Are orbiting objects at rest for light?
mluttgens
with a definite velocity [speed] c which is independent of the state
of motion of the emitting body."
.Also, light velocity c is independent of the state of motion of the
light receiving body.
Thus, logically, for light, the velocity of an orbiting body is zero.
Is it this not paradoxical, as we know that orbiting objects have a
tangential velocity?
Marcel Luttgens
dlzc
On Aug 29, 3:16 pm, mluttgens <mluttg...@orange.fr> wrote:
> According to Einstein, "... light is always
> propagated in empty space with a definite
> velocity [speed] c which is independent of
> the state of motion of the emitting body."
> .Also, light velocity c is independent of
> the state of motion of the light receiving
> body.
Not strictly true. If the receiving body is accelerating, there is no
guarantee light speed will be measured as c.
> Thus, logically, for light, the velocity of
> an orbiting body is zero.
No. We have no basis for saying this, with any form of logic. We
don't know what light measures. The logic of Lorentz transforms
cannot necessarily be applied to light.
> Is it this not paradoxical, as we know that
> orbiting objects have a tangential velocity?
Well, it is your misunderstanding, and heaven knows you've said
paradoxical things before. You just never stick around long enough to
see where you messed up.
David A. Smith
BURT
You can get behind light in space by acceleration to high speed
motion.
You can also leave it behind so that it might take longer to catch up.
In either case there is a relative motion of light ahead or behibd
matter
through space that is not C.
Mitch Raemsch
Androcles
"dlzc" <dl...@cox.net> wrote in message
news:44ddc80c-da2e-41e3...@q21g2000prm.googlegroups.com...
mluttgens:
On Aug 29, 3:16 pm, mluttgens <mluttg...@orange.fr> wrote:
> According to Einstein, "... light is always
> propagated in empty space with a definite
> velocity [speed] c which is independent of
> the state of motion of the emitting body."
> .Also, light velocity c is independent of
> the state of motion of the light receiving
> body.
Not strictly true. If the receiving body is accelerating, there is no
guarantee light speed will be measured as c.
============================================
Ah, right. So Einstein was talking out of his arse since all bodies
everywhere are constantly accelerating.
GogoJF
Yes, I think that Einstein stressed the difference between velocity of
the transmitted medium is independent of the observation of orbiting
objects.
Thus, yes, the orbiting body is zero- or as close to instantaneous as
possible.
BURT
In an ideal circular orbit there is constant speed around the center
of gravity.
There is no longer any strength of gravity changing energy speed in
the round closed curve.
Mitch Raemsch
GogoJF
This all has to do with perpetual motion. These motions happen- and
we hardly have any mention of their motions. Because, in reality, we
are shielded towards this type of motion- this type of logic- its
called very special relativity- or, man's crude point of view. With
special relativity we put ourselves into a calm, relaxed, perceptual,
reference frame- and then create our mathematics.
GogoJF
Sorry, I was just jokin.
Tom Roberts
> mluttgens:
> On Aug 29, 3:16 pm, mluttgens <mluttg...@orange.fr> wrote:
>> According to Einstein, "... light is always
>> propagated in empty space with a definite
>> velocity [speed] c which is independent of
>> the state of motion of the emitting body."
>> .Also, light velocity c is independent of
>> the state of motion of the light receiving
>> body.
>
> Not strictly true. If the receiving body is accelerating, there is no
> guarantee light speed will be measured as c.
Hmmmm. A "receiving body" cannot possibly measure the speed of the received
light -- that requires measuring multiple points along the light's trajectory,
and a single "body" cannot do that.
Moreover, Einstein was only considering measurements in INERTIAL FRAMES when he
wrote that:
Measuring the speed of anything requires observing it at two
(or more) places along its trajectory, using synchronized
clocks. This is equivalent to constructing a coordinate system
and measuring the coordinate speed of the object. Obviously
the value obtained will depend on the details of how the
coordinates were constructed. In this paper, Einstein always
used standard clocks and rulers at rest in an inertial frame,
with the clocks synchronized as he described.
>> Thus, logically, for light, the velocity of
>> an orbiting body is zero.
You use words funny, and in such an unusual manner that this is nonsense.
In SR, light emitted by a moving source travels in vacuum with speed c relative
to any inertial frame. This holds for orbiting sources, but not for orbiting
frames (which are not inertial in the sense of SR).
> No. We have no basis for saying this, with any form of logic. We
> don't know what light measures. The logic of Lorentz transforms
> cannot necessarily be applied to light.
