# My Mind going to blow! Why Conservative field Path independ?

1. Aug 7, 2010

### TMSxPhyFor

Hi

I'm now reading about Vector fields, everything is clear and intuitive for me as curl divergence ..ect , except one simple thing that I'm straggling with for the last 4 days! I searched Internet and a lot of math & physics books but in vain :

Why Conservative field is Path independent? or why work between two point will be path independent, by definition conservative field is the one generated out of gradient of some scalar, and purely mathematically it's very straight forward and clear, but it's not intuitive for me at all, I can't find any geometric explanation for that even so it seems to be very neat , Beautiful and simple result...

Any I ideas? thx in advance.

Last edited: Aug 7, 2010
2. Aug 7, 2010

### timewalker

At first, problem is determining which sentence is prior to the other.
Path-independent line integral vector field => Conservative field?
Or
Conservertive field => path-independent?

I think first one is more better explanation for our intuition. But if we are gonna make the definition of conservative field then may be we can write it like follow
"A conservative field is a differentiable(integrable) vector field if and only if any line integral upon closed countour is zero.

Once we define it like above, next two sentences are equivalent.

"Conservative vector fields are path-independent."
"A vector field that line integrals are path-independent is conservative."

3. Aug 7, 2010

### TMSxPhyFor

Dear timewalker

Tnx for your replay, personally, i think it more logical or abstract to think about it in the second way, field is gradient of a scalar -> it's path independent, then I say: the interesting thing that gradient is always pointing to the max increment of the function, and for "some" magical reasons when we doing inner product of this vector (gradient) by a dr which by the way as i realized after a long mind straggling that it is also a vector field dr=d(xi+yj+zk) and this vector field is going in circles around the origin , so by doing this inner product we actually getting function increment along this circulation (directional derivative) or we getting the projection of the gradient vectors on that circulating vector field, which happens to became get full derivative of some scalar function! that amazing and surprising for me as i can't understand why that happening, and that last thing is always path independent as a consequence!! why? what if after that product we got inexact derivative, what that mean? also it's interesting to note that simplest circulating field r=yi-xj is not path independent! I'm sure I'm missing here something obvious and very important!

Also it's useful to recall that infinitesimal circulation (curl or rotor) of conservative field is zero, it's very clear why if we know that it will be path in-depended.

I just can't fit all this parts together! pls help!

4. Aug 7, 2010

### lugita15

The attached proof isn't mathematically rigorous, but hopefully it will give you an intuitive sense of why conservative vector fields are conservative. Tell me if any of the steps are unclear.

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5. Aug 8, 2010

Have you seen the proof of the gradient theorem, which states for a scalar field $f$ and a path $\gamma$ from $\mathbf r_0$ to $\mathbf r_1$
$$\int_\gamma \nabla f \cdot d\mathbf{r} = f(\mathbf r_1) - f(\mathbf r_0)$$
?

Path independence follows from the fact that the right side depends only on the endpoints.

6. Aug 8, 2010

### quasar987

And the "gradient theorem", as many things in life , is just a facet of the fundamental theorem of calculus:

Let C be a curve joining two points r_0 and r_1 of R³.

$$\int_C\nabla f\cdot dr=\int_{t_0}^{t_1}\sum_{i=1}^3\frac{\partial f}{\partial x^i}(\gamma(t))\frac{d\gamma}{dt}dt=\int_{t_0}^{t_1}\frac{d}{dt}(f\circ\gamma)dt=(f\circ\gamma)(t_1)-(f\circ\gamma)(t_0)=f(r_1)-f(r_0)$$

Explanation: first equality is just the definition of the line integral, where \gamma is a parametrization of the curve C. Second equality is the chain rule. Third equality is the fundamental theorem of calculus.

What the above tells us is that understanding path independence of gradient vector fields, like many things in life, boils down to understanding the FTC.

7. Aug 8, 2010

### TMSxPhyFor

As I said, from pure mathematical manipulations, it's quite clear why is that happening, but again you trying to explain me the answer in that framework, which is not my goal.

I will try to rephrase my question in a more physical sense (and the answer I'm seeking for is in this sense also)

If we had a field, some point is moving on couple predefined paths , we calculating the work by summarizing infinitely many works along the path, we calculating the work on each of those paths, everything till here is cool, then we changing our field a little bit in one or another way, and recalculating works...
Suddenly, in one of our experiments, we got that work along all of this paths obtained the same value! putting in mind that those paths are very different in length, it will be very strange to me how that happens.

