I once explained to a speaker that additions of relativistic speeds were
not done in a common way, and that for example 0.5c+0.5c did not make c.
This Internet user refused to believe me.
For what? Because it is very difficult to give water to a donkey who is
not thirsty, and who categorically refuses to understand or discuss.
I think that this makes most of the speakers smile, because they know A LITTLE realtivity, and if they do not necessarily know the general
formula
for adding relativistic speeds, they at least know the longitudinal
formula that is w=(v +u)/(1+v.u/c²) or here w=0.8c.
But we must go further. Physicists don't make this kind of mistake, but
they do make others. I told Paul B. Andersen that his magnificent
integration formula
<http://news2.nemoweb.net/jntp?EKV4LWfwyF4mvRIpW8X1iiirzQk@jntp/Data.Media:1>
was incorrect PHYSICALLY even though mathematically it was obviously
perfect.
Paul doesn't want to believe me. This confuses him.
However he is wrong and I pointed out to him that if we could integrate
all the proper times, to obtain the sum of the total proper time, we
could
not do it with improper times, the sum of which segment by segment was greater than the total evolution.
A bit like realtivist speed additions where the sum is not equal to the common, mathematical sum.
Paul doesn't want to believe me, because he wasn't taught that way, and
he complains about me.
Why doesn't he complain about those who taught him incorrectly?
R.H.
On Fri, 19 Jul 2024 19:51:32 +0000, Richard Hachel wrote:
formula that is w=(v +u)/(1+v.u/c²) or here w=0.8c.
<http://news2.nemoweb.net/jntp?EKV4LWfwyF4mvRIpW8X1iiirzQk@jntp/Data.Media:1>
That's not a valid comparison, Dr. H.
It seems to me that he complained properly.
On Fri, 19 Jul 2024 19:51:32 +0000, Richard Hachel wrote:
I once explained to a speaker that additions of relativistic speeds were
not done in a common way, and that for example 0.5c+0.5c did not make c.
This Internet user refused to believe me.
For what? Because it is very difficult to give water to a donkey who is
not thirsty, and who categorically refuses to understand or discuss.
I think that this makes most of the speakers smile, because they know A
LITTLE realtivity, and if they do not necessarily know the general
formula
for adding relativistic speeds, they at least know the longitudinal
formula that is w=(v +u)/(1+v.u/c²) or here w=0.8c.
But we must go further. Physicists don't make this kind of mistake, but
they do make others. I told Paul B. Andersen that his magnificent
integration formula
<http://news2.nemoweb.net/jntp?EKV4LWfwyF4mvRIpW8X1iiirzQk@jntp/Data.Media:1>
was incorrect PHYSICALLY even though mathematically it was obviously
perfect.
Paul doesn't want to believe me. This confuses him.
However he is wrong and I pointed out to him that if we could integrate
all the proper times, to obtain the sum of the total proper time, we
could
not do it with improper times, the sum of which segment by segment was
greater than the total evolution.
A bit like realtivist speed additions where the sum is not equal to the
common, mathematical sum.
That's not a valid comparison, Dr. H.
Paul doesn't want to believe me, because he wasn't taught that way, and
he complains about me.
Why doesn't he complain about those who taught him incorrectly?
R.H.
It seems to me that he complained properly. You have made an accusation without a justification. If one can't use "improper" times to integrate
a proper time, what does one use? And if using "improper" times is
wrong in this case, that throws all of calculus in doubt.
Frankly, I found that physicists get into more trouble when they ignore mathematical rules.
I once explained to a speaker that additions of relativistic speeds
were not done in a common way, and that for example 0.5c+0.5c did not
make c.
On 2024-07-19 19:51:32 +0000, Richard Hachel said:
I once explained to a speaker that additions of relativistic speedsThat is a clear indication that you cannot explain.
were not done in a common way, and that for example 0.5c+0.5c did not
make c.
Mikko <mikko.levanto@iki.fi> wrote or quoted:
On 2024-07-19 19:51:32 +0000, Richard Hachel said:
I once explained to a speaker that additions of relativistic speedsThat is a clear indication that you cannot explain.
were not done in a common way, and that for example 0.5c+0.5c did not
make c.
When nobody says otherwise, the symbols in "0.5c+0.5c" mean what
they usually do (from math class back in school), so that sum is
still c. Thing is, it doesn't give you the speed of a guy relative
to Earth in a situation where he's cruising at 0.5c on one of
those moving walkways that's also zipping along at 0.5c itself.
I'm sorry, Dr. H., but Paul's equation makes perfect sense
Le 20/07/2024 à 04:47, hitlong@yahoo.com (gharnagel) a écrit :
I'm sorry, Dr. H., but Paul's equation makes perfect sense
That is what I am saying.
This is why the trap is colossal.
Simply perfect mathematics applied to abstract physics no longer has any practical use.
R.H.
On Sat, 20 Jul 2024 12:23:21 +0000, Richard Hachel wrote:
Le 20/07/2024 à 04:47, hitlong@yahoo.com (gharnagel) a écrit :
I'm sorry, Dr. H., but Paul's equation makes perfect sense
That is what I am saying.
This is why the trap is colossal.
Simply perfect mathematics applied to abstract physics no longer has any
practical use.
R.H.
But you haven't explained "the trap." Inventing equations out of thin
air
is not an explanation. You have TWO points to prove: How to derive
your
equations and how they agree with reality. You have done neither.
You have done neither.
Le 20/07/2024 à 14:55, hitlong@yahoo.com (gharnagel) a écrit :
You have done neither.
Vous plaisantez, j'espère?
R.H.
I told Paul B. Andersen that his magnificent
integration formula
<http://news2.nemoweb.net/jntp?EKV4LWfwyF4mvRIpW8X1iiirzQk@jntp/Data.Media:1>
was incorrect PHYSICALLY even though mathematically it was obviously
perfect.
Paul doesn't want to believe me. This confuses him.
However he is wrong and I pointed out to him that if we could integrate
all the proper times, to obtain the sum of the total proper time, we
could not do it with improper times, the sum of which segment by segment
was greater than the total evolution.
I suppose this means that the equation is incorrect according to
your theory.
So please show the equation for the proper time as a function of
the velocity and the time in the inertial system, which is
correct physically and mathematically according to your theory.
<http://news2.nemoweb.net/jntp?EKV4LWfwyF4mvRIpW8X1iiirzQk@jntp/Data.Media:1>
was incorrect PHYSICALLY even though mathematically it was obviously
perfect.
Of course it is correct physically and mathematically according to SR.
Le 20/07/2024 à 22:05, "Paul.B.Andersen" a écrit :
So please show the equation for the proper time as a function of
the velocity and the time in the inertial system, which is
correct physically and mathematically according to your theory.
I've already given all that away for a long time.
I take the main equations again.
To: time in the laboratory (or terrestrial), observable time.
Tr: proper time, tau.
a: acceleration
x=(1/2)a.Tr²
x=(c²/a)[sqrt(1+a²To²/c²) -1]
To=(x/c)sqrt(1+2c²/ax)
Tr=sqrt(2x/a)
To=Tr.sqrt(1+(1/4)a².Tr²/c²)
R.H.
Den 20.07.2024 23:55, skrev Richard Hachel:
But you know of course that all clocks in the same time zone
are synchronous.
You have said strange statements about synchronisation of clocks.
Le 21/07/2024 à 21:26, "Paul.B.Andersen" a écrit :
Den 20.07.2024 23:55, skrev Richard Hachel:
But you know of course that all clocks in the same time zone
are synchronous.
No, and I've kept telling you that.
Two stationary watches are not synchronous and never will be.
You are, once again, confusing anisochronia and dyschronotropia.
Two watches placed in two different places, stationary between them, obviously beat at the same speed.
It's logic.
Den 20.07.2024 23:55, skrev Richard Hachel:
So: in an inertial system K, a clock C is in inertial motion.
t is the time in K.
A and B are two stationary, synchronous clocks in K.
At t = t₁
C
A B
--|---------------------|------> x
0 L
Clock A is showing t₁, clock C is adjacent to A and is set to zero.
At t = t₂
C
A B
--|---------------------|------> x
0 L
Clock B is showing t₂, clock C is adjacent to B and is showing τ
Let L = 0.0001 light second = 29979 m
Let (t₂ - t₁) = T = 125 μs
v = L/T = 239833966.4 m/s = 0.8c
w = L/τ
What kind of time is T ?
What kind of speed is v ?
What is τ ? (equation, value and type of time)
What kind of speed is w ?
What is the physical significance of w?
Can it say anything about the position of C at the time τ ?
Le 21/07/2024 à 21:26, "Paul.B.Andersen" a écrit :
You have said strange statements about synchronisation of clocks.
Clock synchronization.
I have already told you hundreds of times over the past 40 years that
it is impossible to synchronize two clocks placed in different places.
If I place a very precise atomic clock, one on this bench, one on this
table, the other on the mantelpiece, I could never synchronize them absolutely, because it is impossible and an abstract representation of
the notion of simultaneity of present time.
In the best case of my synchronization, each watch will delay the other
by t=x/c unavoidable, universal, physical delay.
We will say: “How does GPS work?”
GPS works on the idea of an abstract point, located outside our universe, and placed orthogonally and ideally almost to infinity, such
that an impulse coming from it will be returned to it at the same time
by all the components of our universe.
It is on such an “ideal and abstract” point that GPS works, giving the illusion of a real universal present time.
Le 21/07/2024 à 21:26, "Paul.B.Andersen" a écrit :
Den 20.07.2024 23:55, skrev Richard Hachel:
So: in an inertial system K, a clock C is in inertial motion.
t is the time in K.
A and B are two stationary, synchronous clocks in K.
At t = t₁
C
A B
--|---------------------|------> x
0 L
Clock A is showing t₁, clock C is adjacent to A and is set to zero.
