BUT efficiency is actually 100%. As we can see when we build a heat engine where a black hole serves as a heat sink. This engine has 99.9999999 efficiency, or 99.999999999999999999999 with bigger and colder black hole.
When energy goes downhill in an electric power line on a hill side, voltage and amperage increase.
When energy goes downhill in an electric power line on a hill side, voltage and amperage increase.
Interesting. Were does the extra current come from?
No Jartza, I'm not sure how it will express itself but I don't see it as impossible, you need to have really high energies for it to be noticeable I think. And the same arguments as we use for describing 'frames of reference' should be applicable. But you have a really nice argument there. And what I think it is about is.
"What the he* is energy"?
You are climbing up the stairs - you are not doing work.
You are climbing up the stairs carrying a box of photons - you are doing work.
A box of photons is tumbling down the stairs - photons are doing work. (making noise is the work they do)
Here's the correct answer to the quiz:
The solar panel will say: after the black hole moved closer I have been receiving more energy, the photons are more energetic and there are more photons per second.
Yes. when looked at as waves it will blue-shift from the view point of the solar panel, and as far as I understand it will increase its blue-shift the closer we get to the event-horizon. The interesting thing is why it can do it :)
A and B are identical robots with same mass, they are at floor 2.
A is carried to floor 1. B uses the stairs to go to floor 1.
A and B have different masses now. B has bigger mass.
Because A has cool brakes, while B has hot brakes.
Oh here I come lol
Replacing R in first formula with 2GM/c² and simplifying we get: ¼mc²
So they say efficiency is 75%
BUT efficiency is actually 100%. As we can see when we build a heat engine where a black hole serves as a heat sink. This engine has 99.9999999 efficiency, or 99.999999999999999999999 with bigger and colder black hole.
Could someone explain this a little better to me. Where does this efficiency arise? The calculations?
Oh yes.
Jartza, are you thinking of energy converted into mass?
Oh here I come lol
Replacing R in first formula with 2GM/c² and simplifying we get: ¼mc²
So they say efficiency is 75%
BUT efficiency is actually 100%. As we can see when we build a heat engine where a black hole serves as a heat sink. This engine has 99.9999999 efficiency, or 99.999999999999999999999 with bigger and colder black hole.
Could someone explain this a little better to me. Where does this efficiency arise? The calculations?
Expert's not so good efficiency calculation:
Click this link: http://www.scholarpedia.org/article/Bekenstein_bound (http://www.scholarpedia.org/article/Bekenstein_bound)
Find picture where mass is hanging over bigger mass. Find in the text the part that talks about what is going on in the picture.
Jartza's correct efficiency estimation based on this:
http://en.wikipedia.org/wiki/Carnot%27s_theorem_%28thermodynamics%29 (http://en.wikipedia.org/wiki/Carnot%27s_theorem_%28thermodynamics%29)
The Hawking Equation for Black Hole Entropy is a good startE=Rmc^4 /8GM
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fplus.maths.org%2Fissue18%2Ffeatures%2Fhawking%2Fimages%2Fformula.gif&hash=c33232921cba408337fd641e90fbb5c2)
The Hawking Equation for Black Hole Entropy is a good startE=Rmc^4 /8GM
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fplus.maths.org%2Fissue18%2Ffeatures%2Fhawking%2Fimages%2Fformula.gif&hash=c33232921cba408337fd641e90fbb5c2)
This is the swarzschild radius of a black hole with mass M:
R=2GM/c²
Replacing R in first formula with 2GM/c² and simplifying we get: ¼mc²
So they say efficiency is 75%
using the above, the OP simplifies 2GM/c² and simplifying we get: ¼mc²
How is this simplification acheived? Am I missing something obvious? 1/4Mc^2 is quite different to the quantity 2GM/c^2. A bit lost here following the OP's calculations. And who says the black hole is 75% efficient?
I must be doing the calculation wrong somehow, because rereading, the OP is not simplyfing 2GM/c² exactly. The OP is asking me to replace that quantity with R in the first equation - it would be good to read it properly from my behalf. Right... the G's and M's cancel, 2 divides into 8 giving, 4... right fine. Yes, it leaves ¼mc².
I must be doing the calculation wrong somehow, because rereading, the OP is not simplyfing 2GM/c² exactly. The OP is asking me to replace that quantity with R in the first equation - it would be good to read it properly from my behalf. Right... the G's and M's cancel, 2 divides into 8 giving, 4... right fine. Yes, it leaves ¼mc².
I'll be honest, I'm not sure what it means. The OP says:
''The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon''
so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...
I could be wrong.
I'll be honest, I'm not sure what it means. The OP says:
''The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon''
so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...
I could be wrong.
Black Holes are quite different creatures
They apparently seem to violate some fundamental laws of physics
the 1/4mc^2 term relates to the rest mass ENERGY.
The body of mass "m" accelerates towards the speed of light as it enters the black hole event horizon
Is the mass converting to "non-kinematic" energy?
Are you saying that the black hole absorbs 75% of the rest mass ENERGY and 25% is lost?
What is this efficiency term you talk about?
