# Product rule of variations

I am deriving the Einstein equation using the Einstein-Hilbert action:

It is obvious that the variation in the Riemann Tensor is calculated from a variational product rule. What is not obvious to me is why variations obey this rule, and I'll like an explanation.

• Elementary answer: Variations are derivatives of 1-parameter families of functions, thus they follow the product rule. Functional answer: Variations are exterior derivatives on the covariant phase space, and as such they follow the (anti-)product rule. Differential geometric answer: Variations are Lie derivatives on a jet bundle, and as such they follow the product rule. Nov 4, 2019 at 20:22

The easiest way to go about this is to think of variation in terms of the background field method. Here, you write the metric as some background plus perturbation:

$$$$g_{\mu \nu} = \bar{g}_{\mu \nu} + h_{\mu \nu}$$$$

and then you expand every quantity that depends on $$g_{\mu \nu}$$ as a power series in $$h$$:

$$\Gamma = \bar{\Gamma} + \Gamma^{(1)} + \Gamma^{(2)} \cdots \\ R = \bar{R} + R^{(1)} + R^{(2)} \cdots$$

where the superscript $$^{(n)}$$ labels the $$\mathcal{O} (h^n)$$ perturbation. You can then easily see that

$$\Gamma \Gamma = (\bar{\Gamma} + \Gamma^{(1)} + \Gamma^{(2)} \cdots)(\bar{\Gamma} + \Gamma^{(1)} + \Gamma^{(2)} \cdots)$$

Extracting the $$\mathcal{O}(h)$$ variation, like in your problem, we automatically get both the contributions:

$$(\Gamma \Gamma)^{(1)} = \bar{\Gamma} \Gamma^{(1)} + \Gamma^{(1)} \bar{\Gamma}$$

It's a fundamental property of infinitesimal variations $$\delta$$ that they are linear derivations, i.e. they obey Leibniz product rule. Short of a rigorous definition of infinitesimals, the proof is basically to consider variations as 1-parameter families of functions, and then differentiate wrt. the parameter, cf. above comment by Bence Racsko.