Science a GoGo's Home Page Forums General Discussion General Science Discussion Forum mass of photon

I agree that the Standard Model is just a model and that the Higgs hasn't been experimentally confirmed to exist. But you do have to acknowledge that the W and the Z particles and the various relations between their masses were correctly predicted. That's a very strong result.

I think I will start another thread on renormalization some time in the future here. renormalization can be seen as a trick to remove infinities. However, as pointed out by Wilson, it is more natural to consider field theories as an effective theory valid at low energies obtained by integrating out the high energy degrees of freedom.

So, you assume that the unknown ''theory of everything'' reduces to the standard model at low energies, just like magnetism in a solid can be described by a renormalizable phi^(4) field theory.

I agree with this, but this argument can work in many ways. E.g. when we detect particles like electrons or photons what we really 'see'' are effects in macroscopic detectors. The theory describing electrons and photons, Quantum Mechanics, in fact precludes you from observing particles in the same way you can observe classical objects. This means that it is theoretically possible that the Standard Model is just some effective theory and that the particles it describes don't really exist.

Such theories have been proposed by 't Hooft. He has proposed deterministic models. Bell's theorem is evaded because the particles are only effective mathematical objects that appear in intermediary calculations.

Wilson's argument is still just a conjecture. Nonetheless,t we are moving nearer to an understanding. As I have acknowledged, the physics that has been done using quantum field theory should not be thrown out of the window. Just as Bohr's atom did not invalidate the applicability of the Periodic Table of the Elements. There has been extraordinary successes like the W and Z particles; however, I do not believe that these theories are complete theories and their complexity is, to my mind, also an indication that there must be simpler mechanisms behind the symmetries predicted and observed. I believe that the real physics involved might be simpler than we anticipate at present. Sometimes I think that we are trying to fix what do not need fixing; and in the process we are developing wonderful new mathematics but less and less of an understanding of the physics involved.

In turn you do have to acknowledge that on another thread I have shown aspects of superconduction which the BCS model cannot explain; and I have postulated a single mechanism that explains superconduction in all the superconductors that have been discovered to date. Whatever you might think of me subjectively; do you not think that under such circumstances the "experts on superconduction" should evaluate the new model objectively vis-avis the BCS model; instead of just running away in horror while slinging around insults? Is this what scientists have become? Or have they always been the real dogmatists?

I dont agree so easilyi with Johnny argument:

"E=mc2 does not apply to a photon because a photon do not have potential rest energy; it only has kinetic energy given by Planck's relationship"

I think that E=mc2 implies or predicts that no one
particle in the Nature could be massless. Because his energy would be 0, and how to move in with E=0. Isn't impossible?

Yes this is a core argument and I believe very central to making progress in future physics. It is based on the acceptance that the Copenhagen interpretation is correct. I think it is this interpretation that has been leading us around in circles. It is the Copenhagen interpretation that precludes you from observing quantum mechanical "particles" in the same way you can observe classical objects; however, it does not explain why we can observe an electron (which is a quantum "particle") to follow a classical path except by invoking "magic". To say it is because the measurement apparatus is "classical" only avoids the dilemma. You have to bring in the "magic wand" by saying that sometimes you can see them and sometimes you cannot. If you rather accept that everything consists of waves and when we "observe" particles we observe localised waves, and when we do not, the waves are near to each other to superpose or entangle, then I believe that I can re-iterpret Quantum Mechanics without invoking wave-particle duality "magic". You then find that all interactions (even the photo-electric effect) are typically what you expect when waves interact.

This is the nice thing in life; or rather should be the nice thing in life: We can agree to disagree and argue about it, hopefully politely. I have difficulty to follow why E=mc2 must mean that every particle has mass.

If you write down the energy equation derived by Einstein you will find that a photon must have zero mass in order to have the momentum E/c which, within experimental accuracy,it has. experimentally has.

A photon is usually thought of as a massless particle. However we can associate a mass with a photon by using two well known relations. The first is Einstein's theorem on the equivalence of energy and inertia that is usually expressed as E = m c^2. The second, also due to Einstein, is the energy-frequency relation. This is usually written as E= h f, where h is Planck's constant and f is the frequency of the light. By combining these we get m = h f /c^2. This is the mass equivalent to the energy of the photon.