How does it relate to the Hawking radiation of balck holes?
I'll be honest, I'm not sure what it means. The OP says:
''The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon''
so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...
I could be wrong.
Black Holes are quite different creatures
They apparently seem to violate some fundamental laws of physics
the 1/4mc^2 term relates to the rest mass ENERGY.
The body of mass "m" accelerates towards the speed of light as it enters the black hole event horizon
Is the mass converting to "non-kinematic" energy?
Are you saying that the black hole absorbs 75% of the rest mass ENERGY and 25% is lost?
What is this efficiency term you talk about?
How does it relate to the Hawking radiation of balck holes?
Nothing is lost, especially once an object has passed the event horizon, as that would imply it destroys the information paradox. Kinetic energy is simply an energy of a system supplied to its momentum, so its an energy of a system which is in movement. A system can loose kinetic energy if its speed slows down, or gain it, if it gets faster.
Again, I've never seen the term 1/4Mc^2 before, so I am unsure how to approach the question other than saying nothing is fundamentally lost.
I'll be honest, I'm not sure what it means. The OP says:
''The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon''
so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...
I could be wrong.
Black Holes are quite different creatures
They apparently seem to violate some fundamental laws of physics
the 1/4mc^2 term relates to the rest mass ENERGY.
The body of mass "m" accelerates towards the speed of light as it enters the black hole event horizon
Is the mass converting to "non-kinematic" energy?
Are you saying that the black hole absorbs 75% of the rest mass ENERGY and 25% is lost?
What is this efficiency term you talk about?
How does it relate to the Hawking radiation of balck holes?
Nothing is lost, especially once an object has passed the event horizon, as that would imply it destroys the information paradox. Kinetic energy is simply an energy of a system supplied to its momentum, so its an energy of a system which is in movement. A system can loose kinetic energy if its speed slows down, or gain it, if it gets faster.
Again, I've never seen the term 1/4Mc^2 before, so I am unsure how to approach the question other than saying nothing is fundamentally lost.
One of the few things that Hawking contributed was to show how black holes can radiate energy at from the event horizon and would therefore eventually evaporate
If you were to plot the velocity profile from just outside the event horizon and project it inwards towards the central singularity point, what do you end up with?
Remember, according to relativity laws an infinite amount of energy is required to accelerate a body at rest towards the speed the light
So what is happening at and past the event horizon?
Our physical Laws break down
Here we have a ....
And now I'll have too look into that :)
Because assuming that 'photons' have a mass will indeed change the relation.
And of course if the photon did indeed turn out to be NON-massless, then the speed of light cannot be a constant and even the innate beauty of the expression E=mc^2 will have to be re-written in another way.
And of course if the photon did indeed turn out to be NON-massless, then the speed of light cannot be a constant and even the innate beauty of the expression E=mc^2 will have to be re-written in another way.
If the photon is found to be non-massless, we'll need to just change our terminology so that c is the "cosmic speed limit," then E=mc2 still holds, whereas light now acts like other massive particles and can never reach that speed.
However, there is no evidence whatsoever that light has mass.
In special relativity a critical assumption is that there exists a cosmic speed limit. Since light is massless, this is equal to the speed of light. If you wanted to make up a universe where light had mass, you'd still have a cosmic speed limit and light would be like anything else zipping around within it.
If a photon is non-massless, therefore, you could calculate its energy just like you calculate the energy of any moving particle with mass. E2=m2c4+p2c2.
In special relativity a critical assumption is that there exists a cosmic speed limit. Since light is massless, this is equal to the speed of light. If you wanted to make up a universe where light had mass, you'd still have a cosmic speed limit and light would be like anything else zipping around within it.
If a photon is non-massless, therefore, you could calculate its energy just like you calculate the energy of any moving particle with mass. E2=m2c4+p2c2.
Yor_on, why must you stick your head into the box?The policeman receives that energy, only relatively of the cow.
Try to answer this:
A policeman is measuring, with the police radar thing, the speed of a cow that is walking away from the police. Radar wave gets redshifted. Where does the energy go?
JP I think Foolsophy have a valid point there, it's not only to change one expression. You will need to go over all expressions that build on the assumption of bosons for that then, not that I know all there is :) And also all further expressions that build on those assumptions ad infinitum.Special relativity appears to be true based on a lot of tests, and the cosmic speed limit appears to be the speed of light, which also indicates that photons are massless. If they had mass, all the experiments we've done would still be true and the cosmic speed limit wouldn't change, but the speed of light would. I never said only one equation changes. All our theories about photons would have to be modified to account for this mass.
If it was as simple, and if it mattered that little to our universe then I would expected Einstein to already have considered it when creating his theory of relativity.Of course he had good reasons! Light appears to be massless and there is no evidence to the contrary. Of course, we now know that Einstein's theory of relativity doesn't require that light is massless, and that it would still hold if light had a tiny bit of mass.
