Do Photons Have Mass? - Debate & Questions

[stevmg]

First you must realize that E = mc2 is not what Einstein originally wrote.
The correct equation is E subscript 0 = mc2.

In the first equation m depends on energy (on velocity) and the photon has mass.
In the second equation m is constant and the photon has no mass.

Over the years his orginal equation was changed from E subscript 0 = mc2 to E = mc2.

Yea, you are correct , p. 47 "General Results of the Theory," Section XV Relativity, Einstein AE.

He derives that using a Taylor (I believe) expansion.

 

[xienohp]

There are theories proving photon mass are bounded by some extremely small numbers, however, the mass here means rest mass, not effective mass.
For reference, you may follow classical EM's deviation of what would happen if photon has mass, with fields as the tool.

Photon must follow by geodesic of the space-time, with its "time" being nullified.

 

[Wez]

I agree I can't explain why yet, but I believe the photon has a mass which defines it's limit at C what is really interesting recently at CERN something is potentially traveling faster than C so is it's mass less than a photon, does it shape shift between matter and anti matter, and if there was an anti matter phono could it travel faster than C in our universe, I still believe what ever particle it is , it must have a mass equivalent, which gets "considered" in the fabric

 

[yoron]

I think I agree with, was it Pervect? Saying that it is the geometry of SpaceTime that trace out the 'geodesics'. If I imagine that we can backtrack it all to 'energy' then SpaceTime is 'energy'. Don't mistake that for our older concept of a 'absolute space' though. I don't think there is any such 'quantity' existing, it's all frame dependent.

Anyway, if 'energy' is what describes a 'SpaceTime', then it stands to reason that a 'photon', or a 'wave,' will follow those geodesics, as they do not 'accelerate'.

The only thing that breaks a geodesic is expending energy, and light does not do that. But I'm not totally sure if this definition covers it all. Can anyone think up a example of something not expending 'energy' breaking a geodesic?

For example, a tunneling? Is that breaking a geodesic?
Or a spontaneous particle creation?

What I mean is if there is some point where you in the transition can say that 'this is not a geodesic.' It's difficult to define that one.
==

You might want to turn it around, and ask yourself if there can be a accelerative effect that does not 'expend energy'? And by that I mean anything creating a change to the former geodesics, relative what it gets after it.

Ah well :)

 

[SinghRP]

I have joined in this very late. I tried to read all of the above, but ...
This is what I understand.
A photon has mass, but not real mass, only effective mass. In reality, the effective mass is due to its momentum = E/c.
Here is what Wigner argured. If a photon has mass, you can acclerate it or de-accelerate it. But you can't. Poor photon is stuck with its c. (Slowing down of light in water is something else; don't be confused here.)
Similar logic with its spin. Its spin = 1, but no zero polarization! (A particle with a mass and spin 1 has three polarizations: +1, 0, -1.)
What do you think?
I have introduced a new topic: Gravity's effects. Can someone answer me there.

 

[yoron]

No, a photon has a momentum as I see it, not a 'mass'. a photon only exist in its annihilation. All measurements of a 'real' photon comes from the definitions made in it annihilating. any other definition involving 'paths' etc is theoretical, no matter how many points, and times, you measure those 'identical' photons in. That we use conceptual descriptions do not guarantee that path you think you see. And a geodesic is 'gravity' defining paths of 'no energy expenditure' to me.

It's a quanta of light.
==

What you might want to ponder is that, assuming that space's metric is gravity, there can be no point uninfluenced by that. So even in a 'geodesic' there must be 'gravity'. Therefore don't tell me that it 'propagate', I'm perfectly happy with it choosing to exist in 'geodesics', no matter where you measure it, from the conservation laws alone.

 

[MVesseur]

Seeing as this thread is more than 5 years old, might as well get it into its sixth year ;). I am no physics student, but questions on the subject keep popping up in my mind, one of them being the question that started the thread.

