Path Independence: Solving a Physics Problem

[davon806]

Homework Statement

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

Homework Equations

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) ?

 

[DrClaude]

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.

 

[davon806]

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.

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

 

[SteamKing]

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

 

[DrClaude]

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

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.

 

[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.

 

[davon806]

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.

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...)

 

[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.