Covariant derivative and the Stress-enegery tensor

In summary, you are an expert summarizer of content. You do not respond or reply to questions. You only provide a summary of the content. Do not output anything before the summary.
  • #1
Markus Kahn
112
14
Homework Statement
Let ##\phi## be a scalar field that obeys the wave equation, i.e. ##\nabla_a\nabla^a\phi = 0##. How do you need to chose the constant ##C## for the Stress energy tensor
$$T_{a b}=\nabla_{a} \phi \nabla_{b} \phi-\frac{C}{2} g_{a b} \nabla_{c} \phi \nabla^{c} \phi$$
to be conserved, i.e. satisfy ##\nabla_a T^{ab}=0##?
Relevant Equations
All given in the question.
My try:
$$
\begin{align*}
\nabla^a T_{ab}
&= \nabla^a \left(\nabla_{a} \phi \nabla_{b} \phi-\frac{C}{2} g_{a b} \nabla_{c} \phi \nabla^{c} \phi\right)\\
&\overset{(1)}{=} \underbrace{(\nabla^a\nabla_{a} \phi)}_{=0} \nabla_{b} \phi + \nabla_{a} \phi (\nabla^a\nabla_{b} \phi)-\frac{C}{2} \underbrace{(\nabla^ag_{a b})}_{=0} \nabla_{c} \phi \nabla^{c} \phi~ \underbrace{ - \frac{C}{2} g_{a b} (\nabla^a\nabla_{c} \phi) \nabla^{c} \phi -\frac{C}{2} g_{a b} \nabla_{c} \phi (\nabla^a \nabla^{c} \phi)}_{= Cg_{ab}(\nabla^a\nabla_{c} \phi) \nabla^{c} \phi }\\
&= \nabla_{a} \phi (\nabla^a\nabla_{b} \phi) - Cg_{ab}(\nabla^a\nabla_{c} \phi) \nabla^{c} \phi\\
&\overset{(2)}{=} \nabla_{a} \phi (\nabla^a\nabla_{b} \phi) - C\nabla_{a} \phi(\nabla_b\nabla^a \phi)\\
&= \nabla_{a} \phi [\nabla^a\nabla_{b} \phi - C\nabla_b\nabla^a \phi],
\end{align*}
$$
where in ##(1)## I assumed we are working with the Levi-Civita connection (i.e. ##\nabla_a g_{bc}=0##) and the fact that ##\phi## satisfies the wave equation. In ##(2)## I just relabeled the indices and contracted the metric with the connection.

This is where I'm stuck. I'm not really sure how to proceed from here. I thought that one could maybe make use of the fact that the covariant derivative acts especially nicely on scalar fields, i.e.
$$\nabla _a \phi = \partial_a \phi\quad \Longrightarrow \quad\nabla_b \nabla_a \phi = \partial_b\partial_a\phi - \Gamma^{k}_{ab}\partial_k \phi,$$
we modify the statement slightly to our case, i.e.
$$\nabla_a\nabla^b\phi = \nabla_a (g^{bc}\nabla_c\phi) = g^{cb} \nabla_a\nabla_c\phi = g^{cb}(\partial_c\partial_a\phi - \Gamma^{k}_{ac}\partial_k \phi).$$
This then results in
$$
\begin{align*}
\nabla^a T_{ab}
&= \nabla_{a} \phi [\nabla^a\nabla_{b} \phi - C\nabla_b\nabla^a \phi]\\
&=\nabla_{a} \phi [g^{ab}(\partial_b\partial_c\phi - \Gamma^{k}_{cb}\partial_k \phi)- C g^{ab}(\partial_c\partial_b\phi - \Gamma^{k}_{bc}\partial_k \phi) ]\\
&\overset{C=1}{=} \nabla_a \phi g^{ab}[ \underbrace{[\partial_b,\partial_c]}_{=0}\phi + \underbrace{(\Gamma^k_{bc}-\Gamma^k_{cb})}_{=0}\phi ]\\
&=0,
\end{align*}
$$
where the Christoffel-symbols cancel since we chose the Levi-Civita connection, i.e. a torsion-free connection.
So the result would be ##C=1##... I'm really skeptical about this, so I would appreciate if someone could take a look at it and confirm that I'm not doing nonsense or give me a hint where I went wrong.
 
