Are the stings in string theory, considered to be matter?

[lightarrow]

You would agree in that charge is conserved in a reaction, like 0=-e+e, but not mass? My answer is that Yes, the photon has mass. If a container with photon gas has mass, why should not a single photon has mass?

Because mass is not additive...

 

[The Dagda]

Clearly mathematics is not physics, but you must admit that "the power of the equality sign" has been quite successful so far!

1) p=h/\lambda (waves has a momentum) \rightarrow\;\lambda=h/p -> matter are waves!
2) E=mc^2 (mass contains energy) \rightarrow\;m=E/c^2 matter can be PRODUCED by any kind of energy!

2) is the most fruitful result by Einstein, not E=mc^2.

You would agree in that charge is conserved in a reaction, like 0=-e+e, but not mass? My answer is that Yes, the photon has mass. If a container with photon gas has mass, why should not a single photon has mass?

Perhaps I have misunderstood things here, but If it is the Higgs particle that GIVES particles their mass, Its would not exist -> Quite extreme prediction these days... But if it only intermediate mass in some way (like Feynman diagram Kirchoff current of mass) its OK, but why don't predict a "particle" that intermediate momentum and angular momentum at the same time? I here predict the momentum particle in the case you find the Higgs particle.

Well considering that the photon appears to be a special case, and that no one in fact knows for sure if its mass is 0, or it is beneath the Planck scale, or tinier still. Suffice to say without a method of measuring it, photons having particular mass is an entirely hypothetical concern. I doubt it having 1x10^-67kg mass would bother the theory all that much either but there you go.

Mass and energy are equivalent but they are not the same thing exactly, we have different ways of modelling mass particles and waves. Is it possible for something to be pure energy? Yes, is it possible light isn't? Yes, who knows...

 

[malawi_glenn]

The very confusion could perhaps lie in the definition of mass and matter - that is why there is such confusion. We have both quantum physics, general relativity and special relativity.

Trying to be diplomatic here ;-)

 

[Dmitry67]

Because mass is not additive...

Again confusion between relativisitc and rest mass...
Before the post I recommend adding what do you mean by 'mass'
Otherwise it is not a mass, but a mess :)

 

[per.sundqvist]

Because mass is not additive...

Well again, its the relativistic mass I was talking of and you are talking of rest mass?

According to the textbook "Introduction to special relativity" by Rindler we have such mass conservation:

\sum*m=0 (* means - for outgoing + for incoming etc)

but he also mention that E=mc^2, here meaning the equivalence of energy and mass is part of a hypothesis from Einstein, but which so far has been shown to be right. About light he write (chap 27):
If every form of energy has mass, we would expect light to have mass and thus bend in a gravitational field like that of the sun. Indeed, this has been observed. The radiation which the sun itself pours into space represents a mass loss to it of more than four million tons per second!

I agree that its a mess with matter and mass... But assume you create a electron positron pair with light you get: " something=m1+m2", what is something? Mass does not appear out of nowhere, it comes from the photon, but it change its manifestation, from being bound in an EM-field to be bound in two point particles.

 

[lightarrow]

Well again, its the relativistic mass I was talking of and you are talking of rest mass?

You don't need to talk about relativistic mass: in a confined space, it's *invariant* mass of the photons to be non-zero, probably it's this that you and others have missed.

 

[Dmitry67]

*invariant* mass of a photon is non zero?
something really new...
In Standard Model it is 0

 

[lightarrow]

*invariant* mass of a photon is non zero?
something really new...
In Standard Model it is 0

Read well what I wrote: not "one photon", but "photons".
In a confined space, for example a box with perfectly reflecting internal walls, you will have stationary EM waves and so no net wave's momentum. Then put p = 0 in E^2 = (mc^2)^2 + (cp)^2 and you get that the total invariant mass of the EM wave is E/c^2 and so non-zero.

 

[Dmitry67]

If this box is big enough then it is just a photon gas - photons moving in all directions.

If walls are close enough (in comparison with a wavelength) then if fact it can be described as a stationary wave, however, in that case you can not analyze just the wave, you need to analyze the whole system: walls plus wave.

 

[Phrak]

If this box is big enough then it is just a photon gas - photons moving in all directions.

If walls are close enough (in comparison with a wavelength) then if fact it can be described as a stationary wave, however, in that case you can not analyze just the wave, you need to analyze the whole system: walls plus wave.

