# Path independence

1. Mar 24, 2016

### davon806

1. The problem statement, all variables and given/known data
Hi,I am trying to understand the proof attached.
My problem is shown by a red arrow.Can someone explain those 2 steps?
And please answer it as simply as you can...Since I haven't done multivariable calculus..It is a physics course.Thanks

2. Relevant equations

3. The attempt at a solution
Let say the integral of A(r') 。d(r') is L(r),so if you take the definite integral it becomes L(r+dr) - L(r),but in the proof they said dr is infinitesimal so dr = 0,so L(r+dr) - L(r) = 0 =/= A(r) 。d(r) ?

Last edited: Mar 24, 2016
2. Mar 24, 2016

### Staff: Mentor

If the integral was path dependent, then you couldn't combine the two integrals together because you could not guarantee that the integral from r to r + dr would pass through r0.

An infinitesimal dr does not mean dr = 0. It means that it is extremely small, the smallest you can get without being 0. Hence, you can consider A(r') to be constant over the interval r to r+dr and take it out of the integral.

3. Mar 24, 2016

### davon806

Sorry, why if A(r') is constant then it becomes A(r) ?

4. Mar 24, 2016

### SteamKing

Staff Emeritus
If you have a HW question with ∫ or ∂ in it, it's a safe bet it belongs in the Calculus HW forum.

5. Mar 24, 2016

### Staff: Mentor

I mean that it is constant in the interval r to r+dr, so A(r) = A(r+dr) = A(r+dr/2), etc. By convenience, you take the value to be the one at r. A is still a function of r.

6. Mar 24, 2016

### andrewkirk

Hi Davon. The notation in the second last formula with two $i$s is 'Einstein notation', which is used in differential geometry. Also, the expression $d\phi(\underline{r})$ has a rigorous meaning in differential geometry but in many other uses is either a piece of notation that has no stand-alone meaning, or else just a hand-wave. This leads me to guess that the text might be written in a differential geometry context.
Is that correct? The way of interpreting the proof is different in diff geom from what would be needed in other contexts.

7. Mar 24, 2016

### davon806

Hi Andrew,it was actually part of the notes of my physics course,and I am still in my 2nd year of a physics degree so I guess it is irrelevant to the more advanced maths.My lecturer introduced Einstein notation for proving various vector identities(scalar product,vector product,curl,div,etc...)

8. Mar 24, 2016

### andrewkirk

Here's a way you can make the proof rigorous without any high-faluting maths:

Replace $d\phi(\underline{r})$ by $\frac{d\phi(\underline{r}+h\underline{e}_i)}{dh}$ where $\underline{e}_i$ is the $i$th basis vector of $\mathbb{R}^3$ (I'm assuming here that the context is 3D Euclidean space).

Then on the right-hand side, divide by $h$, replace $d\underline{r}$ by $h\underline{e}_i$, and enclose both sides of the equation within $\lim_{h\to 0}\Bigg(...\Bigg)$. Note that the LHS is, by definition, the $i$th component of $\nabla(\phi(\underline{r}))$.

Following steps similar to what's written above, but with these modifications, you should be able to deduce

$$\lim_{h\to 0}\left(\frac{d\phi(\underline{r}+h\underline{e}_i)}{dh}\right)=\underline{A}(\underline{r})\cdot\underline{e}_i$$

See how you go.