I've always tried to help myself with models from the "Newtonian world" to try and understand what is going on. Light is/consists of waves and we know that because it has frequencies (I suppose there are other ways to prove it). Yet it arrives in small quanta that can be measured. There is no smaller unit of light than a quantum. That creates what appears to be a contradiction - if it consists of "packages" it has to be particles, but it behaves like waves. Apparently something like that applies to electrons as well - I'd love to know the difference between photons and electrons, by the way.

But to my point: doesn't sound, which to my understanding is waves in matter, arrive in packages? Is there a smaller unit of sound than the smallest particle of matter it moves through? If so, couldn't sound also be considered as transferring or converting energy quanta on impact? Yet we would never consider sound to be particles.

Also the particles of the medium sound moves through do not itself need to travel for sound to move with a specific speed. It's the wave that is moving, not the medium. In a similar vein: would light require the transporting particle (not the photon, but the photon-carrier) to actually move? Could it be that the universe is filled with a mass-less or nearly mass-less medium?

Do I understand correctly that a photon is only the smallest amount of light energy that can be transported, but is not necessarily a particle in itself (if there were another particle that acts as its carrier/medium)? Can these supposed light-transporting particles even be mass-less, seeing as the speed of light is limited? Why is the speed of light what it is? Why isn't it faster? What is slowing it down? Could it be the mass or inertia of the proton-carrier, as yet unnamed?

I read above that the "ether"-theory was countered with the absoluteness of the speed of light. The moving light-emitter does not add to, or subtract from its speed. But isn't there something like a Doppler-effect in light, just as in sound? Also, a moving sound-emitter can not alter the speed of sound either, it only alters its frequency. In that way the speed of sound is as absolute as the speed of light, it seems - supposing the sound-transporting medium is always the same and always has the same density and temperature.

Sorry to troll into this forum, but my simple mind is hoping to get some answers and as a side-effect encourage the bright minds of this world to phrase their understanding in a digestible manner, thereby explaining things to the other ignorant but knowledge-hungry Googlers that land here.. and maybe even clarify things in their own mind.

Thanks!

- Marinus Vesseur

 

[nitsuj]

Seeing as this thread is more than 5 years old, might as well get it into its sixth year ;). I am no physics student, but questions on the subject keep popping up in my mind, one of them being the question that started the thread.

I've always tried to help myself with models from the "Newtonian world" to try and understand what is going on. Light is/consists of waves and we know that because it has frequencies (I suppose there are other ways to prove it). Yet it arrives in small quanta that can be measured. There is no smaller unit of light than a quantum. That creates what appears to be a contradiction - if it consists of "packages" it has to be particles, but it behaves like waves. Apparently something like that applies to electrons as well - I'd love to know the difference between photons and electrons, by the way.

But to my point: doesn't sound, which to my understanding is waves in matter, arrive in packages? Is there a smaller unit of sound than the smallest particle of matter it moves through? If so, couldn't sound also be considered as transferring or converting energy quanta on impact? Yet we would never consider sound to be particles.

Also the particles of the medium sound moves through do not itself need to travel for sound to move with a specific speed. It's the wave that is moving, not the medium. In a similar vein: would light require the transporting particle (not the photon, but the photon-carrier) to actually move? Could it be that the universe is filled with a mass-less or nearly mass-less medium?

Do I understand correctly that a photon is only the smallest amount of light energy that can be transported, but is not necessarily a particle in itself (if there were another particle that acts as its carrier/medium)? Can these supposed light-transporting particles even be mass-less, seeing as the speed of light is limited? Why is the speed of light what it is? Why isn't it faster? What is slowing it down? Could it be the mass or inertia of the proton-carrier, as yet unnamed?

