recoiling photons evidence?

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...

No particle increases its mass, charged or otherwise. Only
their energy changes.

This is double talk. If by mass you mean the
gravitational
interaction then the mass increases.

How about a little test. What is the mass of an electron
moving at 0.9999c? I say it is 9.1095*10^-31 kg. What do
you say?

I say the mass is greater (1-.9998)^-1/2 about times
9.1095*10^-31 kg.
You say the product of this times .9999c is what is greater
which implies that the mass as observed in mass spectrometers
Kaufmann's experiemtn is greater.
Of course you can play semantic games but they are only
sematic games unless you can say where the increased momentum
resides.
I say it resides inside the charged particle and that the mass
does not increase as the spectrometer says but that the magnetic
interaction in Kaufmann's experiment becomes noticeably non
linear as v approaches c.

If you want to change the
definition so that this change is attributed to an increase
in
energy to make it more compatible with the concept of the
photon
etc,

The energy a particle possesses due to motion has been
known as "kinetic energy" for a very long time, I just
stick to that. As for mass, again from the FAQ:

"Technically, it is the invariant length of the
particle's four-momentum."

You should look at Kaufmann's experiment FAQ

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

How about a little test. What is the mass of an electron
moving at 0.9999c? I say it is 9.1095*10^-31 kg. What do
you say?

I say the mass is greater (1-.9998)^-1/2 about times
9.1095*10^-31 kg.

OK, that's M_r = 6.442*10^-29kg. Now suppose the particle
is moving due east. (Bear with me please, the point I am
trying to illustrate becomes apparent soon

Apply a force of 6.442*10^-29N in an northerly direction
to this particle. What is the acceleration?


Apply a force of 6.442*10^-29N in an easterly direction
to this particle. What is the acceleration?

You say the product of this times .9999c is what is greater
which implies that the mass as observed in mass spectrometers
Kaufmann's experiemtn is greater.

Current physics says the mass is m = 9.1095*10^-31 kg
and the kinetic energy is KE = 5.7075*10^-12 J

Note that your M_r = m + KE/c^2

Of course you can play semantic games but they are only
sematic games unless you can say where the increased momentum
resides.
I say it resides inside the charged particle and that the mass
does not increase as the spectrometer says but that the magnetic
interaction in Kaufmann's experiment becomes noticeably non
linear as v approaches c.

I don't follow, you just said the mass does change above.
I agree the momentum is not linearly related to speed, the
same is also true for the kinetic energy.

You should look at Kaufmann's experiment FAQ

I tried "Kaufmann FAQ" in Google and got a baseball library
and a "rotten tomato forum". Can you give me the URL for
this document?

George

Ralph Sansbury replied to Craig Markwardt:

But there are no physically massless particles because it is
impossible to measure zero mass. Therefore you can only DEFINE the
photon as massless... There is no independent rationale for this.

There are no physically giraffeless particles because it is
impossible to measure zero giraffes. Therefore you can only
DEFINE the photon as giraffeless. There is no independent
rationale for this.

Measurements show that photons have less than 0.1 nanogiraffe,
but it wouldn't make sense to say that photons therefore have
no giraffes at all.

-- Jeff, in Minneapolis

..

Ralph Sansbury replied to Craig Markwardt:

But there are no physically massless particles because it is
impossible to measure zero mass. Therefore you can only DEFINE the
photon as massless... There is no independent rationale for this.

There are no physically giraffeless particles because it is
impossible to measure zero giraffes. Therefore you can only
DEFINE the photon as giraffeless. There is no independent
rationale for this.

Replace "giraffe" with anything you want.

-- Jeff, in Minneapolis

..

Jeff Root wrote about giraffeless photons...

Oops! I didn't mean to post both versions! Heh.

-- Jeff, in Minneapolis

..

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...
I say the mass is greater (1-.9998)^-1/2 about times
9.1095*10^-31 kg.

OK, that's M_r = 6.442*10^-29kg. Now suppose the particle
is moving due east. (Bear with me please, the point I am
trying to illustrate becomes apparent soon

Sorry no Address the point directly. momentum for particles
with zero mass can be defined to be E/c or hv/c while particles
with mass have momentum mv where m is measureable. Yes or no?

Current physics says the mass is m = 9.1095*10^-31 kg
and the kinetic energy is KE = 5.7075*10^-12 J

Note that your M_r = m + KE/c^2

Of course you can play semantic games but they are only
sematic games unless you can say where the increased momentum
resides.
I say it resides inside the charged particle and that the
mass
does not increase as the spectrometer says but that the
magnetic
interaction in Kaufmann's experiment becomes noticeably non
linear as v approaches c.

I don't follow, you just said the mass does change above.
Yes it appears not to curve as much away from the
gravitational
force pulling it down as one would expect from the increased
magnetic
interaction.

I agree the momentum is not linearly related to speed, the
same is also true for the kinetic energy.

The reason for this is not because momentum is not mv but
because the linear magnetic interaction becomes non linear near
c.


