Post by: jeffreyH on 08/03/2015 18:37:14
Post by: jeffreyH on 09/03/2015 01:49:50
1) The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.
2) Meaning, I assume, in free space, far away from all gravitating bodies?
Is this correct?
Post by: Ethos_ on 09/03/2015 21:03:59
I found these two consecutive posts on physicsforums.It's really very simple Jeff. Think of this in terms of the black hole. We are all familiar with the escape velocity of a black hole as being equal to c.
1) The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.
2) Meaning, I assume, in free space, far away from all gravitating bodies?
Is this correct?
Understanding this, the fact that escape velocity for any mass in question is caused by the gravitational influence, and as such is directly proportional to the time dilation associated with that mass. That fact should come as no surprise.
Post by: jeffreyH on 09/03/2015 21:58:38
I found these two consecutive posts on physicsforums.It's really very simple Jeff. Think of this in terms of the black hole. We are all familiar with the escape velocity of a black hole as being equal to c.
1) The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.
2) Meaning, I assume, in free space, far away from all gravitating bodies?
Is this correct?
Understanding this, the fact that escape velocity for any mass in question is caused by the gravitational influence, and as such is directly proportional to the time dilation associated with that mass. That fact should come as no surprise.
Well, that being said, I am asking if the same escape velocity for any size of mass will give the same time dilation. This requires a variation in density.
EDIT: This should relate to black hole entropy.
http://en.wikipedia.org/wiki/Black_hole_thermodynamics
This also involves the Bekenstein Bound
http://en.wikipedia.org/wiki/Bekenstein_bound
Post by: Ethos_ on 09/03/2015 22:47:02
That's very true Jeff but as densities are different for rocky planets like earth, and gas giants like Jupiter, these differences in density also effect where the surface of each is found. And consequently, the radius of each which is used to calculate the escape velocities of those bodies.
This requires a variation in density.
Post by: jeffreyH on 09/03/2015 23:58:32
s = (1/2)gt^2
Since g = GM/r^2
We can say that
S = (1/2)(GM/r^2)t^2
Then 2Sr/t^2 = GM/r
Ve = SQRT(2GM/r)
So:
4Sr/t^2 = 2GM/r
The factor of 4 is now present
SQRT(4Sr/t^2) = SQRT(2GM/r)
Since g = GM/r^2
Then
Ve = SQRT(2gr) for any mass.
Post by: alancalverd on 10/03/2015 00:16:38
Post by: jeffreyH on 10/03/2015 00:25:55
But that all assumes point masses.
Yes and that is a major problem. Which highlights just why GR isn't a piece of cake.
Post by: Bill S on 10/03/2015 00:53:33
Quote from: Ethos
We are all familiar with the escape velocity of a black hole as being equal to c.
If that were the case, wouldn't light escape?
Post by: Ethos_ on 10/03/2015 01:05:43
Yes of course...........I should have said "escape velocity greater than c." But greater by only the smallest degree.Quote from: EthosWe are all familiar with the escape velocity of a black hole as being equal to c.
If that were the case, wouldn't light escape?
Post by: jeffreyH on 10/03/2015 11:59:42
Post by: Ethos_ on 10/03/2015 15:11:26
With Ve = SQRT(2gr) if we hold g constant and vary r as in a change in density then we have a differing value for Ve. So gravitational acceleration and escape velocity are not proportional. This means that while for the earth the value of g at the surface is much less than Ve there can be situations in which g is greater than Ve near very dense objects. In theory this can result in superluminal acceleration.I think you've misunderstood my use of the word proportional Jeff. I was not saying they were equal, Webster's defines proportional as:
"to arrange the parts of (a whole) so as to be harmonious. a ratio"
Post by: jeffreyH on 10/03/2015 17:44:43
With Ve = SQRT(2gr) if we hold g constant and vary r as in a change in density then we have a differing value for Ve. So gravitational acceleration and escape velocity are not proportional. This means that while for the earth the value of g at the surface is much less than Ve there can be situations in which g is greater than Ve near very dense objects. In theory this can result in superluminal acceleration.I think you've misunderstood my use of the word proportional Jeff. I was not saying they were equal, Webster's defines proportional as:
"to arrange the parts of (a whole) so as to be harmonious. a ratio"
Well there is a proportionality and it is balanced but not like the inverse square nature of the gravitational field. Add time dilation to the mix and stir. Can we say time dilation is a function of Ve, g or both?
