Supported Theory: Light cannot "enter" black holes.

kmarinas86 · Sep 21, 2005

Gravity slows down the speed of light, physically: http://en.wikipedia.org/wiki/Shapiro_effect In General relativity, the Shapiro effect, or gravitational time delay, is one of the four classic solar system tests of general relativity. It says that a radar beam (or light beam) which passes near a massive object as it travels from some observer's location to a target and returns to the observer, takes slightly longer to make the round trip (as measured by the observer) than it would if the object were not present. More generally, the "travel time" of any signal moving at the local speed of light can be affected by the gravitational field in regions of spacetime through which it travels. In general relativity (and in most other gravitation theories), the local speed of light is a constant of nature, but the time delay effect implies that the effective global speed of light is path-dependent. The time delay effect was first noticed in 1964, by Irwin I. Shapiro. Shapiro proposed an observational test of his prediction: bounce radar beams off the surface of Venus and Mercury, and measure the round trip travel time. When the Earth, Sun, and Venus are most favorably aligned, Shapiro showed that the expected time delay, due to the presence of the Sun, of a radar signal traveling from the Earth to Venus and back, would be about 200 milliseconds, well within the limitations of 1960s era technology. The first test, using the MIT Haystack radar antenna, was successful, matching the predicted amount of time delay. The experiments have been repeated many times since, with increasing accuracy. Velocity time dilation deals with the clocks of objects with mass. Gravitational time dilation deals with the clock of the space time fabric itself! http://www.google.co</safety_wrapper

kmarinas86 · Sep 21, 2005

I wonder what hapens, then, to that radiation which reaches the event horizon of singularity. Is it split into components, as with a supercollider? A photon would be boundlessly blueshifted as it reaches the blackhole in the same way a photon would be redshifted to a trival existence when leaving from just above the surface of a black hole. The photon approaching the black hole may spontanously form a quark-anti-quark pair if it has an energy equal to the energy of two electrons. And the quark-anti-quark pair may combine to form a photon with equivalent energy.

kmarinas86 · Sep 21, 2005

Mass-Radius Relationship of Black Holes Radius = Mass * 1.48494276 × 10-27 meters / kilogram 10^30 kg = weight of typical star 10^41 kg = weight of spiral galaxy with 100 billion stars 10^53 kg = weight of a trillion galaxies 10^16 m = ~10^0 LY = ~1 light year 10^3 m = ~10^-13 LY = radius of a black hole with the weight of typical star 10^14 m = ~10^-2 LY = radius of a black hole with the weight of spiral galaxy with 100 billion stars 10^26 m = ~10^10 LY = radius of a black hole with the weight of a trillion galaxies = ~10 billion light years

jatslo · Sep 21, 2005

Well now, this is an interesting thread. time loop?

kmarinas86 · Sep 21, 2005

Premises: http://en.wikipedia.org/wiki/Length_contraction A particle has a space-time continuum within it. When this particle speeds up, its space-time continuum (or gravity field) is compressed such that it becomes "more solid". It is compressed because the whole system of this particle is accelerated from the rear. The "solidness" compares to the solidity of really dense matter in the universe such as neutron stars or even black holes. This causes a Gravitational Time Dilation. Conclusion: The state of accelerated particles, such as mesons, which are usually quick to decay, is preserved beyond their usual lifetime, due to gravitational time dilation caused by pressures exerted from the outside which also increase the particles' velocity. You cannot change the value for the local speed of light, but you can change gravitational time dilation through various means discovered elsewhere in the universe. If you can manage to create a tunnel of low gravitational time dilation to get from point A to point B, and point B to point A, one could concievably have a two-way trip between the two points without a large time passing in the universe.

jatslo · Sep 21, 2005

if you can survive, then I bet you will see light all the way through, unless there is a time loop

Saiph · Sep 21, 2005

Gravitational time dilation also prohibits outside observers from ever seeing matter cross the event horizon. However, the concept of reference frames remains intact. The person in the reference frame notices time move normally, and will observe themselves crossing the event horizon (granted, they'll note the outside reference moving at a much different, and faster, pace). <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

jatslo · Sep 21, 2005

Yep, if the speed is so fast relative to rest that the subject travels backwards, then the subject would not reach rest, because of time looping. <--- Speculation

alkalin · Sep 21, 2005

What if we are looking thru an atmosphere which the sun has and therefore are influenced by the sun’s atmospheric index, N?

kmarinas86 · Sep 21, 2005

What if we are looking thru an atmosphere which the sun has and therefore are influenced by the sun’s atmospheric index, N? The sun atmosphere would distort our vision of space like how the earth's atmosphere does.

