Solving Annoying Integrals from Paper63.pdf

[latentcorpse]

I'm on q4 of this paper:
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2008/Paper63.pdf
For the first bit i said the coordinates had ranges $t \in ( - \infty , \infty) , \quad \chi \in [0, 2 \pi)$
Is that correct?

Anyway for the next bit we can take the equation $g_{\mu \nu} u^\mu u^\nu =- \sigma$ where $\sigma = \begin{cases} 1 \quad \text{for timelike} \\ 0 \quad \text{for null} \\ -1 \quad \text{for spacelike} \end{cases}$

So take the null equation, we get $\dot{t} = \cosh^2{t} \dot{\chi}$
which gives $\int d \chi = \int \frac{dt}{\cosh{t}}$

and the timelike one gives $\dot{t}^2 = \cosh^2{t} \dot{\hi}^2 -1 = \sinh^2{t} \dot{\chi}^2$
So this gives $\int d \chi = \int \frac{dt}{\sinh{t}}$

I don't know how to solve either of these and wolfram is giving me a fairly complicated answer! which makes me think i have made a mistake. Can anyone tell me how to solve these?

Thanks!

 

[fzero]

Can you think of any identities for $$\cosh$$ and $$\sinh$$ that would allow you to write these in terms of simpler functions that would lead to an obvious change of variables?

 

[latentcorpse]

Can you think of any identities for $$\cosh$$ and $$\sinh$$ that would allow you to write these in terms of simpler functions that would lead to an obvious change of variables?

Well yeah rewrite in terms of exponentials. The null one gives $\chi=2 \tan^{-1}{e^t}$ as expected.

The timelike one reduces to $\chi = -2 \int \frac{dy}{1-y^2}$ which I'm not sure how to solve. Wolfram says we get a complicated log expression. This will be hard to plot (see next part of question). Have I made a mistake?

Also, the spacelike case is even harder as we get $\dot{t}^2 = \cosh^2{t} \dot{\chi}^2 + 1$ which we have to solve but unlike the timelike case we can't make use of the identity $\cosh^2-\sinh^2=1$ and so we are left with the annoying $+1$ term which is causing me problem when it comes to separation of variables?

Thanks.

 

[fzero]

Well yeah rewrite in terms of exponentials. The null one gives $\chi=2 \tan^{-1}{e^t}$ as expected.

The timelike one reduces to $\chi = -2 \int \frac{dy}{1-y^2}$ which I'm not sure how to solve. Wolfram says we get a complicated log expression. This will be hard to plot (see next part of question). Have I made a mistake?

That integral can be easily done by partial fraction decomposition, but you did make a mistake in the algebra that got you to it. There's a typo in your post #1 that made it hard to spot, but it's simply that

$\cosh^2{t} \dot{\chi}^2 -1 \neq \sinh^2{t} \dot{\chi}^2$

Also, the spacelike case is even harder as we get $\dot{t}^2 = \cosh^2{t} \dot{\chi}^2 + 1$ which we have to solve but unlike the timelike case we can't make use of the identity $\cosh^2-\sinh^2=1$ and so we are left with the annoying $+1$ term which is causing me problem when it comes to separation of variables?

Thanks.

For the spacelike and timelike cases, you have that

$$\dot{t}^2 - \cosh^2t ~ \dot{\chi}^2 = - \sigma.$$

In order to solve this equation, I think you need to use the fact that $$\partial/\partial \chi$$ is a Killing vector, with an associated conserved momentum that you can use to substitute for $$\dot{\chi}$$.

 

[latentcorpse]

That integral can be easily done by partial fraction decomposition, but you did make a mistake in the algebra that got you to it. There's a typo in your post #1 that made it hard to spot, but it's simply that

$\cosh^2{t} \dot{\chi}^2 -1 \neq \sinh^2{t} \dot{\chi}^2$

Right so it should have been $\cosh^2{t} \dot{\chi}^2 -\dot{chi}^2= \sinh^2{t} \dot{\chi}^2$
So your saying that because we have a 1 not a $\dot{\hi}^2$ we can't make this substitution so should proceed as described below?

For the spacelike and timelike cases, you have that

$$\dot{t}^2 - \cosh^2t ~ \dot{\chi}^2 = - \sigma.$$

In order to solve this equation, I think you need to use the fact that $$\partial/\partial \chi$$ is a Killing vector, with an associated conserved momentum that you can use to substitute for $$\dot{\chi}$$.

I find that the conjugate momentum is (which we can also get just by looking at the Euler Lagrange but I guess its best to think of this as arising from a KVF)

$\dot{\chi} = - \frac{C}{2 \cosh^2{t}}$

However, take the timelike case:

$\dot{t}^2 = \cosh^2{t} \dot{\chi}^2 -1$
$\dot{t}^2 = \cosh^2{t} \frac{C^2}{4 \cosh^4{t}} + 1$
$\dot{t}^2 = \frac{C^2}{4 \cosh^2{t}}+1$

We still have the cosh present though?

 

[fzero]

Right so it should have been $\cosh^2{t} \dot{\chi}^2 -\dot{chi}^2= \sinh^2{t} \dot{\chi}^2$
So your saying that because we have a 1 not a $\dot{\hi}^2$ we can't make this substitution so should proceed as described below?



