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The force is often written without an argument as, $$ \mathbf{F}=m\mathbf{a}=m\frac{d\mathbf{v}}{dt}=\frac{d\mathbf{p}}{dt} $$ Does it mean the force is a functions of time, $t$? Is it the same for the momentum?

I mean is the explicit form of the above formula actually

$$ \mathbf{F}(t)=m\mathbf{a}(t)=m\frac{d\mathbf{v}(t)}{dt}=\frac{d\mathbf{p}(t)}{dt}\quad ? $$ So $\mathbf{F}$ is an abbervation for $\mathbf{F}(t)$?

And $\mathbf{p}$ is just an abbervation for $\mathbf{p}(t)=m\mathbf{v}(t)$?

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Force is generally a function of $\mathbf{x}$, $\mathbf{v}$ (or $\mathbf{p}$) and $t$.

For example, Lorentz-force is given by $$ \mathbf{F}(\mathbf{x},\mathbf{v},t)=q(\mathbf{E}(\mathbf{x},t)+\mathbf{v}\times\mathbf{B}(\mathbf{x},t)), $$ where I allowed the electric and magnetic fields to have time-dependence.


More in-depth explanation: (In which I'll attempt to explain the general functional dependencies of classical mechanical variables)

In advanced mechanics, a configuration of a mechanical system is described by $n$ independent coordinates $\mathbf{q}=(q^1,...,q^n)$, and the states of the system are described by the $2n$ independent coordinates $(\mathbf{q,p})=(q^1,...,q^n,p_1,...,p_n)$.

The $q$s are called generalized coordinates, and the $p$s are called generalized momenta. These might be cartesian coordinates of a particle and components of the particle's momentum vector in that cartesian frame, but they also might be something else. Angle variables, parametrizations of rotations matrices etc.

The space spanned by the $q$ coordinates is called configuration space, and I shall denote it by $\mathcal{C}$. The space spanned by the $q$s and $p$s together is called phase space and I shall denote it by $T^*\mathcal{C}$.

It should be clear intuitively, that a single point of $\mathcal{C}$ corresponds to a single "configuration" of a mechanical system, eg. the positions of all construents of the system.

A single point of $T^*\mathcal{C}$ corresponds to a single "state" of the mechanical system, because you know from Newton's laws, that given the forces, the mechanical state of the system is completely specified by the positions and the momenta (velocities).

Now, a priori, neither the $q$ coordinates, nor the $p$ momenta are functions of time. They are just variables.

An observable quantity in classical mechanics, is some function $$ F:T^*\mathcal{C}\rightarrow\mathbf{R}, $$ eg. an observable is something that associates a number to any mechanical state of the system. An observable quantity might also be time-dependent, so $F=F(\mathbf{q},\mathbf{p},t)$, but this time-dependence is not related to any motion of a particle. (Think of time-varying electric fields, not electric fields that seem varying because a particle moves through them!)

Now, the trajectory of a classical mechanical system is a curve through $\mathcal{C}$, which can be represented by giving time-dependence to the coordinates: $\mathbf{q}(t)=(q^1(t),...,q^n(t))$. It might be not trivial from this (maybe I should have used velocity phase space instead of momentum phase space from the get go :/), but a curve through $\mathcal{C}$ will also induce a curve through $T^*\mathcal{C}$, so now your momentum variables will also have time dependence: $$ (\mathbf{q}(t),\mathbf{p}(t))=(q^1(t),...,q^n(t),p_1(t),...,p_n(t)). $$

If you compose the function $F$ with the trajectory, you will get $$ F(\mathbf{q}(t),\mathbf{p}(t),t), $$ which is now solely a function of time.

But, if we do not take any possible explicit time dependence into account, an observable (eg. a force) will not depend on time by itself. It depends on the states of the mechanical system (positions and momenta). However, if we have an explicit trajectory, then the force will become a function of time, because the states (positions and momenta) will change with time.

