answer to YBM's bell problem
rbwinn
transformation equations and no length contraction with frame of
reference B moving in the +x direction relative to frame of reference
A, light is emited at a distance of a either side of the origins of A
and B when they coincide. YBM says there is a bell at the origins
that will ring when light from both sources hits it. Since he did not
specify whether the bell was at the origin of A or the origin of B, we
will work the problem with the bell at the origin of A. In A, the
light is emitted at x=a and x=(-a). In B the light is emitted at
x'=a and x'=(-a). Light emitted at -a has a velocity of c relative
to both frames of reference. Light emitted at a has a velocity of -c
relative to both frames of reference. If the bell is at the origin of
A, it will ring when t=a/c.
The mathematics for this is, for the light from -a, x1=-a, x2=0
x2-x1 = 0-(-a) = a
x=wt=ct =a
t=a/c
For the light from a, x1=a, x2=0 w=(-c)
x2-x1= 0 - a
x=wt=(-c)t = -a
t=a/c
Light from both directions reaches the bell at a time of a/c, and the
bell rings.
To determine what time a clock in B reads when the bell rings,
n'= t(1-v/w)
n'=a/c(1-v/c)
Robert B. Winn
YBM
> YBM proposed the folowing problem.
Here it is :
>> two light rays are emitted from (-a,0,0) and (a,0,0) (a>0)
>> at the same time in A in the direction of the origine
>> of A (so velocities are respectively (c,0,0) and (-c,0,0).
>> When both arrive at O it makes a bell ring.
>>
>> Try to seriously answer these questions :
>> In frame A and in frame B :
>> - What are the coordinates of both photons at a
>> given time ?
>> - What are the coordinates of the event "the bell ring"
>> in A and in B.
>>
>> At every step explain why you choose t, t' or n' (and
>> which n') as time coordinate.
> ... Since he did not
> specify whether the bell was at the origin of A or the origin of B, we
> will work the problem with the bell at the origin of A.
At least you've guess one thing right in your life. The bell is
at rest in A frame. You seem to have guess right, as well, that
both light rays were emitted ą t=t'=0.
> In A, the
> light is emitted at x=a and x=(-a).
> In B the light is emitted at
> x'=a and x'=(-a).
It is what you'd get if you use Galilean transformations.
> Light emitted at -a has a velocity of c relative
> to both frames of reference.
This is NOT what you'd get if you use Galilean transformations.
> Light emitted at a has a velocity of -c
This is NOT what you'd get if you use Galilean transformations.
> relative to both frames of reference. If the bell is at the origin of
> A, it will ring when t=a/c.
>
> The mathematics for this is, for the light from -a, x1=-a, x2=0
> x2-x1 = 0-(-a) = a
> x=wt=ct =a
> t=a/c
>
> For the light from a, x1=a, x2=0 w=(-c)
> x2-x1= 0 - a
> x=wt=(-c)t = -a
> t=a/c
You forget to say that this is in frame A. Then this is ok.
> Light from both directions reaches the bell at a time of a/c, and the
> bell rings.
> To determine what time a clock in B reads when the bell rings,
>
> n'= t(1-v/w)
> n'=a/c(1-v/c)
You have a very fancy way to use your own "thoery". There is
TWO light rays. One with velocity -c, the other one with
velocity c. Here you only use w=c. Is there something
special with the light ray comming from (-a,0,0) ? Moreover
if you compute the speed of the other light ray, using
n'=a/c(1-v/c), you won't get c.
Do you see why I asked you a problem with *two* light rays to consider ?
To point out that this force you to arbitrarely choice one of
the rays to setup your "slower clocks", but then the other light
ray won't propagate at c.
If you had followed your *own words* you'd have used w=c
for the ray emitted at (-a,0,0), giving for you "slower
clock" : n'=a/c(1-v/c), then when considering the
light ray emitted at (a,0,0), at velocity -w you'd have
used n'=a/c(1+v/c).
Then you'd have found that your "theory" does not
provite a unique time for a single event... In other
words, what is an event in frame B, is *two* event
in frame B : both light rays, according to you theory,
do not arrive on the bell at the time... so the bell
does not ring in B. But it rings in A : contradiction !
There is no such absurdities in SR : for SR, the bell
rings in both frames... It just happens that in frame
B they were emitted at coordinates
(-a/sqrt(1-v^2/c^2),0,0) at time va/(c^2*sqrt(1-v^2/c^2))
for the "left" light ray, and :
(a/sqrt(1-v^2/c^2),0,0) at time -va/(c^2*sqrt(1-v^2/c^2))
Next time you'll pretend I've proven something, please
don't lie and write that I've proven Robert Winn's modified
"Galilean Transformation" to be absurd.
YBM
> YBM proposed the folowing problem.
Here it is :
>> two light rays are emitted from (-a,0,0) and (a,0,0) (a>0)
>> at the same time in A in the direction of the origine
>> of A (so velocities are respectively (c,0,0) and (-c,0,0).
>> When both arrive at O it makes a bell ring.
>>
>> Try to seriously answer these questions :
>> In frame A and in frame B :
>> - What are the coordinates of both photons at a
>> given time ?
>> - What are the coordinates of the event "the bell ring"
>> in A and in B.
>>
>> At every step explain why you choose t, t' or n' (and
>> which n') as time coordinate.
> ... Since he did not
> specify whether the bell was at the origin of A or the origin of B, we
> will work the problem with the bell at the origin of A.
At least you've guess one thing right in your life. The bell is
at rest in A frame. You seem to have guess right, as well, that
both light rays were emitted ą t=t'=0.
> In A, the
> light is emitted at x=a and x=(-a).
> In B the light is emitted at
> x'=a and x'=(-a).
It is what you'd get if you use Galilean transformations.
> Light emitted at -a has a velocity of c relative
> to both frames of reference.
This is NOT what you'd get if you use Galilean transformations.
> Light emitted at a has a velocity of -c
> relative to both frames of reference.
This is NOT what you'd get if you use Galilean transformations.
> If the bell is at the origin of
> A, it will ring when t=a/c.
>
> The mathematics for this is, for the light from -a, x1=-a, x2=0
> x2-x1 = 0-(-a) = a
> x=wt=ct =a
> t=a/c
>
> For the light from a, x1=a, x2=0 w=(-c)
> x2-x1= 0 - a
> x=wt=(-c)t = -a
> t=a/c
You forget to say that this is in frame A. Then this is ok.
> Light from both directions reaches the bell at a time of a/c, and the
> bell rings.
> To determine what time a clock in B reads when the bell rings,
>
> n'= t(1-v/w)
> n'=a/c(1-v/c)
You have a very fancy way to use your own "theory". There is
TWO light rays. One with velocity -c, the other one with
velocity c. Here you only use w=c. Is there something
special with the light ray comming from (-a,0,0) ?
Do you see why I asked you a problem with *two* light rays to consider ?
To point out that this force you to arbitrarely choose one of
the rays to setup your "slower clocks", but then the other light
ray won't propagate at c, see :
if n'=t(1-v/c), for the light ray coming from (a,0,0) you get :
x'=x-vn'=-ct-vt, so :
x'/n' = -t(c+v)/(t(1-v/c)) = -(c+v)/(1-v/c) = -c*(c+v)/(c-v) =/= -c !
So here we have a choice of a preferred light ray for absolutely
no reason, and a contradiction with your postulate that both
light rays propagate at c in both frames.
Now, let's see what happen if one tries to apply your "theory"
that a clock in B would behave differently given the light
ray you consider :
If you had followed your *own words* you'd have used w=c
for the ray emitted at (-a,0,0), giving for your "slower
clock" : n'=a/c(1-v/c), and when considering the
light ray emitted at (a,0,0), at velocity -w you'd have
used n'=a/c(1+v/c).
Then you'd have found that your "theory" does not
provite a unique time for a single event... In other
words, what is an event in frame A, is *two* events
in frame B : both light rays, according to you theory,
do not arrive on the bell at the same time... so the bell
does not ring in B. But it rings in A : contradiction !
There is no such absurdities in SR : for SR, the bell
rings in both frames... It just happens that in frame
B they were emitted at coordinates
(-a/sqrt(1-v^2/c^2),0,0) at time va/(c^2*sqrt(1-v^2/c^2))
for the "left" light ray, and :
(a/sqrt(1-v^2/c^2),0,0) at time -va/(c^2*sqrt(1-v^2/c^2))
but both arrives at the time a/(c*sqrt(1-v^2/c^2)) at
coordinates (-va/(c*sqrt(1-v^2/c^2)).0,0).
There is no such absurties in Galilean Relativity either,
but then speed of light rays won't be c but c-v and c+v,
which is refuted by experiments.
rbwinn
> rbwinn a �crit :
>
> > YBM proposed the folowing problem.
>
> Here it is :
>
>
>
>
>
> >> two light rays are emitted from (-a,0,0) and (a,0,0) (a>0)
> >> at the same time in A in the direction of the origine
> >> of A (so velocities are respectively (c,0,0) and (-c,0,0).
> >> When both arrive at O it makes a bell ring.
>
> >> Try to seriously answer these questions :
> >> In frame A and in frame B :
> >> - What are the coordinates of both photons at a
> >> � given time ?
> >> - What are the coordinates of the event "the bell ring"
> >> � in A and in B.
>
> >> At every step explain why you choose t, t' or n' (and
> >> which n') as time coordinate.
> > �... Since he did not
> > specify whether the bell was at the origin of A or the origin of B, we
> > will work the problem with the bell at the origin of A. �
>
> At least you've guess one thing right in your life. The bell is
> at rest in A frame. You seem to have guess right, as well, that
For light from the other direction, you get -c for the velocity of the
light.
> Do you see why I asked you a problem with *two* light rays to consider ?
Well, I can consider both light rays. What is the problem you see?
> To point out that this force you to arbitrarely choice one of
> the rays to setup your "slower clocks", but then the other light
> ray won't propagate at c.
The other light ray has a velocity of -c. That is what w means.
> If you had followed your *own words* you'd have used w=c
> for the ray emitted at (-a,0,0), giving for you "slower
> clock" : n'=a/c(1-v/c), then when considering the
> light ray emitted at (a,0,0), at velocity -w you'd have
> used n'=a/c(1+v/c).
