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What is this in relation to?

 

Are you asking about conservative and non conservative fields or forces?

 

Gravity is conservative, friction is non conservative.

 

Or are you asking about zero v nonzero curl?

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My previous question posted by mistake. I considered that there is a relation between them. Actually I just wanted to know about the line integral. Is every line integral independent of path?

Edited by SF Shawn
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  • 5 months later...

My previous question posted by mistake. I considered that there is a relation between them. Actually I just wanted to know about the line integral. Is every line integral independent of path?

 

I am puzzled by this question. Any text that introduces path integrals will tell you the conditions under which this is true. If a path integral, [math]\int_p^q f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z) dy[/math] is independent of the path, then we could define a function F(x,y,z) by "F(x,y,z) is the integral from (0, 0, 0) to (x, y, z)". But then it would follow that [math]\frac{\partial F}{\partial x}= f[/math], [math]\frac{\partial F}{\partial y}= g[/math], [math]\frac{\partial F}{\partial z}= h[/math]. But then we would have to have [math]\frac{\partial^2 F}{\partial x\partial y}= f_y= g_x= \frac{\partial F}{\partial y\partial x}[/math], [math]\frac{\partial^2 F}{\partial x\partial z}= f_z= h_x= \frac{\partial F}{\partial z\partial x}[/math], and [math]\frac{\partial^2 F}{\partial y\partial z}= g_z= h_y= \frac{\partial F}{\partial z\partial y}[/math]. (A differential for which that is true is called an exact differential.)

 

As an easy example, if we integrate [math]\int y^2dx+ xdy+ 2dz[/math] on the straight line from (0, 0, 0), we can take as parametric equations x= y= z= t so the integral becomes [math]\int_0^1 t^2dt+ tdt+ 2dt= \int_0^1 (t^2+ t+ 2)dt= \left[\frac{1}{3}t^3+ \frac{1}{2}t^2+ 2t\right]_0^1= \frac{1}{3}+ \frac{1}{2}+ 2= \frac{17}{6}[/math].

 

But if we integrate that same integrand from (0, 0, 0) to (1, 1, 1) by taking the path (0, 0, 0) to (1, 0, 0), then to (1, 1, 0), then to (1, 1, 1) we get:

On (0, 0, 0) to (1, 0, 0) take x= t, y= 0, z= 0. The integral becomes [math]\int_0^1 0 dt= 0[/math].

On (1, 0, 1) to (1, 1, 0) take x= 1, y= t. z= 0. The integral becomes [math]\int_0^1 1 dt= 1[/math].

On (1, 1, 0) to (1, 1, 1) take x= 1, y= 1. z= t. The integral becomes [math]\int_0^1 2dt= 2[/math].

The integral from (0, 0, 0) to (1, 1, 1), along that path is the sum 0+ 1+ 2= 3.

 

Here, of course, [math]\frac{\partial y^2}{\partial y}= 2y\ne 1= \frac{\partial x}{\partial x}[/math]

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