Should photons be considered massless?

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  • #1
Jonnyb302
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Hello all, I am asking this question in the context of general relativity.

In general relativity the stress-energy tensor is related to the spacetime metric through the Einstein field equations. The production of a curved spacetime is what creates what we call gravitation. For example a single particle with mass creates a spacetime which can be described by the schwarzschild metric (among others).

Now electromagnetic fields have a well known stress energy tensor, and of course photons are an electromagnetic phenomena. So photons seem like they would produce gravitational effects. But would they be identical to a particle with equivalent energy?

Could the spacetime around a photon be described with the schwarzschild metric? I know that for a massive particle, the mass shows up in the schwarzschild metric, but that mass is of course an experimentally determined thing. If the photon set up a schwarzschild spacetime, we could experimentally assign it a "mass".

So any thoughts? Do photons curve spacetime? What makes them different from what we typically call massive particles? This is a speculative matter, not an experimental one obviously.
 

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  • #2
Drakkith
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Yes, light curves spacetime. What makes them different from massive particles is that they have no mass. But they don't need mass to curve spacetime, as they have energy.
 
  • #3
WannabeNewton
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GR and Einstein's equations are classical, photons are not. In GR, space-time curvature couples to the energy-momentum of the classical electromagnetic field, that's all.
 
  • #4
Drakkith
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GR and Einstein's equations are classical, photons are not. In GR, space-time curvature couples to the energy-momentum of the classical electromagnetic field, that's all.

So an EM wave passing through an area of space does affect the curvature?
 
  • #6
Jonnyb302
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Well Drakkith what I mean is, if photons curve spacetime, then what is the difference between massive and massless particles in a GR setting? How do they behave differently.

Also WannabeNewton I don't understand your post. Are you saying photons don't curve spacetime?

My question is, do photons set up a spacetime distinguishable from what we typically call massive particles? Is a difference between massive and massless particles even possible in gr since mass is not always a definable term?

Why do we call photons massless? What properties warrant that term?
 
  • #7
Dale
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Well Drakkith what I mean is, if photons curve spacetime, then what is the difference between massive and massless particles in a GR setting? How do they behave differently.
Massive particles always travel at less than the speed of light in any local inertial frame, and massless particles always travel at the speed of light in such frames.

My question is, do photons set up a spacetime distinguishable from what we typically call massive particles?
Yes, EM radiation will have a pp wave spacetime or perhaps a null dust spacetime. Massive particles will other spacetimes depending on the distribution of the matter.

Why do we call photons massless? What properties warrants that term?
Mostly the fact that their mass is 0 to the best of our measurements. Properties that are related to that 0 mass include traveling at c, having two polarization states, Coulomb's law, and I am sure there are others.
 
  • #8
WannabeNewton
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Also WannabeNewton I don't understand your post. Are you saying photons don't curve spacetime?
No I'm saying that you are combining classical equations with non-classical objects. Photons are often used semi-classically in GR but as far as the energy-momentum tensor goes, it's just that of the classical electromagnetic field. We speak of space-times describing electromagnetic radiation e.g. a pp-wave space-time.
 
  • #9
Jonnyb302
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First time using the quote system, hopefully I don't screw this up.

Massive particles always travel at less than the speed of light in any local inertial frame, and massless particles always travel at the speed of light in such frames.

Ok that's fair (at least to the best of my knowledge)

Mostly the fact that their mass is 0 to the best of our measurements. Properties that are related to that 0 mass include traveling at c, having two polarization states, Coulomb's law, and I am sure there are others.

Well what I mean is how can you say the mass is zero? If I have a reflective box with an equal amount of anti matter and matter on a triple beam balance, and I let the antimatter and matter react until only light is left bouncing around inside, am I going to read a different mass?

If the answer is no, then ofcourse one answer is to say that is not mass, its energy. And of course energy gravitates so problem solved. But then my question would be well then what is the difference between mass and energy? And if there is no difference between the two, how can we call a particle massless when it has energy?

I suppose what I would really like is a definition of mass, but I know that can get real tricky.
 
