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Closed curve line integrals of conservative vector fields

Showing that the line integral along closed curves of conservative vector fields is zero. Created by Sal Khan.

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  • aqualine ultimate style avatar for user McWilliams, Cameron
    I'm having a hard time with this: Why would a vector field NOT be conservative. The constraint of being conservative seems like a big one, but can't think of a time where a field is not conservative.
    (6 votes)
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    • leaf red style avatar for user Jt wat?
      I couldn't explain this mathematically , but I could using a simple physics example.

      Imagine we're on a planet with no atmosphere, meaning no air drag, and let's throw a ball upwards. As the ball goes upward the velocity we gave it is being converted in to height. Eventually the ball will reach its max height and start coming back downwards, converting its height in to velocity. The mechanical energy (sum of potential and kinetic energy) stays constant, or it's energy is conserved.

      Now let's imagine we're back on Earth (with an atmosphere) and we throw that same ball up, some of its velocity is still converted in to potential energy, but it's also bumping in to air particles and transferring some of its kinetic energy to these particles. The ball has now lost some of its mechanical energy, meaning it must have been acted on by non-conservative forces.

      If mechanical energy is being transferred out of an object, it's typically being acted on by a conservative field (I say typically because I've heard different definitions of mechanical energy).

      Sorry if the example is too simplistic.
      (17 votes)
  • blobby green style avatar for user jackhorkheimer
    A bit confusing because my book (Multivariable Calculus by Stewart) uses F to refer to the vector function and f to refer to the scalar function. So the gradient of f is F.

    Not sure what the standard is, just a heads up
    (3 votes)
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  • blobby green style avatar for user plankjames
    If doing a line integral gives you the surface area above a curve, then how can it be 0 for a closed line integral that's path independent?
    (2 votes)
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    • leaf blue style avatar for user Stefen
      You can think of it like this: there are 3 types of line integrals:
      1) line integrals with respect to arc length (dS)
      2) line integrals with respect to x, and/or y (surface area dxdy)
      3) line integrals of vector fields
      That is to say, a line integral can be over a scalar field or a vector field.
      You are confusing integration over a scalar field (surface area) versus over a vector field (and in this case, a vector field that is conservative).
      (4 votes)
  • blobby green style avatar for user Rateeb
    Can you relate this to the work done on the particle by the vector field f?
    (2 votes)
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  • blobby green style avatar for user Olivia Plumb
    If the line integral on a vector field is equal to the work done, then won't the line integral over a closed curve always be equal to zero, since displacement is zero, whether or not the vector field is conservative?
    (3 votes)
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    • male robot hal style avatar for user Pavan Gaonkar
      vector field being conservative is very important since otherwise the line integral wont be zero.. for a physical example consider a block travelling a closed path on a surface having friction.. the work done will not be zero because friction is not a conservative force.. so there will be energy loss due to friction
      (1 vote)
  • mr pants teal style avatar for user SanFranGiants
    So how do we know when the vector field can be expressed as a gradient?
    (2 votes)
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  • sneak peak green style avatar for user Alex Knauth
    Is the curl of a conservative vector field always zero? From the picture it seems like it should be, but that's just a wild guess. And would that mean that all vector fields with 0 curl are conservative?

    Edit: I looked on Wikipedia, and it says that the curl of the gradient of a scalar field is always 0, which means that the curl of a conservative vector field is always zero.
    But then can you go the other way and say that a vector field is conservative if it has a curl of 0?
    http://en.wikipedia.org/wiki/Vector_calculus_identities#Curl_of_the_gradient

    Further Edit: Yay Stokes' Theorem!
    (2 votes)
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    • blobby green style avatar for user Karan Patel
      Yes that is exactly correct. The other way to prove that a vector field is conservative is to show that the dy of the i component = the dx of the j component, the dx of the k component = dz of the i component, and the dz of the j component = the dy of the k component. Essentially, this is the same thing as finding the curl of the vector field and making sure that the curl gives you 0 vector. If the curl does not give you 0 vector, it is not conservative.
      (2 votes)
  • starky ultimate style avatar for user Darren Leung
    I have a question. How can the area be zero?
    (2 votes)
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  • blobby green style avatar for user periyal.arun
    Is this line integral anyway related to stokes theorem?
    If so do we have to consider the "conservative"ness?
    (1 vote)
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  • blobby green style avatar for user comptonpark
    You mention that "the line integral is path independent between any two points". This statement is only true if Rot(f)=0 right? Or a field can only be conservative if Rot(f)=0?
    (1 vote)
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    • spunky sam blue style avatar for user Ethan Dlugie
      You are correct. Path independence is only valid for conservative vector fields (as the title suggests). This concept is very important in physics when one starts to analyze work and energy transitions. The work done to move an object from one height to another is independent of whatever crazy winding path you took between the two points. Why? Because the vector field representing the force of gravity is non-curling, it can be written as the gradient of some potential scalar function.
      (1 vote)

