Why Does Light Have No Mass?

[Antonio Lao]

It is not necessarily to call your objects of rotation quarks. You can call them kurons.

 

[neutroncount]

I have been looking into spin on the web.Found this quote:
we must not really see the origin of spin for electrons and protons etc. in the rotation of the charged body because from known dimensions, they would have to be spinning such that their surface was rotating faster than the speed of light to give rise to the magnitude of the angular momentum properties present.
How much faster than light? I have calculated a speed for a new electromagnetic wave which moves at 10 ^ 20 m/s !Hence all the fuss I make about changing photons and relativity.Nothing to do with tachyons though.I reckon spin o comes about in my model of the photon because one quark rotates clockwise around two others and another rotates anticlockwise around the remaining two.
Just like the electron and proton.And since an electron-positron pair comes from a photon, they are also made of quarks which must be very close together and make them behave pointlike.

You know they don't really "spin" in the real sense right? At least not like a top or baseball. Spin is an intristic property of a particle that resembles that of angular momentum seen in the macro world. Spin is verifiable through the use of magnetic field interactions.

 

[kurious]

spin

They do spin. I think the quantum world is a reflection of the macroscopic world and that the born interpretation of the wavefunction squared is the wrong interpretation.
Kurions that is what I'll call them when I pick up my nobel prize!

 

[Antonio Lao]

Please remember to share your award monies with the people at Physics Forums who in some way helped you in your achievement. If not then your conscience will bother you for a long, long time.

 

[suyver]

Can all the non-established "physics" please be moved to the Theory Development subforum (or at least be split off from this thread)?


On-topic:
@pmb_phy: regarding the Proca Lagrangian, is this the most accurate way to determine the upper-limit for the photon mass?
Also, for the revised Coulomb law,
$$\Phi(r) = q \frac { e^{-\mu r} }{r}$$,
(assuming mu close to 0) wouldn't this be relatively easy to detect in astronomical EM interactions, where r will be huge and the exponential term will dominate?

 

[Hessam]

Feel free to correct me here: but, i thought that light has to have at least some sort of mass, because it has energy...? blah please help

 

[Antonio Lao]

Photon as the unit particle of light does have a relativistic mass from its energy content. Since photon is always moving in vacuum at the speed of light 186,000 miles/s or 300,000 km/s, its rest mass in vacuum is zero.

 

[humanino]

There is a wonderful discussion in the "Lessons on gravitation" by Feynman, where He tales what happened at a "soiree mondaine" in Paris where a famous scientist asked him an upper bound on the photon mass. He doez NOT use any equation (^_^)...

 

[pmb_phy]

suyver - Accuracy is an experimental question. I don't know of any other way to measure the photons rest mass.

Hessam - When it is said that light has no mass it means that light has zero rest mass. The m in the relation E = mc² is not rest mass. Its relativistic mass.

Antonio Lao - To be more precise a photon has mass from its momentum. But since momentum and energy are proportional my point is a bit nitpicky. But this goes to the definition of mass and not to a derived relationship between mass and energy.

Pete

 

[kurious]

Photons have rest mass - it can't disappear when electrons and positrons come together.Just as a charge moving at constant speed has no apparent magnetic field but shows that it does when it deccelerates, I wonder if photons have rest mass but it is latent and not apparent, being released only on decceleration of the photon - when it collides or is absorbed by something.

 

[jcsd]

Photons have rest mass - it can't disappear when electrons and positrons come together.Just as a charge moving at constant speed has no apparent magnetic field but shows that it does when it deccelerates, I wonder if photons have rest mass but it is latent and not apparent, being released only on decceleration of the photon - when it collides or is absorbed by something.

In relativity conservation of mass, just becomes conservation of energy as E² = m₀²c⁴ + p²c² (this is basically just E₀ = m₀c², with kinetic energy included), so E is the conserved quantity in our refrence frame not m. So if an elerton and postitron anhilate to form two photons:

$$E^2 = 2{m_e}^2 c^4 + ({p_{e-}}^2 + {p_{e+}}^2)c^2 = ({p_{\gamma_1}}^2 + {p_{\gamma_2}}^2 )c^2$$

You can see that the m²c⁴ term has disappeared for the photons as they have no mass as E not m is the quantity conserved.

