Why is the range of ø in spherical coordinates limited to 0 to π?

[Modest Learner]

Homework Statement

In spherical coordinates (ρ,θ,ø); I understood the ranges of ρ, and θ. But ø, still eludes my understanding. Why is ø only from 0 to π, why not 0 to 2π??

 

[Ray Vickson]

Homework Statement

In spherical coordinates (ρ,θ,ø); I understood the ranges of ρ, and θ. But ø, still eludes my understanding. Why is ø only from 0 to π, why not 0 to 2π??

Look at a diagram to see why.

 

[AlephZero]

Values of ø between from 0 to π cover the whole surface of the sphere. On maps of the Earth latitude is measured from -90 to + 90 degrees not 0 and 180, and longitude from -180 to +180 not 0 to 360, but the basic idea is the same.

 

[pbuk]

$\phi = 0$ is directly overhead, $\phi = \pi$ is directly beneath your feet, where would $\phi = 2\pi$ be?

 

[Modest Learner]

Look at a diagram to see why.

If seeing the diagram would have had helped, then I would not have asked the question in the first place.

 

[Modest Learner]

$\phi = 0$ is directly overhead, $\phi = \pi$ is directly beneath your feet, where would $\phi = 2\pi$ be?

Okay, I have uploaded two attachments.

When I view from side, ø = π, covers only half the circle (see the picture). When I try to think of it as a clock, ø = π, covers 12 to 6. Now shouldn't ø = 1.5π cover 12 to 9, and ø = 2π cover the whole circle, and reach the same point as π = 0.

Also, in ø = π, the 3d section appears to me as a hemisphere. Shouldn't it be a total sphere??

Or maybe, I am confusing spherical coordinates with polar or cartesian coordinates??

 

[pbuk]

When I view from side, ø = π, covers only half the circle (see the picture).

So what you have shown is a coloured half-disk. For every point on that disk, Θ = 0. If you vary Θ from 0 to 2π the half-disk will sweep out a complete sphere.

 

[Modest Learner]

Please explain further, I can't seem to understand.

EDIT: Okay, I think, I got a little idea of why ø = π would work. Basically, The up and down thing was exactly right. I think, I confused myself, when I added a sense of left and right. It is a total different axis, and the coordinates seem to do exactly the same thing, by putting θ = π to 2π.

 

[Modest Learner]

So what you have shown is a coloured half-disk. For every point on that disk, Θ = 0. If you vary Θ from 0 to 2π the half-disk will sweep out a complete sphere.

Okay, this is the link,

http://mathinsight.org/spherical_coordinates

 

[HallsofIvy]

Since $\theta$ goes from $0$ to $2\pi$, if we allowed $\phi$ to go also from [itex]0[itex]to $2\pi$ some points would have <b>two</b> descriptions. For example, $\theta= 3\pi/2$, $\phi= \pi/4$ and $\theta= \pi/2$, $\phi= 7\pi/4$, $\rho$ and fixed value, say 1, designate the same point.<br /> <br /> You can see that by converting to Cartesian coordinates: $x= \rho cos(\theta) sin(\phi)$, $y= \rho sin(\theta) sin(\phi)$, $z= \rho cos(\phi)$.<br /> <br /> $\rho= 1$, $\theta= 3\pi/2$, $\phi= \pi/4$ gives $x= 1(0)(\sqrt{2}/2)= 0$, $y= 1(-1)(\sqrt{2}/2)= -\sqrt{2}/2$ and $z= 1(\sqrt{2}/2)= \sqrt{2}/2$.<br /> <br /> $\rho= 1$, $\theta= \pi/2$, $\phi= 7\pi/4$ gives $x= 1(0)(-\sqrt{2}/2)= 0$, $y= 1(1)(-\sqrt{2}/2)= -\sqrt{2}/2$, and $z= 1(\sqrt{2}/2)= \sqrt{2}/2$.[/itex][/itex]

 

[Ray Vickson]

If seeing the diagram would have had helped, then I would not have asked the question in the first place.

If I was a mind-reader I would have known that. I had no way to know what you have, or have not looked at already.

 

[Chestermiller]

Phi is the angle between the axis of the sphere and a line drawn through the center of the sphere to a given latitude, measured from the North Pole. It is equal to 90 degrees (i.e., ∏/2) minus the latitude. So, ø =0 represents a line drawn from the center of the sphere through the North pole, ø = ∏/2 represents a line drawn through the center of the sphere to any point on the equator, and ø =∏ represents a line drawn through the center of the sphere to the South pole.

Chet

 

[Modest Learner]

Thanks, got it.