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15

By the Newtonian definition, the photon wouldn't count due to its zero mass, but this is a relativistic collision, so you need a relativistic definition of the center of mass. Relativistically, the c.m. frame is the one in which the total momentum four-vector of the system is purely timelike. No, this does not coincide with the electron's frame. What you're ...


11

The reference frame of the center of mass is, by definition, the one where the total $3$-momentum vanishes. It exists almost always also for massless particles as I go to discuss. The total $4$-momentum $P$ of a system of $N$ free particles is the sum of their $4$-momenta of the particles, i.e., $$P = \sum_{a=1}^N P_{(a)}\:,$$ where each $P_{(a)}$ is a ...


5

Square the proper time $d \tau = dt / \gamma$, multiply by $c^2$, rearrange, and take the square root: $$ \left(\frac{d\tau}{dt}\right)^2 = \gamma^{-2} = 1 - \left(\frac{\bf{v}}{c}\right)^2 \Rightarrow \sqrt{\left(c \frac{d\tau}{dt}\right)^2 + {\bf{v}}^2} = c $$ The proper time $d\tau$ is the time elapsed in the frame moving with respect to the lab frame, in ...


4

I think I have understood the problem. Since I want to compute the length in the moving reference frame $R'$, I must assume that $t_2' = t_1'$ but not $t_1=t_2$, because I want to measure the position of the two ends of the bar at the same time in the moving frame. The calculation is as follows: $$ x_2-x_1 = \gamma(x_2'+ut_2')-\gamma(x_1+ut_1) = ...


4

Yes it is - well, sort of. The coordinate invariant form of velocity is the four velocity, and the magnitude of any four velocity is always $c$ (or $1$). So even if you are stationary in space in your chosen coordinate system the magnitude of your four velocity is still $c$. Whether moving at the speed of light on the time axis is a good way to state this ...


4

"Relativistic" velocities (velocities in excess of 0.1 c) are not needed. Any velocity difference will do. People use the doppler effect right here on Earth. Some sample uses: Catching speeders. How fast is his fastball? (Very important at this time of year.) Where is that tornado going?


3

I realize that in essence there is no object which can be considered as "not moving in space". No object at all is moving in space if you are taking the point of view of its reference frame! The law of conservation of energy is requiring that the energy of its mass (e = mc2) is "transported through time", or in other words, that time is passing for ...


2

This depends quite a bit on what you're willing to accept as "direct." The magnetic force between two parallel current-carrying wires can be interpreted as being due to Lorentz contraction, and that's quite easy to measure -- you can do it with a battery and some strips of aluminum foil. Some measurements can be interpreted as showing time dilation, but in ...


2

As you clearly ask for a detection criterion of the doppler shift on "a sample light wave", I have taken the liberty of assuming the following scenario. You detect light from a source, emitting light of a known wavelength (the usual case in astronomy for instance) with an instrument that is capable of measuring wavelength with an accuracy (i.e. with an ...


1

Actually it was that maxwellian electromagnetism had no problems, in contrast to the newtonian classical mechanics framework. Theory of Relativity alters the Newtonian framework not the Maxwellian framework. i would say that even if Einstein hadn't invented SR, someone else would (as indeed many others notably Poincare, Lorentz et al) were alredy on the ...


1

The fact that different observers in relative motion can measure the same light ray to move at a speed of c has to do with the fact that each observer defines the "speed" in terms of distance/time on rulers and clocks at rest relative to themselves. It's crucial to understand that different observers use different rulers and clocks to measure speed, because ...


1

There are apparently some extensions of the Standard Model that allow for Lorentz symmetry to be violated, although from what I understand the symmetry is broken by spontaneous symmetry breaking which means the theory would have been symmetric in the era when the forces were unified, and the symmetry was broken by a random decay to different vacuum state. So ...


1

No. My answer is negative, even if I confirm the statements of other answers: "The first thing is almost completely arbitrary, especially in full general relativity. The second thing is an unambiguous result of an experiment."(Jerry Schirmer) "In Einsteinian relativity all observers can still agree on a number of facts, they are just ...



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