# How can space warp if it is a non-thing?

Discussion in 'Physics & Math' started by nicholas1M7, May 27, 2011.

1. ### Magneto_1Super PrincipiaRegistered Senior Member

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295
In the short, the answer is yes! I am not that comfortable "yet" with the term "Gravitons." In my opinion, and based on my study, Physicists are still coming to grips with the "Graviton" concept. I am still waiting for the LIGO experiments to verify the existence of "Gravitons" before I start to use this term in my lingo.

I am currently using the term Aetherons to describe what the LIGO experiments are suppose to measure. Once they verify the "graviton", I will switch my terminology.

If "Gravitons" do exist this would be a major, major breakthrough in science. We currently view the world through photons and the electromagnetic spectrum. This is what the Hubble, Spitzer, Chandra, and other telescopes measure; telescopes measure photons and the electromagnetic spectrum.

If "Gravitons" do exist and are measured, this would be a brand new way of viewing the universe at large. And a graviton spectrum similar to the electromagnetic spectrum would emerge. Who knows what strange things we will see with "Graviton Vision", I can't wait!

The answer to this question is two (2) fold.

First, the actual Math that Einstein used, which was the original mathematics of Riemannian, Gauss, and Ricci predict curvature in and of space.

Second, one of Einstein's main goals in General Relativity was to replace Newton's instantaneous action at a distance model of gravity, with a finite (speed of light transmission) action at a distance model of gravity.

In Newton's model of gravitation planets and matter magically interact through the vacuum of space.

In Einstein's General Relativity (GR) model of gravitation planets and matter interact through an isotropic and homogenous Aether vacuum of space.

And as an addendum to Einstein's General Relativity (GR) model of gravitation he adds that matter adds an interaction to this vacuum such that matter creates a gravitational field that is anisotropic and heterogeneous (non-homogenous) in the vicinity or near the matter that creates the vacuum.

Hence in Einstein's model matter superimposes on the isotropic and homogenous Aether vacuum of space, anisotropic and heterogeneous (non-homogenous) space, near or in the local vicinity of matter. It is the matter that creates this anisotropic and heterogeneous (non-homogenous) space; and this is considered warping of space or spacetime.

Best

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3. ### przyksquishyValued Senior Member

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3,203
What the hell are you talking about? A vector field having curl just means that $\bar{\nabla} \times \bar{F} \neq 0$, which generally implies the field has non-zero circulation over closed loops. Kids in highschool learn about how electric currents induce curl in the magnetic field surrounding a wire (even if they don't learn it's called "curl"). It's anything but being ignored and it has absolutely nothing to do with virtual particles in QED.

I think this yet another case of you looking at the words physicists use without understanding their domain specific meanings. "Curl" in physics almost certainly doesn't mean what you think it does. Same with "virtual", which just means that a particle is off shell, and not that the particle is "not real".

This guy draws the same sorts of conclusions you do only because he considers the same narrow definitions you do. For example:
For starters while $\frac{\mathrm{d}t}{\mathrm{d}t}$ is a pretty useless quantity which always equals 1, we usually parameterise worldlines using proper time, and generally $\frac{\mathrm{d}t}{\mathrm{d}\tau$ in a given inertial coordinate system is not equal to one, or necessarily even constant. It's not meaningless either: it's the time dilation factor. Second, this says nothing about the usual 3-velocity $\frac{\mathrm{d}\bar{x}}{\mathrm{d}t}$, which can be anything, so working in spacetime and thinking in terms of worldlines has no implications about whether or not I can drive a car around. If you think "motion" is impossible in spacetime, it's only by a definition of "motion" that nobody uses anyway.

Seriously, use some common sense when reading stuff like this.

That's funny, because when I interpreted what Einstein said in almost exactly the same way (space or spacetime being inhomogenous means that we have to think of it as having properties), you insisted he actually meant that we had to interpret it as meaning gravity and inhomogenous space were synonymous with one another.

Do you intend to reply in the Inflation and Curvature thread, by the way?

