(Special relativity) A tilted dipole composed of $+q$ and $-q$ point charges moves at a horizontal velocity. Will it rotate? Two point charges $+q$ and $-q$ are connected with a rigid insulating stick:

This dipole moves at a velocity $v$ to the right. Therefore, the charges should exert a magnetic Lorentz force $F_B$ on each other, causing the dipole to rotate.
On the other hand, we can find an inertial frame S in which the charges are at rest from the start. Then, according to the transformation formulas for velocities, how can this dipole rotate (i.e., with the charges having a non-zero vertical component in their velocities) in another inertial frame that moves at a velocity of $v$ to the left with respect to the frame S?
 A: Ricky Tensor has beaten me to the punch, and his excellent answer provides the key insight: in the frame where the stick is moving, the force of the stick on the charges is not parallel to the stick itself. In my answer I'll just provide some intuition for why this should be the case.
Simpler setup, no angles
The necessity of this result can be seen in a much simpler setup. Consider a stretchable string connecting two equal masses, which has a constant tension $T$ in its rest frame. If the masses start off at rest, they will come together and collide due to the string.

Now consider the reference frame where the masses are initially moving with velocity $v$ to the right. As the string accelerates the masses towards each other, the rightward velocity $v_x$ must stay the same, so the speed goes up, and the Lorentz factor $\gamma$ increases. But the rightward momentum is $p_x = \gamma m v_x$, which means $p_x$ must increase, which means there must be a force in the rightward direction. This force must be applied by the string itself, meaning that the force it applies is not parallel to itself.
(There is a second subtlety in this particular setup: if the string is exerting rightward forces on both particles, doesn't that violate conservation of momentum? No, because the string itself carries rightward momentum; it must carry an energy per length of $T$, which means it has a mass per length of $T/c^2$. As the string's length decreases, it gives up its momentum. But this is irrelevant to OP's question, since OP's stick has constant length.)
Defining shear stress
Here's a second way to see that this result is plausible. The reason the non-parallel force looks weird is because it suggests that there is a shear stress in the stick, while no such shear stress exists in the stick's rest frame. This would be a contradiction, since, e.g. we could replace OP's stick with a cylindrical fluid-filled balloon, which can't support shear stress at all.
The problem here is in the definition of shear stress. Demanding the force be along the stick is equivalent to demanding that the off-diagonal elements of the stress tensor vanish, $T_{ij} = 0$ if $i \neq j$. But this condition doesn't remain true under a Lorentz transformation, which mixes up diagonal and off-diagonal elements! The real definition of vanishing shear stress, from the solid's point of view, is that the off-diagonal elements are zero in the solid's rest frame. The constitutive relations of a solid only apply in the rest frame. In a general frame, all bets are off.
A microscopic model
The second explanation above might feel like a cop-out; it just says that there's no reason the stick's force has to be parallel to itself. But you might want a concrete, microscopic model that explains how the stick's force could be non-parallel to itself.
Of course, this would be rather difficult to do for a real solid, but here's a toy model. Let's replace the stick with a hollow column, within which is a single neutral particle. In the stick's rest frame, this particle elastically bounces back and forth, reflecting its velocity every time it hits one of the charges, leading to a time-averaged force along the stick.
Now let's consider the same situation in OP's reference frame. The particle still bounces back and forth, reversing its velocity every time it hits one of the charges. But because the stick itself is moving, the particle's velocity is not along the stick. It needs an additional rightward component to stay inside the stick as the stick moves to the right. This is the origin of the non-parallel force.
In principle, something similar is happening at the microscopic level in a real solid, though this would be very hard to show explicitly.
A: Here I am providing a shorter version of Ricky Tensor's excellent answer to this question.
In the frame where the dipole is at rest, each of the charges feels an electro(magnetic) force along the line connecting the two charges, and a stick force of the same magnitude but in the opposite direction, hence in a state of equilibrium.
In the frame where the dipole is moving horizontally, the horizontal length is contracted because of Lorentz contraction, and the perpendicular force is reduced according to force transformation from a rest frame to a moving frame:
$$
F'_\perp = \frac{1}{\gamma}F_\perp
$$
$$
F'_{||} = F_{||}
$$
This force formula applies not only to the stick force $F_\mathrm{s}$, but also to the electromagnetic force $F_\mathrm{em}$:

Therefore, each charge is again in a state of equilibrium, though now moving at a constant velocity.
A: According to the Lorentz transformation formula for the electric field, we have:
$$
\mathbf E_\perp' = \gamma(\mathbf E_\perp+\mathbf v\times \mathbf B_\perp)
$$
Since $\mathbf B=0$ in the rest frame, in the moving frame $\mathbf E$ becomes stronger by a factor of $\gamma$. We might hope for these effects to cancel out. The size of the magnetic field generated is $|B_\perp| = \gamma v|E_\perp|/c^2$, by a similar Lorentz transformation law.
Lorentz force on the moving charge will be $qv|B_\perp|$, since the field is perpendicular to the direction of motion. So the overall force due to magnetic field is $\gamma |E_\perp| qv^2/c^2$. Meanwhile, the magnitude of the electric force on the charge in the direction perpendicular to motion is $\gamma |E_\perp| q$. Since these are pointing in opposite directions, the net attractive force, perpendicular to the direction of motion, is:
$$
\gamma |E_\perp|q(1-\frac{v^2}{c^2}) = \frac{1}{\sqrt{1-v^2/c^2}}|E_\perp|q(1-v^2/c^2) = |E_\perp|q\sqrt{1-v^2/c^2}
$$
But this is not a complete cancellation, the perpendicular electromagnetic force is still reduced in the moving frame. The only hope now is that maybe in the moving frame, the insulating stick applies a force that isn't parallel to itself.
In the resting frame, the stick transmits a force $F$ between the two charges. The definition of force is $dp/dt$: the rate of flow of momentum. In the resting frame, both charges have 0 momentum at all times. So the net momentum change is 0. Some momentum is flowing through the electric field between the charges at a rate of $F$, and an equal and opposite quantity of momentum is flowing through the insulating stick, also at a rate of $F$.
Let's consider the change in force applied by the stick after our Lorentz transformation. In the rest frame, we have time coordinate $t$. In the moving frame, we have time coordinate $t' = \gamma(t-vx/c^2)$. At rest, $x=0$ at all times, so $t' = \gamma t$. This means that a small interval of time $dt$ in the rest frame corresponds to a slightly longer interval of time in the moving frame, $dt'=\gamma dt$. The momentum transferred in a time $dt$ is $dp = (Fdt\cos\alpha, Fdt\sin\alpha)$ in the rest frame. According to the transformation rule for 4 momentum, perpendicular momentum transferred remains unchanged while parallel momentum transferred scales by $\gamma$. So we have: $dp'=(\gamma Fdt\cos\alpha, Fdt\sin\alpha)$. Then:
$$
F' = \frac{dp'}{dt'} = \frac{\gamma Fdt\cos\alpha, Fdt\sin\alpha}{\gamma dt} = (F\cos\alpha, \frac{1}{\gamma}F\sin\alpha)
$$
So the force exerted by the stick perpendicular to the direction of motion has been scaled by a factor of $1/\gamma$, just like the perpendicular electromagnetic force was, while parallel force remains the same. Mystery solved, but in a very strange way: The moving stick is exerting a force along a different axis than the one it's pointing in. But that's just special relativity for you. (Length contraction doesn't compensate for this, in fact it makes the stick even less parallel to the force it transfers!)
