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Let's hear about it from the master of the theory to begin with instead...............
I could have done without the 40+ minutes of analogies. If he dumbed it down any further his audience would be the Sesame Street crowd.
Using his black hole as an example, would not someone outside of our light horizon see our universe as a two-dimensional representation, as in his hologram analogy? While we, inside the light horizon (similar to being inside the event horizon of the black hole), are three-dimensional objects, like Alice?
Also, if it is so hot on the surface of the event horizon of the black hole, would not that imply that this jumbled-up two-dimensional representation of everything inside the black hole also has mass?
I could have done without the 40+ minutes of analogies. If he dumbed it down any further his audience would be the Sesame Street crowd.
Using his black hole as an example, would not someone outside of our light horizon see our universe as a two-dimensional representation, as in his hologram analogy? While we, inside the light horizon (similar to being inside the event horizon of the black hole), are three-dimensional objects, like Alice?
Also, if it is so hot on the surface of the event horizon of the black hole, would not that imply that this jumbled-up two-dimensional representation of everything inside the black hole also has mass?
That video was made for TVO, not truly representative of his actual lectures on topics.
And in the channel search just type "Leonard" and you'll see actual college level lectures that you might enjoy more, Glitch.
Would someone outside of our light horizon see anything at all? After all, that's our boundary. Hard to speculate on that answer.
That certainly seems to be true. After all, like he said, we each have our own individual light horizon. So it only matters where you are at the time. Someone a billion light years away would have a completely different light horizon. So unlike his two-dimensional representation of the contents of a black hole, there would be nothing to distinguish our light horizon from the rest of space/time.
Quote:
Originally Posted by beninfl
Just because the black hole would have a 2D surface, doesnt imply that it wouldnt have mass. Black holes do in fact have mass.
I am aware that black holes have mass. What I was wondering was whether or not the creation of this two-dimensional surface on the "skin" of the event horizon also had mass, in addition to the mass already contained within the event horizon. The fact that it gets so hot would imply that it does have mass.
So an object passing through the event horizon of a black hole would not only transfer a two-dimensional image of the object to its event horizon, but also some of its mass, as it crosses that line of no return.
I am aware that black holes have mass. What I was wondering was whether or not the creation of this two-dimensional surface on the "skin" of the event horizon also had mass, in addition to the mass already contained within the event horizon. The fact that it gets so hot would imply that it does have mass.
So an object passing through the event horizon of a black hole would not only transfer a two-dimensional image of the object to its event horizon, but also some of its mass, as it crosses that line of no return.
I dont think the 3D versus 2D matters for mass. Because an object is 2D or 3D wouldnt imply it has less or more mass. If the energy dissipated via Hawking Radiation from a black hole was more or less than what it took in, then that might imply differently. But the black holes shrink proportionate to their energy release.
The 2D image of the object is the actual object, transformed like a hologram would be from 3D to 2D. All of the information is there, it's just "flat". And the information carries with it the full mass and energy of the original object.
When the energy is dissipated into the Universe again, the 2D surface area becomes more "pixely" as bits of the information are radiated at different times. Eventually all of the bits are gone, and voooosh, black hole is no longer.
Otherwise we'd be violating conservation of energy and all that jazz.
I dont think the 3D versus 2D matters for mass. Because an object is 2D or 3D wouldnt imply it has less or more mass. If the energy dissipated via Hawking Radiation from a black hole was more or less than what it took in, then that might imply differently. But the black holes shrink proportionate to their energy release.
The 2D image of the object is the actual object, transformed like a hologram would be from 3D to 2D. All of the information is there, it's just "flat". And the information carries with it the full mass and energy of the original object.
When the energy is dissipated into the Universe again, the 2D surface area becomes more "pixely" as bits of the information are radiated at different times. Eventually all of the bits are gone, and voooosh, black hole is no longer.
Otherwise we'd be violating conservation of energy and all that jazz.
That makes absolutely no sense whatsoever. How can an object passing the event horizon of a black hole transfer not only an image of itself but also all of its mass to the event horizon and still have the same mass as it passes the event horizon? That would be doubling the mass of every object that passes the event horizon, and we know that does not happen.
Holograms are merely photons, and photons at rest have no mass. Therefore, the two-dimensional image "imprinted" on the event horizon of a black hole should have no mass. However, he said that the surface of the event horizon would be exceedingly hot. Therefore, something with mass must be generating that heat under the extreme gravitational pressure of the black hole. We know photons cannot have any mass (otherwise they would not be able to travel at 299,792 km/s), therefore if the "skin" of an event horizon is truly as hot has he claims, then something with mass, other than photons, was also transferred to the "skin" of the event horizon as the object passed through into the black hole.
That also means that there is some mass loss from the object to the event horizon, albeit a very small amount, as the object passes the event horizon into the black hole.
