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

  1. Aug 7, 2010 #1
    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. jcsd
  3. Aug 7, 2010 #2
    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."
     
  4. Aug 7, 2010 #3
    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!
     
  5. Aug 7, 2010 #4
    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.
     

    Attached Files:

  6. Aug 8, 2010 #5
    Have you seen the proof of the gradient theorem, which states for a scalar field [itex]f[/itex] and a path [itex]\gamma[/itex] from [itex]\mathbf r_0[/itex] to [itex]\mathbf r_1[/itex]
    [tex]\int_\gamma \nabla f \cdot d\mathbf{r} = f(\mathbf r_1) - f(\mathbf r_0)[/tex]
    ?

    Path independence follows from the fact that the right side depends only on the endpoints.
     
  7. Aug 8, 2010 #6

    quasar987

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    And the "gradient theorem", as many things in life :rolleyes:, 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.

    [tex]\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)[/tex]

    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.
     
  8. Aug 8, 2010 #7
    Dear lugita15,adriank,quasar987 thx for your answers.

    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...
     
  9. Aug 8, 2010 #8
    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.
     
  10. Aug 8, 2010 #9

    quasar987

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    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?
     
  11. Aug 9, 2010 #10
    Dear adriank

    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
    [tex]\int\grave{A(x)}dx[/tex]
    (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!
     
  12. Aug 9, 2010 #11

    quasar987

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    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 [itex]\nabla f(r)[/itex] satisfies

    [tex]\vect{\nabla f}(r)\cdot \vect{\Delta r}\approx f(\vect{r}+\vect{\Delta r})-f(\vect{r})[/tex]

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

    [tex]\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})[/tex]

    Read this carefully before dispatching it as "just another math proof" because it doesn't get any more intuitive than that.
     
  13. Aug 9, 2010 #12
    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 :(
     
  14. Aug 9, 2010 #13

    quasar987

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    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
  15. Aug 10, 2010 #14
    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.
     
  16. Aug 10, 2010 #15

    quasar987

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    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 [itex]G:\mathbb{R}^3\rightarrow \mathbb{R}[/itex]

    [tex]x\mapsto\int_0^xX\cdot dr[/tex]

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

    is well-defined and its gradient is X.

    To see intuitively why this is true, observe that

    [tex]\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[/tex]
    (because if x varies a little in the i direction, then dx^j=0 for [itex]j\neq i[/itex])
    [tex]=X^i(x)[/tex]
    (for essentially the same reason that
    [tex]\frac{d}{dx}\int_0^xf(t)dt=f(x)[/tex]
    (fondamental theorem of calculus)).
     
  17. Aug 12, 2010 #16
    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.
     
  18. Aug 12, 2010 #17
    map = function
     
  19. Aug 12, 2010 #18

    quasar987

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    A "map" is a synonym for "function". As is "transformation" and "application"
     
  20. Aug 14, 2010 #19
    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.
     
  21. Aug 22, 2010 #20
    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!
     
  22. Aug 22, 2010 #21

    quasar987

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    I'm glad to hear it!
     
  23. Aug 22, 2010 #22

    HallsofIvy

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    Well, no, they are not- I think it would be better to say that the "partial derivatives" with respect to x and y are specifically the directional derivatives in the direction of the positive x and y axes. Rather than saying that all direction derivatives are partial derivatives, I would say that the partial derivatives are special directional derivatives.

    I think you would be better served by thinking of the "derivative" of a function as something different from the partial derivatives. There exist simple functions that have partial derivatives at a given point but are not "differentiable" at that point. For real valued functions of several variables, it is better to think "the" derivative as being the gradient vector.

    ?? What do you mean by the "radius vector of the function itself"? If you mean the position vector from the origin to the given point, that is not, in general, the direction the gradient points.

    I'm afraid that doesn't quite make sense to me. It is true that when we are integrating a gradient, it does not matter what the path was- but that was your original question- why does it not matter. I don't know what you mean by saying "we simplydon't know along which path this function's change should be considered". If we are asked to integrate along a given path, then that is the path we need to consider!