Post by: JohnDuffield on 10/03/2015 17:54:08
Quote from: jeffreyH
The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.It's no coincidence. The thing is, GR doesn't actually explain why matter falls down. It doesn't actually say why an object acquires some given speed, which is escape velocity when you flip things round. However it's very easy to understand if you think about the wave nature of matter and stuff electron diffraction and spin. Just simplify the electron to light going round and round, then simplify it further to light going round a square path. The horizontals bend in the gravitational gradient, and the electron falls down. It's similar if you accelerate the electron in gravity-free space. Sadly you never seem to see any texts which combine relativity with the wave nature of matter.
Post by: jeffreyH on 10/03/2015 21:51:05
It isn't a function of g, because that denotes the "local slope" of gravitational potential. Gravitational time dilation denotes the depth of gravitational potential. And to escape it, you need to "take a run at it", wherein your escape velocity is related to the gravitational time dilation.Quote from: jeffreyHThe time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.It's no coincidence. The thing is, GR doesn't actually explain why matter falls down. It doesn't actually say why an object acquires some given speed, which is escape velocity when you flip things round. However it's very easy to understand if you think about the wave nature of matter and stuff electron diffraction and spin. Just simplify the electron to light going round and round, then simplify it further to light going round a square path. The horizontals bend in the gravitational gradient, and the electron falls down. It's similar if you accelerate the electron in gravity-free space. Sadly you never seem to see any texts which combine relativity with the wave nature of matter.
Of course it has to be a function of g. If you are stationary on the earth you are experiencing time dilation. You are not accelerating away from it. The velocity involved in escaping the field is due to kinetic energy. If the gravitational field were absent the time dilation would be due to the velocity. At a stationary position on the earth there is still a potential for acceleration. However there is NO potential for escape unless there is an impetus away from the surface. This will INDUCE a time dilation. Without the impetus there is no dilation due to Ve. Can't you see that? You only have the value of g operating.
Post by: jeffreyH on 10/03/2015 23:34:14
c = SQRT(2gr)
This is then simplified to:
g = c^2/2r
EDIT: Of course this should be:
g = c^2/2rs
This is likely the reason for the survival of the G2 gas cloud.
Post by: jeffreyH on 11/03/2015 00:58:28
a = (GM)/(r^2*SQRT[1-rs/r])
Where the above is:
g = c^2/2rs
In the Schwarzschild form the acceleration goes infinite at the event horizon.
Post by: jeffreyH on 11/03/2015 01:53:06
g = c^2/(2[rs+n]*SQRT(1-rs/[rs+n]))
The value of n can then be increased as needed to plot the gradient of g away from the event horizon.
Post by: JohnDuffield on 11/03/2015 14:11:00
Of course it has to be a function of g.No it isn't. See this depiction of gravitational potential:
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2Fd%2Fd9%2FGravityPotential.jpg%2F300px-GravityPotential.jpg&hash=848f35613bf74f8f41fe54b2817d5567)
CCASA image by AllenMcC, see Wikipedia (http://commons.wikimedia.org/wiki/File:GravityPotential.jpg).
The time dilation is represented by the depth of potential, how low in the plot you are, whilst g is the slope of the plot. Note that there's an inflection, so g at one elevation is the same as g at another. You can find two places on the plot where you can draw the same tangent.
If you are stationary on the earth you are experiencing time dilation.Yes.
You are not accelerating away from it.Yes.
The velocity involved in escaping the field is due to kinetic energy.Yes, and in your ascending rocket going faster and faster you are swapping gravitational time dilation for special-relativistic time dilation.
If the gravitational field were absent the time dilation would be due to the velocity.Yes.
At a stationary position on the earth there is still a potential for acceleration. However there is NO potential for escape unless there is an impetus away from the surface. This will INDUCE a time dilation.Yes. And as you ascend away from the surface you reduce the gravitational time dilation.
Without the impetus there is no dilation due to Ve. Can't you see that?Yes.
You only have the value of g operating.The acceleration due to gravity is g, and it is due to the local slope of the potential. That's like the local slope of the gravitational time dilation. If your clock at the ceiling runs at the same rate as your clock at the floor, your pencil doesn't fall down. If your clock at the ceiling runs faster than your clock at the floor, your pencil falls down. If your clock at the ceiling runs much faster than your clock at the floor, your pencil falls down much faster. Note that your clocks might be light clocks, and that g can be expressed as the local slope of your plot of the coordinate speed of light.