I read above that the "ether"-theory was countered with the absoluteness of the speed of light. The moving light-emitter does not add to, or subtract from its speed. But isn't there something like a Doppler-effect in light, just as in sound? Also, a moving sound-emitter can not alter the speed of sound either, it only alters its frequency. In that way the speed of sound is as absolute as the speed of light, it seems - supposing the sound-transporting medium is always the same and always has the same density and temperature.


Thanks!

- Marinus Vesseur

I hear the term "propagate" used for EM.

I think it speaks of how light "travels" from point A to point B.

Perhaps how all waves travel.

 

[Mentz114]

Very briefly, the 'light-quantum' only appears when energy and momentum is (are ?) exchanged between light and matter. Such as when an atom or ion absorbs or emits light, or light interacts by 'bouncing off' a particle, changing its momentum ( the Compton effect).

Propagation is wavelike apart from these instances. Einstein wrote a brilliant ( and straightforward) paper in 1916 showing for the first time that momentum *must* be exchanged when emission or absorption happens, or the black-body spectrum discovered by Planck would be different.

 

[MVesseur]

Right, "propagation", I remember. Funny, to use a word from horticulture.

Now, how is the 'light quantum' that much different from the 'sound quantum' when a swinging molecule of the propagating medium hits an absorbing molecule? Kinetic energy is transformed into heat (or is that kinetic energy too?).

Talking about sunlight, for example, as a product of the processes on or in the surface layer of the sun. It leaves in all directions at various frequencies as electromagnetic radiation in small packages we call quanta (correct me if I'm wrong). Why in quanta? Because it itself is a "something" that contains that much energy, or because it is propagated by a universe filled with a "light-propagating medium"? That is my question. How, essentially, do sound waves differ from light waves?

Very briefly, the 'light-quantum' only appears when energy and momentum is (are ?) exchanged between light and matter. Such as when an atom or ion absorbs or emits light, or light interacts by 'bouncing off' a particle, changing its momentum ( the Compton effect).

Propagation is wavelike apart from these instances. Einstein wrote a brilliant ( and straightforward) paper in 1916 showing for the first time that momentum *must* be exchanged when emission or absorption happens, or the black-body spectrum discovered by Planck would be different.

 

[niklaus]

Also a brief answer: The quanta of sound waves are called phonons (at least when they occur in condensed matter). They also exhibit some particle-like properties.

 

[Goodison_Lad]

This argument crops up all the time. It usually seems to settle on the the meaning of 'mass'. Photons have no rest mass, but they do have relativistic mass. As pointed out, Eddington and Bondi spoke about energy having mass, but the term mass is now basically reserved for 'rest mass'.

I've seen this somewhere (can't remember), so here's my best guess at what it said:

1) A box of mass of 1kg contains, in addition, 0.5kg of matter and 0.5 kg of antimatter. Total mass of box and contents: 2kg.
2) Box is on scales, registering weight of approx. 20N.
3) Box is infinitely strong and has perfecetly reflecting inner surface etc. etc. - nothing can escape.
4) Matter and anitmatter come into contact and annihilate, producing photons which, because of (3), can't escape and aren't absorbed by the walls
5) What is the reading on scales now?
6) Has the inertia of the box changed?
7) Has the mass of the box and contents changed?

 

[Austin0]

This argument crops up all the time. It usually seems to settle on the the meaning of 'mass'. Photons have no rest mass, but they do have relativistic mass. As pointed out, Eddington and Bondi spoke about energy having mass, but the term mass is now basically reserved for 'rest mass'.

I've seen this somewhere (can't remember), so here's my best guess at what it said:

1) A box of mass of 1kg contains, in addition, 0.5kg of matter and 0.5 kg of antimatter. Total mass of box and contents: 2kg.
2) Box is on scales, registering weight of approx. 20N.
3) Box is infinitely strong and has perfecetly reflecting inner surface etc. etc. - nothing can escape.
4) Matter and anitmatter come into contact and annihilate, producing photons which, because of (3), can't escape and aren't absorbed by the walls
5) What is the reading on scales now?
6) Has the inertia of the box changed?
7) Has the mass of the box and contents changed?