At t = t₂
C
A B
--|---------------------|------> x
0 L
Clock B is showing t₂, clock C is adjacent to B and is showing τ
Let L = 0.0001 light second = 29979 m
Let (t₂ - t₁) = T = 125 μs
v = L/T = 239833966.4 m/s = 0.8c
w = L/τ
What kind of time is T ?
T=To (observable time in the laboratory, time observable but not real
mesured by two anisochronic clocks A and B).
What kind of speed is v ?
Observable speed Vo.
What is τ ? (equation, value and type of time)
Proper time of C (real time).
What kind of speed is w ?
Real speed (Vr)
What is the physical significance of w?
Vr=AB/τ
It is a physical notion, but not very important in Galilean relativistic physics, on the other hand, which becomes fundamental in the physics of accelerated mobiles and which we can no longer do without as soon as we
leave basic relativistic physics.
I recognize that it is quite strange to say that it is in the laboratory
that we measure the real distance traveled, and at the level of the
particle that we measure the real time to travel it.
Can it say anything about the position of C at the time τ ?
x = Vo.To = Vr.Tr = Vapp.Tapp
x=Vr.τ
Le 21/07/2024 à 21:26, "Paul.B.Andersen" a écrit :
You have said strange statements about synchronisation of clocks.
Clock synchronization.
I have already told you hundreds of times over the past 40 years that it
is impossible to synchronize two clocks placed in different places.
If I place a very precise atomic clock, one on this bench, one on this
table, the other on the mantelpiece, I could never synchronize them absolutely, because it is impossible and an abstract representation of
the notion of simultaneity of present time.
In the best case of my synchronization, each watch will delay the other
by t=x/c unavoidable, universal, physical delay.
We will say: “How does GPS work?”
GPS works on the idea of an abstract point, located outside our universe, and placed orthogonally and ideally almost to infinity, such
that an impulse coming from it will be returned to it at the same time
by all the components of our universe.
It is on such an “ideal and abstract” point that GPS works, giving the illusion of a real universal present time.
What kind of speed is w ?
Real speed (Vr)
w = L/τ = 399723277 m/s = (4/3)c
This is the "proper velocity".
But is it "a real speed"?
Den 21.07.2024 22:34, skrev Richard Hachel:
In the best case of my synchronization, each watch will delay theother
by t=x/c unavoidable, universal, physical delay.
It is obviously impossible to make two clocks side by side show
the same with infinite precision, there will always be a difference.
For atomic clocks this difference may be less than 1 ns,
for say - wristwatches it will be less than 1 second.
Clock C is moving with the speed v relative to clock A and B,
and clock A and B are moving with the speed v relative to clock C.
It is simply a distance in an inertial system divided
with the proper time of a clock to travel the distance.
It is a speed, but not the speed of a physical object relative
to another physical object.
Den 21.07.2024 23:06, skrev Richard Hachel:
The "proper velocity" is never used in SR.
What do you call the phenomenon that when you look at the clocks
on your table and on your mantelpiece, they always show the same?
(to within the precision with which you have set the clocks.)
Please address what I wrote:
You know of course that all clocks in the same time zone
are synchronous. In France and Norway clocks are currently
showing GMT + 2 hour, so my clock and your clock are actually
synchronous.
Please explain why our clocks are NOT synchronous.
(To within few seconds)
Den 21.07.2024 23:06, skrev Richard Hachel:
Let's look at the LHC again.
The length of the circuit is L = 27 k, γ = 7460
The real speed of the proton in the lab frame is
v = 0.999999991·c
The real time measured in the lab frame for the proton to go
around the circuit is
T = L/v ≈ 90 μs
The proper time of a proton to go around the circuit is
τ = T/γ ≈ 12 ns
"proper speed" = L/τ = 6947c
This "speed" is not the speed of a proton or anything else.
The proton is moving at the speed 0.999999991·c relative to the lab,
and the lab is moving at the speed 0.999999991·c relative to the proton.
And you thought that the real speed speed of the proton in the lab frame
was 6947c and therefore you "tell them [the physicists at CERN] that
the proton rotates 78 million times per second".
Which is 6933 times the real number, ≈ 11.25 thousand times per second.
It is obviously impossible to make two clocks side by side show
the same with infinite precision, there will always be a difference.
For atomic clocks this difference may be less than 1 ns,
for say - wristwatches it will be less than 1 second.
As long as the difference is less than the precision of your
measurements, the clocks can be considered to be synchronous.
Practical examples:
100 m sprint:
Two synchronous clocks at start and finish line.
The precision of the measurements is 0.01 second
So the clocks must be synchronous to within 10 ms.
Tour de France.
Start and finish line may be ~200 km from each other.
The precision of the measurement is 1 second.
So the clocks at the start and the finish must
be synchronous to within 1 second.
Do you accept this, or are you still insisting that it
is impossible to have clocks synchronous to within
the precision of the actual measurement?
On Mon, 22 Jul 2024 19:37:17 +0000, Paul.B.Andersen wrote:
Den 21.07.2024 22:34, skrev Richard Hachel:
other
In the best case of my synchronization, each watch will delay the
by t=x/c unavoidable, universal, physical delay.
"Dr. H." is confusing the issue by bringing up the time delay of a
remote
clock. In SR, one has observers at every location who records the time
shown on the local clock when a local event occurs. All the records are
then compared, so speed-of-light time delay is NOT an issue.
It is obviously impossible to make two clocks side by side show
the same with infinite precision, there will always be a difference.
For atomic clocks this difference may be less than 1 ns,
for say - wristwatches it will be less than 1 second.
Which is why "absolute" synchronization is a canard.
Den 22.07.2024 21:37, skrev Paul.B.Andersen:
You know of course that all clocks in the same time zone
are synchronous. In France and Norway clocks are currently
showing GMT + 2 hour, so my clock and your clock are actually
synchronous.
Please explain why our clocks are NOT synchronous.
(To within few seconds)
But I keep explaining it to you.
This is a property of space that can be called universal anisochrony.
This does not translate into the idea that the “plan of present time” so dear to physicists does not exist, it is a thought that seems logical to them, but it is an abstract thought.
What do you call the phenomenon that when you look at the clocks
on your table and on your mantelpiece, they always show the same?
(to within the precision with which you have set the clocks.)
If I place myself equidistant from the two watches, and they are correctly adjusted, that is to say that they beat at the same time, with great precision, I will notice that FOR ME (I beg you to do the (effort to understand myself, and this is why Iwrite in capital letters), FOR ME, they always mark the same time at the same moment.
This means that they are perfectly regulated, and that they have an identical chronotropy (because they are in the same stationary frame of reference).
they are inertial, stationary) but with a strange delay of 1 second.
It is obviously impossible to make two clocks side by side show
the same with infinite precision, there will always be a difference.
For atomic clocks this difference may be less than 1 ns,
for say - wristwatches it will be less than 1 second.
As long as the difference is less than the precision of your
measurements, the clocks can be considered to be synchronous.
Practical examples:
100 m sprint:
Two synchronous clocks at start and finish line.
The precision of the measurements is 0.01 second
So the clocks must be synchronous to within 10 ms.
Tour de France.
Start and finish line may be ~200 km from each other.
The precision of the measurement is 1 second.
So the clocks at the start and the finish must
be synchronous to within 1 second.
Do you accept this, or are you still insisting that it
is impossible to have clocks synchronous to within
the precision of the actual measurement?
You don't understand anything I'm saying...
Pfffff...
I'm not talking about technical precision, I'm talking about a real problem linked to the nature of space and time.
This problem is anisochrony, two identical and well-adjusted watches, which we slowly and in the same way separate over a distance of 300,000 km will be irremediably out of tune.
They will always have the same chronotropy (internal speed of the watch mechanism) and for me, who am at the center, they will always mark the same time.
But between them, there will be a real time difference due simply to the distance. This gap, absolutely real and unavoidable due to the nature of space and time, will be one second between these two watches. Each sees the other beat at the same speed (
Physicists do not seem to understand this property, and only understand the internal chronotropy when watches are in reciprocal motion, but that is not enough. We must also understand anisochrony, which is a real phenomenon, of the first degree,unavoidable, and which has nothing to do with the speed of light. Information propagates instantly, BUT the present moments did not correspond to the departure.
All the clocks in the the GPS system (satellite clocks, ground clocks)
are synchronous with the UTC (Coordinated Universal Time)
to within ~1 ns.
Your clock on your table is synchronous with the UTC+2 hours,
to within the precision you have set the clock.
The UTC is universal in the sense that it covers the whole Earth
and the space in its vicinity.
It is coordinated in the sense that it is defined at any point
on the Earth and in the space in Earth's vicinity.
It is real even if it is defined by man. It is no illusion.
All clocks on Earth and in the GPS-, GLONASS- and Galileo-satellites
are synchronous to the UTC or UTC+n hours.
It is a fact that you can synchronise clocks via the GPS.
The GPS receiver determines four entities, the time, altitude,
latitude and longitude. If the spatial position is within 1 m,
the time must be the UTC to within ~2 ns.
Yes, your GPS receiver does indeed determine the time to within few ns
of the UTC, it must do that to determine the position to few metres.
It is obviously no point in displaying the time with this precision
on the screen.
No comment to this, Richard?
Are you insisting that the GPS doesn't work because the satellite clocks can't be synchronous because of the nature of space an time?
Paul
Den 22.07.2024 23:37, skrev Richard Hachel:But why exactly?
But "proper speed" is never used in SR because:That's right.
Le 22/07/2024 à 21:34, "Paul.B.Andersen" a écrit :
Den 21.07.2024 23:06, skrev Richard Hachel:
Let's look at the LHC again.
