# Tag Info

## New answers tagged relativity

2

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 ...

-1

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.

0

(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 ...

0

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 ...

3

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 ...

1

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.

1

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: $\gamma m_0 \frac{v^2}{r}$

0

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 ...

6

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 ...

4

The problem with your idea is that each time the light reflects off the mirror it transfers some of its energy to the mirror (to increase the mirror's kinetic energy) and is red shifted as a result. So the thrust would fade as the light red shifts away to nothing. For obvious reasons the light can only transfer as much energy to the mirrors (in the form of ...

0

Taken from this review of Conway and Smith's "On Quaternions and OCtonions: It follows that the quaternions of norm 1 form a group under multiplication. This group is usually called ${\rm SU}(2)$, because people think of its elements as $2 \times 2$ unitary matrices with determinant 1. However, the quaternionic viewpoint is better adapted to ...

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