kmarinas86 · Sep 22, 2005

Gravitational time dilation also prohibits outside observers from ever seeing matter cross the event horizon. However, the concept of reference frames remains intact. It does, however.... because of the Infinite Gravitational Time Dilation, this needs to be explained in more detail. Assuming that the external universe can exist after infinity years, then the photon will cross the event horizon (after infinity years have been passed in the exterior universe). The person in the reference frame notices time move normally, and will observe themselves crossing the event horizon (granted, they'll note the outside reference moving at a much different, and faster, pace). If you imagine that in say, 10^300 years in the external universe's future, something will come with enough force to rip that mass from the black hole. Something, will inevitably rip it from there. Of course, observing such phenonemon would make more sense given that the observable universe is a repeating fractal (a universe which is ultimately the most understandable, whatever our universe might be). So far, the best candidates I can think for being our observable universe is gluon matter "strong force" exhanged between quarks. I have given a minimal explanation for how this might be possible. A Fractal Universe http://uplink.space.com/showflat.php?Cat=&Board=phenomena&Number=316930&page=0&view=collapsed&sb=5&o=0&fpart=

Saiph · Sep 22, 2005

All right, here's the references: berkley NASA Both say the same thing I did. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

kmarinas86 · Sep 22, 2005

It does show also that there is massive time distortion AFTER a mass/photon passes the Shhwartschild event horizon. But anything can fall into a black hole, contrary to the above title of the thread. Matter can of course, be inside a black hole, this I'm not refuting. After matter is piled on top, eventually, the schwarzschild radius will change (we all know that). I'm just saying that the matter doesn't "enter" the black hole, it only "becomes" a part of it as soon matter is in "proximity" to it. Just because I said it "didn't enter" didn't mean I said it's "never there" - it is there. So in this way, we actually agree, you just think we disagree because you missed the underlying meaning of what I was really saying when I said "enter".

jatslo · Sep 22, 2005

A black hole is to broad a statement, with respect to generalization, but I see your point. Matter and energy entering a quasar would never reach rest, because of length contraction. It is quite plausible that there is just simply not enough time in the universe to accomplish this task. Your argument is a valid one contrary to what stevehw33 says, so keep plugging away. <img src="/images/icons/wink.gif" /> You have a history of providing mathematics, but most people cannot translate the nomenclature, so with respect to feedback, I would like to request that you explain the mathematics philosophically, so that we can reference your frame of mind in the simplest form.

kmarinas86 · Sep 22, 2005

http://cosmology.berkeley.edu/Education/BHfaq.html#q3 What would happen to me if I fell into a black hole? ---------------------------------------------------- Let's suppose that you get into your spaceship and point it straight towards the million-solar-mass black hole in the center of our galaxy. (Actually, there's some debate about whether our galaxy contains a central black hole, but let's assume it does for the moment.) Starting from a long way away from the black hole, you just turn off your rockets and coast in. What happens? At first, you don't feel any gravitational forces at all. Since you're in free fall, every part of your body and your spaceship is being pulled in the same way, and so you feel weightless. (This is exactly the same thing that happens to astronauts in Earth orbit: even though both astronauts and space shuttle are being pulled by the Earth's gravity, they don't feel any gravitational force because everything is being pulled in exactly the same way.) As you get closer and closer to the center of the hole, though, you start to feel "tidal" gravitational forces. Imagine that your feet are closer to the center than your head. The gravitational pull gets stronger as you get closer to the center of the hole, so your feet feel a stronger pull than your head does. As a result you feel "stretched." (This force is called a tidal force because it is exactly like the forces that cause tides on earth.) These tidal forces get more and more intense as you get closer to the center, and eventually they will rip you apart. And, of course, I'm not arguing against this. For a very large black hole like the one you're falling into, the tidal forces are not really noticeable until you get within about 600,000 kilometers of the center. Note that this is after you've crossed the ho

kmarinas86 · Sep 22, 2005

Cf. this thread's title "Light cannot enter black holes." That is mistaken, clearly. The title is mistakenly worded. Let me, unmistakenly, rephrase it. "Light cannot cross the schwarzschild radius" Now, I have no problem if you say that this is mistaken. After all, the fact of the matter is, even though I am "capable" of learning significant parts of Relativity, I have yet to do so. Unfortunately, I do not know how to do second order derivatives yet, so obviously I cannot understand it all right now. I am not forcing anyone to believe me, but anyone is free to correct my errors (or criticize my mistakes). It's a fact that the concept of singularity is somewhat controversial (I'm questioning this concept by preseting an unvalidated guess (spiced up with the title) - I've yet to see validation for the singularity though the mathematics is not necessarily going to keep me from seeing it). I've noticed that "deviations" occur quite often in discussions about singularity, given the following pages at Wikipedia: http://en.wikipedia.org/wiki/Gravitational_time_dilation http://en.wikipedia.org/wiki/Gravitational_singularity Experts seem to be absent from the pages. The real experts, I think, are those who understand the mathematics of (general) relativity, and actually know what it is. I obviously don't. It's interesting how this subject is easily misintepretable.