The length of the circuit is L = 27 k, γ = 7460
The real speed of the proton in the lab frame is
v = 0.999999991·c
The real time measured in the lab frame for the proton to go
around the circuit is
T = L/v ≈ 90 μs
The proper time of a proton to go around the circuit is
τ = T/γ ≈ 12 ns
"proper speed" = L/τ = 6947c
Yes.
This "speed" is not the speed of a proton or anything else.
It is.
Calculate p=mVr and E=mc².sqrt(1+Vr²/c²)
The proton is moving at the speed 0.999999991·c relative to the lab,
and the lab is moving at the speed 0.999999991·c relative to the proton.
That what I say.
There is nothing illogical or abnormal there.
And you thought that the real speed speed of the proton in the lab frame
was 6947c and therefore you "tell them [the physicists at CERN] that
the proton rotates 78 million times per second".
Which is 6933 times the real number, ≈ 11.25 thousand times per second.
But no!
We can't say that if a proton spins once, it spins thousands of times,
that's stupid.
No one will ever tell you that.
Den 22.07.2024 23:37, skrev Richard Hachel:
When calculating velocities the distance and time must be measured
in the same frame of reference.
The distance in one frame and the time in another frame is not
a speed of anything if the frames are moving relative to each other.
Den 22.07.2024 23:37, skrev Richard Hachel:
Calculate p=mVr and E=mc².sqrt(1+Vr²/c²)
In the real word we have p = mvγ E = γmc²
these are extremely well confirmed.
Den 22.07.2024 23:37, skrev Richard Hachel:
Give it up, Richard.
The physicists at CERN measures that the proton 'rotates'
11.25 thousand times per second, you "tell them that the proton
rotates 78 million times per second."
Is the calculation (78 million)/(11.25 thousand) = 6933/1 too hard
for you?
If the proton is passing a point in the ring with the speed v
relative to the point, then the point in the ring is passing
the proton a the speed v relative to the proton.
Le 23/07/2024 à 23:29, "Paul.B.Andersen" a écrit :
Den 22.07.2024 23:37, skrev Richard Hachel:
Le 22/07/2024 à 21:34, "Paul.B.Andersen" a écrit :
Let's look at the LHC again.
The length of the circuit is L = 27 k, γ = 7460
The real speed of the proton in the lab frame is
v = 0.999999991·c
When calculating velocities the distance and time must be measured
in the same frame of reference.
The distance in one frame and the time in another frame is not
a speed of anything if the frames are moving relative to each other.
I know it may seem very strange (relativistic physics is a bit strange but unavoidable) but that's how it happens.
It is the observing frame of reference that has the right distance, but it is the particle or rocket that has the right time.
Observable time being only an anisochronous illusion.
The proton is moving at the speed 0.999999991·c relative to the lab,
and the lab is moving at the speed 0.999999991·c relative to the proton.
That what I say.
No, that is not what you say.
You say the proton is moving at the speed 6947c relative to the lab.
And you thought that the real speed speed of the proton in the lab frame >>>> was 6947c and therefore you "tell them [the physicists at CERN] that
the proton rotates 78 million times per second".
Which is 6933 times the real number, ≈ 11.25 thousand times per second.
Give it up, Richard.
The physicists at CERN measures that the proton 'rotates'
11.25 thousand times per second, you "tell them that the proton
rotates 78 million times per second."
Don't tell me you don't understand that the proton rotates 11.25 million times per second in the laboratory frame but 78 million times per second in the proton frame.
This is called time dilation.
Den 24.07.2024 00:19, skrev Richard Hachel:
Le 23/07/2024 à 23:29, "Paul.B.Andersen" a écrit :
Den 22.07.2024 23:37, skrev Richard Hachel:
Le 22/07/2024 à 21:34, "Paul.B.Andersen" a écrit :
Let's look at the LHC again.
The length of the circuit is L = 27 k, γ = 7460
The real speed of the proton in the lab frame is
v = 0.999999991·c
When calculating velocities the distance and time must be measured
in the same frame of reference.
The distance in one frame and the time in another frame is not
a speed of anything if the frames are moving relative to each other.
I know it may seem very strange (relativistic physics is a bit strange
but unavoidable) but that's how it happens.
It is the observing frame of reference that has the right distance,
but it is the particle or rocket that has the right time.
Observable time being only an anisochronous illusion.
It isn't only "strange", it is idiotic nonsense.
Measured in the lab frame the proton is moving around
the L = 27 km long ring in T = 90.0623065140618 μs.
The very real speed of the proton relative to the lab is
v = L/T = 0.999999991·c
Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :
Measured in the lab frame the proton is moving around
the L = 27 km long ring in T = 90.0623065140618 μs.
The very real speed of the proton relative to the lab is
v = L/T = 0.999999991·c
γ = 7460
Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v = L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
This should be blazingly obvious for anybody but complete morons:
If the proton is passing a point in the ring with the speed v
relative to the point, then the point in the ring is passing
the proton a the speed v relative to the proton.
This is the only interesting sentence in your post.
The rest is just nonsense or tautology.
Indeed, if the proton passes at Vo=0.999991 c (for example) at a point A
of the device, then the laws of physics state that point A passes at Vo=0.999991c.
If we transpose into real speed Vr, we have:
Vr=Vo/sqrt(1-Vo²/c)=235.7c
Likewise, this real speed is reciprocal.
In the proton frame, it is point A which passes near it at Vr=235.7c.
Now what does the global ring look like in the proton frame of
reference, and above all what is the trajectory of point A during one revolution?
This is a good relativistic physics question.
Have fun answering this question...
I hope you have a lot of fun.
Quite.
But your jokes aren't funny the umpteenth time they are told,
It is getting boring.
Paul
Your idiot guru Roberts could explain you that
"measured" doesn't necessarily mean "real"
in the liturgy of your moronic church.
Le 23/07/2024 à 22:04, "Paul.B.Andersen" a écrit :
Den 22.07.2024 23:55, skrev Richard Hachel:
Den 22.07.2024 21:37, skrev Paul.B.Andersen:
All the clocks in the the GPS system (satellite clocks,ground clocks)
are synchronous with the UTC (Coordinated Universal Time)
to within ~1 ns.
Your clock on your table is synchronous with the UTC+2 hours,
to within the precision you have set the clock.
The UTC is universal in the sense that it covers the whole Earth
and the space in its vicinity.
It is coordinated in the sense that it is defined at any point
on the Earth and in the space in Earth's vicinity.
It is real even if it is defined by man. It is no illusion.
All clocks on Earth and in the GPS-, GLONASS- and Galileo-satellites
are synchronous to the UTC or UTC+n hours.
It is a fact that you can synchronise clocks via the GPS.
The GPS receiver determines four entities, the time, altitude,
latitude and longitude. If the spatial position is within 1 m,
the time must be the UTC to within ~2 ns.
Yes,your GPS receiver does indeed determine the time to within few ns
of the UTC, it must do that to determine the position to few metres.
It is obviously no point in displaying the time with this precision
on the screen.
No comment to this, Richard?
Are you insisting that the GPS doesn't work because the satellite clocks
can't be synchronous because of the nature of space an time?
Damn Paul!
I say exactly the opposite.
I say that if GPS works, it is PRECISELY thanks to universal anisochrony.
This is what GPS measures, and it is thanks to this that, converting anisochrony into spatial metrics, they can practically give the position
to the nearest meter.
You know of course that all clocks in the same time zone
are synchronous. In France and Norway clocks are currently
showing GMT + 2 hour, so my clock and your clock are actually
synchronous.
Please explain why our clocks are NOT synchronous.
(To within few seconds)
But I keep explaining it to you.
This is a property of space that can be called universal anisochrony.
This does not translate into the idea that the “plan of present time” so dear to physicists does not exist, it is a thought that seems logical to them, but it is an abstract thought.
I interpret this to mean that watches in Norway and France are
not synchronous even if they both show GMT+2h
I notice that in your examples, you talk about airports or planes. It's absurd. It is clear that the laws of relativity do not apply at such low speeds, where Vr=Vo and Tr=To.
Relativity only exists if we go very far, or if we go very quickly, and
not in the common world.
Le 24/07/2024 à 20:44, "Paul.B.Andersen" a écrit :
Den 24.07.2024 15:08, skrev Richard Hachel
Now what does the global ring look like in the proton frame of reference, and above all what is the trajectory of point A during one revolution?
Irrelevant.
The point A is at any instant adjacent to the proton.
We consider an arbitrary instant I.
Let K(x,t) be an inertial frame of reference which at the instant I
is momentarily comoving with the point A.
The speed of the proton in K is v = dx/dt = L/T = 0.999999991·c
Do you know another definition of the speed of the proton in K
than dx/dt ?
Let K'(x',τ) be an inertial frame of reference which at the instant I
is momentarily comoving with the proton.
The speed of the point A in K' is v' = dx'/dτ = (L/γ)/τ = 0.999999991·c >>
Do you know another definition of the speed of the point A in K'
than dx'/dτ ?
It's not a joke, it's a question.
We take a very large particle accelerator of several kilometers.
On this particle accelerator, we fix a point A, coordinates (0,0,0) and
we ask to draw the trajectory of the proton in R.
We assume z=0.
We then have a large circle.
We then request a change of reference frame, and we request the
trajectory of point A in the proton reference frame.
I wish good luck to whoever answers, it's not high school level.
R.H.
Den 24.07.2024 15:08, skrev Richard Hachel:
Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :
Measured in the lab frame the proton is moving around
the L = 27 km long ring in T = 90.0623065140618 μs.
The very real speed of the proton relative to the lab is
v = L/T = 0.999999991·c
γ = 7460
Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v = L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
This should be blazingly obvious for anybody but complete morons:
If the proton is passing a point in the ring with the speed v
relative to the point, then the point in the ring is passing
the proton a the speed v relative to the proton.
