| Topic: |
Science > Physics |
| User: |
"Tim Margeson" |
| Date: |
19 Aug 2004 01:18:08 AM |
| Object: |
Photon mass between the candle and the observer |
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
Second question:
Assumming the answer to the first question is greater than 0, what is the
mass of all the photons radiated by the lamp (assumming radiation field is
a sphere) over a period of 4.5E9 years?
Thanks.
Tim
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| User: "Old Man" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 02:01:38 AM |
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"Tim Margeson" <posted@mvei.com> wrote in message
news:Xns9549ED0C1F160newspostmveicom@216.168.3.44...
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
For a parallel beam, zero. The detected photons follow
a massless null trajectory. The Beam may have a small
mass due to beam divergence. For each pair of photons,
each of momentum p,
E_pair = M c^2 = 2 (p c) sin(2*theta).
Second question:
Assumming the answer to the first question is greater than 0, what is the
mass of all the photons radiated by the lamp (assumming radiation field is
a sphere) over a period of 4.5E9 years?
The lamp center sits at the center-of-momentum for
the emitted photons. For a light emission rate of P
(watts), the total center-of-mass, M, of emitted
photons is
delta_E = M c^2 = P (delta_t)
Do your own arithmetic.
[Old Man]
Thanks.
Tim
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| User: "Old Man" |
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| Title: Correction: Re: Photon mass between the candle and the observer |
19 Aug 2004 02:35:31 PM |
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"Old Man" <nomail@nomail.net> wrote in message
news:yeydneHs95POzbncRVn-gA@prairiewave.com...
"Tim Margeson" <posted@mvei.com> wrote in message
news:Xns9549ED0C1F160newspostmveicom@216.168.3.44...
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
For a parallel beam, zero. The detected photons follow
a massless null trajectory. The Beam may have a small
mass due to beam divergence. For each pair of photons,
each of momentum p,
E_pair = M c^2 = 2 (p c) sin(2*theta).
This is incorrect. For full divergence angle, theta, between
two photons,
E_pair = M c^2 = 2 (p c) sin(theta / 2).
If "A" is the area of the detector, and "R" is the distance
between lamp (assumed point source) and detector, then
the mass, delta_m, in the diverging beam between source
and detector is given by the solid angle and volume,
( delta_m ) c^2 = ( A / 4 pi R^2 ) P ( R / c )
Where P is the lamp power. delta_t = R / c is the photon
transit time.
[Old Man]
Second question:
Assumming the answer to the first question is greater than 0, what is
the
mass of all the photons radiated by the lamp (assumming radiation field
is
a sphere) over a period of 4.5E9 years?
The lamp center sits at the center-of-momentum for
the emitted photons. For a light emission rate of P
(watts), the total center-of-mass, M, of emitted
photons is
delta_E = M c^2 = P (delta_t)
Do your own arithmetic.
[Old Man]
Thanks.
Tim
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| User: "Gregory L. Hansen" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 09:49:06 AM |
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In article <Xns9549ED0C1F160newspostmveicom@216.168.3.44>,
Tim Margeson <posted@mvei.com> wrote:
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
The mass is the energy in the center of momentum frame of the particles,
and you can't transform to the center of momentum frame of a photon. You
can, however, transform to the center of momentum frame of some photons, a
lamp, and a detector.
Second question:
Assumming the answer to the first question is greater than 0, what is the
mass of all the photons radiated by the lamp (assumming radiation field is
a sphere) over a period of 4.5E9 years?
Thanks.
Tim
--
"The preferred method of entering a building is to use a tank main gun
round, direct fire artillery round, or TOW, Dragon, or Hellfire missile to
clear the first room." -- THE RANGER HANDBOOK U.S. Army, 1992
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| User: "eric gisse" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 02:09:33 AM |
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On Thu, 19 Aug 2004 06:18:08 -0000, Tim Margeson <posted@mvei.com>
wrote:
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
[snip]
0, to a rather large degree of accuracy.
A photon has energy but not mass.
In case you wish to point out E=mc^2, I will pre-emptively point out
that half the story is missing.
This is the full version of the equation:
E^2 = (mc^2)^2 + (pc)^2
An object can have finite energy with no mass.
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| User: "Old Man" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 06:09:30 PM |
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"eric gisse" <fsegg@NOSPAMATuaf.edu> wrote in message
news:dck8i0lj31jnmpsl9s8nck9u4kccbqja8l@4ax.com...
On Thu, 19 Aug 2004 06:18:08 -0000, Tim Margeson <posted@mvei.com>
wrote:
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
[snip]
0, to a rather large degree of accuracy.
A photon has energy but not mass.
In case you wish to point out E=mc^2, I will pre-emptively point out
that half the story is missing.
This is the full version of the equation:
E^2 = (mc^2)^2 + (pc)^2
An object can have finite energy with no mass.
A single photon has no mass. If a pair of photons are
converging to, or diverging from, a point, they collectively
have mass. For full divergence angle, theta, between
two photons of equal momentum, p,
M c^2 = 2 (p c) sin(theta / 2).
