How can something be massless?

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SUMMARY

The discussion centers on the concept of massless particles, specifically photons and gluons, and their existence in the universe. Participants clarify that photons are massless due to the gauge invariance of electrodynamics, which allows them to exist without rest mass while still possessing energy. The conversation also touches on the Higgs mechanism, explaining that while massless particles do not have rest mass, they can exhibit energy and momentum. Key equations such as E=mc² and E=hν are referenced to illustrate the relationship between energy and mass.

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  • Understanding of special relativity and its implications on mass and energy.
  • Familiarity with gauge invariance in electrodynamics.
  • Knowledge of the Higgs mechanism and its role in particle physics.
  • Basic grasp of quantum mechanics, particularly the behavior of photons.
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  • Study the implications of gauge invariance in quantum field theory.
  • Explore the Higgs mechanism in detail, focusing on its impact on particle mass.
  • Learn about the energy-momentum relationship for massless particles.
  • Investigate the principles of special relativity and their mathematical formulations.
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I am confused about a problem that has arose from my reading of the forums.

If something is massless, then how can it exist? I was taught that all objects in the universe have mass, however small, so how can photons and gluons have no mass?
 
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VelociBlade said:
I am confused... I was taught that all objects in the universe have mass, however small,
You were taught wrong.
 
The fact that photons are massless, rises from the Gauge invariance of electrodynamics, which means that the Lagrangian can be a function of the Field tensor only.
Usually if we want a Gauge invariant field theory, then the mediator of the field must be massless.
Mass is actually not quite easy to define...
 
VelociBlade said:
I am confused about a problem that has arose from my reading of the forums.

If something is massless, then how can it exist? I was taught that all objects in the universe have mass, however small, so how can photons and gluons have no mass?
Photons and gluons have no rest mass, but moving at the speed of light allows them to have energy equivalent.
 
It is a hard concept to explain.My teacher was jokingly talking about Newton's infamous low, F = m * a, which he explained as: "F is the force applied, a is the acceleration as a consequence as of that... and well, then there's this proportionality constant that we call m".

And really, that's all that is mass. The ratio of force to acceleration. Light (or any gauge boson) cannot be accelerated by a force. It does travel slower in different media and can seem to make a curve when in fact the space is curved, but neither case has anything to do with "force".
Hope that helps.
 
LennoxLewis said:
Light (or any gauge boson) cannot be accelerated by a force.
Hope that helps.

Actually that's not quite true. In the general case the four-force is not orthogonal to the four-velocity, so there is four-work. Light(photons) is a special case, since in electrodynamics its true, that four-force, doesn't do four-work.
But in a general field-theory it isnt.
And if there is four-work, then the rest mass of the given particle changes. For example, if the force is some four-gradient of a scalar field, then the this field changes the rest mass of the particle, and the particle will have different mass throughout the field. This is basically the Higgs-Mechanism, if the field is the Higgs field, which is constant everywhere, so actually the mass of the mediator, the Higgs boson doesn't change throughout the field.
 
It is ok for something to be proven mathematically to have no mass at rest, because it still has mass until it comes to a rest. If a photon ever came to a rest, then we would have something to talk about!

It is true that it is impossible for something to exist without mass. A photon has mass because it has energy, it has energy because it moves.
 
No that's not right. A photon does not have mass. The only thing it has is energy, which is not mass. If it would have mass, then it would have a mass shell.
It may have something like "equivalent" mass for energy, but that is NOT mass.
 
Thaakisfox said:
Actually that's not quite true. In the general case the four-force is not orthogonal to the four-velocity, so there is four-work. Light(photons) is a special case, since in electrodynamics its true, that four-force, doesn't do four-work.
But in a general field-theory it isnt.
And if there is four-work, then the rest mass of the given particle changes. For example, if the force is some four-gradient of a scalar field, then the this field changes the rest mass of the particle, and the particle will have different mass throughout the field. This is basically the Higgs-Mechanism, if the field is the Higgs field, which is constant everywhere, so actually the mass of the mediator, the Higgs boson doesn't change throughout the field.

