Who is Correct, Tom Roberts or Brian Jones?
Rabbo
messages:
1.
> First - an assumption: I am assuming that the light "beam" from the
> star takes on the rotational characteristics of the star
Tom Roberts ("which way is light going" , 6/1/99):
"If I understand you correctly, then this is false. Once a given light
ray is emitted from the surface of a star, any subsequent motion or
rotation of the star cannot affect the light ray. Moreover, any
particular photon of the ray is emitted instantaneously, so any prior
motion or rotation is also irrelevant.
In synchrotron radiation from charged particles orbiting in a circle,
the radiation can be collimated by multiple apertures in a straight
line -- the light goes straight even though the sorce is circular."
----------------------------
2.
Brian Jones ("What does 'rest mass' imply", 6/1/99):
"All one ever hears is that light's (propagation) speed is source
independent; where was it stated that light's (propagation) direction
is not?
Your misunderstanding of my proof has of course no bearing on its
validity.
When light rays in each frame travel different paths through space at
equal speeds, the only way observers in each frame can obtain the same
(one-way) speed for these rays is if their clocks intrinsically slow
(and slow differently in each frame).
My simple (I thought!) proof showed clearly that light rays emitted at
one and the same point in space must and will travel different spatial
paths if the light sources are in different frames."
-------------------------------------
Can we clear this point up, once and for all. Is the behaviour of
light, emitted at a particular point in space, independent of source
velocity.
Stephen Wells
rr...@Xbatemansbay.com wrote:
The *speed* of light in an inertial frame is always c, regardless of
the motion of the emitter.
The *frequency* of the light can depend on the direction of its motion
and the relative motion between the emitter and the observer; this is
relativistic Doppler shift, or you can work it out using the
energy-momentum 4vector.
The *angular distribution* of light depends on the relative motion of
the emitter and the observer. If, for example, the emitter is emitting
isotropically, then to an observer in motion wrt the emitter, the
distribution will not be isotropic. This is the angle transformation, or
again you can work it out using 4velocities.
This may be what Brian Jones meant about 'different spatial paths.'
However, his conclusion that this requires intrinsic clock slowing is
not valid (Minkowski spacetime is a counterexample). Also I have no idea
what is meant by 'light sources being in different frames'. All events are
in all frames of reference- people using different frames just disagree on
coordinates.
--
Stephen Wells
"Mieulx est de ris que de larmes escrire,
Pour ce que rire est le propre de l'homme."
-Alcofribas Nasier.
Eleaticus
Ra...@the.brink(Rabbo) wrote:
> 1.
> > First - an assumption: I am assuming that the light "beam" from the
> > star takes on the rotational characteristics of the star
path because the emitter surface at the point and time
of emission was following a curved (rotational) tra-
jectory, then I doubt anyone says that.
But the/a major failing of SRian analysis of binary/etc
star light velocity studies as evidence against Galilean
c+v light is that the SRians don't show (in these ngs) under-
standing that the rotational velocity and direction/axis
of the emitting star, and its direction and velocity of
orbit, and the instaneous velocity of any fields or stellar
atmosphere through which the light must pass must be known.
Tom Roberts didn't even show the basics of the necesary
understanding in his sagnac pieces, a much simpler scenario.
> 2.
> Brian Jones ("What does 'rest mass' imply", 6/1/99):
>
> "All one ever hears is that light's (propagation) speed is source
> independent; where was it stated that light's (propagation) direction
> is not?
Move at v=.9c and see if your laser light shines
to the rear or to the side when pointed in the
direction of (forward) movement.
Eleaticus
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Eleaticus Oren C. Webster Thnk...@concentric.net ?
! "Anything and everything that requires or encourages systematic ?
! examination of premises, logic, and conclusions" ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
Tom Roberts
> [...] Is the behaviour of
> light, emitted at a particular point in space, independent of source
> velocity.
I'm not sure just what you are asking.
From your preceeding (omitted) text, I assume you are asking about
direction. Yes, all light emitted in a given direction will go in that
direction regardless of the source velocity (:-)). Note please that
direction is observer dependent. But if one changed "direction" into
"hits this object", then it is observer independent (and remains
independent of source velocity in the sense that all emitted light
which hits the object does indeed hit the object (:-)).
The vagueness of your question is what keeps making me
find tautologies (:-)). It is the observer-dependence of
direction which makes me unable to discuss how the direction
of an emitted light ray might vary with source velocity;
the physical situation must be specified....
The speed of the light is also independent of source velocity, both
in SR and GR, and experimentally.
The frequency and wavelength of the emitted light, however, are not
independent of source velocity, and depend upon the relative velocity
of the source and observer, including the direction of this velocity
3-vector wrt the beam of light. The SR Doppler formula describes this
quite well (i.e. agrees with experimental measurements).
