# Derive General Metric Wave Equation

• A
• mertcan
In summary, Susskind explains that the wave equation looks the same whether you have curved spacetime or curvilinear coordinates in flat spacetime, and that you can derive the form in arbitrary curvilinear coordinates by just doing some calculus.

#### mertcan

hi, I know the wave equation in terms of minkowski metric, and we use ordinary derivative in that equation. I also know this wave equation in terms of general metric form using covariant derivative, but I do not know the derivation of it. Could you spell out the steps of derivation?? How do we derive the general metric form of wave equation?

Your question is unclear. What do you want to derive the wave equation for? The wave equation itself is just the divergence of the covariant derivative. It is not really much to derive.

Ok, I try to make it more clear. In this video (, at 35.13)

You can see a wave equation and it's minkowski metric interpretation at 35.13. Also at 52.00, we have the general metric interpretation of wave equation(leonard suskind said so) with covariant derivatives. So, I wonder: how does this metric transformation yield a covariant derivatives?? ( by the way suskind tells something but I can not fully understand)...Could you spell this situation out?

So many views but I can not received response... I hope my question is explicit...

I watched most of the video, but I don't understand what part you are having trouble with. He's trying to figure out two things about a wave field:

1. How does gravity (metric + Christoffel symbols) affect a wave.
2. How does a wave affect gravity?
The second part is about how to write the stress-energy tensor $T_{\mu \nu}$ for a scalar field $\phi$. It doesn't have much to do with the covariant derivative.

The first part, though, is about the covariant derivative. The wave equation in terms of the covariant derivative is just:

$g^{\mu \nu} \nabla_\mu \nabla_\nu \phi = 0$

where $\nabla_\mu$ is the covariant derivative.

Initially I understand the wave equation with minkowski metric, but my struggle is to derive wave equation in terms of metric tensor and covariant derivative.so How can I derive it?? How can I reach it??

mertcan said:
Initially I understand the wave equation with minkowski metric, but my struggle is to derive wave equation in terms of metric tensor and covariant derivative.so How can I derive it?? How can I reach it??
So we come back to the original problem. That it is unclear what you mean. The wave equation itself is not something you derive, it is just a name you give to a certain equation. The wave equation in curved space-time is its natural generalisation.

Whether or not a physical quantity obeys the wave equation is a diffrent question and the derivation might in general depend on the quantity in question.

mertcan said:
Initially I understand the wave equation with minkowski metric, but my struggle is to derive wave equation in terms of metric tensor and covariant derivative.so How can I derive it?? How can I reach it??

I'm not sure what you mean by deriving it. Deriving it starting from what?

As Susskind says in the video, the wave equation looks the same, whether you have curved spacetime or curvilinear coordinates in flat spacetime. So if you start with the flat spacetime, Cartesian coordinates form for the wave equation, you can derive the form in arbitrary curvilinear coordinates by just doing some calculus:

$\eta^{a b} \partial_a \partial_b \phi = 0$ in Cartesian coordinates. Now let's switch to curvilinear coordinates. I'll use Greek letters for the curvilinear indices, and Roman letters for the Caresian indices.

The relationship between the two kinds of derivatives is: $\partial_b \phi = L^\mu_b \partial_\mu \phi$ where $L^\mu_b = \frac{\partial x^\mu}{\partial x^b}$. When you take a second derivative, you end up derivatives of $L^\mu_b$ in addition to derivatives of $\phi$. The derivatives of $L^\mu_b$ become part of the Christoffel symbols in the curvilinear coordinates.

mertcan
stevendaryl said:
I'm not sure what you mean by deriving it. Deriving it starting from what?

As Susskind says in the video, the wave equation looks the same, whether you have curved spacetime or curvilinear coordinates in flat spacetime. So if you start with the flat spacetime, Cartesian coordinates form for the wave equation, you can derive the form in arbitrary curvilinear coordinates by just doing some calculus:

$\eta^{a b} \partial_a \partial_b \phi = 0$ in Cartesian coordinates. Now let's switch to curvilinear coordinates. I'll use Greek letters for the curvilinear indices, and Roman letters for the Caresian indices.