I stand corrected. I know about the Higgs mechanism, but doesn't this apply only to massive particles? If there is four-work, then the particle is no longer massless, right?
 
  • #10
The Higgs mechanism, is through which particles gain their mass, so it doesn't apply only to massive particles.
Yes, if there is four work then the particle is no longer massless. So for example if a "particle" doesn't have rest mass, and it is but into some field, which does four-work, then the particle will gain mass
 
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  • #11
Thaakisfox said:
No that's not right. A photon does not have mass. The only thing it has is energy, which is not mass. If it would have mass, then it would have a mass shell.
It may have something like "equivalent" mass for energy, but that is NOT mass.

But mass and energy are the same thing, as an object gains energy of any form it is proven that its mass increases.

If an object, even a massless object, has speed it attains mass, so a photon must have mass.
 
  • #12
DeepSeeded said:
But mass and energy are the same thing, as an object gains energy of any form it is proven that its mass increases.

If an object, even a massless object, has speed it attains mass, so a photon must have mass.

seriously, what are you talking about?
 
  • #13
E=MC^2 and C^2 is a constant, so E=M.

E=1/2MV^2 so Velocity is proportional to M, a photon has velocity so it has energy so it has mass.

If that doesn't convince you, a photon also has momentum, and momentum is VM. If M were 0 it would not have any momentum.
 
  • #14
E=m*c^2 isn't the same as E=m . Maybe you could say light and mass are two different states of energy? I picture mass as a 'potential' form of energy, that requires lots of energy to release. (correct me if this doesn't sound right).
 
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  • #15
nuby said:
E=m*c^2 isn't the same as E=m

Sure it is, the c^2 is just there because our SI units, chosen independently, are not going to line up perfectly, for example, Joules and grams were both decided independently and 1 Joule is not going to turn out to be equal to exactly 1 gram.

c itself comes from two constants that were measured through lab experiments, these constants are there only to bridge the difference between our SI units.
 
  • #16
DeepSeeded said:
E=MC^2 and C^2 is a constant, so E=M.

E=1/2MV^2 so Velocity is proportional to M, a photon has velocity so it has energy so it has mass.

If that doesn't convince you, a photon also has momentum, and momentum is VM. If M were 0 it would not have any momentum.

E=mc^2 is only for a particle of mass m at rest, and the speed of light sure ain't zero. For a particle with speed v, basic special relativity gives
E = \frac{mc^2}{\sqrt{1 - v^2 / c^2}}
(checking the v = 0 limit is easy).

Now, the speed of light is v = c. Plug this into the formula above, and you get zero in the denominator. The only way to salvage the situation is to have m = 0 in the numerator. Now the expression is merely an ill-defined 0 / 0. (The appropriate formula for massless particles is E = h\nu, where h is Planck's constant and the Greek letter \nu (nu) is the particle's frequency.)

Photons are massless. So are any other particles that travel at the speed of light. And, going the other direction, all massless particles must travel at the speed of light.

If thinking about massless particles is giving you conniptions, recall that light is a wave, and try thinking about massless waves for a while. (Once that gives you conniptions, try going back to thinking about massless particles. And study some more special relativity in the mean time.)
 
  • #17
DeepSeeded said:
If that doesn't convince you, a photon also has momentum, and momentum is VM. If M were 0 it would not have any momentum.

To follow up on my previous post, the photon's energy is E = h\nu and its energy and momentum are related by E = pc (where p = |\textbf p| is the magnitude of the momentum three-vector). Therefore the momentum of a massless particle is
p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda},
where \lambda is the particle's wavelength. (The direction of \textbf p is the direction of the particle's propagation.)

You can get a deeper understanding of this from Wikipedia, or any modern physics or special relativity book.
 
  • #18
How funny it seems, that it is supposed to be more natural for things to have mass than not to have it, since no one knows where it comes from anyway.
Next someone will pull out Zeno again, and say infinities don't exist.
Just because you can't imagine it, doesn't mean it can't exist.
 
  • #19
DeepSeeded said:
E=MC^2 and C^2 is a constant, so E=M.

E=1/2MV^2 so Velocity is proportional to M, a photon has velocity so it has energy so it has mass.

