travelling at the spead of light

[d.m.falk]

Even the most basic of mathematics will tell you that anything multiplied or divided by zero is zero.

Thanks d.m.f... I figured that anything that was multiplied by zero was zero... I just wasn't sure if the presence of a term that is nonsenical (ie. gamma at lightspeed) rendered the whole equation moot.

However, is something that is DIVIDED by zero, zero?

If you devide by 2, you have two equal products totaling the original value. If you devide by 1, you have a single product, equal to the original value. If you divide by zero, you have no products divided from the original. Anything divided by 0 is 0. There is no product.

GoHeelsWeeeehoooo: will we reach other stars? are they too far? who's to say? even if we can never travel faster than the space shuttle, there is

[snip]

The space shuttle is actually very slow, even compared to the space probes we have out there already. The "New Horizons" Pluto-Kuiper mission is projected to have the fastest probe ever launched from Earth, in which they expect to reach Pluto within 5 years after launch.

The current fastest right now belongs to Voyager 1.

d.m.f.

[d.m.falk]

We know now that all particles have mass, and all forces do not. Even the electron's mass was known well before I was born (of which I'll be 40 in two months). Even quarks have mass. Gluons do not, however, because they're not particles, but rather forces.

(See, it doesn't mean Einstein's theory is wrong, but how it's interpreted that's gotten sticky.)

Is not the photon simply the means by which the electromagnetic force is expressed?

Nope. It's defined as a quantum particle, with the characteristics of both a wave and a particle- Even Einstein realised this some years after the Theory of Special Relatiivity he proposed.

So would that mean, if the statements in the above quote are true, that a photon does NOT have mass?

Photons have observed, measured masses, but the value is far, far less than that of even an electron.

d.m.f.

[BlindedByTheLight]

If you devide by 2, you have two equal products totaling the original value. If you devide by 1, you have a single product, equal to the original value. If you divide by zero, you have no products divided from the original. Anything divided by 0 is 0. There is no product.

Well, I'm no mathematician (I believe, at least, THAT much is clear) but it seems to me that "0" and "no product" are not the same things at all. "Zero" is the absence of any units. "No product" means the entire issue of how many units there are is N.A. (or not applicable) since the question is non-sensical.

In any case, using your analogy, you can see that the result of dividing by zero yields, if anything at all, a number that is as far from zero as possible. If you divide say, 10, by 1, you get 10 again. If you divide 10 by 0.5, you get 20, a larger number. If you divide 10 by a small number yet (say 0.25) you get even a larger number.

As the divisor (bottom number) approaches zero, the result gets ever higher and higher (ie. FURTHER from zero).

As the divisor gets infinitely close to zero, the result gets infinitely large as well. Ultimately, when the divisor reaches zero, the result becomes non-sensical. What it is, may becomes a matter of semantics. What it most certainly is NOT, however - is zero.

Any real mathematicians out there wanna chime in? I hate sounding so confident about something I don't really have any clue about. A reality check would be nice.

TRAVEL TO OTHER STARS:

Yeah, the space shuttle is kinda slow, isn't it? But, my point was, it's the fastest thing we have that can carry people. I like the optimism about the 22nd century, though!

Nope. It's defined as a quantum particle, with the characteristics of both a wave and a particle- Even Einstein realised this some years after the Theory of Special Relatiivity he proposed.

Photons have observed, measured masses, but the value is far, far less than that of even an electron.

d.m.f.

I will happily defer to your expertise on the matter - since I've always been fuzzy on that point. However, whether or not a photon is defined as a quantum particle... doesn't change the fact that it carries the EM force... or it does not. I was under the impression that it does. If I am mistaken, ignore this next line...

If I am NOT mistaken, and the photon (aka quantum particle) DOES carry the EM force, how does that reconcile with your statement:

We know now that all particles have mass, and all forces do not.

...and the photon having mass? Does it carry the EM force or not?

[d.m.falk]

If you devide by 2, you have two equal products totaling the original value. If you devide by 1, you have a single product, equal to the original value. If you divide by zero, you have no products divided from the original. Anything divided by 0 is 0. There is no product.