A bit more precisely put:
In the lab frame the proton is passing a point in the ring with
the speed v = L/T = 0.999999991·c.
In the proton frame the point in the ring is passing the proton with
the speed v = (L/γ)/τ = 0.999999991·c.
This is the only interesting sentence in your post.
The rest is just nonsense or tautology.
Indeed, if the proton passes at Vo=0.999991 c (for example) at a point
A of the device, then the laws of physics state that point A passes at
Vo=0.999991c.
If we transpose into real speed Vr, we have:
Vr=Vo/sqrt(1-Vo²/c)=235.7c
Nothing is moving at the speed L/τ = 235.7c
Likewise, this real speed is reciprocal.
The reciprocal of L/τ is (L/γ)/T = 0.0001340c
Equally meaningless. Not the speed of anything.
In the proton frame, it is point A which passes near it at Vr=235.7c.
In the proton frame the point in the ring is passing the proton with
the speed v = (L/γ)/τ = 0.999999991·c.
Now what does the global ring look like in the proton frame of
reference, and above all what is the trajectory of point A during one
revolution?
Irrelevant.
The point A is at any instant adjacent to the proton.
We consider an arbitrary instant I.
Let K(x,t) be an inertial frame of reference which at the instant I
is momentarily comoving with the point A.
The speed of the proton in K is v = dx/dt = L/T = 0.999999991·c
Do you know another definition of the speed of the proton in K
than dx/dt ?
Let K'(x',τ) be an inertial frame of reference which at the instant I
is momentarily comoving with the proton.
The speed of the point A in K' is v' = dx'/dτ = (L/γ)/τ = 0.999999991·c
Do you know another definition of the speed of the point A in K'
than dx'/dτ ?
This is a good relativistic physics question.
Have fun answering this question...
I hope you have a lot of fun.
Quite.
But your jokes aren't funny the umpteenth time they are told,
It is getting boring.
Le 24/07/2024 à 20:45, Paul.B.Andersen a écrit :
Den 24.07.2024 15:08, skrev Richard Hachel:
Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :
Measured in the lab frame the proton is moving around
the L = 27 km long ring in T = 90.0623065140618 μs.
The very real speed of the proton relative to the lab is
v = L/T = 0.999999991·c
γ = 7460
Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v = L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
This should be blazingly obvious for anybody but complete morons:
If the proton is passing a point in the ring with the speed v
relative to the point, then the point in the ring is passing
the proton a the speed v relative to the proton.
A bit more precisely put:
In the lab frame the proton is passing a point in the ring with
the speed v = L/T = 0.999999991·c.
In the proton frame the point in the ring is passing the proton with
the speed v = (L/γ)/τ = 0.999999991·c.
This is the only interesting sentence in your post.
The rest is just nonsense or tautology.
Indeed, if the proton passes at Vo=0.999991 c (for example) at a
point A of the device, then the laws of physics state that point A
passes at Vo=0.999991c.
If we transpose into real speed Vr, we have:
Vr=Vo/sqrt(1-Vo²/c)=235.7c
Nothing is moving at the speed L/τ = 235.7c
Likewise, this real speed is reciprocal.
The reciprocal of L/τ is (L/γ)/T = 0.0001340c
Equally meaningless. Not the speed of anything.
In the proton frame, it is point A which passes near it at Vr=235.7c.
In the proton frame the point in the ring is passing the proton with
the speed v = (L/γ)/τ = 0.999999991·c.
Now what does the global ring look like in the proton frame of
reference, and above all what is the trajectory of point A during one
revolution?
Irrelevant.
The point A is at any instant adjacent to the proton.
We consider an arbitrary instant I.
Let K(x,t) be an inertial frame of reference which at the instant I
is momentarily comoving with the point A.
The speed of the proton in K is v = dx/dt = L/T = 0.999999991·c
Do you know another definition of the speed of the proton in K
than dx/dt ?
Let K'(x',τ) be an inertial frame of reference which at the instant I
is momentarily comoving with the proton.
The speed of the point A in K' is v' = dx'/dτ = (L/γ)/τ = 0.999999991·c >>
Do you know another definition of the speed of the point A in K'
than dx'/dτ ?
This is a good relativistic physics question.
Have fun answering this question...
I hope you have a lot of fun.
Quite.
But your jokes aren't funny the umpteenth time they are told,
It is getting boring.
We are dealing on fr.sci.* with this idiot for thirty years, go figure!
W dniu 24.07.2024 o 22:27, Python pisze:
Le 24/07/2024 à 20:45, Paul.B.Andersen a écrit :
Den 24.07.2024 15:08, skrev Richard Hachel:
Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :
Measured in the lab frame the proton is moving around
the L = 27 km long ring in T = 90.0623065140618 μs.
The very real speed of the proton relative to the lab is
v = L/T = 0.999999991·c
γ = 7460
Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v = L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
This should be blazingly obvious for anybody but complete morons:
If the proton is passing a point in the ring with the speed v
relative to the point, then the point in the ring is passing
the proton a the speed v relative to the proton.
A bit more precisely put:
In the lab frame the proton is passing a point in the ring with
the speed v = L/T = 0.999999991·c.
In the proton frame the point in the ring is passing the proton with
the speed v = (L/γ)/τ = 0.999999991·c.
This is the only interesting sentence in your post.
The rest is just nonsense or tautology.
Indeed, if the proton passes at Vo=0.999991 c (for example) at a
point A of the device, then the laws of physics state that point A
passes at Vo=0.999991c.
If we transpose into real speed Vr, we have:
Vr=Vo/sqrt(1-Vo²/c)=235.7c
Nothing is moving at the speed L/τ = 235.7c
Likewise, this real speed is reciprocal.
The reciprocal of L/τ is (L/γ)/T = 0.0001340c
Equally meaningless. Not the speed of anything.
In the proton frame, it is point A which passes near it at Vr=235.7c.
In the proton frame the point in the ring is passing the proton with
the speed v = (L/γ)/τ = 0.999999991·c.
Now what does the global ring look like in the proton frame of
reference, and above all what is the trajectory of point A during
one revolution?
Irrelevant.
The point A is at any instant adjacent to the proton.
We consider an arbitrary instant I.
Let K(x,t) be an inertial frame of reference which at the instant I
is momentarily comoving with the point A.
The speed of the proton in K is v = dx/dt = L/T = 0.999999991·c
Do you know another definition of the speed of the proton in K
than dx/dt ?
Let K'(x',τ) be an inertial frame of reference which at the instant I
is momentarily comoving with the proton.
The speed of the point A in K' is v' = dx'/dτ = (L/γ)/τ = 0.999999991·c >>>
Do you know another definition of the speed of the point A in K'
than dx'/dτ ?
This is a good relativistic physics question.
Have fun answering this question...
I hope you have a lot of fun.
Quite.
But your jokes aren't funny the umpteenth time they are told,
It is getting boring.
We are dealing on fr.sci.* with this idiot for thirty years, go figure!
And whatever you say - Poincare
poor stinker
Le 24/07/2024 à 22:41, Maciej Wozniak a écrit :
W dniu 24.07.2024 o 22:27, Python pisze:
Le 24/07/2024 à 20:45, Paul.B.Andersen a écrit :
Den 24.07.2024 15:08, skrev Richard Hachel:
Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :
Measured in the lab frame the proton is moving around
the L = 27 km long ring in T = 90.0623065140618 μs.
The very real speed of the proton relative to the lab is
v = L/T = 0.999999991·c
γ = 7460
Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v = L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
This should be blazingly obvious for anybody but complete morons:
If the proton is passing a point in the ring with the speed v
relative to the point, then the point in the ring is passing
the proton a the speed v relative to the proton.
A bit more precisely put:
In the lab frame the proton is passing a point in the ring with
the speed v = L/T = 0.999999991·c.
In the proton frame the point in the ring is passing the proton with
the speed v = (L/γ)/τ = 0.999999991·c.
This is the only interesting sentence in your post.
The rest is just nonsense or tautology.
Indeed, if the proton passes at Vo=0.999991 c (for example) at a
point A of the device, then the laws of physics state that point A
passes at Vo=0.999991c.
If we transpose into real speed Vr, we have:
Vr=Vo/sqrt(1-Vo²/c)=235.7c
Nothing is moving at the speed L/τ = 235.7c
Likewise, this real speed is reciprocal.
The reciprocal of L/τ is (L/γ)/T = 0.0001340c
Equally meaningless. Not the speed of anything.
In the proton frame the point in the ring is passing the proton with
In the proton frame, it is point A which passes near it at Vr=235.7c. >>>>
the speed v = (L/γ)/τ = 0.999999991·c.
Now what does the global ring look like in the proton frame of
reference, and above all what is the trajectory of point A during
one revolution?
Irrelevant.
The point A is at any instant adjacent to the proton.
We consider an arbitrary instant I.
Let K(x,t) be an inertial frame of reference which at the instant I
is momentarily comoving with the point A.
The speed of the proton in K is v = dx/dt = L/T = 0.999999991·c
Do you know another definition of the speed of the proton in K
than dx/dt ?
Let K'(x',τ) be an inertial frame of reference which at the instant I >>>> is momentarily comoving with the proton.
The speed of the point A in K' is v' = dx'/dτ = (L/γ)/τ = 0.999999991·c
Do you know another definition of the speed of the point A in K'
than dx'/dτ ?
This is a good relativistic physics question.
Have fun answering this question...
I hope you have a lot of fun.
Quite.
But your jokes aren't funny the umpteenth time they are told,
It is getting boring.
We are dealing on fr.sci.* with this idiot for thirty years, go figure!
And whatever you say - Poincare
... would kick your asses: yours and Lengrand's.
