• =?UTF-8?Q?Re=3A_How_Einstein_missed_his_opportunity_to_derive_Loren?= =

    From Paul B. Andersen@21:1/5 to Paul B. Andersen on Mon Feb 10 21:39:39 2025
    Den 09.02.2025 00:40, skrev rhertz:
    I was reading again the English version (1923) of his 1905 paper, and
    got interested in a footnote from the editor:

    QUOTE:
    5The equations of the Lorentz transformation may be more simply deduced directly from the condition that in virtue of those equations the
    relation
    x² + y² + z²  = c² t² shall have as its consequence the second relation ξ² + η² + ζ²  = c²τ².
    END QUOTE

    Right.

    A rather obvious consequence of the postulates of SR is that
    the speed of light is invariant c, that is c in all inertial
    frames of reference.

    That is, the light will use the time t = L/c to travel the distance
    L = √(x²+y+z²) => c²t² = x²+y²+z² and thus c²τ² = ξ²+η²+ζ²

    When Einstein in §3 of "Electrodynamics" writes
    x² + y² + z² = c²t²
    and
    ξ² + η² + ζ² = c²τ²
    it is a mathematical expression of
    "the speed of light in K is c" and
    "the speed of light in k is c"

    You can find the derivation of the LT from this starting
    point in several textbooks, and on the net:

    https://testbook.com/physics/derivation-of-lorentz-transformation https://www.vedantu.com/physics/derivation-of-lorentz-transformation https://people.iith.ac.in/kdmakwana/ep2017/lecture2.pdf (see eqn. (33) https://tinyurl.com/26f3xgrp https://dacollege.org/uploads/stdmat/Physics-derivation-of-Lorentz-Transformation-Equation-dem4.pdf

    And there are several other ways to derive the Lorentz transform,
    common for them all are that they start with the postulates
    of SR, usually in the form "speed of light is invariant c"-

    February 7, Paul B. Andersen wrote
    | It is a _fact_ that the Lorentz transform follows from
    | the postulates of SR.
    |
    | This is proven many times by several physicists.
    | Einstein was the first to do it, but he did it in a rather
    | cumbersome way which, as you have demonstrated, may not
    | be simple to understand.
    | In the more than a century since Einstein did it, it has
    | been done many times in much more elegant ways which are
    | simpler to understand.



    ------------------------------------------------------------------------

    This is the derivation of Lorentz transforms using the above concepts:


    ***********************************************

    He could have derived Lorentz transforms by simply postulating that as

    c² t² - x² - y² - z²  = 0

    and

    c² τ² - ξ² - η² - ζ²  = 0

    then

    c² τ² - ξ² - η² - ζ²  = c² t² - x² - y² - z²

    As y = η  and  z = ζ, this equation reduces to c² τ² - ξ²  = c² t² - x²

    Perfect so far!


    Assuming that ξ and x differs in a factor β from Galilean transform,
    where β = 1, the new transform can be written as

    ξ  = β (x - vt)

    This has nothing to do with the Galilean transform.
    You know the transform can be written on the form above.

    Would you say:
    "Assuming that τ and t differs in a factor β from Galilean transform,
    where β = 1, the new transform can be written as τ = βt " ?

    Obviously not, because you know the transform has nothing to do
    with the Galilean transform and has a different form.

    It is however true that you by reasoning can find that
    the transform must be of the form ξ = β (x - vt).

    But you have made no reasoning, you are using your knowledge
    of what the form of the transform should be to derive the transform.

    But never mind, let it pass.
    It is true that the transform is of the form ξ = β (x - vt)


    and

    x  = β (ξ + v τ)

    then

    x  = β² (x - vt) + β v τ

    β v τ = x (1 - β²) + β² v t

    τ = x (1 - β²)/βv + β  t

    τ = β [(1/β² - 1) x/v +  t]

    Replacing ξ and τ in ξ² = c² τ², it follows that

    β² (x - vt)² = c² β² [x (1/β² - 1) x/v +  t]²

    Factoring the above equation with a little help of algebra, it's
    obtained

    [β² - c²β²/v² (1/β² - 1)] x² - 2 β² [v + c²/v (1/β² - 1)] xt = β² (c² -
    v²) t²

    To verify

    x² + y² + z²  = c² t²

    It's required that, in the previous equation,

    [β² - c²β²/v² (1/β² - 1)] = 1

    [v + c²/v (1/β² - 1)] = 0

    β² (c² - v²) = c²

    From the last equation,

    β²  = c²/(c² - v²)

    or

    β  = 1/√(1 - v²/c²) , which is the modern factor Gamma (γ).