Well, I'm no mathematician (I believe, at least, THAT much is clear) but it seems to me that "0" and "no product" are not the same things at all. "Zero" is the absence of any units. "No product" means the entire issue of how many units there are is N.A. (or not applicable) since the question is non-sensical.

Actually, "no product" is just that, since it's an absolute, and not a relative, statement.

In any case, using your analogy, you can see that the result of dividing by zero yields, if anything at all, a number that is as far from zero as possible. If you divide say, 10, by 1, you get 10 again. If you divide 10 by 0.5, you get 20, a larger number. If you divide 10 by a small number yet (say 0.25) you get even a larger number.

As the divisor (bottom number) approaches zero, the result gets ever higher and higher (ie. FURTHER from zero).

As the divisor gets infinitely close to zero, the result gets infinitely large as well. Ultimately, when the divisor reaches zero, the result becomes non-sensical. What it is, may becomes a matter of semantics. What it most certainly is NOT, however - is zero.

This can be explained by the fact that the ever-larger product is what you need to multiply the fractional divisor (ie; 1/0.25) to equal 1!

(eg: 1/0.25=4)

As zero is NOT a fractional divisor, but the value of null, your product simply does not exist. There is no product! It's that simple.

Nope. It's defined as a quantum particle, with the characteristics of both a wave and a particle- Even Einstein realised this some years after the Theory of Special Relatiivity he proposed.

Photons have observed, measured masses, but the value is far, far less than that of even an electron.

I will happily defer to your expertise on the matter - since I've always been fuzzy on that point. However, whether or not a photon is defined as a quantum particle... doesn't change the fact that it carries the EM force... or it does not. I was under the impression that it does. If I am mistaken, ignore this next line...

If I am NOT mistaken, and the photon (aka quantum particle) DOES carry the EM force, how does that reconcile with your statement:

Short answer: No. Magnetism is a force not carried by any particle, but are emitted by elecvtrons. (Some atoms are more magnetic than others.)

We call it the "electromagnetic spectrum" because photons are the products of the interactions between electrons and their magnetic forces.

d.m.f.

[Cham]

Geez, what a pack of bull s...

[d.m.falk]

Geez, what a pack of bull s...

Thank you for your exposition.

d.m.f.

[t00fri]

...
Photons have observed, measured masses, but the value is far, far less than that of even an electron.

d.m.f.

Please stop telling people all this blunder! Apparently, you have vaguely heard about these things before, but understood almost nothing.

While virtually every second of your statements in this thread is incorrect or strongly misleading, let me just pick your above claims about the photon mass as an example.

The world authoritative experimental knowledge about the photon mass is collected here

http://pdg.lbl.gov/2005/listings/s000.pdf

Please observe the '<' sign in ALL entries of the first column! This means that there has NEVER been any evidence for a non-vanishing photon mass, but rather that all measurements have been compatible within their errors with a vanishing mass. According to the laws of statistics one may then convert the experimental uncertainties into an upper experimental bound for the photon mass. That's what is quoted in the above official reference.
The most serious upper bound for the photon mass is thus

Code: Select all

 m(photon) < 6  10^(-17) eV  (electron Volt)


Experimentally, one can of course NEVER prove that the photon mass is EXACTLY ZERO, since each experiment also involves uncertainties.

But unlike other incorrect claims in this thread, the theoretical evidence for an exactly massless photon (and massless gluons) is overwhelming!

Classical electrodynamics, quantum electrodynamics and the tremendously successful Standard Model of elementary particles (Nobel prize...), rest crucially on the validity of a symmetry, the so-called local gauge symmetry that REQUIRES the exact masslessness of the photon and the gluons! Even a tiny photon (gluon) mass would completely destroy the incredibly successful predictivity of our basic particle theory. A non-vanishing photon mass would destroy the "renormalizability" of the theory, which in simple words means that certain infinities arising at the quantum level would not cancel!