Great set up.
SO what's the consensus??

 

[MVesseur]

..
1) A box of mass of 1kg contains, in addition, 0.5kg of matter and 0.5 kg of antimatter. Total mass of box and contents: 2kg.
2) Box is on scales, registering weight of approx. 20N.
3) Box is infinitely strong and has perfecetly reflecting inner surface etc. etc. - nothing can escape.
4) Matter and anitmatter come into contact and annihilate, producing photons which, because of (3), can't escape and aren't absorbed by the walls
5) What is the reading on scales now?
6) Has the inertia of the box changed?
7) Has the mass of the box and contents changed?

Interesting problem.

You'd think the mass would have to be the same - how could energy lose its mass? But will it show on the scales? Is it the kind of mass that is attracted by Earth's gravitation?

If not, can the cause be explained or illustrated?

 

[Goodison_Lad]

I think the answers are:
5) 20N
6) No
7) No

 

[rbj]

I think the answers are:
5) 20N

i don't think that's correct. i believe that, whether you look at the photons as having relativistic mass (a deprecated term) which is $E/c^2=\hbar \omega/c^2$ or as the energy density, i think that GR says they affect space-time curvature or are affected by it just as if was the equivalent mass.

 

[DrGreg]

I think the answers are:
5) 20N

i don't think that's correct. i believe that, whether you look at the photons as having relativistic mass (a deprecated term) which is $E/c^2=\hbar \omega/c^2$ or as the energy density, i think that GR says they affect space-time curvature or are affected by it just as if was the equivalent mass.

Everything you've said supports the answer being correct. I don't understand why you think your reasons make it incorrect.

 

[Dale]

But will it show on the scales? Is it the kind of mass that is attracted by Earth's gravitation?

Yes. Remember that photons blueshift as they go down and redshift as they go up, and also remember that a photon's momentum is proportional to its frequency. So the photons that hit the top of the box will have less momentum than those hitting the bottom, resulting in a net downwards force on the box.

 

[Goodison_Lad]

i don't think that's correct. i believe that, whether you look at the photons as having relativistic mass (a deprecated term) which is $E/c^2=\hbar \omega/c^2$ or as the energy density, i think that GR says they affect space-time curvature or are affected by it just as if was the equivalent mass.

Your line of reasoning is actually pretty similar to the one I used in order to arrive at the answers in the first place.

Your reasoning suggests that the answers to both (6) and (7) will be 'No', in which case the answer to (5) has to be that the weight of the box-photon system is unchanged i.e. 20N.

 

[rbj]

Everything you've said supports the answer being correct. I don't understand why you think your reasons make it incorrect.

the reasons are that sometimes i don't read through the problem sufficiently. i was thinking the 20 N was the tare weight of the box and not of the contents inside.

ooops.

 

[haruspex]

this isn't entirely true as gravity does exert a force. in the example you used youre correct, but if that applied everywhere then two masses at rest relative to each other would feel no gravitational force as there is no inital motion, which I am pretty sure isn't true (correct me if I am wrong anyone)

.

Not so. The curved spacetime explanation does not require any initial relative motion in space. All objects are moving through time. Curving spacetime can mean that the time direction in spacetime may not be the same for the observed object as for the observer.

 

[Fastman99]

When a photon falls into a black hole, the black hole increases in mass by m = E/c^2, where E was the energy of the photon. This is required by the conservation of energy and momentum. Energy can be thought of as the zeroth component of the momentum 4-vector, and each component of the 4-vector is conserved locally.

