Trying to prove a consequence of harmonic gauge in GR

In summary, the conversation is about the harmonic gauge condition and its equivalence to the condition for linearized gravity. The problem being discussed is how to prove this equivalence and the individual is asking for help and providing their work so far. They are also discussing the reason why the de Donder condition is called the harmonic coordinate condition.
  • #1
hideelo
91
15
So, I am following the PI lecture series by Neil Turok. He starts with the following description of harmonic gauge condition

$$g^{\mu \nu}\Gamma^{\lambda}_{\mu \nu}=0$$
He then claims that for linearized gravity (weak field) i.e.
$$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} $$ with $$ |h_{\mu \nu}| <<1 $$ that harmonic gauge is equivalent to the condition that $$h^{\lambda}_{\nu , \lambda} - \frac{1}{2} h^{\lambda}_{\lambda , \nu} = 0$$

My problem is in proving this is how do I prove this, I've been trying for a few days now with no luck, I really need some pointers. I tried using the definition of the Christoffel symbol to work it out and I think I'm going nowhere. Any help would be appreciated.

TIA
 
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  • #2
If you don't show any of your work, how are we supposed to help? We don't know where exactly you got stuck. For example, did you make sure to work to only first order in the perturbation when computing the connection coefficients?
 
  • #3
hideelo said:
So, I am following the PI lecture series by Neil Turok. He starts with the following description of harmonic gauge condition

$$g^{\mu \nu}\Gamma^{\lambda}_{\mu \nu}=0$$
He then claims that for linearized gravity (weak field) i.e.
$$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} $$ with $$ |h_{\mu \nu}| <<1 $$ that harmonic gauge is equivalent to the condition that $$h^{\lambda}_{\nu , \lambda} - \frac{1}{2} h^{\lambda}_{\lambda , \nu} = 0$$

My problem is in proving this is how do I prove this, I've been trying for a few days now with no luck, I really need some pointers. I tried using the definition of the Christoffel symbol to work it out and I think I'm going nowhere. Any help would be appreciated.

TIA
Use [tex]\nabla_{\mu}(\sqrt{-g}g^{\mu\nu}) = 0,[/tex] and rewrite the Harmonic condition in the form [tex]\partial_{\mu}(\sqrt{-g}g^{\mu\nu}) = 0 .[/tex] Expand and contract with [itex]g_{\nu\tau}[/itex] to obtain [tex]\frac{1}{\sqrt{-g}}\partial_{\tau}\sqrt{-g} + g_{\nu\tau}\partial_{\mu}g^{\mu\nu} = 0 .[/tex] This is the same as [tex]\frac{1}{2} g^{\rho\sigma}\partial_{\tau}g_{\rho\sigma} + g_{\nu\tau}\partial_{\mu}g^{\mu\nu} = 0 .[/tex] Now use [itex]g_{\mu\nu} = \eta_{\mu\nu} + \epsilon \ h_{\mu\nu}[/itex] and ignore the [itex]\mathcal{O}(\epsilon^{2})[/itex] terms.
 
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  • #4
WannabeNewton said:
If you don't show any of your work, how are we supposed to help? We don't know where exactly you got stuck. For example, did you make sure to work to only first order in the perturbation when computing the connection coefficients?

So what I have so far is $$0 = g^{\mu \nu} \Gamma^{\lambda}_{\mu \nu} = \frac{1}{2}g^{\mu \nu}g^{\tau \lambda}(g_{\mu \tau ,\nu} + g_{\nu \tau , \mu} -g_{\mu \nu , \tau}) = \frac{1}{2} g^{\tau \lambda}(g_{\mu \tau} \, ^{, \mu} + g_{\nu \tau} \, ^{, \nu}) - \frac{1}{2}g^{\mu \nu}g_{\mu \nu} \, ^{, \lambda} = g^{\tau \lambda} g_{\mu \tau} \, ^{, \mu} - \frac{1}{2}g^{\mu \nu}g_{\mu \nu} \, ^{, \lambda}$$

Now I do the following
$$g_{\mu \tau} \, ^{,\mu} = \partial_\mu g_{\mu \tau} = \partial_\mu( \eta_{\mu \tau} + h_{\mu \tau}) = \partial_\mu h_{\mu \tau} = h_{\mu \tau} \, ^{,\mu}$$ doing this for both terms gives me $$g^{\tau \lambda} h_{\mu \tau} \, ^{, \mu} - \frac{1}{2}g^{\mu \nu} h_{\mu \nu} \, ^{, \lambda} = 0$$

This is what I got so far.

Edit: What it looks like I should do is use the metric tensor to raise the indices on the derivatives of h in the final line, but I don't think I can do that because the derivatives of h aren't tensors so it would be like trying to raise indices on the christoffel symbol, am I wrong?
 
Last edited:
  • #5
bump?
 
  • #6
hideelo said:
bump?
Why? Didn’t I show you exactly how to solve your problem? Do you even know why the de Donder condition ,[itex]g^{\mu\nu}\Gamma_{\mu\nu}^{\rho}=0[/itex], is called harmonic coordinate condition?
 
  • #7
samalkhaiat said:
Why? Didn’t I show you exactly how to solve your problem? Do you even know why the de Donder condition ,[itex]g^{\mu\nu}\Gamma_{\mu\nu}^{\rho}=0[/itex], is called harmonic coordinate condition?

First of all, thanks for answering.

As I understand it, the reason why its called harmonic gauge is because under this condition the coordinate functions are harmonic i.e. $$\Box x^\mu = 0$$

the bump was in response to this:

WannabeNewton said:
If you don't show any of your work, how are we supposed to help? We don't know where exactly you got stuck. For example, did you make sure to work to only first order in the perturbation when computing the connection coefficients?

who asked me to show the work I had done, but then didnt respond, and I wanted to know if I had gone in the totally wrong direction or if I was almost there.
 
  • #8
Of course. The de Dongle condition. Everybody knows it..
 

FAQ: Trying to prove a consequence of harmonic gauge in GR

1. What is harmonic gauge in general relativity?

Harmonic gauge is a specific choice of coordinate system in general relativity where the metric tensor satisfies the wave equation, also known as the "harmonic condition". This condition simplifies the equations of motion and makes it easier to solve for the metric and other physical quantities.

2. How does harmonic gauge relate to the theory of general relativity?

Harmonic gauge is a mathematical tool used in the theory of general relativity to simplify the equations of motion. By imposing the harmonic condition on the metric tensor, the equations of motion become more manageable and can be solved more easily.

3. What is the significance of proving a consequence of harmonic gauge in general relativity?

Proving a consequence of harmonic gauge in general relativity helps to validate the theory and provides a deeper understanding of the mathematical principles behind it. It also allows for more accurate predictions and interpretations of physical phenomena in the universe.

4. What are some common consequences of harmonic gauge in general relativity?

Some common consequences of harmonic gauge in general relativity include the simplification of the equations of motion, the ability to solve for the metric and other physical quantities, and the validation of the theory itself. It also leads to a better understanding of the relationship between space, time, and gravity.

5. How do scientists go about proving a consequence of harmonic gauge in general relativity?

Scientists use a combination of theoretical and mathematical approaches to prove a consequence of harmonic gauge in general relativity. This may involve manipulating the equations of motion, conducting experiments or observations, and using advanced mathematical techniques to validate the results. It is a collaborative effort among scientists in the field of general relativity.

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