I pick just two of these successful predictions of the Standard Model, the anomalous magnetic moment of the muon and electron , to illustrate the enormous level of theoretical and experimental precision I am talking about:

Code: Select all

Theory:   a_mu = (11659182.8 +-7.3)* 10^(-10)
Experiment: a_mu = (11659208+-5+-4)*10^(-10)


Using the value of the electromagnetic fine structure constant as measured from the Quantum Hall effect,

Code: Select all

alpha =1/137.03599911(46)


the anomalous magnetic moment of the electron is prediced
within Quantum-Electrodynamics as

Code: Select all

a_e theor. = 0.0011596521535(240)


in incredibly good agreement with the experimental precision value

Code: Select all

a_e_exp. = 0.0011596521859(38)


This success and many many others would have to be accidental, if the photon had a tiny non-vanishing mass!

Bye Fridger

[Spaceman Spiff]

I knew it! I leave this topic over night, and it get filled with appalling errors, muddled explanations and mysticism!

Blinded: you've been fed rubbish. I'll post clarity separately.

The Hall of Shame:

First off, me! (oh the shame!):

Particles of light, photons, do travel at the speed of light. They have no kinetic energy because they have no inertial mass, ...

I should have written 'rest' mass, not 'inertial' mass. Bad!

But photons do have mass, with its observable effects- Recoil, decay, gravitationally lensing or bending, and resistance to medium density. (That is, light slows down in an atmosphere, and moreso in fluids and solids.)

The fact that photons are observed to have momentum is not inconsistent with Einstein's Special Relativity theory. It simply holds that phtotns have no rest mess, and that it is the energy of the photon that contains its mass. Gravitational bending of light is stated to be due to light following goedesic curves through warped spacetime, influenced by the mass of a body different from the photon (e.g., a galaxy): it is not the mass of the photon (that is, its energy) that causes the photon to be pulled by gravity. If one says this, one returns to the Newtownian model, and one is back to having the Principle of Equivalence unexplained again, something which Einstein's General Relativity does admirably.

What does 'resistance to medium density' mean? The slowing of light in media is nothing to do with photons having mass. It is to do with an increased electric permittivity of the medium.

If photons have no mass, they cannot have any energy, ...

In which you seem to follow the masses in misunderstanding that only rest mass is the subject of the E = mc?? equation. It is the energy of the photon that has mass.

The speed of light is also not infinite, but actually rather slow; however, our present understanding of physics has difficulties grasping that it isn't, and may, with much reluctance, require revision of standard theories to account for observed anomalies that appear to defy long-held theory.

Speak for yourself! What is the difficulty with modern physics over a finite speed of light? This has been know since R??mer. What are the observed anomalies?

And, oh yes- Before I forget:
http://www.phys.uni.torun.pl/~jkob/physnews/node35.html

If you bother to read and understand what you found, you'd notice that this paper is posting a reduced upper limit to photon rest mass from experimental observations. Fridger has put the exact right notion to you about how to deal with evaluating the theoretical notion of photons having zero rest mass. This link you post is a precise example of that. It does not say that photons are found to have finite (non-vanishing) rest mass.

And I never said his theory was fundamentally wrong; only that the conclusion is incorrect.

What is the conclusion that is incorrect (with Einstein's Special Relativity theory)?

And thus you hit on what I was saying in the first place: The continuous attempts to fit the anomalous nature of photons within Einstein's Theory of Special Relativity, rather than trying to quantise the nature of photons themselves. A better way of putting it is that Einstein was on the right track, but came up with the wrong conclusion. Yes, photons have energy, and thus they have mass. This is observable, and has been measured. However, it conflicts with Einstein, and just as you observed, you can't really have something from nothing- The exchange between mass and energy exists on all levels. A photon's mass is just marginally above that of a quark's. Heck, I'd even say that it's best to state that mass and energy are just two states of matter, which would be accurate, giving the mutual exchange that usually results between both, and the energy for the exchange is subtracted from the total.

You don't list the anomalies. Thus I see no conflict, except your misunderstanding of relativity compared to what Einstein meant. The photon's mass comes from its electrical energy - it is carrying electrical energy from one place to another. The energy has mass. You use Einstein's E = mc?? equation to find out what that is, and that is the mass of the photon. What on Earth you mean by 'A photon's mass is just marginally above that of a quark's.'? There are six quarks and they have extremely different masses. You can make a photon much heavier or lighter simply by increasing or decreasing the photon's frequency, hence energy. It's arbitrary.

Even the most basic of mathematics will tell you that anything multiplied or divided by zero is zero.

