# Vector Field Question

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1. Aug 1, 2015

### amenhotep

Hello,
I'm having problems understanding what exactly is meant by a conservative vector field. So, allow me to explain how I currently understand it.

Let $\overrightarrow{F}$ be some sort of a vector field that is conservative. The stream plot of $\overrightarrow{F}$ is shown below.

From my understanding, for a particle that interacts with field, the streamlines show the direction of flow of the particle. I have colored two streamlines in red to illustrate that. So, any particle placed at any location in the vector field will move along one of the streamlines by itself. In other words the work done on the particle at it moves along the streamline by the vector field is $\int_C{\overrightarrow{F}\cdot d\overrightarrow{r}}$. This makes sense.

So, now let's draw any closed curve (shown in pink) inside the domain where $\overrightarrow{F}$ is defined. According to the calculus, since $\overrightarrow{F}$ is conservative, it turns out that $\oint{\overrightarrow{F}\cdot d\overrightarrow{r}}=0$. I really don't understand two things.

1. How do we get a particle to move along the closed curve we drew even when we know that by itself the particle wants to move along one of the streamlines. Let's say $\overrightarrow{F}$ is some electric field and the particle is a charged particle. How can we make it move along that closed curve. In most calculus books, they just say, "if the particle moves along the closed curve..." But how when it should move along a streamline?

2. Let's assume that somehow the particle decides to move along the closed curve ? How can the work done by the field be 0 ? What caused the particle to move then ?

Please help clarify.

2. Aug 1, 2015

### Staff: Mentor

No, they show the direction of the force. The motion can be in a different direction.
Only in very special cases. In general, they won't.
You can have some additional forces or constraints (like a track). In some cases it will move in closed circles even without external forces. Think of the Kepler problem: planets move around the sun in closed loops (ellipses) and the acceleration is always towards the sun.
It can have started with a non-zero velocity. We just know in total, it does not gain or lose energy from/to the field.

3. Aug 1, 2015

### amenhotep

Would you mind explaining more of this because I seem to have this idea engraved in my head. I always seem to be thinking that a vector field is the same as the direction field of an ordinary differential equations. I always think that things will move in the direction of the force. If you push a box along a table, it will move in the direction of the push.

4. Aug 1, 2015

### Staff: Mentor

Gravity is a conservative force, and acting always downwards on Earth. Do you move downwards all the time, and never horizontally? What happens to a ball thrown in the air at an angle? Does it move vertically downwards?

5. Aug 1, 2015

### amenhotep

Your answer seems to be implying that things don't move along the direction of the force because there's something blocking it.
I already know that it is possible to move in directions other than that of a vector field as long as you can provide the necessary forces to do so.
What I'm trying to say is let's suppose that the field in the figure I presented is that of, say Gravity and there's no restriction in any direction and a particle placed in the field cannot generate any thrust of its own to change direction. How can the particle then move along the pink loop ?

6. Aug 1, 2015

### Staff: Mentor

That is one reason, but not the only one. See the thrown ball.
It cannot, but there are other loops where it can.

7. Aug 2, 2015

### HallsofIvy

If an object, of mass m, is stationary at the origin of a coordinate system in force field $\vec{F}$ then its velocity at time t will be $\frac{1}{m}\int F(t)dt$. Since that is a integral of the force field vector, it is always pointing in the direction of F- the object will move along the integral curves. HOWEVER, if the object has some initial velocity $\vec{v_0}$ then the velocity is give by $\vec{v_0}+ \frac{1}{m}\int F(t)dt$ which, because of the $\vec{v_0}$, will NOT be along the integral curves.

8. Aug 2, 2015

### Staff: Mentor

It does not have to. F can change its direction over the trajectory.
Even more explicit: if the object would always move along the direction of acceleration, then its direction can never change. Which is in conflict with the changing direction of F in the general case.

9. Aug 2, 2015

### HallsofIvy

Perhaps I did not make myself clear. If the initial velocity is 0, then the object will be moving in a curve such that its acceleration vector, at each point, is in the direction of the force vector at that point. I did not mean that it moves in a straight line in the direction of the force vector at the initial point.

10. Aug 2, 2015

### Staff: Mentor

Its acceleration vector is always pointing in the direction of the force vector, independent of the original velocity. That is a trivial statement as F=ma.
Its velocity vector does not have to show in the direction of the force vector, also independent of the original velocity. That is the interesting point.

11. Aug 5, 2015

### orzyszpon

A STATIONARY particle at the given point will move in the direction of the field, but if it is already MOVING, then the force field will DEFLECT (accelerate) the particle in the direction of the field lines. In the extreme case of a very fast moving particle, the motion of the particle will be largely uneffected to the naked eye.

The imaginary curve is just that - an imaginary curve. Some external agent (your finger) can move the particle quasistatically (very slowly, i.e., no kinetic energy added) along the imaginary curve. The work done by the field on the particle along the closed curve will be zero for a conservative field. At the same time, along some parts of the curve, the finger will be doing positive work to "prod the particle along" (force in the direction of displacement) and negative work along other parts of the curve to "hold the particle back" (force opposite to the direction of displacement) The sum total will be zero. I hope this imagery helps.

12. Aug 6, 2015

### Radhakrishnam

When we place a particle with zero velocity at a point on a line of field, the particle moves along the field line. When we put a particle with a non zero velocity, it crosses from one field line to another as it moves. In the process of crossing the field lines it changes (gains or losses) its energy. This change in the energy is equal and opposite to the energy change suffered by the force field.

If we consider a closed loop path for the motion of the particle - the particle regains its original state when it reaches the initial position - and continues on its next cycle. This is how the heavenly bodies move in cycles on their own.

On the other hand, when a body released from rest to move downwards along a field line (say, along an inclined plane), so as to enable it to cross field lines in a constant gravitational field, we need to provide another inclined plane for it to change direction and move upwards. On this second plane it moves to reach a height, through which it had fallen on the first inclined plane, and starts retracing its path, thus moving in cycles.

In a non conservative field, the body left to itself to move, does not reach its original state, and, does not repeat its motion in cycles.

Radhakrishnamurty P.

13. Aug 6, 2015

### Staff: Mentor

In general yes, but it does not have to. A circular orbit around a central object is a counterexample.
Again, this is just the general case. It is possible to get cyclical motion.

14. Aug 18, 2015

### Howardb

Many here have accurately answered your questions, but I think your confusion is in how you wrap your mind around what a vector field is.

Maybe you are thinking of it in terms of velocities, which form streamlines. All velocities are vectors, but not all vectors are velocities. Try not to think of it like fluid flows. The vector field is merely a quantity that has direction and in general varies by location. If you are thinking like fluid flow velocities, then you are not understanding the general concept from that specific case.

And as stated by others in other terms, when one integrates along a path, one is not literally moving a particle on that path, it is a mathematical construct. Some might say imaginary or virtual. It is just mathematical equations. Those equations may or may not have physical interpretations, and if you try to use a physical interpretation to understand the more general math, then you are going about it backwards, because this is a math problem, not a physics problem.

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