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

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1. Dec 22, 2017

### faheemahmed6000

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 I originally made it in the $dx$ method.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:

• ###### #7 Neumann's mutual potential energy of two closed circuits s and s'.pdf
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2. Dec 22, 2017

### BvU

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.

3. Dec 22, 2017

### faheemahmed6000

$I=\dfrac{dm}{dS}$ 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 $dm=I dS$

4. Dec 22, 2017

### BvU

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 $\Delta m = I\, \Delta S \ + {\mathcal O}(\Delta S)^2$ .

5. Dec 22, 2017

### faheemahmed6000

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.

6. Dec 22, 2017

### BvU

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.