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When we calculate the solid angle subtended by half a sphere we simply do it as follows:

$\Omega=\int_{0}^{2\pi}\,d\phi \int_{0}^{\pi/2}\,sin \theta \,d \theta =2\pi$

Here,:

$\theta$ = co-latitude (clockwise taken positive)

$\phi$ = azimuthal angle (anti-clockwise taken positve) enter image description here

Now this can also be calculated as:

$\Omega=\int_{0}^{\pi}\,d\phi \int_{0}^{\pi}\,sin\theta \,d\theta =2\pi$

Here, the azimuthal ring corresponding to a co-latitude just doesn't take a full $\,2\pi$ circle at once but gets added to another equal half from the lower portion in the $\pi/2$ to $\pi$ part of the co-latitude to give precisely the same thing.

The first integral above calculates the solid angle subtended by the upper hemispherical cap surface area while the second one calculates the angle subtended by the surface area of the $0$ to $\pi$ the right of the axis (speaking loosely). Now of course they should match due to the isotropy of the sphere and other obvious things.

The problem I face is this: What part of the convention (of choosing the co-latitude, as I have done) makes, an attempt to calculate the same thing like this:

$$\Omega\int_{0}^{\pi}\,d\phi \int_{-\pi/2}^{\pi/2}\,sin\theta \,d\theta $$ futile ?

Since this i integral is obviously zero.

Now this might sound like a stupid question and it probably is but it bugs me. It is obvious on one hand that since $\sin $ is an odd function about 0 so the area under its curve gets cancelled in the full-wave from $-\pi/2$ to $\pi/2$ thus giving zero.

But on the other hand it seems that nothing of the clockwise-anticlockwise convention seems to stop me from writing the wrong integral and getting the wrong answer.

Probably in the last integral the half azimuthal ring at a particular positive value of the co-latitude gets cancelled by the half azimuthal ring at the negative of that co-latitude value (this explanation doesn't seem ryt to me though).

To put my rambling into context,

This thing came to mind when I was trying to solve a problem where I had to calculate the electric flux through a part of a sphere.

enter image description here

For that I needed to find the solid angle subtended by the shaded surface area. Did it as follows: $\int_{-\pi/2}^{\pi/2}\,d\phi \int_{\pi/2-\alpha}^{\pi/2}\,sin\theta \,d\theta $

which yields

$\pi\sin\alpha$

This was listed as the correct option even when at $\alpha=\pi$ it becomes zero instead of $2$

In think this is again related to the fact that co-latitude changes sign to the left of the axis so probably the integral needs to to be split up or something.

So to sum up,

How would you put the limits for that integral so that it gives $2\pi$ ? (but remember I want the upper part to become the surface that subtends)

Do conventions and the clockwise counter-clockwise messing up is whats confusing me or is it totally unrelated to all of the mumbo-jumbo I said?

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  • $\begingroup$ You are trying to calculate the spherical angle between two circles of latitude, which is not really what you are told to do. You can calculate it as an angle between two meridians, if you say that the north pole is in the front of the figure. Then the result is $2\alpha$. $\endgroup$ – Hume2 Feb 4 at 12:20
  • $\begingroup$ I am told to calculate the solid angle. The integral for that requires. co-latitude. So I don't understand what you mean? $\endgroup$ – Lost Feb 4 at 14:28
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    $\begingroup$ I'm looking at that electric flux diagram and all I see bounding the shaded portion is two half planes that meet along a line through the center of the sphere and make an angle $\alpha$ with each other, and the portion of the sphere's surface in the angle between the planes. That portion of the surface is called a lune, and it is congruent to a part of the sphere between two lines of longitude, whether or not the great circle arcs that bound it happen to be lines of longitude. By simple proportions, since the area of a sphere is $4\pi,$ the area of that lune is $2\alpha$ as @Hume2 says. $\endgroup$ – David K Feb 5 at 4:29
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    $\begingroup$ If we suppose the dashed line is the axis from which $\theta$ is measured, the region of the sphere over which you integrated, $-\pi/2 < \phi < \pi/2$, $\pi/2 - \alpha < \theta < \pi/2$, we get a region bounded by four arcs, not two. That region covers parts of the sphere's surface that are definitely not shaded in the figure. Apparently the person who wrote the problem didn't understand their own diagram, because none of the answers is correct. $\endgroup$ – David K Feb 5 at 4:44
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    $\begingroup$ Here's a thought. Go over to physics.se and ask them whether any of the multiple-choice solutions of that electric flux problem is correct. You can show them your solution as proof that you did your own work on the problem, but just look at the "good" case without bringing up what happens when $\theta$ goes negative. Let's see what they think of the problem and the choices for its answer. $\endgroup$ – David K Feb 5 at 6:37
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When you write an integral in the form $$ \int_{\phi_1}^{\phi_2}\,\mathrm d\phi \int_{\theta_1}^{\theta_2} \sin \theta \,\mathrm d \theta,$$ you gain a tighter, unambiguous visual relationship between the variables of integration and the integral signs they apply to (that is, $\mathrm d\phi$ is sitting right next to the integral sign it applies to, rather than all the way on the other side of $\mathrm d\theta$), but you lose the representation of a vital concept of integration over a region of a surface, namely the area element.

