Change of Variable: Physics Math in GR Problem

In summary, in order to solve a GR problem backward, you need to find a connection coefficient between the variables in terms of the wrong variable. This shortcut method uses Green's functions and is reminiscent of black magic.
  • #1
topsquark
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What I am about to do is write an example of what I call "Physics Math."

I am working on a GR problem backward: I have a motion I want to exhibit and am trying to work out the connection coefficients. The problem is that they are in terms of the wrong variable...The equations are expressed in terms of a parameter \(\displaystyle \tau\) and I want them to be in terms of r and t. (The motion is independent of \(\displaystyle \theta\) and \(\displaystyle \phi\) so I am dropping time derivatives of them.)

So. I saw this in one of my Physics texts and if it works it's a good short-cut for me.

Assume a metric
\(\displaystyle d \tau ^2 = -a(r) dt^2 + b(r) dr^2 + r^2 d \theta ^2 + r^2 ~sin^2(\theta) d \phi ^2\).

Thus:
\(\displaystyle d \tau = \sqrt{ -a(r) dt^2 + b(r) dr^2 + r^2 d \theta ^2 + r^2 ~sin^2(\theta) d \phi ^2}\)

\(\displaystyle \frac{d \tau}{dt} = \sqrt{ -a(r) \left ( \frac{dt}{dt} \right ) ^2 + b(r) \left ( \frac{dr}{dt} \right ) ^2 } = \sqrt{ -a(r) + b(r) \left ( \frac{dr}{dt} \right ) ^2 }\)

and
\(\displaystyle \frac{d \tau}{dr} = \sqrt{ -a(r) \left ( \frac{dt}{dr} \right ) ^2 + b(r) \left ( \frac{dr}{dr} \right ) ^2 } = \sqrt{ -a(r) \left ( \frac{dt}{dr} \right ) ^2 + b(r) } \)

How many rules and I breaking and how badly am I breaking them?

-Dan
 
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  • #2
Hmm, it looks like you're dealing with a Schwarzschild-type metric. In any case, I don't see any "rule breaking" but you may want to think carefully about the meaning of those differentials.
 
  • #3
Euge said:
Hmm, it looks like you're dealing with a Schwarzschild-type metric. In any case, I don't see any "rule breaking" but you may want to think carefully about the meaning of those differentials.
Yes, eventually I will putting the whole mess into a Schwarzschild-like problem. The premise is that I'm going to visit a black hole and turn on a rocket to add a constant acceleration downward. It seems that the metric in this case is going to be time-dependent but so far I'm awash in a sea of indices written in terms of the parameter \(\displaystyle \tau\). There is a much more complex method to change the variables but it's kind of frustrating. So I decided to look at a possible shortcut.

And also yes, GR is rather unforgiving about talking about the wrong variables. I have already determined that I need to set \(\displaystyle \frac{d^2 x}{dt^2} = a\) rather than \(\displaystyle \frac{d^2 x}{d \tau ^2} = a\).

I am a little surprised to find that this method works. This is one of those "trick" methods that strongly reminds me of black magic. Green's functions are another! (Wasntme)

Thanks for the help!

-Dan
 

Related to Change of Variable: Physics Math in GR Problem

1. What is a change of variable in physics and how is it used in GR problems?

A change of variable in physics refers to the process of substituting one variable for another in a mathematical equation. In GR problems, this technique is used to simplify and solve complex equations involving spacetime metrics and tensors. By choosing a different set of variables, the equations can be transformed into a more manageable form, making it easier to solve for the desired quantities.

2. What are the benefits of using a change of variable in GR problems?

The main benefit of using a change of variable in GR problems is that it can simplify the equations and make them easier to solve. This can save time and effort in solving complex problems, and can also provide a deeper understanding of the physical concepts involved.

3. Are there any limitations to using a change of variable in GR problems?

While a change of variable can be a useful tool in solving GR problems, it is not always applicable or appropriate. In some cases, the chosen variables may not lead to a simplified equation, or may even make the problem more complicated. It is important to carefully consider which variables to substitute in order to achieve the desired outcome.

4. How do you choose the right variables to substitute in a change of variable for GR problems?

Choosing the right variables to substitute in a change of variable for GR problems requires a good understanding of the physical concepts involved and the mathematical relationships between the different quantities. It also involves trial and error, as different substitutions may lead to varying levels of simplification.

5. Can a change of variable be used in all GR problems?

No, a change of variable is not always applicable or necessary in GR problems. Some problems may already be in a simplified form, or may not involve complex equations that require a change of variable. It is important to assess each problem individually to determine if a change of variable is needed.

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