Velocity addition equation
Vilas Tamhane
equal quantity of fuel q. Accelerometer in my ship shows that I have
attained speed of 0.1c. I am now in the frame 2, cruising with uniform
speed of 0.1c. I cannot measure this velocity, but I have recorded the
same.
I again burn same quantity of fuel. Accelerometer again shows speed of
0.1c, which I add to my record and assume that in new frame 3,
velocity of my ship, is 0.2c. But according to SR it is lesser than
0.2c.
However, so far as spaceship is concerned, there is no difference in
frame 2 and frame 1. All physical laws in frame 2 are same as those in
frame 1. Even SR has to admit this. Indeed, my accelerometer, however
calibrated shows speed attained as 0.1c. It will always show this gain
in speed, whenever I fire the rockets, irrespective of the uniform
speed at which I start. This is a fundamental law, more fundamental
that SR. So why should I refer to SR for my velocity calculations
rather than relying on real meter in my ship?
Dono.
Because you are an imbecile who doesn't understand SR, yet who thinks
that he's found errors in SR? I'll give you a hint, you are mixing up
proper and coordinate speed. I don't expect you to EVER figure out
what your mistake is. But that's ok, it means that you'll forever be
the same pathetic imbecile.
shuba
> Indeed, my accelerometer, however
> calibrated shows speed attained as 0.1c.
Accelerometers don't measure speed.
---Tim Shuba---
Vilas Tamhane
Of course they don't, but they act on force due to acceleration and
noting time, velcoity can be estimated.
shuba
> On May 6, 9:23 am, shuba <tim.sh...@lycos.ScPoAmM> wrote:
>> Accelerometers don't measure speed.
> Of course they don't, but they act on force due to acceleration and
> noting time, velcoity can be estimated.
Velocity *in which frame*? Show your work.
If you don't know how to evaluate tanh(A+B) given A & B and
why it is relevant to the calculation, I suggest you learn.
---Tim Shuba---
Dono.
There is no chance that this antirelativistic imbecile will ever
figure out hyperbolic motion. It is anmazing that he can hold a job
(at Siemens, no less)
Androcles
"Vilas Tamhane" <vilast...@gmail.com> wrote in message
news:50c7f54c-bd53-40ff...@s41g2000prb.googlegroups.com...
aircraft.
If there were ever a WW III and the other side knocked out the GPS
constellation, the military would again be using IN.
Shuba is a babbling cretin trying to pretend he's smarter than he is.
"Let there be given a stationary rigid rod; and let its length be L as
measured by a measuring-rod which is also stationary. We now imagine
the axis of the rod lying along the axis of x of the stationary system of
co-ordinates, and that a uniform motion of parallel translation with
velocity v along the axis of x in the direction of increasing x is then
imparted to the rod. We now inquire as to the length of the moving rod" --
Einstein
"The length to be discovered by the operation (b) we will call ``the length
of the (moving) rod in the stationary system.''"-- Einstein
"This we shall determine on the basis of our two principles, and we shall
find that it differs from L." -- Einstein.
AND THE ANSWER IS...
"xi = (x-vt)/sqrt(1 - v^2/c^2)" -- Einstein.
Yep, xi differs from L, Greek letters differ from Roman letters.
In agreement with experience we further assume the deranged babbling
incompetent cretin couldn't answer his own inquiry, he was too stupid
to realise xi is greater than L when he wrote 'for v=c all moving
objects--viewed from the "stationary'' system--shrivel up into plane
figures', whereas his own equation shows they stretch to infinity...
sqrt(1-c^2/c^2) = 0.
"But the ray moves relatively to the initial point of k, when measured in
the stationary system, with the velocity c-v" - Einstein
"the velocity of light in our theory plays the part, physically, of an
infinitely great velocity" - Einstein.
"In agreement with experience we further assume the quantity
2AB/(t'A -tA) = c to be a universal constant--the velocity of light in
empty space." -- Einstein
He was right. The distance from A to A divided by the time it takes
to get there is undefined. Anyone that divides by zero is a lunatic.
shuba
> There is no chance that this antirelativistic imbecile will ever
> figure out hyperbolic motion.
You may be right, but it doesn't hurt to plant the seed of sanity. We
can ignore calculus and not do full-blown hyperbolic motion, and just
consider the obvious connection between tanh(A+B) and the colinear
velocity addition formula. This is good enough to get the idea, and
those who follow the analogies between circular and hyperbolic
trigonometry will understand that they are equally consistent. Then,
it's a simple matter of noting which one is supported by observation.
---Tim Shuba---
rotchm
> In the space, I start from some point 1 in a spaceship.
Ok. We will call this "point" inertial frame 1 (F1), a frame that has
been coordinated ( a lattice of rulers and clocks).
> I always burn
> equal quantity of fuel q. Accelerometer in my ship shows that I have
> attained speed of 0.1c.
Ok. Your gauge indicates that you are traveling at speed 0.1c WRT F1.
> I am now in the frame 2, cruising with uniform
> speed of 0.1c. I cannot measure this velocity, but I have recorded the
> same.
Ok. But you can measure this 0.1c wrt F1 if you have the instruments
to do so.
Anw, F2 has speed 0.1c wrt F1.
> I again burn same quantity of fuel. Accelerometer again shows speed of
> 0.1c,
Ok. 0.1c WRT F2.
>which I add to my record and assume that in new frame 3,
> velocity of my ship, is 0.2c.
Dont assume. Predict using the chosen model ( SR, Newt./Gall.,
otehr...). Or *measure* it;
Measure your speed (F3) WRT F1. This has been done and the
experimental result is that it is less than 2* 0.1c and has the value
predicted by SR !
> But according to SR it is lesser than 0.2c.
Correct, AND also supported by actual experiments.
> However, so far as spaceship is concerned, there is no difference in
> frame 2 and frame 1. All physical laws in frame 2 are same as those in
> frame 1. Even SR has to admit this.
Correct.
F1 + 0.1c = F2
F2 + 0.1c = F3
These two 'equations' are symbolically identical.
These do not imply F1 + "0.2c" = F3.
> Indeed, my accelerometer, however
> calibrated shows speed attained as 0.1c.
O.1c WRT F1
0.1c Wrt F2
etc
> It will always show this gain
> in speed, whenever I fire the rockets, irrespective of the uniform
> speed at which I start.
Correct. Btw, you always start with speed 0 wrt yourself:
0 + 0.1c = F2 ( F1 wrt F2)
0 + 0.1c = F3 ( F2 wrt F3)
etc
> This is a fundamental law, more fundamental
> that SR. So why should I refer to SR for my velocity calculations
> rather than relying on real meter in my ship?
Dont mix up speed with closing speed, coordinate acceleration with
proper acceleration etc.
PD
The first problem is declaring that your velocity is 0 to start with.
Your velocity to start with is zero in one and only one reference
frame, we'll call this frame A. In all other reference frames B, C, D,
E, F, ...., your velocity is NOT zero.
After your first burn, your velocity in frame A might be 0.1 c. In
frame B (where you were going at -0.1 c before the burn), you are now
at rest with velocity 0.
After your second burn, your velocity in frame B is now 0.1 c, and
your velocity in frame A is now (surprise!) not 0.2 c but 0.198 c.
Your velocity in frame C is now 0.
You assume that the addition rule for velocities (0.1 + 0.1 = 0.2) is
"more fundamental" than special relativity. That is simply not true.
Sue...
> equal quantity of fuel q. Accelerometer in my ship shows that I have
> attained speed of 0.1c. I am now in the frame 2, cruising with uniform
> speed of 0.1c. I cannot measure this velocity, but I have recorded the
> same.
> I again burn same quantity of fuel. Accelerometer again shows speed of
> 0.1c, which I add to my record and assume that in new frame 3,
> velocity of my ship, is 0.2c. But according to SR it is lesser than
Ahh... No.. That is not according to SR.
<< Einstein's relativity principle states that:
All inertial frames are totally equivalent
for the performance of all physical experiments.
In other words, it is impossible to perform a physical
experiment which differentiates in any fundamental sense
between different inertial frames. By definition, Newton's
laws of motion take the same form in all inertial frames.
Einstein generalized[1] this result in his special theory of
relativity by asserting that all laws of physics take the
same form in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
[1]<< the four-dimensional space-time continuum of the
theory of relativity, in its most essential formal
properties, shows a pronounced relationship to the
three-dimensional continuum of Euclidean geometrical space.
In order to give due prominence to this relationship,
however, we must replace the usual time co-ordinate t by
an imaginary magnitude
sqrt(-1)
ct proportional to it. Under these conditions, the
natural laws satisfying the demands of the (special)
theory of relativity assume mathematical forms, in which
the time co-ordinate plays exactly the same rôle as
the three space co-ordinates. >>
http://www.bartleby.com/173/17.html
<< where epsilon_0 and mu_0 are physical constants which
can be evaluated by performing two simple experiments
which involve measuring the force of attraction between
two fixed charges and two fixed parallel current carrying
wires. According to the relativity principle, these experiments
must yield the same values for epsilon_0 and mu_0 in all
inertial frames. Thus, the speed of light must be the
same in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
> 0.2c.
> However, so far as spaceship is concerned, there is no difference in
> frame 2 and frame 1. All physical laws in frame 2 are same as those in
> frame 1. Even SR has to admit this. Indeed, my accelerometer, however
> calibrated shows speed attained as 0.1c. It will always show this gain
> in speed, whenever I fire the rockets, irrespective of the uniform
> speed at which I start. This is a fundamental law, more fundamental
> that SR. So why should I refer to SR for my velocity calculations
> rather than relying on real meter in my ship?
<<the famous Lorentz transformation. [ It ] ensures
that the velocity of
[particle] light is invariant between different
inertial frames, and also reduces to the more familiar
Galilean transform in the limit $v \ll c$. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node109.html
So unless you have a bottle of light particles
hidden away and they move under the influence of
the inertia, your application is not to reals
but rather to imaginaries.
For a modern derivation that doesn't involve
pseudo-particles, see:
~light clock~
http://meshula.net/wordpress/?p=222
Sue...
Vilas Tamhane
> On May 6, 12:04 pm, Vilas Tamhane <vilastamh...@gmail.com> wrote:
>
> Ahh... No.. That is not according to SR.
>
> << Einstein's relativity principle states that:
>
> All inertial frames are totally equivalent
> for the performance of all physical experiments.
>
same I have used in my arguments
>
> Sue...
Please don't reproduce theory here. Stick to my arguments and tell me
if there is anything wrong.
Vilas Tamhane
> On May 6, 11:04 am, Vilas Tamhane <vilastamh...@gmail.com> wrote:
>
>
>
>
>
> The first problem is declaring that your velocity is 0 to start with.
> Your velocity to start with is zero in one and only one reference
> frame, we'll call this frame A. In all other reference frames B, C, D,
> E, F, ...., your velocity is NOT zero.
simply relying on instrument and not theory. In reference frames B,
C...there is no law which can tell me that my velcoity is other than
zero, for this reason I am using the record.
> After your first burn, your velocity in frame A might be 0.1 c. In
> frame B (where you were going at -0.1 c before the burn), you are now
> at rest with velocity 0.
> After your second burn, your velocity in frame B is now 0.1 c, and
> your velocity in frame A is now (surprise!) not 0.2 c but 0.198 c.
> Your velocity in frame C is now 0.
correct.
> You assume that the addition rule for velocities (0.1 + 0.1 = 0.2) is
I am assuming only one law and that is 'All uniform velcocities are
equivalent'.
PD
> On May 6, 12:52 pm, PD <thedraperfam...@gmail.com> wrote:> On May 6, 11:04 am, Vilas Tamhane <vilastamh...@gmail.com> wrote:
>
> > The first problem is declaring that your velocity is 0 to start with.
> > Your velocity to start with is zero in one and only one reference
> > frame, we'll call this frame A. In all other reference frames B, C, D,
> > E, F, ...., your velocity is NOT zero.
>
> I am not using any reference frame for velocity calculations. I am
> simply relying on instrument and not theory. In reference frames B,
> C...there is no law which can tell me that my velcoity is other than
> zero, for this reason I am using the record.
Record but from what starting state? As I said, you don't know what
your velocity is initially.
Secondly, you assume that you can just ADD the results of the
increment each time. Not so.
> After your first burn, your velocity in frame A might be 0.1 c. In
> > frame B (where you were going at -0.1 c before the burn), you are now
> > at rest with velocity 0.
>
> True. From my point of view.
> After your second burn, your velocity in frame B is now 0.1 c, and
> > your velocity in frame A is now (surprise!) not 0.2 c but 0.198 c.
> > Your velocity in frame C is now 0.
>
> Why are you resorting to velocity addition theorem? It need not be
> correct.
No, this is actually confirmed in *measurement*. That's the point.
>
> > You assume that the addition rule for velocities (0.1 + 0.1 = 0.2) is
> > "more fundamental" than special relativity. That is simply not true.- Hide quoted text -
>
> I am assuming only one law and that is 'All uniform velcocities are
> equivalent'.
And the sum rule. Which is incorrect.
Vilas Tamhane
> On May 6, 12:04 pm, Vilas Tamhane <vilastamh...@gmail.com> wrote:
> > I am now in the frame 2, cruising with uniform
> > speed of 0.1c. I cannot measure this velocity, but I have recorded the
> > same.
>
> Ok. But you can measure this 0.1c wrt F1 if you have the instruments
> to do so.
> Anw, F2 has speed 0.1c wrt F1.
>
measure uniform velocity.
> > I again burn same quantity of fuel. Accelerometer again shows speed of
> > 0.1c,
>
> Ok. 0.1c WRT F2.
> >which I add to my record and assume that in new frame 3,
> > velocity of my ship, is 0.2c.
>
> Dont assume. Predict using the chosen model ( SR, Newt./Gall.,
> otehr...). Or *measure* it;
> Measure your speed (F3) WRT F1. This has been done and the
> experimental result is that it is less than 2* 0.1c and has the value
> predicted by SR !
There is no need for me to choose a model. I am actually making
measurements and recording these. You should point out the mistake in
the procedure. On the contrary, you are assuming VAT of SR.
> > But according to SR it is lesser than 0.2c.
>
> Correct, AND also supported by actual experiments.
These experiments are not valid as these were conducted on charged
particles.
> Correct. Btw, you always start with speed 0 wrt yourself:
> 0 + 0.1c = F2 ( F1 wrt F2)
> 0 + 0.1c = F3 ( F2 wrt F3)
> etc
This is what I stated. Problem is with addition. There is no reason
why I should use SR.
> > This is a fundamental law, more fundamental
> > that SR. So why should I refer to SR for my velocity calculations
> > rather than relying on real meter in my ship?
>
> Dont mix up speed with closing speed, coordinate acceleration with
> proper acceleration etc.
Problem is that you are treating VAT as God given. I am not.
Vilas Tamhane
> On May 6, 8:51 pm, Vilas Tamhane <vilastamh...@gmail.com> wrote:
> > Why are you resorting to velocity addition theorem? It need not be
> > correct.
>
> No, this is actually confirmed in *measurement*. That's the point.
Yes and a good point. Can you point out any experiment that is
conducted on uncharged particles?
Alen
In spite of the orthodox ridicule you are getting,
you are actually correct. The SR velocity addition
formula only works for moving frame velocities
measured from a stationary frame origin by means
of light pulses. In a spacelike sense, however, this
does not apply. If an accelerating force is contained
within a moving frame, simple velocity addition is
possible, even though the stationary frame light pulse
method of velocity measurement will not show this result.
This means that superluminal quasar jets are not
really forbidden by SR. They have tried to save the
orthodox version of SR by explaining these by means
of optical effects, but this doesn't always work.
Alen
rotchm
> measure uniform velocity.
Recording the velocity as in measuring it. You can measure a uniform
velo. in the sense that you repeatedly measure the velo. and if you
always get the same value then its safe to assume its a constant velo.
Anw, this is besides our point
> > Ok. 0.1c WRT F2.
>
> Come out of Relativity.
This has nothing to do with SR. YOU made the statement that the speed
was 0.1c. I just cleared up that you meant 0.1c wrt F2 (and F1 in the
first phase)
> There is no need for me to choose a model.
Correct.
> I am actually making measurements and recording these.
Correct
> You should point out the mistake in the procedure.
What mistake? The measurement procedures are (operational)
definitions. There are no mistakes in definitions.
>On the contrary, you are assuming VAT of SR.
VAT? And at this point in the discussion I have not assumed SR nor
any other model.
> These experiments are not valid as these were conducted on charged
> particles.
Not the exps. I am referring to. Ans so what if they were performed on
charged particles. Your speed measurements are independent of charge.
SR's composition of speed is independent of charge.
> This is what I stated. Problem is with addition. There is no
> reason why I should use SR.
Correct. A priori there is no reason to use SR's algebra nor Galilean
Algebra.
> Problem is that you are treating VAT as God given. I am not.
VAT? And since I am presenting nothing as God given, then I am not
presenting VAT as God given.
rotchm
> I am not using any reference frame for velocity calculations. I am
> simply relying on instrument and not theory.
If you are using instruments to measure speed then you are using
reference frames. Speed measurements require the coordination of a
reference frame. Speed is "x/t", IOW x and t have been "setup", have
been operationally defined.
>In reference frames B, C...there is no law which can tell me that >my velcoity is other than zero,
Speed is frame dependent. When you speak of "speed", you must specify
it wrt a chosen frame. Here, you mean that your speed is zero wrt
yourself (your frame)
> Why are you resorting to velocity addition theorem? It need not be
> correct.
Indeed. But empiric verification supports SR's velo. addition.
> I am assuming only one law and that is 'All uniform velcocities
> are equivalent'.
Ok. And what does this law imply concerning velo. composition?
SR's "laws" imply the Lor.Composition of speed and agrees with
experiments...A very good reason to continue to use it, no?
Jim Greenfield
>
Whoah there partner!
When I pointed out this very fact (in "The astute observer), ie that
there are many FoRs,
and we are never sure which one we are in, as regards to the total
universe, youhowled me down.
I now appreciate your defacto acceptance of the premise that there is
ONE FoR, to which all human
imaginings are subject
Question:
Is this correct? (-0.1)+(+0.1)=0
Jim G
c'=c+v
Vilas Tamhane
>And so what if they were performed on
>charged particles. Your speed measurements are independent of charge.
>SR's composition of speed is independent of charge.
> SR's "laws" imply the Lor.Composition of speed and agrees with
> experiments...A very good reason to continue to use it, no?
Experiments on charged particles are invalid as charged particles
radiate energy during accelerations. No wonder, they behave as if
their mass is increased with velocity and of course they cannot exceed
velocity of light as at this velocity they radiate all the energy they
receive.
I am making a simple argument. In frame A, spaceship has 0 velocity.
In frame B it has 0.1c. In frame C it has 0.1c w.r.t. frame B. So in
frame C it has 0.2c velocity w.r.t. frame A.
According to you, using VAT, my procedure of recording is wrong. Using
VAT, it should be less, say 0.198c. In order to arrive at this
velocity, I must change my procedure of recording. Therefore while
moving from frame B to C, I should add 0.098c to the previous value of
0.1c. In doing so I am violating fundamental principle of equivalence
of frames. If I add velocity 0.098c then I must assume that frame B is
not equivalent to frame A.
Vilas Tamhane
This is correct. A particle moving with velocity -v in a frame O'
moving with a velocity +v with respect to frame O has zero velocity
with respect to frame O. Can you elaborate for me?
> Jim G
> c'=c+v- Hide quoted text -
>
rotchm
> radiate energy during accelerations. No wonder, they behave as if
> their mass is increased with velocity and of course they cannot exceed
> velocity of light as at this velocity they radiate all the energy they
> receive.
Irrelevant to the discussion of speed comparisons.
> I am making a simple argument. In frame A, spaceship has 0 velocity.
Ok
> In frame B it has 0.1c.
Ok.
> In frame C it has 0.1c w.r.t. frame B.
This sentence is unclear. What does "it" refer to?
Do you mean that In frame C, frame B has speed 0.1c
> So in frame C it has 0.2c velocity w.r.t. frame A.
Since the sentence above is unclear, this sentence too becomes
ambiguous.
Also, why have you 'added' the velocities? Who told you that Galilean
composition of speeds (or the algebra of the real numbers) are the
valid models to use?
> According to you, using VAT, my procedure of recording is wrong.
What is VAT?
Position, time and speeds have very specific definitions given by the
bureau standards.
If you procedure does not satisfy these definitions then you are
talking about something else.
> Using VAT, it should be less, say 0.198c. In order to arrive at this
> velocity, I must change my procedure of recording.
If you use the definition of speed (an op. definition) then the
predicted result of "0.198c" agrees with experiment, agrees with the
operational verification of this speed.
> Therefore while
> moving from frame B to C, I should add 0.098c to the previous value of
> 0.1c.
Why should you?
SR says that *if* you measure 0.1c and 0.1c you will get 0.198c. This
is confirmed by experiment.
> In doing so I am violating fundamental principle of equivalence
> of frames.
? This needs to be better explained.
SR says that *if* A --> B --> C = 0.198 then
D-->E-->F = 0.198
They are 'equivalent'.
Tom Roberts
After each burn you only know your speed relative to the inertial frame from
which the burn started. You have no way whatsoever to relate your current speed
to any earlier frame. That is, except for the first burn, the "real meter" in
your ship is COMPLETELY UNABLE to display your velocity relative to your initial
frame.
Because your "real meter" essentially integrates acceleration,
and you IMPLICITLY assumed a relationship between the integral
of acceleration and velocity. Your second and third sentences
above calibrate your meter FOR ONE BURN, but do not do so for
multiple burns.
It is relevant to ask: what formula for the composition of velocities will be
self-consistent in this case, so for a given pair of initial and final frames,
it can be applied to any intermediate frame and always yield the same result?
This is equivalent to asking what is the relationship between the
integral of acceleration and velocity.
Analysis based on group theory shows that there are three formulas that yield
self-consistent results:
[Here c is a constant, v3 is the velocity of final frame wrt
initial frame, v1 is velocity of intermediate frame wrt initial
frame, and v2 is velocity of final frame wrt intermediate frame;
all frames are moving in the same direction. wrt = "with respect
to"]
a) the Galilean formula: v3 = v1 + v2
b) the Lorentzian formula: v3 = (v1 + v2) / (1 + v1*v2/c^2)
c) the Euclidean formula: v3 = (v1 + v2) / (1 - v1*v2/c^2)
The question of which is valid in the world we inhabit is an EXPERIMENTAL
question. Numerous experiments have shown that for velocities far below the
speed of light any of these will give reasonably accurate results, but for
velocities approaching the speed of light only Lorentz's formula is correct,
with c being the vacuum speed of light.
IOW: your implicit assumption that the Galilean formula holds is WRONG for the
world we inhabit.
Tom Roberts
Sue...
Your argument gives false references so that is
impossible.
Perhaps LET the the true object of your criticism?
http://en.wikipedia.org/wiki/Lorentz_ether_theory#Later_activity_and_Current_Status
Sue...
Vilas Tamhane
> SR says that *if* A --> B --> C = 0.198 then
> D-->E-->F = 0.198
>
> They are 'equivalent'.
That is what I said. Now go on adding velocites 0.198c.
Vilas Tamhane
>
> - Show quoted text -
Just let me know logical mistake, if any, in my assumption. I am not
interested in knowing whether SRT velocity addition equation is
correct or Galilean.
SRT equation cannot be directly verified. There is no experiment that
is conducted on uncharged microscopic particles or/and macroscopic
bodies. SRT velocity addition equation is not falsifiable.
Vilas Tamhane
> On May 6, 9:40 pm, Vilas Tamhane <vilastamh...@gmail.com> wrote:
>
>
>
>
>
> > On May 6, 1:09 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:> On May 6, 12:04 pm, Vilas Tamhane <vilastamh...@gmail.com> wrote:
>
> > > Ahh... No.. That is not according to SR.
>
> > It is accordidng to SR after application of velocity addition equation> << Einstein's relativity principle states that:
>
> > > All inertial frames are totally equivalent
> > > for the performance of all physical experiments.
>
> > This is not Einstein's principle. Many people stated it before. The
> > same I have used in my arguments
>
> > > Sue...
>
> > Please don't reproduce theory here. Stick to my arguments and tell me
> > if there is anything wrong.
>
> Your argument gives false references so that is
> impossible.
>
What false references? What is impossible?
All inertial frames, whatever might be their uniform velocity are
equivalent. This is the only principle I refered to. Is it false?
> Perhaps LET the the true object of your criticism?
>
> http://en.wikipedia.org/wiki/Lorentz_ether_theory#Later_activity_and_...
>
> Sue...- Hide quoted text -
Dono.
>
> SRT equation cannot be directly verified. There is no experiment that
> is conducted on uncharged microscopic particles or/and macroscopic
> bodies. SRT velocity addition equation is not falsifiable.
You are an ignorant imbecile. The Fizeau experiment 9and its modern
reenactments) do exactly that.
rotchm
You have not answered the clarifying questions I posed to you.
Do not coward from them and try to continue on a decent conversation.
Dont expect us to answer your questions if you cant give back the same
respect.
Once more, answer the questions I posed to you so that we my further
understand
your topic.
Vilas Tamhane
>
> > radiate energy during accelerations. No wonder, they behave as if
> > their mass is increased with velocity and of course they cannot exceed
> > velocity of light as at this velocity they radiate all the energy they
> > receive.
>
> Irrelevant to the discussion of speed comparisons.
I think it is not only important but relevant.
> > I am making a simple argument. In frame A, spaceship has 0 velocity.
>
> Ok
>
> > In frame B it has 0.1c.
>
> Ok.
>
> > In frame C it has 0.1c w.r.t. frame B.
>
> This sentence is unclear. What does "it" refer to?
> Do you mean that In frame C, frame B has speed 0.1c
In frame C spaceship has speed of 0.1c w.r.t. frame B.
>
> > So in frame C it has 0.2c velocity w.r.t. frame A.
>
> Since the sentence above is unclear, this sentence too becomes
> ambiguous.
It means spaceship.
> Also, why have you 'added' the velocities? Who told you that Galilean
> composition of speeds (or the algebra of the real numbers) are the
> valid models to use?
I am not sticking to any model. Reason why I made Galilean addition is
explicit in the original text.
>
> > According to you, using VAT, my procedure of recording is wrong.
>
> What is VAT?
Velocity addition theorem. May be this abbreviation is not in vogue.
> Position, time and speeds have very specific definitions given by the
> bureau standards.
> If you procedure does not satisfy these definitions then you are
> talking about something else.
I am not deviation from these definitions.
> > Using VAT, it should be less, say 0.198c. In order to arrive at this
> > velocity, I must change my procedure of recording.
>
> If you use the definition of speed (an op. definition) then the
> predicted result of "0.198c" agrees with experiment, agrees with the
> operational verification of this speed.
Do you want to say that there is nothing illogical in my original
statement but I should use SRT equation because this equation is
applicable to actual experiment?
>
> > Therefore while
> > moving from frame B to C, I should add 0.098c to the previous value of
> > 0.1c.
>
> Why should you?
>
> SR says that *if* you measure 0.1c and 0.1c you will get 0.198c. This
> is confirmed by experiment.
As above.
>
> > In doing so I am violating fundamental principle of equivalence
> > of frames.
>
> ? This needs to be better explained.
>
> SR says that *if* A --> B --> C = 0.198 then
> D-->E-->F = 0.198
>
> They are 'equivalent'.
So A-->F= 0.198+0.198
Vilas Tamhane
> > In doing so I am violating fundamental principle of equivalence
> > of frames.
>
> ? This needs to be better explained.
The principle that all inertial frames are equivalent is a fundamental
principle. It is also accepted that I cannot measure my own uniform
speed. During each acceleration, meter in my spaceship will show
velocity installment of 0.1c. When my spaceship reaches frame B, I
record this velocity and assume that the spaceship is traveling at a
speed of 0.1c w.r.t. frame A. With the second boost, my meter will
show that the spaceship has gained velocity of 0.1c. If I must use SRT
equation then my velocity will be say 0.198c with respect to the frame
A. In this case I must make correction to the reading of the meter and
instead of recording 0.1c+0.1c, I must record 0.1c+0.098c. Thus I must
reject reading of the meter. In short I must treat frame B as not
equivalent to frame A.
John Park
>
> Just let me know logical mistake, if any, in my assumption. I am not
> interested in knowing whether SRT velocity addition equation is
> correct or Galilean.
> SRT equation cannot be directly verified. There is no experiment that
> is conducted on uncharged microscopic particles or/and macroscopic
> bodies. SRT velocity addition equation is not falsifiable.
Someone else can provide detailed references, but the predictions of SR
are consistent with the behaviour of electrons, muons and protons in
accelerators. They are charged--all three particles have numerically equal
charges--but they range in mass by a factor of almost 2000, and there is
no evidence of any deviation from expected behaviour over that range of
charge to mass; so you have no prima facie evidence aginst SR.
--John Park
Vilas Tamhane
Charged particles radiate energy during acceleration. They radiate all
the energy they gain at the speed of c and so they cannot exceed speed
of light. In any case, energy loss during radiation makes them
inappropriate candidates for testing SR theories. That is why I said
SR theory is not falsifiable.
John Park
> On May 8, 9:06=A0am, af...@FreeNet.Carleton.CA (John Park) wrote:
>> Vilas Tamhane (vilastamh...@gmail.com) writes:
>>
>> > Just let me know logical mistake, if any, in my assumption. I am not
>> > interested in knowing whether SRT velocity addition equation is
>> > correct or Galilean.
>> > SRT equation cannot be directly verified. There is no experiment that
>> > is conducted on uncharged microscopic particles or/and macroscopic
>> > bodies. SRT velocity addition equation is not falsifiable.
>>
>> Someone else can provide detailed references, but the predictions of SR
>> are consistent with the behaviour of electrons, muons and protons in
> l
>> charges--but they range in mass by a factor of almost 2000, and there is
>> no evidence of any deviation from expected behaviour over that range of
>> charge to mass; so you have no prima facie evidence aginst SR.
>>
>
> Charged particles radiate energy during acceleration. They radiate all
> the energy they gain at the speed of c and so they cannot exceed speed
> of light. In any case, energy loss during radiation makes them
> inappropriate candidates for testing SR theories. That is why I said
> SR theory is not falsifiable.
I know what you said. It's nonsensical and there is no evidence for it.
--John Park
Dono.
>
> Charged particles radiate energy during acceleration. They radiate all
> the energy they gain at the speed of c and so they cannot exceed speed
> of light. In any case, energy loss during radiation makes them
> inappropriate candidates for testing SR theories. That is why I said
> SR theory is not falsifiable.
You ARE an imbecile,SR is the theory that explains radiation during
acceleration. google brehmstallung.
Besides, carged particles in circular motion do NOT radiate, google
Poynting vector.
How does such an idiot survive day to day life? Even nematodes have
better brains than you.
PD
Sure. It's conducted on neutrino production experiments and in direct
photon production experiments. Try E702 at Fermilab.
PD
> On May 7, 7:55 am, rotchm <rot...@gmail.com> wrote:
>
> > > In doing so I am violating fundamental principle of equivalence
> > > of frames.
>
> > ? This needs to be better explained.
>
> The principle that all inertial frames are equivalent is a fundamental
> principle. It is also accepted that I cannot measure my own uniform
> speed. During each acceleration, meter in my spaceship will show
> velocity installment of 0.1c. When my spaceship reaches frame B, I
> record this velocity and assume that the spaceship is traveling at a
> speed of 0.1c w.r.t. frame A. With the second boost, my meter will
> show that the spaceship has gained velocity of 0.1c. If I must use SRT
> equation then my velocity will be say 0.198c with respect to the frame
> A.
Why rely on yourself? Why not have someone with measuring instruments
in frame A actually *measure* your speed and tell you what it is,
rather than you guessing which set of assumptions you should apply to
*derive* your own speed?
This is the whole point of *experimental verification* of the velocity
addition formula, in that it doesn't require you to do any guessing.
You just *measure* to see what the result is.
PD
Which is precisely the point I made to Vilas in my first response.
Read it.
> I now appreciate your defacto acceptance of the premise that there is
> ONE FoR, to which all human
> imaginings are subject
>
> Question:
> Is this correct? (-0.1)+(+0.1)=0
Not with regard to velocities.
You can't say, "OK, I've got a velocity of 0.3 c and another velocity
of 0.3 c, and 0.3 + 0.3 = 0.6, so the result MUST be a velocity of 0.6
c."
>
> Jim G
> c'=c+v
PD
> On May 7, 2:04 am, Vilas Tamhane <vilastamh...@gmail.com> wrote:
>
>
>
> > In the space, I start from some point 1 in a spaceship. I always burn
> > equal quantity of fuel q. Accelerometer in my ship shows that I have
> > attained speed of 0.1c. I am now in the frame 2, cruising with uniform
> > speed of 0.1c. I cannot measure this velocity, but I have recorded the
> > same.
> > I again burn same quantity of fuel. Accelerometer again shows speed of
> > 0.1c, which I add to my record and assume that in new frame 3,
> > velocity of my ship, is 0.2c. But according to SR it is lesser than
> > 0.2c.
> > However, so far as spaceship is concerned, there is no difference in
> > frame 2 and frame 1. All physical laws in frame 2 are same as those in
> > frame 1. Even SR has to admit this. Indeed, my accelerometer, however
> > calibrated shows speed attained as 0.1c. It will always show this gain
> > in speed, whenever I fire the rockets, irrespective of the uniform
> > speed at which I start. This is a fundamental law, more fundamental
> > that SR. So why should I refer to SR for my velocity calculations
> > rather than relying on real meter in my ship?
>
> In spite of the orthodox ridicule you are getting,
> you are actually correct. The SR velocity addition
> formula only works for moving frame velocities
> measured from a stationary frame origin by means
> of light pulses.
Flatly untrue.
Inertial
news:50c7f54c-bd53-40ff...@s41g2000prb.googlegroups.com...
>
>In the space, I start from some point 1 in a spaceship. I always burn
>equal quantity of fuel q. Accelerometer in my ship shows that I have
>attained speed
Accelerometers do not show speed
> of 0.1c. I am now in the frame 2,
You are always in EVREY frame
> cruising with uniform speed of 0.1c.
Relative to what?
> I cannot measure this velocity,
You haven't defined what it means
> but I have recorded the
>same.
>I again burn same quantity of fuel. Accelerometer again shows speed of
> 0.1c,
Accelerometers do not show speed
> which I add to my record and assume that in new frame 3,
>velocity of my ship, is 0.2c. But according to SR it is lesser than
>0.2c.
>However, so far as spaceship is concerned, there is no difference in
>frame 2 and frame 1. All physical laws in frame 2 are same as those in
>frame 1. Even SR has to admit this. Indeed, my accelerometer, however
>calibrated shows speed attained as 0.1c. It will always show this gain
>in speed, whenever I fire the rockets, irrespective of the uniform
>speed at which I start. This is a fundamental law, more fundamental
>that SR. So why should I refer to SR for my velocity calculations
>rather than relying on real meter in my ship?
You are very confused about what frames mean here
Inertial
news:d48f0bba-8cba-405a...@22g2000prx.googlegroups.com...
On May 7, 2:04 am, Vilas Tamhane <vilastamh...@gmail.com> wrote:
> In the space, I start from some point 1 in a spaceship. I always burn
> equal quantity of fuel q. Accelerometer in my ship shows that I have
> attained speed of 0.1c. I am now in the frame 2, cruising with uniform
> speed of 0.1c. I cannot measure this velocity, but I have recorded the
> same.
> I again burn same quantity of fuel. Accelerometer again shows speed of
> 0.1c, which I add to my record and assume that in new frame 3,
> velocity of my ship, is 0.2c. But according to SR it is lesser than
> 0.2c.
> However, so far as spaceship is concerned, there is no difference in
> frame 2 and frame 1. All physical laws in frame 2 are same as those in
> frame 1. Even SR has to admit this. Indeed, my accelerometer, however
> calibrated shows speed attained as 0.1c. It will always show this gain
> in speed, whenever I fire the rockets, irrespective of the uniform
> speed at which I start. This is a fundamental law, more fundamental
> that SR. So why should I refer to SR for my velocity calculations
> rather than relying on real meter in my ship?
>In spite of the orthodox ridicule you are getting,
>you are actually correct. The SR velocity addition
>formula only works for moving frame velocities
>measured from a stationary frame origin
Yes .. by definition of what a velocity addition formula means
> by means
>of light pulses.
It doesn't matter how you measure them.
Inertial
news:a7f26760-e1c4-45df...@u26g2000vby.googlegroups.com...
>On May 7, 2:51 am, Jim Greenfield <jgreenfiel...@gmail.com> wrote:
>> Question:
>> Is this correct? (-0.1)+(+0.1)=0
Of course that is correct.
There are two problems though that you're glossing over.
1) Velocity addition is not the same as adding numbers. Unfortunately (in
English at least) we use the word 'addition' to mean two different things
when we talk about arithmetic addition of numbers and addition (composition)
of velocities. And to make it worse, especially due to the limitations of
ASCII characters, e use the same symbol for it. I often use (+) for
velocity addition, to avoid that confusion.
2) Even though the velocity addition and arithmetic addition happen to give
the same answer in this case (when the value is zero) does NOT mean it is
the same in all case.
So for numbers we have
-0.1 + 0.1 - 0
For velocities we have
-0,1 (+) 0.1 = 0
That does NOT imply that in general
u + v = u (+) v
> Not with regard to velocities.
You fell into his little trap, PD. It IS true for velocities .. for
velocities whose numeric sum is zero
>You can't say, "OK, I've got a velocity of 0.3 c and another velocity
>of 0.3 c, and 0.3 + 0.3 = 0.6, so the result MUST be a velocity of 0.6
>c."
That's correct .. but the example he showed (where the sum is zero) it IS
the same
Of course, as you say, implying anything for other values doesn't work.
That would be like saying 0+0 = 0, 0*0 = 0 .. so therefore we must have a+b
= a*b for all a and b
Tom Roberts
> Just let me know logical mistake, if any, in my assumption.
You are assuming that Galilean relativity holds, at least for the composition of
velocities.
Note this is COMPOSITIOON, not "addition". That is, you are
composing the speed of frame A relative to frame B with the
velocity of an object relative to frame A, to get its velocity
relative to B. Addition only applies when the two velocities
are relative to the same frame; then it is called a "closing
velocity" when they are directed oppositely (other cases have
no specific name, but are not very useful).
> I am not
> interested in knowing whether SRT velocity addition equation is
> correct or Galilean.
Then you are not interested in knowing whether your assumption is correct.
that's pretty stupid.
> SRT equation cannot be directly verified. There is no experiment that
> is conducted on uncharged microscopic particles or/and macroscopic
> bodies. SRT velocity addition equation is not falsifiable.
This is just plain false. Experiments such as Fizeau's are a direct
implementation of it:
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Other_experiments
> The principle that all inertial frames are equivalent is a fundamental
> principle. It is also accepted that I cannot measure my own uniform
> speed. During each acceleration, meter in my spaceship will show
> velocity installment of 0.1c. When my spaceship reaches frame B, I
> record this velocity and assume that the spaceship is traveling at a
> speed of 0.1c w.r.t. frame A.
So far, OK. You have integrated your acceleration from A to B to get 0.1 c wrt A.
> With the second boost, my meter will
> show that the spaceship has gained velocity of 0.1c.
Hmmm. Actually, it only shows that you now have a speed of 0.1 c wrt B (the
frame you were at rest in when the burn started). That is, for this meter "gain"
can only be measured from rest -- that is the only calibration you have.
> If I must use SRT
> equation then my velocity will be say 0.198c with respect to the frame
> A. In this case I must make correction to the reading of the meter and
> instead of recording 0.1c+0.1c, I must record 0.1c+0.098c.
NO! You do have a speed 0.1 c wrt B -- that is ALL your calibration is good for.
Your calibration told you how to integrate the acceleration of ONE burn, not two.
There is no "0.1c+0.098c". There is only "You have a speed 0.1 c wrt B", and YOU
HAVE NOT YET ESTABLISHED WHAT YOUR SPEED IS WRT A. Indeed, you do NOT have
sufficient information to do so.
In particular, your meter is only calibrated for ONE BURN. You need a NEW
CALIBRATION to handle two burns. And that is the rub -- you don't know how to do
that, and must turn to theory. As I said before, there are three possible
theories, but two of them are solidly refuted in the world we inhabit, and the
only one that remains valid is SR.
> Thus I must
> reject reading of the meter.
NO! Your meter is only valid for ONE BURN. Your meter correctly reads 0.1 c wrt
B. You meter is INCAPABLE of indicating speed wrt any frame except the frame you
were at rest in when the burn started.
What you must reject is your assumption of Galilean addition of velocities.
> In short I must treat frame B as not
> equivalent to frame A.
Again NO! Your meter treats all frames the same, and for each SINGLE burn it
always reads 0.1 c wrt the frame you were at rest in when the burn started. That
is ALL YOU KNOW. You need more information to relate that to your speed wrt
earlier frames; in the world we inhabit, the only valid way to do that is to use SR.
> Charged particles radiate energy during acceleration. They radiate all
> the energy they gain at the speed of c and so they cannot exceed speed
> of light.
That is OUTRAGEOUSLY naive, and tantamount to being wrong. Indeed in SR neutral
particles emit no radiation, and still cannot exceed speed c wrt any inertial
frame. This is confirmed experimentally for neutrons, photons, and neutrinos
(all are neutral subatomic particles), at speeds up to 0.99999 c or so.
> In any case, energy loss during radiation makes them
> inappropriate candidates for testing SR theories.
That, too, is OUTRAGEOUSLY naive. The way one tests a theory like SR is to
compute what one should measure in a given experiment, and compare to what is
actually measured. yes, charged particles radiate, but the theory [#] covers
that, and one simply includes radiation in the calculation.
[#] SR plus classical electrodynamics.
> That is why I said
> SR theory is not falsifiable.
SR is falsifiable in myriad ways, but has never been refuted within its domain.
Tom Roberts
Alen
Only in orthodox SR
Alen
Alen
Of course, I already know that answer, but then
I also know that it is a result of the fundamental
mistake in orthodox SR, in which SR really
has nothing to do with light itself.
Alen
Eric Gisse
Then perhaps you should let everyone know what you are arguing about
if not SR.
Eric Gisse
People have been telling you for years that SR is not a theory of
light, Alen.
W-w-wayback machine!
http://groups.google.com/group/sci.physics.relativity/msg/1d4ed80869fb80f1?dmode=source
Three years later, SR continues to not be a theory of light.
Alen
That doesn't mean it really isn't (in nature, that is)
> W-w-wayback machine!
>
> http://groups.google.com/group/sci.physics.relativity/msg/1d4ed80869f...
>
> Three years later, SR continues to not be a theory of light.
That's no big deal. I believe a lot of people initially
resisted quantum theory for years. Einstein apparently
never accepted that probability could be a
fundamental reality. I suspect he might have been
right btw!
Alen
Vilas Tamhane
> Vilas Tamhane wrote:
> > Just let me know logical mistake, if any, in my assumption.
>
> You are assuming that Galilean relativity holds, at least for the composition of
> velocities.
>
> Note this is COMPOSITIOON, not "addition". That is, you are
> composing the speed of frame A relative to frame B with the
> velocity of an object relative to frame A, to get its velocity
> relative to B. Addition only applies when the two velocities
> are relative to the same frame; then it is called a "closing
> velocity" when they are directed oppositely (other cases have
> no specific name, but are not very useful).
>
> > I am not
> > interested in knowing whether SRT velocity addition equation is
> > correct or Galilean.
>
> Then you are not interested in knowing whether your assumption is correct.
> that's pretty stupid.
>
> > SRT equation cannot be directly verified. There is no experiment that
> > is conducted on uncharged microscopic particles or/and macroscopic
> > bodies. SRT velocity addition equation is not falsifiable.
>
> This is just plain false. Experiments such as Fizeau's are a direct
No doubt, quite impressive! But I don’t agree with you. You cannot
have it both ways. You cannot say that the ship acquires speed of 0.1c
from frame B to C; but to satisfy equation (b), you must treat it as
0.0988c. Velocity addition theorem has to do everything with frames.
Only when you use Galilean addition, you treat all frames equivalent.
As soon as you add velocity of 0.0988c and not 0.1c, you are NOT
treating frame B same as frame A. In other words, SR method is true
only if you treat succeeding frames different from preceding frames.
Dono.
>
> No doubt, quite impressive! But I don’t agree with you. You cannot
> have it both ways. You cannot say that the ship acquires speed of 0.1c
> from frame B to C; but to satisfy equation (b), you must treat it as
> 0.0988c. Velocity addition theorem has to do everything with frames.
> Only when you use Galilean addition, you treat all frames equivalent.
> As soon as you add velocity of 0.0988c and not 0.1c, you are NOT
> treating frame B same as frame A. In other words, SR method is true
> only if you treat succeeding frames different from preceding frames.
You can relax, Vilas, your imbecility is not curable, so it will
always follow you, wherever you go. To your death.
PD
And you made a statement about the *SR* velocity addition formula. And
your statement is flatly untrue in SR.
>
> Alen
PD
He proposed the experiment that was performed after his death that
proved him wrong, Alen.
PD
Again you make a mistake in thinking that EVERYTHING must be the same
for all frames. This is simply not the case and not what the principle
of relativity says. It ONLY says that the laws of physics are the same
in all frames. What you do procedurally to analyze something is not a
law of physics.
Tom Roberts
> But I don’t agree with you.
Your problem, not mine. This is not subject to opinion.
> You cannot
> have it both ways.
I don't "have it both ways". Rather, you have it wrong because you implicitly
assume Galilean composition of velocities, and don't recognize you are assuming it.
> You cannot say that the ship acquires speed of 0.1c
> from frame B to C;
RIGHT! But not in the way you think. Speed is not any property that a ship "has"
or "acquires"; ditto for velocity. Rather, each is a RELATIONSHIP TO A SPECIFIC
FRAME.
So a ship never "acquires speed of 0.1c"; the ship can be accelerated to a speed
of 0.1 c wrt frame A, but that is a quite different statement. The ship can also
be accelerated to a speed of 0.1 c wrt frame B, but that is yet another
different statement.
> but to satisfy equation (b), you must treat it as
> 0.0988c.
This is plain not true. For each burn the ship is accelerated to speed 0.1 c
relative to the frame in which it was at rest when the burn started. That is ALL
YOU CAN SAY, because you have not determined how to compose velocities, which is
what is required to relate the speed of the ship wrt frame B to its speed wrt
frame A.
> Velocity addition theorem has to do everything with frames.
Not in the way you think. One can only apply 3-velocity addition when the
3-velocities are measured in A SINGLE FRAME. (You say "velocity" when you really
mean 3-velocity.)
4-velocity addition is something else, however, because 4-velocity
is independent of frame. But let's not go there....
> Only when you use Galilean addition, you treat all frames equivalent.
I repeat: you need to COMPOSE THE VELOCITIES, not add them.
In SR, all frames are equivalent, when one adds velocities referenced to a
single frame, or when one composes velocities measured in different frames to
relate velocities relative to the different frames.
> As soon as you add velocity of 0.0988c and not 0.1c, you are NOT
> treating frame B same as frame A.
BUT ONE NEVER DOES THAT!!! I repat: we do not ADD velocities for this case, we
COMPOSE velocities.
What one does is to compose the velocity 0.1 c of the ship wrt frame B with the
0.1 c of frame B wrt frame A, to obtain 0.1988 c for the speed of the ship wrt
frame A. THIS IS NOT ADDITION! THIS IS COMPOSITION!
Note the important point that the two velocities involved are
of DIFFERENT TYPES: one is the velocity of one frame relative
to another frame, and the other is the velocity of an object
relative to a frame. That difference is why addition does not
apply.
> In other words, SR method is true
> only if you treat succeeding frames different from preceding frames.
Again not true. You are confusing velocity addition with the composition of
velocities. They are different: Addition ONLY applies to a SINGLE FRAME, while
composition applies between DIFFERENT frames.
Tom Roberts
Jim Greenfield
> > equal quantity of fuel q. Accelerometer in my ship shows that I have
> > attained speed of 0.1c. I am now in the frame 2, cruising with uniform
> > speed of 0.1c. I cannot measure this velocity, but I have recorded the
> > same.
> > I again burn same quantity of fuel. Accelerometer again shows speed of
> > 0.1c, which I add to my record and assume that in new frame 3,
> > velocity of my ship, is 0.2c. But according to SR it is lesser than
> > 0.2c.
> > However, so far as spaceship is concerned, there is no difference in
> > frame 2 and frame 1. All physical laws in frame 2 are same as those in
> > frame 1. Even SR has to admit this. Indeed, my accelerometer, however
> > calibrated shows speed attained as 0.1c. It will always show this gain
> > in speed, whenever I fire the rockets, irrespective of the uniform
> > speed at which I start. This is a fundamental law, more fundamental
> > that SR. So why should I refer to SR for my velocity calculations
> > rather than relying on real meter in my ship?
>
> which the burn started. You have no way whatsoever to relate your current speed
> to any earlier frame. That is, except for the first burn, the "real meter" in
> your ship is COMPLETELY UNABLE to display your velocity relative to your initial
> frame.
>
> Because your "real meter" essentially integrates acceleration,
> and you IMPLICITLY assumed a relationship between the integral
> of acceleration and velocity. Your second and third sentences
> above calibrate your meter FOR ONE BURN, but do not do so for
> multiple burns.
>
> It is relevant to ask: what formula for the composition of velocities will be
> self-consistent in this case, so for a given pair of initial and final frames,
> it can be applied to any intermediate frame and always yield the same result?
>
> This is equivalent to asking what is the relationship between the
> integral of acceleration and velocity.
>
> Analysis based on group theory shows that there are three formulas that yield
> self-consistent results:
>
> [Here c is a constant, v3 is the velocity of final frame wrt
> initial frame, v1 is velocity of intermediate frame wrt initial
> frame, and v2 is velocity of final frame wrt intermediate frame;
> all frames are moving in the same direction. wrt = "with respect
> to"]
>
> a) the Galilean formula: v3 = v1 + v2
>
> b) the Lorentzian formula: v3 = (v1 + v2) / (1 + v1*v2/c^2)
>
> c) the Euclidean formula: v3 = (v1 + v2) / (1 - v1*v2/c^2)
>
> The question of which is valid in the world we inhabit is an EXPERIMENTAL
> question. Numerous experiments have shown that for velocities far below the
> speed of light any of these will give reasonably accurate results, but for
> velocities approaching the speed of light only Lorentz's formula is correct,
> with c being the vacuum speed of light.
>
> IOW: your implicit assumption that the Galilean formula holds is WRONG for the
> world we inhabit.
>
> Tom Roberts- Hide quoted text -
>
> - Show quoted text -
Passing strange that AE accolytes can use (c^2) willy-nilly, and claim
perfect validity.........
.........and THEN claim that 300,000kps+300,000kps+300,000kps
300,000 TIMES
which is the exact (though somewhat time-consuming)equivalent .......
gives a WRONG RESULT
If you wish to argue that multiplication is NOT addition, go right
ahead.
I will enjoy quoting you in future.
Jim G
c'=c+v
Androcles
"Jim Greenfield" <jgreen...@gmail.com> wrote in message
news:3cddf0e6-0301-4c58...@k27g2000pri.googlegroups.com...
Jim G
c'=c+v
================================================
Illiterate Humpty Roberts has more gaffes than the crew of the Pequod.
"In SR of course, the closing speed of light and star is c+v or c-v
(depending on
direction of star's motion). That is NOT the speed of light leaving the star
measured in the star's instantaneously comoving inertial frame, which is of
course c." -- Tom Roberts
news:F8qdnQhCgsV...@giganews.com
Eric Gisse
[...]
> If you wish to argue that multiplication is NOT addition, go right
> ahead.
> I will enjoy quoting you in future.
>
> Jim G
> c'=c+v
Yes, quote the correct answer while arguing it isn't. Not the first
time you've made a fool of yourself.
Jim Greenfield
3x3=9
3^ =9
3+3+3=9 ...ALWAYS they are equivalent AND the same
When you, or any other sandbox player thinks that if numbers rely on
other things for their validity, then DIFFERENT answers are correct,
you are in the comoving frame of brainwashed simpletons who applaud
the naked king's magical cloak
Jim G
c'=c+v
Jim Greenfield
wrote:
> "Jim Greenfield" <jgreenfiel...@gmail.com> wrote in message
Love it when self-contradiction goes straight over the head.
Pure entertainment when wriggle room runs out.
Jim G
c'=c+v
>
> "In SR of course, the closing speed of light and star is c+v or c-v
> (depending on
> direction of star's motion). That is NOT the speed of light leaving the star
> measured in the star's instantaneously comoving inertial frame, which is of
> course c." -- Tom Roberts
> news:F8qdnQhCgsV...@giganews.com- Hide quoted text -
Tom Roberts
> 3x3=9
> 3^2 =9
> 3+3+3=9 ...ALWAYS they are equivalent AND the same
Sure. But the problem here is NOT addition.
Tom Roberts
PD
Why would you add them like that? That's not how velocities combine.
PD
Oh, dear, Jim. Oh, good grief.
When you see c^2, you are seeing a *unit conversion*. You're not
combining c velocities of value c.
Man oh man.
Eric Gisse
Jim, why did you come back? You obviously do not accept science, so
why not go back to whatever you were doing previously?
Jim Greenfield
Yeh, I know; the problem is 'religion'
Or is it that a 'mile/hour' is no longer a 'mile/hour'
"Velocities don't add" is a total crock, manufactured to support AE
(and maybe Lorentz/Fitz ?)
Jim G
c'=c+v
Jim Greenfield
>
> - Show quoted text -
So what is 3^ (answer here............)
What is 300,000^ (answer here...................................)
A "unit" is 1 So what is "1" converted INTO?
There is psychobabble attributed to psychologists, but mayby a more
apt word around here would be
"physicsbabble"
Jim G
c'=c+v
PS:
My daughters boyfriend is just about to complete his Masters Degree in
math. He informs me that noone in the faculty has taken Einstein
seriously for years. AE has gone from a joke to an annoying
distraction.
Jim Greenfield
>
> - Show quoted text -
I would if the crap you spout was science.
All I read here is waffle and religion.
The 'Scientific Method' has been abandoned, and any non-believer in
AE, Islam,
JC et al is put to the Sword of Ignorance.
If it looks like shit, smells like shit, and every chemical analysis
shows it to BE SHIT,
then I for one, refuse to slurp it down every day, like you.
Jim G
c'=c+v
Dono.
Does it hurt? Being such a cretin, day in and day out?
Eric Gisse
As an example of slurping shit, let's visit your signature. Every
observation in the past century has disproved ballistic theories of
light, yet you cling on. Why is that?
Jim Greenfield
>
> - Show quoted text -
But every MEASUREMENT (of wave length) has proved c'=c+v
Why is that?
Hint: use STANDARDISED RULERS and CLOCKS
Jim G
c'=c+v
PD
It isn't a crock if it's verified in *experiment*, Jim.
One simply cannot say, "But addition MUST work by COMMON SENSE," if
the results do not agree with direct measurement.
And yes, velocities of the same thing in different reference frames
have been directly measured. And direct addition gets the answer
WRONG.
>
> Jim G
> c'=c+v
PD
No it hasn't, Jim.
Because measurements of the same light and taking DIRECT measurements
of both frequency and wavelength show that the product of those two
shifted quantities is c, not c'=c+v.
PD
Fine, I'll play. You have a box that is 3 ft wide by 3 ft long, and
the bottom of the box then has an area of 9 ft^2. Please tell me where
the 3 + 3 + 3 comes in?
You have a box that is 3 ft wide by 3 ft long and 3 ft tall, and the
volume of the box is then 27 ft^3. Please tell me where the 3 + 3 + 3
+ 3 + 3 + 3 + 3 + 3 + 3 comes in?
Eric Gisse
Since you claim 'every measurement' proves it, perhaps you could name
one that you have done yourself.