The relationship between the two kinds of derivatives is: $\partial_b \phi = L^\mu_b \partial_\mu \phi$ where $L^\mu_b = \frac{\partial x^\mu}{\partial x^b}$. When you take a second derivative, you end up derivatives of $L^\mu_b$ in addition to derivatives of $\phi$. The derivatives of $L^\mu_b$ become part of the Christoffel symbols in the curvilinear coordinates.
Stevendarly thank you I got your last answer, you understand my question truly, by the way I can not reach covariant form of wave equation using the notion that you shared, because I can not compile all the christoffel symbols stuff so, Would you mind explaining your last response in more detail using little bit more mathematical demonstration?

mertcan said:
Stevendarly thank you I got your last answer, you understand my question truly, by the way I can not reach covariant form of wave equation using the notion that you shared, because I can not compile all the christoffel symbols stuff so, Would you mind explaining your last response in more detail using little bit more mathematical demonstration?

Okay. In Cartesian coordinates, there is no difference between partial derivatives and covariant derivatives. So if $V^b$ is a vector field, then
$\nabla_a V^b = \partial_a V^b$

Now, switch to a curvilinear coordinate system. Let $L^\mu_a = \partial_a x^\mu$ and letting $\tilde{L}^b_\nu = \partial_\nu x^b$. $L$ and $\tilde{L}$ are inverses, in the sense that:

$L^\mu_a \tilde{L}^b_\mu = \delta^b_a$
$L^\mu_a \tilde{L}^a_\nu = \delta^\mu_\nu$

Then since $\nabla_a V^b$ is a tensor, we can rewrite it in terms of the curvilinear coordinates:

$\nabla_a V^b = L^\mu_a \tilde{L}^b_\nu \nabla_\mu V^\nu$

We can also write:

$\partial_a V^b = L^\mu_a \partial_\mu (\tilde{L}^b_\nu V^\nu)$
$= L^\mu_a (\partial_\mu \tilde{L}^b_\nu) V^\nu + L^\mu_a \tilde{L}^b_\nu \partial_\mu V^\nu$

Here's where I think we have to do some unmotivated manipulation. We want to pull out a factor of $\tilde{L}^b_\nu$. So we can rewrite:

$(\partial_\mu \tilde{L}^b_\nu) V^\nu$
$= (\partial_\mu \tilde{L}^b_\lambda) V^\lambda$
$= \delta^b_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda$
$= \tilde{L}^b_\nu L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda$

So we can write:
$\partial_a V^b = L^\mu_a \tilde{L}^b_\nu [ L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu]$

Putting the facts together, we have:
$L^\mu_a \tilde{L}^b_\nu \nabla_\mu V^\nu = L^\mu_a \tilde{L}^b_\nu [ L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu]$

which implies that
$\nabla_\mu V^\nu = L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu$

Comparing that with the usual expression for $\nabla_\mu V^\nu$ tells us that:

$\Gamma^\nu_{\mu \lambda} = L^\nu_c (\partial_\mu \tilde{L}^c_\lambda)$

mertcan
stevendaryl said:
Okay. In Cartesian coordinates, there is no difference between partial derivatives and covariant derivatives. So if $V^b$ is a vector field, then
$\nabla_a V^b = \partial_a V^b$

Now, switch to a curvilinear coordinate system. Let $L^\mu_a = \partial_a x^\mu$ and letting $\tilde{L}^b_\nu = \partial_\nu x^b$. $L$ and $\tilde{L}$ are inverses, in the sense that:

$L^\mu_a \tilde{L}^b_\mu = \delta^b_a$
$L^\mu_a \tilde{L}^a_\nu = \delta^\mu_\nu$

Then since $\nabla_a V^b$ is a tensor, we can rewrite it in terms of the curvilinear coordinates:

$\nabla_a V^b = L^\mu_a \tilde{L}^b_\nu \nabla_\mu V^\nu$

We can also write:

$\partial_a V^b = L^\mu_a \partial_\mu (\tilde{L}^b_\nu V^\nu)$
$= L^\mu_a (\partial_\mu \tilde{L}^b_\nu) V^\nu + L^\mu_a \tilde{L}^b_\nu \partial_\mu V^\nu$

Here's where I think we have to do some unmotivated manipulation. We want to pull out a factor of $\tilde{L}^b_\nu$. So we can rewrite:

$(\partial_\mu \tilde{L}^b_\nu) V^\nu$
$= (\partial_\mu \tilde{L}^b_\lambda) V^\lambda$
$= \delta^b_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda$
$= \tilde{L}^b_\nu L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda$

So we can write:
$\partial_a V^b = L^\mu_a \tilde{L}^b_\nu [ L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu]$

Putting the facts together, we have:
$L^\mu_a \tilde{L}^b_\nu \nabla_\mu V^\nu = L^\mu_a \tilde{L}^b_\nu [ L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu]$

which implies that
$\nabla_\mu V^\nu = L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu$

Comparing that with the usual expression for $\nabla_\mu V^\nu$ tells us that:

$\Gamma^\nu_{\mu \lambda} = L^\nu_c (\partial_\mu \tilde{L}^c_\lambda)$
stevendarly thank you for your last return, by the way I tried to merge the information you had given in "post 8" with the information you had given in "post 10", but still I can not entirely switch the wave equation in cartesian coordinates to curvilinear coordinates or covariant form of wave equation. I need your a little bit more support. Could you help me a bit more to obtain wave equation in curvilinear coordinates or covariant form of wave equation ?

mertcan said:
stevendarly thank you for your last return, by the way I tried to merge the information you had given in "post 8" with the information you had given in "post 10", but still I can not entirely switch the wave equation in cartesian coordinates to curvilinear coordinates or covariant form of wave equation. I need your a little bit more support. Could you help me a bit more to obtain wave equation in curvilinear coordinates or covariant form of wave equation ?

It's just $g^{\mu \nu} \nabla_\mu \nabla_\nu \phi = 0$. Expanding in terms of Christoffel symbols, it's:

$g^{\mu \nu} (\partial_\mu \partial_\nu \phi + \Gamma^\lambda_{\mu \nu} \partial_\lambda \phi) = 0$

 I got a sign wrong. The covariant derivative of a covector has a minus sign:

$g^{\mu \nu} (\partial_\mu \partial_\nu \phi - \Gamma^\lambda_{\mu \nu} \partial_\lambda \phi) = 0$

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## 1. What is the Derive General Metric Wave Equation?

The Derive General Metric Wave Equation is a mathematical equation used in the field of general relativity to describe the propagation of gravitational waves through spacetime. It is a more general form of the standard wave equation, taking into account the curvature of spacetime caused by massive objects.

## 2. How is the Derive General Metric Wave Equation derived?

The Derive General Metric Wave Equation is derived from the Einstein field equations and the linearized equations of general relativity. It involves manipulating the equations to isolate the second-order derivatives of the metric tensor, resulting in a wave equation that describes the behavior of gravitational waves.

## 3. What are the applications of the Derive General Metric Wave Equation?

The Derive General Metric Wave Equation is used in the study of gravitational waves, which can provide information about the dynamics of massive objects, such as black holes and neutron stars. It also has applications in cosmology, as gravitational waves can help us understand the early universe and its expansion.

## 4. Can the Derive General Metric Wave Equation be applied to any type of wave?

No, the Derive General Metric Wave Equation is specifically designed to describe the propagation of gravitational waves through curved spacetime. It cannot be applied to other types of waves, such as electromagnetic or sound waves, which have their own equations.

## 5. How does the Derive General Metric Wave Equation relate to the concept of spacetime curvature?

The Derive General Metric Wave Equation takes into account the curvature of spacetime, which is caused by the presence of massive objects. This curvature is what allows gravitational waves to propagate through the universe, and the equation describes how this curvature changes in response to the waves. Essentially, the equation links the behavior of gravitational waves to the concept of spacetime curvature in general relativity.