Yes, that is correct.
> Then you'd have found that your "theory" does not
> provite a unique time for a single event... In other
> words, what is an event in frame B, is *two* event
> in frame B : both light rays, according to you theory,
> do not arrive on the bell at the time... so the bell
> does not ring in B. But it rings in A : contradiction !
The time n' in B is saying that the two light rays meet at the origin
of B, not at the origin of A. That being the case, they cannot be
used to compute the time when the bell rings. The bell rings when the
light meets at the origin of A according to the time in A, which is
t=a/c. The way this relates to time in B is that B has traveled a
distance of vt, and the origin of B is a distance of vt from the
origin of A when the bell rings. So the bell rings in B when the
origin of B is a distance of vt from the origin of A, not when an
observer in A thinks light has reached somewhere. The light is not
controlled by observers.
> There is no such absurdities in SR : for SR, the bell
> rings in both frames... It just happens that in frame
> B they were emitted at coordinates
> (-a/sqrt(1-v^2/c^2),0,0) at time va/(c^2*sqrt(1-v^2/c^2))
> for the "left" light ray, and :
> (a/sqrt(1-v^2/c^2),0,0) at time -va/(c^2*sqrt(1-v^2/c^2))
>
> Next time you'll pretend I've proven something, please
> don't lie and write that I've proven Robert Winn's modified
> "Galilean Transformation" to be absurd.- Hide quoted text -
>
Well, it would have taken me years to have gotten n' from the
equations you derived it from. I have to give you credit for it. So
you do not think the length contraction is an absurdity. I think the
length contraction is an absurdity.
Robert B. winn
YBM
For a single case, you have two clocks, each of them tweaked in
order to give a specific result about part of the experiment. This
has nothing to do with what both Galilean and Lorentz transformations
consider : a simple clock, running well, at rest in B.
>> To point out that this force you to arbitrarely choice one of
>> the rays to setup your "slower clocks", but then the other light
>> ray won't propagate at c.
> The other light ray has a velocity of -c. That is what w means.
>> If you had followed your *own words* you'd have used w=c
>> for the ray emitted at (-a,0,0), giving for you "slower
>> clock" : n'=a/c(1-v/c), then when considering the
>> light ray emitted at (a,0,0), at velocity -w you'd have
>> used n'=a/c(1+v/c).
> Yes, that is correct.
>
>> Then you'd have found that your "theory" does not
>> provite a unique time for a single event... In other
>> words, what is an event in frame B, is *two* event
>> in frame B : both light rays, according to you theory,
>> do not arrive on the bell at the time... so the bell
>> does not ring in B. But it rings in A : contradiction !
> The time n' in B is saying that the two light rays meet at the origin
> of B
There is no "time n'" in B in this case : there is two candidates.
Which one do you select ? Why ?
>, not at the origin of A.
Beside not being what you formulas says, this is absurd : now,
according to you, the two light rays are meeting on the bell
for observers in A, but elsewhere for observers in B. Note that,
then, the bell won't ring neither.
How absurd are you going to be ?
> That being the case, they cannot be
> used to compute the time when the bell rings. The bell rings when the
> light meets at the origin of A according to the time in A, which is
> t=a/c. The way this relates to time in B is that B has traveled a
> distance of vt, and the origin of B is a distance of vt from the
> origin of A when the bell rings. So the bell rings in B when the
> origin of B is a distance of vt from the origin of A, not when an
> observer in A thinks light has reached somewhere. The light is not
> controlled by observers.
Is it a kind of random poetry ? Did you realize that as seen in B,
according to your theory and the variant (without formulas) you
just introduced, the bell doesn't ring ?
It is indeed not in SR and GR, in you "theory" it is worse : what
happens differs according moving observers, clocks differs according
to the event an observer consider, and so on ...
>> There is no such absurdities in SR : for SR, the bell
>> rings in both frames... It just happens that in frame
>> B they were emitted at coordinates
>> (-a/sqrt(1-v^2/c^2),0,0) at time va/(c^2*sqrt(1-v^2/c^2))
>> for the "left" light ray, and :
>> (a/sqrt(1-v^2/c^2),0,0) at time -va/(c^2*sqrt(1-v^2/c^2))
>>
>> Next time you'll pretend I've proven something, please
>> don't lie and write that I've proven Robert Winn's modified
>> "Galilean Transformation" to be absurd.- Hide quoted text -
>>
> Well, it would have taken me years to have gotten n' from the
> equations you derived it from. I have to give you credit for it.
I just reverse two of the meaningless I provide in order to get
the value of n' you didn't provide in a specific case. If you
are so dumb that it would have taken years for you to get your
own meaningless equations from a equivalent form you just provide,
why don't you ask yourself a simple question : "Is it surprising
that everyone says me I don't understand equations of motion,
Galilean transformations, Lorentz transformations, ... ?"
> So
> you do not think the length contraction is an absurdity. I think the
> length contraction is an absurdity.
Given that what you've proposed for 11 years has now being proven
completely absurd, could you go on and try to understand what
lenght contraction is ? Feel free to ask here for help. But
- please - read the numerous FAQ and posts in the archive about
this issue first.
And I strongly insist :
rbwinn
> rbwinn a �crit :
> > The time n' in B is saying that the two light rays meet at the origin
> > of B
>
> There is no "time n'" in B in this case : there is two candidates.
> Which one do you select ? Why ?
>
> >, not at the origin of A. �
>
> Beside not being what you formulas says, this is absurd : now,
> according to you, the two light rays are meeting on the bell
> for observers in A, but elsewhere for observers in B. Note that,
> then, the bell won't ring neither.
> How absurd are you going to be ?
All it takes to ring a bell at the origin of A is for two light rays,
one from each direction to meet at the origin of A at the same time.
I proved that was happening. The bell rings, regardless of anything
else. In the same amount of time, B moves a distance of vt relative
to A. When the origin of B is a distance of vt from the origin of A,
the bell rings, whether an observer in B is hallucinating, dead, or
whatever. That is just the way it works.
> > That being the case, they cannot be
> > used to compute the time when the bell rings. �The bell rings when the
> > light meets at the origin of A according to the time in A, which is
> > t=a/c. �The way this relates to time in B is that B has traveled a
> > distance of vt, and the origin of B is a distance of vt from the
> > origin of A when the bell rings. �So the bell rings in B when the
> > origin of B is a distance of vt from the origin of A, not when an
> > observer in A thinks light has reached somewhere. �The light is not
> > controlled by observers.
>
> Is it a kind of random poetry ? Did you realize that as seen in B,
> according to your theory and the variant (without formulas) you
> just introduced, the bell doesn't ring ?
It absolutely does ring. Light reaches the origin of A from both
directions at the same time, and the bell rings. If it is operating,
it cannot do anything but ring.
Well, You claim to have proven the bell will not ring. It is obvious
to me that it would.
Robert B. Winn
YBM
> On Sep 9, 4:10�pm, YBM <ybm...@nooos.fr> wrote:
>> rbwinn a �crit :
>
>
>>> The time n' in B is saying that the two light rays meet at the origin
>>> of B
>> There is no "time n'" in B in this case : there is two candidates.
>> Which one do you select ? Why ?
>>
>>> , not at the origin of A. �
>> Beside not being what you formulas says, this is absurd : now,
>> according to you, the two light rays are meeting on the bell
>> for observers in A, but elsewhere for observers in B. Note that,
>> then, the bell won't ring neither.
>> How absurd are you going to be ?
>
> All it takes to ring a bell at the origin of A is for two light rays,
> one from each direction to meet at the origin of A at the same time.
> I proved that was happening. The bell rings, regardless of anything
> else. In the same amount of time, B moves a distance of vt relative
> to A. When the origin of B is a distance of vt from the origin of A,
> the bell rings, whether an observer in B is hallucinating, dead, or
> whatever. That is just the way it works.
Let me recall you what you just wrote :
> The time n' in B is saying that the two light rays meet at the origin
> of B
... where you just express clearly how absurd your "theory" is.
...
> Well, You claim to have proven the bell will not ring. It is obvious
> to me that it would.
> Robert B. Winn
I've proven that IN YOUR "THEORY" the bell ring or not according to
different observers.
You happen not to care asserting contradictory statements : the
bell ring for A and for B (what you just wrote), "the bell ring
for A, not for B" (what your theory directly implies).
Next time you'll pretend I've proven something, please
don't lie and write that I've proven Robert Winn's modified
"Galilean Transformation" to be absurd.
You can even provide the link :
http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de79ced
rbwinn
So you are saying that if a bunch of drunk scientists in B do not
believe the bell rings, then it does not ring. I say that if two
light rays meet at the origin of A in frame of A from each direction
of the x axis at a time of a/c, then the bell will ring because those
were the conditions you set forth to have it ring. What the drunk
scientists in B believe does not affect whether or not the bell rings,
just whether two light rays meet in A at the origin. Sure enough, two
light rays meet in A at the origin. The bell rings.
Robert B. Winn
> > Well, You claim to have proven the bell will not ring. �It is obvious
> > to me that it would.
> > Robert B. Winn
>
> I've proven that IN YOUR "THEORY" the bell ring or not according to
> different observers.
>
> You happen not to care asserting contradictory statements : the
> bell ring for A and for B (what you just wrote), "the bell ring
> for A, not for B" (what your theory directly implies).
>
> Next time you'll pretend I've proven something, please
> don't lie and write that I've proven Robert Winn's modified
> "Galilean Transformation" to be absurd.
> You can even provide the link :http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de...- Hide quoted text -
>
> - Show quoted text -
YBM
>> Let me recall you what you just wrote :
>>
>>> The time n' in B is saying that the two light rays meet at the origin
>>> of B
>> ... where you just express clearly how absurd your "theory" is.
>
> So you are saying that if a bunch of drunk scientists in B do not
> believe the bell rings, then it does not ring.
I'm quoting you. Did you forget ?
> I say that if two
> light rays meet at the origin of A in frame of A from each direction
> of the x axis at a time of a/c, then the bell will ring because those
> were the conditions you set forth to have it ring. What the drunk
> scientists in B believe does not affect whether or not the bell rings,
> just whether two light rays meet in A at the origin. Sure enough, two
> light rays meet in A at the origin. The bell rings.