If it's a boson gas, it's still nonzero. But if we knew the position of each phton fairly well, so that they don't overlap too much, the mass would tend to zero. It seems mass is observer dependent.

 

[enotstrebor]

The very confusion could perhaps lie in the definition of mass and matter - that is why there is such confusion. We have both quantum physics, general relativity and special relativity.)

Mass is the property of a resistance to a change in linear velocity (and associated requirement to apply a force to change a massed particles velocity). The massed particle in a vacuum can have any velocity between zero and up to but not including ``c''.

This is true in quantum physics, general relativity and special relativity.

Massed particles also exhibit other properties related to their environment as given by quantum physics, general relativity and special relativity.

The mass values is thus measured via its resistance to a change via a fource (see http://en.wikipedia.org/wiki/Mass" ) ``In physical science, mass refers to the degree of acceleration a body acquires when subject to a force: bodies with greater mass are accelerated less by the same force.''


Matter (see http://simple.wikipedia.org/wiki/Matter" ) ``is the substance or material of which all things are made. Matter has mass. Most matter has volume. All things we can touch, taste or smell are matter.''

``Not Matter'' (see http://simple.wikipedia.org/wiki/Matter" )
``heat is not matter, it is also known as infrared radiation, or another form of light
light is not matter
sound is not matter, it is vibrations of molecules in the air''.

Then there is the photon whose velocity in a vacuum is only ``c''. The photons velocity is only dependent on the media through which it travels. The photon has no invariant mass as it does not have mass.

The fact that the path of a photon around the sun (according to GR) does not mean that it has mass as the path curvature has nothing to do with the property of mass (which is the resistance to a change in linear velocity), but reaction of the photo to the curvature of space. If my memory is correct, the reaction of a massed particle to the curvature of space is also different than the reaction of the photon (but I wouldn't bet my life and can't give a reference).

There are some who want to give the massless photon ``mass'' for the sake of their theoretical beliefs. But there is no physical experimental evidence to indicate the photon has mass.

 

[lightarrow]

If this box is big enough then it is just a photon gas - photons moving in all directions.

If walls are close enough (in comparison with a wavelength) then if fact it can be described as a stationary wave, however, in that case you can not analyze just the wave, you need to analyze the whole system: walls plus wave.

Ok, then let's consider the first case (very big box) and, inside of it, two photons with the same energy E traveling in opposite directions: this system has invariant mass (=2E/c^2). If these two photons will hit the walls of the box an hour later, does it make any difference?

 

[Dmitry67]

Ok, then let's consider the first case (very big box) and, inside of it, two photons with the same energy E traveling in opposite directions: this system has invariant mass (=2E/c^2). If these two photons will hit the walls of the box an hour later, does it make any difference?

Such system with 2 photons has RELATIVISTIC mass of 2E/c^2 (in the frame of the box, as relativistic mass is not invariant). Invariant (rest) mass is 0.

 

[per.sundqvist]

Mass is the property of a resistance to a change in linear velocity (and associated requirement to apply a force to change a massed particles velocity). The massed particle in a vacuum can have any velocity between zero and up to but not including ``c''.

This is true in quantum physics, general relativity and special relativity.

Massed particles also exhibit other properties related to their environment as given by quantum physics, general relativity and special relativity.

The mass values is thus measured via its resistance to a change via a fource (see http://en.wikipedia.org/wiki/Mass" ) ``In physical science, mass refers to the degree of acceleration a body acquires when subject to a force: bodies with greater mass are accelerated less by the same force.''


Matter (see http://simple.wikipedia.org/wiki/Matter" ) ``is the substance or material of which all things are made. Matter has mass. Most matter has volume. All things we can touch, taste or smell are matter.''

``Not Matter'' (see http://simple.wikipedia.org/wiki/Matter" )
``heat is not matter, it is also known as infrared radiation, or another form of light
light is not matter
sound is not matter, it is vibrations of molecules in the air''.

Then there is the photon whose velocity in a vacuum is only ``c''. The photons velocity is only dependent on the media through which it travels. The photon has no invariant mass as it does not have mass.

The fact that the path of a photon around the sun (according to GR) does not mean that it has mass as the path curvature has nothing to do with the property of mass (which is the resistance to a change in linear velocity), but reaction of the photo to the curvature of space. If my memory is correct, the reaction of a massed particle to the curvature of space is also different than the reaction of the photon (but I wouldn't bet my life and can't give a reference).

