Does light have mass?

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Offline Pmb

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Does light have mass?
« Reply #100 on: 30/09/2009 02:19:50 »
Any *fixed* region of space containing an energy E has a mass E/c2.
Hi lightarrow! How goes it? I found this response from you to be unexpected. Normally in the passt you have used the term "mass" to mean proper mass. Here you use it to mean relativistic mass. Is there a reason for this that I'm not aware of? Thanks.

Pete

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Offline lightarrow

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« Reply #101 on: 30/09/2009 07:41:53 »
Any *fixed* region of space containing an energy E has a mass E/c2.
Hi lightarrow! How goes it? I found this response from you to be unexpected. Normally in the passt you have used the term "mass" to mean proper mass. Here you use it to mean relativistic mass. Is there a reason for this that I'm not aware of? Thanks.

Pete
No, it's proper = invariant mass even here. If the region of space is fixed, then the total momentum is zero, so from E2 = (cp)2 + (mc2)2 we can infer that m = E/c2.
« Last Edit: 30/09/2009 07:48:19 by lightarrow »

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Offline Mr. Scientist

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« Reply #102 on: 30/09/2009 18:17:34 »
Any *fixed* region of space containing an energy E has a mass E/c2.
Hi lightarrow! How goes it? I found this response from you to be unexpected. Normally in the passt you have used the term "mass" to mean proper mass. Here you use it to mean relativistic mass. Is there a reason for this that I'm not aware of? Thanks.

Pete
No, it's proper = invariant mass even here. If the region of space is fixed, then the total momentum is zero, so from E2 = (cp)2 + (mc2)2 we can infer that m = E/c2.

You'd make this so much easier if you would just shorthand this to having gamma next to Mc^2.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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Offline lightarrow

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Does light have mass?
« Reply #103 on: 30/09/2009 20:46:28 »
Any *fixed* region of space containing an energy E has a mass E/c2.
Hi lightarrow! How goes it? I found this response from you to be unexpected. Normally in the passt you have used the term "mass" to mean proper mass. Here you use it to mean relativistic mass. Is there a reason for this that I'm not aware of? Thanks.

Pete
No, it's proper = invariant mass even here. If the region of space is fixed, then the total momentum is zero, so from E2 = (cp)2 + (mc2)2 we can infer that m = E/c2.

You'd make this so much easier if you would just shorthand this to having gamma next to Mc^2.
Sorry?

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Offline Mr. Scientist

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« Reply #104 on: 01/10/2009 02:36:44 »
You know, relativistic forumla that come in the form E= \gamma Mc^2. No need for the messy definitions concerning mass.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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Offline lightarrow

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« Reply #105 on: 01/10/2009 13:11:21 »
You know, relativistic forumla that come in the form E= \gamma Mc^2. No need for the messy definitions concerning mass.
But I can't understand what exactly you mean. I proved that a system which is not moving in a specific frame of reference and which has energy, also has invariant mass. Relativistic mass is a different concept, that is, is just energy divided by c2, *always*.

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« Reply #106 on: 01/10/2009 14:18:11 »
You know, relativistic forumla that come in the form E= \gamma Mc^2. No need for the messy definitions concerning mass.
But I can't understand what exactly you mean. I proved that a system which is not moving in a specific frame of reference and which has energy, also has invariant mass. Relativistic mass is a different concept, that is, is just energy divided by c2, *always*.
No, its not. reltivistic mass invokes M= \gamma m. This means that it has zero mass. A reativistic mass is never simply E/c^2=M.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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Offline lightarrow

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« Reply #107 on: 01/10/2009 14:59:22 »
You know, relativistic forumla that come in the form E= \gamma Mc^2. No need for the messy definitions concerning mass.
But I can't understand what exactly you mean. I proved that a system which is not moving in a specific frame of reference and which has energy, also has invariant mass. Relativistic mass is a different concept, that is, is just energy divided by c2, *always*.
No, its not. reltivistic mass invokes M= \gamma m. This means that it has zero mass. A reativistic mass is never simply E/c^2=M.
Which is the energy E of a non-zero mass particle? Which is his relativistic mass? Is it different from E/c2?
Which is a photon's energy E? Which is his relativistic mass? Is it different from E/c2?

http://crib.corepower.com:8080/~relfaq/mass.html
http://w3.atomki.hu/fizmind/mag/photon_mass.html
http://crib.corepower.com:8080/~relfaq/light_mass.html
« Last Edit: 01/10/2009 15:20:06 by lightarrow »

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Offline Mr. Scientist

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« Reply #108 on: 01/10/2009 20:49:33 »
You can either listen or not. But I can assure you one last time; relativistic mass is not correct under E=Mc^2, or any algebraic manipulation.
http://www.youtube.com/watch?v=SZGcNx8nV8U

''God could not have had much time on His hands when he formed the Planck Lengths.''

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Offline Mr. Scientist

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« Reply #109 on: 01/10/2009 20:52:34 »
Look into this: http://en.wikipedia.org/wiki/Mass_in_special_relativity

Relativistic mass is an outdated concept. Read especially : The relativistic mass concept
See also: Special relativity#Mass-energy equivalence
[edit] Early developments: transverse and longitudinal mass
http://www.youtube.com/watch?v=SZGcNx8nV8U

''God could not have had much time on His hands when he formed the Planck Lengths.''