    I think what you are saying is this:

    To say that a differential is "conservative" (Which is actually a physics word. I prefer "exact" differential which you use above, but not, I think in this way) just means that it really is the differential of a function rather than just 'looking' like one.

    For example, I can write 2xdx+ xydy but there is NO function f such that df= 2xdx+ xydy.
    2xdx+ xydy looks like a differential but isn't one, really. That is true because if there were such a function, f, we would have
    [tex]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy= 2xdx+ xydy[/tex] so that
    [tex]\frac{\partial f}{\partial x}= 2x[/tex]
    and
    [tex]\frac{\partial f}{\partial y}= xy[/itex]

    But in that case, we would have
    [tex]\frac{\partial f}{\partial x\partial y}= \frac{\partial}{\partial y}\left(2x\right)= 0[/tex]
    and
    [tex]\frac{\partial f}{\partial y\partial x}= \frac{\partial}{\partial x}\left(xy\right)= x[/itex]

    But that is impossible- as long as the derivatives are constant, the "mixed derivatives" must be equal. There is NO f such that df= 2xdx+ xy dy- it is not an exact differential.

    But your original question was about path integrals when the integrand is exact. In that case, there exist a function, f(x,y), such that df= u(x,y)dx+ v(x,y)dy which means that for any path and any parameterization, using, say, t, as the parameter,
    [tex]\frac{df}{dt}= u(x,y)\frac{dx}{dt}+ v(x,y)\frac{dx}{dt}[/tex]

    In particular, integrating u(x,y)dx+ v(x,y)dy along a specific path, C, from p1 to p2, parameterized, with parameter t, so that we are at p1 when t= t1 and at p2 when t= p2,
    then
    [tex]\int_C u(x,y)du+ v(x,y)dy= \int_{t1}^{t2}\left(u(x(t),y(t))\frac{du}{dt}+ v(x(t),y(t))\frac{dv}{dt}\right)dt[/tex]
    [tex]= \int_{t1}^{t2} \(\frac{df}dt}\)dt= \int_{t1}^{t2} df= f(t2)- f(t1)[/tex]
    by the "Fundamental Theorem of Calculus" and, so, is independent of the particular path.

    In calculus of functions of one variable, as long as u(x) is continuous (and often if it is not- as long as it is bounded and its set of points of discontinuity has measure 0), there exist F such that dF= u(x)dx so we an always use the fundamental theorem of calculus. For functions of several variables, that is not necessarily true.
     
    Last edited by a moderator: Aug 22, 2010
  24. Aug 22, 2010 #23
    Dear HallsofIvy

    Thank you for your extensive repaly, maybe i was not very clear becuase i thought that every body knows what i was speaking about, anyway:

    Suppose we have:
    [tex]\delta f = u(x,y)dx + v(x,y)dy[/tex]
    (here i using Thermodynamics notation of inexact derivative, I believe it's more accurate)
    Now if this a full derivative, it will satisfy the conditions you mentioned in your post, but lets suppose it's not!, then we can write:
    [tex]\delta f = \left\{ {u(x,y),v(x,y)} \right\} \cdot \left\{ {dx,dy} \right\} = \vec E \cdot d\vec r[/tex]
    But now lets recall from the course of deferential equations when we proving that each non exact derivative has a multiplier (which by the way not unique), that can convert it to exact derivative, so lets suppose that we found such a multiplier that will always satisfy the condition:
    [tex]1 \le \left| {w(x,y)} \right|[/tex]
    then we can write that:
    [tex]\begin{array}{l}
    dF = w(x,y) \cdot \delta f = \left\{ {\vec E \cdot w(x,y)} \right\} \cdot d\vec r = \\
    = \frac{{\vec E}}{{\cos \left[ {\theta (x,y)} \right]}} \cdot d\vec r = \frac{{\nabla F \cdot \cos \theta }}{{\cos \theta }} \cdot d\vec r = \nabla F \cdot d\vec r
    \end{array}[/tex]
    of course here i used the idea that actually we can suppose that any inexact derivative is actually a directional derivative (but not along the normal vector which will make it exact).