[Old Man]
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| User: "eric gisse" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 07:03:14 PM |
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On Thu, 19 Aug 2004 18:09:30 -0500, "Old Man" <nomail@nomail.net>
wrote:
"eric gisse" <fsegg@NOSPAMATuaf.edu> wrote in message
news:dck8i0lj31jnmpsl9s8nck9u4kccbqja8l@4ax.com...
On Thu, 19 Aug 2004 06:18:08 -0000, Tim Margeson <posted@mvei.com>
wrote:
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
[snip]
0, to a rather large degree of accuracy.
A photon has energy but not mass.
In case you wish to point out E=mc^2, I will pre-emptively point out
that half the story is missing.
This is the full version of the equation:
E^2 = (mc^2)^2 + (pc)^2
An object can have finite energy with no mass.
A single photon has no mass. If a pair of photons are
converging to, or diverging from, a point, they collectively
have mass. For full divergence angle, theta, between
two photons of equal momentum, p,
M c^2 = 2 (p c) sin(theta / 2).
[Old Man]
Mass-mass or mass-equivalent?
Your post makes me thing of geons.
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| User: "Old Man" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 08:28:52 PM |
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"eric gisse" <fsegg@NOSPAMATuaf.edu> wrote in message
news:ukfai0lh7da2dl9qd0kua6t6sft0i4keqp@4ax.com...
On Thu, 19 Aug 2004 18:09:30 -0500, "Old Man" <nomail@nomail.net>
wrote:
"eric gisse" <fsegg@NOSPAMATuaf.edu> wrote in message
news:dck8i0lj31jnmpsl9s8nck9u4kccbqja8l@4ax.com...
On Thu, 19 Aug 2004 06:18:08 -0000, Tim Margeson <posted@mvei.com>
wrote:
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the
detector?
[snip]
0, to a rather large degree of accuracy.
A photon has energy but not mass.
In case you wish to point out E=mc^2, I will pre-emptively point out
that half the story is missing.
This is the full version of the equation:
E^2 = (mc^2)^2 + (pc)^2
An object can have finite energy with no mass.
A single photon has no mass. If a pair of photons are
converging to, or diverging from, a point, they collectively
have mass. For full divergence angle, theta, between
two photons of equal momentum, p,
M c^2 = 2 (p c) sin(theta / 2).
[Old Man]
Mass-mass or mass-equivalent?
Your post makes me thing of geons.
No difference. Rest Mass Gravitates. Center-of-mass
energy gravitates. In the center-of-momentum frame,
photons (N > 1) of total kinetic energy, E, have rest
mass, M, according to M = E / c^2.
It isn't always possible to define the center-of-momentum,
such as when all of the photons travel in exactly the same
direction. In that case, the rest mass of the beam is zero.
A diverging or converging beam of photons has rest mass
according to Old Man's equation .
[Old Man]
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| User: "" |
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| Title: Re: Photon mass between the candle and the observer |
20 Aug 2004 07:41:12 AM |
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In article <bdqdnQO3wLxIzrjcRVn-pg@prairiewave.com>, "Old Man" <nomail@nomail.net> writes:
"eric gisse" <fsegg@NOSPAMATuaf.edu> wrote in message
news:ukfai0lh7da2dl9qd0kua6t6sft0i4keqp@4ax.com...
Mass-mass or mass-equivalent?
Your post makes me thing of geons.
No difference. Rest Mass Gravitates. Center-of-mass
energy gravitates. In the center-of-momentum frame,
photons (N > 1) of total kinetic energy, E, have rest
mass, M, according to M = E / c^2.
In the laboratory frame, a collimated photon beam of total
kinetic energy E has rest mass zero. In this case there is
no center-of-momentum frame.
In the laboratory frame, a non-collimated photon beam of total
energy E that illuminates a 1 cm^2 target at 1 meter downrange
will have a rest mass much less than E/c^2.
Beam divergence is low and the center-of-mass frame is moving at
a very large fraction of the speed of light with respect to the
lab frame. So the relativistic mass of the beam (in the lab frame)
varies greatly from its rest mass (in any frame).
Hence the question:
Mass-mass or mass-equivalent?
John Briggs
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| User: "Old Man" |
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| Title: Re: Photon mass between the candle and the observer |
20 Aug 2004 09:49:12 AM |
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<briggs@encompasserve.org> wrote in message
news:RB9qi8BG5GiW@eisner.encompasserve.org...
In article <bdqdnQO3wLxIzrjcRVn-pg@prairiewave.com>, "Old Man"
<nomail@nomail.net> writes:
"eric gisse" <fsegg@NOSPAMATuaf.edu> wrote in message
news:ukfai0lh7da2dl9qd0kua6t6sft0i4keqp@4ax.com...
Mass-mass or mass-equivalent?
Your post makes me thing of geons.
No difference. Rest Mass Gravitates. Center-of-mass
energy gravitates. In the center-of-momentum frame,
photons (N > 1) of total kinetic energy, E, have rest
mass, M, according to M = E / c^2.
In the laboratory frame, a collimated photon beam of total
kinetic energy E has rest mass zero. In this case there is
no center-of-momentum frame.