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5. ### przyksquishyValued Senior Member

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Well the classical descriptions of the forces all have in common that they're gauge theories, which means that they're all derived from an "equivalence postulate" of sorts. I can post a simple example that should illustrate the sort of thing a gauge theory is if you want. But in short, the four vector potential $A_{\mu}$ and electromagnetic field $F_{\mu\nu}$ are in some ways analogous to the metric $g_{\mu\nu}$ and Riemann curvature tensor in general relativity. The curvature tensor tells you whether or not you can choose coordinates such that the metric tensor simplifies to the Minkowski tensor everywhere. In the same way, whether or not the electromagnetic field is zero tells you whether or not it's possible to choose a phase convention in which the electromagnetic potential disappears. So I suppose you can also think of the electromagnetic field as a type of "warping", but it's a "warping" in a more abstract parameter space associated with the local phases of charged fields, rather than warping in spacetime. (If that last bit doesn't make much sense to you, don't worry. I'm not convinced it makes too much sense to me either).

It's not really so simple. Conceptually, "quantum field theory" is to "classical field theory" as "quantum mechanics" is to "classical mechanics". In quantum mechanics, instead of a particle just existing at some position $x$, it exists in a superposition described by a wavefunction $\psi$, so that $\psi(x)$ is the "presence" of the particle at the position $x$. One way of thinking about quantum field theory (though you'll rarely if ever see it described this way for practical reasons) is that instead of a field having a definite field configuration[sup]*[/sup] $\phi(x)$, it exists in a superposition of field configurations described by a wave-functional $\Psi$, so that $\Psi[\phi(x)]$ is the "presence" of the field in the configuration $\phi(x)$.

When you do this and impose a particular field equation, the quantum field turns out to have discrete excitation levels that have the properties of quantum particles. In other words, particles emerge out of the derivation of quantum field theories rather than simply being put in by hand. A single particle state is simply one of the states a quantum field can be in, and particle states in general are one way of parameterising a quantum field. (This means you can "decompose" a quantum field onto particle states in the same sort of way that you can decompose a particle's quantum state onto position or momentum states. Being quantum physics of course, nothing about the decomposition is necessarily "definite" in the classical sense of the word. For example, in a quantum description of the light emitted by a laser, the number of photons isn't definite. It's in a superposition that follows a Poisson distribution).

[sup]*[/sup]Examples: the Coulomb field solution and sinusoidal electromangetic wave solutions are field configurations that you might one day find the electromagnetic field in.

It would be complementary to it, not an alternative to it. In a quantum theory of gravity, "gravitons" would just be discrete excitation levels in the metric tensor field with particle-like properties.

Last edited: Jun 10, 2011

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7. ### AlphaNumericFully ionizedRegistered Senior Member

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You seem to fail to realise what I said in your links is entirely separate from the fact you were mistaken about the vector potential in quantum field theory.

You said that the 'curl in electromagnetism' was replaced by virtual particles. Exactly the equations which appear in Maxwell's classical electromagnetism exist within quantum field theory, the vector potential is the central object. Instead it is just treated in a slightly different way because its Fourier mode coefficients now take on the form of operators.

The question of whether or not gravitons exist is separate. The gaping hole in your knowledge was to do with the vector potential. Do keep up. And besides, I find it funny you complain about certain things being fiction when your 'work' cannot model a single thing and has no evidence. Gravitons are simply the application of known quantum field theory to the one remaining non-quantised force. All other forces are carried by particles, demonstrably so.

You clearly don't understand quantum field theory. The 'loops' within Feynman diagrams are formed by pair production and the resultant pairs within the loops are virtual. They do not obey $E^{2} = m^{2}+p^{2}$, thus are virtual, and you have to consider all possible momenta for them. If you suppress the quantum field theoretic ability to pair produce (ie render the quantum field theory into a quantum mechanics model by suppressing relativistic effects) then you remove both these loops and virtual particles. It's essentially fixing the number of particles in the system.

The fact you don't understand quantum field theory, can't admit it and have to resort to simply making up paraphrasings of me is not my fault. Look how far it's gotten you so far......

I provided the explaination here and you failed to understand. You tried to then back peddle but proceeded to show you don't know how the electron and photon are modelled differently. I pointed out your mistakes. You always complain people are supposedly only throwing 'outrage and insults' but the fact is you ignore the details because you don't understand them and can't retort them.

