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Calculate flux from a radiating disk

  1. Aug 25, 2015 #1
    1. The problem statement, all variables and given/known data

    Suppose you have a disk of radius r at x=y=z=0 with its normal pointing up along the z-axis. The disk radiates with specific intensity [itex]I(\theta)[/itex] from its upper surface. Imagine the observer plane is at z=Z, where Z is much greater than r. Let [itex]I(\theta)[/itex] = I = constant. Calculate the flux crossing the observer plane as a function of position. Integrate the flux over the entire plane to show that it is equal to the flux emitted by the source.

    2. Relevant equations

    [tex]F = \int{I(\theta) cos\theta d\Omega}[/tex]

    [tex]d\Omega = sin\theta d\theta d\phi[/tex]

    3. The attempt at a solution
    I feel like I'm either complicating this way more than need be, or I'm oversimplifying. I think I'm really just caught up in the coordinates for some reason. Anyway, here's what I'm thinking.

    First off, because z >> R, we can think of a series of parallel rays coming off of the disk and hitting the observer plane some very large distance away. If we were to project the disk onto the observer plane, the flux should be the same everywhere within that projected region and zero outside of it.

    From our above expression for the flux:

    [tex]F = \int{I(\theta) cos\theta d\Omega} = I\int\int cos\theta sin\theta d\theta d\phi[/tex]

    since we're treating I as a constant.

    At this point, I think I'm getting hung up on the units of integration. [itex]\phi[/itex] should go from 0 to 2pi, and it makes sense that it shouldn't depend on position in [itex]\phi[/itex]. Now, does [itex]\theta[/itex] range from 0 to pi/2 (since we're dealing with the top side of a disk)?

    If that's the case, we get [itex]F = I \pi[/itex].

    I think that's showing the flux emitted by the source and NOT integrating the flux over the entire observing plane, which should evidently be equal. Not sure how to go about that part...

    Now, I'm also confused about the first part of the question, which asks to calculate the flux crossing the plane as a function of position. The only difference then would be that our theta limits go from 0 to theta, right?

    [tex]F = I \pi sin^{2}\theta[/tex]

    Now I need to get theta in terms of x, y, and z, right? In this case, z is fixed, so the flux won't depend on z, while it certainly would if we cared about the flux at different points along Z. On the other hand, if we go to some x and y that are outside of the radius r, there should be no flux. But beyond that, I guess I'm getting lost in coordinates. I would think that [itex]\sin\theta = r^{2}/\sqrt{y^{2}+z^{2}}[/itex] but that's contrary to my idea that it shouldn't depend on z for our plane that is located at z=Z.

    Hopefully I'm at least somewhat on the right track. I feel like I'm headed in the right direction, but I could definitely use some guidance. Thanks!
  2. jcsd
  3. Aug 25, 2015 #2


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    Often, when you see that Z >> r, it implies you can treat the disk as a point source.
    If you have a point source, then the flux is constant over spherical regions.
    However, you are calculating this over a flat, planar region.
    So...I would recommend first writing flux as a function of rho.
    ##\rho = \sqrt{x^2 + y^2 + z^2} ##
    or in cylindrical coordinates,
    ##\rho = \sqrt{R^2 + Z^2} .##
  4. Aug 25, 2015 #3
    Ah, you mean expressing the solid angle in cylindrical coordinates and in turn the flux? So [itex]d\Omega ={z d\phi d\rho}/\rho^{2}[/itex]?

    The problem also says that by assuming Z>>r, you can use the small angle approximation, so in the expression for flux can I let [itex]cos\theta[/itex] = 1, and then F is entirely in terms of [itex]d\rho[/itex] and [itex]d\phi[/itex]?

    Ok, here's what I've got, expressing my solid angle in cylindrical coordinates:

    [tex]F = \int{I cos\theta z d\rho d\phi/ \rho^{2}}= {-2\pi I z} / \rho = {-2\pi I z} / \sqrt{r^2 + z^2}[/tex]

    Not sure I understand the negative sign..
    Last edited: Aug 25, 2015
  5. Aug 26, 2015 #4
    Added some work to my last post. Bumping this in case anyone is able to help further. Thanks!
  6. Aug 26, 2015 #5


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    I don't understand how you get that. Are you using rho etc. in the same way that RUber used them?
    It comes to the same thing, but I think it is a little more natural to consider a point in the observer plane and the solid angle the disc subtends there. If the disc has area A, what is the angle it subtends at a point distance t from the z axis in the observer plane?

    By the way, I suspect that the theta in ##I(\theta)## was intended to represent temperature, but since it's constant it does not matter.
  7. Aug 26, 2015 #6
    In general, it would be [itex]A/t^2[/itex], right?
  8. Aug 26, 2015 #7


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    Draw a diagram in side view. You have a disc width 2r centred at O, a line length z perpendicular to it from O to P, say, and a line length t from P to Q perpendicular to OP. Connect the two edges of the disc, X1, X2, to the point Q. Drop a perpendicular from X1 to QX2. What, approximately, is the length of that? If the disc has area A, what does that tell you about the solid angle the disc subtends at Q?
  9. Aug 27, 2015 #8
    Ok, I think you're essentially asking me to do this (page 32), right? I don't think I fully grasped the concept of solid angle. It's the differential of the projected area (projected onto the direction of the observer) divided by the radius of a sphere squared. So, above, I was regurgitating the formula for the solid angle subtended by a disk, but it definitely makes more sense to get the solid angle by projecting the disk onto a sphere, as done in that example.

