise 7.1, and using the principle of least value, derive an identity? The functional is I = \int[/L d[4][/SUP]]x, where L = -1/(\pi*8*G)*\eta<sup>[/\alpha\beta](\Phi)<sub>[/,\alpha](\Phi)<sup>[/,\beta] - \int m (e^\Phi) \delta(<b>[/x - z](\tau)) d\tau Vary with respect to \Phi. I apologize in advance for notation.<br /> <br /> <br /> <br /> <h2>Homework Equations</h2><br /> Euler- Lagrange Eq&#039;ns. I know variational methods, but this one perplexes me.<br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2></b></sup></sub></sup>
 
Physics news on Phys.org
OK, so the functional integral you mean is:

I = \int \mathcal{L} \textrm{d}^4x

where

\mathcal{L} = -\frac{1}{8\pi G}\eta^{\alpha\beta}\frac{\partial \Phi}{\partial x^\alpha}\frac{\partial \Phi}{\partial x^\beta} -m \int e^{\Phi} \delta^{(4)}(\mathbf{x}-\mathbf{x}(\tau))\textrm{d}^4 x

We want to know what the equations of motions are when we vary \Phi. The easiest way to do this is to use the Euler-Lagrange equations. For fields, these equations read

\partial_\alpha \frac{\delta \mathcal{L}}{\delta(\partial_\alpha \Phi)} -\frac{\delta \mathcal{L}}{\delta \Phi}= 0

The first term (of the E.L.) gives:
\partial_\alpha \frac{\delta \mathcal{L}}{\delta(\partial_\alpha \Phi)} = \frac{1}{4\pi G}\partial^\alpha\partial_\alpha \Phi

The second term gives:

-\frac{\delta \mathcal{L}}{\delta \Phi} = m \int e^{\Phi} \delta^{(4)}(\mathbf{x}-\mathbf{x}(\tau))\textrm{d}^4 x

Which is probably more than enough info you need... I think?
 
Yes, thanks, sorry for all the inconvenience you may have experienced with the horrid notation I used
 
One more question (sorry!): Is there a way to simplify the last integral with the delta function? I'm not sure how you would do it with a 4-d integral.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top