Come on Farsight, look at how badly this attitude has served you for years now. You've been stuck in a rut for more than half a decade because you can't accept the sun doesn't shine out your backside. If you'd listened to people's comments and engaged in honest discussion all those years ago you might have actually gotten somewhere. Instead you're stuck being laughed at on forums. Go you!

So I had to correct you on its role in quantum field theory and the specific notation but you understand its 'true' role? And you know this without knowing anything about specific experiments?

Seriously, ask yourself if that sounds at all believable. You have demonstrated you don't know what role the potentials play, as you couldn't even state them properly. Now proclaiming you have insight others don't, when you don't have even the raw data others do is laughable.

If you have more insight then can you provide a working model which is more accurate, applicable and powerful than anything current? Can you provide any model? Thought not.

Pair production doesn't have to have anything to do with the nucleus. That's an observed experimental fact. This is why I say you make these claims without any of the raw data, you have no idea what actually happens in reality yet you claim you understand it more than anyone else. It's one thing to be naive/ignorant of models you can find in books but at least have skimmed the Wiki page of but you're claiming insight into phenomena huge billion dollar machines work to examine. Since you don't have one of those or the data from one of those you're basing your claims in very little.

Can you cite a paper which sees this phenomenon occur? $\gamma+\gamma \to e^{-}+e^{+}$ is not a tree level process, there is no such interaction quantum quantum electrodynamics. As such it requires that one of the photons first turn into the electron/positron pair and then one of them absorbed the second photon. This is experimentally tested as it is a basic prediction of QED that no such direct interaction occurs.

It's little details like that which show the holes in your understanding.

Way to put your foot in your mouth!

Haven't you learnt by now that whenever you say something specific you end up buggering it up? Like I said, you might be used to bull****ing friends and family (or that nut on that TV show) but on physics forums you'll find actual physicists who know how certain phenomena behave. Or have you gotten so used to just making stuff up and talking like you do on that TV show that you can't even stop here?

I have no embarrassment about that. I think gravitons will eventually be distinguished, as I think all the forces can be quantised. You aren't embarrassing me in that regard. You did back peddle, you did make mistakes and you've made more of them.

You always do and now przyk has commented on some of them too, as he's done in other threads in regards to your nonsense. It is no skin of our noses if you continue spouting BS. It's your time and money to waste and my, haven't you wasted a lot of both. And for what? Nothing. Rather than convincing the Nobel Committee you're a genius you're stuck arguing on forums about your 'genius' and paying for advertisements to shift that laughable book of yours. All the while the rest of us continue on with our educations or jobs, actually contributing something. At least I don't pay to get my ideas to the attention of the right people, they pay me

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That about sums up my thoughts on everything you've ever said.

8. ### Farsight

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3,492
And that's wrong, pryzk. It's the electromagnetic field. Reduce that electric current to a single electron. It's got an electromagnetic field. Now move past it. There's no curl being created or "induced", you're just feeling the "magnetic curl" of the electromagnetic field because you now have relative motion with respect to it. One field, two forces, like Minkowski said.

No it isn't. This is another case of mathematicians having only a schoolboy grasp of electromagnetism.

Don't be specious, I know what curl means.

Now pay attention. See electric field. See where it says The electric field is a vector field ? See magnetic field. See where it says The magnetic field at any given point is specified by both a direction and a magnitude (or strength); as such it is a vector field ? These vector fields are describing "what it does" rather than "what it is". It's a description of the forces, not the electromagnetic field itself.

That's wrong too. The evanescent wave is the physical reality underlying virtual photons. I'm not kidding you about this, and I'm not just making it up. Do the research.

Strike three pryzk. There is no motion in spacetime. Come on, think it through.

That's not the issue. You can make marks on the worldline to denote proper time, but the worldline is still an all-times view. There is no motion up the worldline.

Yes the worldline can be angled. But there's still no motion going on in the mathematical space.

No, it's by a definition of "spacetime" that everybody uses. You don't move through spacetime, it's a fallacy.

