# Conservative vector field

1. Apr 9, 2015

### Calpalned

1. The problem statement, all variables and given/known data

Secondly, is a "conservative vector field" the same thing as a "conservative force" (a force that does not depend on the path taken)? If not what is it?

2. Relevant equations
n/a
3. The attempt at a solution
How was $f(x,y,z)=\frac{mMG}{\sqrt{x^2+y^2+z^2}}$ derived?

Last edited: Apr 9, 2015
2. Apr 9, 2015

### MarcusAgrippa

A conservative (vector) field is the mathematical generalisation of a conservative force field in physics. It is a vector field that can be written as the negative gradient of a scalar function. That is
$\vec{F} = - \nabla \phi$
The negative sign is pure convention, introduced to match the physics definition.

An alternative, equivalent, definition is that it is a vector field $\vec{X}$ that has the property that the path integral
$\int_1^2\vec{F}\cdot d\vec{x}$
has the same value along any path that joins point 1 to point 2.

A necessary consequence of either of the above two definitions is that the vector field has zero curl,
$\nabla \times \vec{F} = 0$
The converse of this result is not necessarily true, except in the case where the space containing points 1 and 2 is simply connected. If you did not understand this last statement, forget that I said it and return to this point in a few years time when you know a bit of topology.

In gravitation and electrostatics, the conservative property is interpreted in terms of the conservation of energy. This is interpretation is not possible (or even relevant) in general, and the term "conservative field" is to be understood to mean the above two mathematical definitions.

Last edited: Apr 9, 2015
3. Apr 9, 2015

4. Apr 9, 2015

### MarcusAgrippa

Either by performing the integration, or else by guessing a function whose negative gradient gives the correct answer. Do you know yet how to perform line integrals?

5. Apr 9, 2015

### Calpalned

My textbook talks about "closed curves". I assume this means curves where the path taken leads to the same end/beginning point. In other words, there is distance but no displacement. I am under the impression that "... is independent of path" if and only if the curve is closed.

6. Apr 9, 2015

### Calpalned

Yes, I can do line integrals. However, my textbook didn't show me how to do a gradient line integral. (assuming that f(x,y,z) is the integral of the gradient)

7. Apr 9, 2015

### MarcusAgrippa

The closed curve property is equivalent to the property that the line integral has the same value for all possible curves that join a given initial point to a given final point. This means that the value of the integral depends only on the initial and the final points and not on the path used to join them.

Every closed path can be considered as made up of two curves, the first is the portion of the chosen closed curve that begins at any point 1 on the closed curve and ending at another arbitrary point 2 on the closed curve, and the second is the curve that runs from point 2 along the other part of the closed curve to point 1. Reverse the direction of this second curve so that it now runs from point 1 to point 2. You now have 2 curves that join point 1 to point 2. When you go in the opposite direction along a curve, you evaluate the same integral, but with the limits reversed so you get the negative of the original integral. So we can write
$\oint_C \vec{X} \cdot d\vec{x} = {_{C_1}\int}_1^2 \vec{X} \cdot d\vec{x} + {_{C_2}\int}_2^1\vec{X} \cdot d\vec{x} = {_{C_1}\int}_1^2 \vec{X} \cdot d\vec{x} - {_{C_2}\int}_1^2\vec{X} \cdot d\vec{x}$
So
$\oint_C \vec{X} \cdot d\vec{x} = 0 \ \ \Leftrightarrow \ \ {_{C_1}\int}_1^2 \vec{X} \cdot d\vec{x} = {_{C_2}\int}_1^2\vec{X} \cdot d\vec{x}$

Last edited: Apr 9, 2015
8. Apr 9, 2015

### MarcusAgrippa

Ok. Then do it by evaluating a line integral. Alternatively, take the answer they give you and calculate its gradient. I see you book has defined f without the unnecessary negative sign.

Note that I have edited #2.

