Variation in the context of symmetries

Question

I’m rephrasing a suggestion as a question because there was an aspect to it where I wanted to know more as well.

I have studied both general relativity and particle physics, though in both cases my experience with theory ended around the time of my general exam in graduate school (because I switched fields, did data analysis, got sick, ended with a masters, various…)

I did a project based on finding an interaction cross section assuming certain symmetries many years ago. I don’t remember exactly what I did, but I think I enforced those symmetries prior to perturbing the first-order Feynman integral.

But I am now looking at the Einstein-Hilbert action and see that if the spacetime background is Schwarszchild, then $R=0$, and the action is zero. Obviously the equations of motion still exist.

What’s the difference here? Why can the relevant relationships between the form factors be derived (if I recall) in particle physics but not in this instance of general relativity, using symmetry constrained variations?

In the E-H action, obviously the geodesic equation can be derived and the Schwarzschild equations of motions computed, but it seems to be necessary to keep $R$ symbolic until the geodesic equation, at which point the Christofel symbols can be evaluated.

Why the difference? Or is my 15 year old memory or past work on the summer particle physics project likely just wrong?

Answer 1

The Einstein-Hilbert action is $$ S = \frac{1}{2\kappa} \int R \sqrt{-g} d^4x$$ with $\kappa$ the dimensionful constants.

If you set $R=0$ at the outset, you have the action $S = 0$, which is a different (and completely uninteresting) action. The whole point of the variational approach is to consider "various" values of the fields, being a bit loose with the language intentionally. You then set the variation, not the action itself, to 0 to find the equations of motion. It does turn out that some solutions have $R = 0$, as you know, but you compute that as part of the process. You don't put it in from the start.

This is directly analogous to taking regular derivatives to find a minimum. If you want to compute the minimum of $f(x) = x^2$, you take the derivative first and set it to 0. So we look for $x$ such that $f'(x) = 2x = 0$, which implies $x = 0$. You can then put that back into the original function to see what value $f$ has at the minimum. The specific value $f(0) = 0$, however, is constant. You never take the derivative of that constant at any point in the process.

Answer 2

It's hard to say for sure, since you don't remember the details, but usually we don't assume anything about the solutions to the equations of motion when deriving them (by computing the variation of the action). If you plug some field configuration into the action (Like setting $R=0$ in the E-H action), you are computing the action of this particular solution, which is something you can always do if you want to, but that has nothing to do with the fact that this particular field configuration happens to sit in a minimum of the action. For that you don't want the value of the action itself, you want the first derivative (the variation $\delta S$), which is what you set to zero to derive the equations of motion. Note that the symmetries of a classical spacetime are usually things like rotational symmetry $x \to Rx$, time translation symmetry $t \to t+a$, or discrete symmetries like time reversal $t \to -t$. These correspond to transformations of physical space. The action is not a function of spacetime, it is a function(al) of the fields, $S=S[g_{\mu \nu}]$ in the case of GR. So when varying the action, you are looking for the minimum in the "space of metrics", not physical space. The particular symmetries one solution might have does not enter this calculation, what chooses the symmetries (and the solution itself) are the initial or boundary conditions.

In quantum field theory the situation is a bit different, since you are not anymore looking for the classical solutions. Usually people want to compute correlation functions, like the following two-point function \begin{equation} \left\langle \phi(x) \phi(y) \right\rangle = \mathcal{N} \int \mathcal{D} \phi e^{iS[\phi]} \phi(x) \phi(y) \end{equation} of some theory with the field $\phi(x)$, where $\mathcal{N}$ is a normalization. The correlation functions are the basic ingredients for the calculation of scattering amplitudes and cross sections, so it is likely that you computed some, in whatever theory you were working with. These objects actually do inherit the symmetries of the action. For instance, if the action is translation-invariant, the same holds for the two-point function above, meaning that it can only depend on the difference $x-y$, and not on the value of each one separately: \begin{equation} \left\langle \phi(x) \phi(y) \right\rangle = f(x-y). \end{equation} Other symmetries can be similarly used to further constrain the correlation functions. Working in momentum space, if your fields have tensor indices and your action is Lorentz invariant, there can be enough symmetry to reduce the unknown part to some scalar "form factors", as you mentioned, each multiplied by some tensor built from combinations of the flat metric and the external momenta. Maybe this is what you were doing. Note however that using the symmetries of the action to simplify the correlation functions (and therefore the scattering amplitude) has nothing to do with plugging the symmetries into the action itself.