Yes, that too. Sort of: it's not "we don't know what light measures", but rather
that light cannot make any measurements. Lorentz transforms can be applied to
the propagation of light rays relative to inertial frames, but cannot be applied
to the "rest frame" of a light ray in vacuum (because such a frame does not
exist in SR).
Tom Roberts
GogoJF
Tom says: You use words funny, and in such an unusual manner that
this is nonsense.
I say: I'm just trying to make a normally boring subject, a little
more interesting.
whoever
news:776112a1-aad2-4388...@t2g2000yqe.googlegroups.com...
>
>According to Einstein, "... light is always propagated in empty space
>with a definite velocity [speed] c which is independent of the state
>of motion of the emitting body."
Yes
>.Also, light velocity c is independent of the state of motion of the
>light receiving body.
Yes
>Thus, logically, for light, the velocity of an orbiting body is zero.
No .. that doesn't follow by any logic at all. You could *perhaps* have
concluded that the velocity of the orbiting body is 'c' relative to the
frame of reference of the light. But light doesn't have an inertial frame
of reference, and talking about it as you would other inertial frames is
incorrect.
>Is it this not paradoxical,
No .. it is just ignorance on your part. Easily resolved by some thought
and study
> as we know that orbiting objects have a
>tangential velocity?
Try again
--- news://freenews.netfront.net/ - complaints: ne...@netfront.net ---
BURT
Relativity is misinterpreted as saying that light moves at C relative
to all motion of matter. But the truth is there is a difference from
C
in terms of light for matter. It is as you said C in empty space
always
with matter slower.
Mitch Raemsch
eric gisse
Trying to understand SR again, Marcel?
Good luck.
dlzc
On Aug 29, 8:12 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> dlzc wrote:
> > mluttgens:
> > On Aug 29, 3:16 pm, mluttgens <mluttg...@orange.fr> wrote:
> >> According to Einstein, "... light is always
> >> propagated in empty space with a definite
> >> velocity [speed] c which is independent of
> >> the state of motion of the emitting body."
> >> .Also, light velocity c is independent of
> >> the state of motion of the light receiving
> >> body.
>
> > Not strictly true. If the receiving body
> > is accelerating, there is no guarantee light
> > speed will be measured as c.
>
> Hmmmm. A "receiving body" cannot possibly
> measure the speed of the received light --
> that requires measuring multiple points along
> the light's trajectory, and a single "body"
> cannot do that.
Two detectors a meter apart, oriented in line with the light source, a
really good clock, and you've got it. Same as any measurement of
light speed would require. Or just one detector and "a" mirror (if
you weren't worried about changing light's speed with it). Given a co-
operative light source (pulses). ;>)
I just don't think you need to struggle so with "measuring light's
speed" with a single detector and no extent. We've never been able to
do that...
> Moreover, Einstein was only considering
> measurements in INERTIAL FRAMES when he
> wrote that:
>
> Measuring the speed of anything requires
> observing it at two (or more) places
> along its trajectory, using synchronized
> clocks. This is equivalent to constructing
> a coordinate system and measuring the
> coordinate speed of the object. Obviously
> the value obtained will depend on the
> details of how the coordinates were
> constructed. In this paper, Einstein always
> used standard clocks and rulers at rest in
> an inertial frame, with the clocks
> synchronized as he described.
David A. Smith
PD
> According to Einstein, "... light is always propagated in empty space
> with a definite velocity [speed] c which is independent of the state
> of motion of the emitting body."
> .Also, light velocity c is independent of the state of motion of the
> light receiving body.
>
> Thus, logically, for light, the velocity of an orbiting body is zero.
No, this is wrong. You have a hidden assumption that you've not shown.
Your assumption is that, to find the speed of light with respect to a
body moving with velocity V, then the result would have to be either c
+V or c-V, and since the result is stated to be c, then this is how
you conclude that for light, V must be = 0.
However, your assumption about how the velocities would combine is
wrong.
BURT
>
> - Show quoted text -
You can move behind light and to you it will then be going ahead
of you slower than it is through space which is a constant limit.
Mitch Raemsch
PD
Nope.
BURT
>
> - Show quoted text -
Matter is therefore not at rest with regard to light.
Mitch Raemsch
PD
Didn't say it was. Marcel did. But Marcel has difficulties.
BURT
>
> - Show quoted text -
Matter moves. Light moves. Matter moves behind light causing light's
relative motion ahead to be less but still at C in space.
Mitch Raemsch
PD
Nope. I told you that already.
>
> Mitch Raemsch
BURT
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Both have different motions through space creating relativity between
the two.
Mitch Raemsch
PD
Relativity doesn't mean "take the difference".
This is the part neither you nor Marcel seem to get.
>
> Mitch Raemsch
mluttgens
OK, thank you!
BURT
One point in Relativity is that it does.
Relativity for co-moving objects is their difference in their motion
through space together.
Mitch Raemsch
PD
Nope. Not so, Mitch.
Not v1-v2.
>
> Mitch Raemsch
Androcles
"PD" <thedrap...@gmail.com> wrote in message
news:323ba02b-e094-45cb...@h19g2000yqb.googlegroups.com...
=============================================
Yep. Is so, Duck.
"But the ray moves relatively to the initial point of k, when measured in
the stationary system, with the velocity c-v" -- your tin god, you stupid
wanker.
BURT
Also light moving in the opposite direction of matter's
movement through space creates a seperation above the speed
of light. Distane inbewteen them is groing in space in two
ways.
Mitch Raemsch
PD
wrote:
> "PD" <thedraperfam...@gmail.com> wrote in message
Nope.
> "But the ray moves relatively to the initial point of k, when measured in
> the stationary system, with the velocity c-v" -- your tin god, you stupid
> wanker.
That's closing speed.
Androcles
Nope.
That's closing speed.
==============================================
Nope, it's opening velocity. Not that a pair of fuckwitted wankers like
you and Einstein would know velocity from Geschwindigkeit anyway.
"I've lost interest. Foam and blather and waste all the time you want.
You're not getting anywhere." -- Phuckwit Duck
(Meaning "I lost that argument, those grapes are sour".)
Ref: d23006a4-4a88-4efb...@c34g2000yqn.googlegroups.com
"You are not entitled to be educated. Someone who insists on
being willfully ignorant does not deserve to be dissuaded.
Nobody owes you anything. Nobody *should* do anything for
you. It's your choice to learn or not to learn."-- Phuckwit Duck
Ref: 571b8ace-cca8-4392...@o28g2000yqh.googlegroups.com
PD
As I said, you are having problems with something very simple.
>
> "I've lost interest. Foam and blather and waste all the time you want.
> You're not getting anywhere." -- Phuckwit Duck
> (Meaning "I lost that argument, those grapes are sour".)
>
> Ref: d23006a4-4a88-4efb-b1f4-12b115399...@c34g2000yqn.googlegroups.com
>
> "You are not entitled to be educated. Someone who insists on
> being willfully ignorant does not deserve to be dissuaded.
> Nobody owes you anything. Nobody *should* do anything for
> you. It's your choice to learn or not to learn."-- Phuckwit Duck
>
> Ref: 571b8ace-cca8-4392-ba69-0a328320a...@o28g2000yqh.googlegroups.com
Androcles
============================================
That description of yourself is apt, simpleton, although I'm not
actually having a problem with you.
mluttgens
> On Aug 29, 5:16 pm, mluttgens <mluttg...@orange.fr> wrote:
>
> > According to Einstein, "... light is always propagated in empty space
> > with a definite velocity [speed] c which is independent of the state
> > of motion of the emitting body."
> > .Also, light velocity c is independent of the state of motion of the
> > light receiving body.
>
> > Thus, logically, for light, the velocity of an orbiting body is zero.
>
> No, this is wrong. You have a hidden assumption that you've not shown.
> Your assumption is that, to find the speed of light with respect to a
> body moving with velocity V, then the result would have to be either c
> +V or c-V, and since the result is stated to be c, then this is how
> you conclude that for light, V must be = 0.
> However, your assumption about how the velocities would combine is
> wrong.
According to Tom Roberts,
"In SR, light emitted by a moving source travels in vacuum with speed
c relative
to any inertial frame. This holds for orbiting sources, but not for
orbiting
frames (which are not inertial in the sense of SR)."
I suppose that this means that SR can't refute my hypothesis that for
light,
the velocity of an orbiting body is zero.
Then why would not the result be either c+V or c-V?
Marcel Luttgens
dlzc
On Sep 1, 7:54 am, mluttgens <mluttg...@orange.fr> wrote:
...
> According to Tom Roberts,
>
> "In SR, light emitted by a moving source travels
> in vacuum with speed c relative to any inertial
> frame. This holds for orbiting sources, but not
> for orbiting frames (which are not inertial in
> the sense of SR)."
>
> I suppose that this means that SR can't refute
> my hypothesis that for light, the velocity of
> an orbiting body is zero.
Right, since SR doesn't do "orbitting". And if you use the Lorentz
tranforms, and let v->c, that velocity could also be assumed to be
infinite.
