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mathman

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Thanks, Mathman.

Does this imply photons with more energy have more gravity, and will 'react' differently to a planet/star's gravity? Has this been observed?

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Does this imply photons with more energy have more gravity, and will 'react' differently to a planet/star's gravity? Has this been observed?

It has been observed that photon's deflection angle does not depend on its energy/momentum (or wavelength). Otherwise, the (displaced) image of the distant star would be seen as a little rainbow. It fact, it remains point-like in spite of the light bending. The deflection angle is (approximately) [tex] \Delta p/p [/tex], where p is photon's momentum and [tex] \Delta p [/tex] is the transversal momentum change in the gravity field. From this you can conclude that [tex] \Delta p [/tex] is proportional to p. So, yes, higher energy photons are deflected more.

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I am not aware of such experimental details. On the other hand, it is plausible that the proportionality between [tex]\Delta p [/tex] and p holds only approximately, so some minor rainbowing may be present in exact theory, even if the chromosphere effect is not present.

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I thought photons just followed a "straight" path through curved space time (or whatever is closest to straight), so why would their energy make any difference? Any photon travels at the same speed and flies through the same curved space-time along the shortest path.I am not aware of such experimental details. On the other hand, it is plausible that the proportionality between [tex]\Delta p [/tex] and p holds only approximately, so some minor rainbowing may be present in exact theory, even if the chromosphere effect is not present.

As for the gravity created by photons: yes, photons don't have mass but they do have energy and any kind of energy (even including gravity itself) causes gravity. In fact, the amount of gravity coming from the photons inside the sun is probably far from negligible. They have so much energy that their pressure is even important in helping to keep the sun from collapsing, so we probably feel their gravity quite noticeably too.

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Wheeler thought that energy created gravity.

http://en.wikipedia.org/wiki/Geon_(physics [Broken])

http://search.yahoo.com/search?p=geons&ei=UTF-8&fr=hp-pvdt

You may have to go to search and click on the link there.

http://en.wikipedia.org/wiki/Geon_(physics [Broken])

http://search.yahoo.com/search?p=geons&ei=UTF-8&fr=hp-pvdt

You may have to go to search and click on the link there.

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As for the gravity created by photons: yes, photons don't have mass but they do have energy and any kind of energy (even including gravity itself) causes gravity. In fact, the amount of gravity coming from the photons inside the sun is probably far from negligible. They have so much energy that their pressure is even important in helping to keep the sun from collapsing, so we probably feel their gravity quite noticeably too.

This brings up another controversial matter though, because according to Einstein energy and mass are equivalent, so saying that photons have no mass would in turn say that they have no energy, and vise versa, in saying they have energy says they have mass, so either Einstein was "wrong" or there is something about photons that we still don't know

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As for the gravity created by photons: yes, photons don't have mass but they do have energy and any kind of energy (even including gravity itself) causes gravity. In fact, the amount of gravity coming from the photons inside the sun is probably far from negligible. They have so much energy that their pressure is even important in helping to keep the sun from collapsing, so we probably feel their gravity quite noticeably too.

"The Photon" has no mass. the mass of "The Photon" can be found to be exactly equal to zero in a table of elementry particles. On the other hand, photons have mass. A culminated beam of photons has virtually no mass. A perfect planar wave has no mass. A bunch of photons distributed in a spherical box curves space exactly as you would expect from a bunch of atoms distributed in the same box.

Specifically, it is not just mass, but the stress energy tensor that curves space in 10 different ways. A perfectly culminated beam of photons curves space differently than, say, a rod of solid matter with velocity along its length. Unfortunately, I don't know the details of the difference.

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light being a wave should refract around massive objects due to gravitational time dilation. didnt some russian guy prove that it could all be accounted for by time dilation?

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This brings up another controversial matter though, because according to Einstein energy and mass are equivalent, so saying that photons have no mass would in turn say that they have no energy, and vise versa, in saying they have energy says they have mass, so either Einstein was "wrong" or there is something about photons that we still don't know

The correct statement of special relativity is that energy (E), mass (m) and momentum (p) are inter-related. They are connected by formula

[tex] E= \sqrt{m^2c^4 + p^2c^2} [/tex]

If a massive body is at rest (p=0), then you'll get the famous energy-mass relationship

[tex] E= mc^2 [/tex]

If a body has zero mass (like photons), then another famous relationship follows

[tex] E= pc [/tex]

So, massless particles can have non-zero energy.