  • #10
Jonnyb302
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No I'm saying that you are combining classical equations with non-classical objects. Photons are often used semi-classically in GR but as far as the energy-momentum tensor goes, it's just that of the classical electromagnetic field. We speak of space-times describing electromagnetic radiation e.g. a pp-wave space-time.

ahh ok thank you. I had never heard of a pp-wave spacetime before now. I have only taken one undergraduate course in gr (a senior level one).

Edit: also to give more perspective that was at a quarter school, ie 10 week classes.
 
  • #11
WannabeNewton
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ahh ok thank you. I had never heard of a pp-wave spacetime before now. I have only taken one undergraduate course in gr (a senior level one).
I'm not sure what textbook you used but if you're interested: I'm 99.99% sure MTW talks about it and I know for sure that d'Inverno talks about it in his GR text. There are also more specialized texts devoted solely to exact solutions in GR which will delve into it.

But the overall idea is that yes the electromagnetic field and the space-time geometry are coupled to each other. What I mean by this is that Maxwell's equations ##\nabla_a F^{ab} = j^b## and ##\nabla_{[a}F_{bc]} = 0## are coupled to ##g_{ab}## because of ##\nabla_{a}## and that ##G_{ab} = \kappa T_{ab}## are coupled to ##F_{ab}## because of the energy-momentum tensor associated with the electromagnetic field.
 
  • #12
rbj
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Why do we call photons massless? What properties warrant that term?

using an anachronistic term, their rest mass is zero. they have momentum. and you can construct a physical quantity of the same dimension of mass by taking their momentum and dividing by their speed, which is the same quantity you get from their energy [itex] E = \hbar \omega[/itex] and dividing by [itex]c^2[/itex].
 
  • #13
Jonnyb302
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using an anachronistic term, their rest mass is zero. they have momentum. and you can construct a physical quantity of the same dimension of mass by taking their momentum and dividing by their speed, which is the same quantity you get from their energy [itex] E = \hbar \omega[/itex] and dividing by [itex]c^2[/itex].

See my reply to DaleSpam above.


How can you say their rest mass is zero? How do you measure the rest mass of a particle opposed to just all the energy stored in it (including rest mass energy)? In a general relativistic setting what is the difference between stored energy and rest mass if any?
 
  • #14
rbj
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How can you say their rest mass is zero? How do you measure the rest mass of a particle opposed to just all the energy stored in it (including rest mass energy)? In a general relativistic setting what is the difference between stored energy and rest mass if any?

i am going to get in trouble, because currently the concept of "rest mass" and "relativistic mass" is deprecated in physics. i found this differentiation helpful, but the physicists here don't like it.

if you were to define the momentum of a particle or body to be simply

[tex] p = m v[/tex]

no matter what, then, if you know the momentum of a particle and its speed, you can always define a quantity that has the same dimension or units of mass:

[tex] m = \frac{p}{v} [/tex]

i would call that [itex]m[/itex] the "inertial mass" of the particle or body. sometimes it's called the "relativistic mass", but that term is deprecated in modern times.

now, all EM waves move in a vacuum at the same speed [itex]c[/itex] no matter who the observer is and what their speed (relative to some other observer) is. so there is no light waves that move slower than [itex]c[/itex] for anyone. but for normal particles or bodies made outa matter that move slower (w.r.t. some observer) than [itex]c[/itex], then the relativistic expression of the particle's momentum is

[tex] p = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} v[/tex]

where [itex]m_0[/itex] is the symbol that i am calling "rest mass" and what is most often called "invariant mass" and is what you would normally think of the mass of that particle if you were holding it in your hand. at speeds much slower than [itex]c[/itex], that momentum is well approximated as [itex] p = m_0 v [/itex], which is the classical Newtonian expression.

so the relationship between this "inertial mass" and "invariant mass" is

[tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

light (or EM radiation) has both wave-like properties (that we describe with Maxwell's equations) and particle-like properties (that we describe as photons with energy [itex]E = h \nu[/itex]).
we normally think of photons as moving at the same speed as the EM wave that is the wave-like dual of the photons. (that's sort of an issue of debate because some people, and i am not one of them, like to think of photons as possibly moving slightly slower than [itex]c[/itex] which would give them some rest mass. but most physicists don't think that photons move slower than the EM wave speed.)

