EoM via varying action - covariant derivative when integrate

In summary, if given an action with an integration factor to keep the integral invariant, integrating by parts with this factor will always result in a covariant derivative, which can be traded for a partial derivative in the case of a scalar density. This is because the covariant divergence of a scalar density is equal to the partial divergence. However, if there are additional indices contracted between the tensors in the integrand, the covariant derivative must be present from the beginning for the expression to be coordinate invariant. The number of indices cannot be the same in the integrand or the integral must be taken over a hypersurface.
  • #1
1,251
11
##\int d^4 x \sqrt {g} ... ##

if I am given an action like this , were the ##\sqrt{\pm g} ## , sign depending on the signature , is to keep the integral factor invariant, when finding an eom via variation of calculus, often one needs to integrate by parts. When you integrate by parts, with this integration factor, is it now, always, the covariant derivative, replacing the partial derivative as in normal calculus when integrating by parts ? ..
 
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  • #2
I'm not sure I get the question .but by definition the covariant divergence of a scalar density equals the partial divergence. So one can trade partial for covariant derivatives and vice versa in that case.
 
  • #3
If I am understanding the question correctly, you are looking for something of the form
$$
\int d^4x \sqrt{g} A^\mu \partial_\mu f,
$$
where you want to do partial integration. Partial integration will always give you
$$
- \int d^4x f \partial_\mu(\sqrt{g} A^\mu),
$$
which you can rewrite
$$
- \int d^4x \sqrt{g} f \left[ A^\mu \frac{1}{\sqrt{g}}\partial_\mu\sqrt{g} + \partial_\mu A^\mu\right].
$$
Noting that ##\Gamma^\nu_{\mu\nu} = \partial_\mu \ln(\sqrt{g})##, it follows that
$$
\int d^4x \sqrt{g} A^\mu \partial_\mu f = - \int d^4x \sqrt{g} f [A^\mu \Gamma^\nu_{\mu\nu} + \partial_\mu A^\mu]
= - \int d^4x \sqrt{g} f \nabla_\mu A^\mu.
$$
If this is what you mean by having to use the covariant derivative (keeping the ##\sqrt{g}## factor out of it), then yes.

Note that if you have other indices contracted inside ##f## or ##A^\mu## they will not matter. If you have additional indices that are contracted between ##f## and ##A##, then you should have the covariant derivative from the beginning or your expression would not be coordinate invariant. The integration by parts for those indices does not change the sign of the corresponding Christoffel symbol - which gives the relative sign between the terms that you need to go from the covariant derivative of one part to the covariant derivative of the other. For example (just writing out the integrand apart from ##\sqrt g##)
$$
A^\mu_\nu \nabla_\mu f^\nu =
A^\mu_\nu (\partial_\mu f^\nu + \Gamma_{\mu\lambda}^\nu f^\lambda) \to
(-\partial_\mu A^\mu_\nu - \Gamma^\mu_{\mu\lambda} A^\lambda_\nu + \Gamma_{\mu\nu}^\lambda A^\mu_\lambda) f^\nu
= -(\nabla_\mu A^\mu_\nu) f^\nu.
$$
As before, the ##\Gamma_{\mu\lambda}^\lambda## came from the differentiation of ##\sqrt{g}## after the partial integration. The second term was not partially integrated (it contains no derivative) and therefore did not switch sign.
 
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Likes binbagsss
  • #4
binbagsss said:
were the ##\sqrt{\pm g}## , sign depending on the signature

If you mean the signature convention you adopt for the metric (timelike ##+---## vs. spacelike ##-+++##), it doesn't. For a Lorentzian spacetime, the determinant ##g## of the metric is always negative, so you always have ##\sqrt{-g}##.
 
  • #5
PeterDonis said:
If you mean the signature convention you adopt for the metric (timelike ##+---## vs. spacelike ##-+++##), it doesn't. For a Lorentzian spacetime, the determinant ##g## of the metric is always negative, so you always have ##\sqrt{-g}##.
To be a little more specific, in an even-dimensional Lorentzian manifold, it does not depend on the signature convention.
 
  • #6
Orodruin said:
Note that if you have other indices contracted inside ##f## or ##A^\mu## they will not matter. If you have additional indices that are contracted between ##f## and ##A##, then you should have the covariant derivative from the beginning or your expression would not be coordinate invariant. .
Thank you I think your post helped alot
But why is this ?
 
