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DrStupid said:
And that means that everything that is not an interaction cannot be a force - including fictitious forces or temperature.
That makes sense now. Forces are caused by interactions, and the Newtonian space-time structure dictates translation invariance and thus energy, momentum, and angular-momentum conservation for closed systems. That determines the action of Newtonian point-particle systems. It does however not restrict everything to two-body interactions only. There can be generic n-body interactions (see #176). I've no clue what anything of this has to do with temperature (at least not within non-relativistic Newtonian mechanics) at all.
"Fictitious forces" (I prefer to call them "inertial forces") are terms that from a principle point of view do not belong to the right-hand side of the equation of motion. They occur from bringing parts of ##m \ddot{\vec{r}}##, where ##\vec{r}## is a vector within an inertial reference frame to the other side in the equation, when expressing the equations of motion in terms of the vector components (sic!) wrt. a non-inertial reference frame. The latter is defined by the origin ##O'## of the non-inertial reference frame, which is moving in an arbitrary way against the origin of the inertial frame, and a right-handed Cartesian basis ##\vec{e}_k'(t)=D_{jk}(t) \vec{e}_j## which can arbitrarily rotate against the inertial Cartesian basis ##\vec{e}_j##, where ##(D_{jk})=\hat{D} \in \mathrm{SO}(3)##. This is the most general form of an accelerated reference frame (accelerated against an inertial frame of course).
Now take the most simple example of a particle moving in some external field (e.g., a gravitational field of a very heavy body, e.g., the Earth's gravitational field acting on a stone or a charged particle in an electrostatic field, neglecting radiation reaction, which is a relativistic effect anyway). Then in the inertial frame of reference you have
$$m \ddot{\vec{r}}=\vec{F}(\vec{r})$$
or in components
$$m \frac{\mathrm{d}^2}{\mathrm{d} t^2} r_j \vec{e}_j=F_j(\vec{r}) \vec{e}_j \; \Rightarrow \; m \ddot{r}_j = F_j(\vec{r}).$$
Now you have
$$\vec{r}=\vec{R}+\vec{x},$$
where ##\vec{R}=\overrightarrow{OO'}## is the position vector of the accelerated reference frame's origin wrt. the inertial frame's origin, and ##\vec{x}## is the position vector wrt. ##O'##.
To get the equations of motion for the components ##\uvec{r}'=(r_1',r_2',r_3')^{\text{T}}## of the position vector ##\vec{r}=\overrightarrow{RP}##, we need the time derivative of an arbitrary vector ##\vec{V}(t)## in terms of it's non-inertial components, i.e.,
$$\dot{\vec{V}}=\mathrm{d}_t (V_k' \vec{e}_k')=\dot{V}_k' \vec{e}_k + V_k' \dot{\vec{e}}_k'.$$
Now we have
$$\vec{e}_k' = D_{jk} \vec{e}_j \; \Rightarrow \; \dot{\vec{e}}_k' = \dot{D}_{jk} \vec{e}_j = D_{jl} \dot{D}_{jk} \vec{e}_l'=\Omega'_{lk} \vec{e}_l'.$$
Next we show that ##\Omega'_{lk}=-\Omega'_{kl}##. This follows from the orthogonality of the rotation matrix ##\hat{D}##, i.e., ##D_{jl} D_{jk}=\delta_{lk}## and thus
$$\Omega_{lk}'=\dot{D}_{jl} D_{jk}=-\dot{D}_{jl} D_{jk}=-\Omega_{kl}'.$$
So we can write
$$\Omega_{lk}'=\epsilon_{klm} \omega_m'=-\epsilon_{lkm} \omega_m'.$$
Then we find
$$\dot{\vec{V}}=\dot{V}_k' \vec{e}_k + V_{k}' \epsilon_{klm} \omega_m' \vec{e}_l'=(\dot{V}_l'+\epsilon_{lmk} \omega_m' V_k') \vec{e}_l'.$$
For simplicity we can now introduce a covariant time-derivative for vector components by
$$\dot{\uvec{V}}'=\dot{\uvec{V}}'+\uvec{\omega}' \times \uvec{V}'.$$
Now the equation of motion for ##\vec{r}## are easily expressed in terms of the non-inertial components. We only need the 2nd covariant time derivative. First we have, as just derived,
$$\mathrm{D}_t{\uvec{r}}'=\dot{\uvec{r}}'+\uvec{\omega}' \times \uvec{r}',$$
and then after some algebra
$$m \mathrm{D}_t{\uvec{r}}'=m[\ddot{\uvec{r}}' + 2 \uvec{\omega} \times \dot{\uvec{r}}' + \dot{\uvec{\omega}}' \times \uvec{r}'+\uvec{\omega}' \times (\uvec{\omega}' \times \uvec{r}')]=\uvec{F}'(\uvec{r}').$$
As you see everything is form-invariant when using the proper time derivatives ##\mathrm{D}_t## for components in the intertial frame. Then there are just the "real forces", i.e., in our case just ##\uvec{F}'## on the right-hand side of the equation, but now it is customary to single out ##m \ddot{\uvec{r}}'## and write the equation of motion as
$$m \ddot{\uvec{r}}' =\uvec{F}'-m[2 \uvec{\omega} \times \dot{\uvec{r}}' + \dot{\uvec{\omega}}' \times \uvec{r}'+\uvec{\omega}' \times (\uvec{\omega}' \times \uvec{r}')]$$
and calls the additional pieces on the right-hand side "fictitious forces" or "inertial forces" and gives them fancy names like "Coriolis force", "centrifugal force", and the part with ##\dot{\uvec{\omega}}'##, for which I don't know whether there's also a specific name.
It's just written the usual equations of motion, which are well-defined in the intertial frame in a non-covariant way in terms of the vector components with respect to the basis fixed in the non-inertial reference frame which can time-dependently rotate against the basis fixed in the inertial frame.