You "theory" predict it will ring in A not in B.
You have some kind of mental disability I've never heard before.
Let me write it clear :
I didn't show "the bell won't ring in B".
I've shown "According to Robert Winn theory the ring will ring in A
and not in B".
Then, if YOU assert both your "theory" AND "the bell ring in A and in
B", you contradict yourself.
By the way, next time you'll pretend I've proven something, please
don't lie and write that I've proven Robert Winn's modified
"Galilean Transformation" to be absurd.
You can even provide the link :
http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de79ced
rbwinn
> ...
> >> Let me recall you what you just wrote :
>
> >>> The time n' in B is saying that the two light rays meet at the origin
> >>> of B
> >> ... where you just express clearly how absurd your "theory" is.
>
> > So you are saying that if a bunch of drunk scientists in B do not
> > believe the bell rings, then it does not ring.
>
> I'm quoting you. Did you forget ?
>
> > �I say that if two
> > light rays meet at the origin of A in frame of A from each direction
> > of the x axis at a time of a/c, then the bell will ring because those
> > were the conditions you set forth to have it ring. �What the drunk
> > scientists in B believe does not affect whether or not the bell rings,
> > just whether two light rays meet in A at the origin. �Sure enough, two
> > light rays meet in A at the origin. �The bell rings.
>
> You "theory" predict it will ring in A not in B.
transformation equations meant?
>
> You have some kind of mental disability I've never heard before.
> Let me write it clear :
>
> I didn't show "the bell won't ring in B".
>
> I've shown "According to Robert Winn theory the ring will ring in A
> and not in B".
event in B.
> Then, if YOU assert both your "theory" AND "the bell ring in A and in
> B", you contradict yourself.
Well, I don't think so. If the bell rings in A, it has no choice but
to ring in B.
> By the way, next time you'll pretend I've proven something, please
> don't lie and write that I've proven Robert Winn's modified
> "Galilean Transformation" to be absurd.
> You can even provide the link :http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de...
Well, I don't know what else I can do to thank you, YBM. I
alreadysaid that if they try to give me the Nobel Prize, I will make
sure that they give it to you.
Robert B. Winn
YBM
> On Sep 9, 5:01�pm, YBM <ybm...@nooos.fr> wrote:
>> rbwinn a �crit :
>>
>>> On Sep 9, 4:36 pm, YBM <ybm...@nooos.fr> wrote:
>> ...
>>>> Let me recall you what you just wrote :
>>>>> The time n' in B is saying that the two light rays meet at the origin
>>>>> of B
>>>> ... where you just express clearly how absurd your "theory" is.
>>> So you are saying that if a bunch of drunk scientists in B do not
>>> believe the bell rings, then it does not ring.
>> I'm quoting you. Did you forget ?
Nothing to say here ? Disgusting hypocrite !
>>> �I say that if two
>>> light rays meet at the origin of A in frame of A from each direction
>>> of the x axis at a time of a/c, then the bell will ring because those
>>> were the conditions you set forth to have it ring. �What the drunk
>>> scientists in B believe does not affect whether or not the bell rings,
>>> just whether two light rays meet in A at the origin. �Sure enough, two
>>> light rays meet in A at the origin. �The bell rings.
>> You "theory" predict it will ring in A not in B.
> If it rings in A, it has to ring in B. What did you think
> transformation equations meant?
>> You have some kind of mental disability I've never heard before.
>> Let me write it clear :
>>
>> I didn't show "the bell won't ring in B".
>>
>> I've shown "According to Robert Winn theory the ring will ring in A
>> and not in B".
> Why wouldn't it ring in B? If it is an event in A, it is also an
> event in B.
Just apply the equations of your "theory", it leads to this
contradiction.
>> Then, if YOU assert both your "theory" AND "the bell ring in A and in
>> B", you contradict yourself.
> Well, I don't think so. If the bell rings in A, it has no choice but
> to ring in B.
>> By the way, next time you'll pretend I've proven something, please
>> don't lie and write that I've proven Robert Winn's modified
>> "Galilean Transformation" to be absurd.
>> You can even provide the link :http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de...
>
> Well, I don't know what else I can do to thank you, YBM. I
> alreadysaid that if they try to give me the Nobel Prize, I will make
> sure that they give it to you.
Just quote me right, and provide the link.
YBM
>> You "theory" predict it will ring in A not in B.
> If it rings in A, it has to ring in B. What did you think
> transformation equations meant?
Right. Then you admit that your set of equations does
not form a transformation and that your "theory" is
meaningless.
So, next time you'll pretend I've proven something, please
rbwinn
No contradiction at all. The transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
What is there about these equations that you do not understand?
If an event happens at the origin of A in A, then it happens at the
origin of A in B. When t'=t.
If a bell rings at the origin of A in A, then it rings at the origin
of A in B. The bell ringing is one event. The drunk scientists
cannot understand from frame of reference B why the bell rings. If
two photons in frame of reference A reach the origin of A as
specified, the bell is going to ring, that is just the way it is.
Sorry you scientists do not like it.
> >> Then, if YOU assert both your "theory" AND "the bell ring in A and in
> >> B", you contradict yourself.
> > Well, I don't think so. �If the bell rings in A, it has no choice but
> > to ring in B.
> >> By the way, next time you'll pretend I've proven something, please
> >> don't lie and write that I've proven Robert Winn's modified
> >> "Galilean Transformation" to be absurd.
> >> You can even provide the link :http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de...
>
> > Well, I don't know what else I can do to thank you, YBM. �I
> > alreadysaid that if they try to give me the Nobel Prize, I will make
> > sure that they give it to you.
>
> Just quote me right, and provide the link.- Hide quoted text -
>
> - Show quoted text -
Well, you said n'=t(1-v/w). I could not have said it better myself.
Robert B. Winn
rbwinn
The transformation equations are
x'=x-vt
y'=y
x'=x
t'=t
If you have a clock showing light to be traveling at c in B, then
light from
-a and a is going to meet at the origin of B if it is emitted
simultaneously.
Since there is no distance contraction, the light is emitted
simultaneously in both frames of reference. The light beams meet at
the origin of B at a time of a/c according to the n' clock.
Robert B. Winn
YBM
>> Just apply the equations of your "theory", it leads to this
>> contradiction.
> No contradiction at all. The transformation equations are
>
> x'=x-vt
> y'=y
> z'=z
> t'=t
>
> What is there about these equations that you do not understand?
This is the equations of the Galilean Transformations, not of your
"theory", as usual you deliberately "forget" the meaningless
n'=t(1-v/w) you add.
I already said you what GT predict : of course the bell ring
in both frame... *but* with GT the speeds of the light rays
are not c in *both* frames.
> Well, you said n'=t(1-v/w). I could not have said it better myself.
Listen, disgusting bastard : in order to proove that any theory
is absurd, one have to play with it, to write down its formulas,
to make some logical deduction from them. The fact that I wrote
some of you stupid equations does not mean that I accept them.
As a matter of fact, I've proven them to be absurd.
So, next time you'll pretend I've proven something, please
don't lie and write that I've proven Robert Winn's modified
"Galilean Transformation" to be absurd.
You can even provide the link :
http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de79ced
Robert Winn's best fumble :
> The time n' in B is saying that the two light rays meet at the origin
> of B
so... the ring bell in B but not in A (the bell is not at the origin
of B whenever the light rays meet !)
rbwinn
> >> contradiction.
> > No contradiction at all. �The transformation equations are
>
> > � � � � � � � � � x'=x-vt
> > � � � � � � � � � y'=y
> > � � � � � � � � � z'=z
> > � � � � � � � � � t'=t
>
> > � �What is there about these equations that you do not understand?
>
> This is the equations of the Galilean Transformations, not of your
> "theory", as usual you deliberately "forget" the meaningless
> n'=t(1-v/w) you add.
>
> I already said you what GT predict : of course the bell ring
> in both frame... *but* with GT the speeds of the light rays
> are not c in *both* frames.
>
> > Well, you said n'=t(1-v/w). �I could not have said it better myself.
>
> Listen, disgusting bastard : in order to proove that any theory
> is absurd, one have to play with it, to write down its formulas,
> to make some logical deduction from them. The fact that I wrote
> some of you stupid equations does not mean that I accept them.
> As a matter of fact, I've proven them to be absurd.
>
> So, next time you'll pretend I've proven something, please
> don't lie and write that I've proven Robert Winn's modified
> "Galilean Transformation" to be absurd.
>
> Robert Winn's best fumble :
> �> The time n' in B is saying that the two light rays meet at the origin
> �> of B
>
> so... the ring bell in B but not in A (the bell is not at the origin
> of B whenever the light rays meet !)
Well, in the frame of reference of the bell, where do you think the
light rays meet?
Robert B. Winn
YBM
This is probably the most stupid question you'll ever ask...
On the bell : the origine of A, (0,0,0) in A frame. Got it ?
(this is what both TL and GL predict, not what predicts
"Robert Winn stupid so-called transformation).
Again : do you have a problem with posts longer that three lines ?
You seem to not even read anything longer that this, since you only
repond to the very end of any post.
If you are mentaly disabled, as it is quite clear now, we can
do the effort to write to you about one problem at a time, with
as few words as possible. Would it help you to read what we wrote ?
rbwinn
>
> - Show quoted text -
I just post the relevant facts. n' is not a transformation equation.
It is time on a clock in B that shows light to be traveling at a speed
of c.
Robert B. Winn
YBM
Right. As I've shown it cannot be, so you'd better drop it.
> It is time on a clock in B that shows light to be traveling at a speed
> of c.
Nope. Read again the setup I proposed : You'd get c of one ligh ray, not
for the other one. Same conclusion : meaningless formula sould be
dropped (and if not it would lead to contradictory conclusion, as I
shown you).
So, next time you'll pretend I've proven something, please
don't lie and write that I've proven Robert Winn's modified
"Galilean Transformation" to be absurd.
You can even provide the link :
http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de79ced
YBM
> I just post the relevant facts. n' is not a transformation equation.
