# Are force vectors members of a vector space?

Vectors in vector spaces depend only on their size and direction.

Force vectors, for example, depend also on their location. Opposite force at different locations, for example, do not annihilate each other but form torque.

Why this anomaly?

The answer is that force is a vector, you just have to be careful about what the vector space is. As a starting point, I think it would be useful to review what exactly a force is. Imagine we've chosen some inertial reference frame $$O$$ and are using it to describe a universe which consists of nothing but a single particle, which has coordinates $$\mathbf{x}(t)$$. As I'm sure you're aware, Newton's second law for this system just asserts that if we have some force law for the particle $$\mathbf{F}(\mathbf{x}, \mathbf{\dot{x}},t)$$, the particle's motion satisfies

$$\frac{d^2\mathbf{x}}{dt^2} = \mathbf{F}(\mathbf{x}, \mathbf{\dot{x}},t)/m$$

where m is the mass of the particle. The important thing to notice about this is that force isn't tied to random points in space-- it's tied to the movement of particles within space. I can't apply a force on an empty piece of vaccuum-- I can only do so to a particle. In this vein, we should then not be thinking of the force as acting on the vector space describing space, but rather the vector space describing the location of our particles.

Now, for one particle, this is somewhat of a moot point because both of these vector spaces are just $$\mathbb{R}^3$$. But what about if we have two particles? There's no "nice" (ie continuous, linear) way to describe the location of two particles in space using one vector in $$\mathbb{R}^3$$, which we need if we're gonna be using vector algebra and calculus to solve stuff. But there is a nice way if we expand our vector space to $$\mathbb{R}^6$$-- just let the first three components describe the location of particle 1, and the second three components describe the location of particle 2! In this formalism, it's clear how a force acting on particle 1 is different from a force acting on particle 2-- they're completely different vectors in $$\mathbb{R}^6$$.

Hopefully the generalization to arbitrary numbers of particles is obvious-- we can describe the location of $$N$$ particles using a vector in $$\mathbb{R}^{3N}$$, and it's in this vector space-- known as configuration space-- that force lives. Using this concept of configuration space is very useful in physics and guides us to the incredibly important concept of lagrangian mechanics. Note-- when I say force "lives" in configuration space, I'm using that as shorthand for saying that it's the co-domain of the force.

To recap-- force isn't tied to arbitrary locations in physical space, it's tied to the location of particles inhabiting that physical space. And these locations are described by the $$3N$$ dimensional configuration space, which is the vector space that force formally lives in.

## An alternate viewpoint

If my description of force as a vector in $$\mathbb{R}^{3N}$$ seems contrived, be aware that another, perhaps more intuitive way of looking at the situation is by saying that each particle has its own force vector associated with it that depends on time and the locations/velocities of itself and all the other particles. In this viewpoint (which is the one taken in JiK's answer), the reason that applying a force to two different particles results in different dynamics of your system is because although both forces are members of $$\mathbb{R}^3$$, they're two completely independent vectors since they're describing the dynamics of different particles. It's this independence that lets us combine the vector spaces of all the different forces in a natural way using the direct sum.

Aside from just doing it because we can, this combination of vector spaces is useful for three main reasons:

1. As mentioned before, it helps lead us to Lagrangian mechanics.
2. On the aesthetic front, it combines $$N$$ different vector differential equations into one.
3. It helps clarify the conditions we need to impose on $$\mathbf{F}$$ to guarantee existence and uniqueness of trajectories through configuration space via the Picard-Lindelöf theorem. Strictly speaking to do so we must first turn the force equation from a $$3N$$ dimensional 2nd order ODE to a $$6N$$ dimensional 1st order ODE, but this is trivial.

## Further discussion

In the comments, Aloizio Macedo brought up a good point, which is that if you read my answer in a certain way it can seem like it's saying that force should only be described in terms of what particles are actually doing. Yet, I also seem to implicitly define force as a function on all of configuration space which is a contradiction because the system doesn't occupy every point of configuration space along its trajectory. So what gives?

