# Equations of motion in curved spacetime

I'm trying to understand how to use covariant actions to derive equations of motion. A simple example would be the free scalar field $$S = \int\;d^4x\; \sqrt{-g} \left( -\frac{1}{2}\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2 \right)$$

Now from classical mechanics we know that the equations of motion pop out when we set the variation of the action to zero. How would we do that in this formulation?

The $$\nabla_\mu$$ s represent covariant derivatives!

What I have tried: Using the $$\partial_\mu \rightarrow\nabla_\mu$$ perscription the lagrangian of the free scalar field will be $$\mathcal{L} = \left( -\frac{1}{2}\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2 \right)$$ Hence the equations of motion will come from the modified E-L equation:

$$\frac{\partial\mathcal{L}}{\partial\phi}=\nabla_\mu\frac{\partial\mathcal{L}}{\partial(\nabla_\mu\phi)}$$ My question is basically how do we arrive to this equation from varying the action?

• Hi fielder, please read the meta on exercises so that you can adjust the question to the Physics SE standards. In particular, we need you to show some independent effort to solve the problem.
– Void
Nov 6 '19 at 19:18
• Although this is just an example I thought, I will add some more context and thoughts of mine. Nov 6 '19 at 19:29
• The Dirac 1975 booklet on basic GR should be helpful. Nov 6 '19 at 19:31
• The covariant derivative of a scalar field is identical to the usual partial derivatives. If the field is a real singlet (instead of a complex multi-field), then there isn't any gauge field coupled to it so $\nabla_{\mu} \phi \equiv \partial_{\mu} \phi$.
– Cham
Nov 6 '19 at 20:22
• Nov 6 '19 at 20:33

From Ref. 1, one sees that the Euler-Lagrange equations generalize to $$\frac{\partial \mathcal{L}}{\partial\phi}=\nabla_\mu\left(\frac{\partial\mathcal{L}}{\partial(\nabla_\mu\phi)}\right)$$ In general, when one does these things, one must make sure of two things: that the index placement is the same, and that the indexes are NOT the same in the derivative and the Lagrangian. Use the metric to raise and lower indexes as needed, and a simple relabeling will take care of the the second. Thus, I change your Lagrangian as such: $$\mathcal{L}=-\frac{1}{2}\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2\to -\frac{1}{2}\nabla_\alpha\phi\,g^{\alpha\beta}\nabla_\beta\phi-\frac{1}{2}m^2\phi^2$$ We also note that for an object with one index, $$\frac{\partial V_\alpha}{\partial V_\beta}=\delta^\beta_\alpha$$. Thus, your scalar Lagrangian becomes \begin{align} -m^2\phi&=-\frac{1}{2}\nabla_\mu\left(\frac{\partial}{\partial(\nabla_\mu\phi)}\left(\nabla_\alpha\phi\,g^{\alpha\beta}\nabla_\beta\phi\right)\right)\\ & =-\frac{1}{2}\nabla_\mu\left(g^{\alpha\beta}\delta^\mu_\alpha\nabla_\beta\phi+g^{\alpha\beta}\nabla_\alpha\phi\delta_\beta^\mu\right) \\ &=-\frac{1}{2}\nabla_\mu\left(\nabla^\mu\phi+\nabla^\mu\phi\right) \end{align} So, finally, we have $$\nabla^\mu\nabla_\mu\phi-m^2\phi=0$$ which is just the flat spacetime version generalized by the usual prescription.
I think it's more straightforward to vary the action with respect to $$\phi$$ and cancel surface terms. Also, check this: Derivation of Klein-Gordon equation in General Relativity