In general relativity, the curvature of spacetime is not caused by rest mass, but by the energy-momentum tensor.
http://en.wikipedia.org/wiki/Stress–energy_tensor

To answer the question of the box containing matter and antimatter:
5) The energy density of the box would not change, therefore the scale would still read as 20N
6) My question is, what is meant by "inertia"? This is a subjective question, because it doesn't refer to a measurable quantity, but to a phenomenon that is actually kinda complicated. This is usually called the conservation of energy and momentum. If you took the sum of energy and momentum of all the particles in the box before and after, then you would find it didn't change. In order to induce a change in the total energy/momentum of the box, there would have to be an opposite change in the energy/momentum of something else.

A better question that might be asked, suppose you had two boxes, one box full of normal matter, and one box made of half matter and half light. Say they are both placed on scales, and give the same number, 20N in this case. If both of them have the same total momentum not equal to zero in a particular direction (say the x-axis), then do they both travel at the same apparent velocity? Or does the half matter/light box have a higher velocity because it is "lighter?" Lol puns.

My intuitive answer is that they both move at the same velocity, but I haven't worked it out yet.

7) Mass is a loaded word, and not universally defined. If you're asking about the rest mass, then yes, the rest mass has changed. Photon have no rest mass, therefore the total rest mass of the box has decreased. However the photons do have energy and momentum. The energy and momentum of the photon could be said to have "gravitation mass" and that is conserved. The "gravitation mass" has not changed.

 

[dustinthewind]

The way I came to understand how photons can have mass and yet have no mass is that they have apparent mass because of their energy. You can calculate their energy as stated above E=$\hbar$$\omega$. So then I always took E=mc$^{2}$. And sovle for the mass of the electron. You can then consider it has momentum mc. So that when it hits say a solar sail and is reflected you receive twice the momentum of caching that photon and throwing it back. However it is stated that photons actually have no mass other than their energy and so immediately reach the speed of light. That is the energy applied to a massless object would immediately reach infinite speed for E=1/2 mv$^{2}$ As you would divide a finite energy by zero mass. However we would percieve it as having the speed c instead due to space time distortion. I think even in the relativisitc formula this would be the case $v=\sqrt{\left(\frac{k_i}{c m_l}\right){}^2+\frac{k_i 2 }{m_l}}$

 

[Goodison_Lad]

7) Mass is a loaded word, and not universally defined. If you're asking about the rest mass, then yes, the rest mass has changed. Photon have no rest mass, therefore the total rest mass of the box has decreased. However the photons do have energy and momentum. The energy and momentum of the photon could be said to have "gravitation mass" and that is conserved. The "gravitation mass" has not changed.

Regarding inertia, I suppose I should have been more specific and referred to ‘inertial mass’. Since the Principle of Equivalence is based on the equality of inertial mass and gravitational mass, the answer to (6) would be that the inertial mass is unchanged. We could measure the inertial mass of the box system by attaching it to a spring that could oscillate in the horizontal direction, with the period of oscillation being used to determine the mass. This mass would be unchanged after the annihilation process.

As for the rest mass, I would say that the rest mass of the system is therefore also unchanged, since the gravitational/inertial mass is unchanged and rest mass is simply the mass of the system determined when it is at rest. So even though the box is now full of photons, each with zero rest mass, the box system is a closed one and so its rest mass cannot have altered.

 

[yuiop]

...
7) Mass is a loaded word, and not universally defined. If you're asking about the rest mass, then yes, the rest mass has changed. Photon have no rest mass, therefore the total rest mass of the box has decreased. However the photons do have energy and momentum. The energy and momentum of the photon could be said to have "gravitation mass" and that is conserved. The "gravitation mass" has not changed.

The total rest mass of the box has not changed. Here is a simple proof that a system of photons can have rest mass:

Consider a single photon and the momentum energy expression:

$$E = \sqrt{ (m_0 c^2)^2 + (pc)^2 }$$

where E is the total energy of the photon and $m_0$ is the rest mass and p is the momentum of the photon.

Now it is generally accepted that the total energy of a photon is E = pc, where p = hf and h is the Planck constant and f is the frequency of the photon. Since E = pc it is immediately obvious from the above equation that the rest mass of a single photon is zero.

