Intuition for conservative field

[aaaa202]

A scalar field is conservative, i.e. the line integral does not depend on the path taken, if it has a gradient.

Now, can someone give me intuition behind why the gradient would have something to do with this? :)

For me the gradient is merely a way of writing up the partial derivates as a vector, which then points in the direction where the field is steepest.

 

[LCKurtz]

A scalar field is conservative, i.e. the line integral does not depend on the path taken,

You mean a vector field is conservative...

if it has a gradient.

If WHAT has a gradient? Any differentiable function has a gradient. And that isn't what you mean anyway. The condition is that the vector field ##\vec F## has a potential function ##\phi## which means the ##\vec F =\nabla \phi##. I don't know if it will help your intuition that much, but you might want to look at the proof where independence of path is assumed and it gives the argument to build a potential function. That might help you see why a force field including friction along the path isn't conservative.

 

[HallsofIvy]

You seem to have a lot of misunderstandings here. I think you are reversing the "scalar field" and its "gradient". It doesn't make sense to talk about "the line integral does not depend on the path taken" for a scalar field. A line integral is something of the form [itex]\int f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz[/itex] which can be interpreted as the dot product of the vector function [itex]<f(x,y,z), g(x,y,z), h(x,y,z)>[/itex] with the "differential of the path", <dx, dy, dz>.

What is true is that the integral (between two given points) of such a vector valued function is independent of the path used if it is a gradient, not if it has one. That's because if there exist some function F(x,y,z) so that [itex]\nabla F= <f(x,y,z), g(x,y,z), h(x,y,z)>[/itex] then [itex]f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz= (f(x,y,z)(dx/dt)+ g(x,y,z)(dy/dt)+ h(x,y,z)(dz/dt))dt= dF/dt[/itex], where t is some parameter for the path.

That means that the integral along any path, from [itex](x_0, y_0, z_0)[/itex] to [itex](x_1, y_1, z_1)[/itex] is [itex]\int f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz= \int (f(x,y,z)(dx/dz)+ g(x,y,z)(dy/dt)+ h(x,y,z)(dz/dt))dt= F(x_1, y_1, z_1)- F(x_0, y_0, z_0)[/itex]. That last part depends only on the endpoints, not the specific path.