View Full Version : Gravity (again)


Just Curious
05-24-10, 04:02 AM
Having watched dozens of Discovery Channel programs on the origins of the Universe I am getting increasingly annoyed and confused with their description of Gravity. They very quickly dismiss the myth that Gravity is a force and describe it more correctly as a warping of space-time caused by massive objects. They even use the infamous rubber sheet model to illustrate this. So OK, we accept this. But later in the programs they start talking of Gravity as a FORCE again. They say at the big bang that all four forces were equal but the Gravity force broke away and became much weaker. They even postulate that it diffused into higher order dimensions which we cannot see. There is also the question of why mass warps space-time. What is the property that mass contains to do this? Have gravitons got anything to do with it or do they only exist at the quantum level. So the basic question remains. Is Gravity a force or not? Once the scientists decide then perhaps the rest of us will understand.

BenTheMan
05-24-10, 12:47 PM
Hi JC---

The point is that, a long distances, the description of gravity as warping of space-time works better. At short distances, it is likely that gravity has a particle interpretation (i.e. the graviton), and thus is more aptly described as a force.

CptBork
05-24-10, 01:29 PM
Hi JC---

The point is that, a long distances, the description of gravity as warping of space-time works better. At short distances, it is likely that gravity has a particle interpretation (i.e. the graviton), and thus is more aptly described as a force.

Ok, I hope I don't complicate the discussion too much, but I've been really curious about a certain aspect of Quantum gravity. Now in quantum field theory describing all the "forces" except gravity, spacetime fundamentally enters the description, but it's the flat spacetime in the absence of gravity, i.e. the Minkowski metric. If you want to describe Quantum physics in the presence of gravity, don't you still have to take the region's spacetime curvature into account? I thought the problem with Quantum Gravity wasn't the ability to incorporate a curved spacetime, but rather difficulties in describing how mass causes that curvature?

Pandaemoni
05-24-10, 03:15 PM
If you want to describe Quantum physics in the presence of gravity, don't you still have to take the region's spacetime curvature into account?

Yes, and there are such things as relativistic quantum field theories which are generalizations from non-relativistic quantum mechanics. The issue with quantum gravity is that nothing th quantum field theory "predicts" the existence of gravity. Rather quantum field theory tends to be a version of quantum mechanics with allowances made specifically to reflect what we happen to know about relativity.

For many real world purposes though, space-time is so close to flat that the relativistic effects don't greatly influence the outcome of experiments. Just like Newton's law of universal gravitation is is good enough for most uses, non-relativistic quantum mechanics can be used as a good approximation in most Earth based experiments.

Relativistic quantum field theory is very successful, but it is a classical theory applied in the context of quantum mechanics, rather than a true quantum theory of gravity.


I thought the problem with Quantum Gravity wasn't the ability to incorporate a curved spacetime, but rather difficulties in describing how mass causes that curvature?

I think the problem is deeper than that. What scientists want is a single theory that predicts all the fundamental forces, including gravity, rather than two separate theories that can be grafted together. It is certainly true that they do not know "why" spacetime curves i the presence of mass, but that is only one of the mysteries. Another is whether the fine structure of spacetime (even where it appears to be flat at our level) is or is not the famous "spacetime foam". Just like a billiard ball appears smooth to the naked eye, and yet is revealed to have pronounced projections and rifts under magnification, seemingly placid spacetime at the level of the Planck length may be a roiling and chaotic soup.

CptBork
05-25-10, 07:15 AM
Relativistic quantum field theory is very successful, but it is a classical theory applied in the context of quantum mechanics, rather than a true quantum theory of gravity.

I understand that. I just wanted to note that there's a very simple, natural way to incorporate curved spacetimes into quantum field theories. Just replace the Minkowski metric with a curved spacetime metric, and replace all derivatives with covariant derivatives (covariant in the GR sense). I believe this is done to some extent for Hawking radiation, i.e. you treat the edge of the event horizon as a locally flat region to do the calculation in QFT, and then convert to the coordinate system of a distant observer to add in the gravitational effects. I also know there are physicists actively studying quantum field theories in curved spacetimes, although I hear there are some strange results coming from this approach.


I think the problem is deeper than that. What scientists want is a single theory that predicts all the fundamental forces, including gravity, rather than two separate theories that can be grafted together. It is certainly true that they do not know "why" spacetime curves i the presence of mass, but that is only one of the mysteries.

Well I don't think the problem is with QFT not "predicting" gravity or curved spacetime- there's a straightforward recipe for taking the "classical" Lagrangian of GR and plugging it into QFT as a field energy term, then adding in kinetic and other interaction terms for the graviton excitations of this field. The problem I keep hearing about (and have seen sketched out mathematically) is that the bread and butter technique of QFT renormalization doesn't work with this recipe, thus perturbation expansions don't work in scattering calculations, and the resulting theory requires an infinite number of parameters.

Just Curious
05-25-10, 07:37 AM
Hi JC---

The point is that, a long distances, the description of gravity as warping of space-time works better. At short distances, it is likely that gravity has a particle interpretation (i.e. the graviton), and thus is more aptly described as a force.

Thanks Ben. So, somewhere along the way between the big bang, where gravity was a force best described using the graviton, to where we are now with the size of the universe where gravity is best described by the warping of space-time, gravity changed. The elusive theory of everything would unite these two proprties of gravity.

kurros
05-25-10, 07:37 AM
Well I don't think the problem is with QFT not "predicting" gravity or curved spacetime- there's a straightforward recipe for taking the "classical" Lagrangian of GR and plugging it into QFT as a field energy term, then adding in kinetic and other interaction terms for the graviton excitations of this field. The problem I keep hearing about (and have seen sketched out mathematically) is that the bread and butter technique of QFT renormalization doesn't work with this recipe, thus perturbation expansions don't work in scattering calculations, and the resulting theory requires an infinite number of parameters.

I also believe it is the case that if consider the resulting field theory to be simply an effective field theory then everything is fine, you can regularise everything with a cutoff at the scale the theory breaks down and off you go. In fact I have heard it said that such a field theory of gravity can be considered to be a much better theory than the standard model, which is probably an effective theory with cutoff somewhere not much higher than the electroweak scale, because the gravity qft cutoff is way way higher, the Planck scale I believe.
The standard model is still nicer though since it is renormalisable, it just starts being wrong above the electroweak scale (or so it seems) while the gravity qft really starts to go to hell above the cutoff. Or something like that, I don't really remember.

