Is the Graviton Really Massless?

[FeDeX_LaTeX]

I was just reading a Wikipedia article about a hypothetical particle known as the graviton. It stated that it was massless -- but how is this possible? I thought that every particle has to have a mass.

 

[torquil]

Yes it would be massless. So is the photon. In the standard model, all particles are massless, but they appear massive because of their interactions with the hypothetical Higgs particle.

Torquil

 

[tiny-tim]

Photons are particles that don't have mass, and that enables them to travel at the speed of light.

 

[FeDeX_LaTeX]

So a particle has to have a mass of 0 to reach the speed of light on the dot?

An electron has a mass of 9.11*10^-31 kg... is that why we say that an electron can get very close to the speed of light, but not exactly to the speed of light?

Also, when we say 'massless', do we mean the mass is so small that it is 0? I am probably wrong...

EDIT: Just did a google search on photon mass and it yields '0'. So if F = ma, then;

F/a = m

F/a = 0

So F/a must be 0? Confused...

http://www.aip.org/pnu/2003/split/625-2.html - this article also seems to be talking about the limit of a photon mass...

 

[torquil]

So a particle has to have a mass of 0 to reach the speed of light on the dot?

An electron has a mass of 9.11*10^-31 kg... is that why we say that an electron can get very close to the speed of light, but not exactly to the speed of light?

It is why it cannot move at the speed of light. But any massive particle can get arbitrarily close to the speed of light. The higher the mass, the more energy is needed to reach a given speed. Massless particles move atthe speed of light in a vaccuum. Massive particles move at any speed less than the speed of light.

Also, when we say 'massless', do we mean the mass is so small that it is 0? I am probably wrong...

EDIT: Just did a google search on photon mass and it yields '0'. So if F = ma, then;

F/a = m

F/a = 0

So F/a must be 0? Confused...

Photon movement is out of the domain of Newtonian dynamics. So it doesn't make sense to apply Newtons law as you have written it to a photon. To describe movements of massless particles, you need to apply Einsteins theory of relativity.

http://www.aip.org/pnu/2003/split/625-2.html - this article also seems to be talking about the limit of a photon mass...

This means that experiments have shown that the poton mass is less than some experimentally observed amount. The accepted theory says that the photon is massless, so theory and experiment are consistent. Everything must be checked experimentally, and experiments always have uncertainties, so it would not show directly that m=0 for a photon. Experiments will never determine that the mass of the photon is exactly zero, because that is impossible.

Torquil

 

[tiny-tim]

So a particle has to have a mass of 0 to reach the speed of light on the dot?

Yes!

(or rather, it doesn't reach the speed of light, it can only be at the speed of light )

Also, when we say 'massless', do we mean the mass is so small that it is 0? I am probably wrong...

No, we mean the mass (the rest-mass, of course) actually is zero.

EDIT: Just did a google search on photon mass and it yields '0'. So if F = ma, then;

F/a = m

F/a = 0

So F/a must be 0? Confused...

Photon movement is out of the domain of Newtonian dynamics. So it doesn't make sense to apply Newtons law as you have written it to a photon. To describe movements of massless particles, you need to apply Einsteins theory of relativity.

No, I disagree.

This is the danger of using the "easy" version of Newton's second law … F = ma

The official version is "force = rate of change of momentum", or F = dp/dt.

For an ordinary particle, m ≠ 0, and so p = mv, and therefore F = dp/dt = ma.

For a photon, m = 0, but the Newtonian F = dp/dt is still valid.

 

[torquil]

No, I disagree.

This is the danger of using the "easy" version of Newton's second law … F = ma

The official version is "force = rate of change of momentum", or F = dp/dt.

For an ordinary particle, m ≠ 0, and so p = mv, and therefore F = dp/dt = ma.

For a photon, m = 0, but the Newtonian F = dp/dt is still valid.

Agreed!

Torquil