# Gravity Waves

Discussion in 'Physics & Math' started by Little Bang, Sep 26, 2015.

1. ### danshawenValued Senior Member

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This is the core issue I've been trying t0 get across for over a year since joining SF in July, 2014.

Do we need a Higgs boson/field to explain inertia or not? Electrons, quarks, electroweak bosons, and their antiparticles all get their inertial mass from Higgs. Gluons don't. Color charge is not involved. Gluons and color charge interactions with quarks make up 98% of the mass of the atom, but without that other 2% -- the whole atomic structure flys apart. And really, who gives a damn whether gluons and color charge interactions are energy or mass, because it is all exactly the SAME THING. Since 1905. Since E=mc^2. What? Do you all just look at that equation, use it every day, and then just glaze over when someone asks you what it means?

Inertia means traveling in a straight line. Photons do that. So does matter. Unless acted upon by an outside force. Matter does that. So do photons. Same thing, as in SAME THING. There is only energy, and time.

Particle physicists seem to be in denial, if not of relativity, of what the foundational particle of the Standard Model means. They are too busy looking for the hidden and broken and super symmetries and the infinite joy of Hilbert spaces and Lie groups to bother about physical bindings to the reality they have no choice other than to inhabit, even if they are determined to ignore exactly what that means. Sometimes, the promethean board needs to be erased instead of publishing more of the same old, same old.

I'm sure the likes of John Dewey and all of academia built on his philosophy of teaching whatever is available to teach would approve. This doesn't mean that I need to. Sooner or later, in my own case MUCH later, you need to decide on your own which things are important, and which you may safely ignore. There simply isn't enough room between your ears to keep both and have time enough left to do anything that is really new with it. Keep E=mc^2, if only because Dewey never really learned it.

Last edited: Oct 21, 2015

3. ### Fednis48Registered Senior Member

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Thanks! Looks like a more in-depth discussion of the proposed experiment I linked above (at least in part). All the topology is a little too dense for me to sift through right now, but I'll try to take another crack at it later. Of course, if you have any insights that you could express in plain language, I'd love to hear those, too.

5. ### krash661[MK6] transitioning scifi to realityValued Senior Member

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i have typed this here before. but recently, i have re-iterated this agian here: https://www.physforum.com/index.php?showtopic=155461&view=findpost&p=841072
also, i have to continue the typing aspect of this. i may do so now since i referenced it here.
i'll notify when the second typing part is complete.
any ways, here's what i typed:
you have to be clear about some fundamental facts. the very first thing is that you must divide up the conception of the physical world; because each existence consist of different layers. lets say for simplicity's sake that it consist of a material illusion and this field of influence. certain physical conditions are associated only with the realm of the material.while other and more complicated conditions are associated only with this sphere of influence of the material world. your conception of the physical world is based upon a simple material illusion. that illusion is further subdivided into three elementary elements, or basic conditions of matter.a fourth and very important condition also exists,which you simply pay attention too more or less as you choose. it is the one bordering on the sphere of influence or plasma realm. for you, the theory for control transformation or an elevation of the frequency of matter in the stabling existence of this fourth aggregate condition of matter is not very common; for it exist at a very primitive level. as on a side, there's simply five states of matter but the post plasma state would really be going too far, and it would only serve to confuse you.besides, it's not necessary for understanding of the basic theory.it is connected with diverse phenomenon which you would characterize as paranormal. now back to the essentials, plasma. now, with plasma i don't mean just "hot gas" (as the concept is generally simplified), by normal people but rather i mean a higher aggregate condition of matter.the plasma state of matter is a special form of matter which lies between its real existence and the sphere of influence; that is, a complete loss of mass and pure accretion of energy of various form whenever matter is "pushed or shoved"[there's no use of a word for 'pushed or shoved'].the fourth state of matter is very important for certain physical conditions which can be used, for example, to to generate antigravity. essentially, in the world of real physics, there are no bipolar forces, but rather only "observer dependent reflective behavior" of a single, large unified force at different levels.