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Offline lightarrow

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« Reply #110 on: 02/10/2009 13:27:25 »
Look into this: http://en.wikipedia.org/wiki/Mass_in_special_relativity

Relativistic mass is an outdated concept.
Good news! I thought to be the only one to say this!

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Offline lightarrow

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« Reply #111 on: 02/10/2009 13:29:00 »
You can either listen or not. But I can assure you one last time; relativistic mass is not correct under E=Mc^2, or any algebraic manipulation.
Ok. So, can you please give me the definition of relativistic mass of a photon?

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Offline Mr. Scientist

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« Reply #112 on: 03/10/2009 01:43:11 »
You can either listen or not. But I can assure you one last time; relativistic mass is not correct under E=Mc^2, or any algebraic manipulation.
Ok. So, can you please give me the definition of relativistic mass of a photon?

The mass of a photon in relativity is zero but has itself, a non-zero energy:

E=M^2c^4+p^2c^2

Plugging in the appropriate values, one finally assumes that for Mc^2, we actually have:

E= \gamma Mc^2

Which reduces its mass total to zero; this is the mass of the photon.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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Offline lightarrow

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« Reply #113 on: 03/10/2009 09:09:00 »
You can either listen or not. But I can assure you one last time; relativistic mass is not correct under E=Mc^2, or any algebraic manipulation.
Ok. So, can you please give me the definition of relativistic mass of a photon?

The mass of a photon in relativity is zero but has itself, a non-zero energy:

E=M^2c^4+p^2c^2
This is correct.

Quote
Plugging in the appropriate values, one finally assumes that for Mc^2, we actually have:

E= \gamma Mc^2
This is *not* correct for photons. What does infinite multiplied by zero means?

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Offline Mr. Scientist

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« Reply #114 on: 03/10/2009 17:17:53 »
You can either listen or not. But I can assure you one last time; relativistic mass is not correct under E=Mc^2, or any algebraic manipulation.
Ok. So, can you please give me the definition of relativistic mass of a photon?

The mass of a photon in relativity is zero but has itself, a non-zero energy:

E=M^2c^4+p^2c^2
This is correct.

Quote
Plugging in the appropriate values, one finally assumes that for Mc^2, we actually have:

E= \gamma Mc^2
This is *not* correct for photons. What does infinite multiplied by zero means?

It is correct for photons, because M is the rest mass, and \gamma makes the value of matter to zero. That is why the rest energy of a photon is given by: E=\gamma Mc^2. You can learn this stuff quite independantly and easily on web sites spralled all over the place.
http://www.youtube.com/watch?v=SZGcNx8nV8U

''God could not have had much time on His hands when he formed the Planck Lengths.''

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Offline Mr. Scientist

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« Reply #115 on: 03/10/2009 17:19:07 »
You can either listen or not. But I can assure you one last time; relativistic mass is not correct under E=Mc^2, or any algebraic manipulation.
Ok. So, can you please give me the definition of relativistic mass of a photon?

The mass of a photon in relativity is zero but has itself, a non-zero energy:

E=M^2c^4+p^2c^2
This is correct.

Quote
Plugging in the appropriate values, one finally assumes that for Mc^2, we actually have:

E= \gamma Mc^2
This is *not* correct for photons. What does infinite multiplied by zero means?

Also, where i have bolded; what infinite value? There is no infinite value in question here. ts a simple case of algebra.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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Offline lightarrow

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« Reply #116 on: 04/10/2009 13:21:29 »
It is correct for photons, because M is the rest mass, and \gamma makes the value of matter to zero. That is why the rest energy of a photon is given by: E=\gamma Mc^2. You can learn this stuff quite independantly and easily on web sites spralled all over the place.

And how much is gamma for a photon?
You shouldn't base your knowledge on internet sites only, you should also go to school, at least...

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Offline Mr. Scientist

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« Reply #117 on: 04/10/2009 14:10:50 »
It is correct for photons, because M is the rest mass, and \gamma makes the value of matter to zero. That is why the rest energy of a photon is given by: E=\gamma Mc^2. You can learn this stuff quite independantly and easily on web sites spralled all over the place.

And how much is gamma for a photon?
You shouldn't base your knowledge on internet sites only, you should also go to school, at least...

Talking about ''infinities'' did nothing for the conversation. And I dont need more school. I've had a shitload of it so far; i have education and certificates in physics too.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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Offline lightarrow

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« Reply #118 on: 04/10/2009 19:06:22 »
Talking about ''infinities'' did nothing for the conversation. And I dont need more school. I've had a shitload of it so far; i have education and certificates in physics too.
Well, so you should know that m*gamma is meaningless for a photon.
At high school they call it "Indeterminate form":
http://en.wikipedia.org/wiki/Indeterminate_form
« Last Edit: 04/10/2009 19:08:34 by lightarrow »

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Offline Mr. Scientist

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« Reply #119 on: 05/10/2009 09:20:10 »
Talking about ''infinities'' did nothing for the conversation. And I dont need more school. I've had a shitload of it so far; i have education and certificates in physics too.
Well, so you should know that m*gamma is meaningless for a photon.
At high school they call it "Indeterminate form":
http://en.wikipedia.org/wiki/Indeterminate_form

M alone is meanngless, not to mention incorrect.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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Offline lightarrow

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« Reply #120 on: 05/10/2009 12:15:14 »
You wrote:

<<E= \gamma Mc^2
Which reduces its mass total to zero; this is the mass of the photon>>

If you didn't mean gamma*Mc^2, what did you mean, then???