W dniu 24.07.2024 o 22:45, Python pisze:
Le 24/07/2024 à 22:41, Maciej Wozniak a écrit :
W dniu 24.07.2024 o 22:27, Python pisze:
Le 24/07/2024 à 20:45, Paul.B.Andersen a écrit :
Den 24.07.2024 15:08, skrev Richard Hachel:
Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :
Measured in the lab frame the proton is moving around
the L = 27 km long ring in T = 90.0623065140618 μs.
The very real speed of the proton relative to the lab is
v = L/T = 0.999999991·c
γ = 7460
Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v = L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
This should be blazingly obvious for anybody but complete morons: >>>>>>>
If the proton is passing a point in the ring with the speed v
relative to the point, then the point in the ring is passing
the proton a the speed v relative to the proton.
A bit more precisely put:
In the lab frame the proton is passing a point in the ring with
the speed v = L/T = 0.999999991·c.
In the proton frame the point in the ring is passing the proton with >>>>> the speed v = (L/γ)/τ = 0.999999991·c.
This is the only interesting sentence in your post.
The rest is just nonsense or tautology.
Indeed, if the proton passes at Vo=0.999991 c (for example) at a
point A of the device, then the laws of physics state that point A >>>>>> passes at Vo=0.999991c.
If we transpose into real speed Vr, we have:
Vr=Vo/sqrt(1-Vo²/c)=235.7c
Nothing is moving at the speed L/τ = 235.7c
Likewise, this real speed is reciprocal.
The reciprocal of L/τ is (L/γ)/T = 0.0001340c
Equally meaningless. Not the speed of anything.
In the proton frame the point in the ring is passing the proton with >>>>> the speed v = (L/γ)/τ = 0.999999991·c.
In the proton frame, it is point A which passes near it at Vr=235.7c. >>>>>
Now what does the global ring look like in the proton frame of
reference, and above all what is the trajectory of point A during
one revolution?
Irrelevant.
The point A is at any instant adjacent to the proton.
We consider an arbitrary instant I.
Let K(x,t) be an inertial frame of reference which at the instant I
is momentarily comoving with the point A.
The speed of the proton in K is v = dx/dt = L/T = 0.999999991·c
Do you know another definition of the speed of the proton in K
than dx/dt ?
Let K'(x',τ) be an inertial frame of reference which at the instant I >>>>> is momentarily comoving with the proton.
The speed of the point A in K' is v' = dx'/dτ = (L/γ)/τ =
0.999999991·c
Do you know another definition of the speed of the point A in K'
than dx'/dτ ?
This is a good relativistic physics question.
Have fun answering this question...
I hope you have a lot of fun.
Quite.
But your jokes aren't funny the umpteenth time they are told,
It is getting boring.
We are dealing on fr.sci.* with this idiot for thirty years, go figure! >>>
And whatever you say - Poincare
... would kick your asses: yours and Lengrand's.
Could be.
Den 24.07.2024 20:55, skrev Richard Hachel:
And your point is?
We are dealing on fr.sci.* with this idiot for thirty years, go figure!
And whatever you say - Poincare had enough wit
to understand how idiotic rejecting Euclid
would be, and he has written it clearly
enough for anyone able to read (even if not
clearly enough for you, poor stinker).
Le 24/07/2024 à 22:41, Maciej Wozniak a écrit :
We are dealing on fr.sci.* with this idiot for thirty years, go figure!
And whatever you say - Poincare had enough wit
to understand how idiotic rejecting Euclid
would be, and he has written it clearly
enough for anyone able to read (even if not
clearly enough for you, poor stinker).
It is not useful to advise poor stinker Python. He doesn't understand
Le 24/07/2024 à 22:55, Maciej Wozniak a écrit :
W dniu 24.07.2024 o 22:45, Python pisze:
Le 24/07/2024 à 22:41, Maciej Wozniak a écrit :
W dniu 24.07.2024 o 22:27, Python pisze:
Le 24/07/2024 à 20:45, Paul.B.Andersen a écrit :
Den 24.07.2024 15:08, skrev Richard Hachel:
Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :
Measured in the lab frame the proton is moving around
the L = 27 km long ring in T = 90.0623065140618 μs.
The very real speed of the proton relative to the lab is
v = L/T = 0.999999991·c
γ = 7460
Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v = L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
This should be blazingly obvious for anybody but complete morons: >>>>>>>>
If the proton is passing a point in the ring with the speed v
relative to the point, then the point in the ring is passing
the proton a the speed v relative to the proton.
A bit more precisely put:
In the lab frame the proton is passing a point in the ring with
the speed v = L/T = 0.999999991·c.
In the proton frame the point in the ring is passing the proton with >>>>>> the speed v = (L/γ)/τ = 0.999999991·c.
This is the only interesting sentence in your post.
The rest is just nonsense or tautology.
Indeed, if the proton passes at Vo=0.999991 c (for example) at a >>>>>>> point A of the device, then the laws of physics state that point >>>>>>> A passes at Vo=0.999991c.
If we transpose into real speed Vr, we have:
Vr=Vo/sqrt(1-Vo²/c)=235.7c
Nothing is moving at the speed L/τ = 235.7c
Likewise, this real speed is reciprocal.
The reciprocal of L/τ is (L/γ)/T = 0.0001340c
Equally meaningless. Not the speed of anything.
In the proton frame, it is point A which passes near it at
Vr=235.7c.
In the proton frame the point in the ring is passing the proton with >>>>>> the speed v = (L/γ)/τ = 0.999999991·c.
Now what does the global ring look like in the proton frame of
reference, and above all what is the trajectory of point A during >>>>>>> one revolution?
Irrelevant.
The point A is at any instant adjacent to the proton.
We consider an arbitrary instant I.
Let K(x,t) be an inertial frame of reference which at the instant I >>>>>> is momentarily comoving with the point A.
The speed of the proton in K is v = dx/dt = L/T = 0.999999991·c >>>>>>
Do you know another definition of the speed of the proton in K
than dx/dt ?
Let K'(x',τ) be an inertial frame of reference which at the instant I >>>>>> is momentarily comoving with the proton.
The speed of the point A in K' is v' = dx'/dτ = (L/γ)/τ =
0.999999991·c
Do you know another definition of the speed of the point A in K'
than dx'/dτ ?
This is a good relativistic physics question.
Have fun answering this question...
I hope you have a lot of fun.
Quite.
But your jokes aren't funny the umpteenth time they are told,
It is getting boring.
We are dealing on fr.sci.* with this idiot for thirty years, go
figure!
And whatever you say - Poincare
... would kick your asses: yours and Lengrand's.
Could be.
No doubt about that. Glad to see you admit it.
And your point is?
When a proton moves around the circuit once, a stationary clock
in the circuit will measure the one round around the circuit to
last the time T = 90.0623 μs
The proton (if it had a clock) will measure the one round around
the circuit to last the time τ = 12.0727 ns
Does this mean that when the proton moves around the circuit once,
then it moves once around the circuit in the lab frame while
it moves T/τ = 7460 times around the circuit in the proton frame?
Le 24/07/2024 à 22:45, "Paul.B.Andersen" a écrit :
And your point is?
Don't tell me you don't understand that the proton rotates
11.25 thousand times per second in the laboratory frame but
78 million times per second in the proton frame.
"La relativité restreinte est mathématiquement très simple, mais elle
est bourré de petits pièges".
En fait, il suffit simplement d'appliquer les bonnes équations
originelles, c'est à dire celles qui ont été données par Poincaré.
C'est très simple.
Si je me déplace dans un nouveau référentiel de gauche à droite:
x'=(x+Vo.To)/sqrt(1-Vo²/c²) les distances et les longueurs se dilatent (DILATENT)! y'=y
z'=z
To'=(To+x.Vo/c²)/sqrt(1-Vo²/c²)
Si je me déplace de droite à gauche, inversion de Vo en -Vo et on
obtient les équations réciproques.
x'=(x-Vo.To)/sqrt(1-Vo²/c²) les distances et les longueurs se dilatent (DILATENT)! y'=y
z'=z
To'=(To-x.Vo/c²)/sqrt(1-Vo²/c²)
Poincaré montre qu'il y a une dilatation réciproque des temps, et une DILATATION réciproque des longueurs et des distances par permutation de référentiel.
Certes, si j'observe une tige de 10 mètres, qui passe devant moi transversalement à Vo=0.8c, elle semblera ne mesurer que 6 mètres. Pourtant, elle est plus grande elle mesure 16.666m, mais ce que je vois
n'est pas dans l'axe de visée perpendiculaire dans le référentiel de la tige, et le numérateur de l'équation va énormément contracter la valeur de l'équation (bien plus que gamma).
Le piège consiste à croire que c'est gamma qui a contracté la tige,
alors que ce facteur la dilate, comme il dilate aussi le temps.
C'est très simple et réciproque, mais il faut bien considérer l'espace-temps et non s'inventer des espace-temps minkowskiens, à la
fois d'une laideur épouvantable et d'une irréalité physique.
Pour les transformations en milieu tournants, c'est également très
simple si l'on a compris. <http://news2.nemoweb.net/jntp?BQ5j0PykzttrIMqyw16zXh2VQVU@jntp/Data.Media:1>
R.H.
Yeah, you might think so, then though to equip the model
where the frames, in the space, are space-frames and frame-spaces,
so that the particle's _space_ besides its _frame_ are moving,
what results that space-contraction in effect, is real, that
the particle brings its space with it.
The linear accelerators are mostly aggregates of quite a large
number of, abstractly, particles, as with regards to energy
input and energy arrived.
The linear accelerators, like SLAC, illustrate that space-contraction
can be observed, affecting the surrounds of the main beam-line as
it were, as if according to a space contraction, and indeed about
the Galilean, inputs and outputs.
In the cosmological setting, the larger body or system being
its own rotational frame altogether, illustrating again that
the space contraction is observable, the Lorentzian in the
rotational, helps explain why theories like MOND have a
physical explanation and not just an algebraic model.