so, assuming photons move at speed [itex]c[/itex] the real physicist and the anachronistic old electrical engineers (like me) agree that photons have momentum of

[tex] p = \frac{E}{c^2}c = \frac{h \nu}{c} [/tex]

now that expression that multiplies velocity [itex]c[/itex] which is [itex]\frac{E}{c^2}[/itex] has dimensional units of mass, and old codgers like me might call that the inertial mass of a photon.

[tex] m = \frac{E}{c^2} = \frac{h \nu}{c^2} [/tex]

dunno what the real physicists want to call that quantity, but they cannot deny that, dimensionally, it is the same dimension of "stuff" that mass is. and this [itex]m[/itex] quantity is real, positive, and finite. it's a number (with dimensional units).

but if you were to relate the "rest mass" back to the "invariant mass" (turning around the equation above), you get:

[tex] m_0 = m \sqrt{1 - \frac{v^2}{c^2}} [/tex]

but, if for photons [itex]v = c[/itex], so as long as [itex]m<\infty[/itex], then the rest mass [itex]m_0=0[/itex] no matter what finite value [itex]m[/itex] is.

that is, from my anachronistic understanding, why photons have zero rest mass or zero invariant mass which is why they are called "massless" particles. if photons did have non-zero rest mass, then if they moved at speed [itex]c[/itex] (and not even a tiny bit slower), they would have infinite inertial mass, infinite momentum, and infinite energy (which they do not have).
 
  • #15
jtbell
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If I have a reflective box with an equal amount of anti matter and matter on a triple beam balance, and I let the antimatter and matter react until only light is left bouncing around inside, am I going to read a different mass?

No. The mass of a system of particles is not, in general, the sum of the masses of the individual particles.

I suppose what I would really like is a definition of mass, but I know that can get real tricky.

For a single particle, the mass is given by
[tex]mc^2 = \sqrt{E^2 - (|\vec p|c)^2}[/tex]
For a system of particles, the mass is given by
[tex]mc^2 = \sqrt{E_{total}^2 - (|\vec p_{total}|c)^2}[/tex]
The "trick" is that the magnitude of the total momentum does not generally equal the sum of the magnitudes of the individual particles' momenta.
 
  • #16
BruceW
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for a single photon, it is simple. The rest mass is zero, but relativistic mass is nonzero. And if we have a collection of photons, then the rest mass of the collection is generally not zero.

And instead, if we think about the electromagnetic stress-energy tensor, the relativistic mass is the 0,0 component (which is generally nonzero). So classical light does have relativistic mass. Now, for the rest mass of the EM stress-energy tensor, uh... that is not so clear to me. The electromagnetic stress-energy tensor is trace-free, so I guess maybe that is the equivalent?

edit: Maybe something to do with the 'stress' part of the stress-energy tensor is closer to the concept of the rest mass.
 
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  • #17
WannabeNewton
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Keep in mind that the term "rest mass" as applied to photons is a misnomer. The concept of rest mass as the mass measured in a Lorentz rest frame is of course meaningless for a photon. It's really the invariant mass (##p_{\mu}p^{\mu}##), which for massive particle is the same as the intuitive concept of rest mass.
 
  • #18
Dale
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Well what I mean is how can you say the mass is zero?
The definition of mass is the norm of the four-momentum, or, in units where c=1, ##m^2=E^2-p^2##. For a photon ##E=p## so ##m^2=p^2-p^2=0##.
 
  • #19
Jonnyb302
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The definition of mass is the norm of the four-momentum, or, in units where c=1, ##m^2=E^2-p^2##. For a photon ##E=p## so ##m^2=p^2-p^2=0##.

Ok that's fair enough, although it seems strange to define mass like that.

Does anyone know how you would actually experimentally measure the rest mass as opposed to the total energy of something?