> It is time on a clock in B that shows light to be traveling at a speed
> of c.
> Robert B. Winn
Given your mental disability which prevet you to understand more that
one question at a time, let's simplify the point since you're clearly
lost from the very beginning :
n'=t(1-v/w)
Well. In frame A, I consider two light rays on the (Ox) line (does it
remind you something ?). One at speed w=c in A, the other one at speed
w=-c in A. How would an observer in B would compute, in the context of
you "theory", using n', the speeds of both light rays simultaneously ?
Is there some kind of magic making its clock change whenever he consider
one light rays or the other one ?
--
By the way, next time you'll pretend I've proven
don't lie and write that I've proven Robert Winn'
"Galilean Transformation" to be absurd.
You can even provide the link :
http://groups.google.com/group/sci.physics.relati
Robert Winn's best fumble :
> The time n' in B is saying that the two light r
> of B
so... the ring bell in B but not in A (the bell i
rbwinn
Speed is the magnitude of velocity. Light has a speed of c from
either direction in either frame of reference. Remember your
equation, YBM?
n'=t(1-v/w). w is velocity of light. If the light is coming from -a,
it has a velocity of c, if it is coming from a, it has a velocity of -
c. Either way it has a speed of c.
Robert B. Winn
>
> So, next time you'll pretend I've proven something, please
> don't lie and write that I've proven Robert Winn's modified
> "Galilean Transformation" to be absurd.
> You can even provide the link :http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de...
>
> Robert Winn's best fumble :
> �> The time n' in B is saying that the two light rays meet at the origin
> �> of B
>
> so... the ring bell in B but not in A (the bell is not at the origin
> of B whenever the light rays meet !)- Hide quoted text -
rbwinn
> rbwinn a �crit :
> one question at a time, let's simplify the point since you're clearly
> lost from the very beginning :
>
> n'=t(1-v/w)
>
> Well. In frame A, I consider two light rays on the (Ox) line (does it
> remind you something ?). One at speed w=c in A, the other one at speed
> w=-c in A. How would an observer in B would compute, in the context of
> you "theory", using n', the speeds of both light rays simultaneously ?
OK, say that the lights are emitted at -a and a in both frames of
reference when the origin of B is at the origin of A, each ray of
light directed at the origins. B is moving in the +x direction
relative to A at a velocity of v. The light ray emitted at -a has a
velocity of w=c in both frames of reference. The light ray emitted at
a has a velocity of w=-c in both frames of reference. The light ray
emitted at -a goes from x=-a to the origin of A in frame of reference
A in a time of t=a/c. The light ray emitted at -a goes from x'=-a to
the origin of B in a time of n'=a/c. The light ray emitted at a
travels from x=a to the origin of A in a time of t=a/c. The light ray
emitted at a travels from x=a to the origin of B in a time of n'=a/c.
The equation for n' is
n'=t(1-v/w)
You remember the equation for n', YBM.
> Is there some kind of magic making its clock change whenever he consider
> one light rays or the other one ?
It is not a transformation equation, YBM. It just tells how far light
has gone in B according to a clock that shows n'.
Robert B. Winn
YBM
> On Sep 10, 7:19�pm, YBM <ybm...@nooos.fr> wrote:
>> rbwinn a �crit :
>
>> one question at a time, let's simplify the point since you're clearly
>> lost from the very beginning :
>>
>> n'=t(1-v/w)
>>
>> Well. In frame A, I consider two light rays on the (Ox) line (does it
>> remind you something ?). One at speed w=c in A, the other one at speed
>> w=-c in A. How would an observer in B would compute, in the context of
>> you "theory", using n', the speeds of both light rays simultaneously ?
>
> OK, say that the lights are emitted at -a and a in both frames of
> reference when the origin of B is at the origin of A, each ray of
> light directed at the origins. B is moving in the +x direction
> relative to A at a velocity of v. The light ray emitted at -a has a
> velocity of w=c in both frames of reference. The light ray emitted at
> a has a velocity of w=-c in both frames of reference.
> The light ray
> emitted at -a goes from x=-a to the origin of A in frame of reference
> A in a time of t=a/c. The light ray emitted at -a goes from x'=-a to
> the origin of B in a time of n'=a/c. The light ray emitted at a
> travels from x=a to the origin of A in a time of t=a/c. The light ray
> emitted at a travels from x=a to the origin of B in a time of n'=a/c.
Funny : the same light rays have a different behaviour in A and in B !
Here is what you pretend:
They meet at the origin of A from the point of view of A (so the bell
ring)
They meet at the origin of B from the point of view of B (so the bell,
which is NOT at the origin of B at this time, doesn't ring)
AGAIN: This is a direct conclusion OF WHAT YOU JUST WROTE, not what
I pretend, not what GT or LT pretend.
> The equation for n' is
>
> n'=t(1-v/w)
>
> You remember the equation for n', YBM.
>
>> Is there some kind of magic making its clock change whenever he consider
>> one light rays or the other one ?
>
> It is not a transformation equation, YBM. It just tells how far light
> has gone in B according to a clock that shows n'.
You way you evade questions is ridiculous...
FIRST : they is two light rays in the experiment, SO there is two "n'" :
n'=t(1-v/c) (for light ray coming from (-a,0,0))
n'=t(1+v/c) (for light ray coming from (a,0,0))
Why should the observer in B use two clocks, both of them
being broken ?
SECOND : You have a double language : You claim that n' give
a time coordinate for obervers of events in B, so the
equation for n' is part of a transformation (the fact
I've proven is that this transformation is absurd), then
you claim that this equation is NOT a part of a transformation,
but then you just support Galilean Transformations were
speed of light rays in frame B is no more c.
By the way, next time you'll pretend I've proven something, please
don't lie and write that I've proven Robert Winn's modified
"Galilean Transformation" to be absurd.
You can even provide the link :
http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de79ced
Robert Winn's best fumble :
> The time n' in B is saying that the two light rays meet at the origin
> of B
so... the ring bell in B but not in A (the bell is not at the origin
YBM
> Speed is the magnitude of velocity. Light has a speed of c from
> either direction in either frame of reference. Remember your
> equation, YBM?
This is yours, stupid and disgusting liar.
> n'=t(1-v/w). w is velocity of light. If the light is coming from -a,
> it has a velocity of c, if it is coming from a, it has a velocity of -
> c. Either way it has a speed of c.
To show how absurd is your "theory", one only have to write down
precisely what is implied by you OWN WORDS :
- in order to find a velocity of (c,0,0) for the light ray coming
from (-a,0,0) an observer in B should use a clock giving a
time n'=t(1-v/c)
- in order to find a velocity of (-c,0,0) for the light ray coming
from (a,0,0) an observer in B should use a clock giving a
time n'=t(1+v/c)
Silly : when timing what happens in any experiment an observer need
only ONE clock.
Whenever you're late on a appointement would you
give the excuse that you're in time because your clock is running
slow ?
By the way, next time you'll pretend I've proven something, please
don't lie and write that I've proven Robert Winn's modified
"Galilean Transformation" to be absurd.
You can even provide the link :
http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de79ced
rbwinn
> Funny : the same light rays have a different behaviour in A and in B !
> Here is what you pretend:
> They meet at the origin of A from the point of view of A (so the bell
> ring)
> They meet at the origin of B from the point of view of B (so the bell,
> which is NOT at the origin of B at this time, doesn't ring)
> AGAIN: This is a direct conclusion OF WHAT YOU JUST WROTE, not what
> I pretend, not what GT or LT pretend.
depends on light meeting at the origin of A. By your definition of
the problem, light meets at the origin of A because light is emitted
at -a and a on the x axis in frame of reference A at the same time.
The light meets at the origin of A in a time of t=a/c. The bell rings
in all frames of reference when that happens.
Now you want to talk about frame of reference B. To make this easier
to see, we will put another bell at the origin of B. The bells are
right next to each other when the light is emitted. Bell A rings when
a time of t=a/c has elapsed in A. The bell in B is now a distance of
vt from the bell in A. Scientists have determined by experiment that
the clock in B is slower than the clock in A. The clock in B
continues to move away from the clock in A. When the clock in B reads
n'=a/c, the bell in B rings.
Now, YBM, explain the same events using the Lorentz equations.
> > � � � The equation for n' is
>
> > � � � � � �n'=t(1-v/w)
>
> > You remember the equation for n', YBM.
>
> >> Is there some kind of magic making its clock change whenever he consider
> >> one light rays or the other one ?
>
> > It is not a transformation equation, YBM. �It just tells how far light
> > has gone in B according to a clock that shows n'.
>
> You way you evade questions is ridiculous...
>
> FIRST : they is two light rays in the experiment, SO there is two "n'" :
> � � � � � �n'=t(1-v/c) (for light ray coming from (-a,0,0))
> � � � � � �n'=t(1+v/c) (for light ray coming from (a,0,0))
> � � � � �Why should the observer in B use two clocks, both of them
> � � � � �being broken ?
Well, according to you, the observer in B only had one clock, which
showed light to be traveling at a speed of c in B. The observer in B
knows that an observer in A has a clock that shows light to be
traveling at a speed of c in A. You specified in the beginning that B
was moving with a velocity of v relative to A. That makes A a
preferred frame of reference because it is not moving. To transform
coordinates, you use a clock in A. To determine where light appears
to be according to the n' clock, you use the n' clock. In B light was
emitted at x'=-a and x'= a at n'=0. The light will meet at the origin
of B at a time of n'=a/c, and the bell in B will ring when that
happens. It will ring in all frames of reference when it rings, not
just in B.
> SECOND : You have a double language : You claim that n' give
> � � � � �a time coordinate for obervers of events in B, so the
> � � � � �equation for n' is part of a transformation (the fact
> � � � � �I've proven is that this transformation is absurd), then
> � � � � �you claim that this equation is NOT a part of a transformation,
> � � � � �but then you just support Galilean Transformations were
> � � � � �speed of light rays in frame B is no more c.
The speed of light in B is definitely c according to the n' clock.
Robert B. Winn
rbwinn
> > either direction in either frame of reference. �Remember your
> > equation, YBM?
>
> This is yours, stupid and disgusting liar.