In the laboratory frame, a non-collimated photon beam of total
energy E that illuminates a 1 cm^2 target at 1 meter downrange
will have a rest mass much less than E/c^2.
Beam divergence is low and the center-of-mass frame is moving at
a very large fraction of the speed of light with respect to the
lab frame. So the relativistic mass of the beam (in the lab frame)
varies greatly from its rest mass (in any frame).
Hence the question:
Mass-mass or mass-equivalent?
John Briggs
Old Man agrees completely, and has said as much in this
thread and elsewhere.
When Old man says "mass" he means "invariant mass",
and nothing else.
For the diverging light beam considered here, the center-of-
momentum (CM) is moving at near the speed of light. In
the CM, the total photon energy, E_cm, is Doppler shifted
such that E_cm << E_lab, where E_lab is the total energy
of the diverging photon beam in the lab:
E_cm / E_lab ~ 1 - ( v_cm / c )
The invariant rest mass, M, of the beam is given by
M = E_cm / c^2 ~ [ E_lab / c^2 ] [ 1 - ( v_cm / c ) ]
For two diverging photons of equal energy with angle
"theta" between their momentum vectors;
v_cm / c = cos( theta / 2 )
for a light point source of total power, P, a distace,
R, from a detector of surface area, A,
E_lab ~ ( A / 4 pi R^2 ) P ( R / c )
Where R / c is the photon transit time between source
and detector.
[Old Man]
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| User: "Old Man" |
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| Title: Correction: Re: Photon mass between the candle and the observer |
20 Aug 2004 05:19:44 PM |
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"Old Man" <nomail@nomail.net> wrote in message
news:haSdncaBy-L0krvcRVn-tw@prairiewave.com...
<briggs@encompasserve.org> wrote in message
news:RB9qi8BG5GiW@eisner.encompasserve.org...
In article <bdqdnQO3wLxIzrjcRVn-pg@prairiewave.com>, "Old Man"
<nomail@nomail.net> writes:
"eric gisse" <fsegg@NOSPAMATuaf.edu> wrote in message
news:ukfai0lh7da2dl9qd0kua6t6sft0i4keqp@4ax.com...
Mass-mass or mass-equivalent?
Your post makes me thing of geons.
No difference. Rest Mass Gravitates. Center-of-mass
energy gravitates. In the center-of-momentum frame,
photons (N > 1) of total kinetic energy, E, have rest
mass, M, according to M = E / c^2.
In the laboratory frame, a collimated photon beam of total
kinetic energy E has rest mass zero. In this case there is
no center-of-momentum frame.
In the laboratory frame, a non-collimated photon beam of total
energy E that illuminates a 1 cm^2 target at 1 meter downrange
will have a rest mass much less than E/c^2.
Beam divergence is low and the center-of-mass frame is moving at
a very large fraction of the speed of light with respect to the
lab frame. So the relativistic mass of the beam (in the lab frame)
varies greatly from its rest mass (in any frame).
Hence the question:
Mass-mass or mass-equivalent?
John Briggs
Old Man agrees completely, and has said as much in this
thread and elsewhere.
When Old man says "mass" he means "invariant mass",
and nothing else.
For the diverging light beam considered here, the center-of-
momentum (CM) is moving at near the speed of light. In
the CM, the total photon energy, E_cm, is Doppler shifted
such that E_cm << E_lab, where E_lab is the total energy
of the diverging photon beam in the lab:
E_cm / E_lab ~ 1 - ( v_cm / c )
Old Man forgot the gamma-factor here. For v_cm -> c
E_cm / E_lab ~ sqrt{ [ 1 - ( v_cm / c ) ] / 2 }
The invariant rest mass, M, of the beam is given by
M = E_cm / c^2 ~ [ E_lab / c^2 ] [ 1 - ( v_cm / c ) ]
Old Man forgot the gamma-factor here. For v_cm -> c
M ~ ( E_lab / c^2 ) sqrt{ [ 1 - ( v_cm / c ) ] / 2 }
For two diverging photons of equal energy with angle
"theta" between their momentum vectors;
v_cm / c = cos( theta / 2 )
for a light point source of total power, P, a distace,
R, from a detector of surface area, A,
E_lab ~ ( A / 4 pi R^2 ) P ( R / c )
Where R / c is the photon transit time between source
and detector.
[Old Man]
[Old Man]
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| User: "eric gisse" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 11:37:53 PM |
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On Thu, 19 Aug 2004 20:28:52 -0500, "Old Man" <nomail@nomail.net>
wrote:
[snip]
A single photon has no mass. If a pair of photons are
converging to, or diverging from, a point, they collectively
have mass. For full divergence angle, theta, between
two photons of equal momentum, p,
M c^2 = 2 (p c) sin(theta / 2).
[Old Man]
Mass-mass or mass-equivalent?
Your post makes me thing of geons.
No difference. Rest Mass Gravitates. Center-of-mass
energy gravitates. In the center-of-momentum frame,
photons (N > 1) of total kinetic energy, E, have rest
mass, M, according to M = E / c^2.