I find that the conjugate momentum is (which we can also get just by looking at the Euler Lagrange but I guess its best to think of this as arising from a KVF)

$\dot{\chi} = - \frac{C}{2 \cosh^2{t}}$

However, take the timelike case:

$\dot{t}^2 = \cosh^2{t} \dot{\chi}^2 -1$
$\dot{t}^2 = \cosh^2{t} \frac{C^2}{4 \cosh^4{t}} + 1$
$\dot{t}^2 = \frac{C^2}{4 \cosh^2{t}}+1$

We still have the cosh present though?

It still looks like the integration (after some intermediate substitution) can be done by a trig substitution.

 

[latentcorpse]

It still looks like the integration (after some intermediate substitution) can be done by a trig substitution.

I can't see it. What would the substitution be?

 

[fzero]

I can't see it. What would the substitution be?

Start with $$y=\sinh t$$.

 

[latentcorpse]

Start with $$y=\sinh t$$.

$y=\sinh{t} \Rightarrow \cosh^2{t}=1+\sinh^2{t}=1+y^2$
$\frac{dy}{dt}=\cosh{t}$
So $\dot{y}=\frac{dy}{d \tau} = \frac{dy}{dt} \frac{dt}{d \tau}$
Therefore $\dot{t}=\frac{dt}{d \tau} = \frac{dt}{dy} \dot{y} = \frac{\dot{y}}{\cosh{t}} = \frac{\dot{y}}{\sqrt{1+y^2}}$

So we get

$\frac{\dot{y}}{\sqrt{1+y^2}} = \frac{C^2}{4(1+y^2)} + 1$
$\sqrt{1+y^2} \dot{y} - 4(1+y^2) = \frac{C^2}{4}$

Now we have that annoying term on the left. I was trying to complete the square or something but I can't get it...

 

[fzero]

$y=\sinh{t} \Rightarrow \cosh^2{t}=1+\sinh^2{t}=1+y^2$
$\frac{dy}{dt}=\cosh{t}$
So $\dot{y}=\frac{dy}{d \tau} = \frac{dy}{dt} \frac{dt}{d \tau}$
Therefore $\dot{t}=\frac{dt}{d \tau} = \frac{dt}{dy} \dot{y} = \frac{\dot{y}}{\cosh{t}} = \frac{\dot{y}}{\sqrt{1+y^2}}$

So we get

$\frac{\dot{y}}{\sqrt{1+y^2}} = \frac{C^2}{4(1+y^2)} + 1$
$\sqrt{1+y^2} \dot{y} - 4(1+y^2) = \frac{C^2}{4}$

Now we have that annoying term on the left. I was trying to complete the square or something but I can't get it...

The equation contains $$\dot{t}^2$$ and you only computed $$\dot{t}$$.

 

[latentcorpse]

The equation contains $$\dot{t}^2$$ and you only computed $$\dot{t}$$.

Ok. So then I get $\dot{t}^2 = ( \frac{dt}{dy})^2 ( \frac{dy}{ d \tau})^2 = \frac{1}{\cosh^2{t}} ( \frac{dy}{d \tau} )^2$

$\Rightarrow ( \frac{dy}{ d \tau})^2 = \frac{C^2}{4} + \cosh^2{t} = \frac{C^2}{4} + 1 + y^2$

$\Rightarrow d \tau = \frac{dy}{\sqrt{1+(\frac{C^2}{4}+y^2)}}$

This still seems horrific! I was thinking of substituting $x^2=\frac{C^2}{4}+y^2$ but then $dx$ has $y$ terms in it?

 

[fzero]

You can rescale y to put that in a form where you can go back to hyperbolic trig functions to do the integration.

 

[latentcorpse]

You can rescale y to put that in a form where you can go back to hyperbolic trig functions to do the integration.

Ok. so I set $y^2=\frac{C^2}{4}x^2 \Rightarrow dy = \frac{C}{2} dx$

$d \tau = \frac{\frac{C}{2} dx}{\frac{C}{2} \sqrt{\frac{4}{C^2} + (1 + x^2)}}$

Now I can't see anything on this page

http://en.wikipedia.org/wiki/Differentiation_rules#Derivatives_of_trigonometric_functions

that will help integrate something in that form?

 

[fzero]

The derivative of $$\sinh^{-1} x$$ is $$1/\sqrt{1+x^2}$$.

 

[latentcorpse]

The derivative of $$\sinh^{-1} x$$ is $$1/\sqrt{1+x^2}$$.

Hey. I don't know what's wrong with the latex at the moment but in my last post i had three terms in the square root. I had the 1+x^2 we want but also a C^2/4. How do we get rid of that?

Thanks.

 

[fzero]

By a linear change of variable.

 

[latentcorpse]

By a linear change of variable.

The best I've managed is

$t^2=x^2+\frac{4}{C^2} \Rightarrow x=\sqrt{t^2-\frac{4}{C^2}}$

so $\frac{dx}{dt} = \frac{1}{2} \frac{2}{\sqrt{t^2-\frac{4}{C^2}}} = \frac{1}{x}$

But then we have $x dx =dt$

But in our last line we only had dx on the numerator so we're going to have an extra factor of x kicking about aren't we?

 

[fzero]

Note that

$\frac{dy}{\sqrt{1+(\frac{C^2}{4}+y^2)}} = \frac{1}{\sqrt{1+\frac{C^2}{4}}} \frac{dy}{\sqrt{1 + \frac{y^2}{1+\frac{C^2}{4}}}},$

therefore we can rescale y to simplify this.