Use Stokes's on Line Integral to Show Path Independence

In summary: But what about a toroidal or cylindrical field?Consider a single-turn copper coil with a mixture of ∂B/∂t of both polarities thruout its area. Curl of the E field = -∂B/∂t could be nonzero everywhere within its planar area but ∫curl E*dA = 0 is entirely possible, so that the circulation of E (the emf) around the coil would be zero too. Of course, what you say is entirely correct for a lamellar field like gravity or electrostatic E field.But what about a toroidal or cylindrical field?The curl of the electric field around a toroidal
  • #1
sikrut
49
1
Use Stokes's theorem to show that line integral of ##\vec{F}(\vec{r})## over an curve ##L##, given by ##\int_L \vec{F}(\vec{r}) d\vec{r}##, depends only on the start and endpoint of ##L##, but not on the trajectory of ##L## between those two points.

Hint: Consider two different curves, ##L## and ##M##, say, which share a common start and endpoint. Construct from them a closed curve ##C## to which Stokes's theorem can be applied.

$$\vec{F}(\vec{r}) \equiv [F_x(\vec{r}), F_y(\vec{r}), F_z(\vec{r})] = [2z^2, 3z^2, (4x+6y)z] \rightarrow where \rightarrow \vec{r} \equiv [x,y,z].$$
 
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  • #2
First, you need to remember that Stoke's theorem says that $$\iint\limits_S curl( \vec{F} ) d \vec{S} = \oint\limits_C \vec{F} d \vec{r}$$ You need to compute ##curl( \vec{F})## and show that it is the zero vector. This would mean that any closed line integral will be zero. This makes the integral over the path ##C## zero, which forces the two line integrals ##\int\limits_L \vec{F} d\vec{r}## and ##\int\limits_M \vec{F} d \vec{r}## to be equal.

Good Luck!
 
  • #3
It's not necessarily the curl of F that needs to be zero, but the integration over the circumscribed area of the curl dotted into an element of normal area.
 
  • #4
rude man said:
It's not necessarily the curl of F that needs to be zero, but the integration over the circumscribed area of the curl dotted into an element of normal area.

this is true, but I think if the curl is nonzero at some point then you could find a surface that makes the surface integral nonzero (you would need some sort of continuity assumption to prove this). Since this problem is talking about an arbitrary curve, L, I believe the only possibility is to show that the curl itself is zero.

I could imagine, however, a problem stated "for an arbitrary path on the surface, S, show that..." in which the curl isn't zero, but the same property holds.
 
  • #5
hapefish said:
this is true, but I think if the curl is nonzero at some point then you could find a surface that makes the surface integral nonzero (you would need some sort of continuity assumption to prove this). Since this problem is talking about an arbitrary curve, L, I believe the only possibility is to show that the curl itself is zero.

I could imagine, however, a problem stated "for an arbitrary path on the surface, S, show that..." in which the curl isn't zero, but the same property holds.

Consider a single-turn copper coil with a mixture of ∂B/∂t of both polarities thruout its area. Curl of the E field = -∂B/∂t could be nonzero everywhere within its planar area but ∫curl E*dA = 0 is entirely possible, so that the circulation of E (the emf) around the coil would be zero too.

Of course, what you say is entirely correct for a lamellar field like gravity or electrostatic E field.
 

What is Stokes's Theorem?

Stokes's Theorem is a mathematical theorem that relates the integral of a vector field over a surface to the line integral of the same vector field along the boundary of the surface.

How is Stokes's Theorem used to show path independence?

Stokes's Theorem can be used to show path independence by showing that the line integral of a vector field along a closed loop is equal to the surface integral of the curl of the same vector field over the enclosed surface.

What is the significance of path independence?

Path independence is important because it allows us to evaluate line integrals without needing to know the path taken. This simplifies calculations and makes it easier to analyze vector fields.

What are the conditions for using Stokes's Theorem?

The conditions for using Stokes's Theorem are that the vector field must be continuous and differentiable, and the surface must be smooth and orientable.

Can Stokes's Theorem be used for any type of surface?

No, Stokes's Theorem can only be used for smooth and orientable surfaces. If the surface is not smooth or not orientable, then the theorem cannot be applied.

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