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I've heard both yes and no.

Is $\mathbf{F}=m\mathbf{a}$ a vector field or just a vector?

I think it's ambiguous, it's always written without an argument.

For sake of clarity: I use the notation $\mathbf{F}(x,y,z)$ or $\mathbf{F}(\mathbf{r})$ for a vector field $\mathbb{R}^3 \rightarrow \mathbb{R}^3$. $\mathbf{F}(t)$ for a vector-valued function $\mathbb{R} \rightarrow \mathbb{R}^3$ and $\mathbf{F}$ for a vector (no argument, just a constant vector).

EDIT:

I don't grasp if $\mathbf{F}=m\mathbf{a}$ can be written explicit as:

A vector field: $$ \mathbf{F}(x,y,z)=m\mathbf{a}(x,y,z) $$

A vector field with time $t$: $$ \mathbf{F}(x,y,z,t)=m\mathbf{a}(x,y,z,t) $$

A vector-valued function: $$ \mathbf{F}(t)=m\mathbf{a}(t) $$

Or if it always is a vector (no function, just a constant vector): $$ \mathbf{F}=m\mathbf{a} $$

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/295419 $\endgroup$ – SRS Dec 23 '16 at 15:26
  • $\begingroup$ Related : physics.stackexchange.com/q/286903 $\endgroup$ – SRS Dec 23 '16 at 15:28
  • $\begingroup$ A vector field $\vec F(\boldsymbol{x})$ is vector valued function on space(-time); it is a rule that associates a vector to each point (event). If the velocity of a particle of mass $m$ is $\vec{v}(t)$, is the acceleration of the particle $\vec{a}(t)=\frac{d}{dt}{\vec{v}}(t)$ a vector field? $\endgroup$ – Alfred Centauri Dec 23 '16 at 15:52
  • $\begingroup$ @AlfredCentauri Either that comment should be an answer, or the question is unclear. $\endgroup$ – DanielSank Dec 23 '16 at 16:18
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    $\begingroup$ @DanielSank, answering is sometimes a process. $\endgroup$ – Alfred Centauri Dec 23 '16 at 16:55
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It depends on the specific situation.

A priori, the equation $F=ma$ is to be read as a differential equation for $a =\ddot{x}$ such that we want to solve for $x(t)$ in $$ F(x(t),\dot{x}(t),t) = m\ddot{x}(t)$$ However, in many physical situations, the function $F(x,\dot{x},t)$ does not depend on $\dot{x}$ or $t$, i.e. the force is indeed just a vector field. This is the case e.g. for all conservative forces, which are the gradient fields of scalar potentials.

On the other hand, many friction forces such as Stokes friction depend on the velocity of the particle, and are therefore no vector fields on space, since they can have different values at the same point in space dependingon what velocity the moving object has there.

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In rigid body mechanics $\bf{F}$ is not a vector field because to accelerate a rigid body (center of mass) with $\bf{a}$ you apply a force $\bf{F}$ regardless of where it is applied. The location of the force does not affect the motion of the center of mass.

On the other hand, the acceleration $\bf{a}$ is a vector field because different parts of a rotating rigid body accelerate differently. But rotational velocity $\boldsymbol{\omega}$ and acceleration $\boldsymbol{\alpha}$ are not because they are shared with the entire rigid body.

In addition, the net torque applied $\bf{T}$ is a vector field because it is (almost) always defined as a force at a distance $\bf{T} = \bf{r} \times \bf{F}$ and therefore the location of the force $\bf{F}$ changes the torque. A pure torque (or a force couple) is not a vector field because its location isn't important.

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The RHS of Newton's equation $\mathbf F=m\ddot {\mathbf x}$ is a vector field along the curve $\mathbf x(t)$. The LHS, depending upon the context, might be:

  1. A constant vector, e.g., $\mathbf F=m\mathbf g.$
  2. A vector field, e.g., $\mathbf F=-\frac{\partial U}{\partial \mathbf x}.$
  3. A vector valued function defined on the tangent space $T\mathbb R ^3\simeq \mathbb R ^3 \times \mathbb R ^3$, e.g., $\mathbf F = q\mathbf E (\mathbf x ,t)+ q\mathbf v \times \mathbf B (\mathbf x ,t).$

There's no difficulty into equating a vector field along a curve like $m\ddot {\mathbf x}$ to any of the objects 1, 2 or 3. That's why physicists usually talk of $\mathbf F$ simply as a "vector". The same loose language is used in special relativity, where one refers to the mass of a particle or to its proper time simply as "scalars". The first item is just a number, while the second is a scalar field along the particle's worldline$^1$.

You can also phrase Newton's equation purely in terms of vector fields, by passing to a "velocity phase space" formulation: $$\dot {\mathbf x}=\mathbf v \\ \dot {\mathbf v} = \mathbf F (\mathbf x ,\mathbf v,t).$$ Now, the "force" $(\mathbf v,\mathbf F)$ is a honest vector field on $\mathbb R ^3 \times \mathbb R ^3$ (which may depend parametrically on time $t$).


$^1$This example also shows that one generally has to distinguish beetween a tensor field along a trajectory and a tensor field along a parametrization. The proper time doesn't depend upon the particular parametrization $x^\mu = x^\mu (\lambda)$ while the Lorentz force ofcourse does.

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  • $\begingroup$ Thanks! I added a clarification of the notation. You write "The RHS of Newton's equation $\mathbf F=m\ddot {\mathbf x}$ is a vector field along the curve $\mathbf{x}(t)$" Isn't the RHS a vector-valued function of $t$, e.g. not a vector field? With the notation I mean a vector field $\mathbf{F}(x,y,z)=m\mathbf{a}(x,y,z)$ or maybe $(x,y,z,t)$. But a vector-valued function is $\mathbf{F}(t)=m\mathbf{a}(t)$. $\endgroup$ – JDoeDoe Dec 23 '16 at 19:18
  • $\begingroup$ Dear JDoeDoe, indeed the acceleration is not a vector field. It is, let me emphasize, a vector field along the curve $\mathbf x = \mathbf x(t)$, which is exactly what you say: a vector valued map $t\mapsto \mathbf a (t)$. On the other hand, the force is, in general, a vector valued map $(\mathbf x , \mathbf v , t)\mapsto \mathbf F (\mathbf x ,\mathbf v ,t)$ (which is neither a vector field because it can depend on $\mathbf v$). $\endgroup$ – pppqqq Dec 23 '16 at 19:29
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It's a vector field. In general, a force (for example the force applied by a eletric field on a charge) is dependent on the position in space (just think of the electric force of a point charge) and therefore $F = F(x,y,z)$ is a field.

There is also no ambiguity here as a vector is just a constant vector field (as a number is just a constant function).

Edit: I think your confusion could entirely come from the word "field", so I want to clarify that a bit further.

A field is nothing more than a function usually from the domain of $\mathbb{R}^n$ for $n \ge 2$, so just a multi-valued function. A vector field refers to a map $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ (which is of course far away from a general definition of the term vector field e.g. as a section of a vector bundle). As this function takes its values in $\mathbb{R}^n$, it transforms automatically as a vector and is therefore a vector for every n-tupel $(x_1,\ldots, x_n) \in \mathbb{R}^n$.

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protected by Qmechanic Dec 23 '16 at 18:03

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