You should look at Kaufmann's experiment FAQ

I tried "Kaufmann FAQ" in Google and got a baseball library
and a "rotten tomato forum". Can you give me the URL for
this document?


Google on Kaufmann mass increase

George

"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...
I say the mass is greater (1-.9998)^-1/2 about times
9.1095*10^-31 kg.

OK, that's M_r = 6.442*10^-29kg. Now suppose the particle
is moving due east. (Bear with me please, the point I am
trying to illustrate becomes apparent soon

Sorry no Address the point directly.

My questions do address it directly. I think they will also
provide an answer to your comments on Kaufmann. I've answered
all your questions for many months so it's your turn now:

OK, that's M_r = 6.442*10^-29kg. Now suppose the particle
is moving due east. (Bear with me please, the point I am
trying to illustrate becomes apparent soon

Apply a force of 6.442*10^-29N in an northerly direction
to this particle. What is the acceleration?


Apply a force of 6.442*10^-29N in an easterly direction
to this particle. What is the acceleration?


You say the product of this times .9999c is what is greater
which implies that the mass as observed in mass spectrometers
Kaufmann's experiemtn is greater.

The only Kaufmann experiment I've found that relates is
from 1901. Is that the one you are reffering to?

momentum for particles
with zero mass can be defined to be E/c or hv/c while particles
with mass have momentum mv where m is measureable. Yes or no?

No. Answer the questions above and I will be able to
make that clearer.

Current physics says the mass is m = 9.1095*10^-31 kg
and the kinetic energy is KE = 5.7075*10^-12 J

Note that your M_r = m + KE/c^2

Of course you can play semantic games but they are only
sematic games unless you can say where the increased momentum
resides.
I say it resides inside the charged particle and that the mass
does not increase as the spectrometer says but that the magnetic
interaction in Kaufmann's experiment becomes noticeably non
linear as v approaches c.

I don't follow, you just said the mass does change above.
Yes it appears not to curve as much away from the gravitational
force pulling it down as one would expect from the increased
magnetic interaction.

Indeed. If that's what you meant, my questions will go a long
way to clarifying the issue.

I agree the momentum is not linearly related to speed, the
same is also true for the kinetic energy.

The reason for this is not because momentum is not mv but
because the linear magnetic interaction becomes non linear near
c.


You should look at Kaufmann's experiment FAQ

I tried "Kaufmann FAQ" in Google and got a baseball library
and a "rotten tomato forum". Can you give me the URL for
this document?


Google on Kaufmann mass increase

I found a couple of pages just giving the usual relativistic
mass increase derivation and referring the experiments on
high-speed electrons done in 1901. I haven't found a FAQ
though. If you know where it is, why can't you just say?

George

"ralph sansbury" writes:
"Craig Markwardt" wrote in
message news
[ ... ]
The "relativistic mass" is an analogy or crutch for
understanding
relativity. It begins by asking the question: *if* we can
express the
momentum by the classical equation p = mv, then what is the
mass m
which makes that equation work under relativity. For massive
particles, the answer is of course m = m0 / sqrt(1-(v/c)^2).
For
massless particles, the premise of the statement is false (we
can't
express momentum as p = mv), therefore any conclusions are
irrelevant.

But there are no physically massless particles because it is i
mpossible to measure zero mass. Therefore you can only DEFINE the
photon as massless.m_p=0, and define its momentum as hf/c but
not (m_p)c. There is no independent rationale for this.

Your logic is flawed. What is or isn't measureable has no bearing on
whether photons are massless. The "rationale" for the theory is
irrelevant, if the theory works.

Once these definitions are in place you show the conservation
of momentum of course but my question is can you extrapolate from
this as you do with ordinary particles. That is,if you have an
isolated photon emitter, will the conservation of momentum work
to predict a recoil of the emitter.

Of course. It is how atom trapping works.

In all of the experiments etc where this occurs and can be
described with the above definitions and the conservation of
momentum, the emitter and reflector/scatterer are near one
another and a quantitative explanation of the phenomena is
possible also in terms of the magnetic force between oscillating
electrons in the source and the receiver see Feynman Lectures etc
on light pressure.

Your distinction is irrelevant. Momentum conservation and light
pressure are different ways to say the same thing. Momentum (density)
conservation also exists for classical optical light waves.

CM

"Craig Markwardt" wrote in
message news

"ralph sansbury" writes:
.. In any case all of
these possible properties for the photon no matter which ones
you
accept make the photon unlike other particles for which the
conservation of momentum has been shown to be applicable.

The "relativistic mass" is an analogy or crutch for
understanding
relativity. It begins by asking the question: *if* we can
express the
momentum by the classical equation p = mv, then what is the
mass m
which makes that equation work under relativity. For massive
particles, the answer is of course m = m0 / sqrt(1-(v/c)^2).
For
massless particles, the premise of the statement is false (we
can't
express momentum as p = mv), therefore any conclusions are
irrelevant.

