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Let $\textbf{u},\textbf{w}$ be the four velocities of observers at rest in the original and primed frames, respectively. Let $\Delta V$ be a small volume in $\textbf{u}$'s frame, and let $\Delta p^{\mu}$ be the total amount of four momentum contained in $\Delta V$. $\Delta p^{\mu}$ is also expressible as $T^{\mu0}\Delta V=T^{\mu\alpha}(u_{\alpha}\Delta V)$ - ...

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A simple argument shows that contraction can only occur in the direction of motion: there is no way to uniquely specify any other axis in space along which an object could contract. If nature is to be consistent, then longitudinal contraction is the only choice.

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" Is it because the contraction happens in the direction the object is going, which is usually only a single direction? If so, are there any cases which would involve 2- or 3-dimensional contraction?" Yes. In order to illustrate the principle, uniform motion (constant velocity) is usually specified as occurring along a single spatial dimension, such as "the ...

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Is it because the contraction happens in the direction the object is going, which is usually only a single direction? Yes, exactly. The dimensions perpendicular to the relative motion of the object and the observer are not affected. And as far as I know, there is no motion in which more than one (spatial) dimension is contracted.

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I think, you are mixing quantum electrodynamics with classical (relativistic) electrodynamics. The equation of motion you stated above is from the classical electrodynamics where particles are considered point charges. In quantum electrodynamics particles are just field excitation, so you cannot ask for their four-velocity but rather for the time evolution ...

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Let the probe move with v <<c, i.e. sub-relativistic thus only first-order in v effect relevant. Order of magnitude of v may be the drifting speed of conduction electrons.

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(This was intended to be a comment but ended up being too long) A related "paradox": Consider the diurnal motion of the stars as they appear to rise and set each night (the non-circumpolar stars at least) traversing circular paths on the celestial sphere. An obvious back-of-the-envelope value for their angular velocity ...

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If you tell me that an object has velocity $\omega r$, it will have velocity $\omega r$. There's nothing stopping me from imagining a nonphysical/theoretical point moving this fast. If you tell me that a physical particle of something, or something that moves at this speed, has this velocity, I will tell you that you're wrong when $\omega>c/r$, and I'll ...

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As Sachin says, there is no limit to acceleration. In fact you can show this by considering an observer hovering at a fixed distance from a black hole. As described in this question, the acceleration required to maintain a fixed distance from a black hole is given by: $$a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}}$$ where $r_s$ is the radius of the ...

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There's no theoretical limit to acceleration as long as you are not accelerating something to $c$. However, there'll be practical challenges as high acceleration would require high amount of energy.

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Since relativistic momentum p = $\gamma m_0 v$, then: $\vec{F} = \frac{d\vec{p}}{dt} = \gamma m_0 \frac{d\vec{v}}{dt}$, which, when solved using vectors is equal to: $\gamma m_0 \frac{v^2}{r}$

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Each carries a clock, which they synchronize to zero time at a moment when they are at the same position on the ring. Bob predicts that when they next meet, Alice's clock will read less than his because of the time dilation arising because she is moving relative to him. What's wrong [...] ? There are two things wrong with Bob's argument as ...

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Both move with constant speed V as measured in that frame. Constant speed but not constant velocity; both Alice and Bob are accelerated, i.e., each observes the other's accelerometer to read non-zero so standard SR reasoning doesn't apply. However, their worldlines, between the initial event and the event they next meet, are congruent thus, the proper ...

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