If you devide by 2, you have two equal products totaling the original value. If you devide by 1, you have a single product, equal to the original value. If you divide by zero, you have no products divided from the original. Anything divided by 0 is 0. There is no product.

Gak! You achieve F-!! A blatant crime against scientific orthodoxy!!! Back to school for indoctrination again!!!!

No. If you divide by 2, you have two equal products, each equaling half the total, totalling the original value. So, if you divide by half, you have half a product equaling twice the value of the original, totalling the original value? You get the total by multiplying the half with the twice, or multiplying the twice with the half? Why allow division by two, not half? So there is a product, because multiplication is the inverse function of division. But, if you divide by zero, you have zero products each equaling infinity times the total. This does not necessarily equal zero.

Don't believe me? So, tell me why does sinc(x) = 1 when x = 0, and not zero ?

Surprise: zero times something only equals zero if something is a number. Infinity is not a number, it's a concept. Zero times infinity is not necessarily zero. Otherwise you can all go back to that Equation 1 of Cham's and be happy that you can set m0 to zero, and pull out zero kinetic energy for all photons. The conclusion is true, but this is not the mathematical proof of it.

And. A number divided by zero may be zero, it may not be zero. All that oddities like this tell you is that you have to use a different way to find the result... Look up sinc(x) = 1 for x = 0.

Actually, "no product" is just that, since it's an absolute, and not a relative, statement.

I bet calculus is beyond you: "so if I divide the area under the curve into strips of zero width, then there's none of them, and I can't add them up into the finite area under the curve? <scratches head>".

Calculus is an excellent mathematical method that divides an area under a curve into an infinite amount of infinitesimally thin strips and adds them up to a finite area. But not necessarily zero.

We know now that all particles have mass, and all forces do not.

Gak! Again! Pointless comment: energy has mass, whatever type it is: kinetic, gravitational potential, electrical, or that 'frozen' kind referred to by the term 'rest mass'. All particles have energy in some form. This energy has mass. All particles have mass. The point is: do all particles have 'rest mass'. Answer: to date: 'NO!'. Example: photon. Has mass (due to its energy), has no rest mass (thus travels at c). Photons are the force carrier particles of electromagnetism. Which brings me to:

Gluons do not, however, because they're not particles, but rather forces.

Gak! Yet again! Gluons are particles: they are the force carrier particles of the strong nuclear force. They have mass! They even have rest mass. That's why they decay and don't get far outside the nucleons: they don't travel at the speed of light, don't have time dilated to zero, and we can catch them decaying.

When Einstein made his special relativity theory, the photon wasn't yet known at that time. Light was only of a wave nature, until Einstein himself suggested its quantum nature.

Nope. It's defined as a quantum particle, with the characteristics of both a wave and a particle- Even Einstein realised this some years after the Theory of Special Relatiivity he proposed.

Nooo! Both wrong!! Einstein published his paper on the Photoelectric Effect in 1905, the same year he published Special Relativity. In it, he proposed light as a quantum particle right there and then. It is that very photoelectric effect paper he was awarded the Nobel Prize for. Please!

Photons have observed, measured masses, but the value is far, far less than that of even an electron.

Gaa-a-ak! Have you not heard of Compton scattering? The measured mass of high energy gamma rays far exceeds the (rest) mass of an electron. Why do you supposed the annihilation of electrons with positrons generates two gamma rays each of energy 511keV? And do you suppose it's a coincidence that the summed mass of the energy of those two gamma rays happens to equal the summed rest masses of the electron and positron?

Short answer: No. Magnetism is a force not carried by any particle, but are emitted by elecvtrons. (Some atoms are more magnetic than others.)

We call it the "electromagnetic spectrum" because photons are the products of the interactions between electrons and their magnetic forces.

&^%$??&*9! NOOOO! Magnetism is predicted by Special Relativity as an apparent reduction of an electron charge relative to its mass (by the Lorentz transform) when it has a relative velocity to other charged particles! Magnetism is therefore just a relativisitic effect of electricity. Maxwell's equations are predicted by Einstein's theory of Special Relativity.

No wonder Fridger has written in CAPITALS LETTERS AGAIN!