That makes absolutely no sense whatsoever. How can an object passing the event horizon of a black hole transfer not only an image of itself but also all of its mass to the event horizon and still have the same mass as it passes the event horizon? That would be doubling the mass of every object that passes the event horizon, and we know that does not happen.
Holograms are merely photons, and photons at rest have no mass. Therefore, the two-dimensional image "imprinted" on the event horizon of a black hole should have no mass. However, he said that the surface of the event horizon would be exceedingly hot. Therefore, something with mass must be generating that heat under the extreme gravitational pressure of the black hole. We know photons cannot have any mass (otherwise they would not be able to travel at 299,792 km/s), therefore if the "skin" of an event horizon is truly as hot has he claims, then something with mass, other than photons, was also transferred to the "skin" of the event horizon as the object passed through into the black hole.
That also means that there is some mass loss from the object to the event horizon, albeit a very small amount, as the object passes the event horizon into the black hole.
When we say the "image" of itself, it means that all of the 3D mass is transformed into 2D mass. It's still fully there, mass, energy, and all.
Holograms is really just the consumer-world analogy that we use to describe the 3D/2D/3D transformation process.
Photons are traditionally said to be massless. This is a figure of speech that we use to describe something about how a photon's particle-like properties are described by the language of special relativity.
The logic can be constructed in many ways, and the following is one such. Take an isolated system (called a "particle") and accelerate it to some velocity v (a vector). Newton defined the "momentum" p of this particle (also a vector), such that p behaves in a simple way when the particle is accelerated, or when it's involved in a collision. For this simple behaviour to hold, it turns out that p must be proportional to v. The proportionality constant is called the particle's "mass" m, so that p = mv.
In special relativity, it turns out that we are still able to define a particle's momentum p such that it behaves in well-defined ways that are an extension of the newtonian case. Although p and v still point in the same direction, it turns out that they are no longer proportional; the best we can do is relate them via the particle's "relativistic mass" mrel. Thus
p = mrelv
When the particle is at rest, its relativistic mass has a minimum value called the "rest mass" mrest. The rest mass is always the same for the same type of particle. For example, all protons have identical rest masses, and so do all electrons, and so do all neutrons; these masses can be looked up in a table. As the particle is accelerated to ever higher speeds, its relativistic mass increases without limit.
It also turns out that in special relativity, we are able to define the concept of "energy" E, such that E has simple and well-defined properties just like those it has in newtonian mechanics. When a particle has been accelerated so that it has some momentum p (the length of the vector p) and relativistic mass mrel, then its energy E turns out to be given by
E = mrelc2 , and also E2 = p2c2 + m2restc4
There are two interesting cases of this last equation:
If the particle is at rest, then p = 0, and E = mrestc2.
If we set the rest mass equal to zero (regardless of whether or not that's a reasonable thing to do), then E = pc.
In classical electromagnetic theory, light turns out to have energy E and momentum p, and these happen to be related by E = pc. Quantum mechanics introduces the idea that light can be viewed as a collection of "particles": photons. Even though these photons cannot be brought to rest, and so the idea of rest mass doesn't really apply to them, we can certainly bring these "particles" of light into the fold of equation (1) by just considering them to have no rest mass. That way, equation (1) gives the correct expression for light, E = pc, and no harm has been done. Equation (1) is now able to be applied to particles of matter and "particles" of light. It can now be used as a fully general equation, and that makes it very useful.
When we say the "image" of itself, it means that all of the 3D mass is transformed into 2D mass. It's still fully there, mass, energy, and all.
Holograms is really just the consumer-world analogy that we use to describe the 3D/2D/3D transformation process.
Photons are traditionally said to be massless. This is a figure of speech that we use to describe something about how a photon's particle-like properties are described by the language of special relativity.
The logic can be constructed in many ways, and the following is one such. Take an isolated system (called a "particle") and accelerate it to some velocity v (a vector). Newton defined the "momentum" p of this particle (also a vector), such that p behaves in a simple way when the particle is accelerated, or when it's involved in a collision. For this simple behaviour to hold, it turns out that p must be proportional to v. The proportionality constant is called the particle's "mass" m, so that p = mv.
In special relativity, it turns out that we are still able to define a particle's momentum p such that it behaves in well-defined ways that are an extension of the newtonian case. Although p and v still point in the same direction, it turns out that they are no longer proportional; the best we can do is relate them via the particle's "relativistic mass" mrel. Thus
p = mrelv
When the particle is at rest, its relativistic mass has a minimum value called the "rest mass" mrest. The rest mass is always the same for the same type of particle. For example, all protons have identical rest masses, and so do all electrons, and so do all neutrons; these masses can be looked up in a table. As the particle is accelerated to ever higher speeds, its relativistic mass increases without limit.
