recoiling photons evidence?

When a bullet and the exploding gas shoots out of the barrel of a
gun, the gun moves in the opposite direction at a lesser velocity
because of its greater mass according to conservation of
momentum..
But the emission of a photon which has no mass except the
ascribed E/c^2 is not quite the same.
Instances of light pressure on reflective surfaces and Compton
like scattering of xrays at lower frequencies with the missing
energy transformed into the linear movement of the electron
suggests a recoil of the electron but not of the photon.
What is the evidence for recoiling photons?

"ralph sansbury" wrote in message
...
When a bullet and the exploding gas shoots out of the barrel of a
gun, the gun moves in the opposite direction at a lesser velocity
because of its greater mass according to conservation of
momentum..
But the emission of a photon which has no mass except the
ascribed E/c^2 is not quite the same.

That is the wrong equation.The correct form is:

E^2 = m^2 * c^4 + p^2 * c^2

hence

m = sqrt(E^2 - p^2 * c^2) / c^2

For a photon that is zero as far as anyone has measured.

http://math.ucr.edu/home/baez/physic...oton_mass.html

Instances of light pressure on reflective surfaces and Compton
like scattering of xrays at lower frequencies with the missing
energy transformed into the linear movement of the electron
suggests a recoil of the electron but not of the photon.
What is the evidence for recoiling photons?

If the photon did not carry momentum, there would be
no effect at all on the electron in the Compton effect.
If the incoming photon carried momentum but the outgoing
photon did not, the angles would be different to what is
observed.

George

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...
When a bullet and the exploding gas shoots out of the barrel
of a
gun, the gun moves in the opposite direction at a lesser
velocity
because of its greater mass according to conservation of
momentum..
But the emission of a photon which has no mass except the
ascribed E/c^2 is not quite the same.

That is the wrong equation.The correct form is:

E^2 = m^2 * c^4 + p^2 * c^2

No this is the wrong equation. The correct form is
E^2 = (m_0)^2 * c^4 + p^2 * c^2
The correct implicit equation I was using was
m=E/c^2 not m_0=E/c^2.
And since p=mv=Ev/c^2
and mc^2=(m_0)c^2/(1-v^2/c^2)^1/2 you get
the corrected equation above.



hence

m = sqrt(E^2 - p^2 * c^2) / c^2


For a photon that is zero as far as anyone has measured.

The rest mass may be zero but it is never at rest and the
difference between the
rest mass and the mass in motion is not zero. A gamma ray photon
impinging upon a nucleus can be converted into the rest mass of
an electron
and a positron and the kinetic energy of these particles.
In the Compton effect you mention the xray photon impinging on
an electron produces the energy of its oscillation at a lower
frequency
plus the kinetic energy of its linear motion.
But lets get back to my question as to the evidence of photon
recoil
in the emission of a photon? When the distance between the source
and the receiver is small as in the Compton effect and various
other cases eg
http://www.aip.org/enews/physnews/19...t/pnu450-1.htm,
The action between the source magnetic force of the oscillating
electrons
on the receiver magnetic property of the oscillating electrons
produces
a repulsion and part of the energy and momentum is this and part
is the
electrical field producing the oscillation.
If however you start an oscillation of charge in the source
by thermally exciting bound electrons in tungsten for example
the oscillating electrons may recoil from one another but the
whole source would not be recoiling against anything.
To say that some equation that works in the above cases
implies that it would work in this case may not be correct.


http://math.ucr.edu/home/baez/physic...clear/photon_m
ass.html

Instances of light pressure on reflective surfaces and
Compton
like scattering of xrays at lower frequencies with the
missing
energy transformed into the linear movement of the electron
suggests a recoil of the electron but not of the photon.
What is the evidence for recoiling photons?

If the photon did not carry momentum, there would be
no effect at all on the electron in the Compton effect.

It is not the momentum of the photon but the magnetic effect
which causes the linear movement at some angle is between
oscillating charges in the xray source and the oscillation of the
receiver electrons and the electric effect is the oscillation at
a slightly lower frequency.



If the incoming photon carried momentum but the outgoing
photon did not, the angles would be different to what is
observed.

In the case I am talking about there is no incoming photon.