If that doesn't convince you, a photon also has momentum, and momentum is VM. If M were 0 it would not have any momentum.

Now I see that you have not done Special Theory of relativity in collage.

The Energy and momentum relation is NOT E = mc^2 but

E^2 = p^2 + m^2

and hence p = E for massless particles, i.e p = mv is for newtionian dynamics, but now you were invoking Einstein, so play by those rules please, no cheating ;-)

daschaich posts are also illuminating, E = gamma * m, if v = c, gamma -> infinity, which is non sense here.
 
  • #20
The big question is, what does the speed of light squared actually represent? Is it two different velocities, if so, what are they for?
 
  • #21
nuby said:
The big question is, what does the speed of light squared actually represent? Is it two different velocities, if so, what are they for?

As DeepSeeded points out, it's a unit conversion factor. Not that big a deal. I don't understand your second sentence.
 
  • #22
nuby said:
The big question is, what does the speed of light squared actually represent? Is it two different velocities, if so, what are they for?

It is not a big question, in SI units it is a conversion factor to get the correct units, but units are arbitrary, you can choose any gauge you want I would say.
 
  • #23
If you have a single (real) proton. How do you turn it into pure energy?
 
  • #24
nuby said:
If you have a single (real) proton. How do you turn it into pure energy?

Define "pure" energy...
 
  • #25
nuby said:
If you have a single (real) proton. How do you turn it into pure energy?

We're discussing photons here, protons are rather different. How do you feel photons differ from "pure energy"?
 
  • #26
1.50e-10 Joules, and the mass goes away.
 
  • #27
nuby said:
1.50e-10 Joules, and the mass goes away.

i) is there an requirement for your definition that energy is not attached to mass? if, why? is a photon "pure" energy? Why should your definition be standard? who cares?

ii) easy, let the proton interact with an antiproton -> two photons of energy 1.50e-10 Joules will be created.

Why is it ok that a photon flies away with 1.5 e-10 joules but not a proton? :S
"Instead of mass", a photon has angular frequency -> I claim this is not pure energy, I want energy not attached to angular frequency of Electric- and Magnetic field.

Can you solve it?
 
  • #28
daschaich said:
E=mc^2 is only for a particle of mass m at rest, and the speed of light sure ain't zero. For a particle with speed v, basic special relativity gives
E = \frac{mc^2}{\sqrt{1 - v^2 / c^2}}
(checking the v = 0 limit is easy).

Now, the speed of light is v = c. Plug this into the formula above, and you get zero in the denominator. The only way to salvage the situation is to have m = 0 in the numerator. Now the expression is merely an ill-defined 0 / 0. (The appropriate formula for massless particles is E = h\nu, where h is Planck's constant and the Greek letter \nu (nu) is the particle's frequency.)

Photons are massless. So are any other particles that travel at the speed of light. And, going the other direction, all massless particles must travel at the speed of light.

If thinking about massless particles is giving you conniptions, recall that light is a wave, and try thinking about massless waves for a while. (Once that gives you conniptions, try going back to thinking about massless particles. And study some more special relativity in the mean time.)

Very good point, I guess I was wrong. I have not taken relativity yet, though I do remember that equation.

Now I have to wonder, if light doesn't have mass... what does? Everything is made of forces, electric, magnetic, strong, weak, etc. If none of these forces have mass then well nothing does.
 
  • #29
DeepSeeded said:
Very good point, I guess I was wrong. I have not taken relativity yet, though I do remember that equation.

Now I have to wonder, if light doesn't have mass... what does? Everything is made of forces, electric, magnetic, strong, weak, etc. If none of these forces have mass then well nothing does.

What kind of nonsense claim is that - that everything is made up of forces? :S

What you have in the standard model is that you have particles and interactions.

Electrons have mass and cannot move at the speed of light in vacuum, but always below that speed. Quarks have mass, and so on.
 
  • #30
malawi_glenn said:
Electrons have mass .

I thought electrons were just the result of a resonance phenomenon of energy, I never thought of them being made of some sort of substance... I guess I am over my head, this post was originally on the classical physics forum.. now I am in quantum physics :redface:
 

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