• Re: Addition of velocity

    From Sylvia Else@21:1/5 to All on Wed Apr 19 15:42:52 2023
    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 ′
    prime ) is reduced accordingly, to a value no greater than 50.

    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 symmetric chosen? Why does the symmetric take precedence over the asymmetric?


    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.

    Sylvia.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From xip14@21:1/5 to All on Tue Apr 18 22:24:59 2023
    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 ′
    prime ) is reduced accordingly, to a value no greater than 50.

    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 symmetric chosen? Why does the symmetric take precedence over the asymmetric?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Sylvia Else@21:1/5 to Maciej Wozniak on Wed Apr 19 16:35:52 2023
    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 ′
    prime ) is reduced accordingly, to a value no greater than 50.

    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
    symmetric chosen? Why does the symmetric take precedence over the asymmetric? >>>
    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.


    And if you're dealing with high velocities, you'll get the wrong answer.

    Sylvia.

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  • From Maciej Wozniak@21:1/5 to Sylvia Else on Tue Apr 18 23:34:16 2023
    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 ′
    prime ) is reduced accordingly, to a value no greater than 50.

    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
    symmetric chosen? Why does the symmetric take precedence over the asymmetric?

    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.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Maciej Wozniak@21:1/5 to Sylvia Else on Tue Apr 18 23:50:50 2023
    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 ′
    prime ) is reduced accordingly, to a value no greater than 50.

    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 symmetric chosen? Why does the symmetric take precedence over the asymmetric?

    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.

    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).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Maciej Wozniak@21:1/5 to Paul B. Andersen on Wed Apr 19 02:06:34 2023
    On Wednesday, 19 April 2023 at 10:29:17 UTC+2, Paul B. Andersen wrote:
    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.

    No.

    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.


    And aren't you asserting that these transformations
    are reflecting the reality?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Paul B. Andersen@21:1/5 to All on Wed Apr 19 10:29:10 2023
    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.

    https://paulba.no/div/LTorigin.pdf
    Chapter 3 p.3


    <snip nonsense>

    --
    Paul

    https://paulba.no

    --- SoupGate-Win32 v1.05
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  • From Sylvia Else@21:1/5 to Maciej Wozniak on Wed Apr 19 19:40:49 2023
    On 19-Apr-23 4:50 pm, Maciej Wozniak wrote:
    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 ′
    prime ) is reduced accordingly, to a value no greater than 50.

    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 symmetric chosen? Why does the symmetric take precedence over the asymmetric?

    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.

    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. Congratulations - more than 180 years
    after the event, and you'd still be getting it wrong.

    Sylvia.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Maciej Wozniak@21:1/5 to Sylvia Else on Wed Apr 19 02:54:11 2023
    On Wednesday, 19 April 2023 at 11:40:52 UTC+2, Sylvia Else wrote:
    On 19-Apr-23 4:50 pm, Maciej Wozniak wrote:
    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 ′
    prime ) is reduced accordingly, to a value no greater than 50.

    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 symmetric chosen? Why does the symmetric take precedence over the asymmetric?

    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.

    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.

    No. You're fabricating. I'm not trying to predict
    what happened almost 200 years ago. And
    whatever You imagine, describing this result
    with GT is no way any real problem.. Samely as
    any other result You have.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From xip14@21:1/5 to All on Wed Apr 19 03:43:33 2023
    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?

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  • From Richard Hachel@21:1/5 to All on Wed Apr 19 22:45:20 2023
    Le 19/04/2023 à 07:25, xip14 a écrit :

    Addition of velocity, 1905: V = ( v + w ) / ( 1 + vw/c² )

    <http://news2.nemoweb.net/jntp?JQL4PSAnOl8zUAIxhmmN8WU8IuQ@jntp/Data.Media:1>

    <http://news2.nemoweb.net/?DataID=JQL4PSAnOl8zUAIxhmmN8WU8IuQ@jntp>

    R.H.

    --- SoupGate-Win32 v1.05
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  • From mitchrae3323@gmail.com@21:1/5 to All on Wed Apr 19 15:22:27 2023
    On Tuesday, April 18, 2023 at 10:25:01 PM UTC-7, 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 ′ prime )
    is reduced accordingly, to a value no greater than 50.

    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
    symmetric chosen? Why does the symmetric take precedence over the asymmetric?

    You can only add to your own velocity.
    Frames have their own motion obeying light speed in space.
    Individual frames obey the speed limit "converging
    or diverging in space" below 2c.
    There are two levels.

    Mitchell Raemsch

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  • From xip14@21:1/5 to All on Wed Apr 19 17:24:16 2023
    R. H.

    https://groups.google.com/g/alt.arts.poetry.comments/c/kNrDf1vv1Yw

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  • From Volney@21:1/5 to All on Thu Apr 20 02:37:55 2023
    On 4/19/2023 6:43 AM, xip14 wrote:
    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?

    Since the speed of the train was already defined to be 0.8c, why are you
    even asking this?

    Symmetric addition: V ′ = 0.8c /1.64 + 0.8c /1.64 ≈ 0.49c + 0.49c = 0.98c

    What the hell is this? Make up your own physics day? And why are you
    using a mixture of nonrelativistic speed addition and the relativistic
    speed combination formulas?