CptBork
05-25-10, 07:52 AM
I also believe it is the case that if consider the resulting field theory to be simply an effective field theory then everything is fine, you can regularise everything with a cutoff at the scale the theory breaks down and off you go. In fact I have heard it said that such a field theory of gravity can be considered to be a much better theory than the standard model, which is probably an effective theory with cutoff somewhere not much higher than the electroweak scale, because the gravity qft cutoff is way way higher, the Planck scale I believe.
The standard model is still nicer though since it is renormalisable, it just starts being wrong above the electroweak scale (or so it seems) while the gravity qft really starts to go to hell above the cutoff. Or something like that, I don't really remember.

If my understanding is correct, the Standard Model is considered by many to be an effective cutoff of an SU(5) Yang-Mills theory. The effective cutoff would make low-energy consequences such as the "strong force" artificially appear to be nonrenormalizable. Similarly, many consider GR to be an effective low-energy field theory and that the correct, complete field theory would hopefully not run into the same difficulties. Still, I don't really see any grounds to call one theory superior over another, regardless of their respective cutoff scales; I would think physicists would call both models flawed and seek them both as consequences of a single unified theory, such as String Theory, but I haven't heard of anyone being able to deduce the Standard Model just by extending GR on its own.

Anyhow what I'm really most curious about is Ben's comment that spacetime curvature might just be a convenient approximation at large scales. I'm just saying by the looks of it, from my own personal readings and understanding, it seems that even if gravity can be modelled by the exchange of gravitons, those gravitons must still have some sort of effect on the spacetime background in which everything is moving. This is what I'm most curious about.

kurros
05-25-10, 08:12 AM
Anyhow what I'm really most curious about is Ben's comment that spacetime curvature might just be a convenient approximation at large scales. I'm just saying by the looks of it, from my own personal readings and understanding, it seems that even if gravity can be modelled by the exchange of gravitons, those gravitons must still have some sort of effect on the spacetime background in which everything is moving. This is what I'm most curious about.

Ahh yes, well that is the most interesting question isn't it? At what point do the little perturbations of spacetime which are gravitons become something which has a noticable influence on the background geometry? Or how does the background geometry arise from them fundamentally?

Gravitons are really just the quanta of a linearised gravitational theory, in which one quantises perturbations about a background metric (gravitational waves) after making the approximation that the perturbations are small enough not to affect the background (linearising the theory). The theory breaks down when you go near the non-linear regime, which is where all the interesting stuff happens. I have some vague picture in my head of lots of gravitons building up into classical gravitational waves which as they become bigger become real changes to the curvature of space, but this dodges the question of what the heck the background is in the linearised theory. Maybe we'll know one day. It's also a pretty crap picture since I guess regular curvature of space has more to do with virtual gravitons than real ones. Ahh who knows.

BenTheMan
05-25-10, 07:58 PM
Thanks Ben. So, somewhere along the way between the big bang, where gravity was a force best described using the graviton, to where we are now with the size of the universe where gravity is best described by the warping of space-time, gravity changed. The elusive theory of everything would unite these two proprties of gravity.

Yes, but I wouldn't say gravity ``changed''. If you look at a single water molecule, what does it mean to talk about temperature? The point is that the same substance can be described by different degrees of freedom depending on whether you're interested in the short distance properties or the long distance properties.

Jack_
05-25-10, 09:34 PM
Thanks Ben. So, somewhere along the way between the big bang, where gravity was a force best described using the graviton, to where we are now with the size of the universe where gravity is best described by the warping of space-time, gravity changed. The elusive theory of everything would unite these two proprties of gravity.

Since gravitation mass and inertial mass are not distinguishable, then there must be some relationship between Higgs and the graviton, if all this junk is true.
Without establishing a model to explain that mass stimulates gravity and gravity operates on mass, we certainly scientifically do not yet have the answer.

In other words, a bijective function of some nature is required between the two.

CptBork
05-26-10, 01:36 AM
Since gravitation mass and inertial mass are not distinguishable, then there must be some relationship between Higgs and the graviton, if all this junk is true.
Without establishing a model to explain that mass stimulates gravity and gravity operates on mass, we certainly scientifically do not yet have the answer.

In other words, a bijective function of some nature is required between the two.

If you had actually known what you were talking about and not making stuff up off the top of your head, you would have observed that the Higgs mechanism only determines the rest masses of the particles. Since gravity depends on the relativistic energy and momentum of the particles, not the rest mass, a particle's gravity has no direct relation with the Higgs. For example, by your reasoning, radiation would produce no gravity of its own, but in fact it actually does.

Just Curious
05-26-10, 07:00 AM
CptBork. I don't want to incur your wrath like Jack but can you explain this statement
"Since gravity depends on the relativistic energy and momentum of the particles, not the rest mass."

CptBork
05-26-10, 08:17 AM
CptBork. I don't want to incur your wrath like Jack but can you explain this statement
"Since gravity depends on the relativistic energy and momentum of the particles, not the rest mass."

Jack has been making all sorts of posts about how conventional physics is totally wrong, that's why I was a little harsh in my response to him. It's ok to make mistakes, just not ok to make mistakes and then call everyone else stupid because they can recognize the mistake.

In General Relativity, the mass and energy present in a system determines the resulting spacetime curvature through a mathematical object called the "Stress-Energy tensor (http://en.wikipedia.org/wiki/Stress-energy_tensor)". The stress-energy tensor has applications in many areas of physics, and the quantities used to calculate it can depend on its usage, but in General Relativity the quantities you use are the relativistic energy density and relativistic momentum density of the particles/gases being considered. The important point is that the rest mass is not what you need to calculate the gravitational field, and the rest mass is all the Higgs mechanism determines. If I take a particle and add energy to it to make it start moving, that extra energy will be seen as an increase in the particle's relativistic mass/energy, and there will be an increased gravitational field as a result.

