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I recently started learning multivariable calculus and I came across some questions regarding conservative vector fields and it's a bit confusing.

If I took the derivative of my potential function (as in $\nabla f(x,y,z)$), it will give me a vector field. But if I integrated my vector field, would it give me the potential function as well, or would it give me a completely different function? If I encounter a problem where I have to find the potential function of a conservative vector field, should I approach the problem as an initial value condition problem except with multiple independent variables?

First of all, I can't imagine the problem graphically. What does it mean for a vector field to be conservative? I don't get why my professor said it has to be a closed path in order to find the potential. What's the meaning of this symbol $\oint$? How does this relate to other real-world applications? I don't really understand much of this at all.

Note: I don't have a good background in physics, so it doesn't really help to say "know your physics" as others have said to me. And I just started learning line integrals, so I would really love some insight that allows to me view the problem more easily.

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Assume that your vector field is for force. It will make the calculations easier. Any other sort of thing differs only be a constant factor.

The amount of work done on an object by the field is defined as $$\int \mathbf{F} \cdot \mathrm{d} \mathbf{r}$$ where the integral is taken over the path the object takes. The important thing to realize is that this is independent of how fast the object moved along this path.

Suppose there existed a closed path $P$ such that this integral were not $0$. Then by going around this loop over and over again, you'd either be able to achieve arbitrary large positive work done on the object (in which case you'd get infinite energy), or arbitrarily large negative work done on you. By going the other direction, you'd be able to attain the former case.

Thus, for all "natural" force fields, every integral around a closed path of the above form should be $0$. Such a vector field is said to be conservative. In a conservative vector field, there is an associated "potential field" (well, technically an infinite number of ones each differing by a constant), found by picking some arbitrary location $J$ as $0$ potential, and setting the potential at a particular point $A$ to be $$-\int \mathbf{F} \cdot \mathrm{d} \mathbf{r}$$ where the integral is taken along any path $J$ to $A$. Because the field is conservative, such a definition is consistent.

Now, if we denote the potential field by $U$, we obtain the very nice looking identity

$$W = \int \mathbf{F} \cdot \mathrm{d} \mathbf{r} = U(A) - U(B)$$

where the integral is taken along any path from $A$ to $B$.

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  • $\begingroup$ While I do get some of the steps in your explanation, when you stated that "every integral around a closed path of the above form should be 0", are you implying that no matter what path the integral takes, the work must always equal 0? So $U(A) = U(B)$? $\endgroup$ – Daniel Lee Jun 5 '18 at 5:02
  • $\begingroup$ No. When I say closed path, I mean one where the start is he same as the end $\endgroup$ – extremeaxe5 Jun 5 '18 at 7:01

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