nexium · Sep 22, 2005

To observers on Earth, (if the theory isn't nonsence) neither light nor mass enter black holes, but to observers orbiting the black hole in the accreation disk, the speed difference is less than c, so time dialation is moderate rather than extreme. Once light is inside the event horizon, we can not know what happens. It may take infinite time for the light photons to be absorbed into the zero surface area of the singularity. For black holes less than ten solar mass; most of the mass inside the Schartzschild is likely converted to photons and free quarks. How can a three dimentional quark with dimentions in pico meters enter a much smaller singularity? I picture the singularity coated with a thick layer of quarks. Neil

rpmath · Sep 23, 2005

To observers on Earth, (if the theory isn't nonsence) neither light nor mass enter black holes, but to observers orbiting the black hole in the accreation disk, the speed difference is less than c, so time dialation is moderate rather than extreme. I disagree about the second... (well I'm used to disagree with everybody about black holes) If time dilation at the event horizon is infinite relative to Earth, it is still infinite for observer orbiting the black hole, even if they have a huge time dilation relative to earth... Infinity divided per whatever huge finite number you choose is still infinity... Only observers falling toward the event horizon can see things crossing it because they will get infinite time dilation too... but I think saying those observers are as valid as other reference frames is going too far for relativity: May be time is what a clock says, but talking about what will happen after a broken clock marks the next our is a nonsense.

kmarinas86 · Sep 23, 2005

Suppose you have a clone. Your clone is near the black hole with a gravitational time dilation of 1000 and a velocity time dilation of 1000. You are not moving. You measure the speed of light to be c far from the black hole. Your clone measures the speed of light near the black hole to be c. Now, say two photons of the same frequency were emitted from the same point, one to your clone, and the other to you. The photon headed towards you will be gravitational redshifted in accordance with gravitational time dilation - not velocity time dilation. If the photon that reaches you observe to be 10 Hz, the clone will observe his photon to be 10 Hz (1000 times gravitational time dilation increases the photon's energy & frequency by 1000 times while doppler redshift will cause the light to be 1000 times longer in wavelength in the eyes of your clone). For every 1000 seconds you pass, your clone passes 1/1000th of a second, and the light passes just 1 second. If your clone is gravitationally time dilated by a factor of 1,000 versus you, and if he also observes the speed of light to be c, then that must mean, by definition, the gravitational time dilation is an altering of space time where the both mass and photons are dilated equally as a far as gravity goes (since the source of gravitational time dilation is the space time fabric itself).

jatslo · Sep 23, 2005

Velocity is the source. Gravity is like an engine that creates force.

kmarinas86 · Sep 24, 2005

Suppose that you were to use a rocket in a region with a time dilation of 1000. The front end of the rocket would less gravitationally time dilated than the rear end. The entire rocket is held together by metallic bonds. Gravitational Time Dilation, approx. T = 1000 = 10^3 = 1/sqrt(1-2GM/rc²) 1/10^6=1-2GM/rc² 1-1/10^6=2GM/rc² c²(1-1/10^6)=2GM/r Escape Velocity=sqrt(2GM/r) Escape Velocity=sqrt(c²(1-1/10^6)) Escape Velocity=c*sqrt(1-1/10^6) This Escape Velocity is the velocity that space craft has to travel to lose grip on the black hole. As the object escapes the black hole, it will accelerate in units of m/s² or m·Hz². The acceleration of the black hole's gravity is also measured in m/s² or m·Hz². The "Hz" is lower for objects with greater velocity time dilation and gravitational time dilation. In deepening gravity, light has a wavelength (m) decreased by "x" and a frequency (of Hz) increased by "x" (caused by a transfer of gravitational potential energy to the photon's massless momentum), however, in addition to this, gravitational time dilation is a decrease in the rate of time (of Hz) by "x" (independent of frequency). Given Gravitational Time Dilation (decreased Hz of Hz²), Hz² does not change, while frequency (increased Hz of Hz²) increases, and wavelength (m) is still shorter. The local speed of light, measured in m·Hz, does not account for gravitational space-time dilation, while the global speed of light, also m·Hz, factors in gravitational time dilation. Let's factor in Time Dilation (TimeDilation /> 1): Escape Velocity=c*sqrt(1-1/T)/TimeDilation and Velocity is measured in (m·Hz). Only a single Hz variable (Hz¹) should be affecting velocity. And this is true. This single Hz is decreased by the product of gravitational time dilation and velocity time dilation (a function of velocity)! Escape Velocity=v=sqrt(2GM/r)*sqrt(1-2GM/rc²)*sqrt(1-v²/c²) Let c = unity Let v b