I am, you aren't. Now sit down and think this one through. You can move through three-dimensional space, we all know that, we can see it. You can add a fourth "dimension" to derive Minkowski spacetime and draw a worldline in it. Typically we shed one of the spatial dimensions to draw Minkowski spacetime as a block. If you're aren't moving through space the worldline is vertical. You aren't moving through space, and you aren't moving through spacetime either. When you move uniformly through space the worldline is angled, but you're moving through space, not through spacetime. There is no motion in an all-times view of the world.

This comes back to your confusion between space and spacetime. You have to learn to make the distinction between space and the mathematical "space" called spacetime.

I've never intended to not reply to you. Alphanumeric yes - it's just hopeless trying to talk to the guy. It fell off my radar, I'll take a look.

9. ### Farsight

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3,492
Sorry for butting in, pryzk, but I think this is important.

It's spatial curvature as opposed to spacetime curvature. A photon goes past you, it's a pulse of $A_{\mu}$. Take a derivative for the sinusoidal electromagnetic field variation. For the field to vary there has to be some form of current, only this is a photon not an electron, and it's displacement current rather than conduction current. It displaces space. It really does. That's why you can use this photon in pair production to make an electron and a positron. The electron is just a self-trapped 511keV photon in a closed path. Set gravity aside and a photon goes in a straight line. But if the space it's going through is all displaced and curved because the photon is going through itself, that straight line can be a closed path.

The particle is a wave.

Don't forget that there's only two static stable particles with mass - the electron and the proton, multiplied by two for their antiparticles.

The sinusoidal electromagnetic wave isn't a field configuration per se. The $A_{\mu}$ pulse is the field configuration. $A_{\mu}$ is "more fundamental". Read this. Pair production alters the configuration. Different configuration, different particle.

Gotta go.

10. ### wellwisherBannedBanned

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5,160
Bending space is a mathematical construct that uses the imagination to help us visualize. This visualization assumption makes it easier to model what we see happening in reality.

In ancient times the stars, planets and constellations used a mental construct based on mythology. This helped organize the collected data, using the imagination, so one could plot and predict the movements of the heavens. It allowed more and more data to be collected, but it was not a good visualization for math. Another visualization was needed and was created based on bending something that does not exist. From a practical POV, if it works, all is good.

Often the biggest funders of science want models that can be used for practical applications. Practical means money within the free market to help recoup the price of the science. It can also mean tools and toys for national defense. Practical does not have to be fully real to work, since practical results are most important.

I remember an old timer who used to spit chewing tobacoo juice into his electroplating tank. His work was always top notch. His spitting tobacoo juice was not real or proven science in terms of a cause and effect for his excellent results. But because of his excellent results, so he could continue spitting all he wanted. If you lose track of practical versus real, eventually spitting is called real due to results.

Much of science funded for practical purposes, is not always about pure science, but it does do the job very well. Often doing the job very well is mistaken for being real, like spitting chewing tobacco.

Bending space and time, with space defined as empty and time defined as just a mental construct, sort of looks like one has lost touch with reality. But it gets the job done. If you try to be more real about such imaginary assumption, it appears like you are out of touch. It has to due to with comparative practical. Go figure?

11. ### przyksquishyValued Senior Member

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3,203
Er actually there is. The electric and magnetic parts of the electromagnetic field are frame-dependent and in a sense rotate between one another as you switch frames. So you can well have no magnetic field in a point charge's rest frame, and a magnetic field with curl surrounding it in another frame.

No, it really isn't. You're confused about things everyone who has learned electromagnetism - from mathematicians to physicists to engineers alike - knows about.

Why are you telling me about basic definitions everyone already knows about? Yes, I know the electric field is a vector field. :bugeye:

Strictly speaking it's an axial vector field.

That's just physics being what it is. Experiments can only test behaviour and properties of objects. That's why everything in physics is essentially defined by its behaviour and interactions. Any statement about what the electromagnetic field or anything else "really is" that doesn't have implifications for its behaviour falls out of the realm of science. That's why we have a much more minimal view of the electromagnetic field than Maxwell originally did: we're comfortable with the notion of a field and we've whittled away all the "window dressing" that doesn't actually contribute to making testable predictions.