9. Apr 9, 2015

### HallsofIvy

No, "independent of path" does NOT apply "if and only if the curve is closed". For example, we if $ds= ydx+ x dy$, then the integral $\int_{p_0}^{p_1} ds$ depends only on the endpoints, $p_1$ and $p_0$, not upon the particular path between the endpoints.

I had never seen the term "conservative vector field". I have, of course, seen "conservative force field" to mean a force field in which the work done by or against the force, between two points, is independent of the path. Personally, for mathematics rather than physics, I prefer to use mathematics terms. A differential, $ds= f(x,y)dx+ g(x,y)dy$ (or $ds= f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz$ in three dimensions) is said to be an "exact differential" if and only if there exist a function F(x,y) such that $dF= f(x,y,z)dx+ g(x,y,z)dy$ (or $dF= f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz$ in three dimensions).

10. Apr 9, 2015

### MarcusAgrippa

I suspect Calpalned was careless with his choice of words.

See Marsden and Tromba, Vector Calculus, 5 ed, p 551.

Your terminology is preferable, but I don't expect that as a beginning student Calpalned will have covered Pfaffian forms or to have available the concepts of closed and exact 1-forms. Too advanced.

Last edited: Apr 9, 2015
11. Apr 9, 2015

### Calpalned

I just found out how was $f(x,y,z)=\frac{mMG}{\sqrt{x^2+y^2+z^2}}$ derived. It's known as a potential. My textbook skipped deriving it, so it was very confusing.

12. Apr 9, 2015

### MarcusAgrippa

Correct. The function f in $\vec{X} = - \nabla f$ is called a potential for the conservative field.

13. Apr 9, 2015

### SammyS

Staff Emeritus
Your book said that it stated that the gradient of this function gives the desired force field in Section 16.1. Maybe it wasn't derived, but it was given.

14. Apr 10, 2015

### vanhees71

Ad #10: In this textbook excerpt there's a subtle mistake. In (iv) you have to add that $\vec{\nabla} \times \vec{F}=0$ implies the existence of a potential only in simply connected domains of $\vec{F}$.

A counter example is the "potential vortex"
$$\vec{F}(\vec{x})=\frac{1}{x^2+y^2} \begin{pmatrix} -y \\ x \\0 \end{pmatrix}$$
You can show that for all $x^2+y^2 \neq 0$
$$\vec{\nabla} \times \vec{F}=0,$$
but that the circulation of this vector field for any curve running around the $z$ axis once is $2 \pi$.

Nevertheless the field has a potential in every simply connected domain, not containing the $z$ axis. The maximal such domains are given by taking out a surface with the $z$ axis as a boundary, which you can choose arbitrary otherwise. The corresponding potential has a jump along that surface.

15. Apr 10, 2015

### MarcusAgrippa

No, the book does not contain a subtle mistake. The book states quite clearly on what domain the vector field must be defined: $\vec{F} \mbox{ is defined on } R^3, \mbox{ not } R^2, \mbox{ or } R^3$ with a line or curve removed. The book state explicitly that the field is assumed to be defined everywhere in R^3 except perhaps at a finite number of points. That makes the domain of the vector field simply connected.

If one works in $R^2$, the the book is still correct provided that the vector field is defined everywhere, since the domain is simply connected. If the field fails to be defined at any point on $R^2$, then the domain fails to be simply connected and the result is true only in simply connected subdomains. The book does not cover this case.

But thank you for making that point. It is an important point of which many are unaware. I made that point earlier in my comments when I said:

"A necessary consequence of either of the above two definitions is that the vector field has zero curl,
$\nabla \times \vec{F} = 0$
The converse of this result is not necessarily true, except in the case where the space containing points 1 and 2 is simply connected. If you did not understand this last statement, forget that I said it and return to this point in a few years time when you know a bit of topology."

Last edited: Apr 10, 2015
16. Apr 10, 2015

### vanhees71

Ah, sorry, I've overlooked the condition given in the very beginning of the quoted text. Of course, in $\mathbb{R}^3$, if the field is well defined everywhere except for a finite number of points, the domain is simply connected and thus the statements correct.