Video transcript

In the last video, we saw that if a vector field can be written as the gradient of a scalar field-- or another way we could say it: this would be equal to the partial of our big f with respect to x times i plus the partial of big f, our scalar field with respect to y times j; and I'm just writing it in multiple ways just so you remember what the gradient is --but we saw that if our vector field is the gradient of a scalar field then we call it conservative. So that tells us that f is a conservative vector field. And it also tells us, and this was the big take away from the last video, that the line integral of f between two points-- let me draw two points here; so let me draw my coordinates just so we know we're on the xy plane. My axes: x-axis, y-axis. Let's say I have the point, I have that point and that point, and I have two different paths between those two points. So I have path 1, that goes something like that, so I'll call that c1 and it goes in that direction. And then I have, maybe in a different shades of green, c2 goes like that. They both start here and go to there. We learned in the last video that the line integral is path independent between any two points. So in this case the line integral along c1 of f dot dr is going to be equal to the line integral of c2, over the path c2, of f dot dr. The line, if we have a potential in a region, and we may be everywhere, then the line integral between any two points is independent of the path. That's the neat thing about a conservative field. Now what I want to do in this video is do a little bit of an extension of the take away of the last video. It's actually a pretty important extension; it might already be obvious to you. I've already written this here; I could rearrange this equation a little bit. So let me do it. So let me a rearrange this. I'll just rewrite this in orange. So the line integral on path c1 dot dr minus-- I'll just go subtract this from both sides --minus the line integral c2 of f dot dr is going to be equal to 0. All I did is I took this take away from the last video and I subtracted this from both sides. Now we learned several videos ago that if we're dealing with a line integral of a vector field-- not a scalar field --with a vector field, the direction of the path is important. We learned that the line integral over, say, c2 of f dot dr, is equal to the negative of the line integral of minus c2 of f dot dr where we denoted minus c2 is the same path as c2, but just in the opposite direction. So for example, minus c2 I would write like this-- so let me do it in a different color --so let's say this is minus c2, it'd be a path just like c2-- I'm going to call this minus c2 --but instead of going in that direction, I'm now going to go in that direction. So ignore the old c2 arrows. We're now starting from there and coming back here. So this is minus c2. Or we could write, we could put, the minus on the other side and we could say that the negative of the c2 line integral along the path of c2 of f dot dr is equal to the line integral over the reverse path of f dot dr. All I did is I switched the negative on the other side; multiplied both sides by negative 1. So let's replace-- in this equation we have the minus of the c2 path; we have that right there, and we have that right there --so we could just replace this with this right there. So let me do that. So I'll write this first part first. So the integral along the curve c1 of f dot dr, instead of minus the line integral along c2, I'm going to say plus the integral along minus c2. This-- let me switch to the green --this we've established is the same thing as this. The negative of this curve, or the line integral along this path, is the same thing as the line integral, the positive of the line integral along the reverse path. So we'll say plus the line integral of minus c2 of f dot dr is equal to 0. Now there's something interesting. Let's look at what the combination of the path of c1 and minus c2 is. c1 starts over here. Let me get a nice, vibrant color. c1 starts over here at this point. It moves from this point along this curve c1 and ends up at this point. And then we do the minus c2. Minus c2 starts at this point and just goes and comes back to the original point; it completes a loop. So this is a closed line integral. So if you combine this, we could rewrite this. Remember, this is just a loop. By reversing this, instead of having two guys starting here and going there, I now can start here, go all the way there, and then come all the way back on this reverse path of c2. So this is equivalent to a closed line integral. So that is the same thing as an integral along a closed path. I mean, we could call the closed path, maybe, c1 plus minus c2, if we wanted to be particular about the closed path. But this could be, I drew c1 and c2 or minus c2 arbitrarily; this could be any closed path where our vector field f has a potential, or where it is the gradient of a scalar field, or where it is conservative. And so this can be written as a closed path of c1 plus the reverse of c2 of f dot dr. That's just a rewriting of that, and so that's going to be equal to 0. And this is our take away for this video. This is, you can view it as a corollary. It's kind of a low-hanging conclusion that you can make after this conclusion. So now we know that if we have a vector field that's the gradient of a scalar field in some region, or maybe over the entire xy plane-- and this is called the potential of f; this is a potential function. Oftentimes it will be the negative of it, but it's easy to mess with negatives --but if we have a vector field that is the gradient of a scalar field, we call that vector field conservative. That tells us that at any point in the region where this is valid, the line integral from one point to another is independent of the path; that's what we got from the last video. And because of that, a closed loop line integral, or a closed line integral, so if we take some other place, if we take any other closed line integral or we take the line integral of the vector field on any closed loop, it will become 0 because it is path independent. So that's the neat take away here, that if you know that this is conservative, if you ever see something like this: if you see this f dot dr and someone asks you to evaluate this given that f is conservative, or given that f is the gradient of another function, or given that f is path independent, you can now immediately say, that is going to be equal to 0, which simplifies the math a good bit.