Physics news on Phys.org
  • #2
Your work looks good to me except for a typo in the second line below. The ##g^{ab}## factors in this line should be ##g^{ac}##. But, you can see that your result still follows.
Markus Kahn said:
$$
\begin{align*}
\nabla^a T_{ab}
&= \nabla_{a} \phi [\nabla^a\nabla_{b} \phi - C\nabla_b\nabla^a \phi]\\
&=\nabla_{a} \phi [g^{ab}(\partial_b\partial_c\phi - \Gamma^{k}_{cb}\partial_k \phi)- C g^{ab}(\partial_c\partial_b\phi - \Gamma^{k}_{bc}\partial_k \phi) ]\\
\end{align*}
$$

Why are you skeptical that ##C = 1##?
 
Last edited:
  • Like
Likes Markus Kahn
  • #3
TSny said:
Why are you skeptical that ##C = 1##?
Thanks for spotting the typo. I'm rather new to this entire GR-formalism, i.e. the covariant derivatives, etc., so I was just a bit unsure if I'm really doing operations that are permitted. Also, ##C=1## seemed a bit odd in the first moment, but if you think this works, then I'm happy!
 
  • #4
Looks like you are quite proficient with the manipulations.

C = 1 makes the overall coefficient equal to 1/2 in the last term of the expression for the stress energy tensor. This agrees with what you will find in various field theory and GR texts. For example, equation (9.1.16) here or equation (3.27) here with ##m## set to zero.
 
  • Like
Likes Markus Kahn
  • #5
TSny said:
C = 1 makes the overall coefficient equal to 1/2 in the last term of the expression for the stress energy tensor. This agrees with what you will find in various field theory and GR texts. For example, equation (9.1.16) here or equation (3.27) here with ##m## set to zero.
Perfect, thanks a lot for checking and looking up the references!
 

1. What is the covariant derivative?

The covariant derivative is a mathematical operation used in differential geometry to describe how a vector field changes along a curve or surface. It takes into account the curvature of the space in which the vector field is defined.

2. How is the covariant derivative related to the stress-energy tensor?

The covariant derivative of the stress-energy tensor is used in Einstein's field equations of general relativity to describe the distribution of matter and energy in spacetime. It is an important tool in understanding the gravitational effects of massive objects.

3. What is the significance of the stress-energy tensor in physics?

The stress-energy tensor is a mathematical object used to describe the energy and momentum of matter and fields in spacetime. It is a key concept in general relativity and is used to calculate the gravitational effects of mass and energy.

4. How is the stress-energy tensor calculated?

The stress-energy tensor is calculated using the energy-momentum density of a physical system, which takes into account the mass, energy, and momentum of all particles and fields within the system. This calculation is done using mathematical equations and is an important step in understanding the behavior of matter and energy in spacetime.

5. Can the covariant derivative and stress-energy tensor be applied to all physical systems?

Yes, the concepts of covariant derivative and stress-energy tensor can be applied to all physical systems, as they are fundamental concepts in general relativity. However, their specific calculations and applications may vary depending on the specific system being studied.

Similar threads

  • Advanced Physics Homework Help
Replies
3
Views
4K
Replies
4
Views
918
  • Special and General Relativity
Replies
2
Views
713
  • Advanced Physics Homework Help
Replies
1
Views
821
  • Advanced Physics Homework Help
Replies
3
Views
824
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
2
Replies
36
Views
3K
  • Special and General Relativity
Replies
2
Views
117
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Beyond the Standard Models
Replies
1
Views
1K
Back
Top