BenTheMan
05-26-10, 02:39 PM
Since gravitation mass and inertial mass are not distinguishable, then there must be some relationship between Higgs and the graviton, if all this junk is true.

Why do you say this?

Jack_
05-26-10, 06:29 PM
If you had actually known what you were talking about and not making stuff up off the top of your head, you would have observed that the Higgs mechanism only determines the rest masses of the particles. Since gravity depends on the relativistic energy and momentum of the particles, not the rest mass, a particle's gravity has no direct relation with the Higgs. For example, by your reasoning, radiation would produce no gravity of its own, but in fact it actually does.

In physics, mass (from Ancient Greek: μᾶζα) commonly refers to any of three properties of matter, which have been shown experimentally to be equivalent: inertial mass, active gravitational mass and passive gravitational mass. In everyday usage, mass is often taken to mean weight, but in scientific use, they refer to different properties.

Inertial mass is a measure of an object's resistance to changing its state of motion when a force is applied. It is determined by applying a force to an object and measuring the acceleration that results from that force. An object with small inertial mass will accelerate more than an object with large inertial mass when acted upon by the same force. One says the body of greater mass has greater inertia.

Active gravitational mass is a measure of the strength of an object’s gravitational flux (gravitational flux is equal to the surface integral of gravitational field over an enclosing surface). Gravitational field can be measured by allowing a small ‘test object’ to freely fall and measuring its free-fall acceleration. For example, an object in free-fall near the Moon will experience less gravitational field, and hence accelerate slower than the same object would if it were in free-fall near the earth. The gravitational field near the Moon is weaker because the Moon has less active gravitational mass.

Passive gravitational mass is a measure of the strength of an object's interaction with a gravitational field. Passive gravitational mass is determined by dividing an object’s weight by its free-fall acceleration. Two objects within the same gravitational field will experience the same acceleration; however, the object with a smaller passive gravitational mass will experience a smaller force (less weight) than the object with a larger passive gravitational mass.


http://en.wikipedia.org/wiki/Inertial_mass#Inertial_mass


Rest mass
A constant intrinsic to a body which determines its inertial and energy-momentum properties. It is a fundamental concept of special relativity, and in particular it determines the internal energy content of a body. It is the same as the inertial mass of classical mechanics. According to the principle of equivalence, the basic physical principle of general relativity, the inertial mass of a body is also equal to its gravitational mass. See Classical mechanics, Gravitation, Relativity

The rest mass or inertial mass of a body, m, is a measure of its resistance to being accelerated at a by a force F ;

http://encyclopedia2.thefreedictionary.com/Rest+mass

Jack_
05-26-10, 06:46 PM
If you had actually known what you were talking about and not making stuff up off the top of your head, you would have observed that the Higgs mechanism only determines the rest masses of the particles. Since gravity depends on the relativistic energy and momentum of the particles, not the rest mass, a particle's gravity has no direct relation with the Higgs. For example, by your reasoning, radiation would produce no gravity of its own, but in fact it actually does.

This is curious, this article tries to unify of GR gravity and Higgs bosons.

http://arxiv.org/PS_cache/arxiv/pdf/1004/1004.4866v1.pdf

Say, based on your strong response, I would like to see your unification of the graviton and Higgs, since I am so wrong in your view.

How long will this take?

If you are unable to do this, my statement that they have not been sucessfully unified stands.

BenTheMan
05-26-10, 07:26 PM
This is curious, this article tries to unify of GR gravity and Higgs bosons.

http://arxiv.org/PS_cache/arxiv/pdf/1004/1004.4866v1.pdf

Say, based on your strong response, I would like to see your unification of the graviton and Higgs, since I am so wrong in your view.

How long will this take?

If you are unable to do this, my statement that they have not been sucessfully unified stands.

See my objections, here (http://www.sciforums.com/showthread.php?t=101783).

Just because it's on the arXiv doesn't mean it's legitimate science.

Jack_
05-26-10, 07:33 PM
See my objections, here (http://www.sciforums.com/showthread.php?t=101783).

Just because it's on the arXiv doesn't mean it's legitimate science.


I read your objections. Since I believe SR is false and have proven it, I believe GR is false also as a consequence. So I do not adhere to this paper anyway.

My objective was to demonstrate the mainstream has not yet proven the logical equivalence of inertial mass and gravitational mass though no experiment has refuted it.

It is mathematically still an open question which was the point of my earlier post.

Billy T
05-26-10, 08:33 PM
... by your {Jack's} reasoning, radiation would produce no gravity of its own, but in fact it actually does. I believe a hot brick makes more gravity than the same brick when it has cooled off. Do you agree?
I believe that photons do not make gravity. Why do you say they do?

Most of this thread is over my head (educated in physic too long ago.) but here is simple argument suggesting that photons do not make gravity:

Look at the night sky (stars) with alternating eyes (only one open at a time) It has the same stellar pattern with either eye. Now consider how long a path the photons traveling to your single observing eyes have had - how close they have been together. If they mutually attract, why have they not coalesced into discrete fine streams? I.e. the right eye should see some stars the the left eye can not and conversely.

If they were hydrogen atoms and only their mutual gravity were present (no radiation pressure or other gravity fields etc.), and traveled so closely together for so long, by the time they arrived at Earth they would have coalesced into hydrogen molecules even neglecting any other mutual forces related to chemical binding.

As stars do have considerable size the photon pairs reaching either eye probably were well separated when leaving the star. Thus, replace the star with much more distant quasar or better still an early supernova - 13B years old event. If photons have mutual attraction would they not when arriving at Earth be coalesced into discrete steams smaller that the separation between my eyes?

Perhaps not true when calculated but that can be relatively easily done as you claim to know their mutual gravity. I.e. calculate how long it takes them to"fall together" under their mutual gravity. For example, start a pair of photons 100,000 meters apart and let them initially have slightly convergent trajectories, aimed towards one point on Earth and travel 13 billion light years. Will they not have a common trajectory long before reaching Earth?

PS my simple rule for when energy makes gravity and when it does not is that it does iff the energy is the same in all references frames, as is the case for the hot brick (Temperature of melting lead is not a function of the frame it is viewed from) but photon energy is frame dependent.

kurros
05-26-10, 10:24 PM
I believe a hot brick makes more gravity than the same brick when it has cooled off. Do you agree?
I believe that photons do not make gravity. Why do you say they do?

