# Delta function in spherical coords

1. Dec 2, 2007

### Lorna

1. The problem statement, all variables and given/known data
If we have a delta function in cartesian coords, how do we convert it into spherical.
for example : delta (r) = delta(x-x0) delta(y-y0) delta(z-z0)

2. Relevant equations

3. The attempt at a solution
I used
delta (r) = delta(r-r0) delta(cos{theta}-cos{theta0}) delta (phi-phi0)/(r sin{theta})^2

How do we find r0,cos{theta0} and phi0, if what im using is the right formula.

do we use:

z= r cos{theta}
y= r sin{theta} cos {phi}
x= r sin {theta} sin {phi}

and then say the delta function is non-zero if x=x0 or x0=r sin {theta} sin {phi}
and so on and then solve fir r, cos{theta} and phi?

thanks

2. Dec 3, 2007

### HallsofIvy

I'm not sure what you mean by writing it in spherical coordinates. The Delta function is 0 if $\rho$ (your r) is non-zero.

3. Dec 3, 2007

### Despondent

If you were working in polar coordinates for example, the 'obvious' thing to do would be to write $$\delta = \delta \left( r \right)$$ but this is incorrect since the delta function would not satisfy all of the required properties. I can't remember exactly off the top of my head but in cylindrical coordinates you define the delta function as something like $$\delta \left( {x,y} \right) = \frac{{\delta \left( r \right)}}{{2\pi r}}$$.

I would imagine that the delta function would be defined in a similar way for spherical coordinates. My guess would be that the 2*pi*r would be replaced by 4*pi*r in spherical coordinates.

4. Dec 3, 2007

5. Dec 3, 2007

### Avodyne

Lorna, your solution is correct, but you should really write the left-hand side as
$$\delta(\vec{r}-\vec{r}_0)$$
This is sometimes written as
$$\delta^3(\vec{r}-\vec{r}_0)$$
to emphasize that this is a 3-dimensional delta function

Last edited: Dec 3, 2007
6. Dec 3, 2007

### Lorna

thanks all

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook