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Understanding the metric transformation under infinitesimal diffeomorphism

Physics Asked by ZacharyC on February 23, 2021

In my general relativity course, we are discussing infinitesimal diffeomorphisms defined by $x^{mu}rightarrow y^{mu}(x) = x^{mu} + xi^{mu}(x)$. We have been examining how different objects transform under this. For example, the metric transformation $g_{munu}(x)dx^{mu}dx^{nu} rightarrow g_{munu}(y) dy^mu dy^nu$ can be found via

$$g_{munu}(y) dy^mu dy^nu = g_{munu}(x+xi)frac{partial(x+xi)^mu}{partial x^alpha} frac{partial (x +xi)^nu}{partial x^beta} dx^alpha dx^beta tag{1} $$

From here, it is my understanding that we Taylor expand the expressions to get:

$$ = left( g_{munu}(x) + xi^alpha partial_alpha g_{munu}(x),+, … right)(delta^mu_alpha + partial_alpha xi^mu ) (delta^nu_beta + partial_beta xi^nu,) mathrm d{x^alpha} mathrm dx^beta tag{2}$$
$$= left( g_{munu}(x) +, xi^sigma partial_sigma g_{munu}(x) +, g_{musigma} partial_nu xi^sigma +, g_{nusigma} partial_mu xi^sigma,+,… right) mathrm dx^mu mathrm dx^nu tag{ 3}$$

where the dots are just higher order terms that we can ignore. From here, one can show that the metric varies with the Lie derivative in the direction of $xi^{mu}$.

I am really struggling to comprehend the mathematical calculation and simplification in this problem. My specific questions are:

Q1) From (1) to (2), how does one taylor expand an expression like $partial(x+xi)^mu / partial x^alpha$? I am having trouble understanding how one gets to the resulting expression.

Q2) From (2) to (3), I am completely lost on what was done. It seems like the kronecker delta’s were applied to $mathrm dx^alpha mathrm dx^beta$. I assume they would need to be factored out of the binomial expressions, but how does that affect the other terms?

Q3) I am often confused when new indices are relabeled. What is the motivation for introducing this sigma index?

Q4) On a more conceptual note, I understand that we want to use this result to show that the action is invariant under infinitesimal diffeomorphisms. But why look at infinitesimal diffeomorphisms rather than diffeomorphisms in general?

I apologize if much of this question is trivially mathematical. I debated putting it in the Math forum but figured I’d rather get a physicist’s answer.

One Answer

Q1) The derivative is a linear operator hence begin{equation} partial_alpha (x+ xi)^mu = partial_alpha x^mu + partial_alpha xi^mu = delta^mu_alpha + partial_alpha xi^mu end{equation} The statement that those 2 factors get Taylor expanded is actually wrong, only the first factor gets Taylor expanded.

Q2) To first order in the derivatives we have something like begin{equation} (g+A)(delta+B)(delta+C) = g delta delta + Adelta delta + gBdelta + gCdelta end{equation} A bit cryptic, but it gives you the four terms considered.

Q3) There isn't really a motivation, just know that you can always relabel whenever there are contraction $xi^alpha xi_alpha = xi^beta xi_beta$. Presumably they got rid of the different $alpha$'s and $beta$'s and replaced it with $sigma$'s.

Q4) For most groups used in physics the general symmetry transformation can be obtained from the infinitesimal transformation, called generators. Therefore, one looks at the transformation under these infinitesimal transformations and want them to be of a certain form which results in an equation to solve for these infinitesimal generators.
I think that the only necessary condition is that the group should be connected to the identity such that for generators $xi^A$ and parameters $epsilon_A$: begin{equation} g = e^{epsilon^A xi_A} end{equation} $A$ is not a spacetime label, but an internal index of the symmetry group. $g$ is the finite group element.

Correct answer by JulianDeV on February 23, 2021

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