# Difference between a vector field and a force field

In mathematics while learning about vector fields, we define a "vector field" as "a function of space whose value at each point is a vector quantity". That is, at each point in space there is a vector quantity attached to it.

Now if we talk about a force field which a three dimensional object is experiencing, the "force at a certain position of three dimensional object" cannot be interpreted as "acting at a point". Instead the force is acting over the whole three dimensional object.

Then how can it be justified that the "force field" is a "vector field" in the sense of above definition of vector field.

• Why force field cannot be interpreted as acting on a point? Even if you have a three dimensional object, you can have a force acting on just a point, for example on its surface. – ndrearu Mar 7 '18 at 14:24
• The force can act on the $c.o.m.$ – Nehal Samee Mar 7 '18 at 14:26
• What shall we do in the case of force on a $3D$ charge due to another charge? – N.G.Tyson Mar 7 '18 at 14:32
• @faheemahmed400 in each point you have a force vector which gives a contribution to the total force acting on the full object. Sorry, maybe I'm lost but I really don't get the point. – ndrearu Mar 7 '18 at 14:38
• Suppose we have a cubic charge in $3D$ space experiencing a force. Now which point in this cube is eligible to take the value of total force acting on the full object. – N.G.Tyson Mar 7 '18 at 14:42

The force acts on every element of a 3d object.

Take for example a 3d charged object. The charges can be distributed any way at all. They could be uniform, or not. One side could be positively charged, the other negative. And so on.

That 3d object has an element of charge existing at every point. (More precisely: in every infinitesimally small volume element.) Each charge element experiences the force as determined by the vector field value at the point. The force on the object is the vector sum of all those contributions.

It is possible to find one point that has the property that the translational motion of the object acts as if the total charge on the object is concentrated at that point. One might call it "center of charge", but as you point out, one does not often hear of such a thing. That is, take that charge at that point (which turns out to be the total charge), and multiply by the value of the electric field vector at that point, and you will obtain the force on the object as a whole.

The fact that there might be one point that can be considered to be "where the force acts" is a consequence of summing, is not fundamental and has some limitations. In a sense it's an illusion. It works only if you can afford to model your object as a point particle. For example, you can't use it to calculate torque.

Imagine that your object is a cube that is uniformly positively charged on its "left" half, and is uniformly positively charged on its "right" half, and has total charge zero. In a uniform electric field there will be no force on the object. The "center of charge" will have $q=0$. The same result obtains by summing the forces on all the charge elements in the cube.

However, the cube will spin! There will be a torque on the cube.

Now, you may not care that it spins. If all you care about is the trajectory of the object, all you care about is the force. In such a case, you can afford to model the object as a point, and the "center of charge" approach will give you what you want. But if you care about rotation, you cannot afford to model the object as a point, and you cannot use the point particle model. If you are interested in total kinetic energy, you'll need to include rotational as well as translational kinetic energy. The "center of charge" approach can't give you that. The point particle model is too simple to account for rotational kinetic energy.

There's a quote attributed to Einstein, but I have my doubts about who might have said it first: "Make your models as simple as possible, but no simpler."

A vector field defines a situation where the magnitude and direction of vectors are a function of location only. We can best understand this on a rotating rigid body where the linear velocity of each particle $\boldsymbol{v}$ depends on its location $\boldsymbol{r}$.

$$\boldsymbol{v} = \boldsymbol{\omega} \times \boldsymbol{r}$$

To describe such vector field we define an "axial vector" $\boldsymbol{\omega}$ which is placed on the origin to describe the rotation of the object. This rotation has the following properties

• Magnitude, $\omega = \| \boldsymbol{\omega} \|$
• Direction, $\boldsymbol{k} = \frac{ \boldsymbol{\omega}}{\omega}$
• Location, the origin.