Most of this thread is over my head (educated in physic too long ago.) but here is simple argument suggesting that photons do not make gravity:

Look at the night sky (stars) with alternating eyes (only one open at a time) It has the same stellar pattern with either eye. Now consider how long a path the photons traveling to your single observing eyes have had - how close they have been together. If they mutually attract, why have they not coalesced into discrete fine streams? I.e. the right eye should see some stars the the left eye can not and conversely.

If they were hydrogen atoms and only their mutual gravity were present (no radiation pressure or other gravity fields etc.), and traveled so closely together for so long, by the time they arrived at Earth they would have coalesced into hydrogen molecules even neglecting any other mutual forces related to chemical binding.

As stars do have considerable size the photon pairs reaching either eye probably were well separated when leaving the star. Thus, replace the star with much more distant quasar or better still an early supernova - 13B years old event. If photons have mutual attraction would they not when arriving at Earth be coalesced into discrete steams smaller that the separation between my eyes?

Perhaps not true when calculated but that can be relatively easily done as you claim to know their mutual gravity. I.e. calculate how long it takes them to"fall together" under their mutual gravity. For example, start a pair of photons 100,000 meters apart and let them initially have slightly convergent trajectories, aimed towards one point on Earth and travel 13 billion light years. Will they not have a common trajectory long before reaching Earth?

PS my simple rule for when energy makes gravity and when it does not is that it does iff the energy is the same in all references frames, as is the case for the hot brick (Temperature of melting lead is not a function of the frame it is viewed from) but photon energy is frame dependent.

Gravity is so weak that none of those things you mentioned will be observable, despite the fact that in principle they do happen. The energy of a single photon, even a super energetic one, is still a lot less than say a neutron, yet if you calculate how long it will take two neutrons, initially say 1 metre apart, to collide in an otherwise empty universe you will come up with a rather big number. Comparable to the age of the universe if I remember correctly. I should do this calculation again...

Also, just because energy is not the same in all reference frames doesn't mean it doesn't gravitate. A planet moving at 0.9999999999c has a lot more mass than one at rest (relative to us, the observer) and will pull on us via gravity more strongly. People do calculations in GR of things like what happens to a 'gas' of photons.

BenTheMan
05-27-10, 08:29 AM
Since I believe SR is false and have proven it, I believe GR is false also as a consequence.

Ah good! That simplifies matters.

If you post anything to that effect in this forum, you will be banned.

Thank you :)

Billy T
05-27-10, 09:27 AM
To BenTheMan:

I am not in favor of banning just because someone holds POV very different for the well accepted POV. (I would be banned as I have posted proofs that the accepted POV of cognitive science about how perception "emerges" is totally wrong and offer an alternative POV, which is slowly winning some converts.)

Instead Jack must tell where he has shown SR is wrong. Then we can tear his argument to shreds. - That is how science is supposed to work, not by edicts issued by "authorities" that the accepted POV cannot be questioned even with rational arguments for an alternative, that also correctly explains the observed facts.

I show that accepted POV about perception is wrong here:
http://www.sciforums.com/showpost.php?p=2502342&postcount=12

And my alternative, with many supporting observations not explained by the accepted POV, is presented here:
http://www.sciforums.com/showpost.php?p=905778&postcount=66

Jack is very likely a crackpot. I am too, but I can and have defended my POV and NO ONE HAS EVER SHOWN IT WRONG OR HOW THE ACCEPTED POV CAN ALSO EXPLAIN AS WELL THE MANY FACTS MY POV DOES NOT ONLY EXPLAIN BUT ALSO PREDICTS MUST BE TRUE FROM ITS BASIC STRUCTURE.

Billy T
05-27-10, 09:45 AM
(1)Gravity is so weak that none of those things you mentioned will be observable, despite the fact that in principle they do happen.... (2) A planet moving at 0.9999999999c has a lot more mass than one at rest (relative to us, the observer) and will pull on us via gravity more strongly. People do calculations in GR of things like what happens to a 'gas' of photons.On (1): That is why I said: "Perhaps not true when calculated..." but as neither you nor I have done the calculation*, we are both just waving our hands.

On(2): Certainly that fast planet (wrt our frame) has greatly increased inertial mass for any force we apply to it, even if applied transverse to its trajectory (as is observed with relativistic particles in a cyclotron ). The question is will it give Earth a greatly increased gravitational impulse as it passes? Applying my simple rule (given in prior post) I would say NO.
What evidence do you have that the correct answer is "yes."

----------------------
*BTW, I suspect that the only way to actually do the calculation is to reverse time and the sign of the mutual gravitation. I.e. start with two photons on parallel trajectories only a Planck distance separated with "gravitation" being repulsive and calculate how long before they are 100,000 meters apart (or separated by so much that the computer rounding error makes the repulsion remain zero for that and all greater distances. If they have mass that makes gravity then surely they have inertia too. So after that end of the calculation computer accuracy limit is reached, they again are on straight but divergent line trajectories.)

kurros
05-27-10, 07:39 PM
On (1): That is why I said: "Perhaps not true when calculated..." but as neither you nor I have done the calculation*, we are both just waving our hands.

Well ok I haven't done the calculations, but one of my colleages it studying solutions to the Einstein-Yang Mills equations for U(1) fields, which is just the fancy way of saying he is looking at what happens to spacetime if there are electromagnetic fields floating around in it. It's not quantum mechanical, so there are no photons, but on the large scale that doesn't matter. There is no matter, it is just pure electromagnetic field. These things still contain energy, and energy is all you need to curve spacetime.



On(2): Certainly that fast planet (wrt our frame) has greatly increased inertial mass for any force we apply to it, even if applied transverse to its trajectory (as is observed with relativistic particles in a cyclotron ). The question is will it give Earth a greatly increased gravitational impulse as it passes? Applying my simple rule (given in prior post) I would say NO.
What evidence do you have that the correct answer is "yes."


Physical evidence? None, I don't know of any experiments in which increased gravitational influence due to increased mass has been observed (there might be some clever astrophyical observations I don't know about though). Certainly inertial mass increase has been observed zillions of times in particle colliders. And if you accept general relativity then the equivalence principle says inertial mass = gravitational mass so it predicts that the relativistic mass creates extra gravitational influence.