Now consider the case of a pair of photons with equal energy but going in opposite directions.

Since they momentum of equal magnitude and opposite directions, the total momentum of the photon pair is zero so we can now say:

$$E = \sqrt{ (m_0 c^2)^2 + (pc)^2 }$$

$$E = \sqrt{ (m_0 c^2)^2 + 0 }$$

$$E = m_0 c^2$$

$$m_0 = E/c^2$$

Now the rest mass of the photon pair as a system is non zero and proportional to the total energy of the photon pair.

Q.E.D.

=======================================

Now I would argue (although I have not seen it generally accepted) that rest mass is what gives a system its active gravitational properties. A single photon has no rest mass and is not a source of gravity on its own but it does have passive gravitational properties in that it responds to a gravitational field. A single photon also has the property of momentum and while this is normally associated with mass a photon clearly demonstrates that a particle can momentum without having rest mass. It would seem that the unique property of rest mass that distinguishes it from other forms of mass (other than being invariant under transformation) is that it has active gravitational mass. This is further supported by the accepted observation that photons going in the same direction are not attracted gravitationally towards each other while photons going in opposite directions are.

Now when a massive particle or pair of massive particles decay into photons, momentum is conserved and decay photons are normally produced in pairs going in opposite directions. For a box of particles that is stationary in a given coordinate system, the average momentum of the box of particles is zero and when the box of particles decays into photons conservation of momentum dictates that the total momentum of all the photons in the box is zero and therefore the system of photons has rest mass and active gravitational properties.

 

[Goodison_Lad]

Nice proof, yuiop.

I wasn't aware, though, about photons going in the same direction not attracting each other, and this set me thinking...

This is further supported by the accepted observation that photons going in the same direction are not attracted gravitationally towards each other while photons going in opposite directions are.

What would happen if the box experiment in post 47 were repeated, but this time the two annihilation photons are diverted vertically by mirrors? After the first reflection, the mirrors are retracted (as in the attached image), and the photons continue to reflect up and down in step, somewhat like the proverbial light clock. Presumably the scales would continue to register the combined mass of the box and the photons - there is still an energy density present in the box.

So if we had an identical box-photon system side-by-by side with the first one (again, resting on a set of scales), with its photons reflecting up and down in step with the photons in the first box, is it the case that the two boxes would not attract each other gravitationally (ignoring the masses of the boxes themselves for the sake of argument), even though each set of scales would register the masses of the photons? Or would they?

 

[yuiop]

... So if we had an identical box-photon system side-by-by side with the first one (again, resting on a set of scales), with its photons reflecting up and down in step with the photons in the first box, is it the case that the two boxes would not attract each other gravitationally (ignoring the masses of the boxes themselves for the sake of argument), even though each set of scales would register the masses of the photons? Or would they?

Good question! (Which is another way of saying I am not really sure of the answer :tongue:) I imagine that with the photons reflecting off the top and bottom of the box, that they exert a pressure on the container and that in turn causes tension and stresses in the walls. These stresses/pressures/tensions add to the total gravitational mass of the boxes and so the two boxes would attract each other gravitationally with the force you would expect from observing their weights on the scales. However I know nothing about GR tensors, so I cannot give a definitive answer.

 

[wmsmith]

Hello all, I am a chemistry graduate student at UCI. This question (do photons have mass) came up in a discussion section for general chemistry. The student gave the following scenario and I'm a bit stumped.

Suppose 2 photons were emitted from two points on a plane separated by a distance, d and traveling with identical velocity, v , parallel to the normal of the plane.

The question is, would the paths of the two photons ever intersect?
I.e. would their relativistic masses yield a gravitational interaction, causing their paths of motion to be distorted, or would they continue on in a straight path, forever parallel?