In the context of the above, I want explanation, not using math, the math you provided is not explaining why especially gradients have such Independence! in other words, why integrating over one variable function is always path independent, in multipliable it depends, and why if the field is gradient it's again don't depend! no math please.. I need intuition! sorry for that...

8. Aug 8, 2010

Before you can make intuitive sense of the gradient theorem, you need to have intuition about what gradients and path integrals mean.

In physics, the typical example is: the scalar field is potential energy, the gradient of potential energy is force (except for opposite sign), and the line integral is work. So when you have a particle in a conservative force field (i.e. it arises from a potential energy field), then the work done on the particle upon moving it doesn't depend on the path it takes.

9. Aug 8, 2010

### quasar987

I understand that you're trying to get an intuitive feel for why the gradient them is true. The point of my post was that if you get an intuitive feel for why each step in the proof is true, then you will have achieved your goal. Don't you agree?

10. Aug 9, 2010

### TMSxPhyFor

So when you have a particle in a conservative force field (i.e. it arises from a potential energy field), then the work done on the particle upon moving it doesn't depend on the path it takes.

You are just saying the same, conservative <=> path independent, I want to dig deeply to understand the reasons behind that.

Dear Quasar

I understand that you're trying to get an intuitive feel for why the gradient them is true. The point of my post was that if you get an intuitive feel for why each step in the proof is true, then you will have achieved your goal. Don't you agree?

I totally agree with the math you posted, but i can't see how it should affect my intuition, in one variable situation we have
$$\int\grave{A(x)}dx$$
(where A(x) is the area function from Newton-Leibenz Theorem) , thus we multiplying the rate of area function increment my our step or movement (dx), the same is with line integrals, we multiplying the rate of increment of potential by dr, but when our field is not conservative, we still multiplying it by dr, but here we also need the path that we are moving on! strange! also we always can think about non conservative field, or inexact derivative (this is my own opinion after long tries to understand what is inexact derivative means) as a directional derivative along some vector (but not parallel to the Maximum increment vector direction or gradient), in other words here we also know along which path we are moving, but still we need additional info on the path! why is that ? there is something missed!

11. Aug 9, 2010

### quasar987

Why is that strange? Suppose you observe a certain phenomenon happening under a certain set of circumstances. If you modify those circumstances, will you find it strange if the phenomenon no longer takes place?!

Here is another intuitive explanation of the phenomenon.

For any point r in R³, the gradient $\nabla f(r)$ satisfies

$$\vect{\nabla f}(r)\cdot \vect{\Delta r}\approx f(\vect{r}+\vect{\Delta r})-f(\vect{r})$$

and the approximation gets better as $\vect{\Delta r}$ gets smaller. Let C be a path joining x_0 to x_1. Also let $\{\vect{r_i}:0\leq i \leq N\}$ be a partition of C with r_0=x_0 and r_N=x_1. Set also $\vect{\Delta r_i}:=\vect{r_{i+1}}-\vect{r_i}$. Then,

$$\int_C\vect{\nabla f}(r)\cdot \vect{\Delta r}=\lim_{\mathrm{max}\vect{\Delta r_i}\rightarrow 0}}\sum_{i=0}^N\vect{\nabla f}(r_i)\cdot \vect{\Delta r_i}\approx \lim_{\mathrm{max}\vect{\Delta r_i}\rightarrow 0}}\sum_{i=0}^N(f(\vect{r_i}+\vect{\Delta r_i})-f(\vect{r_i}))=\lim_{\mathrm{max}\vect{\Delta r_i}\rightarrow 0}}(f(\vect{r_{N}})-f(\vect{r_0}))=f(\vect{x_1})-f(\vect{x_0})$$

Read this carefully before dispatching it as "just another math proof" because it doesn't get any more intuitive than that.

12. Aug 9, 2010

### TMSxPhyFor

Actually I already thought about such an approach, even so it's explaining the situation when our field is gradient, but it doesn't provide satisfactory explanation of the need for additional path in case of inexact derivatives :(

13. Aug 9, 2010

### quasar987

What do you mean "the need for additional path" ?

