Einstein EDoMB-1905-Section §5:prime ) is reduced accordingly, to a value no greater than 50.
https://www.fourmilab.ch/etexts/einstein/specrel/www/
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )
A suitcase version: Two suitcases of weights v = 60 lbs and w = 40 lbs make up a sum of designated V.
V = v + w
V = 60 + 40 = 100
We learn from airline regulation that total luggage weight can be no greater than 50. The sum ( v + w ) can be divided by a number between 1 and 2 inclusive, the divisor being a version of ( 1 + vw/c² ). A new left-hand-side sum ( call it V ′
Equation: V ′ = ( v + w ) / 2the symmetric chosen? Why does the symmetric take precedence over the asymmetric?
50 = ( 60 + 40 ) / 2
The equation is a left ← to ← right assignment operator. The sum of weights 60 + 40 on the right-hand-side is divided by 2 and the new value 50 is assigned to weight variable V ′ on the left-hand-side.
Assignment: 50 ← ( 60 + 40 ) / 2
The quotient ( 60 + 40 ) / 2 is identical in value to 60 / 2 + 40 / 2. When this equality is shown as an equation, it is an identity and the physical situations are irrelevant.
Identity: 60 / 2 + 40 / 2 ≡ ( 60 + 40 ) / 2
Einstein suggests that the assignment operator can be reconfigured so that division by 2 on the right-hand-side is the phenomenon 60 / 2 + 40 / 2.
50 ← 60 / 2 + 40 / 2
Quote: “It is worthy of remark that v and w enter into the expression for the resultant velocity in a symmetrical manner.”
Original: 50 ← ( 60 + 40 ) / 2
Symmetric: 50 ← 60 / 2 + 40 / 2
We get V ′ = 50 by cutting each suitcase weight in half.
Asymmetric: 50 ← 60 / 6 + 40 / 1
We get V ′ = 50 by cutting the heavy suitcase with divisor 6, leaving the lesser suitcase with divisor 1, unchanged.
Einstein is telling us how to repack our bags: divide each by 2. Do the airlines tell us how to repack our bags? No. Just cut total luggage weight in half.
The symmetric and asymmetric divisions are equal as values; also equal to the original. But the symmetric and asymmetric divisions are not identical physical situations. What governs that choice? The airlines don’t care. EDoMB does care. Why is
On Wednesday, 19 April 2023 at 07:42:56 UTC+2, Sylvia Else wrote:prime ) is reduced accordingly, to a value no greater than 50.
On 19-Apr-23 3:24 pm, xip14 wrote:
Einstein EDoMB-1905-Section §5:
https://www.fourmilab.ch/etexts/einstein/specrel/www/
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )
A suitcase version: Two suitcases of weights v = 60 lbs and w = 40 lbs make up a sum of designated V.
V = v + w
V = 60 + 40 = 100
We learn from airline regulation that total luggage weight can be no greater than 50. The sum ( v + w ) can be divided by a number between 1 and 2 inclusive, the divisor being a version of ( 1 + vw/c² ). A new left-hand-side sum ( call it V ′
symmetric chosen? Why does the symmetric take precedence over the asymmetric? >>>
Equation: V ′ = ( v + w ) / 2
50 = ( 60 + 40 ) / 2
The equation is a left ← to ← right assignment operator. The sum of weights 60 + 40 on the right-hand-side is divided by 2 and the new value 50 is assigned to weight variable V ′ on the left-hand-side.
Assignment: 50 ← ( 60 + 40 ) / 2
The quotient ( 60 + 40 ) / 2 is identical in value to 60 / 2 + 40 / 2. When this equality is shown as an equation, it is an identity and the physical situations are irrelevant.
Identity: 60 / 2 + 40 / 2 ≡ ( 60 + 40 ) / 2
Einstein suggests that the assignment operator can be reconfigured so that division by 2 on the right-hand-side is the phenomenon 60 / 2 + 40 / 2.