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Offline Mr. Scientist

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« Reply #121 on: 05/10/2009 14:23:54 »
Let me rephrase this.

E is not the ''energy'' alone. When relativity formulated the equation E=Mc^2 in this specific form referred to the rest mass of a particle. Which means does not include [ in fact - never involved] the description of photons. The photon has a non-zero energy as it is the packet of pure kinetical energy, but this energy is not of a rest form associated to a particle with a mass M.

Instead one needs to reduce to mass to zero, to describe the rest energy of a photon to also be zero in quantity; E=\gamma Mc^2. These are equations used frequently in relativity for the same purposes posted above. Also to clarify, i told you all this because you inferred to the outdated concept of relativistic mass - outdated in the sense that in a qualitative physics course, lecturers usually inform us that the term relativistic mass is hardly ever used nowadays in an academic sense, to cause less confusion; something which i earlier highlighted which you where inexorably conducting.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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Offline lightarrow

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« Reply #122 on: 05/10/2009 20:31:13 »
Let me rephrase this.

E is not the ''energy'' alone. When relativity formulated the equation E=Mc^2 in this specific form referred to the rest mass of a particle. Which means does not include [ in fact - never involved] the description of photons. The photon has a non-zero energy as it is the packet of pure kinetical energy, but this energy is not of a rest form associated to a particle with a mass M.
...and it's exactly for this reason that writing it for a photon, as you did, is meaningless.

Quote
Instead one needs to reduce to mass to zero, to describe the rest energy of a photon to also be zero in quantity; E=\gamma Mc^2. These are equations used frequently in relativity for the same purposes posted above.
I still cannot understand what purposes it can have.
Before your intervention in response of my post, I had used the correct equation E2 = (mc2)2 + (cp)2 and I had never talked of relativistic mass; then you come with the equation E = \gamma Mc2 which is not correct, in general, because is valid only for non-zero mass bodies and which uses a different symbol for the mass (and here it is my erroneous believe that you were talking about relativistic mass).

Then arrives your post with this statement:
<<You'd make this so much easier if you would just shorthand this to having gamma next to Mc^2>>
and I ask you again: what does it mean? It seems meaningless to me.

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Offline yor_on

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« Reply #123 on: 08/10/2009 00:48:21 »
does it?

Not as I see it?

If light had a mass how much heavier would it make a supernova by any chosen magnitude? Should I assume that the supernova before exploding versus after, if able to assemble all its light, to then 'mass' the same?

Light constantly do 'things' mass can't. They are immaterial not able to define except when impacting (photons) like being seen by your eye (photons/waves).

"The definition of the invariant mass of an object is  m = sqrt{E2/c4 - p2/c2}. By this definition a beam of light, is massless like the photons it is composed of. However, if light is trapped in a box with perfect mirrors so the photons are continually reflected back and forth in the box, then the total momentum is zero in the boxes frame of reference but the energy is not. Therefore the light adds a small contribution to the mass of the box. This could be measured - in principle at least - either by an increase in inertia when the box is slowly accelerated or by an increase in its gravitational pull. You might say that the light in the box has mass but it would be more correct to say that the light contributes to the total mass of the box of light. You should not use this to justify the statement that light has mass in general."  http://crib.corepower.com:8080/~relfaq/light_mass.html

To that I would like to add that this 'system' as discussed above may be defined as having no 'momentum' as the light 'bounces' inside the box, and as we define the box itself to be the total 'system'. But this light 'bouncing' still have both a speed and a distance traveled in time as observed by us. ( if we assume that light do 'travel' that is :). If it does so then there will be intervals between its 'bounces' where that box will get no action/reaction. So to prove the concept I think you would need to have the light somehow 'frozen' floating freely inside that box and then weight it. And if you 'freeze' it you are acting on it, introducing a force, and as I see it invalidating the claim of it having a 'restmass' like a particle.
« Last Edit: 08/10/2009 00:51:01 by yor_on »
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Offline Vern

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« Reply #124 on: 08/10/2009 03:32:32 »
yor_on; you are thinking [:)] You can also consider photons of light as mass themselves. They then do not have mass. They are mass. I have made that statement a few times lately and it has not been challenged, but I am sure most folks are not comfortable with it. I arrived at that by just looking at the arithmetic. m = hv / c2. Then just choose the units to eliminate the constants and we are left with m = v; or mass = electromagnetic change. Then restate it simply; mass is electromagnetic change.

Electromagnetic change is any change in the electric and magnetic charge amplitude in a localized area that can be considered as a system.