And believe me, I can understand you feeling like this leaves you with nothing to grasp onto if you've never taken a course in electrodynamics. That still doesn't make the argument that "the equations only describes behaviour" any less specious, since that's physics doing exactly what it's supposed to be doing.

Assuming you're right (I haven't checked), what would that change? What's "not real" about evanescent waves?

I have. I think it's you who's just taking an oversimplistic view of things. I'm not clear on exactly what point you're trying to make anyway. Suppose for argument's sake there is no "motion" in spacetime for some definition of "motion" you have in mind. So what? Does it actually contradict anything we observe?

What are you talking about? When I say something is "moving", I mean it in the simple everyday sense that its position changes over time or, equivalently, it has non-zero 3-velocity. When you say that there is no motion in spacetime, you make it sound like you're saying the notion of spacetime is somehow incompatible with the fact we observe "3-motion". If you just want to say that there is a sense in which there is no "4-motion" in spacetime then fine - depending on exactly how you define it - but then so what?

And...?

Last edited: Jun 11, 2011
12. ### przyksquishyValued Senior Member

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3,203
That interpretation is never going to work. This is obvious when you compare the equations of motion. The equation of motion of a charged particle in a given electromagnetic field is given by the Lorentz force:
$m \frac{\mathrm{d}^{2}\bar{x}}{\mathrm{d}t^{2}} \,=\, q \bigl[ \bar{E} \,+\, \bar{v} \times \bar{B} \bigr] \,. \qquad (1)$​
On the other hand the trajectory of a particle in a curved space is given by the geodesic equation I told you about in the other thread:
$\ddot{x}^{i} \,+\, \Gamma^{i}_{jk} \dot{x}^{j} \dot{x}^{k} \,=\, 0 \,. \qquad (2) [/indent]$
In order to interpret the electromagnetic field as curved space you'd have to show it predicts the right trajectories for charged particles, which means showing that you could interpret equation (1) as a special case of equation (2). But it's clear you can't: these equations describe qualitatively different behaviour that can't be reconciled. For example, in equation (1) the acceleration is an affine (constant + linear) function of the velocity, while in equation (2) the acceleration is a quadratic function of velocity. In equation (1) the acceleration is also different for different particles (it explicitly depends on the charge/mass ratio) which also contradicts equation (2).

Also, I'm sure someone must have told you at some point that the idea of curvature only in space immediately breaks Lorentz invariance, which is a known symmetry of electrodynamics.

And all of this is true because... you say so? This isn't an accurate description of quantum electrodynamics, if that's what you were aiming for.

What does that even mean? In quantum physics the wavefunction is not a wave in a classical field. Part of the point of my brief explanation was to make this clear: in quantum electrodynamics, the whole electromagnetic field gets its own wavefunctional. (Or at least, you can attribute it with a wavefunctional. As I said QFT textbooks don't describe it this way for practical reasons. They use the state/operator formalism instead.)

+1 for the photon (free photons don't decay into anything). +6 for the neutrinos and their antiparticles, which as far as I know are too light to decay into anything else. + possibly a few more I'd have to look up in my old particle physics notes, due to chirality. What's your point?

Er, yes it is, at least in the sense I was using the term "field configuration". The only reason I included that example was to clarify the sense in which I was using the term "field configuration".

I already know about the Aharonov-Bohm effect. You know the effect appears even if you treat the electromagnetic field as classical, right?​

Last edited: Jun 11, 2011
13. ### Farsight

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3,492
A reference frame is an abstract artefact associated with motion. You don't switch frames so much as change your state of motion. Let's say you were motionless, and now you move. Then you start to build up a more complete picture of the electromagnetic field. Think about what you do when you see an unfamiliar object. You move your head from side to side. Likewise you have to move to see the electromagnetic field for what it is. It doesn't have electric and magnetic parts so much as aspects.

Because I don't think you understand the electromagnetic field. If you did you wouldn't have talked about electric currents inducing curl, and you wouldn't have said I'm not convinced it makes too much sense to me either to RJ in post #43. It all makes perfect sense once you understand it.