A photon has no rest mass, as such. It does have energy and energy is equivalent to mass. So light posesses inertia or, one might say, inertial mass. This may sound odd but it is demonstrated by the Compton Effect. In this experiment a photon interacts with an electron in a way taht is like a macroscopic particle collision. In other words the photon acts like a particle just as the electron does.

Dr. R.

P.S.: Another form of the energy relation is used in particle physics is

E^2 = (m0 c^2)^2 + (pc)^2.

In this m0 is the rest mass of the particle in question. Suppose that m0 = 0, as with a photon, then we have

E = pc = (m v)c.

Since we are looking at a photon v = c so that this equation becomes

E = mc^2.

In this expression m is not the proper (rest) mass.

We can also say that

E / c = p = h / L = h f / c.

Rearranging this gives the Einstein frequency relation E = h f. By the way Einstein got this relation from his analysis of the photoelectric effect.

Very well explained dr_rocket. What I have been talking about are particles which have rest masses. All present theories are in accord that a photon cannot have a rest mass.

The omly point a disagree on is to say categorically that the photon act as a particle just like an electron: it only seems as if they act like particles when they "collide". I do not believe that they are REALLY particles.

Still not ok , Johnn.
It seemes to me you established that

"A photon has no rest mass" due to
m= hf/c^2.


But why that m=hf/c^2 isnt a rest mass?
Isn't absurdity that photon hasn't mass?
Why do you talk about photon as a particle if photon hasnt mass ?
Whats the source of his energy?

I do not say that a photon is a "real particle"; I believe that "particles" do not exist. A photon is a light wave that has collapsed to inhabit a small region of space. This happens when the light is being absorbed by an electron; which is also not a "particle". The absorption is akin to a coalescence of the two waves (entanglement). From Galileo's relativity to Newton's laws, "rest mass" relates to the ability of a "body" to be "at rest" within a uniformly moving inertial reference frame. According to Einstein light can never be stationary relative to an inertial reference frame; it musy always move with a speed c. One can, however, assign mass to the kinetic energy that a "body" with rest mass has so the kinetic energy can be written as T=mc^2-m(rest)c^2. where m is larger than m(rest). Similarly, for a light wave (photon?), except that for "a photon" you have that m(rest)=0. I hope this helps.

O yes a lst comment: I believe that in the case of light AND matter the energy relates completely to wave energy. The only difference is that matter can be descibed (by the Schroedinger equation) in terms of stationary time-independent waves; i.e. the wave can be stationary relative to an inertial reference frame; while light cannot be described by stationary time-independent waves (as can be verfified when solving Maxwell's equations).

Thank you JB:

Possibly the last part of my question assumes something that does not exist. The part you did not understand was due to my failure to be more precise in forming the question. I try again.

When our astronomers look into the Galaxy they see pictures that were transported here by light and it is that light which is either the photon itself or the photon is the carrier of that light.

If so the photon has an image that was created not only at the time it left the surface of that foreign star but of what ever else it picked up as it left that distant Galaxy on its way to us. If the photon is pure energy where is this picture stored so it can get here to us to wonder at?

jjw

the photon is the image, what you see is the sum of the photons striking the retinas at the same time from objects around you. they dont 'carry' the image. if there were one photon, all you would see is a short burst of light, with the only info being the frequency of that photon. a group of photons coming from a large area traveling many light years will be spread out and only a few of them will reach you. when they do, they tell you the frequency of all of those that came from that part of the galazy. since the frequency is dependant on how much energy was released in those photons and the energy of the star will be fairly constant, you will see a constant color.

What left the star is probably not a "photon" as it is currently interpreted by the scientific community; i.e. a localised "particle". I believe that it only becomes a localised "photon" once it inertaacts with matter; e.g. our eyes or any other detector. If, on the way, the wave encountered another detector, it would have been absorbed and re-emitted. We will then observe it as a "scatterred photon". While absorbed by the detector, the energy is stored "in the absorber". Afterwards it is released as a (scatterred) wave which collapses into a "photon" when we it reaches our eyes.

