Solving a metric in vacuum

In summary, the problem is to determine the form of the function H for a given metric in order for it to represent a plane gravitational wave propagating in vacuum. The approach involves relating the Einstein equation to the Ricci tensor and solving for the Christoffel symbols, with the goal of showing that the trace of the EFE leads to R_{\mu \nu}=0. This requires working out several Christoffel symbols, with the exception of when a derivative of g_{uu} is needed.
  • #1
joseamck
13
0
Hello,

I'm having problems solving this problem I got in class.
I want to learn the concept and how to approach the solution.

Here it is:

Consider the metric

ds=dx^2+dy^2-dudv+2H(x,y,u)du^2

What form must the function H have for this metric to represent a plane gravitational wave propagating in vacuum?




This is how I'm approaching the problem:

first I let the metric be g_{σβ}=[1,0,0,0;0,1,0,0;0,0,2H,-1;0,0,-1,0]

and I know that in vacuum the Einstein equation in far outside the source's field leads to
T^{σβ} = 0
and I know that a metric representing a plane gravitational wave propagating in vacuum is
g_{σβ=[1,0,0,0;0,1,0,0;0,0,0,-1;0,0,-1,0]

coming from the metric ds^2=-dudv+dx^2+dy^2

Then I don't know how to put the pieces together. I was thinking relating the Einstein equation in terms of the ricci tensor and solving for the Christoffel symbols.

Not sure. I need to understand better the problem and how to approach this problem.
Thanks.
 
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  • #2
Can't you LaTeX this?

If the metric has a -dudv term then I think the matrix should have -0.5 in dudv and dvdu positions
 
  • #3
the metric is

[itex]ds=dx^2+dy^2-dudv+2H(x,y,u)du^2 [/itex]

I know that in vacuum
[itex]T^{σβ} = 0 [/itex]
 
  • #4
joseamck said:
the metric is

[itex]ds=dx^2+dy^2-dudv+2H(x,y,u)du^2 [/itex]

I know that in vacuum
[itex]T^{σβ} = 0 [/itex]

I think you should derive the vacuum field equations.

Consider the trace of the EFE. This should allow you to prove that [itex]R_{\mu \nu}=0[/itex]

So it looks like you're going to have to work out a bunch of Christoffel symbols. Although it looks as though everything will vanish except for the situations when you have to take a derivative of [itex]g_{uu}[/itex]

Hopefully...
 
  • #5


Hello,

Thank you for sharing your problem and thought process. Solving a metric in vacuum can be a complex task, but with the right approach, it can be simplified. First, it is important to understand the concept of a metric and its components. A metric is a mathematical representation of the distance between points in a space, and it is described by a set of coefficients or functions. In this case, the metric is described by four coefficients: dx^2, dy^2, -dudv, and 2H(x,y,u)du^2.

Now, let's focus on the specific problem at hand. The question asks what form the function H must have for the metric to represent a plane gravitational wave propagating in vacuum. To answer this, we need to understand the properties of a gravitational wave. A gravitational wave is a disturbance in the curvature of spacetime, and it is described by a transverse and traceless metric. This means that the metric should only have off-diagonal terms and no trace terms.

Using this information, we can see that the given metric already has two off-diagonal terms (-dudv and 2H(x,y,u)du^2) and one trace term (dx^2+dy^2). Therefore, we need to modify the function H to eliminate the trace term and make the metric transverse and traceless. This can be achieved by setting H=0. With this modification, the metric becomes g_{σβ=[1,0,0,0;0,1,0,0;0,0,0,-1;0,0,-1,0], which represents a plane gravitational wave propagating in vacuum.

To summarize, the function H must be set to zero for the given metric to represent a plane gravitational wave in vacuum. I hope this helps in understanding the problem and approaching the solution. If you have any further questions, please feel free to ask.

Best regards,

 

1. How is a metric solved in a vacuum?

In order to solve a metric in a vacuum, one must use the equations of general relativity, which describe the curvature of spacetime in the presence of matter and energy. These equations can be solved for a vacuum by setting all matter and energy terms to zero.

2. What is a metric in a vacuum?

A metric in a vacuum refers to the mathematical description of the geometry of spacetime in the absence of any matter or energy. It is a fundamental concept in general relativity and is used to calculate the distance between points in space.

3. What is the importance of solving a metric in a vacuum?

Solving a metric in a vacuum allows scientists to understand the fundamental properties of spacetime, such as its curvature and topology. This information is crucial in accurately predicting the behavior of objects in space, as well as in developing theories about the origin and evolution of the universe.

4. How do scientists solve for a metric in a vacuum?

Scientists typically use mathematical techniques, such as tensor calculus and differential geometry, to solve for a metric in a vacuum. These techniques allow them to manipulate the equations of general relativity and find solutions that accurately describe the geometry of spacetime in a vacuum.

5. What are some applications of solving a metric in a vacuum?

Solving a metric in a vacuum has many practical applications, including helping scientists understand the behavior of black holes, predicting the paths of celestial objects, and testing the predictions of general relativity. It is also crucial in developing technologies such as GPS, which rely on accurate measurements of spacetime to function.

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