Question on Phenomenology of Einsteins Field Equations

[robousy]

I am only just starting to realize that there is a correspondence between quantum field theory and Einsteins field equations.

In QFT the approach is to write the Lagrangian and then to solve the Euler Lagrange equation to obtain the equations of motion of the field.

In GR it seems that the starting point is not the Lagrangian but instead the metric - but that the end result, the stress energy tensor, is again, the equations of motion of the field.

Now, in QFT the simplest lagrangian is for a scalar field.

Does one take a similar approach in GR. When one solves the metric what do the eqtns of motion correspond to - ie are they for a field, or a particle or both?

 

[pervect]

You can formulate relativity with a Lagrangian density using the Einstein-Hilbert action.

according to

http://www.lns.cornell.edu/spr/2000-05/msg0024719.html

the answer to your question is the Ricci curvature, plus whatever you need for the Lagrangian density of your "source" fields.

I'll quote the whole article,while it may be guilty of over-explaining, and it might come across as condescending, I think I've been way too terse above.

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Lagrangian formalism in general relativity

* Subject: Lagrangian formalism in general relativity
* From: baez@galaxy.ucr.edu
* Date: 05 May 2000 00:00:00 GMT
* Approved: baez@math.ucr.edu (John Baez)
* Newsgroups: sci.physics.research
* Organization: UCR
* References: <8c01h1$72o$1@nnrp1.deja.com>

>In article <8elmc9$io6$1@pravda.ucr.edu>, John Baez writes
<snip>

Sure. I seem to remember several other threads in which we tried
to explain the Lagrangian formalism to you. If you were just trying
to understand the motion of a single particle, the action would be
the integral of something called the Lagrangian over the worldline
of the particle. In classical mechanics, this Lagrangian is simply
the kinetic energy minus the potential energy! The particle then
cleverly decides to move in such a way that this action is minimized -
subject to the condition that the particle starts here and ends up there.

(More precisely, it finds a stationary point of the action! The particle
is not really smart enough to always find a minimum.)

But if you're talking about a field theory, like Maxwell's equations or
general relativity, the action is the integral over all of spacetime
of something called the Lagrangian density - or just "Lagrangian" when
you get lazy. Again, the fields cleverly wiggle in such a way as to
achieve a stationary point of the action.

Smart folks know that you can figure out everything very efficiently
as soon as you know the formula for the Lagrangian density. So: what
is it in general relativity? Simple: just the Ricci scalar curvature,
plus whatever Lagrangian density you need to describe the matter fields.

(In general relativity, "matter fields" refers to all fields other than
the metric on spacetime.)

What does all this mean? Very crudely speaking, the desire of spacetime
to find a stationary point of the action is very much like the desire of
a taut string to return to its original position when plucked. Spacetime
"likes to be flat" just as the string "likes to be straight". When you
pluck, it bounces back, sending out ripples of gravitational radiation
in all directions. The Lagrangian formalism makes this vague and seemingly
anthropomorphic idea precise and useful.

 

[robousy]

This is fascinating, thank-you.

So, let me check I have this correct - $$R_{\mu\nu}$$ is equivalent to $$L$$

That is the Ricci tensor is the lagrangian density.

...oh, and don't worry about 'over' explaining - that's exactly how I like it explained.

 

[pervect]

This is fascinating, thank-you.

So, let me check I have this correct - $$R_{\mu\nu}$$ is equivalent to $$L$$

That is the Ricci tensor is the lagrangian density.

...oh, and don't worry about 'over' explaining - that's exactly how I like it explained.

Looking this up in my textbook, it looks like it can be even simpler. It's the Ricci scalar that's the Lagrangian density when you define things right.


You get

${\cal{L}} = \sqrt{ -g } R$
You have to do some mucking around with volume elements and tensor densities to get this in this simple form - the sqrt term (which is the sqrt of the determinant of the metric) has to do with the issue of defining a volume element, and all tensor densities have to be proportional to $\sqrt{-g}$ to be tensor densities.

There's another approach called the Palanti action which works the way Baez described, too, but it is probably harder to explain.

Of course the vacuum Einstein equations aren't so interesting, unless you have fields in them to generate gravity. This however, requires more terms in the Lagrangian.

Something like

${\cal{L}} = -\frac{1}{4}\sqrt{-g} g^{ac} g^{bd} F_{ab}F_{cd}$
for the electromagnetic field, for instance.

 

[robousy]

Yes, I recognise the $$\sqrt g$$ term, I believe it corresponds to the Jocobian. Its like multiplying by $$r^2 sin\theta$$ when you work in spherical coordinates.

So I just want to sumamrize:

The Ricci scalar is equivalent to the lagrangian density

the stress energy tensor is equivalent to the equations of motion.

Now, what is the ricci tensor analagous to?

Also, is the stress energy tensor equivalent to the EOM in field theory for a scalar field or some other field?

Thanks!

 

[pervect]

Yes, I recognise the $$\sqrt g$$ term, I believe it corresponds to the Jocobian. Its like multiplying by $$r^2 sin\theta$$ when you work in spherical coordinates.

So I just want to sumamrize:

The Ricci scalar is equivalent to the lagrangian density

the stress energy tensor is equivalent to the equations of motion.

Now, what is the ricci tensor analagous to?

Also, is the stress energy tensor equivalent to the EOM in field theory for a scalar field or some other field?

Thanks!

$g^{ab}$ are your "field" variables, and you get R by taking

$R_{ab} g^{ab} = R^a{}_a = R$, so R_ab determines the mapping of the field variables to the action.

 

[robousy]

when you say field variables - is this equivalent to talking about a scalar field in QFT?

eg $$\phi \rightarrow \phi(x,t)$$ where x and t are you field variables.