Thanks again:

It is a more serious question than I can frame.
Consider the surface of a light source 10,000 light years away from us. Calculate the size of the photon that left that surface and compare it to the size it must achieve at the end of the journey. If the photon expaned constantly over that great distance it would be thousands of times larger that when it started out. If it did not expand then it must be dividing itself to continue to fill up the ever increasing volumn so that no empty space would be left inbetween.

This does not require an answer by any one. It is just one of my curiosities of which I have many to play with. Existing scientific thinking is not always satisfying for me.
jjw

jjw004,
'One photon does not an image make...`
The only 'size` that has much meaning in
relation to a photon is it's wave-length.
It might help you to consider a photon as
as energetic event, frozen in time, fleeing
through the universe at 'C`,
seeking an opportunity to undo the itself,
(the event).
Felt poetic today, but the above is surprisingly
accurate come to think of it.
Pragmatist

the size of the photon is the same as it is when it leave the surface of the sun. the thing is the number of them per square centmeter at the surface or the sphere 100000 light years away is extreamly small. the entire surface of the sun is sending out photons in all directions. some of them will be assorbed by things in the way, some will be deflected, and other will be reflected. when you look at the star, what you are seeing is the sum total of the all the photons thrown off the surface in your direction (or more spacifically, the directions of your eyes) that was not absorb, deflected, or reflected along the way that arrive at that instant.

jjw, why on earth is the size of the photon so important? Don't you think that you are forcing a perspective/interpretation on something that "we" are aware at this time that we cannot have a good pictural image of?

How does this help anything?

Welcome back Pasti. I hope all went well with your endeavours during April-May.

Hi Pasti:

Very nice to hear from you, even with impatience.

I suppose the size of the photon or the "wave" which forms the photon is of no importance to any one that knows it all about photon's.

It just seems to me that something is happening that may be important for us to understand and I broach the subject with a simplistic approach. No one need trouble themselves with what they see as a useless question- but- suppose that 1 degree of a distant stars surface is 30,000 miles. That 1 degree of surface emiting photon's will strech into many millions of miles after traveling many thousands of light years to get to me. The volumn has increased accordingly so photon's that left that star are spread out now.
I am aware that the answers provided are well founded but I remain puzzeled and that is my problem and not the members problem. I think just possibly that the size of the photon may be the reason for a limit on the speed of light.

I may try to restrain myself better. Thanks.
jjw

Thanks for the warm welcomming. And as an added bonus, I still have all my ranking stars. Silence is golden, as usual...

Sorry for the brevity, but I am still swimming in it thick until the end of this month. Anyway, folks, feel free to email me to take it where we left it, and I will get to it sooner or later.

jjw, the issue that is puzzling you has nothing to do with the size of the photon, neither as what we commonly percive as size, nor with the wavelength.

The answer to your question is simply kmown as the intensity law. The intensity of light (i.e. the number of photons per unit time) incident on a transversal surface varies with the inverse squared distance to th4e surface (neglecting scattering and absorption).I~I0/r^2, where I0 is the intensity of the source. Replace intensity with the photon rate, and you got your answer.

If you want to see where this law comes from, there is a nice way to do it. Take a piece of paper, and in its center draw a small blob. that is your source. Take a protractor, and now from your blob draw outwardly every say, 5 degrees, straight lines. Those are your photons emitted by the source. Now, using the blob as the center, take a compass (geometrical copmpass and not a navigational one) and draw concentric circles with radii 1 cm, 2cm, 3cm, 4cm, etc. If you use a letter sized paper, you should get quite a few such concentric circles.

Here is where your imagination must come into place. Imagine these circles as the intersection of concentric spheres with the plane of your paper, and now you have a source of light illuminating the space.

Take now your 1cm radius circle, and chose an arbitrary semicircle out of it. Count how many lines go through this semicircle. If you drew the lines every 5 degrees, you will have 36 lines going through this semicircle.

Now, the length of the semicircle is pi*1cm=3.14 cm. On each of the remaining concentric circles (of increasing radius) measure a length along the circumference equal to 3.14cm, and count how many lines go through this length for each of the circles. Plot the number of lines through this portion of the circumference as a function of the radius, and as a function of the inverse squared radius. Let me know what you get.

Hope this helps.
Cheers.