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This brings up another controversial matter though, because according to Einstein energy and mass are equivalent, so saying that photons have no mass would in turn say that they have no energy, and vise versa, in saying they have energy says they have mass, so either Einstein was "wrong" or there is something about photons that we still don't know

To be a bit clearer

"photon has zero

In general, photon do has mass which is due to its momentum, namely [tex] E=pc [/tex].

With non-zero mass, there should have non-zero gravity created.

Here i have a thought experiment:

Suppose, in a free space, a pair of laser beams (sharp enough of course) are shot out parallel to each other with a distance apart. Will these two beams converge together eventually? it should happen, isn't it?

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ooooops~! i was a bit slower ...

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To be a bit clearer

"photon has zerorestmass, which is the mass we measured in its co-moving frame although it is not possible to do so.

In general, photon do has mass which is due to its momentum, namely [tex] E=pc [/tex].

With non-zero mass, there should have non-zero gravity created.

Here i have a thought experiment:

Suppose, in a free space, a pair of laser beams (sharp enough of course) are shot out parallel to each other with a distance apart. Will these two beams converge together eventually? it should happen, isn't it?

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ooooops~! i was a bit slower ...

May I please ask two questions? 1. When photons travel together are they at rest with respect to each other? 2. Does zero rest mass mean that photons traveling together don't feel gravity towards each other?

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jtbell

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A culminated beam of photons

I think you mean "collimated."

"Culminated" is a genuine word, but it has an entirely different meaning.

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I think you mean "collimated."

"Culminated" is a genuine word, but it has an entirely different meaning.

Oh well. What's an ell between friends...and a u.

I object strongly. It really ought to be 'columnate'. I will rite the editors.

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I thought photons just followed a "straight" path through curved space time (or whatever is closest to straight), so why would their energy make any difference? Any photon travels at the same speed and flies through the same curved space-time along the shortest path.

Photons do follow a straight path, but "straight" depends on the local curvature of spacetime which IS affected by local energies and masses....more energy means more curvature which means more deflection....

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May I please ask two questions? 1. When photons travel together are they at rest with respect to each other? 2. Does zero rest mass mean that photons traveling together don't feel gravity towards each other?

1. yes, like two cars each going along the highway at 55MPH.

2. according to post #13, it looks that way...

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1. From the point of view of a photon,May I please ask two questions? 1. When photons travel together are they at rest with respect to each other? 2. Does zero rest mass mean that photons traveling together don't feel gravity towards each other?

2. No, gravity is caused by energy. Mass is just one kind of energy, but it's so huge that we used to think this was the only cause for gravity. However, other kinds of energy, like pressure for example, cause gravity too. A photon definitely has energy, so it does cause a small amount of gravity.

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jtbell

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1. When photons travel together are they at rest with respect to each other?

Calculate it for yourself using the relativistic "velocity addition" equation:

[tex]u^{\prime} = \frac {u - v} {1 - \frac {uv}{c^2}}[/tex]

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel.html

v is the velocity of "observer B" (one of the photons) with respect to "observer A" (you), which is c, of course. u is the velocity of the other photon with respect to you (which is also c, of course). u' is the velocity of second photon with respect to the first one. What do you get for it?

michelcolman said:2. No, gravity is caused by energy.

And momentum. Remember in GR, the curvature of space-time is determined by the stress-energy tensor, whose components include both energy and the components of momentum.

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So that's 0/0, and is undefined? How can that be explained physically?Calculate it for yourself using the relativistic "velocity addition" equation:

[tex]u^{\prime} = \frac {u - v} {1 - \frac {uv}{c^2}}[/tex]

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jtbell

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[tex]u = \frac {u + v^{\prime}} {1 + \frac {uv^{\prime}}{c^2}}[/tex]

and substitute u = c and v = c as before. You get the equation

[tex]c = \frac {c + v^{\prime}} {1 + \frac {v^{\prime}}{c}}[/tex]

No matter what you substitute for v' (except v' = -c), it satisfies this equation. So v' can be almost anything. There is simply no well-defined answer to the question, "what is the velocity of the second photon, from the 'point of view' of the first photon?"

I think most physicists would say that this means that it doesn't make sense to talk about the "point of view" of a photon, at least in the sense of an inertial reference frame (x, y, z, t coordinate system in which Newton's First Law is valid) in which the photon is at rest.

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Here's another way to think about a photon's gravity. Think of an enormous collection of photons.