It isn't always possible to define the center-of-momentum,
such as when all of the photons travel in exactly the same
direction. In that case, the rest mass of the beam is zero.
A diverging or converging beam of photons has rest mass
according to Old Man's equation .
Interesting. Thanks for the insight.
[Old Man]
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| User: "" |
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| Title: Re: Photon mass between the candle and the observer |
20 Aug 2004 11:30:49 AM |
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In article <ukfai0lh7da2dl9qd0kua6t6sft0i4keqp@4ax.com>, eric gisse <fsegg@NOSPAMATuaf.edu> writes:
On Thu, 19 Aug 2004 18:09:30 -0500, "Old Man" <nomail@nomail.net>
wrote:
"eric gisse" <fsegg@NOSPAMATuaf.edu> wrote in message
news:dck8i0lj31jnmpsl9s8nck9u4kccbqja8l@4ax.com...
On Thu, 19 Aug 2004 06:18:08 -0000, Tim Margeson <posted@mvei.com>
wrote:
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
[snip]
0, to a rather large degree of accuracy.
A photon has energy but not mass.
In case you wish to point out E=mc^2, I will pre-emptively point out
that half the story is missing.
This is the full version of the equation:
E^2 = (mc^2)^2 + (pc)^2
An object can have finite energy with no mass.
A single photon has no mass. If a pair of photons are
converging to, or diverging from, a point, they collectively
have mass. For full divergence angle, theta, between
two photons of equal momentum, p,
M c^2 = 2 (p c) sin(theta / 2).
[Old Man]
Mass-mass or mass-equivalent?
Mass, period. Verify it for yourself by using the full version of the
equation you yourself provided, for a pair of photons.
Mass is not additive, in SR.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Tim Margeson" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 09:50:12 PM |
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Tim Margeson <posted@mvei.com> wrote in
news:Xns9549ED0C1F160newspostmveicom@216.168.3.44:
Thanks to all for the answers.
One further question:
If there's zero mass in this pile of photons, which is what theory says,
how much energy is stored in all the photons that are now traveling
between the galaxies of the universe?
And since we haven't seen them yet, they are a dark energy, correct?
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the
detector?
Second question:
Assumming the answer to the first question is greater than 0, what is
the mass of all the photons radiated by the lamp (assumming radiation
field is a sphere) over a period of 4.5E9 years?
Thanks.
Tim
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Photon mass between the candle and the observer |
20 Aug 2004 03:07:35 AM |
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Tim Margeson wrote:
Tim Margeson <posted@mvei.com> wrote in
news:Xns9549ED0C1F160newspostmveicom@216.168.3.44:
Thanks to all for the answers.
One further question:
If there's zero mass in this pile of photons, which is what theory says,
Did you understand the difference between rest mass (invariant mass)
and relativistic mass?
how much energy is stored in all the photons that are now traveling
between the galaxies of the universe?
1) What has that to do with photons having zero mass?
2) Not very much. I don't know the exact numbers, but I would suspect
that the total energy is far lower (by a factor of at least 1000) than
the mass "contained" in the matter.
And since we haven't seen them yet, they are a dark energy, correct?
Not according to the usual meaning of the word "dark energy" in cosmology.
What is meant there is a type of energy which acts gravitationally
repulsive, and is homogeneously distributed in the universe.
[snip]
BTW: please don't top-post.
Bye,
Bjoern
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| User: "Tim Margeson" |
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| Title: Re: Photon mass between the candle and the observer |
21 Aug 2004 03:46:58 AM |
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Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in
news:cg4bg8$oum$1@news.urz.uni-heidelberg.de:
Tim Margeson wrote:
[.\/.]
[.OO.]
And since we haven't seen them yet, they are a dark energy, correct?
Not according to the usual meaning of the word "dark energy" in
cosmology. What is meant there is a type of energy which acts
gravitationally repulsive, and is homogeneously distributed in the
universe.
[snip]
BTW: please don't top-post.
Bye,
Bjoern
Admittedly I am not a physics person. The questions stem from my point of
view that I haven't seen the photons between stars accounted for in any
tangible form. Yet they are there, in significant numbers. But my grasp
of physics being limited, I have no idea how to ask a question like this
properly: Is the amount of energy/mass of photons (light at all
frequencies) between stars significant in any way? So I built a small
example -- the lamp and detector -- one that made sense to me, and
hopefully to you, to look at the issue and posit a result.
How does current physics account for light that is in transit? Clearly
this light has some energy, or some mass when viewed from the proper
frame, sufficient that it can affect things it hits even to the point of
changing orbits of smaller bodies.
Perhaps this is a question for the sci.astro or similar group.
Thanks to all who have responded!
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| User: "Ken S. Tucker" |
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| Title: Re: Photon mass between the candle and the observer |
21 Aug 2004 12:47:14 PM |
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Tim Margeson <posted@mvei.com> wrote in message news:<Xns954C1225166C2newspostmveicom@216.168.3.44>...