I.e., MOND sort of answers why there is no dark matter,
then there's a sort of inverse-MOND also to explain why
there's no dark energy, that the effects otherwise are
quite simple and holistic, instead of the "missing link"
non-theory of non-science.
In the case you are proposing, there is no contraction of the distances, because the particle is heading TOWARDS its receptor.
The equation is no longer D'=D.sqrt(1-Vo²/c²) and to believe this is to fall into the trap of ease, but D'=D.sqrt[(1+Vo/c)/ (1-Vo/c)] since
cosµ=-1.
For the particle the distance to travel (or rather that the receiver
travels towards it) is extraordinarily greater than in the laboratory reference frame.
R.H.
On Thu, 25 Jul 2024 20:30:09 +0000, Richard Hachel wrote:
In the case you are proposing, there is no contraction of the distances,
because the particle is heading TOWARDS its receptor.
The equation is no longer D'=D.sqrt(1-Vo²/c²) and to believe this is to
fall into the trap of ease, but D'=D.sqrt[(1+Vo/c)/ (1-Vo/c)] since
cosµ=-1.
You are conflating Doppler effect with length contraction. LC is ALWAYS D'=D.sqrt(1-Vo²/c²).
On Thu, 25 Jul 2024 20:30:09 +0000, Richard Hachel wrote:
In the case you are proposing, there is no contraction of the distances,
because the particle is heading TOWARDS its receptor.
The equation is no longer D'=D.sqrt(1-Vo²/c²) and to believe this is to
fall into the trap of ease, but D'=D.sqrt[(1+Vo/c)/ (1-Vo/c)] since
cosµ=-1.
You are conflating Doppler effect with length contraction. LC is ALWAYS D'=D.sqrt(1-Vo²/c²).
For the particle the distance to travel (or rather that the receiver
travels towards it) is extraordinarily greater than in the laboratory
reference frame.
R.H.
Your assertion is in violent disagreement with the LTE:
dx' = gamma(dx - vdt)
dt' = gamma(dt - vdx)
For an object stationary in the unprimed frame, dx = 0:
dx' = gamma(-vdt)
dt' = gamma(dt)
v' = dx'/dt' = -v
For an object moving at v in the unprimed frame, dx' = 0
v = dx/dt = v.
There is no "extraordinarily greater" speed in either frame. This
is true in Galilean motion also. Galileo described it perfectly
with his ship and dock example and blows your assertion out of the
water, so to speak.
Le 26/07/2024 à 01:29, hitlong@yahoo.com (gharnagel) a écrit :
On Thu, 25 Jul 2024 20:30:09 +0000, Richard Hachel wrote:
distances,In the case you are proposing, there is no contraction of the
because the particle is heading TOWARDS its receptor.
toThe equation is no longer D'=D.sqrt(1-Vo²/c²) and to believe this is
fall into the trap of ease, but D'=D.sqrt[(1+Vo/c)/ (1-Vo/c)] since cosµ=-1.
You are conflating Doppler effect with length contraction. LC isALWAYS
D'=D.sqrt(1-Vo²/c²).
laboratoryFor the particle the distance to travel (or rather that the receiver travels towards it) is extraordinarily greater than in the
reference frame.
R.H.
Your assertion is in violent disagreement with the LTE:
dx' = gamma(dx - vdt)
dt' = gamma(dt - vdx)
For an object stationary in the unprimed frame, dx = 0:
dx' = gamma(-vdt)
dt' = gamma(dt)
v' = dx'/dt' = -v
For an object moving at v in the unprimed frame, dx' = 0
v = dx/dt = v.
There is no "extraordinarily greater" speed in either frame. This
is true in Galilean motion also. Galileo described it perfectly
with his ship and dock example and blows your assertion out of the
water, so to speak.
But NO!
WE MUST APPLY POINCARE'S TRANSFORMATIONS!
It took years to find them, and without Poincaré, it is likely that they would have been found only ten or fifteen years later, when they already
had them in 1904.
That is why I am almost certain that Einstein copied them from Poincaré despite his period denials (which he would later contradict by saying
that he had read Poincaré and that he had been captivated by the intellectual
power of this man, considered the best mathematician in the world at
that time).
We must apply Poincaré.
What does Poincaré say?
If an observer moves towards me, at speed Vo=v, and crosses me at
position 0, then for me, he is at (0,0,0,0) and for him, I am at
(0,0,0,0).
But let's assume that it is only a piece of rod 9 cm long
that crosses me, and that the other end has not yet passed.
At what distance will I see the other end of the rod? Let Vo = 0.8c.
Den 25.07.2024 21:50, skrev Richard Hachel:
Le 25/07/2024 à 21:17, "Paul.B.Andersen" a écrit :
I see you have given up responding to my post.
So let us terminate this discussion with the following
demonstration of the geniality of Doctor Richard Hachel:
| Den 24.07.2024 00:19, skrev Richard Hachel:
Don't tell me you don't understand that the proton rotates
11.25 thousand times per second in the laboratory frame but
78 million times per second in the proton frame.
Richard Hachel's statement:
"The proton rotates 11.25 thousand times per second in
the laboratory frame but 78 million times per second
in the proton frame."
is quite genial, because it sums up Richard Hachel's
confusion and stupidity in one single sentence!
Well done, Richard!:-D
When a proton moves around the circuit once, a stationary clock
in the circuit will measure the one round around the circuit to
last the time T = 90.0623 μs
The proton (if it had a clock) will measure the one round around
the circuit to last the time τ = 12.0727 ns
Does this mean that when the proton moves around the circuit once,
then it moves once around the circuit in the lab frame while
it moves T/τ = 7460 times around the circuit in the proton frame?
But no!
It's stupid.
The proton only goes around once, and the time it takes, measured by
the laboratory clock (which is actually TWO clocks A and B combined
into one) is T = 90.0623 μs.
I write To = 90.0623 μs to say that this is the observable time in the
laboratory reference frame.
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
Why do you repeat what is quoted above?
Thus, for the proton, the distance AB (in the laboratory reference
frame) was crossed 7460 times faster.
So we can change the wording of your genial statement above to:
"The proton rotates once per 90.0623 μs in the laboratory frame
but 7460 times per 90.0623 μs in the proton frame."
Even better!
I call this notion the real speed of the proton, even if it sounds
funny when you're not used to seeing things that way.
The speed usually measured, and observed in the laboratory, which is
the distance in the laboratory per laboratory time, I call it v (like
the physicists) or better, Vo, to point out that we only ever observe
one notion of things, and not real things, distorted by the nature of
local space-time, of the local frame of reference.
'nuff said! :-D
A hint:
Measured in the proton frame, the length of the ring is
Le 25/07/2024 à 21:17, "Paul.B.Andersen" a écrit :
I see you have given up responding to my post.
So let us terminate this discussion with the following
demonstration of the geniality of Doctor Richard Hachel:
| Den 24.07.2024 00:19, skrev Richard Hachel:
Don't tell me you don't understand that the proton rotates
11.25 thousand times per second in the laboratory frame but
78 million times per second in the proton frame.
When a proton moves around the circuit once, a stationary clock
in the circuit will measure the one round around the circuit to
last the time T = 90.0623 μs
The proton (if it had a clock) will measure the one round around
the circuit to last the time τ = 12.0727 ns
Does this mean that when the proton moves around the circuit once,
then it moves once around the circuit in the lab frame while
it moves T/τ = 7460 times around the circuit in the proton frame?
But no!
It's stupid.
The proton only goes around once, and the time it takes,
measured by the laboratory clock (which is actually TWO
clocks A and B combined into one) is T = 90.0623 μs.
I write To = 90.0623 μs to say that this is the observable
time in the laboratory reference frame.
But if I measure with the watch that the proton wears on his
left wrist, I will measure a time of τ = 12.0727 ns.
Thus, for the proton, the distance AB (in the laboratory
reference frame) was crossed 7460 times faster.
I call this notion the real speed of the proton, even if
it sounds funny when you're not used to seeing things
that way.
The speed usually measured, and observed in the laboratory,
which is the distance in the laboratory per laboratory time,
I call it v (like the physicists) or better, Vo, to point out
that we only ever observe one notion of things, and not real
things, distorted by the nature of local space-time, of the
local frame of reference.
Le 26/07/2024 à 15:38, Maciej Wozniak a écrit :
W dniu 26.07.2024 o 15:18, Paul.B.Andersen pisze:
Den 25.07.2024 21:50, skrev Richard Hachel:
Le 25/07/2024 à 21:17, "Paul.B.Andersen" a écrit :
I see you have given up responding to my post.
So let us terminate this discussion with the following
demonstration of the geniality of Doctor Richard Hachel:
| Den 24.07.2024 00:19, skrev Richard Hachel:
Don't tell me you don't understand that the proton rotates
11.25 thousand times per second in the laboratory frame but
78 million times per second in the proton frame.
Richard Hachel's statement:
"The proton rotates 11.25 thousand times per second in
the laboratory frame but 78 million times per second
in the proton frame."
is quite genial, because it sums up Richard Hachel's
confusion and stupidity in one single sentence!
Well done, Richard!:-D
When a proton moves around the circuit once, a stationary clock
in the circuit will measure the one round around the circuit to
last the time T = 90.0623 μs
The proton (if it had a clock) will measure the one round around
the circuit to last the time τ = 12.0727 ns
Does this mean that when the proton moves around the circuit once,
then it moves once around the circuit in the lab frame while
it moves T/τ = 7460 times around the circuit in the proton frame?
But no!
It's stupid.
The proton only goes around once, and the time it takes, measured by
the laboratory clock (which is actually TWO clocks A and B combined
into one) is T = 90.0623 μs.
I write To = 90.0623 μs to say that this is the observable time in
the laboratory reference frame.
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
Why do you repeat what is quoted above?