Tom Roberts tjro...@lucent.com
Rabbo
wrote:
>Rabbo wrote:
>> [...] Is the behaviour of
>> light, emitted at a particular point in space, independent of source
>> velocity.
>
>I'm not sure just what you are asking.
Your statement was: 'Moreover, any particular photon of the ray is
emitted instantaneously, so any prior motion or rotation is also
irrelevant.'
Your example of synchrotron radiation sums it up.
Brian Jones, on the other hand says this: "My simple (I thought!)
proof showed clearly that light rays emitted at one and the same point
in space must and will travel different spatial paths if the light
sources are in different frames."
>
>From your preceeding (omitted) text, I assume you are asking about
>direction. Yes, all light emitted in a given direction will go in that
>direction regardless of the source velocity (:-)). Note please that
>direction is observer dependent. But if one changed "direction" into
>"hits this object", then it is observer independent (and remains
>independent of source velocity in the sense that all emitted light
>which hits the object does indeed hit the object (:-)).
>
> The vagueness of your question is what keeps making me
> find tautologies (:-)). It is the observer-dependence of
> direction which makes me unable to discuss how the direction
> of an emitted light ray might vary with source velocity;
> the physical situation must be specified....
My question is quite straightforward.
Take a point in space. Two light sources, travelling at different
speeds, coincidentally pass through that point and emit a spherical
light pulse.
Do the two subsequent spherical(?) wave fronts move out from the point
in exactly the same way or do they not? That is, will any observer,
anywhere, receive light from the two sources simultaneously.
According to you they do - but Brian Jones says that the light path is
dependent on the source reference frame.
Who is correct?
Bob Sanders
>Take a point in space. Two light sources, travelling at different
>speeds, coincidentally pass through that point and emit a spherical
>light pulse. Do the two subsequent spherical(?) wave fronts move
>out from the point in exactly the same way or do they not? That
>is, will any observer, anywhere, receive light from the two sources
All the experimental evidence is that any observer, anywhere, will
receive the two pulses simultaneously. That's the objectively
verifiable part of the answer. As to your parenthetical question,
whether or not the wavefronts are spherical is a matter of convention.
Using the natural system of coordinates in any inertial frame, i.e.,
ordinary clocks and rulers at rest with respect to that frame, the
wave fronts are spherical.
Paul B. Andersen
>
> On Wed, 02 Jun 1999 13:06:21 -0500, Tom Roberts <tjro...@lucent.com>
> wrote:
>
> >Rabbo wrote:
> >> [...] Is the behaviour of
> >> light, emitted at a particular point in space, independent of source
> >> velocity.
> >
> >I'm not sure just what you are asking.
> Your statement was: 'Moreover, any particular photon of the ray is
> emitted instantaneously, so any prior motion or rotation is also
> irrelevant.'
> Your example of synchrotron radiation sums it up.
>
> Brian Jones, on the other hand says this: "My simple (I thought!)
> proof showed clearly that light rays emitted at one and the same point
> in space must and will travel different spatial paths if the light
> sources are in different frames."
Which is a very imprecise statement, open to a number of interpretations.
Taken at "face value", I would say the statement is wrong.
However, having read his "example", I think I understand what he mean.
If you have a light source emitting a narrow beam of light - like a laser,
the _direction_ of the beam obviously depend on the direction of
the source. If the source (laser) is moving perpendicular to it's
length axis in your selected frame of reference, the direction of
the light beam is _not_ parallel to the length axis of the laser,
of obvious reasons if you think about it just a little bit.
Thus, if two parallel lasers are moving relative to each other,
and emits a flash as they pass close by each other, the two light
beams will _not_ be parallel, but have an angle to each other;
they will "travel different spatial paths".
This phenomenon is not very remarkable; it is a example of
aberration, e.g. the phenomenon that the direction of a light beam
(or any velocity) depend on the frame of reference in which it
is observed.
> >From your preceeding (omitted) text, I assume you are asking about
> >direction. Yes, all light emitted in a given direction will go in that
> >direction regardless of the source velocity (:-)). Note please that
> >direction is observer dependent. But if one changed "direction" into
> >"hits this object", then it is observer independent (and remains
> >independent of source velocity in the sense that all emitted light
> >which hits the object does indeed hit the object (:-)).
> >
> > The vagueness of your question is what keeps making me
> > find tautologies (:-)). It is the observer-dependence of
> > direction which makes me unable to discuss how the direction
> > of an emitted light ray might vary with source velocity;
> > the physical situation must be specified....
> My question is quite straightforward.
>
> Take a point in space. Two light sources, travelling at different
> speeds, coincidentally pass through that point and emit a spherical
> light pulse.