Well, I think we have a situation like what happened when Gottfried
Leibniz and Sir Isaac Newton both discovered calculus at the same
time. Who should get credit for it? Well, I am going to have to
defer and let you take credit for
n'=t(1-v/w)
> > n'=t(1-v/w). �w is velocity of light. �If the light is coming from -a,
> > it has a velocity of c, if it is coming from a, it has a velocity of -
> > c. �Either way it has a speed of c.
>
> To show how absurd is your "theory", one only have to write down
> precisely what is implied by you OWN WORDS :
>
> � - in order to find a velocity of (c,0,0) for the light ray coming
> � � from (-a,0,0) an observer in B should use a clock giving a
> � � time n'=t(1-v/c)
> � - in order to find a velocity of (-c,0,0) for the light ray coming
> � � from (a,0,0) an observer in B should use a clock giving a
> � � time n'=t(1+v/c)
You have it exactly right, YBM. This shows the light from -a and a
meeting at the origin of B at n'=a/c as shown by the n' clock.
> Silly : when timing what happens in any experiment an observer need
> only ONE clock.
Right. The observer in A has one clock that shows t, and the observer
in B has one clock that shows n'.
> Whenever you're late on a appointement would you
> give the excuse that you're in time because your clock is running
> slow ?
Well, that is what observers in B usually do.
> By the way, next time you'll pretend I've proven something, please
> don't lie and write that I've proven Robert Winn's modified
> "Galilean Transformation" to be absurd.
> You can even provide the link :http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de...
>
> Robert Winn's best fumble :
> �> The time n' in B is saying that the two light rays meet at the origin
> �> of B
>
> so... the ring bell in B but not in A (the bell is not at the origin
> of B whenever the light rays meet !)
Well, originally we had a bell at the origin of A, which rang when
light beams met at the origin of A. In order to have a bell ring at
the origin of B, you have to put one there.
Robert B. Winn
YBM
> On Sep 11, 6:55�am, YBM <ybm...@nooos.fr> wrote:
>> rbwinn a �crit :
>>
>>> Speed is the magnitude of velocity. �Light has a speed of c from
>>> either direction in either frame of reference. �Remember your
>>> equation, YBM?
>> This is yours, stupid and disgusting liar.
>
> Well, I think we have a situation like what happened when Gottfried
> Leibniz and Sir Isaac Newton both discovered calculus at the same
> time. Who should get credit for it? Well, I am going to have to
> defer and let you take credit for
> n'=t(1-v/w)
>
>>> n'=t(1-v/w). �w is velocity of light. �If the light is coming from -a,
>>> it has a velocity of c, if it is coming from a, it has a velocity of -
>>> c. �Either way it has a speed of c.
>> To show how absurd is your "theory", one only have to write down
>> precisely what is implied by you OWN WORDS :
>>
>> � - in order to find a velocity of (c,0,0) for the light ray coming
>> � � from (-a,0,0) an observer in B should use a clock giving a
>> � � time n'=t(1-v/c)
>> � - in order to find a velocity of (-c,0,0) for the light ray coming
>> � � from (a,0,0) an observer in B should use a clock giving a
>> � � time n'=t(1+v/c)
>
> You have it exactly right, YBM. This shows the light from -a and a
> meeting at the origin of B at n'=a/c as shown by the n' clock.
Which one : n'=t(1-v/c) or n'=t(1+v/c) ?
>> Silly : when timing what happens in any experiment an observer need
>> only ONE clock.
> Right. The observer in A has one clock that shows t, and the observer
> in B has one clock that shows n'.
Which one : n'=t(1-v/c) or n'=t(1+v/c) ?
YBM
> On Sep 11, 6:40�am, YBM <ybm
>> Funny : the same light rays have a different behaviour in A and in B !
>> Here is what you pretend:
>> They meet at the origin of A from the point of view of A (so the bell
>> ring)
>> They meet at the origin of B from the point of view of B (so the bell,
>> which is NOT at the origin of B at this time, doesn't ring)
>> AGAIN: This is a direct conclusion OF WHAT YOU JUST WROTE, not what
>> I pretend, not what GT or LT pretend.
> A bell at the origin of A does not depend on light meeting in B. It
> depends on light meeting at the origin of A.
These sentences are insanely meaningless... The bell at the origin of A
depends on light meeting or not at the origin of A regardless of the
frame you consider this event. It's why any decent theory could not
predict, as yours, that it rings in a frame and not in another one.
> By your definition of
> the problem, light meets at the origin of A because light is emitted
> at -a and a on the x axis in frame of reference A at the same time.
> The light meets at the origin of A in a time of t=a/c.
Right.
> The bell rings
> in all frames of reference when that happens.
It's true, BUT it's not enough to write it down to make your "theory"
says that. As a matter of fact your "theory" predict it won't in
B frame.
> Now you want to talk about frame of reference B. To make this easier
> to see, we will put another bell at the origin of B. The bells are
> right next to each other when the light is emitted. Bell A rings when
> a time of t=a/c has elapsed in A. The bell in B is now a distance of
> vt from the bell in A.
> Scientists have determined by experiment that
> the clock in B is slower than the clock in A.
Nope. What experiment shows is :
- from the point of view of frame B, clocks in A are running slower
- from the point of view of frame A, clocks in B are running slower
Note that this is as well what LTs predicts.
> The clock in B
> continues to move away from the clock in A. When the clock in B reads
> n'=a/c, the bell in B rings.
You're going more insane every day : what you just write implies this :
- For an observer in A, the bell at rest in A rings, the one at rest in
B don't
- For an observer in B, the bell at rest in B rings. the one at rest in
A don't
> Now, YBM, explain the same events using the Lorentz equations.
I did here :
http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de79ced
>> There is no such absurdities in SR : for SR, the bell
>> rings in both frames... It just happens that in frame
>> B they were emitted at coordinates
>> (-a/sqrt(1-v^2/c^2),0,0) at time va/(c^2*sqrt(1-v^2/c^2))
>> for the "left" light ray, and :
>> (a/sqrt(1-v^2/c^2),0,0) at time -va/(c^2*sqrt(1-v^2/c^2))
So it is quite certain that you didn't read.
>>> � � � The equation for n' is
>>> � � � � � �n'=t(1-v/w)
>>> You remember the equation for n', YBM.
>>>> Is there some kind of magic making its clock change whenever he consider
>>>> one light rays or the other one ?
>>> It is not a transformation equation, YBM. �It just tells how far light
>>> has gone in B according to a clock that shows n'.
>> You way you evade questions is ridiculous...
>>
>> FIRST : they is two light rays in the experiment, SO there is two "n'" :
>> � � � � � �n'=t(1-v/c) (for light ray coming from (-a,0,0))
>> � � � � � �n'=t(1+v/c) (for light ray coming from (a,0,0))
>> � � � � �Why should the observer in B use two clocks, both of them
>> � � � � �being broken ?
>
> Well, according to you, the observer in B only had one clock, which
> showed light to be traveling at a speed of c in B. The observer in B
> knows that an observer in A has a clock that shows light to be
> traveling at a speed of c in A. You specified in the beginning that B
> was moving with a velocity of v relative to A. That makes A a
> preferred frame of reference because it is not moving.
There is nothing "preferred" for frame A in the problem I described.
There is indeed something special about A : the bell is at rest in A !
> To transform
> coordinates, you use a clock in A. To determine where light appears
> to be according to the n' clock, you use the n' clock. In B light was
> emitted at x'=-a and x'= a at n'=0. The light will meet at the origin
> of B at a time of n'=a/c, and the bell in B will ring when that
> happens. It will ring in all frames of reference when it rings, not
> just in B.
There is always the problem you refuse to adress : there is two discinct
"n'" clocks in B !
>> SECOND : You have a double language : You claim that n' give
>> � � � � �a time coordinate for obervers of events in B, so the
>> � � � � �equation for n' is part of a transformation (the fact
>> � � � � �I've proven is that this transformation is absurd), then
>> � � � � �you claim that this equation is NOT a part of a transformation,
>> � � � � �but then you just support Galilean Transformations were
>> � � � � �speed of light rays in frame B is no more c.
>
> The speed of light in B is definitely c according to the n' clock.
There is always the problem you refuse to adress : there is two discinct
"n'" clocks in B !
rbwinn
>
> - Show quoted text -
Well, if you are talking about an event in B, the n' values would be
the same, and the two t times wouled be different. You have the
reverse effect you see with n' from A. Not to worry, YBM, you have
really discovered a great principle of physics.
Robert B. Winn
rbwinn
> >> Funny : the same light rays have a different behaviour in A and in B !
> >> Here is what you pretend:
> >> They meet at the origin of A from the point of view of A (so the bell
> >> ring)
> >> They meet at the origin of B from the point of view of B (so the bell,
> >> which is NOT at the origin of B at this time, doesn't ring)
> >> AGAIN: This is a direct conclusion OF WHAT YOU JUST WROTE, not what
> >> I pretend, not what GT or LT pretend.
> > A bell at the origin of A does not depend on light meeting in B. �It
> > depends on light meeting at the origin of A.
>
> These sentences are insanely meaningless... The bell at the origin of A
> depends on light meeting or not at the origin of A regardless of the
> frame you consider this event. It's why any decent theory could not
> predict, as yours, that it rings in a frame and not in another one.
Mine says that if the bell is at the origin of A, and two light rays
meet there as you specified, then the bell will ring in all frames of
reference.
If the bell is at the origin of B, and two light beams meet at the
origin of A, the bell at the origin of B will not ring unless two
beams of light are also meeting at the origin of B.
> > By your definition of
> > the problem, light meets at the origin of A because light is emitted
> > at -a and a on the x axis in frame of reference A at the same time.
> > The light meets at the origin of A in a time of t=a/c. �
>
> Right.
>
> > The bell rings
> > in all frames of reference when that happens.
>
> It's true, BUT it's not enough to write it down to make your "theory"
> says that. As a matter of fact your "theory" predict it won't in
> B frame.
It can't do anything else in B. The mathematics as computed from
either frame of reference shows that in frame of reference A, light
met at the origin of A. The bell has to ring if it is working.