Today, relativity theory strongly suggest (but doesn't prove) that photons should travel at the velocity "c". But the theory can easily accept the other way (v < c) if experiment gives a tiny mass to the photon. It just happened, historically, that Einstein discovered the "c" constant inscribed in vacuo using the properties of light, but this is just an happy coincidence.

No no no! This is completely back to front. Einstein took it as the starting assumption to the theory of Special Relativity that light would always appear to travel at the same speed in a vacuum, c. It does not predict it, it assumes it. Special Relativity makes predictions from the assumption. Details about Cherenkov radiation for when matter exceeds the local speed of light in media is irrelevant. The value of c comes from the values for the electrical permittivity and magnetic permeability of free space: c = 1 / ??? ( ?µ0??0 ). Nothing more, nothing less. Einstein, Maxwell and others were well aware of this.

Equation (1) DEFINES what is "relativistic mass" of a particle.

It really means this: if you observe and weigh a particle 'at rest' (relative to yourself*) and find it to have a mass m0 (the 'rest mass'), then if you observe and weigh it again when it travels at speed v relative to you, it will appear to have a total mass ('relativisitic mass') of m, which is m0 modified by the Lorentz transform. This is because according to Special Relativity, the kinetic energy Ek shoved into the particle with non-zero rest mass (a 'massive particle') also has mass: m = Ek/c??.

Equation (2) ... WHEN APPLIED TO A MASSIVE PARTICLE such an electron.

No. It applies to all forms of energy and mass. A photon has a quantum of energy, E, the energy has a mass of m = E/c??. That gives the photon its apparent mass, and momentum.

Equation (3) is exact for any massive particle. It doesnt apply to a photon, because of equation (1) !

Equation 3 does apply to a photon, and Equation 6 proves it. Equate Equations 2 and 5, rearrange for m, substitute into Eq. 3 while setting v = c, you get Eq. 6. QED.

Equation 4: you're making this up!

Equations 5 and 6, OK.

Equation 7. No, the equation is correct, especially since you do as Einstein suggests and use the Lorentz transform: substitute ??m0 for m. It would have been wrong if you put just m0 in there instead. Bear in mind this equation is a definition, not a 'law of physics': Newton's first and third laws of motion are just definitions, only the second is a 'law'.

The only error is the interpretation of m_0. It isn't "Mass initial of something". It's the "proper mass", "true mass", "intrinsic mass" of the particle.

No. m0 does have a very precise physical meaning: it is the special case of what a particle's mass is measured to be by you when it has zero relative velocity to you. For that reason, it has been called the 'rest mass' even though you could not tell whether you and it are truly at absolute rest. It's also the massof an amoutn of energy one would think is tied up in a non-kinetic form in a particle. Yet, kinetic energy and rest mass energy actually change when both the particle and measurer are travelling together at a different speed. If a laboratory was rushing at 87% the speed of light from we on Earth, and both the lab and we measured the mass of an electron, the lab would find the mass of the electron to be the same as its published rest mass, because as far as the lab is concerned, the electron is at rest. We would find the electron to be twice as heavy, because the kinetic energy of that electron now has a mass equal to the rest mass of an electron.

Other matters I should tidy up...

Photons DO have kinetic energy : K = h f, where "f" is the frequency and "h" is Planck's constant. However, The relativistic Newton equation do not apply to them as photons do not have an intrinsic mass.

No, this energy is not kinetic, it is electromagnetic. A photon carries an amount of electromagnetic force, hence electromagnetic energy.

As for the 'relativistic Newtonian' equation for kinetic energy, is does not apply to photons simply because the Lorentz transform term: 1 / ??? ( 1 - v?? / c?? ) is undefined when v = c. That is all.

The photon interference is a pure quantum phenomenon and has nothing to do with special relativity (to which photons are just particles).

Are you thinking of interference patterns with Young's slits? What is it about quantum mechanics that explains why photons can still interfere when the light level is turned down so low that there can only be one photon at a time between the slits and the image? I'm curious about knowing whether the 'relativisitic' explanation of timelessness has any feasibility behind it.

We know it's not just the electrical charge, it's not the spin, it's ... something else.

Great! How do we know this?

Spiff.

