Recent content by Sagittarius A-Star

The explanation of aberration in the light-clock scenario does not only work with classical particles, but also with EM waves. The reason is relativity of simultaneity. Assume, a horizontal electromagnetic wavefront segment of vertically moving light crosses the horizontal x-axis. In the...

Do you mean the case ##v=0##? You get ##2\arccos{} 0= \pi##, but not ##2\pi##. It seems that you want to describe the headlight effect. https://demonstrations.wolfram.com/TheHeadlightEffect/

In the restframe of the train, the bridge-length maybe even smaller than the train-length due to length contraction, if the bridge moves fast enough in the train's rest-frame. Your scenario has some similarities to the "length contraction paradox", published in 1961 by W. Rindler...

The crater depth will not be different.

A change in kinetic energy is ##\int \vec F \cdot d\vec s##. A force to decelerate the moving body by mechanical contact is in the end an electric force on a moving charge. But the time-component of the 4-current (charge density) depends on the clock synchronization, therefore also the electric...

I think this must be multiplied with ##m## to get the four-momentum.

This means for proper time squared: ##(d\tau)^2 = (dT + \kappa dX)^2 - dX^2 - dY^2 - dZ^2\ \ \ \ \ (1)## Multiplying equation ##(1)## with ##(m/d\tau)^2## yields the Anderson energy-momentum equation:$$m^2 = (\tilde E + \kappa \tilde p_x)^2 - \tilde {p_x}^2 - \tilde {p_y}^2 - \tilde {p_z}^2\ \...

But I think it is correct: The reason is, that the term ##m ({dx \over d\tau}, {dy \over d\tau}, {dz \over d\tau})## does not contain the coordinate-time and the momentum is conserved. Additional argument: The relativistic 3-momentum ##\mathbf v = ({dx \over d\tau}, {dy \over d\tau}, {dz \over...

Conserved energy is the time-component of the 4-momentum ##p_\mu=m {dx_\mu \over d\tau}##. ##E = m {dt \over d\tau}##. The conserved energy of a moving particle depends on the clock-synchronization scheme, but the conserved 3-momentum ##m ({dx \over d\tau}, {dy \over d\tau}, {dz \over d\tau}...

Please see A. Einstein: "The Meaning of Relativity"/Lecture 2, equations (39) and (42) https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_2 https://www.amazon.com/-/de/dp/B0F8RBNSRF?tag=pfamazon01-20

Einstein modified Newton's definition ##p_x=mv_x## to ##p_x=m {dx \over d\tau}##. This does not depend on the clock-synchronization scheme. Newton's laws do hold in an inertial coordinate system based on anisotropic one-way-speeds, if the underlying definitions are modified to the relativistic...

To my understanding this is valid for oberserver ##p_1##, but not for ##p_2## described in the same coordinates.

I think you cannot integrate over 100 revolutions, because the used Born coordinates are only defined in the range ##-\pi < \phi < +\pi##. Source: https://en.wikipedia.org/wiki/Born_coordinates#Transforming_to_the_Born_chart

I think you are only calculating in the simple non-rotating frame and use the time-dilation factors of SR. BTW.: From this I conclude: ##\omega_2 = { d\over dt} (\phi-\omega_1 t~) = -\omega_1## This doesn't answer the question of the OP for a calculation in the rotating frame. The Born...

I think you can't use Galilean (angular) velocity addition for ##\omega_1## an ##\omega_2##.