Well, in my mind, this comes down to a chicken and the egg type situation-- is the path of the particle through configuration space what's given and a force law deduced from that? Or is it the other way around and the force law is what's fundamental and what determines the path through configuration space? When I was writing my answer, I had the second one in mind, with force being given as some function $$\mathbf{F}: \mathbb{R}^{3N} \times \mathbb{R}^{3N} \times \mathbb{R} \rightarrow \mathbb{R}^{3N}$$ satisfying some continuity requirements.

However, as Aloizio said, it's possible to read my second paragraph as saying that forces should only be defined when acting on particles, which necessitates the first viewpoint and a different formalism that makes the path through configuration space $$\gamma$$ the fundamental thing describing our universe and derives force through that.

This wasn't my intention with the second paragraph. I just wanted to emphasize for the OP that force is something that fundamentally has to do with how particles move in space, rather than with space itself-- a statement that's true regardless of which of the above viewpoints you subscribe to. Hopefully that clears things up and addresses the concerns people may have had with my description of force.

• Another way to look at it: for many forces, such as gravitational force, you can describe the vector space of what the force would be if there were a particle (of a certain mass/potential/etc) there. We call that vector space a field. – BlueRaja - Danny Pflughoeft Mar 25 '20 at 22:20
• @BlueRaja-DannyPflughoeft I think your comment makes a valuable contribution, but there's some issue with the terminology. A vector space is a set together with some operations (like $\mathbb{R}^3$ with $+$ and multiplication by scalars). The concept you're describing, an association of a vector to each point in space, is a vector field (hence the shortening to just "field"). Don't mean to be pedantic, but since the OP and this answer are both particular about what vector space they're in, it's worth distinguishing. – jawheele Mar 26 '20 at 6:55
• This answer seems confused. If you require forces to be defined only where they are acting upon something, then forces don't live in the configuration space, they should probably live in something like the space of sections of the pull-back bundle $\gamma^*TM$, where $M$ is the configuration space and $\gamma$ is the path of the particle(s). But your conclusion and final paragraphs imply that forces are vector fields on the configuration space, directly contradicting the first part. – Aloizio Macedo Mar 26 '20 at 14:42
• @bemjanim The force vectors are now six dimensional-- the first 3 components correspond to the force acting on the 1st particle, and the second 3 components correspond to the force acting on the second particle. If you prefer, you can think of them as two different forces acting on each of the particles, but combining them into one vector has some mathematical and aesthetic advantages. – el duderino Mar 28 '20 at 14:52
• @AloizioMacedo After thinking about it for a while, I think I understand what you were getting at and it's a valid point. I think the confusion stems from my second paragraph coming across differently than I intended. I added a note at the bottom discussing this-- hopefully it adequately addresses your concerns. – el duderino Mar 28 '20 at 17:50

Is mass a scalar? If I spread $$10^{57}$$ hydrogen atoms around the universe, it barely has any effect on anything. But if I concentrate them into a small volume, it creates distinguishable gravity and planets have closed orbits around them. So clearly, the position of a mass also matters, not only the amount of the mass.

Is the amount of money a scalar? If I have 1000 € in my bank account and pay a 100 € restaurant bill, I end up having 1000€+(-100€) on my bank account. But if instead John pays the bill, I end up having 1000€ in the bank account. So clearly the owner of the money also matters.

Are real numbers scalars? If a question has a $$2$$ and a $$3$$, then you can't know whether $$5$$ is the correct answer to the question without knowing other things.

Force vectors, for example, depend also on their location. Opposite force at different locations, for example, do not annihilate each other but form torque.

Why this anomaly?

There is no anomaly. Force vectors are vectors. But when you want to find out what happens in a physical system involving forces, you can't just take the vectors and do stuff with them without thinking about where they apply.

Like you can't just take masses and sum them to find gravity without thinking about where the masses are. Like you can't just take amounts of money and sum them without thinking about who owns the money. Like you can't just take real numbers and sum them without thinking about what the problem actually is.

In addition to @elduderino's very good answer, there is a second way to address your question. It can be hard to see what I'm talking about in a Euclidean space (like $$\Bbb{R}^2$$ or $$\Bbb{R}^3$$) so let's talk about point particles constrained to move on the surface of a sphere.