* How can you tell if you really are stationary? You can't. I suspect that the reason why the term 'rest mass' is being phased out is because you cannot prove you are at (absolute) rest, so it's misleading in that sense. Yet, no matter how fast you and the particle travel (but you always travel at less than c), so long as your relative velocity is zero, its measured mass will always appear to you to be that rest mass. Remember, v is relative to your own reference frame, not absolute stop! That's why the theory is called relativity theory.

[selden]

How can you tell if you really are stationary? You can't.

I think perhaps you've forgotten about the cosmic microwave background. You can tell if you're stationary relative to it. This has many subtle implications that I don't understand.

[Spaceman Spiff]

Selden,

I haven't forgotten about the CMB, and I specifically have it in mind when I talk of knowing whether one is absolutely stationary: it doesn't qualify.

Before relativity there was the notion of an absolute reference frame (which the 'ether' was supposed to reveal). As you say, you could tell if you are stationary relative to the CMB, which we are not. Even then, the CMB is actually moving away from us in all directions, or equally we're moving from it, so the CMB is redshifted to microwaves.

And even then "the Earth moves, all the same". We expect to appear stationary from the Cosmological Principle: that this CMB redshift must appear anisotropic. So, we trim out all the special motions, but in the end really we just create a tautology.

So, this is by no means a substitute for the ether or an absolute reference frame, because we cannot tell if the entire observable universe also has a residual lateral motion within a larger as yet unobservable universe...

What subtle implications come to mind?

Spiff.

[Cham]

Spaceman Spiff,

there are so much misunderstanding and confusion in your response, I will not go into all the details here. This is just too discouraging.

[t00fri]

Perhaps one slightly more general sentence about my general attitude in this game:

I am very happy and always ready to spend all my time (and more) in trying to help and explain, if I read physics posts from people

--that convey their true interest in these (admittedly tricky) matters
--that do not pretend expertise if there is none.

Conversely, I tend to get VERY nervous if I have to read lots of posts, written in an authoritative manner, where I can see immediately as a very experienced professional that any understanding and thorough knowledge is actually lacking...

Just some such examples:

alot of the misconception about the status of photonic mass is due to the scientific reluctance to say that Einstein's Theory of Special Relativity is incorrect on the matter.

Einstein was on the right track, but came up with the wrong conclusion. Yes, photons have energy, and thus they have mass. This is observable, and has been measured.

Even the most basic of mathematics will tell you that anything multiplied or divided by zero is zero.

We know now that all particles have mass, and all forces do not.
...
Gluons do not, however, because they're not particles, but rather forces.

etc...

Bye Fridger

[t00fri]

Cham,

just came accross this one:

But the suggestion comes more strongly from Maxwell theory (which is fully relativistic) and its conformal (scale) invariance. Conformal invariance is a very nice mathematical symetry, but it would be totally ruined if the photon had a mass.

Although you are right in stating that Maxwell's theory is conformally invariant if the photon mass is exactly zero, this is unfortunately not a good reason for its masslessness.

Conformal invariance is necessarily broken at the quantum level through the process of regularization of quantum divergences and the subsequent process of renormalization. As you may know, this requires the introduction of a regulator mass which breaks scale invariance. [aside: The fact that it is arbitrary, i.e. only a mathematical tool without physical significance , is then expressed through the so-called "renormalization group" (equations).]

As I already pointed out above: the real theoretical reason for the masslessness of the photon is local gauge symmetry. The latter is also present already in the classical Maxwell theory, but survives at the quantum level! (unlike scale invariance). Moreover all sectors of the Standard Model of particle physics exhibit such a local gauge symmetry.

That's why also the 8 gluons of Quantum Chromodynamics (the theory of strong interactions) have to be strictly massless.

Bye Fridger

PS: just found this one, too:

If we trully find photons to have a tiny mass, guess what ? It wont even put the relativity theory into trouble, like the public believe. We will just have to adjust the basic evolution equations (Maxwell equations) by adding a parameter in them (the value of m_0). According to relativity theory, if the photon has a small mass, it could not travel at exactly the velocity v = c. it would have a velocity v < c and then gamma isn't infinite. The product gamma TIMES m_0 will be legitimate then.