[part 2 next]

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7. ### danshawenValued Senior Member

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High energy plasma dynamics are where all particles of bound energy come from.

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8. ### The GodValued Senior Member

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My paragraph reading ability just scanned it in one shot and concluded that its a great stuff from you. Wow, you could type so much, so what if it was done offline and then pasted afterwords......I am eagerly looking forward to part 2.

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9. ### krash661[MK6] transitioning scifi to realityValued Senior Member

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no. this is actual work. i'm only typing it from my mind. this is why i have not done the second part. i hate writing/typing. i'm lazy. but i'll do it; i'm receiving request on both sites to do so. i'll do it later today/night.

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10. ### Farsight

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No. The photon in the box demonstrates E=mc² nicely. Open the box and a radiating body loses mass. Electron-positron annihilation is like opening one box with another, whereupon each is a body that loses mass. All of it. To be blunt, the Higgs mechanism contradicts E=mc².

Not when they're in the box. Then they go round and round. Think of photon momentum as resistance to change-in-motion for a wave moving linearly at c. Think of electron mass as resistance to change-in-motion for a wave going round and round at c. It's that simple.

11. ### Farsight

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It isn't nuts, and nor is it an objection to conditional probability. He said what he said, you're reading way too much into it.

Scott Aaronson spent a whole pile of time badmouthing Christian, and when you look carefully for his case, it's all smoke and mirrors.

12. ### krash661[MK6] transitioning scifi to realityValued Senior Member

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you're simply pathetically hilarious.

13. ### Farsight

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The photon has a non-zero inertial mass and a non-zero active gravitational mass. It has no rest mass because it isn't at rest, and you can't slow it down like you can slow down an electron. But when you catch it in a mirror-box, it increases the mass of the system. It's still going at c, but it's going round and round at c, so it's effectively at rest. See Light is Heavy by van der Mark and (not the Nobel) 't Hooft: http://arxiv.org/abs/1508.06478 and think of the electron as a 511keV photon in a box of its own making.

If you've got a 511keV photon going round and round at c, you cannot also make the whole thing move linearly at c as well. Because to do that light would have to go faster than light.

Formation of matter is pair-production. Think of it as making light go round and round. A gravitational field slows it down. That's why it curves. See Einstein talking about that :

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14. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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Where does the energy come from that speeds the light back up to c when it leaves the gravitational field?

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I think we'll start calling this Farsight nonsense the "Gospel of Farsight"
Total fairy tale it be.
As I said in post 148
Let me state some other facts for you......
Photons certainly as I said, have no rest mass.
But they do have momentum, and solar sails are an example of that momentum being put to use.
Photons also by definition never slow down.They are always moving at "c"
The "deflection" of a light ray passing the Sun for example, is simply due to the fact that light/photons travel in geodesics....ie they follow the curved/warped path of spacetime created by the massive object present.

http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

What is the mass of a photon?
This question falls into two parts:

Does the photon have mass? After all, it has energy and energy is equivalent to mass.
Photons are traditionally said to be massless. This is a figure of speech that physicists use to describe something about how a photon's particle-like properties are described by the language of special relativity.

The logic can be constructed in many ways, and the following is one such. Take an isolated system (called a "particle") and accelerate it to some velocity v (a vector). Newton defined the "momentum" p of this particle (also a vector), such that p behaves in a simple way when the particle is accelerated, or when it's involved in a collision. For this simple behaviour to hold, it turns out that p must be proportional to v. The proportionality constant is called the particle's "mass" m, so that p = mv.

In special relativity, it turns out that we are still able to define a particle's momentum p such that it behaves in well-defined ways that are an extension of the newtonian case. Although p and v still point in the same direction, it turns out that they are no longer proportional; the best we can do is relate them via the particle's "relativistic mass" mrel. Thus

p = mrelv .
When the particle is at rest, its relativistic mass has a minimum value called the "rest mass" mrest. The rest mass is always the same for the same type of particle. For example, all protons have identical rest masses, and so do all electrons, and so do all neutrons; these masses can be looked up in a table. As the particle is accelerated to ever higher speeds, its relativistic mass increases without limit.