50 ← 60 / 2 + 40 / 2
Quote: “It is worthy of remark that v and w enter into the expression for the resultant velocity in a symmetrical manner.”
Original: 50 ← ( 60 + 40 ) / 2
Symmetric: 50 ← 60 / 2 + 40 / 2
We get V ′ = 50 by cutting each suitcase weight in half.
Asymmetric: 50 ← 60 / 6 + 40 / 1
We get V ′ = 50 by cutting the heavy suitcase with divisor 6, leaving the lesser suitcase with divisor 1, unchanged.
Einstein is telling us how to repack our bags: divide each by 2. Do the airlines tell us how to repack our bags? No. Just cut total luggage weight in half.
The symmetric and asymmetric divisions are equal as values; also equal to the original. But the symmetric and asymmetric divisions are not identical physical situations. What governs that choice? The airlines don’t care. EDoMB does care. Why is the
It really isn't clear what you're trying to say, but I think what you're
getting at is that Einstein somehow modified the addition rule.
The true situation is that there was never a valid reason to think that
velocities would combine by addition, and special relativity tells us
that they do not. So Einstein isn't modifying the addition rule, he's
saying that the addition rule is not what you should use to combine
velocities. Instead you should use a rule that, for small velocities, is
very similar to the addition rule, but for larger velocities is nothing
like it.
Your insane gurus can tell us what we should do to
follow THE BEST WAY we're FORCED to. And we can
ignore them. We do and we will.
On 19-Apr-23 3:24 pm, xip14 wrote:prime ) is reduced accordingly, to a value no greater than 50.
Einstein EDoMB-1905-Section §5:
https://www.fourmilab.ch/etexts/einstein/specrel/www/
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )
A suitcase version: Two suitcases of weights v = 60 lbs and w = 40 lbs make up a sum of designated V.
V = v + w
V = 60 + 40 = 100
We learn from airline regulation that total luggage weight can be no greater than 50. The sum ( v + w ) can be divided by a number between 1 and 2 inclusive, the divisor being a version of ( 1 + vw/c² ). A new left-hand-side sum ( call it V ′
symmetric chosen? Why does the symmetric take precedence over the asymmetric?Equation: V ′ = ( v + w ) / 2
50 = ( 60 + 40 ) / 2
The equation is a left ← to ← right assignment operator. The sum of weights 60 + 40 on the right-hand-side is divided by 2 and the new value 50 is assigned to weight variable V ′ on the left-hand-side.
Assignment: 50 ← ( 60 + 40 ) / 2
The quotient ( 60 + 40 ) / 2 is identical in value to 60 / 2 + 40 / 2. When this equality is shown as an equation, it is an identity and the physical situations are irrelevant.
Identity: 60 / 2 + 40 / 2 ≡ ( 60 + 40 ) / 2
Einstein suggests that the assignment operator can be reconfigured so that division by 2 on the right-hand-side is the phenomenon 60 / 2 + 40 / 2.
50 ← 60 / 2 + 40 / 2
Quote: “It is worthy of remark that v and w enter into the expression for the resultant velocity in a symmetrical manner.”
Original: 50 ← ( 60 + 40 ) / 2
Symmetric: 50 ← 60 / 2 + 40 / 2
We get V ′ = 50 by cutting each suitcase weight in half.
Asymmetric: 50 ← 60 / 6 + 40 / 1
We get V ′ = 50 by cutting the heavy suitcase with divisor 6, leaving the lesser suitcase with divisor 1, unchanged.
Einstein is telling us how to repack our bags: divide each by 2. Do the airlines tell us how to repack our bags? No. Just cut total luggage weight in half.
The symmetric and asymmetric divisions are equal as values; also equal to the original. But the symmetric and asymmetric divisions are not identical physical situations. What governs that choice? The airlines don’t care. EDoMB does care. Why is the
It really isn't clear what you're trying to say, but I think what you're getting at is that Einstein somehow modified the addition rule.
