I Forces between point particles: Always towards or away?

greypilgrim
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Hi.

I've read the following argument why the torques generated by internal forces add up to zero: Since by Newton's third law all forces come in pairs, so do all internal torques, adding them we get
$$\vec{\tau}_1+\vec{\tau}_2=\vec{r}_1\times\vec{F}_1+\vec{r}_2\times\vec{F}_2=\vec{r}_1\times\vec{F}_1-\vec{r}_2\times\vec{F}_1=(\vec{r}_1-\vec{r}_2)\times\vec{F}_1$$
and apparently this is zero since ##(\vec{r}_1-\vec{r}_2)\parallel\vec{F}_1## .

This argument uses that the forces between two point particles always point towards or away from each other. Is this true in general? I don't quite see why it should be, but I also can't find a counterexample within the fundamental forces (although maybe magnetism, but I don't know how magnetic forces work between point particles).
 
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The internal forces between two point particles do not necessarily point along the line joining the particles if the particles are point magnetic dipoles. To get a handle on this you can calculate the force on dipole ##\vec{\mu}_1## due to the presence of dipole ##\vec \mu_2## using ##\vec F_{2~on~1}=-\vec{\nabla}(\vec{\mu}_1\cdot \vec{B}_2)## where ##\vec B_2## is the magnetic dipole field set up by ##\vec \mu_2##. Knowing the force, you can find the torque using ##\vec \tau = (\vec r_1 - \vec r_2) \times \vec{F}_{2~on~1}##.
 
Does that mean magnetic internal forces actually can lead to a nonzero net torque? But doesn't that violate conservation of angular momentum?
 
Not if you take into consideration the angular momentum of the field itself :-)
 
Oudeis Eimi said:
Not if you take into consideration the angular momentum of the field itself :-)
If we define fields as the force that would act on a hypothetical unit charge, then the field is just a mathematical tool and there should be a field-free description with forces only. Where's the angular momentum now?
 
I don't think we can do that, not in a relativistic analysis anyway (which is required for electrodynamics). Since covariance (in classical theories at least) implies locality, the field must be regarded as the basic object, and (classical) point particles taken as a limiting case of a very concentrated charge distribution (so effectively another field, of charge density in this case).
 
So can we somehow conclude from this that "non-relativistic forces" (if there are any) between point particles always have to point along the line joining the particles? Everything else seems to necessarily include fields and SR.
 
greypilgrim said:
So can we somehow conclude from this that "non-relativistic forces" (if there are any) between point particles always have to point along the line joining the particles? Everything else seems to necessarily include fields and SR.

This is a good question, and the most straightforward answer I can give is "I don't really know" :-)

However, allowing for something less straightforward: instant forces at a distance, as they are used in Newtonian mechanics, must comply to the known conservation laws for linear and angular momentum, assuming the system has the appropriate symmetries. So forces that intrinsically don't conserve angular momentum are ruled out, as long as we model them as direct forces at a distance between point particles. Under these constraints, then it seems inevitable that said forces must act along the line joining two particles. Which agrees with your above conclusion.
 
greypilgrim said:
So can we somehow conclude from this that "non-relativistic forces" (if there are any) between point particles always have to point along the line joining the particles? Everything else seems to necessarily include fields and SR.
It depends on how strong assumptions you make concerning the forces between your point particles. The standard way is to consider a many-body system that is (a) describable as a Hamiltonian system, i.e., the equations of motion are derivable from Hamilton's action principle and is (b) closed, i.e., taking into account all particles all 10 conservation laws due to Galilei symmetry are fulfilled or equivalently the action is Galilei invariant and (c) there are only two-body forces. Then from symmetry considerations you can show that the Lagrangian is of the form
$$L=\sum_{j=1}^N \frac{m_j}{2} \dot{\vec{x}}_j^2 - \frac{1}{2} \sum_{j,k; \, j \neq k} V_{jk}(|\vec{x}_j-\vec{x}_k|).$$
In other words the forces are derivable from central potentials for particle pairs, which implies that the forces are always along the line joining the particles:
$$\vec{F}_{j,k}=-\vec{\nabla}_j V_{jk}(|\vec{x}_j-\vec{x}_k|)=-\frac{\vec{x}_j-\vec{x}_k}{|\vec{x}_j-\vec{x}_k|} V_{jk}'(|\vec{x}_j-\vec{x}_k|).$$
An example are Newton's gravitational forces, for which
$$V_{jk}=-\frac{\gamma m_j m_k}{|\vec{x}_j-\vec{x}_k|}.$$
 
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