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Force is generally a function of $\mathbf{r}(t)$, $\mathbf{v}(t)$ and $t$. $$1-)\begin{cases} \mathbf F: \mathbb R^3\times\mathbb R^3 \times \mathbb R \ \rightarrow \mathbb R^3\\ (\mathbf r, \mathbf v, t) \longmapsto \mathbf F(\mathbf r,\mathbf v,t) \end{cases}$$

But I do't understand it . Because function from $\mathbb{R^7}$ to $\mathbb{R^3}$ I don't know how is that .

Can we conclude that $\mathbf F(\mathbf r(t),\mathbf v(t),t)=\mathbf F(t)$ ??

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    $\begingroup$ Possible duplicate of What is function of force $F$ and work $W$? by OP himself. $\endgroup$
    – Sahand Tabatabaei
    Commented Nov 16, 2018 at 16:14
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    $\begingroup$ You are confusing composite functions with the functions used to make the composite function itself. If we have two functions $f,g$ which are able to be composed as so $f\circ g = f(g(x))$. Do not confuse the composition $f(g(x))$ with $f$. These are completely different functions (what makes functions different? What makes a function in the first place? Answer: Domain elements and mappings). Writing $f(g(x))$ is very different than $f$ (because of not making this distinction - the chain rule is often taught wrong). You can't overload $f$ to mean both the composite function and the function f $\endgroup$
    – DWade64
    Commented Nov 16, 2018 at 16:57
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    $\begingroup$ But yes, the composition function $\vec{F}(\vec{r}(t), \vec{v}(t), t)$ is a single variable function of $t$. Also, your pink text looks good but $\vec{F}(\vec{r},\vec{v},t)$ I understand to be very different than $\vec{F}(\vec{r}(t), \vec{v}(t), t)$. The first is just the outside function (that's how I read it), while the second is a composite function $\endgroup$
    – DWade64
    Commented Nov 16, 2018 at 17:01
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    $\begingroup$ (Also, the argument notation () is not needed when you are talking about functions. It's just to show extra information about inputs. To talk about functions, you only need to know the name of the function, then look up the definition because you now know the name. So if you want to talk about $\vec{F}$, then just say $\vec{F}$. Giving the extra information $\vec{F}(\vec{r},\vec{v},t)$ is not necessary because we are assuming since we have a function name, the definition is somewhere out there already defined showing the domain, naming for domain elements, and codomain) $\endgroup$
    – DWade64
    Commented Nov 16, 2018 at 17:37
  • $\begingroup$ It's a little strange though. If $\vec{r}$ is a vector-valued function taking $t \mapsto \vec{r}(t)$, likewise if $\vec{v}$ is a vector-valued function, then writing $\vec{F}(\vec{r},\vec{v},t)$ does look like a composite function. Because $\vec{r}$ and $\vec{v}$ are functions and are now the inside functions. Notice that when I write $f \circ g$, I just write $f \circ g$. I think this is sloppy notation in physics. If $\vec{r}$ is a vector-valued function, you can't also write that $\vec{r} \in \mathbb{R}^3$. This is overloading the symbol once again. Is it a function? Or is it an element $\endgroup$
    – DWade64
    Commented Nov 18, 2018 at 13:30

2 Answers 2

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In order to tell what force a particle experiences at any given moment in time, you must specify a particle's position and velocity in general, and perhaps also a moment of time.

To understand, consider the Lorentz-force acting on a particle of charge $q$: $$ \mathbf{F}(\mathbf{x},\mathbf{v})=q(\mathbf{E}(\mathbf{x})+\mathbf{v}\times\mathbf{B}(\mathbf x)). $$

As you can see the question "What is the value of the Lorentz force on a particle" can only be answered if you know the particle's position (as the E and B fields depend on position), and the particle's velocity, since the magnetic force is velocity-dependent.

Here the Lorentz force is not time-dependent explicitly, because I have assumed the E and B fields are constant in time. However if we allow time dependence there, then $$ \mathbf{F}(\mathbf{x},\mathbf{v},t)=q(\mathbf{E}(\mathbf x,t)+\mathbf v\times\mathbf{B}(\mathbf x,t)), $$ and now you also need to specify a moment of time to know the value of the force.

This is different compared to the time dependence of a force due to motion of the particle. Assume a particle's path is described by the curve $\mathbf x=\boldsymbol{\gamma}(t)$, where, unlike usual conventions in mechanics, I have decided to use a separate symbol, $\gamma$, for the curve function.

Then at time $t$, the particle's position is $\mathbf x=\boldsymbol{\gamma}(t)$, its velocity is $\mathbf v=\dot{\boldsymbol{\gamma}}(t)$ and so the function that describes the forces the moving particle feels over its motion is $$ \mathbf{F}(t)=\mathbf{F}(\boldsymbol{\gamma}(t),\dot{\boldsymbol{\gamma}}(t),t). $$

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No, only if you are talking about a specific solution trajectory $t\mapsto {\bf r}(t)$ (which in turn depends on an initial position ${\bf r}_0\in\mathbb{R}^3$ and an initial velocity ${\bf v}_0\in\mathbb{R}^3$).

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