Admittedly I am not a physics person. The questions stem from my point of
view that I haven't seen the photons between stars accounted for in any
tangible form. Yet they are there, in significant numbers. But my grasp
of physics being limited, I have no idea how to ask a question like this
properly: Is the amount of energy/mass of photons (light at all
frequencies) between stars significant in any way? So I built a small
example -- the lamp and detector -- one that made sense to me, and
hopefully to you, to look at the issue and posit a result.
How does current physics account for light that is in transit? Clearly
this light has some energy, or some mass when viewed from the proper
frame, sufficient that it can affect things it hits even to the point of
changing orbits of smaller bodies.
Perhaps this is a question for the sci.astro or similar group.
Thanks to all who have responded!
I did a crude calculation and estimated the photon
energy in transit within the volume of our Milky
Way galaxy is equivalent to 5 solar masses, quite
insignificant. Choose a random location within the
galaxy in interstellar space and the radiation will
get you heated slightly above absolute zero.
Ken
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| User: "Tim Margeson" |
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| Title: Re: Photon mass between the candle and the observer |
21 Aug 2004 08:39:49 PM |
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(Ken S. Tucker) wrote in
news:2202379a.0408210947.5417dc37@posting.google.com:
Tim Margeson <posted@mvei.com> wrote in message
news:<Xns954C1225166C2newspostmveicom@216.168.3.44>...
[xx]
I did a crude calculation and estimated the photon
energy in transit within the volume of our Milky
Way galaxy is equivalent to 5 solar masses, quite
insignificant. Choose a random location within the
galaxy in interstellar space and the radiation will
get you heated slightly above absolute zero.
Ken
I beleive your calculations are correct for instantaneous heating - after
all you'd just turn around and reradiate the heat buld up. Just intuitively
it seems to me that there is so much volume full of photons that have been
emitted for so long that they'd be more significant than 5 solar masses.
Time for me to get out my books and prove to myself I'm once again
misguided by intuition.
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| User: "Ken S. Tucker" |
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| Title: Re: Photon mass between the candle and the observer |
23 Aug 2004 04:55:37 AM |
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Tim Margeson <posted@mvei.com> wrote in message news:<Xns954CBDDE7159Enewspostmveicom@216.168.3.44>...
dynamics@vianet.on.ca (Ken S. Tucker) wrote in
news:2202379a.0408210947.5417dc37@posting.google.com:
Tim Margeson <posted@mvei.com> wrote in message
news:<Xns954C1225166C2newspostmveicom@216.168.3.44>...
[xx]
I did a crude calculation and estimated the photon
energy in transit within the volume of our Milky
Way galaxy is equivalent to 5 solar masses, quite
insignificant. Choose a random location within the
galaxy in interstellar space and the radiation will
get you heated slightly above absolute zero.
Ken
I beleive your calculations are correct for instantaneous heating - after
all you'd just turn around and reradiate the heat buld up. Just intuitively
it seems to me that there is so much volume full of photons that have been
emitted for so long that they'd be more significant than 5 solar masses.
Time for me to get out my books and prove to myself I'm once again
misguided by intuition.
Hi Tim, I checked a thread entitled
"Mass of all the light" and found the figure
I calculated was 4,000 solar masses (back in
2002), not the 5 I suggested above, oops...
You might check out that thread, I got
some good replies.
Regards
Ken S. Tucker
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Photon mass between the candle and the observer |
23 Aug 2004 03:32:18 AM |
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Tim Margeson wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in
news:cg4bg8$oum$1@news.urz.uni-heidelberg.de:
Tim Margeson wrote:
[.\/.]
[.OO.]
And since we haven't seen them yet, they are a dark energy, correct?
Not according to the usual meaning of the word "dark energy" in
cosmology. What is meant there is a type of energy which acts
gravitationally repulsive, and is homogeneously distributed in the
universe.
[snip]
BTW: please don't top-post.
Bye,
Bjoern
Admittedly I am not a physics person. The questions stem from my point of
view that I haven't seen the photons between stars accounted for in any
tangible form.
That you do not have seen this does not change the fact that
cosmologists indeed *do* account for them.
Yet they are there, in significant numbers. But my grasp
of physics being limited, I have no idea how to ask a question like this
properly: Is the amount of energy/mass of photons (light at all
frequencies) between stars significant in any way?
It isn't, as I pointed out in my last reply.
One can make quite reliable estimates of the total energy contained
in these photons. It turns out that the total amount is negligible.
Try to see it this way: the light which stars emits comes from the
nuclear fusion processes in the cores of the stars. But in these
nuclear fusion processes, the energy which is released in only about
1/1000 of the total energy contained in the masses of the nuclei
which fuse there! So, obviously, the total energy of all light produced
by stars in the universe can be at most 1/1000 of the energy contained
in the normal matter of the universe (actually even far less, since
only a *very* small percentage of that normal matter undergoes nuclear
fusion and releases light).
Taking into account light released by other sources, e.g. by accretion
discs around black holes, does not change this argument much.
So I built a small
example -- the lamp and detector -- one that made sense to me, and
hopefully to you, to look at the issue and posit a result.