> > Now you want to talk about frame of reference B. �To make this easier
> > to see, we will put another bell at the origin of B. �The bells are
> > right next to each other when the light is emitted. �Bell A rings when
> > a time of t=a/c has elapsed in A. �The bell in B is now a distance of
> > vt from the bell in A. �
> > Scientists have determined by experiment that
> > the clock in B is slower than the clock in A.
>
> Nope. What experiment shows is :
> - from the point of view of frame B, clocks in A are running slower
> - from the point of view of frame A, clocks in B are running slower
Well, you specified in the beginning that B was moving relative to A,
so thus far, we are only looking at it from the point of view of frame
A.
> Note that this is as well what LTs predicts.
>
> > The clock in B
> > continues to move away from the clock in A. �When the clock in B reads
> > n'=a/c, the bell in B rings.
>
> You're going more insane every day : what you just write implies this :
> - For an observer in A, the bell at rest in A rings, the one at rest in
> � �B don't
> - For an observer in B, the bell at rest in B rings. the one at rest in
> � �A don't
Well, if you are using the Lorentz equations, only one bell is going
to ring because of relativity of simultaneity. The rays of light will
not meet at the origins of both frames of reference. So from A, an
observer will only observe the bell in A to ring, from B an observer
will only observe the bell in B to ring.
With my equations, the bell in A will ring first, then the bell in
B. You will hear both bells in both frames of reference.
> > � � Now, YBM, explain the same events using the Lorentz equations.
>
> I did here :http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de...
>
> >> There is no such absurdities in SR : for SR, the bell
> >> rings in both frames... It just happens that in frame
> >> B they were emitted at coordinates
> >> (-a/sqrt(1-v^2/c^2),0,0) at time va/(c^2*sqrt(1-v^2/c^2))
> >> for the "left" light ray, and :
> >> (a/sqrt(1-v^2/c^2),0,0) at time -va/(c^2*sqrt(1-v^2/c^2))
Uh huh. But if you put a bell at the origin of each frame of
reference, what will happen?
Robert B. Winn
>
> So it is quite certain that you didn't read.
>
No, I read it, YBM. So explain what happens if there are two bells,
one at the origin of each frame of reference.
Robert B. Winn
> "n'" clocks in B !- Hide quoted text -
YBM
> the same, and the two t times wouled be different.
Wrong : t(1-v/c) and t(1+v/c) are different except at t=0 (or if v=0).
AGAIN : which n'=t(1-v/c) or n'=t(1+v/c) do you choose ?
> You have the
> reverse effect you see with n' from A. Not to worry, YBM, you have
> really discovered a great principle of physics.
I just discover how insane is Robert Winn, this is not a principle
of physics.
YBM
> On Sep 11, 8:55�am, YBM <ybm...@nooos.fr> wrote:
>> rbwinn a �crit :
>>
>>> On Sep 11, 6:40 am, YBM <ybm
>>>> Funny : the same light rays have a different behaviour in A and in B !
>>>> Here is what you pretend:
>>>> They meet at the origin of A from the point of view of A (so the bell
>>>> ring)
>>>> They meet at the origin of B from the point of view of B (so the bell,
>>>> which is NOT at the origin of B at this time, doesn't ring)
>>>> AGAIN: This is a direct conclusion OF WHAT YOU JUST WROTE, not what
>>>> I pretend, not what GT or LT pretend.
>>> A bell at the origin of A does not depend on light meeting in B. �It
>>> depends on light meeting at the origin of A.
>> These sentences are insanely meaningless... The bell at the origin of A
>> depends on light meeting or not at the origin of A regardless of the
>> frame you consider this event. It's why any decent theory could not
>> predict, as yours, that it rings in a frame and not in another one.
>
> Mine says that if the bell is at the origin of A, and two light rays
> meet there as you specified, then the bell will ring in all frames of
> reference.
So you change your mind, remember that YOU wrote :
> The time n' in B is saying that the two light rays meet at the origin
> of B
But now you are saying that the bell will ring in both frames, so
the light rays won't meet at the origin of B. The problem is
that YOUR math implies that the bell won't ring from the point
of view of B. You contradict you OWN "theory".
> If the bell is at the origin of B, and two light beams meet at the
> origin of A, the bell at the origin of B will not ring unless two
> beams of light are also meeting at the origin of B.
Of course, but this is irrelevant since the light rays won't
meet at the origin of B.
You are trying to obfuscate the issue in order to evade the
fact that you "theory" has been proven absurd, aren't you ?
>>> By your definition of
>>> the problem, light meets at the origin of A because light is emitted
>>> at -a and a on the x axis in frame of reference A at the same time.
>>> The light meets at the origin of A in a time of t=a/c. �
>> Right.
>>
>>> The bell rings
>>> in all frames of reference when that happens.
>> It's true, BUT it's not enough to write it down to make your "theory"
>> says that. As a matter of fact your "theory" predict it won't in
>> B frame.
>
> It can't do anything else in B. The mathematics as computed from
> either frame of reference shows that in frame of reference A, light
> met at the origin of A. The bell has to ring if it is working.
So the mathematics of your "theory" have no consistence. Not that
this is a surprise to anyone.
>>> Now you want to talk about frame of reference B. �To make this easier
>>> to see, we will put another bell at the origin of B. �The bells are
>>> right next to each other when the light is emitted. �Bell A rings when
>>> a time of t=a/c has elapsed in A. �The bell in B is now a distance of
>>> vt from the bell in A. �
>>> Scientists have determined by experiment that
>>> the clock in B is slower than the clock in A.
>> Nope. What experiment shows is :
>> - from the point of view of frame B, clocks in A are running slower
>> - from the point of view of frame A, clocks in B are running slower
> Well, you specified in the beginning that B was moving relative to A,
> so thus far, we are only looking at it from the point of view of frame
> A.
Sigh... Obfuscating and evading again... The whole point of
transformations is to study what is observed in a frame (here the B one)
given what is observed in another one (here the A one).
>> Note that this is as well what LTs predicts.
>>
>>> The clock in B
>>> continues to move away from the clock in A. �When the clock in B reads
>>> n'=a/c, the bell in B rings.
>> You're going more insane every day : what you just write implies this :
>> - For an observer in A, the bell at rest in A rings, the one at rest in
>> � �B don't
>> - For an observer in B, the bell at rest in B rings. the one at rest in
>> � �A don't
> Well, if you are using the Lorentz equations, only one bell is going
> to ring because of relativity of simultaneity.
Wrong. I've shown you EXACTLY what Lorentz equations show.
> The rays of light will
> not meet at the origins of both frames of reference. So from A, an
> observer will only observe the bell in A to ring, from B an observer
> will only observe the bell in B to ring.
> With my equations, the bell in A will ring first, then the bell in
> B. You will hear both bells in both frames of reference.
>>> � � Now, YBM, explain the same events using the Lorentz equations.
>> I did here :http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de...
>>
>>>> There is no such absurdities in SR : for SR, the bell
>>>> rings in both frames... It just happens that in frame
>>>> B they were emitted at coordinates
>>>> (-a/sqrt(1-v^2/c^2),0,0) at time va/(c^2*sqrt(1-v^2/c^2))
>>>> for the "left" light ray, and :
>>>> (a/sqrt(1-v^2/c^2),0,0) at time -va/(c^2*sqrt(1-v^2/c^2))
>
> Uh huh.
Is there something you don't understand above ?
> But if you put a bell at the origin of each frame of
> reference, what will happen?
> Robert B. Winn
>> So it is quite certain that you didn't read.
>>
> No, I read it, YBM. So explain what happens if there are two bells,
> one at the origin of each frame of reference.
Look carefully at the coordinates of events I've provided
above. Then figure out yourself :
- at what time in B does both light rays have been emitted ?
are they the same ?
- are the positions of the points of emission in frame B symetric
with respect to the origin of B ?
- given that velocities of the light rays are c and -c in B
are they going to meet at the origin of B ?
rbwinn
compared to a clock in B which shows light to be traveling at a speed
of c, and the other shows where light from x'=a is as compared to a
clock in B which shows light to be traveling at a speed of c if both
beams of light are traveling toward the origin of B. Sorry you don't
like where this light is, YBM. The light will meet at the origin of B
at a time of n'=a/c.
Now getting back to the two bells that ring when light from both
directions reaches them, you will notice that from frame of reference
A, the Lorentz equations show only the bell at the origin of A
ringing. According to your rules, the bell at the origin of B will
not ring.
I say both bells will ring.
Robert B. Winn
rbwinn
require, the beams of light are not emitted at the same time in B.
Therefore, according to you only the bell at the origin of A will ring
as observed by the observer in A.
> - are the positions of the points of emission in frame B symetric
> � �with respect to the origin of B ?
> - given that velocities of the light rays are c and -c in B
I say they will, but the Lorentz equations say they will not. If you
have a bell at the origin of A and a bell at the origin of B, only one
of the bells will ring.
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Robert B. Winn
YBM
>>>>> Right. The observer in A has one clock that shows t, and the observer
>>>>> in B has one clock that shows n'.
>>>> Which one : n'=t(1-v/c) or n'=t(1+v/c) ?- Hide quoted text -
>>> Well, if you are talking about an event in B, the n' values would be
>>> the same, and the two t times wouled be different. �
>> Wrong : t(1-v/c) and t(1+v/c) are different except at t=0 (or if v=0).
>> AGAIN : which n'=t(1-v/c) or n'=t(1+v/c) do you choose ?
>>
> They are both correct. One shows where light from x'= -a is as
> compared to a clock in B which shows light to be traveling at a speed
> of c, and the other shows where light from x'=a is as compared to a
> clock in B which shows light to be traveling at a speed of c if both
> beams of light are traveling toward the origin of B.
If they are "both correct" it means that from the point of view
of B, the events "left light ray arrives at the origin of A" and
"right light ray arrives at the origine of A" have different
time coordinates
n'=a/c(1-v/c) for the right one
n'=a/c(1+v/c) for the left one
so they don't arrive on A at the same instant (let's forget
for a moment how stupid is to compare events with coordinates
build from clocks being broken first, and broken on a different
way then), so for B, the bell in A don't ring.
