Find Distance Between Spherical Polar & Cartesian Points

[MathematicalPhysics]

I have two points, one given in spherical polar coordinates and the other in cartesian coordinates. If I want to work out the distance inbetween these points do I need to convert the cartesian into spherical polars?

 

[matt grime]

Unless you happen to be able to visualize these things in your head, then yes, writing the point in sphericals in cartesian coords would seem like a good step, especially as distance is usually calculated in terms of cartesion coordinates.

 

[StatusX]

the formula for the distance between two points in spherical coordinates is:

d^2=r_1^2+r_2^2-2 r_1 r_2(sin\theta_1 sin\theta_2 cos\phi_1 cos\phi_2 + sin\theta_1 sin\theta_2 sin\phi_1 sin\phi_2 + cos\theta_1 cos\theta_2)

obviously, the cartesian formula is a lot simpler.

 

[MathematicalPhysics]

Okay so I have point P with spherical polar coordinates
(R, \theta , \phi)
and point A with cartesian coordinates
(0, 0, a)

so A is just the z axis with length a? so in spherical polars:

A is (a, 0, 0) ? or maybe (a, \theta, \phi) ?

Im getting confused!

 

[arildno]

With \theta being the angle in the horizontal plane, whereas \phi the angle to the z-axis, a given point at distance abs(a) ("a" itself either positive or negative) on the z-axis has the polar representation:
(abs(a),\theta,\frac{\pi}{2}(1-sign(a))
where sign(a) is 1 when a is positive or -1 when a is negative.

 

[MathematicalPhysics]

Thanks, I've now worked out:

x = r\sin\theta\cos\phi , y = r\sin\theta\sin\phi, z = r\cos\theta

I'll see how far I can get now!

 

[arildno]

Note:
In your notation, the angles are interchanged from how I've used them!

 

[MathematicalPhysics]

Yeah, sorry forgot to mention that I had interchanged them.

So now I have two points in cartesian coordinates:

P (R\sin\theta\cos\theta , R\sin\theta\sin\phi , R\cos\theta)

A (0, 0, a)

In order to work out the distance AP do I need to square that awful looking thing?!

 

[arildno]

Yep, that's what you need to do.

 

[MathematicalPhysics]

Okay so I need:

(R\sin\theta\cos\phi)^2 + (R\sin\theta\sin\phi)^2 + (R\cos\theta)^2

How can I simplify this? In a book I have it just sets a "similar" expression equal to 1 with no intermediate steps:

(\sin\theta\cos\phi)^2 + (\sin\theta\sin\phi)^2 + (\cos\theta)^2 = 1

Also this has no 'R' in it.

edit: sorry had theta where there should have been phi

 

[HallsofIvy]

Okay so I need:

(R\sin\theta\cos\phi)^2 + (R\sin\theta\sin\phi)^2 + (R\cos\theta)^2

How can I simplify this? In a book I have it just sets a "similar" expression equal to 1 with no intermediate steps:

(\sin\theta\cos\phi)^2 + (\sin\theta\sin\phi)^2 + (\cos\theta)^2 = 1

Also this has no 'R' in it.

edit: sorry had theta where there should have been phi

Did you notice that
(R\sin\theta\cos\phi)^2 + (R\sin\theta\sin\phi)^2 + (R\cos\theta)^2
=R^2(\sin\theta\cos\phi)^2 + R^2(\sin\theta\sin\phi)^2 + R^2(\cos\theta)^2
= R^2((\sin\theta\cos\phi)^2 + (\sin\theta\sin\phi)^2 + (\cos\theta)^2)
?

 

[MathematicalPhysics]

Yeah, I've got the distance AP to be

(R^2 - 2aR\cos\theta + a^2)^(1/2)

which I think is correct?

Now writing the triple integral of AP over the sphere R less than or equal to a in terms of spherical polar coords gives:

\iiint {\sqrt(R^2 - 2aR\cos\theta + a^2)} R^2 \sin\theta\,dR\,d\theta\,d\phi

with the integration over R between 0 and a
the integration over \theta between 0 and pi
the integration over \phi between 0 and 2pi

correct so far?