> Then why would not the result be either c+V or
> c-V?
You've been told many times. Will one more time do the trick, or will
you forget this time too?
Maxwell requires that the speed of light be always c, for all inertial
observers. And GR, which makes your question unaskable, will allow
you to know what an orbitting observer would measure.
Since light cannot measure anything, which you seem to ignore, what
exactly is your question? "Why can't light measure anything, so that
I can imagine some sort of problem for SR?"
David A. Smith
PD
SR doesn't refute things. Observations do. SR is a theory whose
success is based on its compatibility with observations where it
claims to apply. SR doesn't refute crazy assertions about nature where
it does not claim to apply. But observations will.
>
> Then why would not the result be either c+V or c-V?
The principle reason is that such a hypothesis is inconsistent with
measurements made in real experiments.
Though it is fine to guess that velocities should combine like v1+v2,
the fact is that this guess does not get the right answer that matches
experimental observation.
The guess that DOES match experimental observation is that velocities
combine like this:
(v1 + v2)/(1 + v1*v2/c^2) or (v1 - v2)/(1 - v1*v2/c^2). This works for
all observations of combined velocities, regardless of the values of
v1 and v2.
I'll let you work out the result of this when v1=c and v2 = V:
(c + V)/(1 + c*V/c^2) = ?
Androcles
================================
SR is a failure based on its idiocy, and so are you. Using aether or
relativity,
explain this real observation:
http://www.britastro.org/vss/gifc/00918-ck.gif
mluttgens
Do you realize that c+V, according to such SR formula, which matches
with
experimental observations, is (c+V)/(1+V/c), meaning that for
physically realistic
velocities like orbital ones, one is entitled to consider that c
combines with V
like c+V or c-V?
Marcel Luttgens
>
PD
Be careful, please. You meant to say that "Do you realize that the
combination of c and V, according to such SR formula..." The
combination of c and V is not c+V. It is (c + V)/(1 + c*V/c^2). These
are certainly different, as I'm sure you can tell.
> according to such SR formula, which matches
> with
> experimental observations, is (c+V)/(1+V/c),
Please continue to simplify this expression. You're not done.
If you need a hint, factor out a 'c' from the first term in the
expression above.
Remember, the objective here is to see if the expression above yields
a number that is different than c. Just because you see a (c+V) in it
somewhere does not mean that.
BURT
>
>
> - Show quoted text -
Orbiting in the direction of light flow causes the relativity to be
a slower light flow ahead. Or faster if orbiting in the other dirction
to its flow.
Mitch Raemsch
mluttgens
You are right, Sum(c+v) = c, according to such formula.
Marcel Luttgens
PD
And again, I just want to reiterate that you are *combining*
velocities, not summing them. Velocities do not combine by summation,
no matter how much you want to believe they should.
But to summarize, what you learn from this is several things:
1. Velocities never combine to give a combined velocity greater than
c.
2. Velocities that are less than c will change value in a different
reference frame, but this ONLY applies to velocities less than c.
3. Light speed doesn't change when looked at from different reference
frames.
Now perhaps you can understand why your conclusion that the speed of
the orbiting satellite from the perspective of light must be zero, is
a nonsensical one, based wholly on an assumption that turns out to be
inappropriate.
mluttgens
Let's consider the relativistic Doppler:
Let v the velocity of the source of light of frequency fs relative to
the observer
(v is negative if the source moves toward the observer, otherwise, it
is positive)
The observed frequency f0 = fs * sqrt((c-v)/c+v)
According to you, c+v = c-v = c, hence f0 = fs !
Marcel Luttgens
Unified_Perspective
> According to Einstein, "... light is always propagated in empty space
> with a definite velocity [speed] c which is independent of the state
> of motion of the emitting body."
> .Also, light velocity c is independent of the state of motion of the
> light receiving body.
>
> Thus, logically, for light, the velocity of an orbiting body is zero.
> tangential velocity?
>
> Marcel Luttgens
In their own frame of reference circularly orbiting bodies are not
accelerated.
They only appear to have a tangential velocity when viewed by an
observer outside their frame of reference.
Review Kepler's Laws of motion and Newton's conservation of momentum
for circular and non-circular orbits if you seek a deeper
understanding of these issues.
The independence of light from mechanical motions is quite real, but
not absolute.
There is a red-shift called transverse red-shift that is dependent on
the motion of the emitting body. If that velocity is high enough to be
a noticeable percentage of the speed of light.
Also the observed pulsation of pulsars arises almost completely from
the effect of their relative motion.
