I have been trying to derive the Einstein equation from the Einstein-Hilbert action $$ S[g_{\mu \nu}] = \frac{1}{16 \pi} \int_M \text{d}^4x \sqrt{-g}R $$ The standard derivation states that the variation $\delta S =0$ when we vary the metric components. In this derivation, we use the fact that $\delta (\sqrt{-g}R)= \delta \sqrt{-g} R + \sqrt{-g}\delta R$. To me this seems quite obvious but I thought I would try and prove this.

My understanding is that if we have an action $$ S[q] = \int \text{d}t L(q,\dot{q},t)$$ we vary it as $$ \delta S[q] = \int \text{d}t \delta L(q,\dot{q},t)$$ so $$ \delta L= \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q} \\ = \left( \frac{\partial L}{\partial q} - \frac{\text{d}}{\text{d}t} \frac{\partial L}{\partial \dot{q}} \right) \delta q + \frac{\text{d}}{\text{d}t} \left( \frac{\partial L}{\partial \dot{q}} \delta q \right)$$ As we integrate over $\delta L$, we ignore the last term as this produces a boundary term which we can take to vanish, therefore I say

$$ \delta L = \left( \frac{\partial L}{\partial q} - \frac{\text{d}}{\text{d}t} \frac{\partial L}{\partial \dot{q}} \right) \delta q $$

Okay, now let's say my Lagrangian is a product: $L(q,\dot{q},t) = f(q,\dot{q},t)g(q,\dot{q},t)$. Plugging this into the above formula for the variation, I have

$$ \frac{\partial L}{\partial q} = \frac{\partial f}{\partial q} g + \frac{\partial g }{\partial q} f $$

$$ \frac{\partial L}{\partial \dot{q}} = \frac{\partial f}{\partial \dot{q}}g + \frac{\partial g}{\partial \dot{q}} f$$

$$ \frac{\text{d}}{\text{d}t} \frac{\partial L}{\partial \dot{q}} = \left( \frac{\text{d}}{\text{d}t} \frac{\partial f}{\partial \dot{q}} \right) g + \left( \frac{\text{d}}{\text{d}t}\frac{\partial g}{\partial \dot{q}} \right) f + \frac{\partial f}{\partial \dot{q}} \frac{\text{d}g}{\text{d}t} + \frac{\partial q}{\partial \dot{q}} \frac{\text{d}f}{\text{d}t}$$

so the variation of this Lagrangian is

$$ \delta L = \left( \frac{\partial f}{\partial q} - \frac{\text{d}}{\text{d}t} \frac{\partial f}{\partial \dot{q}} \right)g \delta q + \left( \frac{\partial g}{\partial q} - \frac{\text{d}}{\text{d}t} \frac{\partial g}{\partial \dot{q}} \right)f \delta q - \frac{\partial f}{\partial \dot{q}} \frac{\text{d}g}{\text{d}t}\delta q - \frac{\partial q}{\partial \dot{q}} \frac{\text{d}f}{\text{d}t}\delta q$$


$$ \delta L = g \delta f + f \delta g - \frac{\partial f}{\partial \dot{q}} \frac{\text{d}g}{\text{d}t}\delta q - \frac{\partial g}{\partial \dot{q}} \frac{\text{d}f}{\text{d}t}\delta q $$

Now I can't seem to get rid of those horrible extra terms - I can't see how they would produce a boundary term when integrated. Maybe my understanding was incorrect and variations do not obey the product rule? Many standard resources suggest that varying the Einstein-Hilbert action obeys the product rule... is this just an exception?

My question:

How can I show that variations obey $\delta (fg) = f \delta g + g \delta f$


It is true that $$ \delta L = f \delta g + g \delta f,$$ but the expressions for variations are: \begin{align} \delta f&=\frac{\partial f}{\partial q} \delta q+ \frac{\partial f}{\partial \dot q}\delta \dot q = \left[ \frac{\partial f}{\partial q} - \frac{d}{dt}\left(\frac{\partial f}{\partial \dot q} \right) \right] \delta q \ + \ \frac{d}{dt}\left( \frac{\partial f}{\partial \dot q} \delta q\right)\\ \delta g&=\frac{\partial g}{\partial q} \delta q+ \frac{\partial g}{\partial \dot q}\delta \dot q = \left[ \frac{\partial g}{\partial q} - \frac{d}{dt}\left(\frac{\partial g}{\partial \dot q} \right) \right] \delta q \ + \ \frac{d}{dt}\left( \frac{\partial g}{\partial \dot q} \delta q\right)\\\delta L&=\frac{\partial L}{\partial q} \delta q+ \frac{\partial L}{\partial \dot q}\delta \dot q = \left[ \frac{\partial f}{\partial q}g+\frac{\partial g}{\partial q}f - \frac{d}{dt}\left(\frac{\partial f}{\partial \dot q}g + \frac{\partial g}{\partial \dot q}f \right) \right] \delta q \ + \ \frac{d}{dt}\left( \frac{\partial f}{\partial \dot q}g \delta q + \frac{\partial g}{\partial \dot q}f \delta q\right)\end{align} In each case, the variation is equal to the Euler-Lagrange equation times $\delta q$, plus a total derivative. The total derivatives are important!

  • $\begingroup$ The OP wants to know why $\delta(fg) = f\delta g + g\delta f$. You just cannot start with "it is true that $\delta(fg) = f\delta g + g\delta f$" $\endgroup$ – caverac Dec 27 '18 at 23:59
  • $\begingroup$ @caverac I think the OP wants to verify that $\delta(fg) = f\delta g + g \delta f$ holds in the example given. And this is manifestly true from the expressions for $\delta f$, $\delta g$ and $\delta L$ that I wrote down. $\endgroup$ – Kenny Wong Dec 28 '18 at 0:01
  • $\begingroup$ @KennyWong Yes this is what I was looking for, thank you. $\endgroup$ – Matt0410 Dec 28 '18 at 0:12

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