Post by: jeffreyH on 11/03/2015 14:56:45
Of course it has to be a function of g.No it isn't. See this depiction of gravitational potential:
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2Fd%2Fd9%2FGravityPotential.jpg%2F300px-GravityPotential.jpg&hash=848f35613bf74f8f41fe54b2817d5567)
CCASA image by AllenMcC, see Wikipedia (http://commons.wikimedia.org/wiki/File:GravityPotential.jpg).
The time dilation is represented by the depth of potential, how low in the plot you are, whilst g is the slope of the plot. Note that there's an inflection, so g at one elevation is the same as g at another. You can find two places on the plot where you can draw the same tangent.If you are stationary on the earth you are experiencing time dilation.Yes.You are not accelerating away from it.Yes.The velocity involved in escaping the field is due to kinetic energy.Yes, and in your ascending rocket going faster and faster you are swapping gravitational time dilation for special-relativistic time dilation.If the gravitational field were absent the time dilation would be due to the velocity.Yes.At a stationary position on the earth there is still a potential for acceleration. However there is NO potential for escape unless there is an impetus away from the surface. This will INDUCE a time dilation.Yes. And as you ascend away from the surface you reduce the gravitational time dilation.Without the impetus there is no dilation due to Ve. Can't you see that?Yes.You only have the value of g operating.The acceleration due to gravity is g, and it is due to the local slope of the potential. That's like the local slope of the gravitational time dilation. If your clock at the ceiling runs at the same rate as your clock at the floor, your pencil doesn't fall down. If your clock at the ceiling runs faster than your clock at the floor, your pencil falls down. If your clock at the ceiling runs much faster than your clock at the floor, your pencil falls down much faster. Note that your clocks might be light clocks, and that g can be expressed as the local slope of your plot of the coordinate speed of light.
You have just contradicted yourself and described time dilation as a function of g.
Post by: David Cooper on 11/03/2015 18:41:23
It's no coincidence. The thing is, GR doesn't actually explain why matter falls down. It doesn't actually say why an object acquires some given speed, which is escape velocity when you flip things round. However it's very easy to understand if you think about the wave nature of matter and stuff electron diffraction and spin. Just simplify the electron to light going round and round, then simplify it further to light going round a square path. The horizontals bend in the gravitational gradient, and the electron falls down. It's similar if you accelerate the electron in gravity-free space. Sadly you never seem to see any texts which combine relativity with the wave nature of matter.
That's a useful way of looking at things. Where do you get your knowledge from, because I'd like to explore the source.
Post by: JohnDuffield on 12/03/2015 17:18:43
That's a useful way of looking at things. Where do you get your knowledge from, because I'd like to explore the source.A whole rack of different places, ranging from old papers by the likes of Maxwell and Einstein, to newer papers and articles that tend not to get much publicity. See for example http://arxiv.org/abs/physics/0512265 and S A Goudsmit talking about the discovery of electron spin (http://lorentz.leidenuniv.nl/history/spin/goudsmit.html): "When the day came I had to tell Uhlenbeck about the Pauli principle - of course using my own quantum numbers - then he said to me: "But don't you see what this implies? It means that there is a fourth degree of freedom for the electron. It means that the electron has a spin, that it rotates". Also see magnetic moment (http://en.wikipedia.org/wiki/Electron_magnetic_moment#Magnetic_moment_of_an_electron) and the Einstein-de Haas effect (http://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect) which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". However if you were to ask a contemporary particle physicist about this, he might say the electron is a point particle, that its spin is not a real rotation, and that gravity is all down to gravitons flitting back and forth.
NB: note that only the horizontals bend downwards, which is why light is deflected twice as much as matter (http://www.astro.ucla.edu/~wright/deflection-delay.html).