Ralph



George

"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...
When a bullet and the exploding gas shoots out of the barrel
of a
gun, the gun moves in the opposite direction at a lesser
velocity
because of its greater mass according to conservation of
momentum..
But the emission of a photon which has no mass except the
ascribed E/c^2 is not quite the same.

That is the wrong equation.The correct form is:

E^2 = m^2 * c^4 + p^2 * c^2

No this is the wrong equation. The correct form is
E^2 = (m_0)^2 * c^4 + p^2 * c^2

Physics these days has generally dropped the old 'relativistic
mass' presentation as anachronistic and confusing. "Mass" means
invariant mass.

The correct implicit equation I was using was
m=E/c^2 not m_0=E/c^2.
And since p=mv=Ev/c^2
and mc^2=(m_0)c^2/(1-v^2/c^2)^1/2 you get
the corrected equation above.

The equation you used implies non-zero mass while the
(invariant) mass of a photon is zero, hence the possible
confusion. If you want to use "relativistic mass", it is
best to use that phrase so everyone understands you.

hence

m = sqrt(E^2 - p^2 * c^2) / c^2


For a photon that is zero as far as anyone has measured.

The rest mass may be zero but it is never at rest

And that comment is exactly the reason the these terms have
been dropped. The "rest mass" is actually invariant so is
a fundamental property that applies regardless of motion.
The "rest mass" of a photon is zero at all speeds.

and the
difference between the
rest mass and the mass in motion is not zero.

The difference is the kinetic energy, it is not mass of any
form. Do you see why these terms are considered confusing?

A gamma ray photon
impinging upon a nucleus can be converted into the rest mass of
an electron
and a positron and the kinetic energy of these particles.

Yes, total energy and total momentum are both conserved.

In the Compton effect you mention the xray photon impinging on
an electron produces the energy of its oscillation at a lower
frequency
plus the kinetic energy of its linear motion.

It is not clear what "its" means in that sentence. The frequency
(hence energy) of the photon is reduced while the kinetic energy
of the electron is increased. The momentum is similarly
redistributed but still conserved.

But lets get back to my question as to the evidence of photon recoil
in the emission of a photon? When the distance between the source
and the receiver is small as in the Compton effect and various
other cases eg
http://www.aip.org/enews/physnews/19...t/pnu450-1.htm,

You need a subscription to read the article but from the abstract
is seems clear they are using radiation pressure to cancel the
motion of the mirror. Since the momentum of the mirror is changed
by the reflection of the photons, there is your evidence.

The action between the source magnetic force of the oscillating electrons
on the receiver magnetic property of the oscillating electrons produces
a repulsion and part of the energy and momentum is this and part
is the electrical field producing the oscillation.

The mechanism is not relevant to your question, you only asked
for evidence that photons carry momentum when they recoil.

If however you start an oscillation of charge in the source
by thermally exciting bound electrons in tungsten for example
the oscillating electrons may recoil from one another but the
whole source would not be recoiling against anything.

In thermal agitation in tungsten, there are many particles
being accelerated in random directions so the mean cancels.

To say that some equation that works in the above cases
implies that it would work in this case may not be correct.

There is no reason to suspect it would differ, and since
the same model is successful at the particle level and for
macroscopic measures such as the mirror you cite, it is a
usable model.

http://math.ucr.edu/home/baez/physic...clear/photon_m
ass.html

Instances of light pressure on reflective surfaces and Compton
like scattering of xrays at lower frequencies with the missing
energy transformed into the linear movement of the electron
suggests a recoil of the electron but not of the photon.
What is the evidence for recoiling photons?

If the photon did not carry momentum, there would be
no effect at all on the electron in the Compton effect.

It is not the momentum of the photon but the magnetic effect
which causes the linear movement at some angle is between
oscillating charges in the xray source and the oscillation of the
receiver electrons and the electric effect is the oscillation at
a slightly lower frequency.

However you describe the mechanism, the effect is the same,
momentum is transferred and the amount is quantised.

If the incoming photon carried momentum but the outgoing
photon did not, the angles would be different to what is
observed.

In the case I am talking about there is no incoming photon.

You only asked:
When a bullet and the exploding gas shoots out of the barrel of a
gun, the gun moves in the opposite direction at a lesser velocity
because of its greater mass according to conservation of
momentum..
What is the evidence for recoiling photons?