NO. You always tend to forget the crucial effects of the respective quantum theory! While classical Maxwell theory could indeed be modified with a small photon mass, the latter would screw the Quantum Maxwell theory = Quantum Electrodynamics completely!

Since then /local gauge invariance/ is broken, the Quantum Maxwell theory would loose its predictivity, be ill-defined and useless.

Reason: the loss of local gauge invariance makes the theory NON-RENORMALIZABLE.

[ajtribick]

Ok, looking up the term "gluon" I find this. Looks like the term "strong force" has got a confusing history.

So if I understand that article correctly, gluons are massless, and have nothing to do with holding the nucleus together - they only operate inside the nucleons. Does the gluon-caused force have infinite range, and if so why is it not felt on large scales (is this to do with the colour-neutrality?)

So this leaves something else to hold the nucleus together - I came across the term "residual strong force" - never heard this term before, though it seems to be what I understood as "strong force"... the terms change so it seems. Is this the force that holds the nucleus together, and are the exchange particles of the residual strong force massless as well?

[t00fri]

Ok, looking up the term "gluon" I find this. Looks like the term "strong force" has got a confusing history.

So if I understand that article correctly, gluons are massless, and have nothing to do with holding the nucleus together - they only operate inside the nucleons. Does the gluon-caused force have infinite range, and if so why is it not felt on large scales (is this to do with the colour-neutrality?)

So this leaves something else to hold the nucleus together - I came across the term "residual strong force" - never heard this term before, though it seems to be what I understood as "strong force"... the terms change so it seems. Is this the force that holds the nucleus together, and are the exchange particles of the residual strong force massless as well?

Gluons are particles in exactly the same sense as the photon is a particle. The difference is that gluons carry a charge = COLOR, while the photon is chargeless (its electrical charge is zero).

Photons and Gluons are special particles, however. Their exchange between (electrically charged) electrons and color-charged quarks, respectively, gives rise to the electromagnetic and strong forces, respectively. Their special role in connection with the forces is also expressed in their required masslessness.

++++++++++++++
An intuitive picture is this: consider two kids (<-> electrons, quarks) sitting each in a boat floating on a lake. If they start throwing a ball (<-> photon , gluons) between them, the two boats experience a force due to energy-momentum conservation and start floating away in various directions. (Note the ball carries energy/momentum over to the other boat).
++++++++++++++


The gluons are permanently confined in the protons (unlike photons). It would take an infinite amount of energy to kick them out!This phenomenon of confinement in strong interactions is very intriguing and unfortunately not so easy to explain.

The interactions of the protons in a nucleus are so-called Van der Waals forces , meaning due to a leftover tail of the gluon-force dropping to zero very fast outside each proton due to the phenomenon of color confinement (=> residual forces!) . That's why the strong interaction among protons is of very short range, while the underlying color interaction among quarks responsible for it, has formally infinite range (<-> gluon mass=0).

Bye Fridger

[Cham]

There's only one point I want to make perfectly clear, about a strong source of historical confusion :

MASS DO NOT INCREASE WITH VELOCITY.

The next formula (equation (1)), which is the definition of the so called "relativistic mass" : DO NOT HAVE ANY PHYSICAL MEANING. IT IS USELESS AND A STRONG SOURCE OF CONFUSION. It should simply be banished from all discussions on relativity theory and especially from vulgarisation articles. That quantity "m" ISN'T OBSERVABLE. What is increasing with velocity is the Kinetic Energy, not mass. As I have shown previously, the "m" may be eliminated from all equations. And it even don't shows without ambiguity in the relativistic Newton equation :



The "relativistic mass" appears ONLY in the special case of rotational motion (equation (10)). In the special case of rectilinear motion, it's not "gamma TIMES m0", it is "gamma-cube TIMES m0". In the general case (equation (11)), it's more complicated and acceleration isn't even parallel to the net force applied.

In any experiment, you cannot measure the particle's mass, while the particle is in motion. IT IS ITS KINETIC ENERGY. The only kind of mass you can measure, is the "rest mass" (or "proper mass", or whatever) while the particle is instantaneously at rest relative to the observer. In this case, v = 0 for that observer and equation (11) reduces to the classical newton equation :



The observer can then measure the mass by a simple comparison between acceleration and the net force applied. If he had done the same while the particle is in motion, he will find a variable, NON-ISOTROPIC MASS, according to the general case (equation (11)), which is very weird.