Thus, for the proton, the distance AB (in the laboratory reference
frame) was crossed 7460 times faster.
So we can change the wording of your genial statement above to:
"The proton rotates once per 90.0623 μs in the laboratory frame
but 7460 times per 90.0623 μs in the proton frame."
Even better!
I call this notion the real speed of the proton, even if it sounds
funny when you're not used to seeing things that way.
The speed usually measured, and observed in the laboratory, which is
the distance in the laboratory per laboratory time, I call it v
(like the physicists) or better, Vo, to point out that we only ever
observe one notion of things, and not real things, distorted by the
nature of local space-time, of the local frame of reference.
'nuff said! :-D
A hint:
Measured in the proton frame, the length of the ring is
A lie, of course, as expected from
a relativistic piece of shit. No
measurements were ever performed
"in proton frame"; what even worse -
according to his moronic physics
(its quantum part) there is no
such thing as "the proton frame".
Quantum mechanics is moronic also,
W dniu 26.07.2024 o 15:18, Paul.B.Andersen pisze:
Den 25.07.2024 21:50, skrev Richard Hachel:
Le 25/07/2024 à 21:17, "Paul.B.Andersen" a écrit :
I see you have given up responding to my post.
So let us terminate this discussion with the following
demonstration of the geniality of Doctor Richard Hachel:
| Den 24.07.2024 00:19, skrev Richard Hachel:
Don't tell me you don't understand that the proton rotates
11.25 thousand times per second in the laboratory frame but
78 million times per second in the proton frame.
Richard Hachel's statement:
"The proton rotates 11.25 thousand times per second in
the laboratory frame but 78 million times per second
in the proton frame."
is quite genial, because it sums up Richard Hachel's
confusion and stupidity in one single sentence!
Well done, Richard!:-D
When a proton moves around the circuit once, a stationary clock
in the circuit will measure the one round around the circuit to
last the time T = 90.0623 μs
The proton (if it had a clock) will measure the one round around
the circuit to last the time τ = 12.0727 ns
Does this mean that when the proton moves around the circuit once,
then it moves once around the circuit in the lab frame while
it moves T/τ = 7460 times around the circuit in the proton frame?
But no!
It's stupid.
The proton only goes around once, and the time it takes, measured by
the laboratory clock (which is actually TWO clocks A and B combined
into one) is T = 90.0623 μs.
I write To = 90.0623 μs to say that this is the observable time in
the laboratory reference frame.
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
Why do you repeat what is quoted above?
Thus, for the proton, the distance AB (in the laboratory reference
frame) was crossed 7460 times faster.
So we can change the wording of your genial statement above to:
"The proton rotates once per 90.0623 μs in the laboratory frame
but 7460 times per 90.0623 μs in the proton frame."
Even better!
I call this notion the real speed of the proton, even if it sounds
funny when you're not used to seeing things that way.
The speed usually measured, and observed in the laboratory, which is
the distance in the laboratory per laboratory time, I call it v (like
the physicists) or better, Vo, to point out that we only ever observe
one notion of things, and not real things, distorted by the nature of
local space-time, of the local frame of reference.
'nuff said! :-D
A hint:
Measured in the proton frame, the length of the ring is
A lie, of course, as expected from
a relativistic piece of shit. No
measurements were ever performed
"in proton frame"; what even worse -
according to his moronic physics
(its quantum part) there is no
such thing as "the proton frame".
W dniu 26.07.2024 o 17:19, Python pisze:
Le 26/07/2024 à 15:38, Maciej Wozniak a écrit :
W dniu 26.07.2024 o 15:18, Paul.B.Andersen pisze:
Den 25.07.2024 21:50, skrev Richard Hachel:
Le 25/07/2024 à 21:17, "Paul.B.Andersen" a écrit :
I see you have given up responding to my post.
So let us terminate this discussion with the following
demonstration of the geniality of Doctor Richard Hachel:
| Den 24.07.2024 00:19, skrev Richard Hachel:
Don't tell me you don't understand that the proton rotates
11.25 thousand times per second in the laboratory frame but
78 million times per second in the proton frame.
Richard Hachel's statement:
"The proton rotates 11.25 thousand times per second in
the laboratory frame but 78 million times per second
in the proton frame."
is quite genial, because it sums up Richard Hachel's
confusion and stupidity in one single sentence!
Well done, Richard!:-D
When a proton moves around the circuit once, a stationary clock
in the circuit will measure the one round around the circuit to
last the time T = 90.0623 μs
The proton (if it had a clock) will measure the one round around
the circuit to last the time τ = 12.0727 ns
Does this mean that when the proton moves around the circuit once, >>>>>> then it moves once around the circuit in the lab frame while
it moves T/τ = 7460 times around the circuit in the proton frame?
But no!
It's stupid.
The proton only goes around once, and the time it takes, measured
by the laboratory clock (which is actually TWO clocks A and B
combined into one) is T = 90.0623 μs.
I write To = 90.0623 μs to say that this is the observable time in
the laboratory reference frame.
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
Why do you repeat what is quoted above?
Thus, for the proton, the distance AB (in the laboratory reference
frame) was crossed 7460 times faster.
So we can change the wording of your genial statement above to:
"The proton rotates once per 90.0623 μs in the laboratory frame
but 7460 times per 90.0623 μs in the proton frame."
Even better!
I call this notion the real speed of the proton, even if it sounds
funny when you're not used to seeing things that way.
The speed usually measured, and observed in the laboratory, which
is the distance in the laboratory per laboratory time, I call it v
(like the physicists) or better, Vo, to point out that we only ever
observe one notion of things, and not real things, distorted by the
nature of local space-time, of the local frame of reference.
'nuff said! :-D
A hint:
Measured in the proton frame, the length of the ring is
A lie, of course, as expected from
a relativistic piece of shit. No
measurements were ever performed
"in proton frame"; what even worse -
according to his moronic physics
(its quantum part) there is no
such thing as "the proton frame".
Quantum mechanics is moronic also,
Not as much as The Shit - at least it
works - but obviously.
And nothing like
"the proton frame" according to it.
Le 26/07/2024 à 17:32, Maciej Wozniak a écrit :
W dniu 26.07.2024 o 17:19, Python pisze:
Le 26/07/2024 à 15:38, Maciej Wozniak a écrit :
W dniu 26.07.2024 o 15:18, Paul.B.Andersen pisze:
Den 25.07.2024 21:50, skrev Richard Hachel:
Le 25/07/2024 à 21:17, "Paul.B.Andersen" a écrit :
I see you have given up responding to my post.
So let us terminate this discussion with the following
demonstration of the geniality of Doctor Richard Hachel:
| Den 24.07.2024 00:19, skrev Richard Hachel:
Don't tell me you don't understand that the proton rotates
11.25 thousand times per second in the laboratory frame but
78 million times per second in the proton frame.
Richard Hachel's statement:
"The proton rotates 11.25 thousand times per second in
the laboratory frame but 78 million times per second
in the proton frame."
is quite genial, because it sums up Richard Hachel's
confusion and stupidity in one single sentence!
Well done, Richard!:-D
But no!
When a proton moves around the circuit once, a stationary clock
in the circuit will measure the one round around the circuit to
last the time T = 90.0623 μs
The proton (if it had a clock) will measure the one round around >>>>>>> the circuit to last the time τ = 12.0727 ns
Does this mean that when the proton moves around the circuit once, >>>>>>> then it moves once around the circuit in the lab frame while
it moves T/τ = 7460 times around the circuit in the proton frame? >>>>>>
It's stupid.
The proton only goes around once, and the time it takes, measured
by the laboratory clock (which is actually TWO clocks A and B
combined into one) is T = 90.0623 μs.
I write To = 90.0623 μs to say that this is the observable time in >>>>>> the laboratory reference frame.
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
Why do you repeat what is quoted above?
Thus, for the proton, the distance AB (in the laboratory reference >>>>>> frame) was crossed 7460 times faster.
So we can change the wording of your genial statement above to:
"The proton rotates once per 90.0623 μs in the laboratory frame
but 7460 times per 90.0623 μs in the proton frame."
Even better!
I call this notion the real speed of the proton, even if it sounds >>>>>> funny when you're not used to seeing things that way.
The speed usually measured, and observed in the laboratory, which
is the distance in the laboratory per laboratory time, I call it v >>>>>> (like the physicists) or better, Vo, to point out that we only
ever observe one notion of things, and not real things, distorted
by the nature of local space-time, of the local frame of reference. >>>>>>
'nuff said! :-D
A hint:
Measured in the proton frame, the length of the ring is
A lie, of course, as expected from
a relativistic piece of shit. No
measurements were ever performed
"in proton frame"; what even worse -
according to his moronic physics
(its quantum part) there is no
such thing as "the proton frame".
Quantum mechanics is moronic also,
Not as much as The Shit - at least it
works - but obviously.
"it works - but obviously" (while being somewhat "moronic"),
"information engineer (ahah)" Wozniak. What does that seriously *mean*?
And nothing like
"the proton frame" according to it.
The link I provided shows you wrong.
A hint:
Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v = L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
Paul
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
Right again.
This is a trivial fact, disputed by no one.
Den 26.07.2024 18:14, skrev Richard Hachel:
Right.
So the proton crossed the distance L = 27 km in the laboratory
reference frame in T = 90.0623 μs, and the speed in
the laboratory reference frame is v = L/T = 0.999999991·c
Le 26/07/2024 à 15:18, "Paul.B.Andersen" a écrit :
Den 25.07.2024 21:50, skrev Richard Hachel:
Le 25/07/2024 à 21:17, "Paul.B.Andersen" a écrit :
| Den 24.07.2024 00:19, skrev Richard Hachel:
Don't tell me you don't understand that the proton rotates
11.25 thousand times per second in the laboratory frame but
78 million times per second in the proton frame.