    OK. So that all makes sense. But now the problem is that we're not interested in the angle subtended by a point, but rather a plane. I feel like that would change everything?
    Last edited: Aug 27, 2015
  10. Aug 27, 2015 #9


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    The link (page 32) is concerned with getting the exact solid angle subtended at a point on the disc's axis. In the present context, we don't have to worry about the distinction between a small disc and a small spherical cap. But we do have to take into account that the arbitrary point on the observer plane is not on the disc's axis. From the observer view, the disc appears as an ellipse.
    This is needed to determine the intensity of light received at that point, and this is what the question asks for ("function of position"). Later, we can integrate that across the plane and check it gives the right total.
  11. Aug 27, 2015 #10

    Ok, fair enough. That being said, I don't think I understand the drawing. I'm not sure what that length (perpendicular from X1 to QX2 is?

    Thanks for the help by the way!
  12. Aug 27, 2015 #11


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    We need to find how much the apparent size of the disc is shrunk by viewing it on the skew. If you view a circle area A from an angle of theta to the normal, what is its apparent size?
  13. Aug 27, 2015 #12
    Ah, should just be the projected area, right? A*cos(theta)?
  14. Aug 27, 2015 #13


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  15. Aug 27, 2015 #14
    Ok...so is what I did above in calculating the flux not the same? The cos term arrises from the projection into the line of sight direction. I am incredibly confused.
  16. Aug 27, 2015 #15


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    In post #3? I couldn't follow your reasoning there, but the result looked wrong. There should be a factor like ##\rho^{-3}##.
  17. Aug 29, 2015 #16
    Hm, ok. Let's see...

    [tex]F = \int{I \cos{\theta} d\Omega}[/tex]
    [tex]d\Omega = dAcos{\theta}/\rho^2[/tex]
    [tex]F = I \int{cos{\theta} dA cos{\theta}/\rho^2}[/tex]
    [tex]cos{\theta} = z/\rho[/tex]
    [tex]F = I \int{(z/\rho)^2 dA/\rho^2} = I \int{(z/\rho)^2 2\pi r dr/\rho^2} = 2\pi I z^2 \int{\rho d\rho / \rho^4} = -\pi I z^2/\rho^2[/tex]

    Not sure where I'm going wrong?
  18. Aug 29, 2015 #17


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    How did you change ##dr## to ##d\rho##? I would convert all variables (r, rho) to functions of theta.
    But there's also a problem right at the start. If we take the intensity received at the point in the plane to be ∫IcosθdΩ, we end up with an extra factor of pi.
    Thought experiment 1:
    Suppose the disc is at the centre of a sphere radius rho. If the illumination at a point on the surface is ##\frac{IA\cos(\theta)}{\rho^2}##, when we integrate that we get ##\pi IA##, not IA.
    Thought experiment 2:
    Consider a spherical source, radius r, intensity I, at the centre of a sphere radius R. Clearly the intensity of illumination in the larger sphere is ##I\frac{r^2}{R^2}##. To each point of the outer sphere, the inner sphere looks the same as a disc radius r and uniform intensity I. So an actual disc radius r intensity I would illuminate a point at distance R on its axis the same. In terms of the area of the disc, ##A=\pi r^2##, the illumination at that point would have intensity ##I\frac{A}{\pi R^2}##. So that's where the factor ##\frac 1{\pi}## that we need comes from.

    Note that this means you cannot treat a flat source as a point source, no matter how far away compared with the diameter of the source.
  19. Aug 31, 2015 #18
    [tex]\rho^2 = r^2 + z^2[/tex]
    [tex]2\rho d\rho = 2rdr[/tex]

    since z is constant.

    I think I understand most of your reasoning, but I'm still not sure how to go about the problem. A disk as viewed from some point P that is off axis will appear as an ellipse -- ok, that much makes sense. So that means the projected area of the disk at P will gain the cos(theta) term. But I have no idea how to deal with the fact that we have a plane. I don't see how we get a [itex]1/\pi[/itex] since the disk is not radiating onto a sphere.
    Last edited: Aug 31, 2015
  20. Aug 31, 2015 #19


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    Ok, I see how you switched to r, but as I posted, I think it is easier if you convert all to functions of the angle to the normal. Please try that.
    I don't know how else to convince of the need for the 1/pi factor. It had not occurred to me until I tackled the problem, and what I posted is how I proved it to myself.
    But it cannot matter whether the point of the observer in that analysis is on a spherical surface or a flat one. All that matters there is the angle of the tangent plane.
  21. Aug 31, 2015 #20
    It seems like we disagree right off the bat with the equation for flux? Isn't this how it's defined in radiative processes? Where [itex]d\Omega = dA/r^2[/itex]. dA is an "infinitesimal amount of surface area that is located a distance r from the source and oriented perpendicular to the position vector r." Now of course in our case, dA is not perpendicular to r (because it's a plane at z=Z), and hence we get [itex]d\Omega = dAcos\theta /\rho^2[/itex], right? But since this cos term is just for [itex]d\Omega[/itex], shouldn't we still have another cos term from our definition for flux?

    Thanks for the patience, by the way. Can't quite crack this one yet.
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