In the general case, path independance does not hold. To see it, it suffices to produce a counter-example. Consider for instance the vector field on R² - 0 given by (x,y)-->(-y,x). This vector field circulates counter clockwise around the origin. (Calculate its vector product product with a radial vector field if you're not convinced.)

It is easy to find two path in this field that have different line integrals.

Last edited: Aug 9, 2010
14. Aug 10, 2010

### TMSxPhyFor

Dear Quasar

Thank you for your efforts, I meant that the above mentioned equations can lead us to explain things clearly and intuitively when the field is gradient, but if it's not, it's still gives no clear reason for the need of additional info on path of integration, also the example you advices is very clear and known, but it still gives you no evidence for why that should happens (the need for the path of integration) in general when the field is not gradient.

15. Aug 10, 2010

### quasar987

Ok, so if the field is not gradient, then independance of path fail. To see this intuitively is the same as to understand intuitively why "X is path independant ==> X is gradient".

The reason for this implication is that if X is path independant, then the map $G:\mathbb{R}^3\rightarrow \mathbb{R}$

$$x\mapsto\int_0^xX\cdot dr$$

(where $\int_0^x:=\int_{\gamma}$ for $\gamma:[0,1]\rightarrow\mathbb{R}^3$ any smooth curve with $\gamma(0)=0, \ \gamma(1)=x$)

is well-defined and its gradient is X.

To see intuitively why this is true, observe that

$$\frac{\partial G}{\partial x^i}=\frac{\partial }{\partial x^i}\int_0^x X\cdot dr=\frac{\partial }{\partial x^i}\int_0^xX^1dx^1+X^2dx^2+X^3dx^3 = \frac{\partial }{\partial x^i}\int_0^xX^idx^i$$
(because if x varies a little in the i direction, then dx^j=0 for $j\neq i$)
$$=X^i(x)$$
(for essentially the same reason that
$$\frac{d}{dx}\int_0^xf(t)dt=f(x)$$
(fondamental theorem of calculus)).

16. Aug 12, 2010

### TMSxPhyFor

Dear Quasar

Thank you again and sorry for the delay..
Unfortunately I'm not well familiar with maps, can you please revise your equations without using maps? thank you in advance.

17. Aug 12, 2010

map = function

18. Aug 12, 2010

### quasar987

A "map" is a synonym for "function". As is "transformation" and "application"

19. Aug 14, 2010

### Curl

He's not asking for a mathematical proof, rather a logical explanation why it makes sense.

First consider the vector field in R^2: f(x,y)= 4i, sort of like a constant, steady wind.

If you move through the wind, it will do work on you while moving in the positive X, and no work when you move in the y direction. If you move back, it will do negative work. So the only thing that matters is the start and end points.

Use logic to convince yourself of other vector fields, try to look at graphs and see if you can find a path that you can take and have work done one way but not an equal and negative work done when you come back another way: it's impossible if the field is conservative.

20. Aug 22, 2010

### TMSxPhyFor

Eureka!! :!!) (sorry but I'm so excited!)
Sorry for the delay, and thank you all for help especially quasar987 for his efforts even so he went far away :-D
After a long straggling, Finlay I found a note in a Russian book (I love Russian books! they much deeper than English!) that lead me to the answer, and I finally closed this naughty gap in mind! (the book is "Curse of High Mathematics", V.I.Smirnov tom 2, page 354)

The basic trick is that (generally speaking) actually all directional derivatives are actually partial derivatives! non of the of books that I checked ever noted that, I'm also was stupid enough to not note that even that I'm using it everyday! but the question is why?

When we are defining it as function change over a length change on some vector (as mentioned in all books that i checked), and that is what not TRUE, actually it's not on a vector, it's on some curve! we divided in our definition only on the length of the arc, but we didn't included any information about this arc it self! this is why it's partial! the result is also depending on the carve along which is not specified!

But the nice thing is in full or Exact derivative (or gradient), we actually moving along the radios vector of the considered function it self, so the information about the path are already self included!

This is why when we integrating a gradient we don't need more info! when we integrating non gradient, we actually integrating a non-full derivative, so we simply doesn't know along which path this function's change should be considered!

That it! very simple! I glad that my intuition still working :-D, and now I believe I'm understanding Universe much better now!