7
$\begingroup$

Consider a system of particles (index $i$). Let the force acting on each particle be $\mathbf{F}_i = \mathbf{F}_i^{e}+\sum_{j, j \neq i}\mathbf{f}_{ji}$, where $\mathbf{F}_i^{e}$ denotes the external force on the particle and $\mathbf{f}_{ji}$ denotes the internal force on particle $i$ due to particle $j$. The time rate of change of the total momentum $\mathbf{P}$ is given by $$\dot{\mathbf{P}}=\frac{d}{dt}\left(\sum_{i}m_i\mathbf{v}_i \right)=\sum_{i}m_i\dot{\mathbf{v}}_i = \sum_{i}\mathbf{F}_i^{e}+\sum_{i,j\neq i}\mathbf{f}_{ji}=\mathbf{F}_{\mathrm{tot}}^{e}+\sum_{i,j>i}(\mathbf{f}_{ji}+\mathbf{f}_{ij}) $$ where $\mathbf{F}_{\mathrm{tot}}^{e}$ denotes the sum of the external forces. In case the (weak) law of action and reaction holds, $\mathbf{f}_{ji}+\mathbf{f}_{ij}=\mathbf{0}$, and we recover Euler's first law.

My question is if, and if so why, this assumption is realistic for bodies in practice. As we know, the electromagnetic force does in general not satisfy the law of action and reaction, since the fields carry both linear and angular momentum. Given that electromagnetic forces ultimately are responsible for all internal forces, why can we in general expect the condition $\mathbf{f}_{ji}+\mathbf{f}_{ij}=\mathbf{0}$ to hold?

$\endgroup$

2 Answers 2

9
$\begingroup$

You're right that in general, the right way to think about electromagnetic interactions isn't between charges at all: instead charges each individually act on the field, which intervenes between them. Newton's third law and its strong form just boil down to overall conservation of linear and angular momentum for both the charges and field together.

However, in most situations where we talk about introductory mechanics, the change in (angular) momentum of the field is negligible. This generally holds as long as particles are not accelerating significantly, and are moving slowly compared to the speed of light.

This can be established heuristically in a few cases. For example, consider two charged particles separated by a distance $r$, with charge $q$ and speed $v$, neglecting emission of radiation. The ordinary electrostatic force between them, which does obey the strong form of Newton's third law, is $$F_e \sim q E \sim \frac{q^2}{\epsilon_0 r^2}.$$ Meanwhile, the magnetic force between them, which does not obey Newton's third law, is $$F_m \sim q v B \sim q v \left(\frac{\mu_0 q v}{r^2}\right) \sim \frac{\mu_0 q^2 v^2}{r^2}.$$ The ratio of these forces is $$\frac{F_m}{F_e} \sim \mu_o \epsilon_0 v^2 \sim \frac{v^2}{c^2}$$ which is indeed small when the charges move nonrelativistically. (Incidentally, the same analysis holds for particles interacting gravitationally, via gravitoelectromagnetism.). To check this, we can also estimate the field momentum. The field momentum density is $$\mathcal{P} \sim \frac{1}{c^2} \frac{E B}{\mu_0}.$$ The right $E$ and $B$ to use here are the electric field of one particle and the magnetic field of the other. (Taking the same fields for both particles would just give the momentum carried by a particle in isolation, which can be absorbed into the definition of the particle's mass.) The product $E B$ is hence non-singular and significant over a volume of order $r^3$, giving an electromagnetic field momentum $$P_{\text{em}} \sim r^3 \mathcal{P} \sim r^3 \, \frac{1}{\mu_0 c^2} \frac{q}{\epsilon_0 r^2} \frac{\mu_0 q v}{r^2} \sim \frac{\mu_0 q^2 v}{r}.$$ What matters is the rate of change of this momentum, which is $$\frac{dP_{\text{em}}}{dt} \sim \frac{\mu_0 q^2 v^2}{r^2}$$ which is precisely the order of $F_m$, i.e. the violation of Newton's third law. So everything checks out; the field picks up the "missing" momentum.

That is precisely why Newton's third law gets mentioned less and less as one continues in the physics curriculum. It's ultimately just an approximation, that ends up replaced with the deeper ideas of momentum and angular momentum conservation.

$\endgroup$
3
  • $\begingroup$ Thanks! So in conclusion, would it be accurate to say that under conditions such that the Newtonian description is applicable, violations of the law of action and reaction, are always negligible? $\endgroup$ Jan 14, 2020 at 9:51
  • $\begingroup$ @ÉtienneBézout Yup, if they weren't, we wouldn't use the language of Newtonian point particle mechanics. $\endgroup$
    – knzhou
    Jan 14, 2020 at 21:08
  • $\begingroup$ Alright, thanks! $\endgroup$ Jan 15, 2020 at 0:23
2
$\begingroup$

In this case you work only with point-like particles, that interact through paired interaction, that satisfy condition: $$ f_{ij} + f_{ji} =0 $$ This condition is essential.

If you wanna to describe interactions of this system through fields, you need be more accurate. For correct description of field you need use special relativity. For details I refer you to

Mansuripur's paradox

Mansuripur's Paradox 1 Problem - Princeton Physics

Trouble with the Lorentz law of force: Incompatibility with special relativity and momentum conservation

Comment on "Trouble with the Lorentz Law of Force: Incompatibility with Special Relativity and Momentum Conservation"

$\endgroup$
1
  • 1
    $\begingroup$ Thank you for the links! $\endgroup$ Jan 14, 2020 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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