It also turns out that in special relativity, we are able to define the concept of "energy" E, such that E has simple and well-defined properties just like those it has in newtonian mechanics. When a particle has been accelerated so that it has some momentum p (the length of the vector p) and relativistic mass mrel, then its energy E turns out to be given by

E = mrelc2 , and also E2 = p2c2 + m2restc4 . (1)
There are two interesting cases of this last equation:

1. If the particle is at rest, then p = 0, and E = mrestc2.
2. If we set the rest mass equal to zero (regardless of whether or not that's a reasonable thing to do), thenE = pc.

In classical electromagnetic theory, light turns out to have energy E and momentum p, and these happen to be related by E = pc. Quantum mechanics introduces the idea that light can be viewed as a collection of "particles": photons. Even though these photons cannot be brought to rest, and so the idea of rest mass doesn't really apply to them, we can certainly bring these "particles" of light into the fold of equation (1) by just considering them to have no rest mass. That way, equation (1) gives the correct expression for light, E = pc, and no harm has been done. Equation (1) is now able to be applied to particles of matter and "particles" of light. It can now be used as a fully general equation, and that makes it very useful.

Is there any experimental evidence that the photon has zero rest mass?
Alternative theories of the photon include a term that behaves like a mass, and this gives rise to the very advanced idea of a "massive photon". If the rest mass of the photon were non-zero, the theory of quantum electrodynamics would be "in trouble" primarily through loss of gauge invariance, which would make it non-renormalisable; also, charge conservation would no longer be absolutely guaranteed, as it is if photons have zero rest mass. But regardless of what any theory might predict, it is still necessary to check this prediction by doing an experiment.

It is almost certainly impossible to do any experiment that would establish the photon rest mass to be exactly zero. The best we can hope to do is place limits on it. A non-zero rest mass would introduce a small damping factor in the inverse square Coulomb law of electrostatic forces. That means the electrostatic force would be weaker over very large distances.

Likewise, the behavior of static magnetic fields would be modified. An upper limit to the photon mass can be inferred through satellite measurements of planetary magnetic fields. The Charge Composition Explorer spacecraft was used to derive an upper limit of 6 × 10−16 eV with high certainty. This was slightly improved in 1998 by Roderic Lakes in a laboratory experiment that looked for anomalous forces on a Cavendish balance. The new limit is 7 × 10−17 eV. Studies of galactic magnetic fields suggest a much better limit of less than 3 × 10−27 eV, but there is some doubt about the validity of this method.
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Q: How can photons have energy and momentum, but no mass?
Posted on September 8, 2010 by The Physicist
Physicist: Classically (according to Newton) kinetic energy is given by

and the momentum is given by

, where m is the mass and v is the velocity. But if you plug in the mass and velocity for light you get

. But that’s no good. If light didn’t carry energy, it wouldn’t be able to heat stuff up.

The difficulty comes from the fact that Newton’s laws paint an incomplete (and ultimately incorrect) picture. When relativity came along it was revealed that there’s a fundamental difference in the physics of the massive and the massless. Relativity makes the (experimentally backed) assumptions that: #1) it doesn’t matter whether, or how fast, you’re moving (all physical laws stay the same) and #2) the speed of light is invariant (always the same to everyone).

Any object with mass travels slower than light and so may as well be stationary (#1).

Anything with zero mass always travels at the speed of light. But since the speed-of-light is always the speed-of-light to everyone (#2) there’s no way for these objects to ever be stationary (unlike massive stuff). Vive la différence des lois! It’s not important here, but things (like light) that travel at the speed of light never experience the passage of time. Isn’t that awesome?

