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Homework Help: Continous charge or mass distributions

  1. Apr 12, 2012 #1
    Often you want to calculate the electric field of a lot of charges or the center of mass for a lot of small masses.

    When we have a rigid body or something similar in the world of electrodynamics, my book tells me always to calculate the cm or total field as an integral, because "the distributions are continous". Now I have always speculated about this line "they are continous". Because in this world everything is discrete. So isn't it just an approximation when you replace the sum of a hell of alot of elements with an integral?
  2. jcsd
  3. Apr 12, 2012 #2


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    Staff: Mentor

    What if it is an approximation? Physics is about tractable models of the world. Often such models are based upon continuous manifolds, continuous and differentiable functions, conservation principles, and other such niceties where warranted. Intractable models have limited appeal.

    Every particle has an uncertainty of position which is further blurred over even brief time periods. How would you set up a calculation in, say, fluid dynamics, if each time you had to account for every atom's location? Could you ever complete the calculation in time to be relevant?

    Feel free employ the machinery of sigma notation rather than integral calculus whenever you can pinpoint the location of every discrete entity for your calculations :smile:

    Which decimal place are you worried about and why? :devil:
  4. Apr 12, 2012 #3
    It's a matter of scale as always. If I spin a bike tire there's only certain discrete values of angular momentum it can have (scaled by h), but on the scale I'm looking at I can't even hope to detect the discrete values so it's not really even an approximation to say that it's continuous.

    If charges are only spaced out at atomic levels then you're even approaching the limit of self energy, and the point where classical electrodynamics breaks. So, looking at the charge distribution from further out, where electrodynamics is more valid, you can call the distribution continuous. Of course, if you charge up some pith balls and place them in a volume, maybe you shouldn't be calling them continuous.
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