The Compton effect demonstrates that both the incoming and
outgoing photons must carry momentum proportional to their
frequency.

George

"George Dishman" wrote in message
...
"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...
That is the wrong equation.The correct form is:

E^2 = m^2 * c^4 + p^2 * c^2

No this is the wrong equation. The correct form is
E^2 = (m_0)^2 * c^4 + p^2 * c^2

Physics these days has generally dropped the old 'relativistic
mass' presentation as anachronistic and confusing.

I hope not.
It is confusing not to make the distinction. You also
misunterstood my question. Namely, when you start an oscillation
of charge in a distant isolated source
by thermally exciting bound electrons in tungsten for example,
the oscillating electrons may recoil from one another as you
suggest in a random and cancelling manner but the whole source
would not be recoiling against anything.
My point then is that you cannot use a mathematical equation
that
applies to source and reflector together over a small distance,
as evidence for photon recoil causing a motion of eg an 8 watt
transmitter on a distant spacecraft as photons are emitted.

My point is that the evidence for photon recoil in a receiver
and
reflector when the source is close cannot be used as evidence for
the case when the source is distant and the energy of the emitted
photons
is large enough but the energy of the received photons is
miniscule.






The correct implicit equation I was using was
m=E/c^2 not m_0=E/c^2.
And since p=mv=Ev/c^2
and mc^2=(m_0)c^2/(1-v^2/c^2)^1/2 you get
the corrected equation above.

The equation you used implies non-zero mass while the
(invariant) mass of a photon is zero, hence the possible
confusion.

The whole concept of the photon is confused and full of
contradictions. There
is no such thing as a photon at rest so the mass of the photon
is never zero!!!

If you want to use "relativistic mass", it is
best to use that phrase so everyone understands you.

I was using mass in the normative sense or total mass
which can be analysed into the rest mass and the relativistic
mass
or whatever else you might desire. You were using mass
in the specialized sense of "rest" mass.


Ralph

"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...
"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...
That is the wrong equation.The correct form is:

E^2 = m^2 * c^4 + p^2 * c^2

No this is the wrong equation. The correct form is
E^2 = (m_0)^2 * c^4 + p^2 * c^2

Physics these days has generally dropped the old 'relativistic
mass' presentation as anachronistic and confusing.

I hope not.
It is confusing not to make the distinction.

I suggest you read the FAQ pages:

http://math.ucr.edu/home/baez/physic...y/SR/mass.html

http://math.ucr.edu/home/baez/physic...y/SR/mass.html

You also
misunterstood my question. Namely, when you start an oscillation
of charge in a distant isolated source
by thermally exciting bound electrons in tungsten for example,
the oscillating electrons may recoil from one another as you
suggest in a random and cancelling manner but the whole source
would not be recoiling against anything.

That depends on how you produce the change of temperature.
If you heat up the tungsten coil of a light bulb by passing
a current through it, obviously there is no recoil. You need
to explain how you heat up "distant isolated" material
thermally.

My point then is that you cannot use a mathematical equation
that
applies to source and reflector together over a small distance,
as evidence for photon recoil causing a motion of eg an 8 watt
transmitter on a distant spacecraft as photons are emitted.

Why not? A photon is just a photon. Either it carries
momentum or is doesn't.

My point is that the evidence for photon recoil in a receiver and
reflector when the source is close cannot be used as evidence for
the case when the source is distant and the energy of the emitted
photons is large enough but the energy of the received photons is
miniscule.

Solar sails are accelerated by recoil from photons from
the Sun.

The correct implicit equation I was using was
m=E/c^2 not m_0=E/c^2.
And since p=mv=Ev/c^2
and mc^2=(m_0)c^2/(1-v^2/c^2)^1/2 you get

Note this gives

m_r * c^2 = 0 * c^2 / sqrt(0)

or

m_r = 0/0

which is undefined.

the corrected equation above.

The equation you used implies non-zero mass while the
(invariant) mass of a photon is zero, hence the possible
confusion.

The whole concept of the photon is confused and full of
contradictions. There
is no such thing as a photon at rest so the mass of the photon
is never zero!!!

