How do I convert an integral from LHS to RHS?

[Ratzinger]

I have a problem with a integral. Could someone tell me how to get from the LHS to RHS? It must be very easy, but as so often I don't get it.

Since I have not mastered Latex yet, I attached a pdf file.

thanks

 

[HallsofIvy]

In LaTex what you have is
$$\frac{1}{k^2+ m^2}\int e^{i\vec{k}\dot\vec{r}}d\vec{k}= \int k d\phi \intk sin(\theta)d\theta \int \frac{e^{ik|r|cos(/theta)}}{k^2+ m^2}dk$$
(Click on that for a pop-up box showing the code.)

I had to stare at that a while until I realized- those blasted engineers have mixed up $\theta$ and [/itex]\phi[/itex] again! Also, it is using "k" for the length of $\vec{k}$. I take it that "$\vec{k}$" is the position vector of a point in the space being integrated over (here all of R³).

The right hand side is in spherical coordinates, using k instead of the more standard $\rho$. In spherical coordinates (mathematics version!)
$$dV= \rho^2 sin(\phi)d\rho d\phi d\theta$$
In "engineer speak", as here, it is
$$dV= \k^2 sin(\theta)dk d\theta d\phi$$

That has been separated into
$$(k d\phi)(k sin(\theta)d\theta)(dk)$$
Of course, since $\vec{u}\dot\vec{v}= |u||v|cos(\theta)$, that
[itex]i\vec{k}\dot\vec{r}= ik|r|cos(\theta)[/tex]<br /> <br /> It's not clear to me that breaking it up that way is valid, because clearly k is a variable (dk in one integral) and so putting k into the $d\theta$ and $d\phi$ integrals doesn't make sense.[/itex]

 

[Galileo]

Basically, all that's happened is writing the integral over all space in spherical coordinates $d^3\vec k=k^2\sin \theta dk d\theta d\phi$ and using $\vec k \cdot \vec r=|r|k\cos \theta$.
But the way it's written it looks like a product of three integrals, which it is not!
And unless $\vec r$ points along the z-axis you must not confuse the angle between $\vec k$ and $\vec r$ with the angle $\theta$ over which you are integrating.

 

[Ratzinger]

Thanks HallsofIvy, thanks Galileo!

Good to know that this time it was not all my fault.