Richard Hachel's statement:
"The proton rotates 11.25 thousand times per second in
the laboratory frame but 78 million times per second
in the proton frame."
is quite genial, because it sums up Richard Hachel's
confusion and stupidity in one single sentence!
The proton only goes around once, and the time it takes, measured by
the laboratory clock (which is actually TWO clocks A and B combined
into one) is T = 90.0623 μs.
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
!!!!!! :-D
Thus, for the proton, the distance AB (in the laboratory reference
frame) was crossed 7460 times faster
A hint:
Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v = L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
Paul
Tu mélanges tout, et tu ne comprends plus rien.
Cela me montre (ce n'est pas de ta faute) la déconnexion totale des médecins avec la théorie de la relativité telle qu'elle devrait être enseignée.
Respire et inhale un peu, Paul, et prends un peu le temps (sauf si on te menace de mort) de lire ce que j'écris, et de considérer que tout se
tient théoriquement (les maths sont les mêmes pour tout le monde) et
qu'il ne reste qu'à l'expérimentation de prouver que j'ai raison (ou de l'infirmer, mais ça m'étonnerait, tant ce que je dis est cohérent et ce que les médecins disent ne l'est pas , même dans un simple Langevin à vitesses apparentes. Ils sont incapables d'expliquer pourquoi je dis
qu'en un temps propre de 9 ans, une personne qui voit revenir un objet
sur elle à Vapp=4c, voit cet objet se déplacer sur 36. al. C'est
tellement cohérent, qu'on est obligé de me cracher dessus pour me faire vivre dans un monde de haute débilité. on connait la circonférence dans
le référentiel terrestre.
On note cette circonférence C. Tu me dis que la circonférence dans le référentiel du proton, cela va être C'=C.sqrt(1-Vo²/c²)?
Mais tu dis n'importe quoi, et là encore on voit que les médecins relativistes n'y comprennent que pouic.
Dans le référentiel du proton, le tunnel va prendre l'aspect non d'un
petit tunnel cylindrique régulier ou contracté, mais au contraire d'un gigantesque ovoïde dévié dans le sens horaire, si le proton tourne dans
le sens trigonométrique (anti-horaire ).
Il va y avoir une dilatation très importante et absolue du disque pour
le proton (il se passe l'inverse que quand c'est un disque qui tourne
dans le labo, le disque se rétrécit de façon absolue).
J'ai donné les transformations pour les référentiels tournants, et si tu les étudiais un peu, tu serais moins moqueur.
R.H.
Den 26.07.2024 18:14, skrev Richard Hachel:
Le 26/07/2024 à 15:18, "Paul.B.Andersen" a écrit :
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
Right again.
This is a trivial fact, disputed by no one.
Den 26.07.2024 18:14, skrev Richard Hachel:
How do you think that the proton can have two different speeds
in the laboratory frame?
Of course it can't in the real world.
Den 26.07.2024 18:14, skrev Richard Hachel:
Only an ignoramus like you will fail to understand that if
the speed of the proton in the lab frame is 0.999999991·c,
then the speed of the lab in the proton frame is 0.999999991·c.
On 07/25/2024 01:30 PM, Richard Hachel wrote:
You mean the distance _in_ the space _in_ the frame?
Le 26/07/2024 à 21:54, "Paul.B.Andersen" a écrit :
Den 26.07.2024 18:14, skrev Richard Hachel:
Right.
So the proton crossed the distance L = 27 km in the laboratory
reference frame in T = 90.0623 μs, and the speed in
the laboratory reference frame is v = L/T = 0.999999991·c
There you go.
Your sentence is almost perfect.
It is perfect in traditional relativity, and almost perfect in RH-style relativity.
I just add the word "observable" not because I like to brag, but because
I find it useful for the song.
So the proton crossed the distance L = 27 km in the laboratory
reference frame in T = 90.0623 μs, and the observable speed in
the laboratory reference frame is v = L/T = 0.999999991·c
Le 26/07/2024 à 21:54, "Paul.B.Andersen" a écrit : >>>> Den 25.07.2024 21:50, skrev Richard Hachel:
The proton only goes around once, and the time it takes, measured by >>>>> the laboratory clock (which is actually TWO clocks A and B combined
into one) is T = 90.0623 μs.
Right.
So the proton crossed the distance L = 27 km in the laboratory
reference frame in T = 90.0623 μs, and the speed in
the laboratory reference frame is v = L/T = 0.999999991·c
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
Right again.
This is a trivial fact, disputed by no one.
It is not a question of discussing what is of rare evidence in both theories. We must remain simple.
We have here two theories, and it is infinitely probable that one of the two is correct.
The problem is that on certain concepts, we do not agree. It is therefore certain that if there is an error, the error is there.
It cannot be found where we agree as here, or when each one poses: To=(x/c).sqrt(1+2c²/ax) to determine the observable time of an accelerated object.
Thus, for the proton, the distance AB (in the laboratory reference
frame) was crossed 7460 times faster
!!!!!! :-D
You say that the proton crossed the distance L = 27 km in
_the laboratory reference frame_ in τ = 12.0727 ns, thus is
the speed in the laboratory reference frame v = L/τ ≈ 7460·c
How do you think that the proton can have two different speeds
in the laboratory frame?
Of course it can't in the real world.
We must say simple things, and we must say true things.
It is very difficult in special relativity because of the frequent conceptual errors. Sometimes when I read certain things here or elsewhere, I have the impression that everything is sinking into horror.
We must be careful about the confusion of words.
You say, a body can only have one speed, and you seem to think that I am an idiot.
But no, I am not an idiot, and it is precisely because of morons like Python that I can pass for an idiot.
Do you think that I am so stupid to say that a moving body can have two different speeds at the same time?
When I say that a body can have, in the same frame of reference, many different speeds, that is OBVIOUSLY not what I am talking about.
Let us assume a speed Vo=0.8c.
It is quite obvious that I cannot have at the same time, at the risk of being absurd, I who claim to describe the most beautiful, the simplest and the most logical theory, a life that is Vo=0.8/c, Vo=0.9c, Vo=0.5c and Vo=0.999c.
It would be absurd, and it would be dishonest to make me say what I did not say.
Now, I can still write Vo=0.8c, Vr=(4/3)c, Vapp'=0.4444c and Vapp"=4c.
You just have to understand what I write, why I write it, and validate it without spitting on it.
Only an ignoramus like you will fail to understand that if
the speed of the proton in the lab frame is 0.999999991·c,
then the speed of the lab in the proton frame is 0.999999991·c.
C'est ce que je dis.
Den 26.07.2024 22:36, skrev Richard Hachel:
Le 26/07/2024 à 21:54, "Paul.B.Andersen" a écrit : >>>> Den 25.07.2024
21:50, skrev Richard Hachel:
The proton only goes around once, and the time it takes, measured
by the laboratory clock (which is actually TWO clocks A and B
combined into one) is T = 90.0623 μs.
Right.
So the proton crossed the distance L = 27 km in the laboratory
reference frame in T = 90.0623 μs, and the speed in
the laboratory reference frame is v = L/T = 0.999999991·c
But if I measure with the watch that the proton wears on his left
wrist, I will measure a time of τ = 12.0727 ns.
Right again.
This is a trivial fact, disputed by no one.
This is experimentally confirmed in the real world.
(With satellites and aeroplanes.)
It is not a question of discussing what is of rare evidence in both
theories.
We must remain simple.
We have here two theories, and it is infinitely probable that one of
the two is correct.
The problem is that on certain concepts, we do not agree. It is
therefore certain that if there is an error, the error is there.
It cannot be found where we agree as here, or when each one poses:
To=(x/c).sqrt(1+2c²/ax) to determine the observable time of an
accelerated object.
What are you talking about? :-D
Thus, for the proton, the distance AB (in the laboratory reference >>>>>> frame) was crossed 7460 times faster
!!!!!! :-D
You say that the proton crossed the distance L = 27 km in
_the laboratory reference frame_ in τ = 12.0727 ns, thus is
the speed in the laboratory reference frame v = L/τ ≈ 7460·c
How do you think that the proton can have two different speeds
in the laboratory frame?
Of course it can't in the real world.
We must say simple things, and we must say true things.
It is very difficult in special relativity because of the frequent
conceptual errors. Sometimes when I read certain things here or
elsewhere, I have the impression that everything is sinking into horror.
We must be careful about the confusion of words.
You say, a body can only have one speed, and you seem to think that I
am an idiot.
But no, I am not an idiot, and it is precisely because of morons like
Python that I can pass for an idiot.
Do you think that I am so stupid to say that a moving body can have
two different speeds at the same time?
I did think so, but now you have corrected me.
So you know that the proton can have only one speed in the lab frame.
When I say that a body can have, in the same frame of reference, many
different speeds, that is OBVIOUSLY not what I am talking about.
Let us assume a speed Vo=0.8c.
It is quite obvious that I cannot have at the same time, at the risk
of being absurd, I who claim to describe the most beautiful, the
simplest and the most logical theory, a life that is Vo=0.8/c,
Vo=0.9c, Vo=0.5c and Vo=0.999c.
It would be absurd, and it would be dishonest to make me say what I
did not say.
Now, I can still write Vo=0.8c, Vr=(4/3)c, Vapp'=0.4444c and Vapp"=4c.
You just have to understand what I write, why I write it, and validate
it without spitting on it.
Why did you use so many words to say that you agree:
"The proton can only have one speed in the lab frame."
Sysop: | Keyop |
---|---|
Location: | Huddersfield, West Yorkshire, UK |
Users: | 366 |
Nodes: | 16 (2 / 14) |
Uptime: | 06:16:57 |
Calls: | 7,825 |
Calls today: | 8 |
Files: | 12,930 |
Messages: | 5,769,091 |
Posted today: | 1 |