« Last Edit: 08/10/2009 03:36:44 by Vern »

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Offline Pmb

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« Reply #125 on: 08/10/2009 05:50:02 »
You know, relativistic forumla that come in the form E= \gamma Mc^2. No need for the messy definitions concerning mass.
But I can't understand what exactly you mean. I proved that a system which is not moving in a specific frame of reference and which has energy, also has invariant mass. Relativistic mass is a different concept, that is, is just energy divided by c2, *always*.
That's not true. Relativistic mass is the ration m = p/v. If a body is under stress then m does not equal E/c^2.

Here is an example: http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

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Offline Pmb

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« Reply #126 on: 08/10/2009 05:52:55 »
Any *fixed* region of space containing an energy E has a mass E/c2.
Hi lightarrow! How goes it? I found this response from you to be unexpected. Normally in the passt you have used the term "mass" to mean proper mass. Here you use it to mean relativistic mass. Is there a reason for this that I'm not aware of? Thanks.

Pete
No, it's proper = invariant mass even here. If the region of space is fixed, then the total momentum is zero, so from E2 = (cp)2 + (mc2)2 we can infer that m = E/c2.
Perhaps I'm not clearon what you mean by "region of space is fixed". What does that mean? Thanks.

Pete

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Offline Pmb

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« Reply #127 on: 08/10/2009 05:57:29 »
Look into this: http://en.wikipedia.org/wiki/Mass_in_special_relativity

Relativistic mass is an outdated concept.
Good news! I thought to be the only one to say this!
That is not correct. Relativistic mass is not an outdated concept. It's used in Cosmology a lot. A survey of recent relativity literature was done by Gary Oas which showed that it's widely used in modern textbooks. I've seen in used in the American Journal of Physics too. It's a very meaningful concept and you can get into trouble if you try to use invariant mass as the definition of mass when you venture outside its use in particle physics

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Offline lightarrow

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« Reply #128 on: 08/10/2009 13:26:58 »
You know, relativistic forumla that come in the form E= \gamma Mc^2. No need for the messy definitions concerning mass.
But I can't understand what exactly you mean. I proved that a system which is not moving in a specific frame of reference and which has energy, also has invariant mass. Relativistic mass is a different concept, that is, is just energy divided by c2, *always*.
That's not true. Relativistic mass is the ration m = p/v. If a body is under stress then m does not equal E/c^2.

Here is an example: http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm
Yes, I had seen that page. Unfortunately I couldn't understand it. [:-'(]

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Offline lightarrow

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« Reply #129 on: 08/10/2009 13:32:28 »
No, it's proper = invariant mass even here. If the region of space is fixed, then the total momentum is zero, so from E2 = (cp)2 + (mc2)2 we can infer that m = E/c2.
Perhaps I'm not clear what you mean by "region of space is fixed". What does that mean? Thanks.
Pete
It mean stationary in the frame of reference considered. IMHO, if you consider that region of space as system, as long as a beam of light (for example) crosses that region, the system acquires (proper) mass, even if the light beam itself hasn't. I know it sounds a bit...esoteric [:)]  but I cannot see how it could be wrong.
« Last Edit: 08/10/2009 13:33:59 by lightarrow »

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« Reply #130 on: 08/10/2009 13:34:55 »
Look into this: http://en.wikipedia.org/wiki/Mass_in_special_relativity

Relativistic mass is an outdated concept.
Good news! I thought to be the only one to say this!
That is not correct. Relativistic mass is not an outdated concept. It's used in Cosmology a lot. A survey of recent relativity literature was done by Gary Oas which showed that it's widely used in modern textbooks. I've seen in used in the American Journal of Physics too. It's a very meaningful concept and you can get into trouble if you try to use invariant mass as the definition of mass when you venture outside its use in particle physics
For example?

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« Reply #131 on: 08/10/2009 21:14:12 »
yor_on; you are thinking [:)] You can also consider photons of light as mass themselves. They then do not have mass. They are mass. I have made that statement a few times lately and it has not been challenged, but I am sure most folks are not comfortable with it. I arrived at that by just looking at the arithmetic. m = hv / c2. Then just choose the units to eliminate the constants and we are left with m = v; or mass = electromagnetic change. Then restate it simply; mass is electromagnetic change.

Electromagnetic change is any change in the electric and magnetic charge amplitude in a localized area that can be considered as a system.

It sound as if we're born under the same full moon here Vern :)

Electromagnetic charge?
How much would Earth and the moon need to have to attract each other?
Or the solarsystem as a whole, and why don't we notice it?
« Last Edit: 09/10/2009 01:18:12 by yor_on »
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« Reply #132 on: 09/10/2009 00:45:11 »
Well; the earth and the moon have as much as they have in the form of the accumulations of their matter. It works just fine. In the innards of the matter the electromagnetic change is taking place giving substance to the matter.

I don't see a problem there.