Example: Instead of looking at forces, let's look at kinetic energy, for a single particle moving in 3D space: $$ T=\frac{\mathbf{p}^2}{2m}. $$ So you see, kinetic energy associates a number to a mechanical state (and this state function does not depend on position, only velocity/momentum). It is not time dependent. But if we have a particle with trajectory $(\mathbf{q}(t),\mathbf{p}(t))$ through phase space, then $T$ will become a function of time, $T(t)=\mathbf{p(t)}^2/(2m)$, as the particle moves through states with possibly different momenta.

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  • $\begingroup$ Attendum: The second part of this post does not strictly answer the question, but I feel that sometimes functional dependencies in CM (what does it mean for something to be time-dependent, total time derivative vs. partial time derivative, on-shell vs off-shell etc.) can be really confusing for a novice so I wanted to clear this up. $\endgroup$ – Bence Racskó May 4 '17 at 14:47
  • $\begingroup$ I should suppose from the way the question has been asked and further investigation on the user's profile that it did not require such an answer. Probably Jaywalker's answer suffices the need. $\endgroup$ – Aaron John Sabu May 4 '17 at 14:53
  • $\begingroup$ @AaronJohnSabu Well, me 4 years ago would have greatly appreciated this explanation, so I suppose this answer should be useful for somebody. With that said, the first part directly answers the question with an example. $\endgroup$ – Bence Racskó May 4 '17 at 14:57
  • $\begingroup$ Hi! In the Lorentz force, isn't the velocity a function of time? I mean $\mathbf{F}(\mathbf{x},\mathbf{v},t)=q(\mathbf{E}(\mathbf{x},t)+\mathbf{v}(t)\times\mathbf{B}(\mathbf{x},t))?$ $\endgroup$ – Donsert Aug 21 '17 at 13:51
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It's not necessary for the force to be a function of time. It could even be constant, i.e, not changing with time. If the velocity(v) of a body is a linear function of time, then 'mv' or momentum is also a linear function of time but the derivative would be a constant - which gives the force.

And Yes. You can write F as F(t) but that doesn't have to mean F has to change with time.

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$F=ma$ does not tell you anything about how force varies with time. It only tells you that force is proportional to acceleration. If acceleration is zero, $F$ is zero. If acceleration is constant, $F$ is constant. If acceleration is increasing, $F$ is increasing.

You have to look at other information about the problem to decide if $F$ is varying with time.

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In general yes, but it depends on the nature of the velocity's dependence on time whether the term $t$ actually appears in the equation for the force. If the velocity has a linear dependence on time for example

$v(t) = 4t + 7$

then the derivative of the function becomes 4 and the force is a constant. If however the velocity changed as the cube of $t$ for example

$v(t) = 3t^3 + 5t^2 + 9$

then the force becomes time dependent as

$F(t) = 9t^2 + 10t$

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  • $\begingroup$ Could you give an example where the welocity does not have a linear dependence with time? $\endgroup$ – Chappy May 4 '17 at 15:23
  • $\begingroup$ So if the velocity changes linearly with time meaning that the acceleration is costant ie for every second of time the velocity increases by the same constant the force is time independent but if the acceleration is not constant ie for every succeeding second the velocity change is different then the acceleration is not constant and the force is also changing with time ie it is time dependent? $\endgroup$ – Chappy May 4 '17 at 16:22
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    $\begingroup$ @Chappy, v = 2(t^2) $\endgroup$ – Aaron John Sabu May 4 '17 at 16:57
  • $\begingroup$ Velovity in a circle of radius R v = R(cos(t), sin(t),0). Non linear dependence and Newtons equation provides the centripetal force, whose components depends on t. $\endgroup$ – Alejandro Menaya May 4 '17 at 17:30
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When two bodies collide exchange of momentum occurs with one body losing and the other body gaining the same amount of momentum.The force experienced by each is of the same magnitude although in opposite directions and the magnitude depends on how fast this exchange of momentum occurs.The shorter the time in which this occurs the more is the force experienced .

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