The point is: light and ordinary matter are very different, and the laws that govern them are just as different.
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and again.......
http://www.ibiblio.org/lunar/school/library/massphot.html

What Is The Mass Of A Photon?
From the sci.physics FAQ

What is the Mass of a Photon?

updated 24-JUL-1992 by SIC

original by Matt Austern

Or, "Does the mass of an object depend on its velocity?"

This question usually comes up in the context of wondering whether photons are really "massless," since, after all, they have nonzero energy. The problem is simply that people are using two different definitions of mass. The overwhelming consensus among physicists today is to say that photons are massless. However, it is possible to assign a "relativistic mass" to a photon which depends upon its wavelength. This is based upon an old usage of the word "mass" which, though not strictly wrong, is not used much today.

Oh, and back to photons: people sometimes wonder whether it makes sense to talk about the "rest mass" of a particle that can never be at rest. The answer, again, is that "rest mass" is really a misnomer, and it is not necessary for a particle to be at rest for the concept of mass to make sense. Technically, it is the invariant length of the particle's four-momentum. (You can see this from Eq. (4).) For all photons this is zero. On the other hand, the "relativistic mass" of photons is frequency dependent. UV photons are more energetic than visible photons, and so are more "massive" in this sense, a statement which obscures more than it elucidates.
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Finally......
https://www.quora.com/How-can-photons-have-no-mass-and-yet-still-have-energy-given-that-E-mc-2

How can photons have no mass and yet still have energy given that E=mc2?

That's correct. m is the rest mass, the one you'd observe if you were in the same reference frame. A photon has no reference frame; it has zero rest mass.

The formula E=mc2 is actually an abbreviation for objects that aren't moving, i.e. objects with no momentum. This is sometimes written as E0, the rest energy, and for clarity emphasize that we mean the rest mass with m0:E0=m0c2.

While photons have no rest mass, they do nonetheless have momentum, given by p=h/λ. The momentum of a photon is kind of an unusual concept, since you're used to thinking of momentum as simply mv, which would be zero. Let's just say that photons have an inherent momentum that's different from the kind of momentum that massive objects have. (It's really exactly the same, but different from the kind you're familiar with in classical Newtonian physics). This is well-confirmed by experiment: it's what makes solar sails work.

For the full energy of a moving particle, with or without mass:

E=m20c4+p2c2−−−−−−−−−−√

For non-moving objects, p=0, and you can see that it easily reduces to the form you're used to. For photons, with m=0, it reduces E=pc, and if you fiddle with that and the definition for p above you'll see that you get the equation you listed in your question.

For objects with mass that are moving close enough to the speed of light for their momentum to make a difference, you do need the full form of the equation.

18. ### quantum_waveContemplating the "as yet" unknownValued Senior Member

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Last edited: Oct 23, 2015

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21. ### danshawenValued Senior Member

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In what way is this a contradiction, Farsight? There is no guarantee that ANY particle (electron, positron, what-have-you) will remain indefinitely in a state of bound energy. Higgs giving inertial mass to both the particle and anti-particle (as well as itself) means only that both particles have mass, which makes it impossible for either of them to exceed the speed of light until or unless they annihilate each other. If either of them could move at the speed of light, they would not be able to briefly form an atom of positronium, would they? Also, they would likely have no charge either, but that's another story.

22. ### danshawenValued Senior Member

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Some photons NEVER leave the gravitational field of a black hole event horizon. I doubt it "slows the photon down" very much in terms of "proper" time.

For photons that do not achieve such orbits, the energy of the photon will be slightly increased (Doppler shifted) as a result of the interaction.

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From any remote FoR the photon appears redshifted to infinity, and is never actually seen to cross the horizon. From a local frame with the photon at just this side of the EH, it will arc back to succumb to the EH, except when emitted directly radially away, when it will hover forever just above the EH, never quite escaping and never succumbing to the EH.
This can best be explained in the water fall model of the BH, [sometimes called the river model] which sees the spacetime coinciding with the EH "falling in" at "c" while the photon is moving away at "c"
http://arxiv.org/pdf/gr-qc/0411060.pdf
also explained here.......