What I mean is, say you put an object on a scale, you would actually be measuring its total energy not its rest mass right?

So say we measured a charged battery vs. an uncharged battery, the charged battery would "weigh" more on our scale because it contains more energy? (realizing its a very small difference).

edit: Energy scaled by c squared
 
  • #20
PAllen
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Ok that's fair enough, although it seems strange to define mass like that.

Does anyone know how you would actually experimentally measure the rest mass as opposed to the total energy of something?

What I mean is, say you put an object on a scale, you would actually be measuring its total energy not its rest mass right?

So say we measured a charged battery vs. an uncharged battery, the charged battery would "weigh" more on our scale because it contains more energy? (realizing its a very small difference).

edit: Energy scaled by c squared

For cases where you can't actually bring something to rest (for either practical or theoretical reasons), a good way to look at Energy versus (invariant) mass is what part of total energy cannot be accounted for by kinetic energy? What is left over is mass. Then E^2 - p^2 exactly answers this question: for all particles it describes what is not accounted for by KE; for massless particles is shows nothing is left over - there is only KE; and this is what characterized massless particles.

For a system of particles, you have to distinguish between analyzing each particle versus the system. Each particle has a rest mass mass, separable from KE. However, the system treated as a whole has an 'rest mass' measurable in its center of mass frame. This system rest mass will include KE of all particles as measured in that COM frame. This is actually all encompassed in the 4-momentum formalism: you add 4-momentum vectors for all the particles, getting a total 4-momentum. In the COM frame, by definition, the total spatial components of 4-momentum will be zero. The resulting norm of the total 4-momentum will be determined just by sum of total energy of particles measured in this frame (energies are all additive, no negative energies). If you analyze the system in a different frame, you will find that you have net momentum and different total energy, but the E^2 - p^2 for the system will match the total energy measured in the COM frame. To sum: for a system of particles, system rest mass is the total energy measured in the COM frame, with positive contributions from particle rest mass and KE, and negative contributions from any relevant binding energy (PE).

In this context, a cavity of light or radio waves will have system rest mass contribution from the KE of the EM radiation, just as it does from the KE of particles.
 
  • #21
Nugatory
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So say we measured a charged battery vs. an uncharged battery, the charged battery would "weigh" more on our scale because it contains more energy? (realizing its a very small difference).

You're right about the difference being very small. There's a thread somewhere here in which someone calculated that the mass difference between a fully charged lead-acid auto battery and a fully discharged one is measured in nanograms. So the quick answer is that because the difference is smaller than the resolution of most reasonable measurements, we just don't care.

That's the quick answer. A more complete answer is that the distinction you're trying to make, between the rest mass of an object when you do and don't count the internal energy that may be hidden inside the object (charge of the battery; kinetic energy of photons bouncing around inside a sealed mirror-box that you're weighing) isn't very useful. Even in the uncharged battery, much of the total mass comes from that internal energy; if you add up the rest masses of every quark and electron in the uncharged battery you'll get something appreciably less than the rest mass of the battery.
 
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  • #22
PAllen
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Another way to look at this is to note that when an unknown particle interacts with known particles in known states of motion, you can separately deduce the E and p of the unknown particle. From this, you can determine it's rest mass without having it at rest via E^2-p^2. For photons, all experiments of this type are consistent with zero rest mass for the photon, with upper bounds exceedingly small.
 
  • #23
Jonnyb302
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PAllen, thank you for your replies they have been very helpful.I have an additional question, would it be fair then to say that it is the total energy which resists motion? So for example the mass term m in Newtons second law could be replaced with E/c^2 ? I realize that for many situations the majority of the energy is going to be mass.

Would it also be appropriate to replace the mass terms in Newtonian gravity with E/c^2?

I recognize that we have much better theories than this small correction to Newtonian gravity, I am asking to bridge the gap between those theories.


Nugatory I disagree that the distinction is not important. If you think about thermal physics, sometimes we would like to measuring something, and find out how many particles are in that system. If much of what we measure is internal energy and not rest mass energy then we could incorrectly predict how many particles are in that system.

