Recent content by Sagittarius A-Star

Yes. In this case, the cross product must be replaced by the wedge product. ##\mathbf L = \mathbf r \wedge \mathbf p##

They try to avoid the use of thrusters and use control moment gyroscopes most of the time, to not disturbe micro-gravity experiments. Source: https://www.quora.com/How-does-the-ISS-keep-its-orientation/answer/Robert-Frost-1?ch=99&oid=42053686&share=6bbd3c52&srid=5Gqkk&target_type=answer

The gyroscope in the video is spinning about it's axis (even while it's housing is not spinning), but it is difficult to see. How this specific gyroscope works, can be seen in the video from 01:20 on:

It's the ISS, which rotated, not the gyroscope axis: Source: https://www.forbes.com/sites/quora/2017/10/03/how-does-the-iss-travel-around-the-earth/

In theses notes is also mentioned, that Johann Georg von Soldner calculated the gravitational deflection of light as it passes the sun. He did this in 1801 (published in 1804), based on Newton's theory...

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