Read the FAQ. The term 'rest mass' comes about because
rest is the condition when the total energy equals the
invariant mass. The contradiction appears only because
you have been confused by the terminology, and that is
precisely why it is being dropped.

If you want to use "relativistic mass", it is
best to use that phrase so everyone understands you.

I was using mass in the normative sense or total mass
which can be analysed into the rest mass and the relativistic
mass
or whatever else you might desire. You were using mass
in the specialized sense of "rest" mass.

No, I am using it in the standard physical sense of
invariant mass whereas you are trying to use it in
the 19th century style of the longitudinal component
of a quantity that has different values in different
directions but can take the place of Newtonian mass
in many equations. It is actually a mathematical term
consisting of the product of the invariant mass and a
speed-dependent factor.

George

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...
"ralph sansbury" wrote in message
...

"George Dishman" wrote in
message
...

"ralph sansbury" wrote in message
...
That is the wrong equation.The correct form is:

E^2 = m^2 * c^4 + p^2 * c^2

No this is the wrong equation. The correct form is
E^2 = (m_0)^2 * c^4 + p^2 * c^2

Physics these days has generally dropped the old
'relativistic
mass' presentation as anachronistic and confusing.

I hope not.
It is confusing not to make the distinction.

I suggest you read the FAQ pages:

Interesting and there does seem to be an acknowledged split.
I side with the traditional Rindler crowd and against the Baez
crowd for obvious reasons that a distinction should be made.
Also if you look at some common physics texts you will see that
the meaning of mass in most text books used today is the total
mass not the relativistic mass or the rest mass. And if this
shows there is something wrong with the concept of the photon
then so be it.
Re a solar sail, this involves much stronger magnetic forces
than the absorption of radiation by a distant spacecraft antenna.
.. Re the emission of a photon eg by a distant 8Watt transmitter
not produced after the absorption of photons by a reflective
surface, between what and what is the momentum conserved? Between
what and what is the magnetic force associated with light
pressure etc.?

Ralph

http://math.ucr.edu/home/baez/physic...y/SR/mass.html

http://math.ucr.edu/home/baez/physic...y/SR/mass.html

You also
misunterstood my question. Namely, when you start an
oscillation
of charge in a distant isolated source
by thermally exciting bound electrons in tungsten for
example,
the oscillating electrons may recoil from one another as you
suggest in a random and cancelling manner but the whole
source
would not be recoiling against anything.

That depends on how you produce the change of temperature.
If you heat up the tungsten coil of a light bulb by passing
a current through it, obviously there is no recoil. You need
to explain how you heat up "distant isolated" material
thermally.

My point then is that you cannot use a mathematical
equation
that
applies to source and reflector together over a small
distance,
as evidence for photon recoil causing a motion of eg an 8
watt
transmitter on a distant spacecraft as photons are emitted.

Why not? A photon is just a photon. Either it carries
momentum or is doesn't.

My point is that the evidence for photon recoil in a
receiver and
reflector when the source is close cannot be used as evidence
for
the case when the source is distant and the energy of the
emitted
photons is large enough but the energy of the received
photons is
miniscule.

Solar sails are accelerated by recoil from photons from
the Sun.

The correct implicit equation I was using was
m=E/c^2 not m_0=E/c^2.
And since p=mv=Ev/c^2
and mc^2=(m_0)c^2/(1-v^2/c^2)^1/2 you get

Note this gives

m_r * c^2 = 0 * c^2 / sqrt(0)

or

m_r = 0/0

which is undefined.

the corrected equation above.

The equation you used implies non-zero mass while the
(invariant) mass of a photon is zero, hence the possible
confusion.

The whole concept of the photon is confused and full of
contradictions. There
is no such thing as a photon at rest so the mass of the
photon
is never zero!!!

Read the FAQ. The term 'rest mass' comes about because
rest is the condition when the total energy equals the
invariant mass. The contradiction appears only because
you have been confused by the terminology, and that is
precisely why it is being dropped.

If you want to use "relativistic mass", it is
best to use that phrase so everyone understands you.

I was using mass in the normative sense or total mass
which can be analysed into the rest mass and the
relativistic
mass
or whatever else you might desire. You were using mass
in the specialized sense of "rest" mass.