In particular (and I only know this because it was my senior capstone) dealing with the number density in a neutron star, that could be problematic as the density with which neutrons are packed is important.
 
  • #24
Dale
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Does anyone know how you would actually experimentally measure the rest mass as opposed to the total energy of something?

What I mean is, say you put an object on a scale, you would actually be measuring its total energy not its rest mass right?
If you put an object on a scale then p=0 wrt the scale so E=m according to the definition of invariant mass given above.
 
  • #25
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In particular (and I only know this because it was my senior capstone) dealing with the number density in a neutron star, that could be problematic as the density with which neutrons are packed is important.

For that calculation you must include the internal energy of the neutrons as well as the rest of mass of their constituent particles, or you will get an absurdly wrong answer for the number of neutrons and hence the density. (Although I may be misunderstanding the problem you were working).
 
  • #26
Jonnyb302
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If you put an object on a scale then p=0 wrt the scale so E=m according to the definition above.

That seems problematic, imagine if I build a scale out of glass which has a lid and closed box. And I fill the scale with light, you would then claim that the light had mass using your definition. This is why it seems difficult to distinguish between mass and energy in any way. And hence why it seems contradictory to me to call a photon massless when it has energy.
 
  • #27
ZapperZ
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That seems problematic, imagine if I build a scale out of glass which has a lid and closed box. And I fill the scale with light, you would then claim that the light had mass using your definition. This is why it seems difficult to distinguish between mass and energy in any way. And hence why it seems contradictory to me to call a photon massless when it has energy.

This makes no sense either.

Your whole argument here seems to be based on a faulty understanding of E=mc^2. If you want to argue that E and m are the "same thing", and thus, ALL energy must be mass, then why simply stop at that equation?

Then look at Hooke's Law, F=kx. If you simply want to apply an equality weely neely, then you are saying that all forces must have some sort of an extension. Look at F=ma. All forces are nothing more than acceleration, regardless if there is an equilibrium or not.

We can play this game ad nauseum.

Photons doesn't have mass is not just based on this, and the consequences of photons having a mass are plenty (longitudinal polarization, etc.), all of which have not been detected. You simply cannot be blind to all these consequences.

Zz.
 
  • #28
Dale
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That seems problematic, imagine if I build a scale out of glass which has a lid and closed box. And I fill the scale with light, you would then claim that the light had mass using your definition. This is why it seems difficult to distinguish between mass and energy in any way. And hence why it seems contradictory to me to call a photon massless when it has energy.
There is nothing problematic or contradictory with that. A single photon is massless, as shown above and as shown experimentally to great precision. A system of multiple photons may be massive.

Suppose that we have one photon of unit energy traveling in the x direction, then its 4-momentum is ##(E,p_x,p_y,p_z)=(1,1,0,0)## its mass is clearly 0. Now, suppose we add a second photon also of unit energy traveling in the -x direction, then its 4-momentum is ##(E,p_x,p_y,p_z)=(1,-1,0,0)## with a mass also 0. The system's 4-momentum is given by ##(E,p_x,p_y,p_z)=(1,1,0,0)+(1,-1,0,0)=(2,0,0,0)## so the system's mass is 2.

The mass of a system of multiple particles is always greater than or equal to the sum of the masses of the individual particles. This is the relativistic equivalent of the triangle inequality.
 
  • #29
Jonnyb302
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This makes no sense either.

Your whole argument here seems to be based on a faulty understanding of E=mc^2. If you want to argue that E and m are the "same thing", and thus, ALL energy must be mass, then why simply stop at that equation?

Then look at Hooke's Law, F=kx. If you simply want to apply an equality weely neely, then you are saying that all forces must have some sort of an extension. Look at F=ma. All forces are nothing more than acceleration, regardless if there is an equilibrium or not.

We can play this game ad nauseum.

Photons doesn't have mass is not just based on this, and the consequences of photons having a mass are plenty (longitudinal polarization, etc.), all of which have not been detected. You simply cannot be blind to all these consequences.

Zz.

