# How can something be massless?

1. Jan 7, 2009

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?

2. Jan 7, 2009

### clem

You were taught wrong.

3. Jan 7, 2009

### Thaakisfox

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...

4. Jan 7, 2009

### mathman

Photons and gluons have no rest mass, but moving at the speed of light allows them to have energy equivalent.

5. Jan 7, 2009

### LennoxLewis

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.

6. Jan 7, 2009

### Thaakisfox

Actually thats 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, doesnt 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 doesnt change throughout the field.

7. Jan 7, 2009

### DeepSeeded

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.

8. Jan 8, 2009

### Thaakisfox

No thats 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.

9. Jan 8, 2009

### LennoxLewis

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. Jan 8, 2009

### Thaakisfox

The Higgs mechanism, is through which particles gain their mass, so it doesnt apply only to massive particles.
Yes, if there is four work then the particle is no longer massless. So for example if a "particle" doesnt have rest mass, and it is but into some field, which does four-work, then the particle will gain mass

Last edited: Jan 8, 2009
11. Jan 8, 2009

### DeepSeeded

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. Jan 8, 2009

### malawi_glenn

seriously, what are you talking about?

13. Jan 8, 2009

### DeepSeeded

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. Jan 8, 2009

### nuby

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).

Last edited: Jan 8, 2009
15. Jan 8, 2009

### DeepSeeded

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. Jan 8, 2009

### daschaich

$$E=mc^2$$ 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. Jan 8, 2009

### daschaich

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. Jan 8, 2009

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. Jan 9, 2009

### malawi_glenn

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. Jan 9, 2009

### nuby

The big question is, what does the speed of light squared actually represent? Is it two different velocities, if so, what are they for?