Post by: David Cooper on 12/03/2015 19:43:58
Post by: JohnDuffield on 12/03/2015 22:21:21
Post by: jeffreyH on 12/03/2015 22:37:25
Read, enjoy and maybe provide all the proofs required.
http://www.quora.com/What-do-physicists-think-of-the-hubius-helix-model-of-electron-structure
Post by: jeffreyH on 12/03/2015 23:00:04
Post by: jeffreyH on 12/03/2015 23:18:45
Post by: jeffreyH on 12/03/2015 23:45:21
Post by: jeffreyH on 12/03/2015 23:48:51
http://background.uchicago.edu/~whu/intermediate/Polarization/polar6.html
Post by: jeffreyH on 13/03/2015 00:16:00
g = c^2/(2[rs+n]*SQRT(1-rs/[rs+n]))
is how would this relate to the Bekenstein bound
http://en.wikipedia.org/wiki/Bekenstein_bound
and the chandrasekhar limit
http://en.wikipedia.org/wiki/Chandrasekhar_limit
If a connection to the black hole entropy can be found that would be a positive step. With the chandrasekhar limit it may be possible to put a lower and upper bound on possible black hole masses. In respect of the big bang this could be very important if the idea of a singularity is to hold.
Post by: JohnDuffield on 13/03/2015 01:01:02
"In contrast, the Mobius strip is a non-orientable surface, because a right-handed figure, moved continuously around the loop until arrive back at its starting point, becomes left-handed. An object must be translated around the loop twice in order to be restored to its original position and chirality. In this sense a Mobius strip is reminiscent of spin-1/2 particles in quantum mechanics, since such particles must be rotated through two complete rotations in order to be restored to their original state...
I'm John Duffield. I don't know any Stuart J Duffield. I'm not claiming to be an authority on spinors or group theory, I'm just telling you about things that aren't in the popscience/student books/articles/etc you've been reading. Do not think that something is "spurious information" or isn't "established physics" just because you don't know about it. The vast bulk of what I tell you about is established physics. I didn't invent the word spinor, and I didn't write the Wikipedia article or put this picture on it:
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2Fd%2Fd7%2FSpinor_on_the_circle.pdf%2Fpage1-330px-Spinor_on_the_circle.pdf.jpg&hash=e2c85ff6701c5649601cd7b3cafd7152)
Quote
Last night I watch the Horizon programme on the BBC about the search for the gravitational waves from the big bang. Alan Guth was one of the physicists included. He propose the theory of inflation right at the start of his career. Finding those waves will tell us a lot more about the big bang and validate Alan's theory. The guys set up telescopes at the pole and spent 3 years recording CMBR from a dark spot in the sky. Then after 3 years they found the data wasn't accurate enough. They then went away and built a better telescope and spent several more years looking for the b mode signal of the early gravitational waves. They found something that looked like the signal and spent more time analyzing it and ruling things out until they were confident in announcing it. Then it was shown that at least 50% of it and maybe 100% of the signal was due to dust. That is physics.The Horizon producers pulled their punches, and didn't say anything about the orchestrated hype that put people's back up. See Physicist Paul Steinhardt Slams Inflation, Cosmic Theory He Helped Conceive (http://blogs.scientificamerican.com/cross-check/2014/12/01/physicist-paul-steinhardt-slams-inflation-cosmic-theory-he-helped-conceive/). And note this: that "first light" dates from 300,000 years after the big bang. How can that tell you anything about the first nanosecond after the big bang? The universe was a seething maelstrom of plasma for a third of a million years. It would be like putting those sonic standing waves into a blender.
Post by: jeffreyH on 13/03/2015 01:18:29
Bispinor on wikipedia.
http://en.wikipedia.org/wiki/Bispinor
Have you even looked at the equations in there? You have group mapping, transformations and the SO(3;1) group all mentioned.
You can find information on orthogonal groups here:
http://en.wikipedia.org/wiki/Orthogonal_group
Try reading it.
"Equivalently, it is the group of n×n orthogonal matrices, where the group operation is given by matrix multiplication, and an orthogonal matrix is a real matrix whose inverse equals its transpose."
"The determinant of an orthogonal matrix being either 1 or −1, an important subgroup of O(n) is the special orthogonal group, denoted SO(n), of the orthogonal matrices of determinant 1. This group is also called the rotation group, because, in dimensions 2 and 3, its elements are the usual rotations around a point (in dimension 2) or a line (in dimension 3). In low dimension, these groups have been widely studied, see SO(2), SO(3) and SO(4)."
What about symmetries? Does that ring any bells?