> Sorry you don't
> like where this light is, YBM. The light will meet at the origin of B
> at a time of n'=a/c.
> Now getting back to the two bells that ring when light from both
> directions reaches them, you will notice that from frame of reference
> A, the Lorentz equations show only the bell at the origin of A
> ringing. According to your rules, the bell at the origin of B will
> not ring.
> I say both bells will ring.
????!!!!! You'd better think before writing such absurdities...
rbwinn
> ...
> >>>>> Right. The observer in A has one clock that shows t, and the observer
> >>>>> in B has one clock that shows n'.
> >>>> Which one : n'=t(1-v/c) or n'=t(1+v/c) ?- Hide quoted text -
> >>> Well, if you are talking about an event in B, the n' values would be
> >>> the same, and the two t times wouled be different.
> >> Wrong : t(1-v/c) and t(1+v/c) are different except at t=0 (or if v=0).
> >> AGAIN : which n'=t(1-v/c) or n'=t(1+v/c) do you choose ?
>
> > They are both correct. � One shows where light from x'= -a is as
> > compared to a clock in B which shows light to be traveling at a speed
> > of c, and the other shows where light from x'=a is as compared to a
> > clock in B which shows light to be traveling at a speed of c if both
> > beams of light are traveling toward the origin of B. �
>
> If they are "both correct" it means that from the point of view
> of B, the events "left light ray arrives at the origin of A" and
> "right light ray arrives at the origine of A" have different
> time coordinates
> � � n'=a/c(1-v/c) for the right one
> � � n'=a/c(1+v/c) for the left one
> so they don't arrive on A at the same instant (let's forget
> for a moment how stupid is to compare events with coordinates
> build from clocks being broken first, and broken on a different
> way then), so for B, the bell in A don't ring.
Sorry, but it does. This can be proven from the fact that the bell
rings when the origin of B is a distance of vt from the origin of A.
From frame of reference B, when the origin of B is a distance of vt
from the origin of A, the bell at the origin of A rings. That is just
what happens.
> > Sorry you don't
> > like where this light is, YBM. �The light will meet at the origin of B
> > at a time of n'=a/c.
> > � �Now getting back to the two bells that ring when light from both
> > directions reaches them, you will notice that from frame of reference
> > A, the Lorentz equations show only the bell at the origin of A
> > ringing. �According to your rules, the bell at the origin of B will
> > not ring.
> > I say both bells will ring.
>
> ????!!!!! You'd better think before writing such absurdities...
Explain how the Lorentz equation show the light turning on the bell at
the origin of B as seen by an observer in A. According to relativity
of simultaneity, the rays of light are not emitted simultaneously in
B. They do not reach the origin of B at the same time. By your own
logic, the bell at the origin of B will not ring.
Robert B. Winn
YBM
This is not what your formulas say. We now have two different
theories from you, both absurd :
The one with formulas, saying that A won't ring as seen in frame B.
The one without formulas, saying that both bells will ring (as seen
by both frames).
>>> Sorry you don't
>>> like where this light is, YBM. �The light will meet at the origin of B
>>> at a time of n'=a/c.
>>> � �Now getting back to the two bells that ring when light from both
>>> directions reaches them, you will notice that from frame of reference
>>> A, the Lorentz equations show only the bell at the origin of A
>>> ringing. �According to your rules, the bell at the origin of B will
>>> not ring.
>>> I say both bells will ring.
>> ????!!!!! You'd better think before writing such absurdities...
>
> Explain how the Lorentz equation show the light turning on the bell at
> the origin of B as seen by an observer in A. According to relativity
> of simultaneity, the rays of light are not emitted simultaneously in
> B. They do not reach the origin of B at the same time. By your own
> logic, the bell at the origin of B will not ring.
right. The absurdity is "both bells will ring". BTW, explaining why
it is absurd to say that the ring at the origin of B will ring is
not a job for a physicist but for a physician (or psychiatrist).
YBM
right.
> Therefore, according to you only the bell at the origin of A will ring
> as observed by the observer in A.
>> - are the positions of the points of emission in frame B symetric
>> � �with respect to the origin of B ?
> The points of emission are equal distances from the origin of B.
right.
>> - given that velocities of the light rays are c and -c in B
>> � �are they going to meet at the origin of B ?
> I say they will, but the Lorentz equations say they will not.
At least you should notice that the Lorentz equations are
perfectly coherent.
> If you
> have a bell at the origin of A and a bell at the origin of B, only one
> of the bells will ring.
You mean according to LT, right ?
So, now you "theory" says : both bells will ring, right ?
(even if this not what your formulas implies).
This is sooooo utterly absurd that I suggest you to take
a rest and think a bit about it.
rbwinn
>
Well, if you scientists do not believe the Galilean transformation
equations, then there is no way to convince you that the bells ring,
but I will make one more effort. First we consider the two bells from
the frame of reference of A. Light is emitted at -a and a. the light
reaches the origin of A at t=t'=a/c, and the bell at the origin of A
rings. According to t'=t, the light reaches the origin of A at t'=a/c
from both directions. An observer in B using t'=t will observe the
light to reach the origin of A and ring the bell. This was the
interpretation that scientists had of the Galilean transformation
equations until 1887. It is good enough for what we are doing here.
Using t'=t, the observer in B has no way of determining that the bell
at the origin of B will ring.
Now we will consider the two frames of reference from B. The
Galilean transformation equations for this observation are
x=x'-v't'
y=y'
z=z'
t=t'
B is the frame of reference at rest and A is the frame of
reference in motion. A is moving relative to B with a velocity of
v'=(-v). So now we consider what the light does in each frame of
reference according to these equations. In B, the light from the -a
goes from x'=-a to the origin of B. The light from a goes from x'=a
to the origin of B. The light reaches the origin of B at a time of
t'=a/c, and the bell at the origin of B rings. An observer in A using
a t=t' clock will see the light from both directions reach the origin
of B at t=t'=a/c and will observe the bell in B ring, but has no means
of observing the light reach the clock at the origin of A using t=t',
so will not observe that bell to ring.
This is the same basic reasoning you are using with the Lorentz
equations. If you can prove that both bells ring from one frame of
reference with the Lorentz equations, I want to see the proof.
Otherwise, I say you are just using the same reasoning I used above
with the old interpretation of the Galilean transformation equations.
The only thing I did different was to say that A is a preferred frame
of reference when observed from A, B is a preferred frame of reference
when observed from B. In Galilean ether theory, A was always the
preferred frame of reference.
All n' does is show that from B, light is observed to meet at the
origin of B.
Robert B. Winn
rbwinn
>
Well, I have thought about it. The Lorentz equations say both bells
will ring, but only if you consider the problem from both frames of
reference, the same way I did with the Galilean transformation
equations, treating each frame in turn as a preferred frame of
reference.
Both bells will ring by anyone's equations. But the objections you
make to my interpretation of the Galilean transformation equations can
also be applied to the Lorentz equations because under their current
interpretation, an observer in one frame of reference cannot see both
bells ring without a consideration from the other frame of reference.
Robert B. Winn
Robert B. Winn
YBM
>>>>> Sorry you don't
>>>>> like where this light is, YBM. The light will meet at the origin of B
>>>>> at a time of n'=a/c.
>>>>> Now getting back to the two bells that ring when light from both
>>>>> directions reaches them, you will notice that from frame of reference
>>>>> A, the Lorentz equations show only the bell at the origin of A
>>>>> ringing. According to your rules, the bell at the origin of B will
>>>>> not ring.
>>>>> I say both bells will ring.
>>>> ????!!!!! You'd better think before writing such absurdities...
>>> Explain how the Lorentz equation show the light turning on the bell at
>>> the origin of B as seen by an observer in A. �According to relativity
>>> of simultaneity, the rays of light are not emitted simultaneously in
>>> B. �They do not reach the origin of B at the same time. �By your own
>>> logic, the bell at the origin of B will not ring.
>> right. The absurdity is "both bells will ring". BTW, explaining why
>> it is absurd to say that the ring at the origin of B will ring is
>> not a job for a physicist but for a physician (or psychiatrist).- Hide quoted text -
>>
> Well, if you scientists do not believe the Galilean transformation
> equations, then there is no way to convince you that the bells ring,
Galilean transformations don't shows that idiocy that both the bells
would ring.
> but I will make one more effort. First we consider the two bells from
> the frame of reference of A. Light is emitted at -a and a. the light
> reaches the origin of A at t=t'=a/c, and the bell at the origin of A
> rings. According to t'=t, the light reaches the origin of A at t'=a/c
> from both directions. An observer in B using t'=t will observe the
> light to reach the origin of A and ring the bell. This was the
> interpretation that scientists had of the Galilean transformation
> equations until 1887. It is good enough for what we are doing here.
> Using t'=t, the observer in B has no way of determining that the bell
> at the origin of B will ring.
He has even a way to determine that the ring at the origin of B won't
ring.
> Now we will consider the two frames of reference from B. The
> Galilean transformation equations for this observation are
>
> x=x'-v't'
> y=y'
> z=z'
> t=t'
>
> B is the frame of reference at rest and A is the frame of
> reference in motion. A is moving relative to B with a velocity of
> v'=(-v). So now we consider what the light does in each frame of
> reference according to these equations. In B, the light from the -a
> goes from x'=-a to the origin of B. The light from a goes from x'=a
> to the origin of B. The light reaches the origin of B at a time of
> t'=a/c, and the bell at the origin of B rings. An observer in A using
> a t=t' clock will see the light from both directions reach the origin
> of B at t=t'=a/c and will observe the bell in B ring, but has no means
> of observing the light reach the clock at the origin of A using t=t',
> so will not observe that bell to ring.
This does not describe the same setup : here you've made the two
light rays to travel at speed c in B... BUT the Galilean Transformation
says that if the two light rays travel at speed c in A (which is a
*given* information of the problem I propose) *then* they will
travel at speed c-v and c+v in frame B.
> This is the same basic reasoning you are using with the Lorentz
> equations. If you can prove that both bells ring from one frame of
> reference with the Lorentz equations, I want to see the proof.
How stupid are you ? How many times should I write that :
- SAYING THAT BOTH BELLS WILL RING IS ABSURD WHATEVER TRANSFORMATION
YOU USE !
