Q-reeus said:
The usual 'low velocity' expression (2) is in fact exact and good for any velocities whatsoever. Provided it is applied as intended. Forces, fields and velocities all evaluated in the one frame they appear. Notice I used dp/dt = F in (2) - not dp/dtau. Important difference. No wiggling out that non-covariant expression (2) was clearly the form being discussed earlier!
If I interpret the two equations as you've just said, they say the same thing:
dp/dtau = dt/dtau dp/dt = gamma dp/dt
Which is fine; if by "force" you just mean the "coordinate force" dp/dt, then you're correct, the equation you wrote is always valid, because it's just a different way of writing the equation I wrote, a way which makes the relativistic covariance more obscure.
The question is, what do we actually measure? The answer is, we actually measure dp/dtau. That's what, for example, a strain gauge attached to the arm that is holding the charge q in place would read. And that observable has to be invariant; it can't be frame dependent, because it's a direct observable. The reading on the strain gauge attached to q can't depend on which frame we decide to calculate it in. I thought that by "force" you meant the actual measured force, the force the strain gauge would read, which is why I thought your version of the force law was incorrect. You're right, it's not incorrect; it just doesn't mean what you think it means.
Q-reeus said:
So what is the correct usage for expression (1)? To be honest, not being conversant with covariant formulations, I'm at a loss.
Indeed. Try the most general covariant version, using the EM field tensor, which appears in the Wiki article (I'm giving it in a slightly different form which will be easier to work with):
\frac{dp^{a}}{d\tau} = q F^{a}_{b} u^{b}
This equation is valid in any frame, so you can just read off the frame-dependent versions by transforming the tensor components and the 4-velocity appropriately.
For example, in the rest frame of the capacitor, we have F^{0}_{1} = F^{1}_{0} = E, with all other EM field tensor components zero. We also have for the 4-velocity of the charge q u^{b} = (\gamma, 0, \gamma v, 0). We have oriented the spatial axes so the capacitor's field is in the "1" direction and the charge is moving at velocity v in the "2" direction (perpendicular to the capacitor field, as was my understanding).
The only nonzero component of the above equation is the "1" component of the 4-acceleration:
\frac{dp^{1}}{d\tau} = q F^{1}_{0} u^{0} = q E \gamma
which is what I wrote. And, of course, we can obtain what you wrote by just writing dp/dt = dtau/dt dp/dtau, which removes the factor of gamma.
Now do the same thing in the rest frame of the charge q. Obviously the 4-velocity is just u^{0} = 1 in this frame, all other components zero. But perhaps it's worth writing out the Lorentz transform of the EM field tensor explicitly, since you actually have to transform twice because the tensor has two indexes. And because we're using the "mixed" form of the tensor, with one upper and one lower index, we need two versions of the transform, one for each kind of index. I haven't found a good online reference for all this, but it's in the textbooks (I'm working from MTW).
A given EM field tensor component transforms like this (indexes are primed or unprimed according to which frame they refer to):
F^{a'}_{b'} = F^{c}_{d} \Lambda^{a'}_{c} \Lambda^{d}_{b'}
where repeated indexes are summed over, and the two transformation matrices are (you do have to keep careful track of which indexes are primed and which are unprimed):
"Forward" matrix (primed upper index):
\Lambda^{0'}_{0} = \Lambda^{2'}_{2} = \gamma
\Lambda^{0'}_{2} = \Lambda^{2'}_{0} = - \gamma v
\Lambda^{1'}_{1} = \Lambda^{3'}_{3} = 1
"Inverse" matrix (unprimed upper index):
\Lambda^{0}_{0'} = \Lambda^{2}_{2'} = \gamma
\Lambda^{0}_{2'} = \Lambda^{2}_{0'} = \gamma v
\Lambda^{1}_{1'} = \Lambda^{3}_{3'} = 1
If you take each of the above matrices individually, and work out how it acts on a vector, you will see that it's saying the same thing as you're used to with Lorentz transformations: x' = gamma (x - vt), t' = gamma(t - vx), etc., with the sign of the v terms switched for the inverse matrix.
Now we need to apply the matrices to the tensor components as shown above. This looks like a lot, but if you work through the possible combinations of indexes, you will see that there are only four nonzero components of the primed EM field tensor: F^{0'}_{1'} = F^{1'}_{0'} = \gamma E, and F^{1'}_{2'} = - F^{2'}_{1'} = \gamma v E. The first two terms correspond to an electric field in the "1" direction, as before, but now you can see that it is stronger, in "coordinate" terms, than it was in the rest frame of the capacitor, by a factor gamma. The last two terms correspond to a magnetic field in the "3" direction, i.e., perpendicular to both the electric field and the relative velocity of the charge and the capacitor. This is what you're used to seeing written out, in 3-vector form, as v x E.
Now read off the force in the primed frame from the index equation above; you will see that there is again only one nonzero component, that for the "1" component of the 4-acceleration:
\frac{dp^{1'}}{d\tau} = q F^{1'}_{0'} u^{0'} = q \gamma E
which is the same number as in the other frame. So the observable quantity is the same in both frames, as required.
But you could "interpret" the two equations differently. You could say that, in the rest frame of the capacitor, the "E field" is just E, and the "coordinate force" on the charge q is qE, but to obtain the observable number, the "4-force", you have to multiply by gamma to convert dp/dt to dp/dtau. Whereas, in the rest frame of the charge, the "E field " is already gamma E, so dp/dt and dp/dtau are the same (because t = tau for the charge in its own rest frame). This is all just "interpretation" and doesn't change the physics.
What you *can't* do is say that somehow the "real force" is different in the two frames, because that leads you to a false prediction: it leads you to predict that the reading on the strain gauge would differ depending on which frame you calculated it in.
We both agreed earlier that there is legitimately a reduction in mass/energy when matter is lowered into a potential well. And it shows remotely - the gravitational contribution felt 'out there' is not that owing to plugging in the locally observed proper mass but the redshifted coordinate value. Agreed?
