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Differential with respect to area or volume

  1. Aug 10, 2010 #1
    This is annoying me beyond comprehension.

    What precisely does one mean by writing:

    - [tex]\frac{dP}{dA}[/tex] for a power transfered via unit area of the surface (P is power here)

    - [tex]\frac{dQ}{dV}[/tex] for a energy density (Q is energy here)

    and how using those formulas can one calculate the defined quantities?
    Last edited: Aug 10, 2010
  2. jcsd
  3. Aug 10, 2010 #2


    Staff: Mentor

    This derivative (it's not a differential) is the derivative of the power with respect to area. It represents the amount of power per square meter, square foot, or whatever the area units are.
    This is the derivative of charge with respect to volume. The units would probably be coulombs per cubic meter, or whatever the volume units are.
  4. Aug 10, 2010 #3
    Thanks for the answer.
    You are right - I meant derivative. This issue aside, I was hoping for a more "mathematical" answer - what do you mean by taking derivative with respect to area/volume? What is the definition here?
  5. Aug 10, 2010 #4


    Staff: Mentor

    If P = f(A) = A3, the dP/dA = 3A2.

    The "with respect to" part tells you which variable is the independent variable. Otherwise, a derivative is a derivative.

    [tex]\frac{dP}{dA} = \lim_{\Delta A \to 0} \frac{f(A + \Delta A) - f(A)}{\Delta A}[/tex]
  6. Aug 10, 2010 #5
    Great! Now I'm starting to get it!

    So in principal to compute [tex]\frac{dQ}{dV}[/tex] I need to express energy as a function of volume and use the definition above.

    I'll ask one last thing though. The main reason why I'm trying to understand this is because a lot of quantities in radiometry are defined this way and a imaginary situation that follows what was bugging me:
    I have w weird lightbulb - a hemisphere with radius r centered at (0,0,0) that emits P watts of light. The light is emitted only straight upward, but a same amount at every point.

    What I'm trying to get around my head is how to compute amount of power per unit area in this situation. This will obviously differ at each point, and the integral over the whole hemisphere has to be the power given.
    But working from other side how do I express power as a function of area here?
    Last edited: Aug 10, 2010
  7. Aug 10, 2010 #6


    Staff: Mentor

    Oftentimes you won't need to compute dQ/dV or whatever: it will be given to you as, say, coulombs/m3.

    The situation you describe seems to involve a point light source at the origin, and a hemispheric (or more likely, parabolic) reflector that reflects the light upward. If the reflector is r units in radius at its top, the horizontal cross-sectional area there is pi*r^2. The power flux (I think that's the right word) would be P/A, where A = pi*r^2.
    Last edited: Aug 10, 2010
  8. Aug 10, 2010 #7
    Thanks. Again cleared a bit on whats going on. However I meant something a bit different - the hemisphere I though of was suppose to be the one described by [tex]\{ (x,y,z) : x^2 + y^2 + z^2 = r^2 \textrm{ and } y \geq 0\}[/tex]. The whole idea was to get a situation where the flux will be non-uniform over the surface so I can see how to compute this in case where I cannot just divide power via area.

    Sorry for making this whole topic so difficult, but I really want to get hang on things.
  9. Aug 10, 2010 #8


    Staff: Mentor

    I'm having trouble understanding how the hemisphere could emit light so that it is traveling in one direction only. The light can't come from a point source and be reflected off the hemisphere, because the reflected light wouldn't go in just one direction. Are there light sources all over the inside surface of the hemisphere?
  10. Aug 10, 2010 #9
    Sorry for making this so weird and thank you for all the effort so far. You answered my original question so don't feel to obligated to answer this one if you think I'm making this too complicated.

    As for the light source I described - I had pulled it out of my head. The idea was to get a light source that had two properties:
    - The flux is non-uniform over the surface
    - Direction of the light is not perpendicular to the surface (at least not everywhere)
    So I guess I overcomplicated this.

    Maybe you can tell me how to compute the flux in following two situations (which sum up to the one above):
    - For the non-uniform case - Imagine a light that is not a point but a circle with radius r on the plane xy and emits light upward. It has the property that emission is non-uniform - its strongest in the center and falls down to zero on the boundary. Mathematically speaking we have got some kind of density function over the circle that has a maximum in the center and falls down to zero on the edge, such that integral of that function over the circle is 1. How would I go about computing the power flux in this case?
    - For the light non perpendicular to the surface - Imagine a area light (say 1x1 square on plane) that emits light in the direction of 45 degrees to its surface. This time uniformly. How to compute the flux now? (I guess this will be somehow dependent on the normal f the area).
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