- LORENTZ TRANSFORMATION DOES NOT PREDICT THAT BOTH BELLS RING !
> Otherwise, I say you are just using the same reasoning I used above
> with the old interpretation of the Galilean transformation equations.
> The only thing I did different was to say that A is a preferred frame
> of reference when observed from A, B is a preferred frame of reference
> when observed from B. In Galilean ether theory, A was always the
> preferred frame of reference.
"Galilean ether theory" ?? What kind of stupidity are you, again,
making up in order to evade how absurd you are ?
> All n' does is show that from B, light is observed to meet at the
> origin of B.
So two impulsions of light will meet in two distinct places and time,
right ?
Isn't that ... stupid to say the least ?
But, wait a minute, it is even worse than that :
... v being unspecified, your idiocy that light rays will meet in two
places "origin of A" and "origin of B", it true for any v one could
choose to consider... so, according to you, it's not only at two
different places that the two light rays will meet, but in infinite
number of different places... Just set up an infinite number of
bells at appropriate speed and only two light pulses will be enough
to make them ALL ring.
Stupidity really can be infinite, Einstein was right !
YBM
DEFINITELY NOT !
This is really amazing ! So far :
- you lied about what Galilean Transformations say
- you lied about what your formulas say
- you lied about what LTs say
> but only if you consider the problem from both frames of
> reference, the same way I did with the Galilean transformation
> equations, treating each frame in turn as a preferred frame of
> reference.
there is no preferred frame of reference in GT.
> Both bells will ring by anyone's equations. But the objections you
> make to my interpretation of the Galilean transformation equations can
> also be applied to the Lorentz equations because under their current
> interpretation, an observer in one frame of reference cannot see both
> bells ring without a consideration from the other frame of reference.
Wrong, silly, stupid, meaningless, ill, all of that is so few words !
rbwinn
I would expect that this is one experiment that scientists are not
going to conduct.
> > Otherwise, I say you are just using the same reasoning I used above
> > with the old interpretation of the Galilean transformation equations.
> > The only thing I did different was to say that A is a preferred frame
> > of reference when observed from A, B is a preferred frame of reference
> > when observed from B. �In Galilean ether theory, A was always the
> > preferred frame of reference.
>
> "Galilean ether theory" ?? What kind of stupidity are you, again,
> making up in order to evade how absurd you are ?
>
> > � � �All n' does is show that from B, light is observed to meet at the
> > origin of B.
>
> So two impulsions of light will meet in two distinct places and time,
> right ?
Two waves of light will meet in two different frames of reference. As
I understand it, you do not like relativity of time.
> Isn't that ... stupid to say the least ?
It does not seem stupid to me. Why do you think it is stupid?
> But, wait a minute, it is even worse than that :
>
> ... v being unspecified, your idiocy that light rays will meet in two
> places "origin of A" and "origin of B", it true for any v one could
> choose to consider... so, according to you, it's not only at two
> different places that the two light rays will meet, but in infinite
> number of different places... Just set up an infinite number of
> bells at appropriate speed and only two light pulses will be enough
> to make them ALL ring.
How do you figure that? The beams of light will meet at the midpoint
between them in A, and will meet at the midpoint between them in B.
What did you think they did?
> Stupidity really can be infinite, Einstein was right !- Hide quoted text -
Einstein would probably know. He was the one who thought that there
was a length contraction.
Robert B. Winn
rbwinn
>
> - Show quoted text -
I forgot something about the Lorentz equations. OK, they show that
only the bell at the origin of A would ring. The ether theory of the
Galilean transformation equations shows that A would be a preferred
frame of reference, regardless of what you believe about it. My
question is, why don't you run the experiment?
Why go around in doubt? I say both bells would ring, you say only
the one in A would ring. So run the experiment and find out.
Robert B. Winn
YBM
>> So two impulsions of light will meet in two distinct places and time,
>> right ?
> Two waves of light will meet in two different frames of reference. As
> I understand it, you do not like relativity of time.
Relativity of space (which is present in GT and LT) means that
two events which happen at the same spacial coordinates in
a frame won't necessarily be at the same spacial coordinates in
another frame.
Relativity of time (which is not in GT but in LT) means that
two events that are simultaneous in a frame, won't be necessarily
be simultaneous in another frame.
A typical example of the former is the light rays emissions
in my example.
>> Isn't that ... stupid to say the least ?
>
> It does not seem stupid to me. Why do you think it is stupid?
You should ask a psychiatrist, you have a cognitive dissonance
at the roots of human rationality (not only human, btw, even animals
would represent this kind of situation correctly or they wouldn't
survive).
Things are moving in space, in straight lines, they meet once, at a
given place, then continue (or not) their journey. What you are saying
is they meet at several times at different places (as a matter of
fact, an infinity of places).
Take two coins, throw them against each other on a table. Count
down the number of places they meet.
>> But, wait a minute, it is even worse than that :
>>
>> ... v being unspecified, your idiocy that light rays will meet in two
>> places "origin of A" and "origin of B", it true for any v one could
>> choose to consider... so, according to you, it's not only at two
>> different places that the two light rays will meet, but in infinite
>> number of different places... Just set up an infinite number of
>> bells at appropriate speed and only two light pulses will be enough
>> to make them ALL ring.
>
> How do you figure that? The beams of light will meet at the midpoint
> between them in A, and will meet at the midpoint between them in B.
Don't forget that what you assert for the B frame should then be true
for a lot of different B frames : just consider different values for
"v"! The light pulse then would meet in a infinite number of places,
at a infinite number of times. Quite an achievement !
> What did you think they did?
They meet at the origin of A. Period.
>> Stupidity really can be infinite, Einstein was right !- Hide quoted text -
>
> Einstein would probably know. He was the one who thought that there
> was a length contraction.
I don't think that someone able to assert that two light rays should
meet at several places is especially the guy to ask for about stupidity.
Well he is quite certainly a guy to study for someone interested by
stupidity as a state of mind.
YBM
> I forgot something about the Lorentz equations. OK, they show that
> only the bell at the origin of A would ring. The ether theory of the
> Galilean transformation equations shows that A would be a preferred
> frame of reference, regardless of what you believe about it. My
> question is, why don't you run the experiment?
Ask your dog, if you have one. Throw something so that he brings it
back to you, proving that he know that linear trajectories only
intersect once.
Believe me : this is the first time I meet someone less clever than
all animals on Earth, including insects, fish and Acarinas.
I would have believed it, if I hadn't read you.
I've read some books about incredible mental disability, I've seen
nothing there about yours. You should definitely contact a department
of psychology in an University near you : you'd be a fascinating
subject of study, and you'd help them to improve science.
> Why go around in doubt? I say both bells would ring, you say only
> the one in A would ring. So run the experiment and find out.
You should ask a psychiatrist, you have a cognitive dissonance
rbwinn
So you are saying that coins are light. I saw a man on television one
time who claimed to be a witch who said something identical. He said
that all material objects are made out of light.
So I guess if they can contract in length, they can do all sorts of
things when you witches need them to.
Anyway, it all comes down to the original problem. You had two frames
of reference, A and B, light was being emitted at -a and +a. Now you
are introducing relativity of simultaneity and length contraction.
OK, we can talk about those things. With regard to where we agree, we
seem to agree that if light is emitted at x=-a and x=+a at t=0, then
the light will meet at the origin of A and a special bell will ring.
That is all very special. Now what seems curious to me is that with
the Lorentz equations, if we say that light is emitted at -a and a in
A, what does it mean in B? We could also say that light is emitted at
-a and a in A when the origin of B is at the origin of A. So when the
origin of B is at A is not the same time that light is emitted at two
points which are not -a and a in B because you say that the length of
B is contracted.
But we can also say that in frame of reference B, light is
emitted at -a and a in B, which rings a bell at the origin of B. In
this case the light is emitted at two points in A which are not -a and
a at times that are not the time when the origin of B is at the origin
of A.
However, at low velocities, the Lorentz equations give
essentially the same values that I showed with the old interpretation
of the Galilean transformation equations. So if the velocity of B
relative to A is zero, both bells should ring according to either set
of equations. If B is moving at all, the Lorentz equations say that
the bell at the origin of B will not ring. So why not do the
experiment?
If you can get the bells to ring when they are sitting still, then
it should be impossible to get them both to ring with light if one of
them is moving. That seems like a fairly clear cut experiment to me.
Why do I have the feeling that we will never hear of it being
performed?
Robert B. Winn
rbwinn
Well, light is energy. I don't pretend to know everything about
energy. Some people do. I saw a man on television one time who
claimed to be a witch. He seemed to agree with you about light. He
had a paper cone which he said was made out of light. You have two
coins made out of light. Where did you study witchcraft?
Robert B. Winn
PD
:>)
And here Bobby pulls his left hand above table level just long enough
for you to see the chain he's yanking. Perhaps he also winked a little
at you. Did you see it?
PD
rbwinn
>
> - Show quoted text -
Quiet, PD, YBM was getting ready to tell us about the distance
contraction.
Robert B. Winn
YBM
No. I'm saying that coins and light (well, to be precise : intersection
of a light front with a radial line) have something in common : a
trajectory. A trajectory is a continuous fonction associating a
position to a instant in time.
What you pretend is that two linear trajectories intersect two
times, even an infinite number of times.
PD is right : there is definitely no way to make you understand the
most basic issues in physics. You mental disability is beyond everything
I've never seen. I'm not a psychiatrist : I give up on your case.
rbwinn
>
> - Show quoted text -
Well, I definitely do not believe in the length contraction. What you
are saying is that anyone who does not believe in the length
contraction is insane. So what do you want to do now?
I say it is easier to resolve the mathematics with energy than to try
to crush a train the way Einstein said he was doing.
I am sorry that you do not like relativity of time, but you are free
to believe anything you want to believe.
That is the difference between how I do things and how you do things.
If you find a psychiatrist who wants to talk to me, send him by. The
only people I enjoy talking to more than psychiatrists are lawyers.
Robert B. Winn
rbwinn
>
> - Show quoted text -
Well, as I said, just run the experiment. If only one bell rings,
then you are right.
Robert B. Winn