Post by: jeffreyH on 13/03/2015 01:22:30
Post by: jeffreyH on 13/03/2015 20:13:04
Jeffrey: David Simmons-Duffin was being overly hostile. Take a look at the picture on the quora page. Then note that the Dirac spinor (http://en.wikipedia.org/wiki/Dirac_spinor) is a bispinor, and have a look at Dirac's belt (http://www.mathpages.com/home/kmath619/kmath619.htm) on Mathspages:
"In contrast, the Mobius strip is a non-orientable surface, because a right-handed figure, moved continuously around the loop until arrive back at its starting point, becomes left-handed. An object must be translated around the loop twice in order to be restored to its original position and chirality. In this sense a Mobius strip is reminiscent of spin-1/2 particles in quantum mechanics, since such particles must be rotated through two complete rotations in order to be restored to their original state...
I'm John Duffield. I don't know any Stuart J Duffield. I'm not claiming to be an authority on spinors or group theory, I'm just telling you about things that aren't in the popscience/student books/articles/etc you've been reading. Do not think that something is "spurious information" or isn't "established physics" just because you don't know about it. The vast bulk of what I tell you about is established physics. I didn't invent the word spinor, and I didn't write the Wikipedia article or put this picture on it:
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2Fd%2Fd7%2FSpinor_on_the_circle.pdf%2Fpage1-330px-Spinor_on_the_circle.pdf.jpg&hash=e2c85ff6701c5649601cd7b3cafd7152)QuoteLast night I watch the Horizon programme on the BBC about the search for the gravitational waves from the big bang. Alan Guth was one of the physicists included. He propose the theory of inflation right at the start of his career. Finding those waves will tell us a lot more about the big bang and validate Alan's theory. The guys set up telescopes at the pole and spent 3 years recording CMBR from a dark spot in the sky. Then after 3 years they found the data wasn't accurate enough. They then went away and built a better telescope and spent several more years looking for the b mode signal of the early gravitational waves. They found something that looked like the signal and spent more time analyzing it and ruling things out until they were confident in announcing it. Then it was shown that at least 50% of it and maybe 100% of the signal was due to dust. That is physics.The Horizon producers pulled their punches, and didn't say anything about the orchestrated hype that put people's back up. See Physicist Paul Steinhardt Slams Inflation, Cosmic Theory He Helped Conceive (http://blogs.scientificamerican.com/cross-check/2014/12/01/physicist-paul-steinhardt-slams-inflation-cosmic-theory-he-helped-conceive/). And note this: that "first light" dates from 300,000 years after the big bang. How can that tell you anything about the first nanosecond after the big bang? The universe was a seething maelstrom of plasma for a third of a million years. It would be like putting those sonic standing waves into a blender.
That article correctly states that Paul Steinhardt helped refine the theory years after Alan Guth conceived it. Don't put your own slant on things and try to rewrite history. So what if he objects to it. HE has a right to. Someone may well come up with a better theory. That is what physics is all about.
Post by: jeffreyH on 13/03/2015 20:46:40
g = c^2/[2rs+L/2
]The value for g that we then get is 299788351 which just about equals the speed of light. To find the lower limit of a feasible black hole we need to vary the parameters until this acceleration is as near to the surface of the event horizon as it is possible to get. This may or may not equal 1 Planck length.
CORRECTION: c/2 has been modified to L/2 to represent the distance traveled by light in one 1/2 second. You just can't add a velocity to a distance.
Post by: jeffreyH on 15/03/2015 20:01:26
Post by: SorryDnoodle on 15/03/2015 20:36:32
Yes of course...........I should have said "escape velocity greater than c." But greater by only the smallest degree.Quote from: EthosWe are all familiar with the escape velocity of a black hole as being equal to c.
If that were the case, wouldn't light escape?
Would't escape velocity be proportional to the mass of the black hole, in other words; A very small black hole being very close to c but a super massive black hole very much great than c?
If not, why?
Post by: jeffreyH on 15/03/2015 21:04:31
Yes of course...........I should have said "escape velocity greater than c." But greater by only the smallest degree.Quote from: EthosWe are all familiar with the escape velocity of a black hole as being equal to c.
If that were the case, wouldn't light escape?
Would't escape velocity be proportional to the mass of the black hole, in other words; A very small black hole being very close to c but a super massive black hole very much great than c?
If not, why?
A black hole traps light at its event horizon. That is why it is called an event horizon because we cannot retrieve any information beyond the horizon. As the radius of the event horizon increases so does the mass but the density of the gravitational field decreases proportionally. If the escape velocity at the event horizon exceeded c then it would no longer be the event horizon. In that case the event horizon would be larger in radius.
You can find details here.
http://en.wikipedia.org/wiki/Schwarzschild_radius