There are some who want to give the massless photon ``mass'' for the sake of their theoretical beliefs. But there is no physical experimental evidence to indicate the photon has mass.

It seems you have a classical view of mass (similar to m=F/a). In this sense m=0 since v=c and cannot change. This corresponds to the rest mass m0, but not with m=E/c^2.

Deflection of light can be predicted by Newtons theory assuming the Force interaction
F=G*m1*m2/r^2 and letting v=c. GR correct the deflection angle by a factor of 2.

All relies on that there, in the Newton case, m1>0 (the photon), but also in the GR-case since there would be no space-time curvature for an object that does not have mass! Even if the mass of the object don't appear in the final result, the very extreme exact case m=0 (not the limit m->0) would lead to a straight line (no bending)!

 

[jtbell]

Such system with 2 photons [moving in opposite directions] has RELATIVISTIC mass of 2E/c^2 (in the frame of the box, as relativistic mass is not invariant). Invariant (rest) mass is 0.

No, the invariant mass is not zero. The invariant mass of a system is

m c^2 = \sqrt {E_{total}^2 - (\vec p_{total} c)^2}

For two photons with equal energy E, moving in opposite directions, the total energy is 2E and the total momentum is zero. Therefore, for this system, m c^2 = 2E.

More generally, for a "gas" of many photons in a stationary box,

m c^2 = \sqrt {(E_{photons} + E_{box})^2 - (\vec p_{photons} c + \vec p_{box} c)^2}

m c^2 = \sqrt {(E_{photons} + m_{box} c^2)^2 - (0 + 0)^2}

m c^2 = E_{photons} + m_{box} c^2

because the total vector momentum of the photons is negligible for any realistic number of photons moving in random directions.

(I use m instead of m_0 for invariant mass here to simplify the notation.)

 

[lightarrow]

No, the invariant mass is not zero. The invariant mass of a system is

m c^2 = \sqrt {E_{total}^2 - (\vec p_{total} c)^2}

For two photons with equal energy E, moving in opposite directions, the total energy is 2E and the total momentum is zero. Therefore, for this system, m c^2 = 2E.

More generally, for a "gas" of many photons in a stationary box,

m c^2 = \sqrt {(E_{photons} + E_{box})^2 - (\vec p_{photons} c + \vec p_{box} c)^2}

m c^2 = \sqrt {(E_{photons} + m_{box} c^2)^2 - (0 + 0)^2}

m c^2 = E_{photons} + m_{box} c^2

because the total vector momentum of the photons is negligible for any realistic number of photons moving in random directions.

(I use m instead of m_0 for invariant mass here to simplify the notation.)

Perfect, Jtbell

 

[enotstrebor]

It seems you have a classical view of mass (similar to m=F/a).

That is the definition and how it is measured. It is not a view of mass it is the definition (period). You seem to want some other definition, so just let's all make up our own.

In this sense m=0 since v=c and cannot change. This corresponds to the rest mass m0, but not with m=E/c^2.

m=E/c^2 is not a definition of physic(s)al equivalence. Mathematic equivalence is not physics equivalence.

m=E/c^2.

I have two energy forms. One a two ton object rotating with rotational energy E_R. One a two ton object traveling with a velocity and translational energy E_L.

The rotational and linear velocities are such that:

E_R=E_L

I will stand in front of the rotating velocity object and you in front of and along the direction of the linear velocity object.

The difference between mathematics and physics will (hopefully) become clear to you before you die.

 

[Phrak]

It seems you have a classical view of mass (similar to m=F/a). In this sense m=0 since v=c and cannot change. This corresponds to the rest mass m0, but not with m=E/c^2.

It may be classical, but it is correct. F/a is a perfectly valid measure of mass, where abstract equations become an experimental science, as you can actually F/a.

Deflection of light can be predicted by Newtons theory assuming the Force interaction F=G*m1*m2/r^2 and letting v=c. GR correct the deflection angle by a factor of 2.
All relies on that there, in the Newton case, m1>0 (the photon), but also in the GR-case
since there would be no space-time curvature for an object that does not have mass! Even if the mass of the object don't appear in the final result, the very extreme exact case m=0 (not the limit m->0) would lead to a straight line (no bending)!

As you've noted, Newtonian gravity may predict the deflection of a photon, but it predicts it incorrectly. So, it lends no supporting evidence to photon mass, one way or the other.