The true situation is that there was never a valid reason to think that velocities would combine by addition, and special relativity tells us
that they do not. So Einstein isn't modifying the addition rule, he's
saying that the addition rule is not what you should use to combine velocities. Instead you should use a rule that, for small velocities, is very similar to the addition rule, but for larger velocities is nothing
like it.
On 19-Apr-23 4:34 pm, Maciej Wozniak wrote:prime ) is reduced accordingly, to a value no greater than 50.
On Wednesday, 19 April 2023 at 07:42:56 UTC+2, Sylvia Else wrote:
On 19-Apr-23 3:24 pm, xip14 wrote:
Einstein EDoMB-1905-Section §5:
https://www.fourmilab.ch/etexts/einstein/specrel/www/
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )
A suitcase version: Two suitcases of weights v = 60 lbs and w = 40 lbs make up a sum of designated V.
V = v + w
V = 60 + 40 = 100
We learn from airline regulation that total luggage weight can be no greater than 50. The sum ( v + w ) can be divided by a number between 1 and 2 inclusive, the divisor being a version of ( 1 + vw/c² ). A new left-hand-side sum ( call it V ′
the symmetric chosen? Why does the symmetric take precedence over the asymmetric?
Equation: V ′ = ( v + w ) / 2
50 = ( 60 + 40 ) / 2
The equation is a left ← to ← right assignment operator. The sum of weights 60 + 40 on the right-hand-side is divided by 2 and the new value 50 is assigned to weight variable V ′ on the left-hand-side.
Assignment: 50 ← ( 60 + 40 ) / 2
The quotient ( 60 + 40 ) / 2 is identical in value to 60 / 2 + 40 / 2. When this equality is shown as an equation, it is an identity and the physical situations are irrelevant.
Identity: 60 / 2 + 40 / 2 ≡ ( 60 + 40 ) / 2
Einstein suggests that the assignment operator can be reconfigured so that division by 2 on the right-hand-side is the phenomenon 60 / 2 + 40 / 2.
50 ← 60 / 2 + 40 / 2
Quote: “It is worthy of remark that v and w enter into the expression for the resultant velocity in a symmetrical manner.”
Original: 50 ← ( 60 + 40 ) / 2
Symmetric: 50 ← 60 / 2 + 40 / 2
We get V ′ = 50 by cutting each suitcase weight in half.
Asymmetric: 50 ← 60 / 6 + 40 / 1
We get V ′ = 50 by cutting the heavy suitcase with divisor 6, leaving the lesser suitcase with divisor 1, unchanged.
Einstein is telling us how to repack our bags: divide each by 2. Do the airlines tell us how to repack our bags? No. Just cut total luggage weight in half.
The symmetric and asymmetric divisions are equal as values; also equal to the original. But the symmetric and asymmetric divisions are not identical physical situations. What governs that choice? The airlines don’t care. EDoMB does care. Why is
It really isn't clear what you're trying to say, but I think what you're >> getting at is that Einstein somehow modified the addition rule.
The true situation is that there was never a valid reason to think that >> velocities would combine by addition, and special relativity tells us
that they do not. So Einstein isn't modifying the addition rule, he's
saying that the addition rule is not what you should use to combine
velocities. Instead you should use a rule that, for small velocities, is >> very similar to the addition rule, but for larger velocities is nothing >> like it.
Your insane gurus can tell us what we should do to
followm t THE BEST WAY we're FORCED to. And we can
ignore them. We do and we will.
And if you're dealing with high velocities, you'll get the wrong answer.
Den 19.04.2023 07:24, skrev xip14:
Einstein EDoMB-1905-Section §5:
https://www.fourmilab.ch/etexts/einstein/specrel/www/
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )The "velocity addition formula" is a misnomer.
It is Lorentz transformation of the velocity v
in one inertial frame to the velocity V in another
frame of reference moving with the speed w relative
to the former.
Einstein EDoMB-1905-Section §5:
https://www.fourmilab.ch/etexts/einstein/specrel/www/
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )
<snip nonsense>
On Wednesday, 19 April 2023 at 08:35:56 UTC+2, Sylvia Else wrote:prime ) is reduced accordingly, to a value no greater than 50.
On 19-Apr-23 4:34 pm, Maciej Wozniak wrote:
On Wednesday, 19 April 2023 at 07:42:56 UTC+2, Sylvia Else wrote:
On 19-Apr-23 3:24 pm, xip14 wrote:
Einstein EDoMB-1905-Section §5:
https://www.fourmilab.ch/etexts/einstein/specrel/www/
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )
A suitcase version: Two suitcases of weights v = 60 lbs and w = 40 lbs make up a sum of designated V.
V = v + w
V = 60 + 40 = 100
We learn from airline regulation that total luggage weight can be no greater than 50. The sum ( v + w ) can be divided by a number between 1 and 2 inclusive, the divisor being a version of ( 1 + vw/c² ). A new left-hand-side sum ( call it V ′
the symmetric chosen? Why does the symmetric take precedence over the asymmetric?
Equation: V ′ = ( v + w ) / 2
50 = ( 60 + 40 ) / 2
The equation is a left ← to ← right assignment operator. The sum of weights 60 + 40 on the right-hand-side is divided by 2 and the new value 50 is assigned to weight variable V ′ on the left-hand-side.
Assignment: 50 ← ( 60 + 40 ) / 2
The quotient ( 60 + 40 ) / 2 is identical in value to 60 / 2 + 40 / 2. When this equality is shown as an equation, it is an identity and the physical situations are irrelevant.
Identity: 60 / 2 + 40 / 2 ≡ ( 60 + 40 ) / 2
Einstein suggests that the assignment operator can be reconfigured so that division by 2 on the right-hand-side is the phenomenon 60 / 2 + 40 / 2.
50 ← 60 / 2 + 40 / 2
Quote: “It is worthy of remark that v and w enter into the expression for the resultant velocity in a symmetrical manner.”
Original: 50 ← ( 60 + 40 ) / 2
Symmetric: 50 ← 60 / 2 + 40 / 2
We get V ′ = 50 by cutting each suitcase weight in half.
Asymmetric: 50 ← 60 / 6 + 40 / 1
We get V ′ = 50 by cutting the heavy suitcase with divisor 6, leaving the lesser suitcase with divisor 1, unchanged.
Einstein is telling us how to repack our bags: divide each by 2. Do the airlines tell us how to repack our bags? No. Just cut total luggage weight in half.
The symmetric and asymmetric divisions are equal as values; also equal to the original. But the symmetric and asymmetric divisions are not identical physical situations. What governs that choice? The airlines don’t care. EDoMB does care. Why is
And if you're dealing with high velocities, you'll get the wrong answer.It really isn't clear what you're trying to say, but I think what you're >>>> getting at is that Einstein somehow modified the addition rule.
The true situation is that there was never a valid reason to think that >>>> velocities would combine by addition, and special relativity tells us
that they do not. So Einstein isn't modifying the addition rule, he's
saying that the addition rule is not what you should use to combine
velocities. Instead you should use a rule that, for small velocities, is >>>> very similar to the addition rule, but for larger velocities is nothing >>>> like it.
Your insane gurus can tell us what we should do to
followm t THE BEST WAY we're FORCED to. And we can
ignore them. We do and we will.
I don't feel the urge to have my answers blessed by
Your insane gurus; I prefer them to be reliable and usable
(some criteria they have no clue about).
On 19-Apr-23 4:50 pm, Maciej Wozniak wrote:prime ) is reduced accordingly, to a value no greater than 50.
On Wednesday, 19 April 2023 at 08:35:56 UTC+2, Sylvia Else wrote:
On 19-Apr-23 4:34 pm, Maciej Wozniak wrote:
On Wednesday, 19 April 2023 at 07:42:56 UTC+2, Sylvia Else wrote:
On 19-Apr-23 3:24 pm, xip14 wrote:
Einstein EDoMB-1905-Section §5:
https://www.fourmilab.ch/etexts/einstein/specrel/www/
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )
A suitcase version: Two suitcases of weights v = 60 lbs and w = 40 lbs make up a sum of designated V.