    Now lets get back to my previous post, actually, acording to the mentioned book, the right defination for the directionl deravative will be:
    [tex]\frac{{\partial f}}{{\partial u}} = \mathop {\lim }\limits_{{M_2} \to {M_2}} \frac{{f({M_2}) - f({M_1})}}{{\widehat{{M_1}{M_2}}}}[/tex]
    (Above M1M2 is Arc sign)
    That means that we actually taking a ration between our function change to the length of the path that we are going along from point M1 to M2, it is NOT just the distance between two points M1M2 as other books states! anyway, on the infinitely small scales, that don't mater us at all, as we always can consider this infinitely small peace of arc as a straight line that represented by the tangent vector, this way most people doesn't notice the difference, and this why we writing the directional derivative as a partial derivative not full, because actually the change in our function is not only relates to the length of the path that we going along, but also to the path it self, but when we integrating, we summarizing infinite number of this very small changes along very small lines, but actually those lines are not arranged in a straight line, but according to some path, that we in general can not know unless it been specified explicitly!
    Now I think it's clear how it differs from exact derivative, gradient by definition means that we taking the directional derivative along the normal vector of the function itself, thus why we writing :
    [tex]df = \nabla f \cdot \overrightarrow {dr} = \left\| {\nabla f} \right\| \cdot ds[/tex]
    which is exact derivative, because here we already included information along which path we getting function change, it is along the curve it self! and that will be the answer to:
    Sorry but I can't see why we can't say the opposite too.

    Yes i faced such a statement in many references, but unfortunately non of them gives an example of how that could be? do you have any? thanx in advance.

    Sorry it's wrong, I should to say we are moving along the carve itself that we calculating the derivative on it.
     
  25. Aug 22, 2010 #24
    A function that has all partial derivatives but is not differentiable at (0, 0) is f(x, y) = xy2 / (x2 + y2), f(0, 0) = 0. Both partial derivatives ∂f/∂x and ∂f/∂y at (0, 0) are zero, but the directional derivative at (0, 0) in the direction of (1, 1) is 1/2, which is not ∂f/∂x + ∂f/∂y, which it would be if f was differentiable at 0. Thus, the gradient doesn't actually exist at (0, 0).
     
  26. Aug 23, 2010 #25
    I think perhaps you should write down your definition of "conservative vector field." When I learned this, we defined it as a vector field which is the gradient of function. Thus, if the path [itex]C[/itex] starts at point [itex]P[/itex] and ends at [itex]Q[/itex], and [itex]\vec{v}=\nabla \phi [/itex] is a conservative (i.e. gradient) vector field, then we get the equivalent of a "Fundamental Theorem of Caclulus" but for vector fields.

    [tex]\int_C \vec{v} \cdot d\vec{r} = \phi(Q) - \phi(P)[/tex]

    Since the line integral does not depend on the path itself -- only the initial and end points, it is path-independent. On the other hand, if you do not have a gradient vector field, then it is possible that different paths (with the same starting and ending points) do not give the same results.

    The easiest example I can think of is to just consider two arbitrary points [itex]P[/itex] and [itex]Q[/itex] and then two paths: [itex]C_1[/itex] starting at [itex]P[/itex] and ending at [itex]Q[/itex]; [itex]C_2[/itex] at [itex]Q[/itex] and ending at [itex]P[/itex]. If we take the line integral over the closed loop [itex]C=C_1+C_2[/itex] then we should (intuitively?) expect the result to be 0 for conservative fields. (If you think of line integrals as just integrals which compute the work done by a potential field over the paths, then a conservative force should conserve energy).

    [tex]\int_{C_1} \vec{v} \cdot d\vec{r} + \int_{C_2} \vec{v} \cdot d\vec{r} = 0[/tex]

    So the line integral over each path has the same magnitude. Accounting for direction, it should be (intuitvely?) clear that two line integrals are equal if the two paths start and end at the same point, and the vector field is conservative.
     
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