Jack_
05-27-10, 07:41 PM
Ah good! That simplifies matters.

If you post anything to that effect in this forum, you will be banned.

Thank you :)


Sorry to offend you.

Perhaps you can come here.
http://www.sciforums.com/showthread.php?t=101058&page=12

I always love to run into a superior mind.

In that thread perhaps you can teach me why relativity is an absolute fact.

CptBork
05-27-10, 08:19 PM
Observing the direct effect of radiation on gravity would be extremely difficult even at macroscopic scales. But that's one of the nice things about cosmology, you can look at the fossils left over from events orders of magnitude higher in energy than anything we could even fathom in our minds. Chapter 8 of Carroll's book discusses the different effects of radiation, dust and gas density on the rate of the universe's expansion.

I presume one can look at clusters of galaxies, get a good estimate of their matter and radiation content and the relationships between density and pressure, use this to predict how the clusters evolve over time from gravity, and compare it to data both from the present as well as estimates of how those galaxies looked in the past. As long as you're choosing sufficiently small scales, dark energy shouldn't have any substantial effect, so I don't see why you couldn't isolate the effects of radiation with careful measurements. In any case, GR predicts that light causes gravity just as light is affected by it.

Jack_
05-27-10, 08:47 PM
Observing the direct effect of radiation on gravity would be extremely difficult even at macroscopic scales. But that's one of the nice things about cosmology, you can look at the fossils left over from events orders of magnitude higher in energy than anything we could even fathom in our minds. Chapter 8 of Carroll's book discusses the different effects of radiation, dust and gas density on the rate of the universe's expansion.

I presume one can look at clusters of galaxies, get a good estimate of their matter and radiation content and the relationships between density and pressure, use this to predict how the clusters evolve over time from gravity, and compare it to data both from the present as well as estimates of how those galaxies looked in the past. As long as you're choosing sufficiently small scales, dark energy shouldn't have any substantial effect, so I don't see why you couldn't isolate the effects of radiation with careful measurements. In any case, GR predicts that light causes gravity just as light is affected by it.

Can you show me the equations to prove light stimulates gravity?

kurros
05-27-10, 09:26 PM
R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = \frac{8 \pi}{c^4}T_{\mu\nu}

These are the Einstein equations, they ARE general relativity. The left hand side is a bunch of geometry stuff which describes the shape of spacetime, the right hand side describes the energy content of the spacetime. The object T_{\mu}{\nu} is called the "energy-momentum tensor", it describes the energy and momentum content of the spacetime in a coordinate independent (observer independent) way. The indices run from 0 to 3, describing the space and time dimensions, so it can be though of as a 4x4 matrix (but it has extra properties to regular matrices, properties which describe how it transforms as you move from frame to frame).

Wikipedia has a good picture of what the components of the energy momentum tensor represent:

http://upload.wikimedia.org/wikipedia/commons/thumb/3/37/StressEnergyTensor.svg/250px-StressEnergyTensor.svg.png

As you see, some of the components describe the energy distribution in the spacetime, others the momentum distribution. Energy and momentum are not observer independent so these components of the matrix change depending on what frame you look at it from.

Anyway the point is the equations in fact never explicitly mention MASS anywhere. They care about energy and momentum, of which light carries both.

Billy T
05-27-10, 09:48 PM
... it is just pure electromagnetic field. These things still contain energy, and energy is all you need to curve spacetime. ...I agree that energy can curve space time or in older terminology, "make gravity." Recall I said that a hot brick made more gravity than same brick when cold. I also said that only if the energy was the SAME in all frames did it curve space time (make gravity).

Do you think one part of space time can have curvature C1 and also have different curvature C2 because in one frame a flux of photons passing thru that spacetime region has energy E1 and in another frame the same flux of photons has energy E2? - As this part of space time can be observed by many observers in many different frames that does not seem reasonable, to me.

All observers in all frames will agree that the hot brick has temperature T but not on the total photon energy. That is why I like the simple rule I invented long ago, which was:

Energy makes gravity IFF it is the same in all frames.

Billy T
05-27-10, 09:55 PM
... GR predicts that light causes gravity just as light is affected by it.Certainly photons traveling thru spacetime are effected by the curvature of space time (gravity to an old time physicist, like me) The famous test of GR with the appearant shift of star's loction when its rays passed near the eclipsed sun shows that.

I don't think that shows in anyway that the photons were "making gravity." They were just going on the least action path, as they always do.

kurros
05-27-10, 10:22 PM
I agree that energy can curve space time or in older terminology, "make gravity." Recall I said that a hot brick made more gravity than same brick when cold. I also said that only if the energy was the SAME in all frames did it curve space time. Do you think one part of space time can have curvature C1 and also have different curvature C2 because in one frame a flux of photons has energy E1 and in another frame the same flux of photons has energy E2? - As this part of space time can be observed by many observers in many different frames that does not seem reasonable, to me.

All observers in all frames will agree that the hot brick has temperature T but not on the total photon energy. That is why I like the simple rule I invented long ago, which was:

Energy makes gravity IFF it is the same in all frames.

You can have your rule if you like but that isn't what the Einstein equations say. Energy is frame dependent, sure, but the energy-momentum tensor is not. However, it is just as valid to calculate what happens in one frame as in another. The spacetime metric does not look the same to all observers; it is a tensor, so you if you know it in one frame you can calculate what it is in another frame, but it will look different. For instance, you can calculate what the spacetime created by a spherical body of uniform mass is as seen from a stationary observer far away (this is the Schwarzschild spacetime, in Schwarzschild coordinates). You can also calculate what it looks like to an observer who is free-falling towards the body (these are called the "normal" coordinates I think), and it will look a lot different.

Maybe if I go back to by super-fast moving planet example I can make this make sense. There is no such thing as the absolute velocity of the planet, so such a concept affects nothing. The planet was only moving at a sizeable fraction of the speed of light relative to Earth (say). It doesn't matter if we observe from Earth, or the other planet, or some centre of momentum frame, we will still see two planets hurtle past each other. This configuration of objects will cause them to pull on each other more strongly than if they moved past each other slowly. Even though it is observer dependent which objects have which energy, there is a well defined sense in which the overall energy of the configuration of objects can be described independently of this. This is the energy momentum tensor in the Einstein equations.