V = v + w
V = 60 + 40 = 100
We learn from airline regulation that total luggage weight can be no greater than 50. The sum ( v + w ) can be divided by a number between 1 and 2 inclusive, the divisor being a version of ( 1 + vw/c² ). A new left-hand-side sum ( call it V ′
the symmetric chosen? Why does the symmetric take precedence over the asymmetric?
Equation: V ′ = ( v + w ) / 2
50 = ( 60 + 40 ) / 2
The equation is a left ← to ← right assignment operator. The sum of weights 60 + 40 on the right-hand-side is divided by 2 and the new value 50 is assigned to weight variable V ′ on the left-hand-side.
Assignment: 50 ← ( 60 + 40 ) / 2
The quotient ( 60 + 40 ) / 2 is identical in value to 60 / 2 + 40 / 2. When this equality is shown as an equation, it is an identity and the physical situations are irrelevant.
Identity: 60 / 2 + 40 / 2 ≡ ( 60 + 40 ) / 2
Einstein suggests that the assignment operator can be reconfigured so that division by 2 on the right-hand-side is the phenomenon 60 / 2 + 40 / 2.
50 ← 60 / 2 + 40 / 2
Quote: “It is worthy of remark that v and w enter into the expression for the resultant velocity in a symmetrical manner.”
Original: 50 ← ( 60 + 40 ) / 2
Symmetric: 50 ← 60 / 2 + 40 / 2
We get V ′ = 50 by cutting each suitcase weight in half.
Asymmetric: 50 ← 60 / 6 + 40 / 1
We get V ′ = 50 by cutting the heavy suitcase with divisor 6, leaving the lesser suitcase with divisor 1, unchanged.
Einstein is telling us how to repack our bags: divide each by 2. Do the airlines tell us how to repack our bags? No. Just cut total luggage weight in half.
The symmetric and asymmetric divisions are equal as values; also equal to the original. But the symmetric and asymmetric divisions are not identical physical situations. What governs that choice? The airlines don’t care. EDoMB does care. Why is
And if you're dealing with high velocities, you'll get the wrong answer.It really isn't clear what you're trying to say, but I think what you're
getting at is that Einstein somehow modified the addition rule.
The true situation is that there was never a valid reason to think that >>>> velocities would combine by addition, and special relativity tells us >>>> that they do not. So Einstein isn't modifying the addition rule, he's >>>> saying that the addition rule is not what you should use to combine >>>> velocities. Instead you should use a rule that, for small velocities, is
very similar to the addition rule, but for larger velocities is nothing >>>> like it.
Your insane gurus can tell us what we should do to
followm t THE BEST WAY we're FORCED to. And we can
ignore them. We do and we will.
I don't feel the urge to have my answers blessed by
Your insane gurus; I prefer them to be reliable and usable
(some criteria they have no clue about).
Your predictions of the results of the Fizeau experiment, performed in
1851, would be in error, then.
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )
Einstein EDoMB-1905-Section §5:is reduced accordingly, to a value no greater than 50.
https://www.fourmilab.ch/etexts/einstein/specrel/www/
Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )
A suitcase version: Two suitcases of weights v = 60 lbs and w = 40 lbs make up a sum of designated V.
V = v + w
V = 60 + 40 = 100
We learn from airline regulation that total luggage weight can be no greater than 50. The sum ( v + w ) can be divided by a number between 1 and 2 inclusive, the divisor being a version of ( 1 + vw/c² ). A new left-hand-side sum ( call it V ′ prime )
Equation: V ′ = ( v + w ) / 2symmetric chosen? Why does the symmetric take precedence over the asymmetric?
50 = ( 60 + 40 ) / 2
The equation is a left ← to ← right assignment operator. The sum of weights 60 + 40 on the right-hand-side is divided by 2 and the new value 50 is assigned to weight variable V ′ on the left-hand-side.