A particle constrained to move on the surface of a sphere cannot react to the component of a force vector along a radius of the sphere, so we might as well restrict our force vectors to have no component along the radius. So the set of allowed force vectors lie in a 2-dimensional real vector space (equivalent to a plane) where we imagine the origin of the vector space is at the particle and the plane is tangent to the sphere. (When we start moving the origins of vector spaces around, we get affine spaces.) At each point of the sphere, we have a different vector space because we have a different tangent plane. (Many pairs of tangent planes seem to intersect in lines -- these are not physically meaningful. We should think of each plane as being tagged by its point on the sphere and we should think of that tag as being an extra coordinate, so these intersections are projection artifacts of ignoring that extra coordinate.)

Because these spaces are tangent to our sphere, they are called tangent spaces. The collection of all of these on our sphere is called the tangent bundle of our sphere. There is much math of tangent spaces and tangent bundles. One thing I want to make very clear is that we have not gone far from @elduderino's answer -- these tangent spaces also have straightforward descriptions in $$\Bbb{R}^{3n}$$-dimensional configuration spaces. Notice that these tangent spaces capture something we know about motion on a sphere -- velocities are parallel to the surface of the sphere, accelerations are parallel to the surface of the sphere, and (as $$F = ma$$ suggests) so are force vectors. (If you imagine some sort of generalized force vector that does not lie in the tangent plane for its particle, then project it into the component that does and discard the other component -- the other component is attempting to violate or geometric constraint, so does nothing.)

Let's circle back to my claim about it being hard to see what's going on in the plane or in 3-space. At each point of the plane, the tangent space is a 2-dimensional real affine space, so it looks exactly like the plane we started with. This can be confusing. At each point in space, the tangent space is a 3-dimensional real affine space, so it looks exactly like the space we start with. This can be confusing. So to explain this, start with a space where the tangent spaces to various points are not just the space you start out with -- many people are familiar with moving around on the surface of an approximate sphere, so start there.

Yes, they are. The vector space is isomorphic to $$\mathbb{R}^3$$, as you might expect.

In at least the Euclidean case, I find that the most helpful way to understand the relationship between points and vectors is to have the idea of an affine space along with a vector space. An affine space is a space where that points are all "homogeneous" in that you cannot discern, using the operations available to you in the space, any point from any other point. This contrasts with a vector space, where you can always obtain the zero vector $$\mathbf{0}$$ by multiplying any given vector by the scalar 0.

In affine space, you can do two things:

1. Subtract two points to get a vector: if $$P$$ and $$Q$$ are two points, then $$Q - P$$ is the vector representing the displacement from $$P$$ to $$Q$$,
2. Add a vector to a point to get another point: if $$P$$ is a point and $$\mathbf{r}$$ a displacement vector, then $$P + \mathbf{r}$$ is the point you get to by following that displacement.

Note that in neither of these does any particular point in the space have any significance as none is referenced, while the zero vector is both explicitly referenced and obtainable in the axiom set for vector spaces. And it is the second idea above - adding a vector to a point - that relates to the idea of forces "having a point of application". In particular, the force $$\mathbf{F}$$ generates an acceleration $$\mathbf{a}$$ which describes change in the rate of displacement $$\mathbf{v}$$, and the object's position finally undergoes small displacements

$$P' = P + (\mathbf{v}\ dt)$$

at each moment of time: here is where we have the affine vector-to-point addition function at work, but the force vector $$\mathbf{F}$$ does not directly add to a point.

However, I believe the notion of "point of application" most properly belongs to the study of spatially extended objects, where the above discussion applies to point-like objects only, because for a point-like object, you cannot intuitively "apply a force" anywhere else but at the single point it occupies, otherwise you're just "pushing at empty space". Describing extended-body movements is a little different. A solid body can be mathematically represented by a suitably-"nice" set of points drawn from space, representing all the locations it occupies, which we may call $$I$$ (for "image"). Then we describe the forces slightly differently, as a function

$$\mathbf{F}(P)$$

that tags each point $$P \in I$$ with a force vector which may be zero. These points on the object where force vectors are associated is are "points of application" for those forces. The "application point" is not a property of the vector, again, but a property of the mapping that tags the points with vectors: it is the point that was tagged with that given vector. (Note that this case subsumes the previous one by allowing $$I$$ to consist of only the single point that a point particle occupies.)