When you integrate over the surface of a unit sphere in spherical coordinates, as long as you keep $\theta$ in the range $0 \leq \theta \leq \pi,$ the area element has the form of a "rectangle" of height $\mathrm d\theta$ and width $\sin\theta\,\mathrm d\phi.$ The "area" of this "rectangle" is therefore $\sin\theta\, \mathrm d\theta\, \mathrm d\phi.$

You can integrate a function $f$ over a suitable region on the sphere by integrating $$ \int_{\phi_1}^{\phi_2} \int_{\theta_1}^{\theta_2} f(\phi,\theta) \sin \theta \,\mathrm d \theta\,\mathrm d\phi. $$ You can turn this integral into a simple area measurement by setting $f(\phi,\theta) = 1$ everywhere. The term $\sin\theta$ is not a function you integrate over the surface of the sphere; it is an essential part of how you integrate anything else over the surface of a sphere.

Conventionally, when doing integrations like these people almost always stay within the bounds of $0 \leq \theta \leq \pi$; after all, there is no point on the sphere that you can't identify using spherical coordinates with $0 \leq \theta \leq \pi$. So writing $\sin \theta \,\mathrm d \theta\,\mathrm d\phi$ works fine.

If you take $\theta$ outside the range $0 \leq \theta \leq \pi,$ you can find values of $\theta$ where $\sin\theta$ is negative. But the area elements on the surface of the sphere don't suddenly have negative areas just because you decided to use some unusual values of $\theta.$ If you want to be able to use arbitrary $\theta$ coordinates, the width of an area element actually is $\lvert\sin\theta\rvert\,\mathrm d\phi,$ and the area element is $\lvert\sin\theta\rvert \,\mathrm d \theta\,\mathrm d\phi$.

In the language of coordinate transformations, where the coordinates $(\theta,\phi,\rho)$ represent a transformation from the local coordinates $(u,v,w)$ over the orthonormal basis $e_\theta,e_\phi,e_\rho,$ the area element in spherical coordinates is derived from the orthonormal area element by the rule $$ \mathrm du \,\mathrm dv = \lvert \det(J)\rvert \,\mathrm d \theta\,\mathrm d\phi$$ where $J$ is the Jacobian of the transformation. Note that for ordinary area (no "wedge products"), we always take the absolute value of the determinant of the Jacobian. For this particular transformation, the determinant of the Jacobian is $\sin\theta$, so the correct area element is technically always $$ \lvert\sin\theta\rvert \,\mathrm d \theta\,\mathrm d\phi. $$ The reason why people write $\sin\theta$ in the integral rather than $\lvert\sin\theta\rvert$ is that their plan is that they will never ever in their lives use this expression in an integral where $\theta$ ventures outside the range $0 \leq \theta \leq \pi.$

The solution to your dilemma is to write the integral in a form that is correct for all $\theta,$ not just for $0 \leq \theta \leq \pi.$ That is, in order to integrate over a region $\phi_1 \leq \phi \leq \phi_2,$ $\theta_1 \leq \theta \leq \theta_2,$ write the area-measuring integral in the form $$ \int_{\phi_1}^{\phi_2} \int_{\theta_1}^{\theta_2} \lvert\sin\theta\rvert \,\mathrm d \theta\,\mathrm d\phi, $$ or if you like, $$ \int_{\phi_1}^{\phi_2} \,\mathrm d\phi \int_{\theta_1}^{\theta_2} \lvert\sin\theta\rvert \,\mathrm d \theta. $$ Using this always-correct area element for the unit sphere, you will not see cancellation effects when you allow negative values of $\phi_1.$ For example, $$ \int_0^\pi \,\mathrm d\phi \int_{-\pi/2}^{\pi/2} \lvert\sin\theta\rvert \,\mathrm d \theta = 2\pi, $$ which is the correct area of the region integrated.