 

[pmb_phy]

In relativity conservation of mass, just becomes conservation of energy as E² = m₀²c⁴ + p²c² (this is basically just E₀ = m₀c², with kinetic energy included), so E is the conserved quantity in our refrence frame not m. So if an elerton and postitron anhilate to form two photons:

$$E^2 = 2{m_e}^2 c^4 + ({p_{e-}}^2 + {p_{e+}}^2)c^2 = ({p_{\gamma_1}}^2 + {p_{\gamma_2}}^2 )c^2$$

You can see that the m²c⁴ term has disappeared for the photons as they have no mass as E not m is the quantity conserved.

Conservation can be a tricky thing if not stated properly. However if taken literally, mass is a conserved quantity if by "mass" you mean relativistic mass, i.e. the "m" in p = mv.

Pete

 

[jcsd]

Well in orddre to try and avoid this issue I used m₀ and m_e which unambigously refer to rest mass. Of course mass is usually defined as rest mass (though you could argue that as the formula I've used is frame depednmt ther'e np reason not to take ion a defintion of mass that's not frame dependent).

 

[kurious]

You can see that the m2c4 term has disappeared for the photons as they have no mass as E not m is the quantity conserved.

This does not prove that REST mass is not conserved.

As far as I am concerned from debates on this forum and elsewhere, the problem of
how photons can become rest masses and vice-versa is unsolved and
possibly one of the most important issues in physics.

 

[jcsd]

You can see that the m2c4 term has disappeared for the photons as they have no mass as E not m is the quantity conserved.

This does not prove that REST mass is not conserved.

As far as I am concerned from debates on this forum and elsewhere, the problem of
how photons can become rest masses and vice-versa is unsolved and
possibly one of the most important issues in physics.

Of course doesn't in itself prove that rexsst mass is not conserved (observational evidence shows us that rest mzass is not consrved), but there's no reaspon to suppose that rest mass is conserved anymore.

Remember just because you refuse to accept the explanation doesn't mean that the problem is unsolved or indeed that there was even a problem in the first place. Can you give me a single rreason why you think rest mass isn't conserved (bearing in mind that this is not the TD forum).

 

[kurious]

I think rest mass is conserved because momentum still exists for a photon as it does for rest masses.When a photon is reflected off a wall it exerts a force on the wall just like any other particle does.You might say this force is a relativistic momentum change but
in general relativity which deals with decceleration and acceleration forces as such don't exist.This seems to me inconsistent.

 

[jcsd]

I think rest mass is conserved because momentum still exists for a photon as it does for rest masses.When a photon is reflected off a wall it exerts a force on the wall just like any other particle does.You might say this force is a relativistic momentum change but
in general relativity which deals with decceleration and acceleration forces as such don't exist.This seems to me inconsistent.

Okay let's work out the momentum of a photon using the formula for a relatvistic particle with rest mass: $p = \gamma mv$ (note m is rest mass). So for a photon we substitute in v = c:

$$p = \frac{mc}{\sqrt{1-\frac{c^2}{c^2}}} = \frac{mc}{0}$$

clearly we have a problem here, as if m is non-zero then we can interpret the momentum as infinite (if we set m as zero we could say that the momentum is indeterminate which is at least more physically realistic as indeed a photon can have have any value for p in special relatvity). So photons with rest masses just don't work in relativity, so the formula we should be using to find the momentum of a photon which can me derived from the formula for relativistic momentum-energy shown in my previous posts is $p = \frac{E}{c}$ so a phtoon's (relativistic) momentum is dependent on it's energy not on it's mass (which is zero).

 

[jtolliver]

This does not prove that REST mass is not conserved.

The only way both restmass and energy-momentum can both be conserved is if all particles have the same four-velocity, which experimental evidence(actually everyday experience) shows is not the case.

 

[reilly]

The proof that a photon has no mass follows directly from the quantum conditions -- as in momentum p=h(1/wavelength), while energy E = h(frequency) -- so E=pc, and that's all she wrote.
Regards,
Reilly Atkinson

(You want details? They will be found in most any first-year physics text.)