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« Reply #133 on: 10/10/2009 07:01:20 »
Look into this: http://en.wikipedia.org/wiki/Mass_in_special_relativity

Relativistic mass is an outdated concept.
Good news! I thought to be the only one to say this!
That is not correct. Relativistic mass is not an outdated concept. It's used in Cosmology a lot. A survey of recent relativity literature was done by Gary Oas which showed that it's widely used in modern textbooks. I've seen in used in the American Journal of Physics too. It's a very meaningful concept and you can get into trouble if you try to use invariant mass as the definition of mass when you venture outside its use in particle physics
For example?
I constantly see people make the following mistakes

(1) The weight of a body does not depend on its speed
(2) The gravitational field of a body does not depend on its speed
(3) The mass density of radiation is zero
(4) Light cannot generate a gravitational field
(5) The ratio p/v is always equal to E/c^2
(6) Relativistic mass is not conserved in nuclear reactions
etc

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« Reply #134 on: 10/10/2009 13:02:33 »
I constantly see people make the following mistakes

(1) The weight of a body does not depend on its speed
(2) The gravitational field of a body does not depend on its speed
(3) The mass density of radiation is zero
(4) Light cannot generate a gravitational field

Maybe it is not necessary to talk about relativistic mass in these cases, if we start from my previous considerations.

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« Reply #135 on: 10/10/2009 13:27:08 »
Quote from: PMB
(4) Light cannot generate a gravitational field
I suspect that this statement is wrong. How do photons attract each other gravitationally if this is so?

Edit: Maybe i misunderstood. What is the mistake? Are the statements mistakenly correct or mistakenly incorrect?
« Last Edit: 10/10/2009 14:18:17 by Vern »

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« Reply #136 on: 10/10/2009 15:05:49 »
Yup he is wrong.

A photon generates a gravitational curvature as it accelerates through spacetime. In fact, curvature, gravitational waves and acceleration are all part and parcel the same thing.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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« Reply #137 on: 10/10/2009 15:12:00 »
Quote from: PMB
(4) Light cannot generate a gravitational field
I suspect that this statement is wrong. How do photons attract each other gravitationally if this is so?

Edit: Maybe i misunderstood. What is the mistake? Are the statements mistakenly correct or mistakenly incorrect?
PMB wrote that those statements are *incorrect*.

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« Reply #138 on: 10/10/2009 15:15:09 »
If that's the case; the first statement is true. So the author has made one mistake or another.
http://www.youtube.com/watch?v=SZGcNx8nV8U

''God could not have had much time on His hands when he formed the Planck Lengths.''

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« Reply #139 on: 10/10/2009 15:16:03 »
In fact, so are the rest apart from the one mentioned.
http://www.youtube.com/watch?v=SZGcNx8nV8U

''God could not have had much time on His hands when he formed the Planck Lengths.''

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« Reply #140 on: 10/10/2009 17:31:20 »
You know, relativistic forumla that come in the form E= \gamma Mc^2. No need for the messy definitions concerning mass.
But I can't understand what exactly you mean. I proved that a system which is not moving in a specific frame of reference and which has energy, also has invariant mass. Relativistic mass is a different concept, that is, is just energy divided by c2, *always*.
That's not true. Relativistic mass is the ration m = p/v. If a body is under stress then m does not equal E/c^2.

Here is an example: http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

Geocities, by the way, is not really a place to be reciting physics papers. In fact, its pretty much the place for the dregs of science society.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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« Reply #141 on: 11/10/2009 01:19:47 »
Quote from: PMB
(4) Light cannot generate a gravitational field
I suspect that this statement is wrong. How do photons attract each other gravitationally if this is so?

Edit: Maybe i misunderstood. What is the mistake? Are the statements mistakenly correct or mistakenly incorrect?
All of those statements are incorrect. I posted them as common errors I've seen people make over and over and over again over the last ten years I've been discussing the subject.

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« Reply #142 on: 11/10/2009 01:21:35 »
Geocities, by the way, is not really a place to be reciting physics papers. In fact, its pretty much the place for the dregs of science society.
I used it only to place calculations I've made so that other people can read the proofs I've constructed. So basically I only reference my own web pages and only then to post proofs I've worked out the math to.

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« Reply #143 on: 11/10/2009 01:24:16 »
Yup he is wrong.

A photon generates a gravitational curvature as it accelerates through spacetime. In fact, curvature, gravitational waves and acceleration are all part and parcel the same thing.
Please reread my post where I started it with I constantly see people make the following mistakes.

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« Reply #144 on: 14/10/2009 23:53:10 »
It seems obvious that the techical side of this arguement will need to be.. well.. argued.
http://www.youtube.com/watch?v=SZGcNx8nV8U

''God could not have had much time on His hands when he formed the Planck Lengths.''

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« Reply #145 on: 16/10/2009 20:14:41 »
Pmb  this is what you see as mistakes right?

(1) The weight of a body does not depend on its speed

So what you mean is that the weight of a body has a direct relation to its speed?
I know that the 'momentum' goes up, but, consider something (object) accelerating and then traveling uniformly. Knowing that you will notice no extra 'jiggling' from the objects atoms traveling, why will it weight more? Are you assuming its gravitational field to grow as its speed grows and therefore by definition also 'weight' more? As a larger gravitational field either should crave more 'relative mass/momentum' or 'invariant mass'

It's an interesting thought
And you define it as weight which makes sense here, not mass.

(2) The gravitational field of a body does not depend on its speed

Seems in that case a variation of the first Q. right? As we now defined speed as creating greater gravitational ripples in SpaceTime.