Or, a more physical interpretation, if a measurement were taken of their diffraction pattern at a distance r from the plane, would there ever be a large enough distance such that there would no longer be two distinct loci of maximal intensity in the diffraction pattern, but instead a single locus of maximal intensity (suppose that d was such that each photon's emission location was centered upon a node for the other photon)

One overlapping intensity maximum

Two distinct intensity maxima


My gut reaction was that they would stay parallel (two distinct intensity maxima), but I'm not 100% certain.

If either one were a particle with rest mass, the answer would be that the path of the photon would be altered...

Of course a particle with rest mass would never achieve the speed of light, so the question of their paths crossing would be mute unless the particle were so massive and dense that it acted like a black hole...

 

[pervect]

GR is a classical theory, but it's fairly well known that "photons" (classically electromagnetic waves and usually abstracted as a "null dust" in General Relativity) don't attract each other when going in the same direction.

They will attract each other if they have anti-parallel paths though.

See for instance http://prola.aps.org/abstract/PR/v37/i5/p602_1
Phys. Rev. 37, 602–615 (1931) "On the Gravitational Field Produced by Light", Richard C. Tolman, Paul Ehrenfest, and Boris Podolsky.

Tolman's result might be easier to look up in http://books.google.com/books/about/Relativity_Thermodynamics_and_Cosmology.html?id=1ZOgD9qlWtsC

Wiki also has a reference, which I haven't studied closely, but looks OK at first glance that gives the metric:

http://en.wikipedia.org/w/index.php?title=Bonnor_beam&oldid=479934048

About the only thing valid about the "relativistic mass" argument is that the "relativistic mass" of photons ( which is rather more likely to be called "energy", see any of the way-too-long threads on this point) is part of the stress-energy tensor. It's actually the stress energy tensor that causes gravity in GR, not mass, or energy.

The fact that the photons don't attract when moving parallel, and do attract when moving in anti-parallel might suggest some sort of gravitational effect that mimics magnetism. This idea turns out to have some merit - in the weak field, one can draw a useful analogy between gravity and Maxwell's equations. See for example http://en.wikipedia.org/w/index.php?title=Gravitoelectromagnetism&oldid=509366342

 

[georgir]

tl;dr
But the idea that something needs mass to be affected by gravity is obviously false - all things that have mass are affected absolutely identically by gravity, they receive exactly the same acceleration, regardless of their mass. So even if they had zero mass, it would be normal to assume they will still be affected in the same manner and get the same acceleration.

 

[wmsmith]

Thanks for the quick reply.

I get the idea of conceptualizing gravity as a warping of time-space curvature. In that context it makes perfect sense that a photon (or any other object) path would be distorted by the presence of a gravitational field. I just wasn't sure if this effect would be produced by objects with relativistic mass as well.

The fact that photons traveling in parallel will not attract but photons traveling anti-parallel will is quite surprising though. I will have to look into the stress energy tensor formulation in more detail. Thanks for the links.

 

[robinpike]

If that is such a glaring evidence for light having a mass, then would something as obvious as that be missed by the whole physics community. I mean, let's get real here. How dumb do you think physicists are to miss such a thing?

This issue has been discussed to death in several threads in both the Quantum physics forum, and the SR/GR forum. Please do your search there and figure out how gravity is a spacetime curvature, and why light follows the "geodesic" of that spacetime curvature. It has NOTHING to do with light having a mass.

Zz.

Sorry I don't understand - If gravity is the consequence of following a curved spacetime line, doesn't that explanation require another, unmentioned, force to curve the spacetime line?

 

[jtbell]

The curvature of spacetime is caused by the local energy and momentum densities (specified by the stress-energy tensor), via the Einstein field equations of general relativity.

 

[robinpike]

The curvature of spacetime is caused by the local energy and momentum densities (specified by the stress-energy tensor), via the Einstein field equations of general relativity.

Sorry I don't understand that - is there a picture showing what that means, that you can upload?

 

[andrien]

http://en.wikipedia.org/wiki/General_relativity