I doubt he missed the inherent 'mysticism' in having bosons and 'point particles' interacting without 'existing' in SpaceTime. So if he never even considered giving light an invariant mass I'm sure he had good reasons.I'm not sure this is true. Bosons do exist in space-time when they interact. Quantum electrodynamics describes how they do so. Photons and gluons are zero mass Bosons, but W and Z bosons have mass and are bosons. The Higgs particle is predicted to be a massive Boson as well. Also, all particles have invariant mass. Photons are just special because their invariant mass is zero.
Photons we have experimental evidence for, gluons? I don't know any experimental proofs myself, isn't they theoretical 'particles' still?We can't see them directly, but we can and have measured their decay products in high energy experiments. The observations match the theoretical objects we call gluons and we don't have another explanation for them.
Are you stating that Einstein didn't 'know' that his theory would hold if he allowed for a slight invariant mass? Maybe, I'm not sure there, he seems to have looked after the simplest explanations that made sense to him, and us too possibly :) I would have expected him to want them to have a certain invariant mass, if he thought he could get away with it as they still are mysterious things, no matter that we know them to interact.I'm sure he knew, but why would you want to postulate that light has an invariant mass when there's no evidence that it does? We'd also have to tweak Maxwell's equations and quantum electrodynamics to account for this and they'd no longer be quite so elegant (though QED wasn't around yet when Einstein proposed special relativity). The elegance of Maxwell's equations alone is enough of a reason why it's natural to leave light massless.
That's a really nice question btw, why didn't he allow for a slight invariant mass if it now would make no difference? To me that would make a world of difference, as Jaztra's ideas for example, mass/energy?
An escaping mirror and a motionless mirror receive various energies from photons.An escaping mirror takes more energy.
I'll be honest, I'm not sure what it means. The OP says:
''The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon''
so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...
I could be wrong.
Black Holes are quite different creatures
They apparently seem to violate some fundamental laws of physics
the 1/4mc^2 term relates to the rest mass ENERGY.
The body of mass "m" accelerates towards the speed of light as it enters the black hole event horizon
Is the mass converting to "non-kinematic" energy?
Are you saying that the black hole absorbs 75% of the rest mass ENERGY and 25% is lost?
What is this efficiency term you talk about?
How does it relate to the Hawking radiation of balck holes?
I'll be honest, I'm not sure what it means. The OP says:
''The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon''
so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...
I could be wrong.
Black Holes are quite different creatures
They apparently seem to violate some fundamental laws of physics
the 1/4mc^2 term relates to the rest mass ENERGY.
The body of mass "m" accelerates towards the speed of light as it enters the black hole event horizon
Is the mass converting to "non-kinematic" energy?
Are you saying that the black hole absorbs 75% of the rest mass ENERGY and 25% is lost?
What is this efficiency term you talk about?
How does it relate to the Hawking radiation of balck holes?
Don't let the subject line confuse you.
Sun has 0.7% efficiency.
An efficient real quasar has 20 % efficiency.
An ideal quasar has 75% efficiency.
A photon rocket has 50% efficiency at velocity 0.5 c
A diesel engine has 40% efficiency, but not really, it's more like 0.000001%
An engine where photon gas expands in a cylinder has ideally 100% efficiency.
Next frame.
Same room, but our rocket is now accelerating non-uniformly, changing acceleration constantly. Will the light still behave the same as on Earth? Inside that room. Assume the room to be 10 light seconds. Put the light bulb at its far end, and in the direction of the rockets motion with you sitting near the rear. And this one is tricky, I expect people to have different opinions here, I have them too :) But I lean towards one though.
Now, let's go back to the first room. Let the room be 10 light seconds long. light-bulb at the front, you at the rear. Will the light be blue shifted reaching you?
The last room. Make it constantly uniformly moving, that is being in a 'free fall' with you being weightless inside it. Make it 10 light seconds long, light-bulb in front, you in the rear. Will the light 'blue/red shift' reaching you?
Now, is that true?
Yep.
In both my first examples I expended energy to make this acceleration.
In the last example I might, or might not have expended energy, but it's indeterminable from inside that room. So let us look at light falling in to a Black Hole again. Where does that light 'expend energy'? Nowhere as i see it.
Does the Black Holes gravity expend 'energy'?
==
'Indeterminable inside that room' as I'm not doing it (expending any energy) as we measure. And that's also what I mean by us looking at what really happens, not considering the 'what ifs'.
Now, is that true?
Yep.
In both my first examples I expended energy to make this acceleration.
In the last example I might, or might not have expended energy, but it's indeterminable from inside that room. So let us look at light falling in to a Black Hole again. Where does that light 'expend energy'? Nowhere as i see it.
Does the Black Holes gravity expend 'energy'?
And yes, according to how I see it, the 'line of sight' will introduce the 'relation' making it possible for that photon to behave differently. Put two observers inside, the 'photon' in the middle 'moving' towards one observer. We will find two definitions of that photons 'energy. and then also a third, the 'invariant light-quanta' we expect a photon to represent in 'itself'.
==
So, does it make sense?
No, even if it was possible to use the momentum that way, you could only use it once. No matter how you define 'propagation' there is only one interaction per photon. You need another argument for that I think?
But what you are thinking of is the blue shift we expect when it getting deeper into a gravity well, right?