(3) The mass density of radiation is zero.
Which then should read as 'non-zero'.

Radiation and 'mass'? Are we referring to 'invariant mass'?
There are different views there, and some might agree but I beg to differ. It depends on your definitions. Me, I prefer the concept of momentum which I see as different from 'invariant mass'. And radiation is a 'mass less' thing as defined in main stream physics..

http://www.weburbia.com/physics/light_mass.html
http://arxiv.org/pdf/hep-ph/9509415v1


(4) Light cannot generate a gravitational field
Well, light can 'push' and they have what I see as 'momentum'. To be able to create a gravitational field it seems to me then that you are saying that momentum can have/create a gravitational field.

"In the "complete" version of GR there is (i) Einstein field equation which relates spacetime curvature and the stress-energy-momentum tensor (ii) Maxwell and other equations which describe the electromagnetic field, charged particles and other "stuff" (iii) A prescription for assigning stress-energy-momentum to stuff. In that language, it is the stress-energy-momentum of stuff which causes spacetime curvature which is gravity. However, it is also often said that "gravity" causes gravity, by which it is meant that the Einstein field equations are nonlinear in the spacetime metric. In an "approximate" version of the theory in which gravity waves propagate on flat spacetime, it is possible to assign gravity some energy. However, this energy is not the same energy as the stress-energy-momentum tensor of the "complete" theory."

So I'm not sure if momentum could create a 'gravitational field'. it also depends on how you view gravitation and momentum/'relative mass'. As 'forces' or as expedients expressing SpaceTimes curvature and the 'disturbance of equilibrium' I see it as. Momentum as an effect won't show up until you 'change' the systems parameters as I understands it, like impacting. So? How could something 'not there / measurable' until then create a gravitational field.

The other two you will have to expand on.

---

thinking some more of your first idea :)
The only thing you speed could be is uniform, am I right?
And if so we fall back to momentum again.
As we can't define uniform speeds as I understands it without comparing?
And with momentum it is like in (4)

Perhaps you meant something different?

----

To be fair I saw this though?

"Theoretical physicist and Nobel laureate Sheldon Glashow explains for PBS NOVA online (Einstein's Big Idea, October 11, 2005) that, "When an object emits light, say, a flashlight, it gets lighter.""

Which seems to support your ideas although I saw no explanation as to how this mass was thought to be lost?
« Last Edit: 16/10/2009 22:57:51 by yor_on »
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« Reply #146 on: 20/10/2009 01:42:51 »
Quote from: yor_on
(1) The weight of a body does not depend on its speed
...

So what you mean is that the weight of a body has a direct relation to its speed?
No. Let me be clear here for people who haven't read all my responses above; The assertion The weight of a body does not depend on its speed is wrong. I posted it as example of an error I see quite often all over the internet. It was even made by a a man who teaches GR at MIT when I first asked him that question. I explained why I thought he was wrong. He later sat down and calculated it and found I was correct. He then made it an homework problem. :)  This is one of the problems with saying "mass doesn't depend on speed" because it can lead you to false assumptions. One can always do the calculations but someone wouldn't do a calcuation if he was sure the answer was zero to begin with. This is one reason I object to people trying to wash the concept of relativsitic mass from the minds of students. Relativists in the early 20th century knew all this and why they defined relativistic mass to begin with. They didn't make an error or were being sloppy or anything like that. They had very sound reasons for what they did.

You might ask what exactly it means for a moving body to weigh more. This is as simple as it sounds. Let me give you an example; Let S be a frame of reference in which the metric has the value

ds^2 = (1 + gz/c^2)^2 dt^2 - dx^2 - dy^2 - dz^2

Then the frame is one in which there is present a uniform gravitational field. The coordinate clock is located at z = 0. Let there be a car sitting at rest at x = y = z = 0 on a huge scale (i.e. a pressure plate/spring scale etc. which measures the weight of whatever is sitting on it). Let W_0 represent the weight of the car when it is at rest. Then W_0 = mg where m is the car's proper mass (aka "rest mass") and g is the locally measured gravitational acceleration at z = 0). Now let the car be moving at a speed v on the scale in the z = 0 plane. Then the weight, W, measured by the scale would be W = mg/sqrt(1 - v^2/c^2) = Mg where M = m/sqrt(1 - v^2/c^2) is the relativistic mass of the car. M > 0. That means that a moving body weighs more than the same object at rest. For proof of this see
http://www.geocities.com/physics_world/gr/weight_moving_body.htm

I provided two methods to arrive at this result. One is a heuristic approach while the other is a straight forward derivation using the principles and techniques of general relativity

Notice that M is also the transverse mass of the car. That this being the measured value was first hinted at in Einstein's first paper on relativity, i.e. On the Electrodynamics of Moving Bodies in section 10 where he noted
Quote
(This force might be measured, for example, by a spring balance at rest in the last-mentioned system.)
Quote from: yor_on
I know that the 'momentum' goes up, but, consider something (object) accelerating and then traveling uniformly. Knowing that you will notice no extra 'jiggling' from the objects atoms traveling, why will it weight more?
It's due to the properties of spacetime. Looking at the metric one can deduce that it is a direct result of time dilation. Also please note that the "jiggling" of the atoms in the body that you mentioned would be different than if the body is at rest. E.g. if the atoms in the body were only moving at cm/s in the rest frame and the car was moving at near the speed of light then as measure in frame S the atoms would also be meaning at near the speed of light.