Fine, but don't ever forget what Minkowski said:

"Then in the description of the field produced by the electron we see that the separation of the field into electric and magnetic force is a relative one with regard to the underlying time axis; the most perspicious way of describing the two forces together is on a certain analogy with the wrench in mechanics, though the analogy is not complete".

The important point is that the vector "fields" are mapping out the forces, not the electromagnetic field itself. People confuse them.

The evanescent wave is perfectly real. What it changes is exchange particles. It means there aren't any. There are still vacuum fluctuations as per the Casimir effect, but that's not the reason an electron and a positron move together. You end up with the photon being more fundamental than the electromagnetic field. Bit of a mess I know, but that's the size of it.

No. It doesn't contradict what we observe, we observe what we observe. It does however contradict what people say happens. It contradicts their understanding.

No problem with that.

I didn't mean to.

I just want to say that things don't move through spacetime. They move through space, over time, and we derive the mathemtaical "space" called spacetime from this.

Light moves through space, not spacetime. In a gravitational field, it's the spacetime that's curved. Not the space. In an electromagnetic field, it's the space that's curved.

Look at the time. I have to go.

14. ### przyksquishyValued Senior Member

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3,203
Er, a reference frame is just a coordinate system in some state of motion. You don't literally need to be moving to work things out in a moving reference frame.

You don't need to do that. You just need to move a test charge around. The force of the electromagnetic field on a charge is given by the Lorentz force I posted above:
$\bar{F} \,=\, q \bigl[ \bar{E} \,+\, \bar{v} \times \bar{B} \bigr] \,.$​
If you measure the force on a charge for a few different (in theory arbitrarily small) velocities, you can work out both $\bar{E}$ and $\bar{B}$.

Er, this was one of the early discoveries in the history of electrodynamics.

I was talking about an abstract point in the derivation of electrodynamics as a gauge theory that I had certain expectations about by analogy with general relativity, but otherwise haven't had much time to think about. This was a comment about material I personally doubt you've ever seen in your life. I certainly was not talking about anything as simple as curvature in space. I already know what that looks like and I'd recognise it pretty easily.

I'm not sure what point Minkowski was trying to make with the wrench, but the first part of that quote is perfectly compatible with what I've already said: the electric and magnetic fields are frame-dependent but together form an object called the electromagnetic field which is covariant. It's just how much of the electromagnetic field is "electric" and how much is "magnetic" that changes when you switch from one frame to another.

Er, not really. If you want to be pedantic about it, the vector fields $\bar{E}$ and $\bar{B}$ are one mathematical representation of all the measurable information about the electromagnetic field. That's why physicists tend not to make much of an (at least explicit) distinction between entities in reality and their mathematical representation in a theory: physics is limited by what is (at least in principle) measurable, so the vector fields contain all the information about the electromagnetic field that is relevant to the physics of electromagnetism.

What's the connection? The way I understand it, virtual particles are just short lived particles that aren't constrained to obey E[sup]2[/sup] = p[sup]2[/sup] + m[sup]2[/sup]. I don't recall the details but as I remember it this is just down to the uncertainty principle: a short lived particle can't have a well defined energy. Evanescent waves at the quantum level are associated with particles penetrating potential barriers via quantum tunneling, which is something different.

You're not making much sense here. As long as what a theory says doesn't conflict with what we observe, where's the problem?

Why should that make spacetime only "mathematical"?

You know that the reason the notion of "spacetime" is so associated with relativity is that the "space" and "time" parts alone are frame dependent? When you change reference frames in relativity the relation between the coordinates is described by a Lorentz transformation:
\begin{align} t' \,&=\, \gamma \bigl( t - \frac{v}{c^{2}} x \bigr) \\ x' \,&=\, \gamma \bigl( x - vt \bigr) \,. \end{align}​
The key point is that the space and time coordinates in one reference frame are a mixture of the space and time coordinates in another reference frame. In a universe where all the laws of physics are relativistic, you have no way of making any "absolute" distinction between space and time.

This is one of the points I was making in my previous post regarding interpreting electromagnetism as curvature in space: we already know that the Lorentz transformation above is a symmetry of electromagnetism, and "space" isn't invariant under that transformation. The spatial plane $x = \text{constant}$ gets mapped to $x' + vt' = \text{constant} / \gamma$, which is a plane in spacetime that isn't horizontal on a Minkowski diagram.

15. ### AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,702
Because each and every time you make a mistake (which is every post) I point it out. It's hopeless for you, yes, you can't go a post without having your BS exposed. When I start pointing out how much time and money and effort you've wasted you stop wanting to play and you go off in a huff. When I point out your 'work' can't model anything you don't want to talk about it, else you'd have to face up to the hypocrisy you show when you talk about string theory. At least it can provide some working models of phenomena.

You're wrapped up in your own little delusional world of ego and you're so daft you think no one can see it.

16. ### CptBorkValued Senior Member

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6,418
Sorry, but you don't seem to have any idea what LIGO is supposed to measure. It's not looking for gravitons, nor would making it 1000000X more precise give us any hope in hell of detecting one. Sure, you can say it's looking for enormous bunches of gravitons, but that's like saying we detect an enormous bunch of electrons when we use a household ammeter to measure electric current. LIGO is for testing General Relativity, not quantum gravity.

17. ### Magneto_1Super PrincipiaRegistered Senior Member

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295
Actually, believe it or not, earlier this year, I was a candidate for the Executive Director Job for the LIGO Lab at Caltech. Unfortunately, I did not get the job; and I am very, very disappointed about it, but life goes on!

Your Avatar reads "Grad Student", so when and if; and that is a big if, you have what it takes to get a PhD; only then do you earn the right to say to me, "Sorry, but you don't seem to have any idea what LIGO is supposed to measure"

Until then, I believe that you are asking me a question concerning "Gravitons." which I will entertain!

Mr. CptBork -Grad Student - keep studying hard!!

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Your statement about individual Gravitons is somewhat correct in that LIGO is measuring "Gravitational Waves" and not individual "Gravitons" specifically.

Graviton - Experimental Observation

"Unambiguous detection of individual gravitons, though not prohibited by any fundamental law, is impossible with any physically reasonable detector. The reason is the extremely low cross section for the interaction of gravitons with matter. For example, a detector with the mass of Jupiter and 100% efficiency, placed in close orbit around a neutron star, would only be expected to observe one graviton every 10 years, even under the most favorable conditions. It would be impossible to discriminate these events from the background of neutrinos, since the dimensions of the required neutrino shield would ensure collapse into a black hole.

However, experiments to detect gravitational waves, which may be viewed as coherent states of many gravitons, are underway (e.g. LIGO and VIRGO). Although these experiments cannot detect individual gravitons, they might provide information about certain properties of the graviton. For example, if gravitational waves were observed to propagate slower than c (the speed of light in a vacuum), that would imply that the graviton has mass"​

The concept of "Gravitons" is still an emerging class of study; and the big questions that are now emerging, ask, whether this is a totally "Quantum Mechanical" effect, or is this mainly a cosmological "General Relativistic" effect. I would say that Gravitational Waves are both a "Quantum Mechanical" effect and a cosmological "General Relativistic." This is why the term "Quantum Gravity" will need to be defined very clearly.

However, whether "Gravity Waves" are a "Quantum Mechanical" effect or mainly a cosmological "General Relativistic" effect, the source of the waves is the "Black Hole" Event Horizon and the Schwarzschild Radius.

And the Schwarzschild radius is discussed both in "Quantum Mechanical" and cosmological "General Relativistic" considerations.

Advanced LIGO - short discussion

"The gravitational wave "sky" is entirely unexplored. Since many prospective gravitational wave sources have no corresponding electromagnetic signature (e.g., black hole interactions), there are good reasons to believe that the gravitational-wave sky will be substantially different from the electromagnetic one. Mapping the gravitational-wave sky will provide an understanding of the universe in a way that electromagnetic observations cannot. As a new field of astrophysics it is quite likely that gravitational wave observations will uncover new classes of sources not anticipated in our current thinking."​

The hope is that "Gravitational Waves" travel at "Light Speed." And if this is true that "Gravitational Waves" travel at light speed, and then there would be a direct correlation between the photon energy and the "Planck Electromagnetic" constant. What, this would mean, is that there is a "Planck Graviton" constant.