Now, lets flip our perspective around in order to draw a parallel with forces later. Consider a rigid body rotating with a vector $\boldsymbol{\omega}$, about a specified location $\boldsymbol{r}$ and we measure the linear velocity $\boldsymbol{v}$ at the origin.

$$\boldsymbol{v} = \boldsymbol{r} \times \boldsymbol{\omega}$$

Does this describe a vector field? Yes, since we have not changed the nature of the problem, but shifted our perspective. Even though we are moving around the axial vector $\boldsymbol{\omega}$ at different locations we measure the effect at the origin. Here $\boldsymbol{v}$ still represents a vector field, and $\boldsymbol{\omega}$ is the axial vector. The properties are now

• Magnitude of rotation, $\omega = \| \boldsymbol{\omega} \|$
• Direction of rotation, $\boldsymbol{k} = \frac{ \boldsymbol{\omega}}{\omega}$
• Location of axis (recovered from velocity), $\boldsymbol{r} = \frac{ \boldsymbol{\omega} \times \boldsymbol{v}}{ \omega^2 }$

Proof: $\require{cancel} \frac{ \boldsymbol{\omega} \times \boldsymbol{v}}{ \omega^2 } = \frac{ \boldsymbol{\omega} \times (\boldsymbol{r} \times \boldsymbol{\omega} )}{ \omega^2 } = \frac{ \boldsymbol{r} \omega^2 - \boldsymbol{\omega} \cancel{(\boldsymbol{r}\cdot \boldsymbol{\omega})}}{\omega^2} = \boldsymbol{r}$ with the rule that $\boldsymbol{r}$ is the location on the axis closest to the origin.

Now let us look at the entire similar situation and consider a rigid body with a force $\boldsymbol{F}$ applied through a location $\boldsymbol{r}$ and we measure the equipollent torque $\boldsymbol{\tau}$ at the origin.

$$\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$$

Does this describe a vector field? Yes, for the same reason(s) as above. Here $\boldsymbol{\tau}$ still represents a vector field, and $\boldsymbol{F}$ is the axial vector. The properties are now

• Magnitude of force, $F = \| \boldsymbol{F} \|$
• Direction of force, $\boldsymbol{e} = \frac{ \boldsymbol{F}}{F}$
• Location of axis (recovered from troque), $\boldsymbol{r} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau}}{ F^2 }$

So the geometry of mechanics dictates the following definitions.

$$\begin{array}{l|c|c} \mbox{quantitity} & \mbox{axial vector} & \mbox{moment vector} \\ \hline \mbox{motion} & \boldsymbol{\omega} & \boldsymbol{v} = \boldsymbol{r}\times \boldsymbol{\omega} \\ \hline \mbox{momentum} & \boldsymbol{p} & \boldsymbol{L} = \boldsymbol{r}\times \boldsymbol{p} \\ \hline \mbox{loading} & \boldsymbol{F} & \boldsymbol{\tau} = \boldsymbol{r}\times \boldsymbol{F} \end{array}$$

Note: I prefer the term moment vector from vector field because it is more descriptive of the specific situation.

There is some room for confusion here because of how liner momentum is defined.

Linear momentum (the axial vector) is defined from the velocity (the vector field) at a specific point (the center of mass).

$$\boldsymbol{p} = m ( \boldsymbol{\omega} \times \boldsymbol{r}_{\rm com} )$$

The equations of motion relate the force (axial vector $\boldsymbol{F}$) with the rate of change of momentum (axial vector $\boldsymbol{p}$). The rotational equations relate torque at the center of mass (vector field $\boldsymbol{\tau}_{\rm com}$) to rate of change of angular momentum at the center of mass (vector field $\boldsymbol{L}_{\rm com}$).

As you can see the equations of motion are consistent with the geometrical interpretation of mechanics.

\begin{aligned} \boldsymbol{F} & = m (\boldsymbol{\alpha} \times \boldsymbol{r}_{\rm com}) + \boldsymbol{\omega} \times \boldsymbol{p} = m \boldsymbol{a}_{\rm com} \\ \boldsymbol{\tau}_{\rm com} & = \mathtt{I}_{\rm com} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \boldsymbol{L}_{\rm com} = \mathtt{I}_{\rm com} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathtt{I}_{\rm com} \boldsymbol{\omega} \end{aligned}