Quote from: yor_on
Are you assuming its gravitational field to grow as its speed grows and therefore by definition also 'weight' more?
No. The gravitational field of the car is ignored. This effect is desccribed as saying that the passive gravitational mass of the body has increased with its speed.

Quote from: yor_on
It's an interesting thought
And you define it as weight which makes sense here, not mass.
The reason I mentioned weight is that weight is defined in terms of mass. Therefore when speaking of mass one should consider areas of physics where mass is applied. Note that weight is different than how force is defined - All weights are forces but not all forces are weights. In particular weight essentially defines the term passive gravitational mass. The equivalence principle from GR states that passive gravitational mass equals inertial mass which is exactly what relativistic mass is.
Quote from: yor_on
(2) The gravitational field of a body does not depend on its speed

Seems in that case a variation of the first Q. right? As we now defined speed as creating greater gravitational ripples in SpaceTime.
No. In the first case I spoke of a gravitational field without considering its source, just the properties of the spacetime of the field.
Quote from: yor_on
The mass density of radiation is zero.
Which then should read as 'non-zero'.
Yes. In that case that is a true statement in general.

Quote from: yor_on
Radiation and 'mass'? Are we referring to 'invariant mass'?
When speaking about densities associated with the stress-energy-momentum of a matter distribution then there is no associated invariant quantity that I'm aware. There is a special quantity which one might refer to as proper mass density. That would be the mass density of the matter as measured in the frame of reference in which the momentum density is zero.

Let us consider a gas of radiation which is unboundend spatially. It's possible that this radiation consists of photons all traveling in the same direction. The invariant mass of that system is zero. Now let us consider a frame of reference in which there is disordered radiation, i.e. photons flying in random directions but the total momentum of all the photons is zero. This frame is the zero momentum frame. The invariant mass of such radiation is non-zero.

Quote from: yor_on
Me, I prefer the concept of momentum which I see as different from 'invariant mass'. And radiation is a 'mass less' thing as defined in main stream physics..
In my opinion it is misleading to say that what you say is how its defined in mass stream physics. Main stream physics includes cosmology, does it not? In cosmology they employ the concepts of relativistic mass. If you were to


Quote from: yor_on
http://www.weburbia.com/physics/light_mass.html
See
http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
Quote from: yor_on
http://arxiv.org/pdf/hep-ph/9509415v1
See
http://arxiv.org/abs/physics/0504110

Notice the percentage of modern general relativity texts which employ the concept of relativistic mass.

Quote from: yor_on
(4) Light cannot generate a gravitational field
Well, light can 'push' and they have what I see as 'momentum'. To be able to create a gravitational field it seems to me then that you are saying that momentum can have/create a gravitational field.
That is correct. For an example please see
http://www.geocities.com/physics_world/gr/grav_light.htm

Quote from: yor_on
The other two you will have to expand on.
Number 5 reads) The ratio p/v is always equal to E/c^2

The ratio p/v is the actual definition of relativistic mass. Let's calll that M as above. The relation ship is then E = Mc^2. Anti-relativistic mass proponents assert that M is just E in different units and conclude that since we have E then we don't need M, since it's superfluos. However M doees not always equal E/c^2. So that argument is flawed. In fact that claim that relativistic mass and energy are the same thing is also flawed since they have different definitions. Asserting that because they are numerically proportional they are the same thing is a flawed argument. This was pointed out by others as well, e.g. see the journal article (which I can send/make available to you)

On the Inertial Mass Concept in Special and General Relativity, by Mendel Sachs, Foundations of Physics Letters, Vol. 1, No. 2, 1988
Quote from: yor_on
(6) Relativistic mass is not conserved in nuclear reactions
etc
That relativistic mass is trule conserved in Nuclear reactions is trivial to prove. I have worked out an example here
http://www.geocities.com/physics_world/sr/nuclear_fission.htm

Quote from: yor_on
To be fair I saw this though?

"Theoretical physicist and Nobel laureate Sheldon Glashow explains for PBS NOVA online (Einstein's Big Idea, October 11, 2005) that, "When an object emits light, say, a flashlight, it gets lighter.""
Nobody denies that. Both sides of the relativistic mass/rest mass debate agree that this is true.

Just to be sure that we're on the same page (somone was confused about this) you do understand, don't you, that when I wrote this list

(1) The weight of a body does not depend on its speed
(2) The gravitational field of a body does not depend on its speed
(3) The mass density of radiation is zero
(4) Light cannot generate a gravitational field
(5) The ratio p/v is always equal to E/c^2
(6) Relativistic mass is not conserved in nuclear reactions
etc

I did state that I constantly see people make the following mistakes
. That means that what is stated in items 1->6 are definitely wrong and are mistakes. So we're all clear on that, right? :)

Pete

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« Reply #147 on: 23/10/2009 00:27:27 »
Well I tried to turn them around as I presumed you meant?
So I'm not sure why you state a 'no' at the beginning of your answer?

