Recent content by timmdeeg

As an aside here is another attempt to get rid of Dark Matter in Galaxies. "Exact solutions for differentially rotating galaxies in general relativity" / https://arxiv.org/pdf/2406.14157 abstract: "Whether or not abundant collisionless dark matter exists, the new solutions very strongly...

The more it is astonishing that these physicists seem to overlook this factor. Thanks for answering.

https://www.mdpi.com/2571-712X/8/3/65 In short, if the dynamics of spiral galaxies are dependent on galactic morphology in the way predicted by GR-SI, and the morphology of galaxies are correlated with their number of satellite galaxies (and more generally, their environment), then we expect to...

Thanks to both of you for clarifying my question.

The decoherence time is extremely short for macroscopic objects like a detector. Does this mean that the outcome of a quantum measurement which happens instantaneous can be understood as being caused by decoherence?

I don't know if its the minimum. I took this picture end of 2021 with my Sony A7III using a 200 mm telephoto lens together with the Omegon MiniTrack LX3, which is mechanically guiding. This picture was my motivation to learn Astrofotografie later.

First light for my RedCat 51 at the Astro-Farm Kiripotib in Namibia in May 2025. The conditions have been overwhelming, 10 hours dark sky each night, SQM around 21.6 and while walking to the Astro-Camp the bright Milky way already at half past 7 pm. Setup Teleskop: RedCat P 51/250 f/4,9...

Thanks for clarifying.

How to calculate the mass within the Hubble Sphere and its time dependence, assuming the L-CDM model?

Do you mean by this - perhaps - that you are in the center of the respective observable universe at any arbitrary place in the universe?

Ah, yes, thanks. I have confused eigenstate and eigenvalue. So, until B is measured it has an eigenstate of spin. Correct?

In more detail, supposed A has been measured, but B not yet. In this case B has an eigenvalue, but depending on the interpretation not an eigenstate. In this case if B is measured thereafter B will have an eigenstate associated with said eigenvalue.

It means, that in the case of B's measurement B has an eigenstate associated with that eigenvalue.

Do you say that if we don't measure B then it isn't spin-down? It has an eigenvalue but not an eigenstate associated with it. Is it wrong to say that if we measure A we effectively measure the system consisting of the entangled pair of particles A and B? But if true wouldn't it mean that the...

Thanks!