    Train-on-ground speed: 0.49c

    Asymmetric addition: V ′ = 0.8c + 0.8c / 4.444 = 0.8c + 0.18c = 0.98c

    More make-up-your-own-physics?

    Train-on-ground speed: 0.8c

    Which is it?

    Since you started off with train-on-ground speed: 0.8c, I'd say 0.8c.
    And shitcan those bizarre attempts at physics.

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  • From Maciej Wozniak@21:1/5 to Richard Hachel on Thu Apr 20 02:29:19 2023
    On Thursday, 20 April 2023 at 11:24:35 UTC+2, Richard Hachel wrote:
    Le 19/04/2023 à 11:06, Maciej Wozniak a écrit :

    And aren't you asserting that these transformations
    are reflecting the reality?
    Poincaré-Lorentz transformations reflect the reality of things.

    Not quite, but enough to make Paul's
    "The "velocity addition formula" is a misnomer."
    mistaken.

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  • From Richard Hachel@21:1/5 to All on Thu Apr 20 09:24:32 2023
    Le 19/04/2023 à 11:06, Maciej Wozniak a écrit :

    And aren't you asserting that these transformations
    are reflecting the reality?

    Poincaré-Lorentz transformations reflect the reality of things.

    They lead to an equation called general addition of velocities
    (relativistic or not).

    If we take an addition of longitudinal velocities, we get:
    w=(v+u)/(1+vu/c²)

    If we take an addition of transverse velocities, we obtain: w=sqrt(v²+u²-v²u²/c²)

    And so on, if we take the right formula.

    <http://news2.nemoweb.net/jntp?ncEkajjMLN0vt4CIU-2YXxP9ZzU@jntp/Data.Media:1>

    Note that it also works for small measurable speeds in our everyday world.

    Good day to all.

    R.H.

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  • From Richard Hachel@21:1/5 to All on Thu Apr 20 09:40:46 2023
    Le 20/04/2023 à 11:29, Maciej Wozniak a écrit :
    Not quite, but enough to make Paul's
    "The "velocity addition formula" is a misnomer."
    mistaken.

    No, only the speed addition formula is correct, but in addition, there is
    no language error by Paul.

    It is indeed an addition of speeds, except that this addition is not
    simple.

    We add a speed phenomenon to another speed phenomenon with a very precise
    law.

    I validate the equation, and I validate the term that I used (so on that,
    I am 100% in agreement with Paul).


    R.H.

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  • From Richard Hachel@21:1/5 to All on Thu Apr 20 10:44:53 2023
    Le 20/04/2023 à 12:28, xip14 a écrit :
    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?

    If the universe is covariant, the effects must be symmetric.

    What an observer sees from the end of his relativistic telescope, the
    other opposite must see the same effects.

    That's how I was able to solve the Langevin paradox that no one had ever
    solved before me.

    It's a pity that the whole universe refuses to listen to me on everything.

    For a long time I believed in an anomaly, but no, the fact that people
    don't listen to me is one of the most beautiful things about healing.

    This discredits all those who show off by believing themselves to be
    geniuses because they invariably repeat the recitations they learned at
    school, and who spit on everything around them.

    This gives evidence of their incapacity.

    R.H.

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  • From xip14@21:1/5 to All on Thu Apr 20 03:28:56 2023
    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?

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  • From xip14@21:1/5 to All on Thu Apr 20 06:24:07 2023
    The idea here seems to be, we “transform” a speed, only one speed, bullet speed-w, not train speed-v. The rifleman on the train maintains bullet speed w = 0.8c . Transformation of speed-w is such that, in the ground based frame of reference, the
    stationmaster sees something else. What has changed? Does the ground guy see bullet-on-train speed different from w = 0.8c ? Or does he see bullet-on-ground speed different from V = 1.6c ? In either case, he hangs onto train-on-ground speed v = 0.8c .

    I don’t think this is Einstein’s approach.

    Stackexchange couldn’t deal with it and I was given the boot.

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  • From Dono.@21:1/5 to All on Thu Apr 20 07:28:49 2023
    On Thursday, April 20, 2023 at 6:24:09 AM UTC-7, xip14 wrote:

    Stackexchange couldn’t deal with it and I was given the boot.

    They don't suffer cretins like you much

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  • From xip14@21:1/5 to All on Tue Aug 22 12:51:56 2023
    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.

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  • From Richard Hachel@21:1/5 to All on Tue Aug 22 20:18:51 2023
    Le 22/08/2023 à 21:51, xip14 a écrit :
    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.

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  • From xip14@21:1/5 to All on Wed Aug 23 06:00:27 2023
    Je ne comprends pas ce que tu essayes de faire.

    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.

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  • From Richard Hachel@21:1/5 to All on Wed Aug 23 14:45:15 2023
    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.

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  • From xip14@21:1/5 to All on Wed Aug 23 10:09:18 2023
    You’re right. It’s wrong.