It also turns out that in special relativity, we are able to define the concept of "energy" E, such that E has simple and well-defined properties just like those it has in newtonian mechanics. When a particle has been accelerated so that it has some momentum p (the length of the vector p) and relativistic mass mrel, then its energy E turns out to be given by
E = mrelc2 , and also E2 = p2c2 + m2restc4
There are two interesting cases of this last equation:
If the particle is at rest, then p = 0, and E = mrestc2.
If we set the rest mass equal to zero (regardless of whether or not that's a reasonable thing to do), then E = pc.
In classical electromagnetic theory, light turns out to have energy E and momentum p, and these happen to be related by E = pc. Quantum mechanics introduces the idea that light can be viewed as a collection of "particles": photons. Even though these photons cannot be brought to rest, and so the idea of rest mass doesn't really apply to them, we can certainly bring these "particles" of light into the fold of equation (1) by just considering them to have no rest mass. That way, equation (1) gives the correct expression for light, E = pc, and no harm has been done. Equation (1) is now able to be applied to particles of matter and "particles" of light. It can now be used as a fully general equation, and that makes it very useful.
Thanks, that makes much more sense. Photons have relativistic mass, and combined with their momentum, are able to impart a significant amount of energy upon a collision.
I encountered an interesting paper published by the Soviet Physics Uspekhi entitled, "Astrophysical upper limits on the photon rest mass." They suggest the upper limit on the photon rest mass is less than ~3.0E-60 g, which is considerable smaller than the Planck constant. Unfortunately, you have to be a paid subscriber to read the paper, and I am not. My first instinct, however, is to distrust anything smaller than the Planck constant, because we are really just guessing at that point.
It looks like I will have to brush up on the theory of quantum electrodynamics.
When we say the "image" of itself, it means that all of the 3D mass is transformed into 2D mass. It's still fully there, mass, energy, and all.
Holograms is really just the consumer-world analogy that we use to describe the 3D/2D/3D transformation process.
Photons are traditionally said to be massless. This is a figure of speech that we use to describe something about how a photon's particle-like properties are described by the language of special relativity.
The logic can be constructed in many ways, and the following is one such. Take an isolated system (called a "particle") and accelerate it to some velocity v (a vector). Newton defined the "momentum" p of this particle (also a vector), such that p behaves in a simple way when the particle is accelerated, or when it's involved in a collision. For this simple behaviour to hold, it turns out that p must be proportional to v. The proportionality constant is called the particle's "mass" m, so that p = mv.
In special relativity, it turns out that we are still able to define a particle's momentum p such that it behaves in well-defined ways that are an extension of the newtonian case. Although p and v still point in the same direction, it turns out that they are no longer proportional; the best we can do is relate them via the particle's "relativistic mass" mrel. Thus
p = mrelv
When the particle is at rest, its relativistic mass has a minimum value called the "rest mass" mrest. The rest mass is always the same for the same type of particle. For example, all protons have identical rest masses, and so do all electrons, and so do all neutrons; these masses can be looked up in a table. As the particle is accelerated to ever higher speeds, its relativistic mass increases without limit.
It also turns out that in special relativity, we are able to define the concept of "energy" E, such that E has simple and well-defined properties just like those it has in newtonian mechanics. When a particle has been accelerated so that it has some momentum p (the length of the vector p) and relativistic mass mrel, then its energy E turns out to be given by
E = mrelc2 , and also E2 = p2c2 + m2restc4
There are two interesting cases of this last equation:
If the particle is at rest, then p = 0, and E = mrestc2.
If we set the rest mass equal to zero (regardless of whether or not that's a reasonable thing to do), then E = pc.
In classical electromagnetic theory, light turns out to have energy E and momentum p, and these happen to be related by E = pc. Quantum mechanics introduces the idea that light can be viewed as a collection of "particles": photons. Even though these photons cannot be brought to rest, and so the idea of rest mass doesn't really apply to them, we can certainly bring these "particles" of light into the fold of equation (1) by just considering them to have no rest mass. That way, equation (1) gives the correct expression for light, E = pc, and no harm has been done. Equation (1) is now able to be applied to particles of matter and "particles" of light. It can now be used as a fully general equation, and that makes it very useful.
Yes, sorry for the math. Glitch is one of the very few on the forums here who makes sense of the concepts when explained with the actual formula and equations. I try to keep plain language as much as humanly possible.
I don't understand escape velocity. Why do we need a rocket to travel 25,000 mph to get into space? Why not a plane with wings and rocket engines that gradually climbs higher and higher till it leaves the atmosphere like this: https://www.youtube.com/watch?v=_OMx8Im16Sw
Speaking of gravity, if the earth's gravity is strong enough to hold the moon in orbit why do astronauts feel zero G just a few miles above earth. And I assume gravity doesn't just go from 1 to 0 at some threshold but decreases gradually. So, If I'm in an airplane at 40k' I weight less than at sea level?
Last edited by geos; 01-13-2014 at 06:00 PM..
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