# Is force the current of momentum?

Momentum is the conserved quantity associated to space translations via Noether's theorem. The momentum density $P_i$ satisfies the continuity equation $$\frac{\partial P_i}{\partial t} + \frac{T_{ij}}{\partial x^j} = 0 \tag 1$$
where $T_{ij}$ is called the *stress tensor* and a sum over $j$ is understood.

Charge is a scalar, so its flow can be described by a vector. Since momentum is a vector quantity, its flow is described by a rank two tensor; $T_{ij}$ is the flow of $i$-momentum in the $j$-direction. (This is explained much better by Misner, Thorne, and Wheeler in *Gravitation* in the appropriate chapter.)

Of course, the above is for a closed system. Looking at only a subsystem, we will find instead of the continuity equation $$ \frac{\partial P_i^1}{\partial t} + \frac{\partial T^1_{ij}}{\partial x^j} = F_i^1 \tag 2 $$ and can identify $F_i^1$ as a force density. For example, consider Maxwell's equations in vacuum. Then the stress tensor is Maxwell's stress tensor $$\sigma_{ij} = \epsilon_0 E_i E_j + \frac{1}{\mu_0} B_i B_j - \frac{1}{2} \big(\epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \big) \delta_{ij}$$ and the momentum density is the Poynting vector $S_i = \frac{1}{\mu_0} (\mathbf E\times \mathbf B)_i$. In vacuum, these satisfy the continuity equation (1). If the sources of the electromagnetic field are a charge density $\rho = qn$ and a current density $\mathbf j = qn\mathbf v$ for some charge $q$, number density $n$ and velocity field $\mathbf v$, on the other hand, we have $$\frac{\partial S_i}{\partial t} + \frac{\partial \sigma_{ij}}{\partial x^j} = -qn(\mathbf E + \mathbf v \times \mathbf B)_i$$ and we recognize the right-hand side as the negative of the Lorentz force (density). If we consider also the momentum density of the charge carriers, $P_i = mnv_i$, then $S_i + P_i$ along with $\sigma_{ij} + T_{ij}$ for an appropriate particle stress tensor $T_{ij}$ will satisfy (1).

The continuity equation in electromagnetism is:

$$ \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j} = 0 $$

If we identify $\rho:=|\mathbf{p}|$ and $\mathbf{j}:=\mathbf{F}$ we obtain an equation that is false, so the momentum-current must be definend differently and then I don't see how this identification can be useful.

If you follow this paper argument, you can identify the Force as the momentum-current in this sense:

$$\frac{dq}{dt}-I_q=0 \quad \text{for definition}$$ $$\frac{d\mathbf{p}}{dt}-\mathbf{I}_p=0 \quad \text{as analogy}$$

Then it follows that $\mathbf{I}_p=\mathbf{F}$. It's up to you to decide if this one is an useful analogy.

The momentum is given by:

$$ p = \int_{t_0}^{t_1} dt F(t) $$

The change of momentum $\dot{p}(t)$ is therefore related to the force $F(t)$. If you want you can say that in some sense your statement is true. Does this answer your question or what exactly do you understand by "current"?