    A stationary point such as x′ = 1 or x′ = 2 can certainly exist on axis X′. It’s just that those stationary points cannot be transformed into points-of-speed on axis X, whereas x′ = 0 of speed-0 can be transformed into speed +v on axis X when
    axis X itself has speed –v on axis X′.

    More: when the full gamma transform is used, stationary point x′ = 1 on axis X′ can be transformed into stationary point-x on axis X.

    x = gamma × ( x′ + vt′ )

    x = gamma × ( 1 + vt′ )

    t′ = 0

    x = gamma × 1

    For example, gamma = 1.5

    x = 1.5 × 1 = 1.5

    Point-x has transformed value 1.5

    It is a vexatious business described in 1905-Section §4. A moving meter stick of length 1 meter does not appear to be something in motion with length 1.5 meters. It’s the reverse. The moving meter stick “appears” to have length 2 / 3 meters.

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  • From mitchrae3323@gmail.com@21:1/5 to Richard Hachel on Wed Aug 23 10:11:41 2023
    On Wednesday, August 23, 2023 at 7:45:18 AM UTC-7, Richard Hachel wrote:
    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.


    Frames can converge on each other in space.
    They have their own velocities below c.
    The atom can move at near light speed toward itself.
    If two atoms are they are converging below 2c.

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  • From xip14@21:1/5 to All on Thu Aug 24 06:27:15 2023
    Section §4: time dilation and Lorentz Contraction

    The ALGEBRA of Section §4 allows the moving guy to say, “My clock is slower than your clock,” and “My meter stick is shorter than your meter stick.” But Special Relativity won’t let the moving guy be the observer. So the stationary guy says,
    “Your clock is slower than my clock,” and “Your meter stick is shorter than my meter stick.”

    Velocity-addition is likewise Lorentz Transformation. The ALGEBRA allows the moving guy to say, “My clock is faster than your clock, so my rifle bullet is slower than your rifle bullet.” But Special Relativity won’t let the moving guy be the
    observer. The stationary guy must be the observer, saying to the moving guy, “Your rifle bullet is not up to speed.” The velocity-addition-formula cannot function as intended on that basis.

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  • From Dono.@21:1/5 to All on Thu Aug 24 07:50:11 2023
    On Thursday, August 24, 2023 at 6:27:17 AM UTC-7, xip14 wrote:
    But Special Relativity won’t let the moving guy be the observer.
    You should stop lying

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  • From Tom Roberts@21:1/5 to All on Thu Aug 24 14:30:40 2023
    On 8/24/23 8:27 AM, xip14 wrote:
    [...] But Special Relativity won’t let the moving guy be the
    observer.

    This is not true. Go back and read Einstein's 1905 paper. In it he
    starts out
    "Let us take a system of co-ordinates in which the
    equations of Newtonian mechanics hold good. In order
    to render our presentation more precise and to
    distinguish this system of co-ordinates verbally from
    others which will be introduced hereafter, we call it
    the “stationary system.”

    This starts with an English translation of the German phrase used by mathematicians to select an ARBITRARY system of coordinates that satisfy
    the given condition. Moreover, his words are quite clear that
    "stationary system" is merely a LABEL for whichever coordinates one
    chooses to give that label.

    So whichever guy you call "moving", anyone else can call "stationary".
    Given Einstein's first postulate, ANY inertial frame can be so chosen.

    [... further mistakes omitted]

    Tom Roberts

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  • From xip14@21:1/5 to All on Sat Aug 26 08:09:14 2023
    Time dilation has been around for 120 years and nobody has been able to ask a question: What is time dilation?

    I ask a question:

    V = ( 8 + 12 ) / 2 = 20 / 2 = 10

    Option-A: V = ( 8 + 12 ) / 2 = 8 / 2 + 12 / 2 = 4 + 6 = 10

    Option-B: V = ( 8 + 12 ) / 2 = 8 / 1 + 12 / 6 = 8 + 2 = 10

    Which is it?

    Einstein says it is A and the Large Hadron Collider says it is B.

    https://dibdeck.blogspot.com/

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  • From Tom Roberts@21:1/5 to All on Sat Aug 26 15:57:45 2023
    On 8/26/23 10:09 AM, xip14 wrote:
    Time dilation has been around for 120 years and nobody has been able
    to ask a question: What is time dilation?

    Nonsense. That question has been asked and answered for those 120 years:
    "time dilation" is simply a geometric projection. Ditto for
    "length contraction".

    Tom Roberts

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  • From mitchrae3323@gmail.com@21:1/5 to Tom Roberts on Sat Aug 26 20:54:27 2023
    On Saturday, August 26, 2023 at 1:57:57 PM UTC-7, Tom Roberts wrote:
    On 8/26/23 10:09 AM, xip14 wrote:
    Time dilation has been around for 120 years and nobody has been able
    to ask a question: What is time dilation?
    Nonsense. That question has been asked and answered for those 120 years: "time dilation" is simply a geometric projection. Ditto for
    Why does force and motion slow time. How do they project?
    "length contraction".

    Tom Roberts
    You can only add to your own velocity.
    Your movement does not move other frames...
    All forces are geometric. And gravity is not an open curve metric.
    Instead of a parabola it is a Riemann sphere ...

    Mitchell Raemsch

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