Specifically, in the first 20,000 years or so after the Big Bang the universe is believed to have been radiation dominated, meaning that any large region of the universe contained far more mass-energy in the form of free radiation than in the form of matter. The gravity of this radiation caused the original expansion of the universe to decelerate dramatically. In fact, because radiation has positive "pressure" equal to 1/3 of its mass-energy (w=1/3), free radiation causes twice as much gravity (density + 3*pressure) as would the same amount of mass-energy in the form of matter. Relative to its total mass-energy, the early universe decelerated gravitationally at twice the rate it would have if it were matter-dominated.

Each photon's individual mass-energy and pressure decay through redshift, at a rate proportional to the expansion of the universe, 1/a. Which is why the universe is no longer radiation-dominated; by comparison, the mass-energy of an atom of matter does not decay in this way.

Specifically, in the first 20,000 years or so after the Big Bang the universe is believed to have been radiation dominated, meaning that any large region of the universe contained far more mass-energy in the form of free radiation than in the form of matter. The gravity of this radiation caused the original expansion of the universe to decelerate dramatically. In fact, because radiation has positive "pressure" equal to 1/3 of its mass-energy (w=1/3), free radiation causes twice as much gravity (density + 3*pressure) as would the same amount of mass-energy in the form of matter. Relative to its total mass-energy, the early universe decelerated gravitationally at twice the rate it would have if it were matter-dominated.

Each photon's individual mass-energy and pressure decay through redshift, at a rate proportional to the expansion of the universe, 1/a. Which is why the universe is no longer radiation-dominated; by comparison, the mass-energy of an atom of matter does not decay in this way.

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DrGreg

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Einstein's 2nd postulate implies the speed of any photon is alwaysI think most physicists would say that this means that it doesn't make sense to talk about the "point of view" of a photon, at least in the sense of an inertial reference frame (x, y, z, t coordinate system in which Newton's First Law is valid) in which the photon is at rest.

The fact the equations quoted early give an indeterminate answer (0/0) is the mathematics' way of telling you "you can't use that equation here".

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Photons have a gravitational force as they interact by not escaping the event horizon of a black hole, so yes photons have gravitational force ,be it very small

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Makes sense, thanks!Einstein's 2nd postulate implies the speed of any photon is alwayscin any inertial frame, therefore it's impossible to find an inertial frame in which a photon is at rest.

The fact the equations quoted early give an indeterminate answer (0/0) is the mathematics' way of telling you "you can't use that equation here".

I thought the reason why photons don't escape the event horizon was because of the huge density of the singularity which creates a huge curvature in space time. I would think the curvature caused by photons can be neglected in comparison to that... the photons themselves are not the reason they stay inside of the event horizon..?Photons have a gravitational force as they interact by not escaping the event horizon of a black hole, so yes photons have gravitational force ,be it very small

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isnt time dilation within (or rather at) the event horizon infinite. (time dilation being proportional to gravitational potential). therefore a photon escaping the event horizon would be infinitely red shifted. i.e. it would have zero energy.

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once a photon enters the event horizon it cannot escape ,looking from outside time would stand still from our point of view, so i am also curious to know if the photon inside the event horizon would be redshifted as anything else would appear stationary, as time stands still to the outside veiwerisnt time dilation within (or rather at) the event horizon infinite. (time dilation being proportional to gravitational potential). therefore a photon escaping the event horizon would be infinitely red shifted. i.e. it would have zero energy.

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I guessed that equivalent things that have more (internal?) pressure would have greater gravity. Thanks for whoever provided that info.

Off my head I would have made a stab that free radiation had a whole lot more than just twice the gravity of mass-energy in matter; but it is only twice as much? Can someone possibly help me with why it is just twice as much?Here's another way to think about a photon's gravity. Think of an enormous collection of photons.

Specifically, in the first 20,000 years or so after the Big Bang the universe is believed to have been radiation dominated, meaning that any large region of the universe contained far more mass-energy in the form of free radiation than in the form of matter. The gravity of this radiation caused the original expansion of the universe to decelerate dramatically. In fact, because radiation has positive "pressure" equal to 1/3 of its mass-energy (w=1/3), free radiation causes twice as much gravity (density + 3*pressure) as would the same amount of mass-energy in the form of matter. Relative to its total mass-energy, the early universe decelerated gravitationally at twice the rate it would have if it were matter-dominated.

Each photon's individual mass-energy and pressure decay through redshift, at a rate proportional to the expansion of the universe, 1/a. Which is why the universe is no longer radiation-dominated; by comparison, the mass-energy of an atom of matter does not decay in this way.

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