The example of the lamp does not help much. Looking at this problem
from the viewpoint of nuclear fusion (see above) yields the result
far more easily.
How does current physics account for light that is in transit? Clearly
this light has some energy, or some mass when viewed from the proper
frame,
If light has mass or not has nothing to do with the "proper frame".
This only depends on what one *means* by "mass" (rest or relativistic
mass).
sufficient that it can affect things it hits even to the point of
changing orbits of smaller bodies.
Well, physics does indeed account for such effects. E.g. the tails
of comets are caused by this effect.
Perhaps this is a question for the sci.astro or similar group.
Good luck.
Thanks to all who have responded!
Hope I could help a bit.
Bye,
Bjoern
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| User: "Sam Wormley" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 08:07:43 AM |
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Tim Margeson wrote:
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
Second question:
Assumming the answer to the first question is greater than 0, what is the
mass of all the photons radiated by the lamp (assumming radiation field is
a sphere) over a period of 4.5E9 years?
Zero, Tim, here's a bigger picture.....
Ref: http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
Physics FAQ] - [Copyright]
Addendum added by DK 2002.
Updated by Jim Carr 1998.
Original by Philip Gibbs 1997.
Does mass change with velocity?
There is sometimes confusion surrounding the subject of mass in
relativity. This is because there are two separate uses of the term.
Sometimes people say "mass" when they mean "relativistic mass",
mr but at other times they say "mass" when they mean "invariant
mass", m0. These two meanings are not the same. The invariant
mass of a particle is independent of its velocity v, whereas
relativistic mass increases with velocity and tends to infinity as the
velocity approaches the speed of light c. They can be defined as
follows:
m_r = E/c^2
m_0 = sqrt(E^2/c^4 - p^2/c^2)
where E is energy, p is momentum and c is the speed of light in a
vacuum. The velocity dependent relation between the two is
m_r = m_0 /sqrt(1 - v^2/c^2)
Of the two, the definition of invariant mass is much
preferred over the definition of relativistic mass. These
days, when physicists talk about mass in their research,
they always mean invariant mass. The symbol m for
invariant mass is used without the subscript 0. Although
the idea of relativistic mass is not wrong, it often leads to
confusion, and is less useful in advanced applications such
as quantum field theory and general relativity. Using the
word "mass" unqualified to mean relativistic mass is
wrong because the word on its own will usually be taken to
mean invariant mass. For example, when physicists quote
a value for "the mass of the electron" they mean its
invariant mass.
At zero speed, the relativistic mass is equal to the invariant mass.
The invariant mass is therefore often called the "rest mass". This
latter terminology reflects the fact that historically it was relativistic
mass which was often regarded as the correct concept of mass in the
early years of relativity. In 1905 Einstein wrote a paper entitled
Does the inertia of a body depend upon its energy content?, to
which his answer was "yes". The first record of the relationship of
mass and energy explicitly in the form E = mc2 was written by
Einstein in a review of relativity in 1907. If this formula is taken to
include kinetic energy, then it is only valid for relativistic mass, but it
can also be taken as valid in the rest frame for invariant mass.
Einstein's conventions and interpretations were sometimes
ambivalent and varied a little over the years; however an
examination of his papers and books on relativity shows that he
almost never used relativistic mass himself. Whenever the symbol
m for mass appears in his equations it is always invariant mass. He
did not introduce the notion that the mass of a body increases with
velocity--just that it increases with energy content. The equation E =
mc2 was only meant to be applied in the rest frame of the particle.
Perhaps Einstein's only definite reference to mass increasing with
kinetic energy is in his "autobiographical notes".
To find the real origin of the concept of relativistic mass, you have to
look back to the earlier papers of Lorentz. In 1904 Lorentz wrote a
paper Electromagnetic Phenomena in a System Moving With Any
Velocity Less Than That of Light. There he introduced the
"longitudinal" and "transverse" electromagnetic masses of the
electron. With these he could write the equations of motion for an
electron in an electromagnetic field in the Newtonian form F = ma
where m increases with mass. Between 1905 and 1909 Planck,
Lewis and Tolman developed the relativistic theory of force,
momentum and energy. A single mass dependence could be used
for any acceleration if F = d/dt(mv) is used instead of F = ma. This
introduced the concept of relativistic mass which can be used in the
equation E = mc2 even for moving objects. It seems to have been
Lewis who introduced the appropriate velocity dependence of mass
in 1908, but the term "relativistic mass" appeared later. [Gilbert
Lewis was a chemist whose other claim to fame in physics was
naming the photon in 1926.]
Relativistic mass came into common usage in the relativity text
books of the early 1920s written by Pauli, Eddington and Born. As
particle physics became more important to physicists in the 1950s,
the invariant mass of particles became more significant, and
inevitably people started to use the term "mass" to mean invariant
mass. Gradually this took over as the normal convention, and the
concept of relativistic mass increasing with velocity was played
down.