CptBork
05-28-10, 01:30 AM
Certainly photons traveling thru spacetime are effected by the curvature of space time (gravity to an old time physicist, like me) The famous test of GR with the appearant shift of star's loction when its rays passed near the eclipsed sun shows that.

I don't think that shows in anyway that the photons were "making gravity." They were just going on the least action path, as they always do.

That's why I gave the cosmological example, it's the only one I can personally think of. The same energy density will have different pressure properties depending on whether it's dust, gas or radiation. Interestingly enough, according to GR the pressure properties of the system under consideration will actually have a noticeable impact on the gravitational field. Furthermore, if space stretches so the distance between any two points increases by a factor a, the energy density of dust and gas will fall as 1/a^3, whereas the energy density of radiation will fall off as 1/a^4 because there's an additional redshift factor to consider.

Janus58
05-28-10, 07:01 AM
Certainly photons traveling thru spacetime are effected by the curvature of space time (gravity to an old time physicist, like me) The famous test of GR with the appearant shift of star's loction when its rays passed near the eclipsed sun shows that.

I don't think that shows in anyway that the photons were "making gravity." They were just going on the least action path, as they always do.

If they didn't, then their passage by the star would not have any effect on the star. But their paths do bend which means a change in momentum for the photons. Without a corresponding change in the star, you would have a conservation of momentum violation.

Billy T
05-28-10, 07:24 AM
If they didn't, then their passage by the star would not have any effect on the star. But their paths do bend which means a change in momentum for the photons. Without a corresponding change in the star, you would have a conservation of momentum violation.That is a good agrument (as all of yours are) but I am not sure it is valid because it (I think) relies on our non-GR idea that the light path was bent / momentum is a vector, etc. thus changed.

I think that one can claim / argue that the path did NOT bend as light always travels in a straight line (in vacuum or any uniform transparent media). I.e. I suspect (but way above my head to show) that if one described the momentum in proper GR terms, it did not change as the photons passed by the sun. If that is true then there is also no momentum change in the sun and your argument collapses.

Billy T
05-28-10, 07:50 AM
To Kurros:

Thanks for post 29 and 31. They helped me understand better. Is there a reason why the Greek letter row and column coordinates of the R, which is next to the equality sign in post 29, are on the left side of the R instead of on the right? I don’t clearly understand but suspect they should be on the other side of R.

If given all 16 components of R, I assume one has a complete description of the curvature of space time. (“gravity” to old timer me) Is this correct?

I am at a loss to understand how thermal energy*, which is the same in all frames can mix with photon energy which differs in different frames is inserted into the Einstein GR equation. You said that there are rules, much like 3D coordinate transforms (but different of course), for transforming the 16 components of the energy momentum tensor into a different coordinate / frame system. The problem I see is with these FIXED transformation rules is that if most of the total energy, E, were thermal, Et, instead of photons (for example) Ep then total in the new frame, E’ should be essentially the same as E, but if E is mainly Ep then when transformed E’ should be very different from E. How can this be if the transforms rules are fixed?

------------------
*Thermal energy is the random KE. If you add a velocity V to every atom (to change to another frame) the random KE is not changed. I.e. Thermal KE and temperature are invariants with frames. (must be true if lead melts at same temperature in all frames and that must be true as the same mass of lead cannot be a pouring liquid when viewed from frame A and yet a solid brick when viewed from frame B.

PS I am quite ignorant about GR, but not dumb. Long ago, when teaching, I noticed how much an ignorant but not dumb student can stimulate your own understanding. I hope I am doing this for you in repayment for your teaching me.

PS1 CptBork has posted some comments about my merger and modification of several ideas to claim dark energy and gravity are two sides of the same coin. If you have time please take a look at my reply to him here: http://www.sciforums.com/showpost.php?p=2538468&postcount=17

I state my idea more clearly there than earlier. Also perhaps you will look at the OP reference, now that thread has a convenient link (in post 18) to what it must be nearly an exact copy and comment on it. I can not evaluate it except to note that anything published is Phys Rev. Letters is very unlikely to be nonsense. Can you follow their argument?

Is it correct that they are claiming the vacuum polarization energy transforms into Dark Energy in the intense field (gradient?) near neutron star etc. That is what I understood to be their point. If not, my idea is nonsense too.

PS2 To CptBork:Mainly for the benefit of others as you know, I note that your inverse alpha to the fourth applies ONLY to EM radiation, not to particles with energy. Thus my concern above about E ---> E' under fixed transformation rules may concern your observation too. I.e. the scaling of energy density with expanding universe is probably approaching inverse alpha cube as the fraction of the total energy in universe is shifting towards KE as the EM component decreases with the red shift. Does this shift to KE "flatten out" the curvature of the universe? Was it more curved earlier? etc.??? Is there anything interesting to idea that initially (big bang) there were no particles and then the universe was so highly curved that it was a point?

kurros
05-28-10, 06:42 PM
To Kurros:

Thanks for post 29 and 31. They helped me understand better. Is there a reason why the Greek letter row and column coordinates of the R, which is next to the equality sign in post 29, are on the left side of the R instead of on the right? I don’t clearly understand but suspect they should be on the other side of R.

Oh, np. Ahh, that second set of indices is actually attached to the g, as in g_{\mu\nu}, which is the metric tensor. This object actually describes the spacetime fully on its own, it tells you the distance between different spacetime points. Eg. in flat space (in an inertial reference frame! Since we are doing GR we could just as legitimately use an accelerated frame, which would make the metric tensor look more strange) it is

g_{\mu\nu} = \left[ \begin{array}{ccc}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right]

Spacetime intervals are calculated from the metric:

ds^2 = g_{\mu\nu}dx^\mu dx^\nu

The repeated indices mean you sum over their values, so the above flat metric tensor gives

ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

(where dx^0 = dt etc.), which is kind of pythagoras' theorem.

