guenter

Could you explain this idea? Thanks.

So am I, Migl. Much simplified Susskind's point seems, that for a hypothetical observer at the EH, the information is somehow (keyword holographic principle) encoded in the EH. And as the EH shrinks the corresponding information is set free to the oustside. In other words, the encoded information decreases proportional to the loss of mass. So, the total information is conserved. The underlying mathematical concept (based on string theory) is very demanding but is taken serious by physicists. I guess you talk about tidal forces. These are stretching things radially and squashing them perpendicular to that direction during their fall towards a mass. At the event horizon of a static black hole of mass M these forces are proportional to 1/M². So, if the BH is large enough, you may not even notice anything, while crossing the EH. Also, this dependence of the tidal forces on the BH's mass is the reason why the Hawking radiation increases with decreasing mass. According to theory black holes can be deformed by external fields. This happens dramatically during the merger of two black holes, creating gravitational waves thereby. If the infalling mass is small, this effect is tiny accordingly. Im am not sure but doubt that the dynamic is fully understood in such detail, as you are questioning. But for sure the Schwarzschild radius rs grows according to rs = 2M.

IM Egdall, one way to answer this question, might come from considering the lightcones. Perhaps you are familiar with that. They are vertical in flat space, getting tipped close to the Black Hole, are tangent at the horizon and are tipped inwards inside it. Now imagine events in spacetime on a surface of constant r. Outside the horizon 2 events on this surface are related timelike, because one event is within the lightcone of the other one. At the horizon 2 events are separated lightlike (lightcone is tangent there). Inside the horizon events with constant r are related spacelike as according to the lightcones being tipped inwards. so, I guess the switching of coordinates results in how 2 events in spacetime are related to each other. A similar consideration should hold for surfaces of constant time. But my knowledge regarding your question is very limited. Hopefully one of the experts can tell more or, in case correct me.

That sounds very natural, Migl. The Hawking radiation which is a black body radiation doesn't contain the missing information. But one should remember Leonard Susskind and his string theoretical calculations. According to that, the information is conserved at the horizon surface. I am not familiar with this stuff, but am sceptical. And - who knows whether string theory is more than a wonderful mathematical building.

Yes, here. Frankly, I prefer the curvature factor in the form (1-2M/r), because it shows the influence of the Mass on the metric. The angle term is zero in case of a radial fall. The point is that for r > 2M the sign of dt² is positiv and that of dr² negativ. Beyond the event horizon its just vice versa. So, dt² and dr² change role at r = 2M. Its a consequence of the coordinate singularity at r = 2M. It might be of interest that the mentioned coordinate singularity vanishes, if the Schwarzschild coordinates are transformed into Kruskal-Szekeres coordinates. Then the t- and r-coordinates remain timelike and spacelike.

Sorry, there is another misunderstanding: One can find this wording in some popular literature, but it is not correct to say so. The wristwatch of the freefaller shows still the flow of the time after he has crossed the event horizon. What happens in Schwarzschild coordinates is that the curvature factor is < 0 for r < 2M, which makes the t-coordinate spacelike and vice versa the r-coordinate timelike. So, these radial coordinates do interchange, not space and time.

Migl, I recommend to careful distinguish Free-fall coordinates, whereby the time coordinate is proper time, from any other coordinates, otherwise there is confusion. The picture of the frozen star, you mentioned, refers to the far distant Schwarzschild observer.

Be careful, using "Anthony" as a reference, see the comment of DrRocket! The proper time for an object to fall from the event horizon of a black hole with 3 billion sun masses e.g. to its center (the singularity) is about 13 hours.

Stars consist of atoms. I guess the .5% relate to their radiation, which gravitates attractive, like mass. The planets contribute only negligible to the mass of star systems.

No, matter - whether hot or cold - exerts attractive gravitation whereas DE works repulsive.

That seems not compelling. This catastrophic factor was calculated from QFT as the sum of zero-point energies. They could gravitate as well. On the other side this factor isn't a real problem, because you can get rid of it by renormalization. What remains to be argued are differences of energy, a good example for which is the Casimir-effect. Some hope comes from LQG to at least model the inflationary period of expansion.