About the photon : I have to insist, the formula E = m c^2 has a clear physical meaning ONLY IN THE CASE OF MATTER (massive particles and objects). IT IS AMBIGUOUS for photons. You may use it, if you wish, to say that a photon has a mass if you write m = E/c^2, but this is an abuse of language and has no physical meaning. If m0 = 0, THERE IS NO RELATIVISTIC NEWTON EQUATION FOR A PHOTON. So you can't measure its "mass" without gravity. Even with gravity, the "m" does not have a clear physical meaning for a photon.


Spaceman Spiff,

You wrote :

Cham wrote:

Photons DO have kinetic energy : K = h f, where "f" is the frequency and "h" is Planck's constant. However, The relativistic Newton equation do not apply to them as photons do not have an intrinsic mass.

No, this energy is not kinetic, it is electromagnetic. A photon carries an amount of electromagnetic force, hence electromagnetic energy.

What the hell is "electromagnetic energy" anyway ? From Maxwell theory (which I know enough to teach it all year long), the total energy of some electromagnetic wave is given by the volumic integral of its electric and magnetic density over all volume of space (I will not write the equations here). IT IS KINEMATICAL AND POTENTIAL ENERGY. So because photons are traveling particles without any possibility of rest, their energy E = h f is kinematical, and nothing else.


You wrote :

Are you thinking of interference patterns with Young's slits? What is it about quantum mechanics that explains why photons can still interfere when the light level is turned down so low that there can only be one photon at a time between the slits and the image? I'm curious about knowing whether the 'relativisitic' explanation of timelessness has any feasibility behind it.

Timelessness has NOTHING to do with interference pattern. Photons (and electrons) are intrinsic quantum entities, and they manifest a double nature : particle and wave. I can explain the exact mathematical meaning of all this (I made all my graduate studies on this kind of stuff), but this is off topic and would be too long. It's not worth it here.

[t00fri]

How can you tell if you really are stationary? You can't.

I think perhaps you've forgotten about the cosmic microwave background. You can tell if you're stationary relative to it. This has many subtle implications that I don't understand.

Right Selden,

that's why a favorite coordinate system for defining galaxy recession velocities is the so-called CMB frame rather than the more traditional frames: galactic center (our galaxy of course) or heliocentric.

Bye Fridger

[Spaceman Spiff]

Gluons are particles: they are the force carrier particles of the strong nuclear force. They have mass! They even have rest mass.

That's why also the 8 gluons of Quantum Chromodynamics (the theory of strong interactions) have to be strictly massless.

Well, I am prepared to admit what I get wrong (which is a requirement of being scientific, by the way). I defer to Fridger's better understanding, and so that last sentence I wrote above is wrong: Gluons do not have rest mass. Wikipedia mentions a few MeV experiemental upper limit for rest mass for them. This doesn't change my prior posts posts, though.

Spiff.

[Spaceman Spiff]

++++++++++++++
An intuitive picture is this: consider two kids (<-> electrons, quarks) sitting each in a boat floating on a lake. If they start throwing a ball (<-> photon , gluons) between them, the two boats experience a force due to energy-momentum conservation and start floating away in various directions. (Note the ball carries energy/momentum over to the other boat).
++++++++++++++

Ah! Fridger, perhaps you could explain as I always wondered about this one: I can see how this analogy explains a repulsive force, but how does it explain an attractive force?

Spiff.

[d.m.falk]

I don't think I'll comment any further, but I'll leave this thread with a couple thoughts.

1: I have heard of one theorum that states that all matter has no real mass, but that this is a perception of force, instead.

2: Similarly, the Universe doesn't exist of its own volition, but instead, is the interference between two intersecting planes. Everything we understand, effectively, does not exist, and what does, is simply beyond our experience.

And lastly, I, for one, will not resort to expletives or name-calling to express disagreement. To do so is childish, even though it seems the prevelant way to debate and conduct discourse on the internet. I've been online now for 14 years, and I've seen too much of such human behaviour, that I have become increasingly cynical and disenfranchised.

Have at what you will.

d.m.f.