> Do the two subsequent spherical(?) wave fronts move out from the point
> in exactly the same way or do they not?
They move out in exactly the same way.
> That is, will any observer,
> anywhere, receive light from the two sources simultaneously.
Yes.
> According to you they do - but Brian Jones says that the light path is
> dependent on the source reference frame.
> Who is correct?
Brian Jones' statement is so imprecise that it is bound to be
misinterpreted. (Or maybe it would be better to say that he has
made an unjustified generalization from his "example".)
However, the direction of a narrow light beam depend on the direction
of the source. In a frame of reference in which the source
is moving, the direction of the light beam is not parallel
to the length axis of the source.
This is not in conflict with what Tom said.
> >The speed of the light is also independent of source velocity, both
> >in SR and GR, and experimentally.
> >
> >The frequency and wavelength of the emitted light, however, are not
> >independent of source velocity, and depend upon the relative velocity
> >of the source and observer, including the direction of this velocity
> >3-vector wrt the beam of light. The SR Doppler formula describes this
> >quite well (i.e. agrees with experimental measurements).
> >
> >
> >Tom Roberts tjro...@lucent.com
Paul
Gerald L. O'Barr
sa...@deathtospam.hermes.cam.ac.uk (Stephen Wells) wrote:
In article <37550e34...@nsw.nnrp.telstra.net>,
rr...@Xbatemansbay.com wrote:
>
>
O'Barr comments:
Near the end of this post, Wells said this:
Stephen Wells wrote: . . . .
. All events are in all frames of reference- people using
different frames just disagree on coordinates.
O'Barr comments:
Now of course Wells and I do not agree on much, but to
me, this statement of Wells is most perfect, and it is
correct. We all live and we all exist in the exact same
reality. Or in Well's words, `All events are in all
frames of reference.' For all frames of references are
the same frame, at their base. No object can escape any
other object or any other event, no matter what its local
clock or extended clocks might do or might not do. In
other words, no changes in any spacetime relationship, in
any frame, under any circumstances, can cause any object
to miss any other object or event. All these `existences'
within a common existence cannot be violated, and this
would help put `physics' over the `math,' as SR math is
applied to our reality!
Thank you Wells for stating such an important concept.
It is a concept that clearly supports the ether. It puts
physics above the math!!!!!!!
--
Gerald L. O'Barr fl...@access1.net
Read: http://www.access1.net/flaco
Read Pete Brown's Aether FAQ at:
http://magna.com.au/~prfbrown/aeth_faq.htm
Read Jan 99 issue of Physics Today about the ether!
Claude Tydtgat
>1.
>> First - an assumption: I am assuming that the light "beam" from the
>> star takes on the rotational characteristics of the star
>
The light beam will follow a null geodesic. The null geodesics can be
calculated from the geometry of space-time. This is fully dependant of the
matter energy distribution. So if the rotating star gives rise to a Kerr
metric (rotating black hole metric), then the rotation of the star has
influence on the light beam.
Claude Tydtgat.
and...@ibm.net
[snip]
In order of most crediblity
in speaking about relativity:
Tom Roberts
Tom Jones (the singer)
Brian Jones
Rabbo
The significance of this point is, I believe, not fully understood or
appreciated.
(For one thing, the same principle should surely apply to moving
observers passing through the same point at the same time. As Tom
Roberts said, emission of light is instantaneous. Should not the
absorption of light be instantaneous, too, and therefore observer
independent? That's just by the way.)
Let's get philosophical, here. It is generally agreed that all light
pulses emitted simultaneously, by any number of moving sources, at a
given point in space, move away from that point in a single spherical
wavefront. It is probably logical to assume that the sphere is
perfect, relative to that point.
Taking this further, we could say that at any instant, each and every
point in space has, passing through it, an amount of EM radiation that
moves away from the point, in a perfectly spherical wavefront. This is
happening continually, whether or not any OBSERVERS are involved.
The light front MOVES outwards and moves outwards at an apparently
constant rate, as defined by the fact that its distance from the point
will be directly proportional to any subsequent time value.
The question I now want to ask is, "in the absence of observers, what
is the rate of movement of this wave front?"
A second related question involves two such points in space. Now I
think it is fair to assume that, by definition, points in space cannot
move relative to eachother (although aetherists may not agree).
If wave fronts were generated simultaneously (yes, synchronised from
the midpoint) at both points, then how long would it take for the
wavefronts to meet?
The standard answer will be, "it will depend on the distance between
the points, as measured by an observer in the same inertial frame as
the points. Divide that by c and you have the time".
That is not what I want to hear. I want to know the answer in the
absence of observers. You will no doubt say that no rate can be
calculated unless there are observers who have pre-defined standards
of length and time.