No, vice versa. Some researchers discuss the possibility that we are in the center of a low matter density region. Imagine the galaxies beyond the visible universe to be spherically symmetric much more dense, then the galaxies within the visible universe would recede accelerated as observed. However having the copernican principle in mind this sounds not too trustful. Assuming vacuum outside one would rather expect deacceleration or even contraction.

michel123456, if you want to show a 'look back time' vs. distance diagram, it's convenient to choose the distance expressed in lightyears and the time axis in years. Then you obtain a straight line under 45°. Choosing miles instead of lightyears the line remains straight, but the angle changes. However, perhaps you want to show something else.

Justin, the universe expands everywhere at the same rate and consequently double distance means double velocity. It's easy to see that, remembering the dotted balloon. Imagine 3 dots in one line and equally spaced, us on A, B 1 cm away from us and C 1 cm away from B, but 2 cm from us. Now we let the balloon expand to double it's radius. Then B is 2 cm away from A = us and also C is 2 cm away from B, but 4 cm from us. Thus from our viewpoint B receded 1 cm and C receded 2 cm within the same time. So, C's velocity is twice as much as B's velocity viewed from A, us. This is Hubble's law, v = d*Ho, the receding velocity is proportional to the distance, as mentioned already in this Thread.

Yes, according to the Hubble law, the receding velocity v is proportional to the distance d, v = d*H.

Good question, I would love to know the answer. To my knowledge the experimental results regarding neutrino oscillations offer an explanation how they acquire mass, but not why the lepton number isn't conserved. The conservation of this number is restricted to the SM. In the frame of most of the GUT-theories the lepton number is not conserved, e.g. not with the predicted proton decay. In these theories only the difference of baryon- and leptonnumber is conserved. However the proton decay isn't detected yet, so it's speculation. But inspite of this the neutrino oscillations indicate that the SM will have to be modified. Perhaps and hopefully the LHC will tell more within some years.

The point is that in case the neutrino is massless one can define only observers who see it from "one side". From this perspective neutrinos are lefthanded and antineutrinos are righthanded, as according to the standard model. If however they possess mass, then neutrinos having a velocity < c can be overtaken and thus look righthanded. If left/right handed depends on a choosen frame of reference, then the left handed neutrino is identical with the righthanded neutrino. This at least is my understanding.

That is plausible, provided both, we and our outer colleague, agree on the current standard cosmology, based on the cosmological principle. It is also helpful to think of the expanding dotted balloon.

You talke about the Hubble sphere, which can be thought as a shell with radius c/Ho (Ho is the Hubble constant, c/Ho is called the Hubble lenght) around us. Galaxies on the Hubble sphere recede with c, hence the distance between the photons emitted by such a galaxy towards us and our position is constant. In that sense one can indeed say these photons stay there but should keep in mind that they travel with c locally. Note, that the Hubble constant decreases over time.

As stated here neutrinos and antineutrinos annihilate via virtual Z-Bosons into charged particle-antiparticle pairs (leptons, quarks). Thus finally- at least theoretically - photons could be produced.

No, annihilation creates gamma rays. That isn't observed. Inspite of this there is no thermal equilibrium of matter/antimatter/radiation at temperartures less than GUT scale. Happy new year to everybody!

Scorcerer, imagine the 3000 K proton-electron plasma being in thermal equilibrium with radiation (photons) 380000 years after the big bang. Now, after cooling by expansion protons and electrons combine to hydrogen atoms, there is no more charge scattering and the universe is transparent. After expansion of the universe we detect these photons at a temperature of 2.73 K. (The energy of photons can be expressed as wavelenght or frequency but also corresponds to a temperature). Now imagine "islands" of antimatter plasma (antiprotons and positrons) here and there in the 3000 K plasma then. The unavoidable matter-antimatter annihilation would have created photons of much higher energy (>> 3000 K) and thus their temperature would exceed 2.73 K today by far. The CMB doesn't show such photons.

Oh, interesting. So annihilation isn't restricted to electric charges. But something else must distinguish then neutrinos and antineutrinos to offer the possibility to annihilate, their chirality? And what is the reaction product? I am not at all familiar with neutrino physics. In case the search for Majorana neutrinos is positiv, would it mean the neutrino is its own antineutrino, like it is true for photons?

Sorry, to avoid misunderstanding: to rule what out?

Neutron and antineutron do annihilate, because they consist of charged particles. The criterion is the existence of opposite charges, not the question whether or not the particles are fundamental.