First off F = -kx lol, sorry I had to.

Second, I think there is a misunderstanding. Are you aware that all energy gravitates? At the beginning of this thread I specifically said this is a discussion in a general relativistic setting.

What I am getting at, is what is the functional difference between energy and mass. What does mass do that energy doesn't and vice versa. Why are photons considered massless? What does mass even mean in a general relativistic sense?

Your statement of longitudinal polarization is something I have never heard of before so I appreciate it. That seems like something that actually distinguishes mass and energy.
 
  • #30
Jonnyb302
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There is nothing problematic or contradictory with that. A single photon is massless, as shown above and as shown experimentally to great precision. A system of multiple photons may be massive.

Suppose that we have one photon of unit energy traveling in the x direction, then its 4-momentum is ##(E,p_x,p_y,p_z)=(1,1,0,0)## its mass is clearly 0. Now, suppose we add a second photon also of unit energy traveling in the -x direction, then its 4-momentum is ##(E,p_x,p_y,p_z)=(1,-1,0,0)## with a mass also 0. The system's 4-momentum is given by ##(E,p_x,p_y,p_z)=(1,1,0,0)+(1,-1,0,0)=(2,0,0,0)## so the system's mass is 2.

The mass of a system of multiple particles is always greater than or equal to the sum of the masses of the individual particles. This is the relativistic equivalent of the triangle inequality.

That is very interesting, I can't believe I never heard about this. Thanks for that good example.
 
  • #31
Dale
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Why are photons considered massless?
I already answered that above. They are considered massless because their invariant mass (formula also given above) is 0 to high precision.
 
  • #32
Dale
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That is very interesting, I can't believe I never heard about this. Thanks for that good example.
You are welcome. I hope that helps.
 
  • #33
PAllen
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Would it also be appropriate to replace the mass terms in Newtonian gravity with E/c^2?

I recognize that we have much better theories than this small correction to Newtonian gravity, I am asking to bridge the gap between those theories.

This will lead you to nothing but grief and error. You will get corrections that are off by factors of 2 or 4, for example. If you want to deal with first order corrections to Newtonian gravity, use what the people who calculate the Ephemeris use: the Einstein-Infeld-Hoffman equations:

http://en.wikipedia.org/wiki/Einstein-Infeld-Hoffmann_equation

(I have checked that the equations given here by Wiki match those incorporated in the NASA documents).
 
  • #34
jtbell
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Does anyone know how you would actually experimentally measure the rest mass as opposed to the total energy of something?

In particle physics experiments it's common to measure the momentum (e.g. by measuring the radius of the curved path in a magnetic field) and the energy (e.g. in a calorimeter) and then use Dale's and my definition to calculate the mass.
 
  • #35
ZapperZ
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First off F = -kx lol, sorry I had to.

If you want to nitpick, I can easily point out that that is a scalar equation, i.e. it signifies the amplitude only. So the negative sign is meaningless in this case.

Second, I think there is a misunderstanding. Are you aware that all energy gravitates? At the beginning of this thread I specifically said this is a discussion in a general relativistic setting.

What I am getting at, is what is the functional difference between energy and mass. What does mass do that energy doesn't and vice versa. Why are photons considered massless? What does mass even mean in a general relativistic sense?

But that's my point. You only care about this from ONE point of view, i.e. the mass-energy equivalence. Instead of looking at it as a conversation factor, you are assigning a physical meaning of one being the same as the other. This is absurd if we apply that same logic to other well-known equality!

Your statement of longitudinal polarization is something I have never heard of before so I appreciate it. That seems like something that actually distinguishes mass and energy.

But really, this should not be a surprise. For example, an electron has mass. If you are saying that this is nothing more than a photon with energy E=mc^2, then you're missing a bunch of things and violating a number of conservations laws, namely charge and spin!

As I've stated, if photons have mass, a number of consequences should occur, including the fact that it must also decay.

http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.111.021801

So assign a mass to photons at your own risk, because the rest of the experiments that have tried to detect it have produced nothing of that magnitude (i.e. using that conversion).

Zz.
 

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