
#1
Jan1411, 03:31 AM

P: 20

Hey guys,
I'm following a course on vector calculus and I'm having some trouble connecting things. Suppose we have a threedimensionale vectorfield F(x,y,z)=(F1,F2,F3) and suppose we have a potential phi for this. So: F=grad(phi). The field lines of a vector field are defined as d(r)/dt = lamba(t)F(r(t)). In words: the derivative of the field line (which is parametrized bij r) is parallel to F(r(t)). Now for my question. Suppose we have equipotential lines, so phi=c with c a constant. What's the connection between these equipotential lines, the field lines and the vector field in terms of being parallel or rightangled? I think the answer should be that the field lines are rightangled with the equipotential lines, but I don't know how to deduce this. Thanks in advance! 



#2
Jan1411, 10:10 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

Let f(x,y,z) be the value of the potential field at each point (x,y,z). Then the vector [itex]grad f= \nabla f[/itex] points in the direction of the "field lines", the lines of fastest increase of the function f. Further, the rate of increase of f in the direction of unit vector [itex]\vec{v}[/itex] is given by [itex]\nabla f\cdot \vec{v}[/itex]. That is, the direction in which the derivative is 0, the "equipotential lines" (strictly speaking, in three dimensions, they would be equipotential surfaces) is exactly the direction in which that dot product is 0, the direction in which the vector [itex]\vec{v}[/itex] is perpendicular to [itex]\nabla f[/itex] and so perpendicular to the field lines.



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