In the case of General relativity, the particle or photon appearing deflected by the influence of a planet is not a result of the curvature of spacetime about the object by the object, but the curvature of spacetime by the planet. Massive and massless objects behave in the same way; following geodesics. Again, no evidence for or against. The difference in the case of gravity is that light follows "null geodesics," but this is the same in Minkowski space.

 

[Phrak]

No, the invariant mass is not zero. The invariant mass of a system is

m c^2 = \sqrt {E_{total}^2 - (\vec p_{total} c)^2}

For two photons with equal energy E, moving in opposite directions, the total energy is 2E and the total momentum is zero. Therefore, for this system, m c^2 = 2E.

More generally, for a "gas" of many photons in a stationary box,

m c^2 = \sqrt {(E_{photons} + E_{box})^2 - (\vec p_{photons} c + \vec p_{box} c)^2}

m c^2 = \sqrt {(E_{photons} + m_{box} c^2)^2 - (0 + 0)^2}

m c^2 = E_{photons} + m_{box} c^2

because the total vector momentum of the photons is negligible for any realistic number of photons moving in random directions.

(I use m instead of m_0 for invariant mass here to simplify the notation.)

I was preparing to argue with you as I assumed you buried a factor or 2 somewhere, as your results differ from the Komar mass which would double the result. After some quick edgikation, I see that general relativity throws a new curve into this mass vs. mass business, even where the total mass of an object is very small on human scales.

The results from the general theory differ from the special theory.

Unless I'm mistaken, we can treat the light in a laser cavity as a one dimensional box. The norm of the energy momentum vector of light contributes an amount of mass equal to what you derrived. The internal pressure of the light contributes an identical amount, doubling what we should expect between the mirrors. And the mechanical arrangment holding the two mirrors together loses this same amount due to the resultant tension.

The total mass is the same, it's location has changed. ...but somehow, I think you already knew this...

 

[enotstrebor]

No, the invariant mass is not zero. The invariant mass of a system is

m c^2 = \sqrt {E_{total}^2 - (\vec p_{total} c)^2}

...

m c^2 = \sqrt {(E_{photons} + E_{box})^2 - (\vec p_{photons} c + \vec p_{box} c)^2}

m c^2 = \sqrt {(E_{photons} + m_{box} c^2)^2 - (0 + 0)^2}

m c^2 = E_{photons} + m_{box} c^2

because the total vector momentum of the photons is negligible for any realistic number of photons moving in random directions.

(I use m instead of m_0 for invariant mass here to simplify the notation.)

What does it take for you to understand that photon energy is not mass!

Therefore photon energy can never contribute to the invariant mass of a system!

The equation you keep using is a statement about the energy value equivalence of a massed object, it is not a physics equivalence.

See my last posting, though it appears you will die first before recognition of the differnce between mathematical equivalence and physics equivalence.

 

[lightarrow]

...

The equation you keep using is a statement about the energy value equivalence of a massed object, it is not a physics equivalence.

When you compute the four-momentum square modulus of a particle, for example, you say that what you get is not mass because it's just mathematics? Weird

 

[per.sundqvist]

It may be classical, but it is correct. F/a is a perfectly valid measure of mass, where abstract equations become an experimental science, as you can actually F/a.



As you've noted, Newtonian gravity may predict the deflection of a photon, but it predicts it incorrectly. So, it lends no supporting evidence to photon mass, one way or the other.

In the case of General relativity, the particle or photon appearing deflected by the influence of a planet is not a result of the curvature of spacetime about the object by the object, but the curvature of spacetime by the planet. Massive and massless objects behave in the same way; following geodesics. Again, no evidence for or against. The difference in the case of gravity is that light follows "null geodesics," but this is the same in Minkowski space.

Here the planet or sun have a huge mass comparable to the object (photon). I can think of an experiment where you have a gamma-ray photon package (Gaussian like blob), like 1Gev which is hitted closely by a photon of visible light 1eV. Would that photon bend in presence of the other? If yes -> photon has mass since it CAUSE gravity. If not not m=E/c^2 is wrong. But deflection is so ultra small and I guess impossible to measure even if you could calculate it, so ...

Also, do you septics believe Rindler was wrong in the statement: "All forms of energy has mass"? Also where origin the mass when you create a electron-positron pair? You can't use E=mc^2 because its just mathematics?