The real fly in the ointment here is not what happens for negative $\theta.$ It's the unfortunate fact that if you take the dashed line in the figure in that electric flux problem as the axis of your spherical coordinates, that is, if $\theta=0$ you are on the dashed line, then the shaded region on the surface of the sphere is not a region with bounds of the form $\phi_1 \leq \phi \leq \phi_2,$ $\theta_1 \leq \theta \leq \theta_2.$ If you say that $-\frac\pi2 \leq \phi \leq \frac\pi2$ on that region, then when $\phi = 0$ the minimum value of $\theta$ is $\frac\pi2 - \alpha,$ but when $\lvert\phi\rvert$ is near $\frac\pi2$ the minimum $\theta$ value is much closer to $\frac\pi2.$ The area of the region was never going to be computed by $$ \int_0^\pi \,\mathrm d\phi \int_{\pi/2-\alpha}^{\pi/2} \lvert\sin\theta\rvert \,\mathrm d \theta, $$ even for small positive values of $\alpha.$

This is not an easy region to integrate using the dashed line as the axis of your spherical coordinates. The choice of axis of spherical coordinates is an arbitrary choice in a problem like this, so you're better off choosing an axis along the line where the two shaded half-discs meet.


By the way, clockwise vs. anticlockwise depends on the location from which you view the rotation. Conventionally, when thinking about rotation in the $x,y$ plane, we look at it from a point of view far up the positive $z$ axis in a right-hand coordinate system, so the direction of $\phi$ in your figure is anti-clockwise. And this same fixed point of view on the $z$ axis gives the same anticlockwise result for the direction of increasing $\phi$ at every $\phi$ and every $\theta$.

But consider the point $Q$ at spherical coordinates $(\theta, \phi, \rho) = \left(-\frac\pi6,\frac\pi2,1\right)$, which also has coordinates $\left(\frac\pi6,-\frac\pi2,1\right).$ No matter what coordinates you use, the point $Q$ is at the end of a great-circle arc of length $\frac\pi6$ that starts at the pole, whose coordinates can be written $\left(0,\frac\pi2,1\right)$ or $\left(0,-\frac\pi2,1\right).$ When $\phi =\frac\pi2,$ we measure the angle of this arc "backwards" from the pole, but when $\phi = -\frac\pi2$ we measure it "forwards". That is, the exact same arc from $P$ to $Q$ goes in a "negative" $\theta$ direction for one value of $\phi$ and a "positive" $\theta$ direction for another value of $\phi.$ There is no place where we can stand where the positive direction of $\theta$ will always be clockwise, just as there is no place where we can stand where it will always be anti-clockwise.

The usual convention is that $\theta$ is measured in a positive direction away from the positive $z$ axis and toward the negative $z$ axis, but that only works because we usually limit the values of $\theta$ to the range $0 \leq \theta \leq \pi.$ If you have to give a meaning of $\theta$ for values outside that range, you can use the conventional definition for the range $0 \leq \theta \leq \pi$ and simply extend it around the same great circle in both directions from the points at $\theta = 0$ and $\theta = \pi.$ I still would not call this either clockwise or counterclockwise unless you also have a continually moving point of view as $\phi$ changes, and then it depends on whether your point of view "leads" $\phi$ or "trails" $\phi.$ In the figure in the question showing an example of an angle $\phi,$ the point of view "trails" $\phi$, but that's an accident of that particular figure.

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  • $\begingroup$ Great. So, the problem lied with the definition of the integral and of course it makes sense cause area element cannot be negative. Sad they don't usually mention this point in usual curriculum books. That is cleared now. $\endgroup$ – Lost Feb 5 at 6:48
  • $\begingroup$ Though, regarding your second part. The integral limits I have put clearly cover the required region. I should mention that the x-axis in the above posted figure was not what I considered in the sphere of the problem. My x-axis is where $alpha$ is subtended, so when I go from $-\pi/2$ to + pi/2 . I cover the shaded portion. $\endgroup$ – Lost Feb 5 at 6:51
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    $\begingroup$ Yes, you cover the shaded area, but you also cover a significant part of the sphere that is not shaded. One "edge" of the covered region runs along the line of longitude $\phi=-\pi/2$ from $\theta = \pi/2-\alpha$ to $\theta = \pi/2.$ Try plotting that arc on the figure. Does it border the shaded area? How about the arc along the line of longitude $\phi = \pi/2$ from $\theta = \pi/2-\alpha$ to $\theta = \pi/2$? $\endgroup$ – David K Feb 5 at 6:57
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    $\begingroup$ If it is any comfort, I am almost certain that the person who wrote the answer choices for this problem did the same integration you did, or something the same except shifting the starting and ending $\phi$ around the circle, because there's no other obvious way I can see to get any of the choices they offered. $\endgroup$ – David K Feb 5 at 7:00
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    $\begingroup$ Yes, at $\alpha = 0$ you're fine, for $0<\alpha<\pi/2$ you're covering too much, at $\alpha=\pi/2$ you're correct again, and I don't even want to think about $\alpha>\pi/2$ :). People make mistakes, even people who write textbooks and exams. But it isn't fun when you're stuck with the task of choosing the answer for a question when all the answers are wrong. I feel your pain. $\endgroup$ – David K Feb 5 at 7:32

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