My prediction is that "Gravitational Waves" travel at "Light Speed" and there is a "Planck Graviton" constant, which will have to be determined.

If the "Gravitational Waves" travel at speeds "slower" than "Light Speed", then this would mean that "Gravitational Waves" behave more like sound waves.

Mr. CptBork - Grad Student - Will you do me a favor? The next time you take a physics class, try and convince "Farsight - John Duffield" to join you!!

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Last edited: Jun 14, 2011
18. ### CptBorkValued Senior Member

Messages:
6,418
So what? I'm head chief of the US Atomic Energy Agency, and Subdirector of NASA's top secret program studying extra-terrestrial hallucinogens.

When you don't have an argument, you can always appeal to your non-existent authority. Classy. What you said about LIGO is a heap of nonsense, I don't need a Ph.D. to spell it out for you.

There you have it, your own link says it's preposterous to talk about detecting gravitons. The rest, like speed of gravitational waves, can be treated as a test of GR. The results would have implications for the properties gravitons would be required to possess, but still wouldn't produce any evidence for their actual existence.

More crap still. I could wave my hand and it would still produce gravitational waves. Doesn't have to be a black hole.

In what sense?

I'd actually prefer to convince you to take a physics class, it would help you get a sense of perspective on what physicists are actually doing.

19. ### Magneto_1Super PrincipiaRegistered Senior Member

Messages:
295
This I do not believe.

This I totally believe!!

This is highly, highly speculative. We have a much better chance measuring "Gravity Waves" if we know the source of the wave; which according to General Relativity is a "Black Hole" Event Horizon, with a Schwarschild radius.

Now if you say that your hand has a characteristic mass which we can assign a Scharzschild radius, then yes the waving of your hand will produce "Gravity Waves."

In the sense that if the "Gravitational Waves" travel at speeds "slower" than "Light Speed", then the waves would behave similar to a mechanical wave that is an oscillation of pressure transmitted through a solid, liquid, or gas. This would mean that there is a relationship between density and pressure of spacetime. This relationship, would also be affected by temperature, which would determine the speed of the "Gravity Wave" within spacetime.

Here, Here, get back to your studies, before you forget something that you thought you learned!!

20. ### TachBannedBanned

Messages:
5,265
It's not gonna happen

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21. ### James RJust this guy, you know?Staff Member

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37,808
So, what you're saying is that you sent in an application letter and you received back a rejection letter.

No surprises there.

22. ### AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,702
You sent in an application. Anyone can do that. I could apply for head of the World Bank or suggest myself for a Nobel Prize but that doesn't mean I'm in the running. Writing a letter doesn't count for anything unless you were in with a serious chance of getting it. Which brings me to what you said next :

You don't have a PhD. You don't have any research papers published. You don't have any right at all to talk down to Cpt for 'only' being a grad student. Remember how I kept calling you Mr Kemp after you tried a similar tactic with me (except I already have my PhD).

Except that Cpt was right, LIGO isn't searching for gravitons and even if it were a billion times more sensitive it couldn't search for them. This only shows how delusional your initial comment about being in the running for the director's job was, you don't even understand the basic physics and reason behind the experiment, let alone be in a sufficiently senior position in the academic/research community to be in with a chance of getting that job. I have more of a chance than you and I have zero chance.

At least he's studying. Better that than resorting to lying and trying to deceive people on internet forums into spending money to buy books he's written on subjects he doesn't understand, which are full of elementary mistakes and which couldn't pass peer review at any competent journal. Better studying than doing as you do.

Cpt could teach a class to you and Farsight. Though it would have to be a very basic one, both you and Farsight don't understand stuff I'd expect of a 1st year. Neither of you understand basic index notation and vector calculus, though you both profess otherwise (the evidence is against you though).

23. ### CptBorkValued Senior Member

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6,418
I've TA'd some electronics and electromagnetism, pretty fun stuff... nice mix of theory and hands-on application. I thought it was funny to see Magneto throw Ricci in there with the mix of "major contributors" to the theory, I presume that's because of the Ricci tensor and Ricci scalar. Yet if it's about throwing names around to pretend one understands the actual substance of their work, it's a shame he missed out on Christoffel and Bianchi.