I will partially agree, -- BUT! -- MOVEMENT apparently occurs. It
occurs without observers and independently of any relationship between
an observer and the points in question, so what is going on?
Is this what can be construed as some kind of absolute space and time?
In fact, does light actually MOVE?
I'm beginning to think that the whole of SPACE and TIME, as we know
it, is very much a psychological construction, put together to
mentally display our interpretation of all the 4D information that
reaches our sensory organs.
That is why relativity is so physically hard to imagine.
>
>> According to you they do - but Brian Jones says that the light path is
>> dependent on the source reference frame.
>> Who is correct?
>
>Brian Jones' statement is so imprecise that it is bound to be
>misinterpreted. (Or maybe it would be better to say that he has
>made an unjustified generalization from his "example".)
>
>However, the direction of a narrow light beam depend on the direction
>of the source. In a frame of reference in which the source
>is moving, the direction of the light beam is not parallel
>to the length axis of the source.
>
>This is not in conflict with what Tom said.
>
>> >The speed of the light is also independent of source velocity, both
>> >in SR and GR, and experimentally.
>> >
>> >The frequency and wavelength of the emitted light, however, are not
>> >independent of source velocity, and depend upon the relative velocity
>> >of the source and observer, including the direction of this velocity
>> >3-vector wrt the beam of light. The SR Doppler formula describes this
>> >quite well (i.e. agrees with experimental measurements).
>> >
>> >
>> >Tom Roberts tjro...@lucent.com
>
>Paul
You could be right about Brian Jones statement although it was Tom
Roberts who was talking about rotating light sources and not Brian.
Jim Carr
>
>But the/a major failing of SRian analysis of binary/etc
>star light velocity studies as evidence against Galilean
>c+v light is that the SRians don't show (in these ngs) under-
>standing that the rotational velocity and direction/axis
>of the emitting star, and its direction and velocity of
>orbit, and the instaneous velocity of any fields or stellar
>atmosphere through which the light must pass must be known.
Curious statement. What "SRians" don't show in these
newsgroups is irrlevant to what is shown in the literature.
What you seem unaware of is the importance of extinction.
The best evidence against c+v is in lab experiments.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
Eleaticus
> The best evidence against c+v is in lab experiments.
You mean like MMX and KTX?
Eleaticus
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Eleaticus Oren C. Webster Thnk...@concentric.net ?
! "Anything and everything that requires or encourages systematic ?
! examination of premises, logic, and conclusions" ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
Subject: 18. What about Michelson-Morley and Kennedy-Thorndike?
The supposed failure of a Galilean model
to fit the Michelson-Morley experimental
results is one commonly asserted reason
that Special Relativity is required.
A C+V Model For The Michelson-Morley and
Kennedy-Thorndike Experiments===========
========================================
Assume that not just emission, but also
refraction and reflection evidence a
c+v effect.
Let A be the 3-space vector describing
an arbitrary arm of a Michelson-Morley
device.
C=cA/|A| is thus the velocity vector
of light directed down the arm of the
MM device.
Let V be an arbitrary velocity vector of
the MM device in an arbitrary inertial
frame.
Because the target end of the arm is
approaching/receding, on the outgoing
light trip we have A+tV as the location
of the arm's far end when the light
reaches it, and t(C+V) as the point the
light reaches at the end of the outgoing
trip, and:
A+tV = t(C+V). (15)
And thus: tC = A, which is to say, the
arbitrary velocity of the device is im-
material. Whatever the length and dir-
ection of the arm, the time it takes is
dependent only on arm length; tc=|A|.
With the arbitrary apparatus speed and
direction being immaterial in the one
direction, it should be clear that it is
also immaterial on the return trip.
Further, this result also holds for any
other arm of the MM apparatus, at right
angles to the first arm or not.
It should be noted this result shows us
that should there be such a thing as
a universal frame, it would not be de-
tectable by means of this device.
Further, since this model assumes a con-
stant, c, for the speed of light depart-
ing any reflective or refractive device,
relative to the refractive or reflective
device, starlight that impinges on such
a device [an MM device renamed a Kennedy
-Thorndike device] has any c+w velocity
of impingement immediately reduced to
c+v, and thus is not a possible device
for detecting effects of differences in
incoming light velocity.
Which is to say that Kennedy-Thorndike
is an exercise in futility in a c+v uni-
verse.
Note that the vector addition of veloc-
ity shows up in the above formula in two
ways: the explicit (c+v), and in setting
up the direction of light as the same as
the direction of the arm; the only way
light could truly reach the end of an
arm if truly aimed down the length of
the arm is if the arm's velocity was
added to the light velocity, with vector
addition.