# A Equation governing the evolution of the scalar field

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1. Apr 11, 2016

### valesdn

I'm looking for a demonstration of the equation governing the evolution of the scalar field:

$\Box \phi = \frac{1}{\sqrt{g}} \frac{ \partial}{\partial x^{\mu}} \sqrt(g)g^{(\mu)(\nu)} \frac{\partial}{\partial x^{\nu}} \phi=0$

I used the lagrangian for a scalar field: $L = \nabla_{\mu}\phi \nabla_{\nu} \phi g^(\mu \nu)$

The density lagrangian is $L= \sqrt(-g) R$, where R is the curvature scalar.
The variation of the Einstein-Hilbert action can be write as follows:

$\delta S= \delta \int (\sqrt (-g) R d^4 x= \int d^4 x \delta (\sqrt(-g)g^{ab}R_{ab}= \int d^4 x \sqrt(-g)g^{ab} \delta R_{ab} + \int d^4 x \sqrt(-g) R_{ab} \delta g^{ab} + \int d^4 x R \delta (\sqrt(-g))$

I've calculated the single terms, respectively.
I found:

- for the first term $\int d^4 x \sqrt(-g)g^{ab} \delta R_{ab}$:

$\int d^4 x \sqrt(-g)g^{ab} \delta R_(ab)=$ $\int d^4 x \sqrt(-g) \nabla_c$ $[g^{ab}$ $\delta (\Gamma^c)_{ab} - g^{ac} \delta$ $(\Gamma ^{b})$$_{ab}$]

I can't calculate the second term $\int d^4 x \sqrt(-g) R_{ab} \delta g^{ab}$.

The third term $\int d^4 x R \delta (\sqrt(-g))$ is equal to $- \frac{1}{2} \int d^4 x R \sqrt(-g) g_{ab} \delta (g^{ab})$.

So, I have $\delta S =$ $\int d^4 x \sqrt(-g) \nabla_c$ $[g^{ab}$ $\delta (\Gamma^c)_{ab} - g^{ac} \delta$ $(\Gamma ^{b})$$_{ab}$] + $\int d^4 x \sqrt(-g) R_{ab} \delta g^{ab}$ + $- \frac{1}{2} \int d^4 x R \sqrt(-g) g_{ab} \delta (g^{ab})$

Is it correct this calculations to find the equation for a scalar field? Is there an other way to find it? How can I demonstrate this equation?
Thank you in advance for your help.

2. Apr 14, 2016

### haushofer

If you vary the action wrt the metric, you will find the EOM for the metric, i.e. the Einstein equations. That's what you derived now, modulo a total divergence. If you want to derive the EOM of a scalar field, you should use your Lagrangian density of the scalar field, not the Einstein-Hilbert Lagrangian. The coupling between the metric and the scalar field will give you, upon varying wrt the metric, the energy-momentum tensor for the scalar field.

I have the idea that you are very confused about what you're doing in the first place. Maybe you should first read a decent text about the whole idea of the action formalism. See e.g. Zee's GR book.

3. Apr 14, 2016

### vanhees71

Indeed, you have to vary the action with respect to the scalar field to get the equations of motion for the field. The variation with respect to $g_{\mu \nu}$ gives (in addition to the Einstein tensor from the Einstein-Hilbert action) the Belinfante energy-momentum tensor of the scalar field as a source for the pseudo-metric (i.e., the gravitational field). Together you have a self-consistent set of equations of motion for the gravitational field and the scalar field.

4. Apr 17, 2016

### valesdn

I just want to thank you, haushofer and vanhees71, for your help. To answer you, haushofer: yes, I am "just a little bit" confused!
I'm trying to demonstrate the equation of motion for a scalar field in this way ( as you have suggested): I wrote the lagrangian for a scalar field $$\phi$$

$$L= \frac{1}{2} g^{\mu \nu} (\nabla_{\nu} \phi) (\nabla_{\mu} \phi)$$

However, for a scalar field, I have: $$\nabla -> \partial$$
The action for a scalar field is $$S = \int d^4 x \sqrt {-g} L$$
I need to calculate the variation of this action to find the equation governing the evolution of a scalar field. This means

$$\delta S = \delta \int d^4 x \sqrt (-g) g^{ \mu \nu} \partial _ {\mu} \phi \partial_{\nu} \phi$$

The principle of minimal coupling states that

$$\delta S = 0$$

It implies that $$\partial _{\mu} \frac{ \partial \sqrt{-g}L}{\partial (\partial_{\mu} \phi)} - \frac{\partial \sqrt{-g}L}{\partial \phi}=0$$
So

$$\partial_{\mu} \frac{\partial \sqrt{-g}[ \frac{1}{2} g^{\mu \nu}(\partial_{\nu} \phi) (\partial_{\mu} \phi)]}{\partial (\partial_{\mu} \phi)}- \frac{\partial \sqrt{-g} \frac{1}{2} g^{\mu \nu}(\partial_{\nu} \phi)(\partial_{\mu} \phi)}{\partial \phi}=0$$

I find
$$\partial_{\mu} \sqrt{-g}[ \frac{1}{2} g^{\mu \nu}(\partial_{\nu} \phi)] =0$$

But $$\partial_{\mu} \sqrt{-g}[ \frac{1}{2} g^{\mu \nu}(\partial_{\nu} \phi)= (\partial_{\mu} \sqrt{-g}) g^{\mu \nu}(\nabla_{\nu} \phi)+\sqrt(g)[(\partial_{\mu}g^{\mu \nu})(\nabla_{\nu}\phi)]= \Gamma ^{\rho}_{\rho \mu} \sqrt{-g} g^{\mu \nu} (\nabla_{\nu} \phi)- \Gamma^{\rho}_{\rho \mu} \sqrt{g} g^{\mu \nu} (\nabla_{\nu} \phi)= \sqrt{-g} \nabla_{\mu} [g^{\mu \nu}(\nabla_{\nu} \phi)= \nabla_{\mu}[\sqrt{-g}g^{\mu \nu}(\nabla_{\nu} \phi)]= \sqrt{-g}g^{\mu \nu} \nabla_{\mu}(\nabla_{\nu} \phi)= \sqrt{-g} \Box \phi$$

So, finally, I have

$$\Box \phi=0$$

Is it correct?
Thank you in advance for your help and suggestions.

Last edited: Apr 17, 2016
5. Apr 17, 2016

### ChrisVer

Hmmm I can't see why you can write the covariant derivaive as partial derivative for the scalar field...but I think you can do that [it's a scalar afterall].
the result is right....
Also, does it help you to carry so many indices in the action? I found it always easier to keep a Lagrangian until reaching the EoM...

Last edited: Apr 17, 2016
6. Apr 17, 2016

### JorisL

In Carroll's book the fact that the covariant derivative reduces to a partial derivative is one of the assumptions made when introducing covariant derivatives. (Relevant chapter of online notes)
A corollary of this assumption together with the fact that the derivative commutes with a contraction is that in that case the connection is the same for both kinds of indices (upper and lower).

The second assumption can be replaced by metric compatibility if I'm not overlooking anything.