So yes, force vectors are members of a vector space, but the "point of application" is not a property of the force vector itself, so much as it is a property of the relation between the force vector and the object.

tl;dr A force vector at a location can be described as part of a vector field. You probably mean to refer to a vector field rather than a vector space.

The tl;dr probably answers the question, if I understand it correctly.

So, just to clarify some more vocabulary...

# Reference: Vocabulary terms.

A temperature can be quantified with a scalar. We can describe scalar values at different points in space as a scalar field, e.g. a temperature field.

A force can be quantified with a vector. We can describe vector values at different points in space as a vector field, e.g. a force field.

Scalars and vectors are both tensors. Scalar fields and vector fields are both tensor fields.

You do add the forces applied at different locations in order to calculate the net force and determine, whether the body is being accelerated as a whole. The difference is that in this case one thinks of the object as one point with the forces applied to it, whereas when analyzing the rotation, one looks at different parts of the body as if they were different points/objects.

Opposite force at different locations, for example, do not annihilate each other but form torque.

Maybe. Maybe not.

For example, you can have two equal and opposite forces applied at different locations that do not form a torque if the forces have the same line of action. Such forces are called "collinear forces". An example is two people pulling on each end of a rope (tug of war) with equal forces. No torque is involved.

On the other hand, a system of two forces that are equal in magnitude, opposite in direction, and parallel to each other acting on an extended (as opposed to a point mass) object will result in a moment, or torque. Such as system is called a "couple" (a.k.a force-couple).

You can also have a system of forces whose lines of action (the direction in which a force acts) meet at a common point. Such forces are called "concurrent force" systems. Such forces can be added up to get a resultant force at their common point. Since they intersect at a common point, there is no torque involved. An example is a point mass suspended by two strings. Three concurrent forces act on the mass: gravity and the tensions in the two strings. No torque is involved.

For a general system of $$n$$ forces (non concurrent, non parallel system) applied to an extended object, equilibrium (no net force and no net torque) will depend on whether or not $$\sum F_{n}=0$$ and $$\sum M_{n}=0$$.

Hope this helps.

• I know. It still means that locations CAN matter. That's not something that happens on normal vectors – user4951 Mar 25 '20 at 15:28
• @user4951 Yes it means locations can matter. I'm just saying it's not ONLY the location that matters but also the nature of the force systems involved. – Bob D Mar 25 '20 at 15:30

Force vectors do not depend on their location. What depends on the location of a force vector is the effect to which that vector contributes to the motion of a body, or to else to considerations of its equilibrium.

Of course, mathematically, vectors are always part of a vector field, regardless of their interpretation in a physical theory!

We consider a body in space, and locate some arbitrary point as the origin. A given force acting on a point on the body creates a moment with regard to the chosen origin. The moment is a combination (dot product) of two vectors: the position vector and the force vector. Moment is a generalization of torque; torque is the magnitude of moment, which is a vector quantity.

The moment is not the only effect of the vector, of course, and it is arbitrarily determined with regard to a chosen point.

We can look at all the forces acting on an body, and consider whether their moments about a given point add to zero. If they add to zero about some point, they add to zero about any other point: the moments are balanced, and the object will not rotate. But the forces might not be balanced, and so the object might accelerate.

It's not that the two forces don't "annihilate" but that the two torques don't cancel out, and that's a different thing.

There are rules on how to move around and add up vectors on Euclidean space and are of course consistent with general rules for vector spaces. One generally can't just simply add together vectors from different points because they usually belong to a strictly different vector space. But if you can move one vector to another point(different vector space)and while doing so being true to its original direction and magnitude, than you can add it up normally. So forces of course are members of a vector space, but not all vector spaces can be used to represent forces.