Assignment: 50 ← ( 60 + 40 ) / 2
The quotient ( 60 + 40 ) / 2 is identical in value to 60 / 2 + 40 / 2. When this equality is shown as an equation, it is an identity and the physical situations are irrelevant.
Identity: 60 / 2 + 40 / 2 ≡ ( 60 + 40 ) / 2
Einstein suggests that the assignment operator can be reconfigured so that division by 2 on the right-hand-side is the phenomenon 60 / 2 + 40 / 2.
50 ← 60 / 2 + 40 / 2
Quote: “It is worthy of remark that v and w enter into the expression for the resultant velocity in a symmetrical manner.”
Original: 50 ← ( 60 + 40 ) / 2
Symmetric: 50 ← 60 / 2 + 40 / 2
We get V ′ = 50 by cutting each suitcase weight in half.
Asymmetric: 50 ← 60 / 6 + 40 / 1
We get V ′ = 50 by cutting the heavy suitcase with divisor 6, leaving the lesser suitcase with divisor 1, unchanged.
Einstein is telling us how to repack our bags: divide each by 2. Do the airlines tell us how to repack our bags? No. Just cut total luggage weight in half.
The symmetric and asymmetric divisions are equal as values; also equal to the original. But the symmetric and asymmetric divisions are not identical physical situations. What governs that choice? The airlines don’t care. EDoMB does care. Why is the
I got into this answering a question of February 9 on Stackexchange.
https://physics.stackexchange.com/questions/749402/velocity-addition-in-special-relativity
Stack’s numbers come directly from Paul A. Tipler vol 3, which is why I answered the question.
A train is on the track with speed v = 0.8c. A rifle on the train fires off a bullet with speed w = 0.8c, speed judged on the moving train deck, not on the ground.
v = 0.8c and w = 0.8c
Velocity V or V ′ ( prime ) is bullet velocity as judged on the ground.
Classical velocity addition: V = v + w
Classical: V = 0.8c + 0.8c = 1.6c
Relativistic addition: V ′ ( prime ) = ( v + w ) / ( 1 + vw/c² )
Relativity: V ′ = ( 0.8c + 0.8c ) / ( 1 + 0.8c × 0.8c /c² )
V ′ = 1.6c / ( 1 + 0.64 )
V ′ = 1.6c / 1.64
V ′ = 0.98c
My answer on Stack is circuitous. Revised:
Bullet speed is to be judged on the ground. Granted, this bullet speed is V ′ = 0.98c, not V = 1.6c. What is train-on-ground speed?
Symmetric addition: V ′ = 0.8c /1.64 + 0.8c /1.64 ≈ 0.49c + 0.49c = 0.98c
Train-on-ground speed: 0.49c
Asymmetric addition: V ′ = 0.8c + 0.8c / 4.444 = 0.8c + 0.18c = 0.98c
Train-on-ground speed: 0.8c
Which is it?
Le 19/04/2023 à 11:06, Maciej Wozniak a écrit :
And aren't you asserting that these transformationsPoincaré-Lorentz transformations reflect the reality of things.
are reflecting the reality?
And aren't you asserting that these transformations
are reflecting the reality?
Not quite, but enough to make Paul's
"The "velocity addition formula" is a misnomer."
mistaken.
Einstein 1905: It is worthy of remark that v and w enter into the expression for
the resultant velocity in a symmetrical manner.
Why is it worthy of remark? What is symmetrical? Is Einstein blinkered benighted purblind and myopic?
Stackexchange couldn’t deal with it and I was given the boot.
