20
$\begingroup$

I have met the expression that force is the current of momentum. At Google I have found only a few papers where force is described that way.

Is this a valid, useful definition?

$\endgroup$
  • 2
    $\begingroup$ Being related to momentum using the same equation as charge is related to current, does not make force a current of it. $\endgroup$ – Aaron John Sabu Mar 5 '17 at 10:16
  • $\begingroup$ I edited my answer, I hope to have indentified the point of your question. $\endgroup$ – Alessandro Zunino Mar 5 '17 at 11:35
  • $\begingroup$ I don't understand the close-vote here - this is perfectly well-posed. That said, @veronika, the expression isn't quite accurate in the details, so it would be good to point to the source where you read it, as the context may change the answer a bit. $\endgroup$ – Emilio Pisanty Mar 5 '17 at 19:10
  • $\begingroup$ @EmilioPisanty It is the Karlshrue physics course, notoriously not mainstream physics and this is the reason I diden't mention it. Also answers of user Maimon at this site. I don't remember which, though. Maybe this definition is useful for teaching momentum, third law of Newton to children. I am a layman, I can't pose the question accurately. $\endgroup$ – veronika Mar 5 '17 at 19:23
  • 1
    $\begingroup$ @sammygerbil My dear friend Sammy I cited my source at the comments."If force is current, what is momentum?" You can always correct my wrong assumptions with an answer. It is useful, I think, for teaching mechanics to children. I would like to write from the beginning pshychoanalysis, not physics. If you explain me why this definition re-writes physics, I will delete my question. $\endgroup$ – veronika Mar 6 '17 at 7:08
21
$\begingroup$

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).

$\endgroup$
  • 4
    $\begingroup$ This is a very good answer, but you're answering a slightly different question, i.e. "can the conservation of momentun be written as a continuity equation?" $\endgroup$ – Alessandro Zunino Mar 5 '17 at 17:13
  • $\begingroup$ Note that if "current" only applies to some "stuff" in the world, there are at least three levels at which you might define "stuff": the weakest sense is that a stuff is a conserved quantity: so every reference frame agrees that an increase of stuff in a box comes at the expense of a decrease in other boxes. A stronger sense would be that everyone also agrees on the magnitude of the changes in the stuff. The strongest would be that also everyone agrees on the amounts of stuff in the boxes. Energy/momentum only fulfill maybe the first two of these, not the third. $\endgroup$ – CR Drost Mar 6 '17 at 16:42
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ However if we identify $\rho$ as momentum and $\mathbf{j}$ as a potential function, potential seems to be the current of momentum, since $-\nabla V = \mathbf{F}$. $\endgroup$ – Udit Dey Mar 5 '17 at 13:59
  • $\begingroup$ Yes at KPK i found this definition $\endgroup$ – veronika Mar 5 '17 at 14:04
  • 1
    $\begingroup$ But the relation $\frac{d\mathbf{p}}{dt}=-\nabla V$ involves the gradient operator, not the divergence, so it has nothing to do with the continuity equation. $\endgroup$ – Alessandro Zunino Mar 5 '17 at 14:11
3
$\begingroup$

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"?

$\endgroup$
1
$\begingroup$

$F=\frac{dp}{dt}$
$I=\frac{dq}{dt}$
$P=\frac{dE}{dt}$
etc

There is an obvious analogy here : force, current and power are the rate of flow of something wrt time. In each case the analogy can be developed.

Generally, finding the structural similarities between two different phenomena can be enlightening (see note below), such as between flow of electrical current in a wire and the flow of fluid in a pipe. Maybe the analogy is useful, maybe it is confusing. If force=current does momentum=charge? Is resistance=mass? It takes effort to identify what quantity corresponds with what other quantity. What happens when we consider forces with an electrical origin? It gets confusing, especially as a teaching aid to those who are new to physics. At some point the analogy breaks down.

These are analogies, not definitions. If the analogy is taken as a definition then it cannot break down. Departures from the analogy must be seen as new physical phenomena, requiring new concepts or new laws of nature to be postulated.


Note: see Conclusion in Analogy between Mechanics and Electricity as cited by Alexandro Zuninio)

$\endgroup$

protected by ACuriousMind Mar 5 '17 at 22:23

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.