----
1) The weight of a body does not depend on its speed (yours)
...
So what you mean is that the weight of a body has a direct relation to its speed?
(Mine understanding of it reversed)
-------

As for relativistic mass or momentum :)
Einstein preferred the last one, didn't he?

http://en.wikipedia.org/wiki/Mass_in_special_relativity#Early_developments:_transverse_and_longitudinal_mass

-----------------------

"Passive gravitational mass is a measure of the strength of an object's interaction with a gravitational field. Passive gravitational mass is determined by dividing an object’s weight by its free-fall acceleration. Two objects within the same gravitational field will experience the same acceleration; however, the object with a smaller passive gravitational mass will experience a smaller force (less weight) than the object with a larger passive gravitational mass."

Do you mean that the weight will differ proportionally to what they weighted before if you let two weights of different mass fall into a black hole 'traveling' side by side?
With them keeping their relative weight-differences although now larger.

That makes sense to me

Looks to me that you are arguing that the only mass worth considering is the 'relativistic mass' as being the most 'exact' is that correct?

-----------

As for your writing.
"Also please note that the "jiggling" of the atoms in the body that you mentioned would be different than if the body is at rest. E.g. if the atoms in the body were only moving at cm/s in the rest frame and the car was moving at near the speed of light then as measure in frame S the atoms would also be meaning at near the speed of light."

You got me stymied there :)

Are you saying that those atoms traveling near light in a uniform motion, will jiggle more as measured inside that frame than when they measure them while at rest?

And that should mean that they will radiate too right?
So then we might see our uniform motion in 'radiation' from those atoms.
Which then should mean that I'm wrong believing that we can't define a uniform motion other than relative another frame?

Can you present any experimental proof for that concept?
I would really like to see it


---
As for momentum and relative mass

Momentum http://www.physicsforums.com/library.php?do=view_item&itemid=53

And relative mass seems to me, somewhat loosely expressed, an inertial property defined by its unwillingness to react to changes in velocity and gravity dynamically changing with gravity/acceleration inside its own frame and motion as compared from another frame. And invariant mass is what you have left when all gravitational and accelerating/moving forces are neutralized when measuring the objects 'unwillingness' to change energy level while moving in a straight line with a steady velocity.


" Since the center of mass of an isolated system moves in a straight line with a steady velocity, an observer can always move along with it. In this frame, the center of momentum frame, the total momentum is zero, the system as a whole may be thought of as being "at rest" (though in a disconnected system parts may be moving relative to each other), and the invariant mass of the system is equal to the total system energy divided by c2."

This might be of interest?
http://www.physicsforums.com/showthread.php?t=60204

-----------

And as for if light have mass.
Wish I knew, don't think so myself.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/blahol.html#c2
But they are talking about gravitational potential energy
Not an 'eigen-mass' of a photon a. k. a. invariant mass

Do you have experimental proofs for that it has?
That light bends is to me due to SpaceTimes geodesics?

----

You also write.
"Let us consider a gas of radiation which is unboundend spatially. It's possible that this radiation consists of photons all traveling in the same direction. The invariant mass of that system is zero. Now let us consider a frame of reference in which there is disordered radiation, i.e. photons flying in random directions but the total momentum of all the photons is zero. This frame is the zero momentum frame. The invariant mass of such radiation is non-zero."

That seems to have to do with how you define your frame of reference right?
If seen as a 'whole system'  there will be a zero 'invariant mass' but if you go into that frame and pick two photons traveling you have another frame of reference in which you may see them as producing a 'mass' just as in that perfectly reflecting box where you enclose light. If I remember right Lightarrow had an example of that too.

But to me there are other explanations to why light bend.
Take two parallel light beams traveling f ex.

"The curvature of spacetime is related to the stress energy (the Ricci tensor). Light would contribute to this. But the contribution is mind bogglingly small, so two parallel rays would not be drawn together in the scale of this universe."

That's not the same as what you propose. But it will make light bend as the stress energy momentum curves SpaceTime  and as I see it, velocity gains momentum. And with that momentum you will get a stress energy tensor which in its turn means a curvature. So photons will curve SpaceTime too.


And now I've argued from a mainstream view, sort of :)
But when we come to my wacky theories it will be downright strange ::))
« Last Edit: 23/10/2009 04:51:59 by yor_on »
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Offline yor_on

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Does light have mass?
« Reply #148 on: 23/10/2009 03:56:57 »
To me Inertia makes most sense when I'm considering SpaceTime to be a field. Then that field will interact with you, wherever you are. If you consider it as gravitons then they will have to be just as 'thick' a soup as that field is?

But hey, what about this then :)
http://www.geocities.com/area51/shadowlands/6583/project289.html

( And my new war cry will be. . . Electrons do not 'orbit' :)
« Last Edit: 23/10/2009 03:59:25 by yor_on »
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Offline litespeed

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Does light have mass?
« Reply #149 on: 01/11/2009 19:34:22 »
Photons do not have mass, but they do have wide varience in energy that is often expressed as wave length. Apparently this means they can create mass when the reflect off something that reduces their wave length to a lower number, thus creating acceleration and mass in the reflective object.