Recent content by Sagittarius A-Star

No, you can't. With reference to the traveler's restframe, the distance between Earth and Alpha Centauri is length-contracted, which explains for him the short proper travel-time.

Some reasons are discussed: Source: https://www.space.com/space-exploration/launches-spacecraft/bad-weather-delays-launch-of-chinas-historic-change-7-mission-to-moons-south-pole According to this video, windows for this mission are also limited by availability of light at the moon south pole...

China's plan for Chang'e 7 and a moon base:

Source: https://www.space.com/astronomy/moon/chinas-change-7-moon-probe-will-launch-this-weekend-on-the-most-ambitious-lunar-mission-in-history

Article about the Russian internet satellite constellation Rasvet: https://www.techtimes.com/articles/323875/20260811/russias-rassvet-clears-ukraine-coverage-threshold-ahead-schedule-gur-warns.htm

I think in the free theory vacuum, the refraction index ##n = \sqrt{\varepsilon_r \mu_r}## must be ##1##. If this were not the case, then light would move with ##c/n## relative to a certain reference frame. But the free theory vacuum does not have a certain state of motion like a medium. If the...

To come closer back to the thread-topic: I found the following Wikipedia article about QED vacuum. Source: https://en.wikipedia.org/wiki/QED_vacuum#Electromagnetic_properties They say that the QED vacuum can be considered a material medium with ##\epsilon_r >1## and ##\mu_r <1##. Is this...

Applying the Hedge dual in this way switches the interior derivative of the field between electric and magnetic sources (zero, because magnetic monopoles do not exist). In 4D-spacetime, there is a common electromagnetic bivector field with 6 components. They split into two sets (E & B) of three...

Consider the inhomogeneous Maxwell equations in (3+1) notation (Gaussian CGS units) with electric charge-density and electric 3-current-density on the RHS. ##\nabla \cdot \mathbf{E} = 4\pi\rho_{el}## ##\nabla \times \mathbf{B} - \frac{\partial \mathbf{E}}{\partial {(ct)}}=...

That's correct. However, for certain scenarios the electric force on a test-charge in it's rest-frame can transform into a pure magnetic force in a specific other frame: https://physics.weber.edu/schroeder/mrr/MRRtalk.html The first tensor in posting #119 is the EM field, the second is the dual...

That's one aspect. Another is, that 4D-notation makes the relativistic structure of Maxwells equations obvious. This is also valid for the Lorentz force, which is in Gaussian units: ## \frac{dp^\mu}{d\tau} = \frac{q}{c}F^\mu{}_{\nu}u^\nu ##. The EM field is here a tensorial coefficient between...

Derivation of the EM wave function (a linear, second-order partial differential equation) from Maxwells equations in (3+1) notation, Gaussian CGS units (see posting #107) Write the ##curl## of Faradays law: ## \nabla \times (\nabla \times \mathbf{E}) = \nabla \times \left( - \frac{\partial...

Yes. The RHS of the wave function contains a 2nd derivative for ##ct##. In SR, ##t## is multiplied with ##c## as conversion factor. You will see SR in Maxwells equations, if you write them in 4-tensor notation. Greek indices 1...4 stand for the spacetime coordinates: ##x^\mu = (x^1, x^2, x^3...

In 4D-spacetime ##(ct, x, y, z)##, the EM field is a bivector. It has 6 components (3 electric + 3 magnetic), which are projections on the 6 planes (xt, yt, zt, yz, zx, xy) of the 4D-coordinate system. Usually, an antisymmetric 4-tensor of grade 2 is used as notation for the EM field...

That 1) and 2) are equivalent can be seen in Gaussian CGS units. Maxwells equations contain partial derivatives for ##ct## of the fields. ##\nabla \cdot \mathbf{E} = 4\pi\rho## ##\nabla \cdot \mathbf{B} = 0## ##\nabla \times \mathbf{E} = -\frac{1}{c}\frac{\partial \mathbf{B}}{\partial t}##...