The case of photons and other particles that move at the speed of
light is special. From the formula relating relativistic mass to
invariant mass, it follows that the invariant mass of a photon must
be zero, but its relativistic mass need not be. The phrase "The rest
mass of a photon is zero" might sound nonsensical because the
photon can never be at rest; but this is just a side effect of the
terminology, since by making this statement, we can bring photons
into the same mathematical formalism as the everyday particles
that do have rest mass. In modern physics texts, the term mass
when unqualified means invariant mass, and photons are said to be
"massless" (see Physics FAQ What is the mass of the photon?).
Teaching experience shows that this avoids most sources of
confusion.
Despite the general usage of invariant mass in the scientific
literature, the use of the word mass to mean relativistic mass is still
found in many popular science books. For example, Stephen
Hawking in A Brief History of Time writes "Because of the
equivalence of energy and mass, the energy which an
object has due to its motion will add to its mass." and
Richard Feynman in The Character of Physical Law wrote "The
energy associated with motion appears as an extra mass,
so things get heavier when they move." Evidently, Hawking
and Feynman and many others use this terminology because it is
intuitive and useful when you want to explain things without using
too much mathematics. The standard convention followed by some
physicists seems to be: use invariant mass when doing research and
writing papers for other physicists but use relativistic mass when
writing for non-physicists. It is a curious dichotomy of terminology
which inevitably leads to confusion. A common example is the
mistaken belief that a fast moving particle must form a black hole
because of its increase in mass (see relativity FAQ article If you go
too fast do you become a black hole?).
Looking more deeply into what is going on, we find that there are
two equivalent ways of formulating special relativity. Einstein's
original mechanical formalism is described in terms of inertial
reference frames, velocities, forces, length contraction and time
dilation. Relativistic mass fits naturally into this mechanical
framework, but it is not essential. If relativistic mass is used, it is
easier to form a correspondence with Newtonian mechanics, since
some Newtonian equations remain valid:
F = dp / dt
p = m_r v
Also, in this picture mass is conserved along with energy.
The second formulation is the more mathematical one introduced a
year later by Minkowski. It is described in terms of spacetime,
energy-momentum four vectors, world lines, light cones, proper
time and invariant mass. This version is harder to relate to ordinary
intuition because force and velocity are less useful in their
four-vector forms. On the other hand, it is much easier to generalise
this formalism to the curved spacetime of general relativity where
global inertial frames do not usually exist.
It may seem that Einstein's original mechanical formalism should be
easier to learn, because it retains many equations from the familiar
Newtonian mechanics. In Minkowski's geometric formalism, simple
concepts such as velocity and force are replaced with world lines
and four vectors. Yet the mechanical formalism often proves harder
to swallow, and is at the root of many people's failure to get over the
paradoxes that are so often discussed. Once students have been
taught about Minkowski space, they invariably see things more
clearly. The paradoxes are revealed for what they are and
calculations also become simpler. But it is debatable whether or not
the relativistic mechanical formalism should be avoided altogether.
It can still provide the correspondence between the new physics and
the old, which is important to grasp at the early stages. The step
from the mechanical formalism to the geometric can then be easier.
An alternative modern teaching method is to translate Newtonian
mechanics into a geometric formalism, using Galileian relativity in
four dimensional spacetime, and then modify the geometric picture
to Minkowski space.
The preference for invariant mass is stressed and justified in the
classic relativity textbook Spacetime Physics by Taylor and Wheeler
who write
"Ouch! The concept of `relativistic mass' is subject to
misunderstanding. That's why we don't use it. First, it
applies the name mass--belonging to the magnitude of a
four-vector--to a very different concept, the time
component of a four-vector. Second, it makes increase of
energy of an object with velocity or momentum appear to
be connected with some change in internal structure of
the object. In reality, the increase of energy with velocity
originates not in the object but in the geometric properties
of space-time itself."
In the final analysis the issue is a debate over whether or not
relativistic mass should be used, and is a matter of semantics and
teaching methods. The concept of relativistic mass is not wrong: it
can have its uses in special relativity at an elementary level. This
debate surfaced in Physics Today in 1989 when Lev Okun wrote an
article urging that relativistic mass should no longer be taught (42
#6, June 1989, pg 31). Wolfgang Rindler responded with a letter to
the editors to defend its continued use. (43 #5, May 1990, pg 13).
The experience of answering confused questions in the news groups
suggests that the use of relativistic mass in popular books and
elementary texts is not helpful. The fact that relativistic mass is
virtually never used in contemporary scientific research literature is
a strong argument against teaching it to students who will go on to
more advanced levels. Invariant mass proves to be more
fundamental in Minkowski's geometric approach to special
relativity, and relativistic mass is of no use at all in general
relativity. It is possible to avoid relativistic mass from the outset by
talking of energy instead. Judging by usage in modern text books,
the consensus is that relativistic mass is an outdated concept which
is best avoided. There are people who still want to use relativistic
mass, and it is not easy to settle an argument over semantic issues
because there is no absolute right or wrong; just conventions of
terminology. There will always be those who post questions using
terms in which mass increases with velocity. It is unhelpful to just
tell them that what they read or heard on cable TV is wrong, but it
might reduce confusion for them in the longer term if they can be
persuaded to think in terms of invariant mass instead of relativistic
mass.