The "analogy" line of reasoning begs the question. Relativity
is not
classical physics, and so p = mv is not a requirement. Under
relativity, it is understood that mass is always the rest mass
(=invariant mass), and the relativistic momentum is redefined
to be p
= sqrt((E/c)^2 - (m0 c)^2), which is m0 v / sqrt(1-(v/c)^2) for
massive particles, but E/c for massless particles.

But this obviously implies a value for mass for non
zero mass particles of m0/sqrt(1-(v/c)).
It is not complicated George. This says that
momentum in this case can be written as p=mv as well as
p=sqrt((E/c)^2)-(m0 c)^2. Thus you can say that the energy E
increases or that the mass increases. The equations say that
both formulations are mathematically possible.
Granted your formulation is more general in taking into
account
the constructed photon particle with zero defined mass as well as
the non zero mass of particles with the usual properties.
But that does not mean that other formulations are not
accurate
also.
Furthermore the trajectory of masses in mass spectrometers
as in Kaufmann's original experiment shows that the either the
mass of the charged particles increases or the magnetic
interaction becomes non linear as the speed of the particles
increase starting at about .1c.
That is it requires a greater magnetic field to push the
rapidly moving charged
particle upward against the force of gravity than a slower moving
charged particle.
The conservation of momentum is not helpful here. And in the
case of
the Compton effect and light pressure it is not helpful either
since it obscures
the physical mechanism which is the magnetic interaction between
oscillating charges



When one says that the relativistic mass increases with speed,
one
really means,.....
Both meanings are real and since a photon is not a real
particle, the meaning
that grew out of the definition of a photon is most
questionable.
Ralph

"ralph sansbury" writes:
"Craig Markwardt" wrote in
message news

"ralph sansbury" writes:
. In any case all of
these possible properties for the photon no matter which ones
you
accept make the photon unlike other particles for which the
conservation of momentum has been shown to be applicable.

The "relativistic mass" is an analogy or crutch for
understanding
relativity. It begins by asking the question: *if* we can
express the
momentum by the classical equation p = mv, then what is the
mass m
which makes that equation work under relativity. For massive
particles, the answer is of course m = m0 / sqrt(1-(v/c)^2).
For
massless particles, the premise of the statement is false (we
can't
express momentum as p = mv), therefore any conclusions are
irrelevant.

The "analogy" line of reasoning begs the question. Relativity
is not
classical physics, and so p = mv is not a requirement. Under
relativity, it is understood that mass is always the rest mass
(=invariant mass), and the relativistic momentum is redefined
to be p
= sqrt((E/c)^2 - (m0 c)^2), which is m0 v / sqrt(1-(v/c)^2) for
massive particles, but E/c for massless particles.

But this obviously implies a value for mass for non
zero mass particles of m0/sqrt(1-(v/c)).
It is not complicated George. This says that
momentum in this case can be written as p=mv as well as
p=sqrt((E/c)^2)-(m0 c)^2. Thus you can say that the energy E
increases or that the mass increases. The equations say that
both formulations are mathematically possible.

First, I am not George.

Second. There is nothing innately correct about p = mv. Under
relativity, there is definitionally a different expression for
momentum than classical mechanics.

Granted your formulation is more general in taking into
account
the constructed photon particle with zero defined mass as well as
the non zero mass of particles with the usual properties.
But that does not mean that other formulations are not
accurate
also.

There are probably an infinite number of alternatives to relativity.
The job of science is to rule out as many of those alternatives as
possible. Relativity defines certain behaviors and expressions for
massive and massless particles, which to date have been borne out by
observations and experiment.


Furthermore the trajectory of masses in mass spectrometers
as in Kaufmann's original experiment shows that the either the
mass of the charged particles increases or the magnetic
interaction becomes non linear as the speed of the particles
increase starting at about .1c.

In the early days of relativity research, it was just as easy to say
the relativistic mass changed at different velocities. However, the
theory has advanced. Now it is understood that certain quantities are
invariant under Lorentz transformations: charge, rest mass, and
four-momentum. Note I said four-momentum, which is the combination of
energy and momentum. Alone, neither is invariant, but combined into a
four-vector, they are. Under these interpretation, it is not the mass
which changes non-linearly with velocity, but the momentum. I invite
you to research this further yourself.


That is it requires a greater magnetic field to push the
rapidly moving charged
particle upward against the force of gravity than a slower moving
charged particle.

First, you appear to be disregarding the electric part of the
electromagnetic field.

Second, when interpretted as a packet of oscillating electric and
magnetic fields, a photon can naturally fill the role you describe.


The conservation of momentum is not helpful here. And in the
case of
the Compton effect and light pressure it is not helpful either
since it obscures
the physical mechanism which is the magnetic interaction between
oscillating charges

Irrelevant. Magnetic fields embody a pressure (B^2/(8*pi)). Pressure
in turn carries a momentum density. Conservation of momentum is
always a consideration.

CM