  • #7
binbagsss said:
Thank you I think your post helped alot
But why is this ?
Why is what? The case of indices contracted between ##f## and ##A^\mu##? I thought my example with a single index contracted between them would address that. More indices contracted between them are just writing more terms with Christoffel symbols.
 
  • #8
Orodruin said:
Why is what? The case of indices contracted between ##f## and ##A^\mu##? I thought my example with a single index contracted between them would address that. More indices contracted between them are just writing more terms with Christoffel symbols.
no , sorry, the coordinate invariant part
 
  • #9
PeterDonis said:
If you mean the signature convention you adopt for the metric (timelike ##+---## vs. spacelike ##-+++##), it doesn't. For a Lorentzian spacetime, the determinant ##g## of the metric is always negative, so you always have ##\sqrt{-g}##.
haha ofc, apologies !
 
  • #10
That is no more difficult than the partial derivative of a tensor component not having the correct transformation properties. You need the covariant derivative for that.
 
  • #11
Orodruin said:
$$
A^\mu_\nu \nabla_\mu f^\nu =
A^\mu_\nu (\partial_\mu f^\nu + \Gamma_{\mu\lambda}^\nu f^\lambda) \to
(-\partial_\mu A^\mu_\nu - \Gamma^\mu_{\mu\lambda} A^\lambda_\nu + \Gamma_{\mu\nu}^\lambda A^\mu_\lambda) f^\nu
= -(\nabla_\mu A^\mu_\nu) f^\nu.
$$

I just came across this and it was extremely useful but I'm having an issue when the number of indices on the two tensors is the same. For example, (in the integral)
\begin{align*}
\sqrt{g}A^{\mu}\nabla_{\nu}B^{\rho} &= \sqrt{g}A^{\mu}\partial_{\nu}B^{\rho} + \sqrt{g}A^{\mu}\Gamma_{\nu\sigma}^{\sigma}B^{\rho} \\
&= -\sqrt{g}\left( \frac{1}{\sqrt{g}}\partial_{\nu}\sqrt{g}A^{\mu} + \partial_{\nu}A^{\mu} - A^{\mu}\Gamma^{\sigma}_{\nu\sigma} \right)B^{\rho}\\
&= -\sqrt{g}\left( A^{\mu}\Gamma^{\sigma}_{\nu\sigma} + \partial_{\nu}A^{\mu} - A^{\mu}\Gamma^{\sigma}_{\nu\sigma} \right)B^{\rho}\\
&= -\sqrt{g}(\partial_{\nu}A^{\mu})B^{\rho}
\end{align*}

When, of course, I would expect

$$
\sqrt{g}A^{\mu}\nabla_{\nu}B^{\rho} = -\sqrt{g}(\nabla_{\nu}A^{\mu})B^{\rho}
$$

Can you please explain what I'm missing or doing wrong?
 
  • #12
You cannot have the same number of indices as your integrand is not a scalar then. Alternatively you have to integrate over a hypersurface instead but it is not clear to me exactly what you want to do.
 
  • #13
Orodruin said:
You cannot have the same number of indices as your integrand is not a scalar then. Alternatively you have to integrate over a hypersurface instead but it is not clear to me exactly what you want to do.
That makes a lot of sense, thanks. I was just thinking about it in a general sense and honestly just failed to consider that, but I now realize that was a foolish oversight. Thanks a lot!
 

1. What is the EoM (equation of motion) via varying action?

The EoM via varying action is a mathematical framework used to describe the behavior of a physical system. It is based on the principle of least action, which states that the true path of a system is the one that minimizes the action (a mathematical quantity representing the energy of the system) over a given period of time.

2. What is a covariant derivative?

A covariant derivative is a mathematical operation used in differential geometry to describe how a vector field changes as it moves along a curved surface. It takes into account the effects of the curvature of the surface on the vector field.

3. How is the covariant derivative used in the EoM via varying action?

The covariant derivative is used in the EoM via varying action to account for the effects of the curvature of spacetime on the behavior of a physical system. It allows for the formulation of a covariant equation of motion that is valid in any coordinate system.

4. What is the process of integration in the EoM via varying action?

The process of integration in the EoM via varying action involves finding the path of least action for a given system by varying the action with respect to the system's variables. This results in a set of differential equations that can be solved to obtain the equations of motion for the system.

5. How does the EoM via varying action differ from other approaches to describing physical systems?

The EoM via varying action differs from other approaches, such as the Lagrangian or Hamiltonian formalisms, in that it takes into account the effects of the curvature of spacetime on the behavior of a system. It also allows for a more general and covariant description of the system, making it applicable in any coordinate system.

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