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I Please suggest whether I should use delta or dx method.

  1. Dec 22, 2017 #1
    I previously made a derivation of Neumann potential. It can be found in the pdf file below. I originally made it in the ##dx## method. It involved equations like ##dm=I dS##. My maths teacher told that such an expression has no meaning, at least in elementary calculus. However I argued that my physics textbook uses such expressions countless times. Anyway I listened to my maths teacher and changed all ##d## with ##\Delta##. The problem with this was that my final results were not perfect but approximations.

    Then I heard about differentials and non standard analysis where expressions like ##dm=I dS## has meaning. However as a graduate student in Physics, I have no understanding in these topics.

    I think these differentials and non-standard analysis strongly suggest that I can go back to my previous ##dx## method.

    Right now I have these two methods. Please suggest which of them is more proper and appropriate in my derivation of Neumann potential. Thanks in advance for your advice.
     

    Attached Files:

  2. jcsd
  3. Dec 22, 2017 #2

    BvU

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    Your pdf uses ##\Delta## in the text, but ## \;d\ ## in the illustrations.
    In physics, as long as you remember that ## \;d\ ## stands for a limit (like in ## \;{dy\over dx} = {\displaystyle \lim_{h\downarrow 0}}\;{y(x+h)-y(x)\over h} \ ##), you should be just fine.
     
  4. Dec 22, 2017 #3
    ##I=\dfrac{dm}{dS}## is a derivative. Here ##d## indeed stands for limit. I don't have problem there.

    I am having problem with the expression: ##dm=I dS##. My math teacher said in terms of elementary calculus, this expression has no meaning. Yet, I think I can use it as in my Physics books. Or will it be a logical fallacy if I do it? Is it proper to replace ##\Delta m=I \Delta S## in my pdf with ##dm=I dS##
     
  5. Dec 22, 2017 #4

    BvU

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    I agree with your math teacher, but I am a physicist and as a physicist I have no problem at all with ##dm=I\, dS##. Much more readable than ##\Delta m \approx I\, \Delta S \ ## or ##\Delta m = I\, \Delta S \ + {\mathcal O}(\Delta S)^2## .
     
  6. Dec 22, 2017 #5
    Then is it necessary for me to learn advanced calculus in order to understand the meaning of ##dm=I dS##. Or is there any simpler way to understand this expression.
     
  7. Dec 22, 2017 #6

    BvU

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    It's not really advanced calculus...
    For a physicist ## dm=I \;dS ## simply means that a small change in ##S## means a change in ##m## that is ##I## times as big.
     
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