spherical co-ordinates

[CPL.Luke]

why is the spherical volume element r^2 sin phi dr d phi d theta ? and not r^2 dr dphi d theta ?

 

 

it doesn' make sense to me as r theta is the arc length in a circle and so r^2 theta x phi should give the area element for a sphere, and then why not just allow for an infinitesimal amount of depth with dr and then declare that the volume element?

 

also why is the angle phi drawn off the z axis and not the x axis like theta is?

[D H]

An easy way to visualize why [math]dV = r^2\, \sin\phi\,dr\,d\phi\,d\theta[/math] is to look at a globe of the Earth. A line of latitude is the set of points with constant r and phi. The arc length of a line of latitude varies with the latitude:

[math]l=2\pi\,R_e \sin(\frac{\pi}2-\phi^\prime)[/math]

where [math]\phi^\prime[/math] is the latitude.

Denoting [math]\phi=\frac{\pi}2-\phi^\prime[/math] yields

[math]l=2\pi\,R_e\,\sin\phi[/math]

A small portion of this arc has length

[math]dl=R_e sin\phi\,d\theta[/math]

 

Replace the line of latitude with a thin band with angular deflection [math]d\phi[/math]. The area of a small chunk of this band is

[math]dA=R_e^{\,2} sin\phi\,d\theta\,d\phi[/math]

 

Now extend the band to the center of the Earth. The volume of a small chunk

of this conical volume is

[math]dV=r^2 sin \phi \, d \theta \, d\phi \, dr[/math]

[CPL.Luke]

but wouldn't rsin phi be a line parrelel to the x axis going from some point p to the z axis?

 

I think my main problem with this is that for arc length I think theta * R, and so if you think of the plane formed by the z and x axeses, then the arc length should be given by r phi.

[D H]

but wouldn't rsin phi be a line parrelel to the x axis going from some point p to the z axis?

I don't know quite what you mean by this.

 

I think my main problem with this is that for arc length I think theta * R, and so if you think of the plane formed by the z and x axeses, then the arc length should be given by r phi.

That works in two dimensions, but not in three. What is the radius [math]r[/math]?

 

Look at a globe of the Earth (or do it in your head if you don't have a globe). Each line of constant latitude is a circle. The circles are tiny near the poles and become a great circle at the equator. What is the radius of the circle at latitude [math]\delta[/math] (colatitude [math]\phi=90^{\circ}-\delta[/math])?

[CPL.Luke]

That works in two dimensions, but not in three. What is the radius ?

 

Look at a globe of the Earth (or do it in your head if you don't have a globe). Each line of constant latitude is a circle. The circles are tiny near the poles and become a great circle at the equator. What is the radius of the circle at latitude (colatitude )?

 

ah now I get it thank you.

 

that also explains why the angle is drawn off the z axis towards the xy plane instead of from the xy plane towards the z axis. awesome thankyou.

[ajb]

Calculate the Jacobian factor when changing from cartesian coordinates to polar coordinates and you will get the factors you desire.