Here are “plus-sign” Lorentz Transforms, axis X′ ( primed ) being the pavement and axis X ( unprimed ) being the moving axis, moving with negative speed-v. The transforms reverse the sign of speed-v to +v.
x = gamma × ( x′ + vt′ )
t = gamma × ( t′ + vx′ / c² )
Here is the Galilean form shown in textbooks.
x = x′ + vt′
t = t′
Example numbers: the origin x = 0 of axis X moves negatively at speed v = – 3
miles per hour. This motion continues for time t′ = 1 hour. Point-x′ is fixed value x′ = 0. It is always at the origin of axis X′. Speed-v reversed
to positive speed-v.
x = 0 miles + 3 mph × 1 hour = 3 miles
t = 1 hour
Speed of point-x on axis X is x / t:
x / t = 3 miles / 1 hour = 3 miles per hour
The origin of axis X moves negatively at 3 mph away from the origin [ x′ = 0 ]
of axis X′. Point-x′ is at origin point [ x′ = 0 ] appears on retreating
axis X to move positively for 1 hour, beginning at [ x = 0 ] and ending up at x =
3 miles, hence 3 mph.
One new number: the origin x = 0 of axis X moves negatively at speed v = – 3
miles per hour. This motion continues for time t′ = 1 hour. Point-x′ is fixed value x′ = 1 on axis X′.
x = 1 mile + 3 mph × 1 hour = 4 miles
t = 1 hour
Speed of point-x on axis X is x / t:
x / t = 4 miles / 1 hour = 4 miles per hour
Wolfgang Pauli: “It’s not even wrong.”
Stationary point x′ = 1 appears on negatively moving axis X to move from x = 1
to x = 4 over the course of an hour. Distance equals 3 miles.
3 miles / 1 hour = 3 miles per hour
Necromancers believe algebra can turn base metal into gold. Not always.
Je ne comprends pas ce que tu essayes de faire.
R.H.
R.H.
The LT’s transform point-x′ into point-x. Point-x′ must start out as x′
= 0. It can remain there or move on to x′ = wt′ ( velocity addition ) or x′
= – vt′ ( time dilation ). A stationary point such as x′ = 1 or x′ = 2 is
a falsehood.
C’est seulement merde.
Le 23/08/2023 à 15:01, xip14 a écrit :
R.H.
The LT’s transform point-x′ into point-x. Point-x′ must start out as x′
= 0. It can remain there or move on to x′ = wt′ ( velocity addition ) or x′
= – vt′ ( time dilation ). A stationary point such as x′ = 1 or x′ = 2 is
a falsehood.
C’est seulement merde.Mais non, c'est pas merde.
C'est toi qui ne comprends pas de quoi on parle et qui croit que c'est merde.
Poincaré transformations are used for this:
We have an event that occurs in an R repository,
in a given position and in a given time.
At the MOMENT, when the event is perceived, and not when it occurs, the observer O starts his watch.
Observer O then says, there is an event E that happened in
x,y,z), there is now a time To.
Attention, To is ALWAYS negative in principle. We never perceive an event that will occur in the future.
He sets for example E=(12,9,0,-15)
To=-sqrt(x²+y²+z²)/c
Another observer O' who crosses O at this precise moment, and at speed Vo=0.8c, also recovers the photons on his retina.
But he will not see the event in the same place (as in the aberration of
the aposition of the stars at the zenith).
Nor will he design the event that took place at the same time.
It is necessary to use the transformers of Poincaré to know all that, and all becomes very simple.
E=(40,9,0,-41).
Here is a perfect example illustrating the transformations of Poincaré
and their utility.
I will never understand how on the physics forums, loads of individuals
find it beautiful and fulfilling to spit on me while I explain things to them with an unprecedented clarity.
This is part of human bullshit, which is unfathomable.
R.H.
But Special Relativity won’t let the moving guy be the observer.You should stop lying
[...] But Special Relativity won’t let the moving guy be the
observer.
[... further mistakes omitted]
Time dilation has been around for 120 years and nobody has been able
to ask a question: What is time dilation?
On 8/26/23 10:09 AM, xip14 wrote:Why does force and motion slow time. How do they project?
Time dilation has been around for 120 years and nobody has been ableNonsense. That question has been asked and answered for those 120 years: "time dilation" is simply a geometric projection. Ditto for
to ask a question: What is time dilation?
"length contraction".You can only add to your own velocity.
Tom Roberts
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