jatslo · Sep 26, 2005

Right, by the time you got out of there, you'd be traveling pretty fast indeed. Invisible fast; however, I have a hypothecation that radiation; I think, radiation is super fast light, so go figure. <img src="/images/icons/wink.gif" /> I would imagine that matter that escapes would be annihilated into invisible alpha and beta particles. I would imagine that a star could get slingshot past “C” relative to Earth under the right circumstances. I would imagine that we could see it’s shockwave; similar to “Cherenkov Radiation”. I would be far more concerned with invisible matter and energy, than visible. The scopes are beginning to detect the invisibles, so new advancements in that arena are sure to follow. Light-Boom is relative to velocity <img src="/images/icons/wink.gif" /> With respect to color, I mean. Opaque would be peak performance, I think. Hawking will report soon, and we will see.

daniko · Sep 27, 2005

Hi everyone ! I'm sorry that I missed the begining of the discussion, because I've been thinking a lot on that theme. I almost agree with kmarinas86that light cannot "enter" black hole It's just a little incomplete. I think the following is more complete: Light cannot "enter" black hole in our Universe My point is that: If any kind of Matter (including light) entering a black hole can "see" the end of Time of our Universe -- /> it cannot be considered as part of our Universe. With other words: if anything happens after all the Time in our Universe had gone it's not part of our Universe. I think that the Matter that dips under the event horizont of a black hole forms a separate Universe. I also got another reason for talking of separate Universe. I think that the tale of the "stretched cosmonaut" who tryed to enter a black hole is wrong. No, he won't be stretched by a tidal forces of gravitation. The reason I believe so is the lenght contraction effect. All we know that if an object passes by on speeds close to [c] we will observe how it's time is slowed and its length is shortened. The same effect will appear with the descending cosmonaut in strong gravitation. As he descends into the black hole the tidal forces will rise. It means that the difference of the gravitational force between two points with different heights respective to the black hole will raise. On the other hand we will observe (from outside the hole) how the cosmonaut becomes shorter. In result the tidal force between the head and the foots of the poor man will remain constant. So he won't feel anything stretchy while free falling. From the view point of the outer observer everything that goes down the black hole contracts its height. If we shift view points, this contraction will appear to the desc

jatslo · Sep 27, 2005

If any kind of Matter (including light) entering a black hole can "see" the end of Time of our Universe, it cannot be considered as part of our Universe. What? You are talking about the invisibles, and the invisibles are part of the universe. They are invisible because they are traveling beyond the scope of our current visual frame of reference. They are either traveling slower than we are, or faster than we are, and we should be able to detect them regardless. I think radiation is evidence of this phenomenon. You think that the Matter that dips under the event horizon of a black hole forms a separate Universe. What? Velocity will open doors to the invisibles, but I can hardly agree that it constitutes a new universe; it only expands on the known universe in my opinion. Now if time is an issue, then we are talking time-loop and plausible annihilation; however, time dilation may make it possible for matter to travel backwards, but that still leaves time looping. Hypothecation: "Stretched Cosmonaut”? Gravity will stretch this individual until his or her eyes pop out, rest assured; However, in terms of velocity, one might be able to exist closer rest provided the G forces on Length contraction = equilibrium. The universe has a constant G which should flux as you approach rest; however, length contraction from velocity could work in your favor, because there is less of your to affect. <blockquote>In reply to:<hr />My point is that: If any kind of Matter (including light) entering a black hole can "see" the end of Time of our Universe --> it cannot be considered as part of our Universe. With other words: if anything happens after all the Time in our Universe had gone it's not part of our Universe. I think that the Matter that dips under the event horizon of a black hole forms a separate Universe. I also got another reason for talking of separate Universe. I think that the tale of the "stre</blockquote>

gammarayburst · Nov 3, 2005

I don't agree with the idea that light cannot enter a black hole "star". Matter is slower than light and matter can be detected entering the event horizon. So if matter can be seen entering a black "stars" grasp then light should be able to enter. Gamma ray bursts, xrays, and other frequencies have been observed coming from black "stars" at their magnetic poles. Does this mean that there is no gravity there? No! It means that the energy can escape the magnetic field. Seeing that energy can escape a black "stars" poles would mean that gravity does not effect light. This would mean it's the magnetic field that effects the light. I would have to conclude that if light could not enter a black star then it would be because of the magnetic field not gravity. < deductive conclusion

Supported Theory: Light cannot "enter" black holes.

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