Anyway the others bits of the LHS of the Einstein equations are the Ricci tensor and scalar, and yes they tell you stuff about the curvature of space, but they are basically just derivatives of the metric tensor. It basically the same as how the second derivative of just a regular 1-D function gives you the curvature of that function. So they don't contain extra information to g_{\mu\nu}.



If given all 16 components of R, I assume one has a complete description of the curvature of space time. (“gravity” to old timer me) Is this correct?

Right idea, except yeah the object you want is g_{\mu\nu}.



I am at a loss to understand how thermal energy*, which is the same in all frames can mix with photon energy which differs in different frames is inserted into the Einstein GR equation. You said that there are rules, much like 3D coordinate transforms (but different of course), for transforming the 16 components of the energy momentum tensor into a different coordinate / frame system. The problem I see is with these FIXED transformation rules is that if most of the total energy, E, were thermal, Et, instead of photons (for example) Ep then total in the new frame, E’ should be essentially the same as E, but if E is mainly Ep then when transformed E’ should be very different from E. How can this be if the transforms rules are fixed?

Hmm I haven't tried to do anything with thermal energy, but yes it should come into it somehow. I guess the energy momentum tensor has components which take pressure energy into account, which is related to temperature through ideal gas law type stuff (I am thinking if we have a big gas cloud at some temperature floating in space). Probably these other components take these forms of energy into account, but sorry I'm not really sure how.



------------------
*Thermal energy is the random KE. If you add a velocity V to every atom (to change to another frame) the random KE is not changed. I.e. Thermal KE and temperature are invariants with frames. (must be true if lead melts at same temperature in all frames and that must be true as the same mass of lead cannot be a pouring liquid when viewed from frame A and yet a solid brick when viewed from frame B.

PS I am quite ignorant about GR, but not dumb. Long ago, when teaching, I noticed how much an ignorant but not dumb student can stimulate your own understanding. I hope I am doing this for you in repayment for your teaching me.

PS1 CptBork has posted some comments about my merger and modification of several ideas to claim dark energy and gravity are two sides of the same coin. If you have time please take a look at my reply to him here: http://www.sciforums.com/showpost.php?p=2538468&postcount=17

I state my idea more clearly there than earlier. Also perhaps you will look at the OP reference, now that thread has a convenient link (in post 18) to what it must be nearly an exact copy and comment on it. I can not evaluate it except to note that anything published is Phys Rev. Letters is very unlikely to be nonsense. Can you follow their argument?

Is it correct that they are claiming the vacuum polarization energy transforms into Dark Energy in the intense field (gradient?) near neutron star etc. That is what I understood to be their point. If not, my idea is nonsense too.

PS2 To CptBork:Mainly for the benefit of others as you know, I note that your inverse alpha to the fourth applies ONLY to EM radiation, not to particles with energy. Thus my concern above about E ---> E' under fixed transformation rules may concern your observation too. I.e. the scaling of energy density with expanding universe is probably approaching inverse alpha cube as the fraction of the total energy in universe is shifting towards KE as the EM component decreases with the red shift. Does this shift to KE "flatten out" the curvature of the universe? Was it more curved earlier? etc.??? Is there anything interesting to idea that initially (big bang) there were no particles and then the universe was so highly curved that it was a point?

Ok I'll have to get back to you on the rest, things to do!

Janus58
05-28-10, 10:51 PM
That is a good agrument (as all of yours are) but I am not sure it is valid because it (I think) relies on our non-GR idea that the light path was bent / momentum is a vector, etc. thus changed.

I think that one can claim / argue that the path did NOT bend as light always travels in a straight line (in vacuum or any uniform transparent media). I.e. I suspect (but way above my head to show) that if one described the momentum in proper GR terms, it did not change as the photons passed by the sun. If that is true then there is also no momentum change in the sun and your argument collapses.

Look at it this way. It is possible to shoot a photon at a black hole such that it bends around the black hole and returns to the sender. Since the sender will recoil due to emitting the photon, and then gain momentum in the same direction in reabsorbing the photon, the sender will gain momentum in one direction in the process, and something has to gain the opposite momentum to compensate. The only other "something" involved in the exchange is the black hole and the only thing it interacts with is the photon.

Billy T
05-29-10, 07:18 AM
Look at it this way. It is possible to shoot a photon at a black hole such that it bends around the black hole and returns to the sender. Since the sender will recoil due to emitting the photon, and then gain momentum in the same direction in reabsorbing the photon, the sender will gain momentum in one direction in the process, and something has to gain the opposite momentum to compensate. The only other "something" involved in the exchange is the black hole and the only thing it interacts with is the photon.Dam you are good. I can't think of any counter to that.

---------------
It is interesting, at least to me, to describe this “180 non-contact mirror” event in terms of the alternative model of gravity described here:
http://www.sciforums.com/showpost.php?p=2538468&postcount=17
Because it requires that even photons cast a “DE flux shadow.” It also increases my desire to modify that post 17 model, slightly to state that the DE flux is never absorbed only scattered. I have been leaning towards this modification for some weeks as it helps Dark Energy to be increasing with time. As you are exceptionally clever, I hope you will take a look at post 17 and comment in that thread.

In the DE flux model of gravity, when your photon is in the “back side” of the loop around the black hole, its DE flux shadow is falling on the side of the BH far from Earth. Thus the DE flux falling on the near Earth side of the BH is slightly stronger. This net excess of front side scattering makes an increase in the BH’s momentum away from the Earth (equal and opposite to that given to the Earth, which of course is twice the momentum carried by the photon. The photon is, of course in the DE flux model, being accelerated in its “U turn” by the shadow of the BH.

Perhaps, the “seeds of destruction” of the DE flux model of gravity are in your simple Gedanken too. I.e. if a tiny BH is compared to a large BH, the photon will spend much less time behind the tiny BH; However, the “darkness” of its scattering shadow will be greater as it "turn radius" is smaller making it closer to the EH of the BH. One should be able, in principle to test if the total impulse given to the BH is independent of the size of the black hole. I.e. does the “greater darkness” exactly compensate for the shorter shadowing time? It must for the DE model to not be destroyed. This calculation does not seem easy to do and does require (or allows some freedom in) some assumed scattering model, but someone as clever as you may be able to find a simple way to show that it can not be independent of the BH mass.