Simple Simon
> Fine, I'll play. You have a box that is 3 ft wide by 3 ft long, and
> the bottom of the box then has an area of 9 ft^2. Please tell me where
> the 3 + 3 + 3 comes in?
are 1 ft wide by 1 ft long. There will be 3 of them in each column. 3
in the first column + 3 in the second column + 3 in the third column
(3 rows): 3 + 3 + 3.
> You have a box that is 3 ft wide by 3 ft long and 3 ft tall, and the
> volume of the box is then 27 ft^3. Please tell me where the 3 + 3 + 3
> + 3 + 3 + 3 + 3 + 3 + 3 comes in?
>
Same as above, but use unit cubes.
Did I win?
PD
> On May 11, 3:23 am, PD <thedraperfam...@gmail.com> wrote:> Fine, I'll play. You have a box that is 3 ft wide by 3 ft long, and
> > the bottom of the box then has an area of 9 ft^2. Please tell me where
> > the 3 + 3 + 3 comes in?
>
> Ooh! Even I know this one. It can be visualized by using boxes that
> are 1 ft wide by 1 ft long.
Whoh there. Where did the 1 ft LONG come in when dealing with the 3 ft
WIDE side of the box? There is no LONG in the WIDE.
Jim Greenfield
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
So c^ is square kilometers/sec/sec ?????????
(hint: linear one dimensional)
Jim Greenfield
You would if you knew the difference between 3 squared and 3 cubed
Bye
Jim G
c'=c+v
(How "simple"?)
PD
c^2 is km^2/s^2, yes. Does this surprise you, Jim? Really?
Here's another one: kinetic energy has units kilograms * (square
meters) / (seconds squared).
Stunning, eh?
>
> (hint: linear one dimensional)
Eric Gisse
[...]
> My daughters boyfriend is just about to complete his Masters Degree in
> math. He informs me that noone in the faculty has taken Einstein
> seriously for years. AE has gone from a joke to an annoying
> distraction.
On one hand, I wonder if the boyfriend is merely placating the insane
father of his girlfriend.
On the other, Einstein has nothing to do with pure mathematics given
his theory uses tools that were previously developed by folks like
Riemann so I'm wondering what the fuck he'd be talking about.
Jim Greenfield
It was Lorentz/Fitz who put the pea under the walnut shell.
They kept the velocity of the train at v(+) at all times, but when
light was considered,
they use c(+) in the direction of the trains motion, but NOT c(-) in
the other.
Hint: -(-c)=(+c)
Whether dumb Albert failed to notice, or was taken in by the math
sleight of hand, it has lead to a lot of dupes beleiving all the SR
crap etc which has followed
Jim G
c'=c+vscessen
Androcles
==========================================
When you see a magician perform a trick, it is still entertaining
and will continue to be until you know how it is done.
So instead of saying it's crap, tell us how the trick is done.
Dono.
> On May 10, 9:37 pm, Jim Greenfield <jgreenfiel...@gmail.com> wrote:
> [...]
>
> > My daughters boyfriend is just about to complete his Masters Degree in
> > math. He informs me that noone in the faculty has taken Einstein
> > seriously for years. AE has gone from a joke to an annoying
> > distraction.
>
> On one hand, I wonder if the boyfriend is merely placating the insane
> father of his girlfriend.
>
Jim Greenfield
wrote:
> "Jim Greenfield" <jgreenfiel...@gmail.com> wrote in message
>
> news:4ed4819e-2c17-4186...@18g2000prd.googlegroups.com...
> On May 13, 3:03 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On May 10, 9:37 pm, Jim Greenfield <jgreenfiel...@gmail.com> wrote:
> > [...]
>
> > > My daughters boyfriend is just about to complete his Masters Degree in
> > > math. He informs me that noone in the faculty has taken Einstein
> > > seriously for years. AE has gone from a joke to an annoying
> > > distraction.
>
> > On one hand, I wonder if the boyfriend is merely placating the insane
> > father of his girlfriend.
>
> > his theory uses tools that were previously developed by folks like
> > Riemann so I'm wondering what the fuck he'd be talking about.
>
> It was Lorentz/Fitz who put the pea under the walnut shell.
> They kept the velocity of the train at v(+) at all times, but when
> light was considered,
> they use c(+) in the direction of the trains motion, but NOT c(-) in
> the other.
> Hint: -(-c)=(+c)
>
> Whether dumb Albert failed to notice, or was taken in by the math
> sleight of hand, it has lead to a lot of dupes beleiving all the SR
> crap etc which has followed
> ==========================================
> When you see a magician perform a trick, it is still entertaining
> and will continue to be until you know how it is done.
> So instead of saying it's crap, tell us how the trick is done.
As above to PD:
Lorentz adds all train v in one direction, but does NOT treat light
the same.
The equation has a situation where v-(-c) should be v+c but isn't. He
confuses direction with magnitude. Stuffed if I know how it has been
overlooked
Jim G
c'=c+v
Androcles
===========================================
Knew it!!!!! We all know it's a crappy trick, but you aren't astute
enough to say how the trick is done so you presumptuously and
blindly drool over a unary minus operator!
Jim Greenfield
>
> - Show quoted text -
Isn't it simply a matter of Lorentz confusing (mating) negative and
positive as directional indicators, with (-) and (+) increase/
decrease.
AE didn't spot the mistake, and deliberately or otherwise has taken
generations of naked-king worshippers for a religious outing.
Jim G
c'=c+v
Androcles
=============================================
Nope. I'll tell you.
This is the pea under the shell:
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img6.gif
These are the shells:
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img11.gif
This is the shell shuffle:
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
through to
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img51.gif
This is the glass pea that the audience think has become a diamond:
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img53.gif
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img54.gif
The two shells, rAB/(c-v) and rAB/(c+v), were different way back
in img11, but the foolish professors study the shuffle instead of the
shells because Einstein told them the shells were identical.
Then they pass the glass bead on to the students and call it "physics".
It's actually a very clever piece of misdirection and sleight-of-hand,
done on paper before your very eyes.
Jim Greenfield
>
> This is the glass pea that the audience think has become a diamond:
> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img53.gif
> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img54.gif
>
> The two shells, rAB/(c-v) and rAB/(c+v), were different way back
> in img11, but the foolish professors study the shuffle instead of the
> shells because Einstein told them the shells were identical.
> Then they pass the glass bead on to the students and call it "physics".
>
> It's actually a very clever piece of misdirection and sleight-of-hand,
>
> - Show quoted text -
Isn't this just a more complicated version of my contention- that Lor/
Fitz
mixed signage for direction, with that of magnitude?
PS:
Did Michaelson/M COMPARE (v+c)/(v-c) ie differently moving light
sources, or merely
measure c (in air)?
It seems unlikely to me that a difference of (say) 100ft/sec could be
detected in 186,000 mps.
Jim G
c'=c+v
Androcles
=====================================
It has nothing to do with Lor/Fitz or anything else.
PS:
Did Michaelson/M COMPARE (v+c)/(v-c) ie differently moving light
sources, or merely
measure c (in air)?
======================================
The name is 'Michelson', no 'a' in it.
Here's the paper:
http://www.aip.org/history/gap/PDF/michelson.pdf
Read for yourself what he did.
Note the first line.
Here's the modern version of the same experiment:
http://www.youtube.com/watch?v=7T0d7o8X2-E
It seems unlikely to me that a difference of (say) 100ft/sec could be
detected in 186,000 mps.
==========================================
It can. It is so sensitive that Grusenick fails when tipped on its side,
the beam splitter bends under gravity.
http://www.youtube.com/watch?v=aNEryiOKkrc
Jim Greenfield
>
> - Show quoted text -
OK. thanks
I was looking for the light in the train tunnel measurement
Jim G
c'=c+v
Dono.
>
> PS:
> Did Michaelson/M COMPARE (v+c)/(v-c) ie differently moving light
> sources, or merely
> measure c (in air)?
Neither, imbecile
> It seems unlikely to me that a difference of (say) 100ft/sec could be
> detected in 186,000 mps.
>
No one gives a shit about what a stupid old fart like you "thinks"
Androcles
============================================
SR was written to explain MMX, by a con artist.