Multi variable integration?

**kev p** · December 22nd 2009, 09:36 PM

I am trying to integrate

$\frac{d\tau}{dp} = \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3}$

where $r_g$ is a constant. I am trying to obtain an expression that can be plotted on a $\tau$ versus p chart.

Attempt at a solution:

My naive attempt was to simply integrate wrt p and I obtained

$\tau = \int \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3} dp \ = \ \left(\frac{3(p-\tau)}{2}\right)^{2/3} r_g^{1/3} +C$

which then solves to

$p = \tau \pm \frac{2 \ (\tau-C)^{3/2}}{3\sqrt{r_g}}$

Unfortunately when plotted it seems unphysical, and I think I may have done the integration wrong. I suspect the problem is that the two variables p and $\tau$ are not independent and multivariable integration is required, but I have no experience of that and no idea where to start. Can anyone help?

Background:

References http://www.phys.au.dk/~fedorov/GTR/09/note11.pdf and Lemaitre metric - Wikipedia, the free encyclopedia

I am using the simplified version of the Lemaitre metric

$ds^{2} = d\tau^{2} - \frac{r_{g}}{r} dp^{2}$

which ignores angular motion and I am trying to obtain an expression for light paths which always have ds=0, so that for light:

$dp/d\tau = \sqrt{ \frac{r}{r_g}}$

and from the references

$r = \left(\frac{3}{2}(p-\tau)\right)^{2/3}\ r_g^{1/3}$

which I substituted into the dp/d $\tau$ equation to obtain the expression that I am trying to integrate. If it helps any, the value 2 can be substituted for $r_g$ . Any help would be greatly appreciated.

**simplependulum** · December 22nd 2009, 10:57 PM

There is a mistake in your solution

$\tau = \int \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3} dp$
It is wrong because $(p - \tau)$ you can find it from the integrand is not just a function of $p$ , it is a function of $p$ and $\tau$

To solve this differential equation ,

$<br /> \frac{d\tau}{dp} = \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3}<br />$

, substitute $p - \tau = u$

differentiate both sides with respect to $p$

$1 - \frac{d\tau}{dp} = \frac{du}{dp}$

$\frac{d\tau}{dp} = 1 - \frac{du}{dp}$

therefore ,

$1 - \frac{du}{dp} = \left(\frac{2\ r_g}{3(u)}\right)^{1/3}$

$\left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] = \frac{du}{dp}$

Then solve it by separating the variables ..

**kev p** · December 23rd 2009, 01:16 AM

Originally Posted by simplependulum

$\left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] = \frac{du}{dp}$

Then solve it by separating the variables ..

so

$du = \left[ 1 - \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3} \right ] dp$

and then

$u = (p-\tau) = \int \left[ 1 - \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3} \right ] dp$

or am I on the wrong track?

**kev p** · December 23rd 2009, 02:10 PM

Originally Posted by simplependulum

To solve this differential equation ,

$<br /> \frac{d\tau}{dp} = \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3}<br />$

, substitute $p - \tau = u$

differentiate both sides with respect to $p$

$1 - \frac{d\tau}{dp} = \frac{du}{dp}$
..

Could you flesh this part out a bit? I am still not getting it...

**dedust** · December 23rd 2009, 03:14 PM

Originally Posted by kev p

Could you flesh this part out a bit? I am still not getting it...

simplependulum said it clearly, differentiate both sides with respect to $p$
you need to simplify $\frac{d}{dp}\{p - \tau \} = \frac{du}{dp}$

**kev p** · December 23rd 2009, 05:35 PM

Originally Posted by dedust

...
you need to simplify $\frac{d}{dp}\{p - \tau \} = \frac{du}{dp}$

Ah! I get it now. For some reason I read

Originally Posted by simplependulum

$\frac{d\tau}{dp} = \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3}$

, substitute $p - \tau = u$

differentiate both sides with respect to $p$

as substitute $p - \tau = u$ (into the equation above it, and then) differentiate both sides (of the equation above it) with respect to $p$ rather than just differentiating both sides of $p - \tau = u$ . Sorry, my mistake. I hope you will bear with me as it is a very long time since I have done any serious calculus.

O.K. so now

$\frac{du}{dp} = \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ]$

$\frac{dp}{du} = \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] ^{-1}$

$p = \int \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] ^{-1} du$

Is that the correct way to go?

**simplependulum** · December 23rd 2009, 07:30 PM

Originally Posted by kev p

$p = \int \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] ^{-1} du$

Is that the correct way to go?

I think it is correct

**kev p** · December 30th 2009, 08:20 PM

$p = \int \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] ^{-1} du$

When I carry out the above indefinite integration, the result is quite a large equation in which the u variable appears numerous times. It seams to be impossible to solve the equation for p or $\tau$ after substituting $(p-\tau)$ back in. Well, Mathematica and Maple seem unable to solve it anyway.

Would it be acceptable to perform the whole operation again, but this time obtain an expression in terms of $\tau$ and u and then use u as a parametric parameter in order to plot p versus $\tau$ ?

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