No, I am using it in the standard physical sense of
invariant mass whereas you are trying to use it in
the 19th century style of the longitudinal component
of a quantity that has different values in different
directions but can take the place of Newtonian mass
in many equations. It is actually a mathematical term
consisting of the product of the invariant mass and a
speed-dependent factor.

George

"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...

I hope not.
It is confusing not to make the distinction.

I suggest you read the FAQ pages:

Interesting and there does seem to be an acknowledged split.
I side with the traditional Rindler crowd and against the Baez
crowd for obvious reasons that a distinction should be made.

It is not a split but a change. It will take a long time
for old ways to die out but there is no debate about it.

Also if you look at some common physics texts you will see that
the meaning of mass in most text books used today is the total
mass not the relativistic mass or the rest mass.

"Relativistic mass" is the combination of invariant (rest) mass
and kinetic energy. I don't know what you mean by "total mass"
but most of the books I have deal only with the modern meaning
of mass and have sidenotes to explain what the old "relativistic
mass" means. The only exception is one printed in the 1950's.

And if this
shows there is something wrong with the concept of the photon
then so be it.

It doesn't. As the FAQ pages say, one of the reasons the term
"relativistic mass" is being dropped is because of these
confusions that it creates.

Re a solar sail, this involves much stronger magnetic forces
than the absorption of radiation by a distant spacecraft antenna.

I'm not aware of any magnetic effect on solar sails. Do you
mean drag like using cables in the Earth's magnetic field?

. Re the emission of a photon eg by a distant 8Watt transmitter
not produced after the absorption of photons by a reflective
surface, between what and what is the momentum conserved?

The photons and the craft. You started this thread by saying

When a bullet and the exploding gas shoots out of the barrel of a
gun, the gun moves in the opposite direction at a lesser velocity
because of its greater mass according to conservation of
momentum..

The photons are like the bullets and the spacecraft
behaves like the gun.

Between
what and what is the magnetic force associated with light
pressure etc.?

I am talking only about radiation pressure, not magnetic effects.

George

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...



RS Baez FAQ says that some people want to change the
notation of
E^2=(m_0)^2c^4+p^2c^2 where p=mv=Ev/c^2. Even in the more
recent texts the distinction between rest mass and the mass
increase
due to relativity of a mass in motion is made in the description
of the formula if not in the notation.

GD It is not a split but a change. It will take a long time
for old ways to die out but there is no debate about it.
As the FAQ pages say, one of the reasons the term
"relativistic mass" is being dropped is because of these
confusions that it creates.

RS The confusion is in the inadequate concept. The rest mass
of a photon is zero, its mass in motion due to the relativity
increase
is infinity so (1/2)mv^2 =(1/2)mc^4 =infinity times c^4 but also
hf where f is the frequency is equal to the kinetic energy of the
moving photon (1/2)mv^2. This says
that infinity is equal to a finite number. Because of such
confusion in the concept I am questioning conclusions like you're
here based on its use.
Re a solar sail, this involves much stronger magnetic
forces than the absorption of radiation by a distant spacecraft
antenna.

GD I'm not aware of any magnetic effect on solar sails.

RS I am aware of the light momentum photon energy argument
"that when light is emitted from a source there is a recoil
effect".
But I am suggesting that there is no evidence for cases like the
8Watt transmitter on a distant spacecraft and that solar sails
involve much greater Wattage and that light pressure on mirrors
etc involve oscillations in the source and the receiver/reflector
acting on each other. That is the mechanism(see Feynman Lectures
p34-10), Bvq, on oscillating charge moving up and down at
velocity v in the direction of the propagation of the fields E
and B causes a driving pressure in the direction of the light
beam which is called light pressure.
Do you know any evidence besides the blanket application of
the momentum argument showing
such pressure effects in cases where the emission of photons
occurs in an isolated source like the spacecraft?
Ralph

In article , says...
Do you know any evidence besides the blanket application of
the momentum argument showing
such pressure effects in cases where the emission of photons
occurs in an isolated source like the spacecraft?
Ralph

I would think the onus is on you to provide some evidence to the
contrary. We've done the experiments and backed up our theories and
applied them. You're the one who needs to say why they don't apply to a
remote spacecraft. What is it about "remoteness" that changes much?

At least, I would think something like that, but you're a complete
crank. The normal processes of thought and experiment don't apply to
you.

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