In a 1948 letter to Lincoln Barnett, Einstein wrote
"It is not good to introduce the concept of the mass M =
m/(1-v^2/c^2)^1/2 of a body for which no clear definition can be
given. It is better to introduce no other mass than `the
rest mass' m. Instead of introducing M, it is better to
mention the expression for the momentum and energy of
a body in motion."
The viewpoint above, emphasising the distinction between mass,
momentum, and energy, is certainly the modern view. Fifty years
later, can relativistic mass be laid to rest?
Addendum: What is the relativistic version
of F = ma ?
For this last section, we'll write down the relativistic version of
Newton's second law, F = ma. In Newton's mechanics, this
equation relates vectors F and a (hence the bold script) via the mass
m of the object being accelerated, which is invariant in Newton's
theory. Because m is just a number, in Newton's theory the force on
a mass is always parallel to the resulting acceleration.
The corresponding equation in special relativity is a little more
complicated. It turns out that the force F is not always parallel to
the acceleration a! To express this fact, we need to use matrix
notation. Letting m be the invariant mass, v be the velocity as a
column vector, and 1 be the 3 x 3 identity matrix, the actual result
turns out to be
F = gamma m (1 + gamma^2 v v^t) a
and
a = (1 - v v^t) F / (gamma m)
(As an aside, there's a nice correspondence to the one dimensional
case here. Just as gamma^2 can be written as 1 + gamma^2 v^2, and its
inverse (i.e. reciprocal) is 1 - v^2, so the matrix 1 + gamma^2 v v^t has
inverse 1 - v v^t, as well as determinant gamma^2.)
Looking at this relativistic version of F = ma, we might say that
when the (invariant) mass m appears, it's accompanied by a factor
of gamma, so that really it is the relativistic mass that's appearing.
Isn't this then, a good reason why we might want to give the notion
of relativistic mass more credence? Not really. Notice that now the
acceleration is not necessarily parallel to the force that produced it.
It's not hard to see from the above equations that it's easier to
accelerate a mass sideways to its motion, than it is to accelerate it in
the direction of its motion. So now, if we still want to maintain some
meaning for relativistic mass, then we'll have to realise that it has a
directional dependence--as if the object somehow has more mass in
the direction of its motion, than it has sideways. Evidently the idea
of relativistic mass is becoming a little more complicated than at first
we might have hoped! And this is another reason why, in the end,
it's so much easier to just take the mass to be the invariant quantity
m, and to put any directional information into a separate, matrix,
factor.
References:
Arguments against the term "relativistic mass" are given in the
classic relativity text book Space-Time Physics by Taylor and
Wheeler, 2nd edition, Freeman Press (1992).
The article Does mass really depend on velocity, dad? by Carl
Adler, American Journal of Physics 55, 739 (1987) also discusses
this subject and includes the above quote from Einstein against the
use of relativistic mass.
Einstein's original papers can be found in English translation in The
Principle of Relativity by Einstein and others, Dover Press.
Some other historical details can be found in Concepts of mass by
Max Jammer and Einstein's Revolution by Elie Zahar.
Special Relativity
http://scienceworld.wolfram.com/physics/SpecialRelativity.html
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 09:19:33 AM |
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Sam Wormley wrote:
Tim Margeson wrote:
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
What's the mass of the photons in route from the candle to the detector?
Second question:
Assumming the answer to the first question is greater than 0, what is the
mass of all the photons radiated by the lamp (assumming radiation field is
a sphere) over a period of 4.5E9 years?
Zero, Tim, here's a bigger picture.....
Ref: http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html
Well, *if* you have to provide a link to the FAQ, I think this one
is more appropriate:
http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/light_mass.html
Bye,
Bjoern
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| User: "Sam Wormley" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 10:37:53 AM |
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Bjoern Feuerbacher wrote:
Well, *if* you have to provide a link to the FAQ, I think this one
is more appropriate:
http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/light_mass.html
Bye,
Bjoern
Yes--you are right Bjoern. Thanks that is better.
-Sam
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Photon mass between the candle and the observer |
19 Aug 2004 03:01:49 AM |
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Tim Margeson wrote:
Easy question for physics buffs.
First question:
Conditions of Experiment:
1: Lamp globe 1 meter away from a 1cm^2 detector.
2: Energy hitting detector is 1w/cm^2.
That's not the energy, that's the power per area.
What's the mass of the photons in route from the candle to the detector?
I'm assuming that you mean the total mass here, not the mass of a single
photon (since that would depend on the frequency of the light).
There isno rest mass; I assume that you mean the relativistic mass,
which is m=E/c^2. Since every second, 1 J is received at the detector,
that means that the total relativistic mass of the photons which hit the
detector every second is 1 J/c^2.
Second question:
Assumming the answer to the first question is greater than 0, what is the
mass of all the photons radiated by the lamp (assumming radiation field is
a sphere) over a period of 4.5E9 years?
Simply multiply the answer above (1 J/c^2/s) with the time (4.5E9 years).
What's your point?
Bye,
Bjoern
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