Note also my prior argument about the night sky being the same pattern for either eye only shows that when photons are not in a shadow they do not make any gravity. This of course follows directly from the DE flux model. Like the tango, it takes two to make gravity in the DE flux model.

Jack_
06-02-10, 10:29 PM
Look at it this way. It is possible to shoot a photon at a black hole such that it bends around the black hole and returns to the sender. Since the sender will recoil due to emitting the photon, and then gain momentum in the same direction in reabsorbing the photon, the sender will gain momentum in one direction in the process, and something has to gain the opposite momentum to compensate. The only other "something" involved in the exchange is the black hole and the only thing it interacts with is the photon.

The black hole is not a necessary condition.

Any gravity that would return light would produce a profit.

Good idea.

Hence, gravity is an energy source.

Billy T
06-03-10, 10:47 AM
Any gravity that would return light would produce a profit. Good idea. Hence, gravity is an energy source.No Jack. The light is the energy source (or what ever made he photons really) Gravity is never and energy source, but one can let it do work and get energy from that work by lowering the gravitational potential energy.

For exampe, nuclear fusion in the sun is an energy source than makes rain on Earth to fill dams. Then humans can get hydro-electric energy by lowering the potental energy of the water stored behind the dam.

Again Gravity is not a source of energy, but it may help make energy for man's use but there was always some other source of energy we are tapping.

Jack_
06-03-10, 09:22 PM
No Jack. The light is the energy source (or what ever made he photons really) Gravity is never and energy source, but one can let it do work and get energy from that work by lowering the gravitational potential energy.

For exampe, nuclear fusion in the sun is an energy source than makes rain on Earth to fill dams. Then humans can get hydro-electric energy by lowering the potental energy of the water stored behind the dam.

Again Gravity is not a source of energy, but it may help make energy for man's use but there was always some other source of energy we are tapping.

Oh, I was talking about a dfifferent model from you.

Let's see, light is an energy source.

Does light come from the sun?

Gravity creates the proximity function such that the strong nuclear force is implemented.

Thus, we have light.

Did you say light is an energy source?

Where does it come from?

If you can implement cold fusion then you are correct, otherwise, I am.

BenTheMan
06-04-10, 09:32 AM
Gravity creates the proximity function such that the strong nuclear force is implemented.

Prove this statement.

Jack_
06-04-10, 07:19 PM
Prove this statement.

Sure, the basic problem to be solved is how you get the nuclei together such that it overcomes the proton repulsion and pushes the nuclei together where the strong nuclear force is implemented.

You can do this in the lab with lasers/heat where the atoms bounce around so much and with such activity that the proton repulsion is overcome and the strong nuclear force is implemented/takes over.

On the other hand, gravity pushes or attracts in your way of thinking, such that the proton repulsion is overcome and the nuclei are close enough together such that the strong nuclear force is implemented and hence E=mc^2 at least up to tungsten.

The problem in logic now is to equate the laser in the lab to gravity.

freziggity
06-06-10, 03:53 AM
If every particle of energy were reassembled what would there be? I think there would be an absolute. this would mean, a perfectly balanced immeasurable density of perfect symmetry. It would have no charge, it would be very hot. It would be glowing white with energy. Space outside of it would be a perfect vacuum. There would be no perferation of the mass of energy. There could not be found the slightest space between particles. it is an incomprehensible thing. The complete whole from which every particle originated. Outside of that is what could be referred to as the father of all blackholes, The event horizion of creation/evolution. A superconductive cold that causes energy to behave differently than what we're used to in our dimension. As gravity draws like charges together in the superconductive void, instead of resisting one another, energies bond with no net loss of energy (like an ideal gas). As the temp rises, the force of electromagnetism goes to work. causing attraction and repulsion. Gravity acts on energy, but is not inherent in matter. Gravity is inherent in vacant space. When Quarks break down, the process of atomic decay begins, but this process is constant over time, as the atom burns away. The atom gradually destabilizes losing power to the vacant cold, the tiniest bits of energy float around until they establish equilibrium and balance in a neutral energy circuit. These types of matter/energy are unstable, they are always breaking down. Where the temperature of space is warmer there is a higher concentration of this energy, where space is cold there is a lower concentration of energy. When blackholes form energy has been completely broken down beyond the atomic and sub atomic levels, perhaps even beyond the quark level. It sinks to a depth beneath this dimension, behind and around the sphere of this energy circuit. As the hole warms, it is filled with this same kind of energy until it disappears. Of course this is all conjecture, but I believe the laws of Physics hold true at every level. I also believe that gravitation is the Master of all forces. I do not believe there needs to be a dividing line between Classical and Quantum physics. So I created this theory to attempt to unify the ideas. I'm no Einstein, but I am faithful in Physical Law. There should be no contradictions in the law and it should be universal. Honestly, none of us know what gravity is. It is the strangest thing in the universe (unless God is out there too). This idea has plagued me for months and months. I'd appreciate it if anyone posting here would shoot it down for me.

Billy T
06-06-10, 12:24 PM
...This idea has plagued me for months and months. I'd appreciate it if anyone posting here would shoot it down for me.Then state the idea clearly so we can tell what it is.
For example, to pick just one brief statement out of your post 45 mess: "gravity acts on energy"

What type of energy? Kinetic energy? Thermal energy? Electromagnetic energy? Chemical energy? Potential energy?

BenTheMan
06-07-10, 10:33 AM
Sure, the basic problem to be solved is how you get the nuclei together such that it overcomes the proton repulsion and pushes the nuclei together where the strong nuclear force is implemented.

You can do this in the lab with lasers/heat where the atoms bounce around so much and with such activity that the proton repulsion is overcome and the strong nuclear force is implemented/takes over.

On the other hand, gravity pushes or attracts in your way of thinking, such that the proton repulsion is overcome and the nuclei are close enough together such that the strong nuclear force is implemented and hence E=mc^2 at least up to tungsten.

The problem in logic now is to equate the laser in the lab to gravity.

Pretty much nothing in this response addresses the actual issue.

For example, how is a laser at all like gravity?

Your statement was definitive, however, you know backpedal. Your explanation is contingent upon an unproven and wrong analogy.

The strong force is very well understood as something completely different from gravity, and has been understood for more than 40 years.