The frequency and intensity of the observed light varies because of
the source, when viewed from our frame of reference, is moving at a
rate high enough to significantly approach the speed of the light it
emits.
If an observer were on a rotating or orbiting pulsar, it would not
appear to pulse. The frequency and amplitude of the emitted light
would appear to be constant. Because, just as you say, "Thus,
logically, for light, the velocity of an orbiting body is zero. Is it
this not paradoxical," ? The answer is yes, it does seem paradoxical.
But resolving this paradox is what Lorentz and Einstein have
substantially accomplished.
Similarly, if an observer has a high velocity with respect to a
constant velocity light stream the observed change "current" of the
stream can be measured by the observer.
However, relative motion of the either the emitter or observer does
not actually change the speed of the light stream. That stream
propagates at a substantially constant velocity. Just as Einstein
stated.
Mr. Gee
PD
No, that is NOT according to me. It is not true that c+v = c.
What is according to me is that the combination of velocities, as what
you would do in a transformation between reference frames, is
according to the rule I gave, which is NOT v1+ v2. What is it instead
is (v1 + v2)/(1 + v1*v2/c^2).
What you keep confusing is identifying "combination of velocities"
with c+v or c-v.
Furthermore, what this means is that you CANNOT substitute "c" every
time you see a formula "c-v" or "c+v". The reason is that "c-v" or "c
+v" does NOT mean "combination of velocities" -- those two are not
synonymous.
PD
mluttgens
You just claimed that velocities never combine to give a velocity
greater than c,
and directly afterward, that c+/-v are not a relativist sum of
velocities(what you call
a combinaison of velocities).
Then what represent c+v or c-v in the relativistic Doppler formula, if
not a "combination of velocities"?
Marcel Luttgens
mluttgens
> On Aug 29, 6:16 pm, mluttgens <mluttg...@orange.fr> wrote:
>
> > According to Einstein, "... light is always propagated in empty space
> > with a definite velocity [speed] c which is independent of the state
> > of motion of the emitting body."
> > .Also, light velocity c is independent of the state of motion of the
> > light receiving body.
>
> > Thus, logically, for light, the velocity of an orbiting body is zero.
> > Is it this not paradoxical, as we know that orbiting objects have a
> > tangential velocity?
>
> > Marcel Luttgens
>
> In their own frame of reference circularly orbiting bodies are not
> accelerated.
>
> They only appear to have a tangential velocity when viewed by an
> observer outside their frame of reference.
>
> Review Kepler's Laws of motion and Newton's conservation of momentum
> for circular and non-circular orbits if you seek a deeper
> understanding of these issues.
>
> The independence of light from mechanical motions is quite real, but
> not absolute.
It is clearly not absolute in the relativistic Doppler formula.
This is another SR paradox, due to above cited postulates leading to
the derivation of the SR formulas.
Simply claiming that c +/- v in the relativistic Dopple formula don't
correspond to what PD calls
a SR combination of velocities looks to me like a mere opinion.
Marcel Luttgens
BURT
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
For the Doppler of light fast flowing matter could absorb light
sideways
and would not energy shift the light. But what about inbetween angles
of light absorption for ahead and behind?
If there is maximum from directly ahead and behind one elevating
light's energy and one lowering light's energy what about the
inbetween angles and perfectly sideways?
Mitch Raemsch
PD
That's correct. Remember (c+v) is not how velocities combine. Nor is
it the case that when you see (c+v) or (c-v) that you are seeing a
combination of velocities.
Velocities combine according to a strict rule: (v1 + v2)/(1 + v1*v2/
c^2). That is how velocities combine. The result of that is NEVER
greater than c.
"So," you must ask, "when I see (c+v), is that not a combination of
velocities?" The answer is NO. That is an algebraic sum involving two
velocities, but it is NOT the combination of velocities.
> and directly afterward, that c+/-v are not a relativist sum of
> velocities(what you call
> a combinaison of velocities).
No, those are NOT combinations of velocities.
>
> Then what represent c+v or c-v in the relativistic Doppler formula, if
> not a "combination of velocities"?
They are simply an algebraic sum that appears in some mathematical
expressions. c+v does NOT represent a physical velocity of ANYTHING as
measured in any reference frame.
What you do is a common freshman physics mistake. You see a term in a
mathematical expression, and you believe that this term must have some
measurable physical significance. That is NOT the case.
c has measurable physical significance. v has measurable physical
significance. f has measurable physical significance. (c+v) and (c-v)
do NOT.
PD
No, sir, it is not mere opinion.
BURT
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
I don't deal in opinion.
Mitch Raemsch