    With this result of β  (γ), the transforms are

    ξ  = β (x - vt) = (x - vt)/√(1 - v²/c²)

    τ = β [(1/β² - 1) x/v +  t] = β (t - vx/c²) = (t - vx/c²)/√(1 - v²/c²)


    OK, well done.

    You have now derived the Lorentz transform from
    the principle "speed of light is invariant c"
    which follows from the postulates of SR.

    So now you agree:
    It is a _fact_ that the Lorentz transform
    follows from the postulates of SR.

    Or don't you?

    --
    Paul

    https://paulba.no/

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From The Starmaker@21:1/5 to Paul B. Andersen on Mon Feb 10 15:10:13 2025
    Paul B. Andersen wrote:

    Den 09.02.2025 00:40, skrev rhertz:
    I was reading again the English version (1923) of his 1905 paper, and
    got interested in a footnote from the editor:

    QUOTE:
    5The equations of the Lorentz transformation may be more simply deduced directly from the condition that in virtue of those equations the
    relation
    x² + y² + z² = c² t² shall have as its consequence the second relation
    ξ² + η² + ζ² = c²τ².
    END QUOTE

    Right.

    A rather obvious consequence of the postulates of SR is that
    the speed of light is invariant c, that is c in all inertial
    frames of reference.

    That is, the light will use the time t = L/c to travel the distance
    L = √(x²+y+z²) => c²t² = x²+y²+z² and thus c²τ² = ξ²+η²+ζ²

    When Einstein in §3 of "Electrodynamics" writes
    x² + y² + z² = c²t²
    and
    ξ² + η² + ζ² = c²τ²
    it is a mathematical expression of
    "the speed of light in K is c" and
    "the speed of light in k is c"

    You can find the derivation of the LT from this starting
    point in several textbooks, and on the net:

    https://testbook.com/physics/derivation-of-lorentz-transformation https://www.vedantu.com/physics/derivation-of-lorentz-transformation https://people.iith.ac.in/kdmakwana/ep2017/lecture2.pdf (see eqn. (33) https://tinyurl.com/26f3xgrp https://dacollege.org/uploads/stdmat/Physics-derivation-of-Lorentz-Transformation-Equation-dem4.pdf

    And there are several other ways to derive the Lorentz transform,
    common for them all are that they start with the postulates
    of SR, usually in the form "speed of light is invariant c"-

    February 7, Paul B. Andersen wrote
    | It is a _fact_ that the Lorentz transform follows from
    | the postulates of SR.
    |
    | This is proven many times by several physicists.
    | Einstein was the first to do it, but he did it in a rather
    | cumbersome way which, as you have demonstrated, may not
    | be simple to understand.
    | In the more than a century since Einstein did it, it has
    | been done many times in much more elegant ways which are
    | simpler to understand.

    ------------------------------------------------------------------------

    This is the derivation of Lorentz transforms using the above concepts:


    ***********************************************

    He could have derived Lorentz transforms by simply postulating that as

    c² t² - x² - y² - z² = 0

    and

    c² τ² - ξ² - η² - ζ² = 0

    then

    c² τ² - ξ² - η² - ζ² = c² t² - x² - y² - z²

    As y = η and z = ζ, this equation reduces to c² τ² - ξ² = c² t² - x²

    Perfect so far!


    Assuming that ξ and x differs in a factor β from Galilean transform, where β = 1, the new transform can be written as

    ξ = β (x - vt)

    This has nothing to do with the Galilean transform.
    You know the transform can be written on the form above.

    Would you say:
    "Assuming that τ and t differs in a factor β from Galilean transform,
    where β = 1, the new transform can be written as τ = βt " ?

    Obviously not, because you know the transform has nothing to do
    with the Galilean transform and has a different form.

    It is however true that you by reasoning can find that
    the transform must be of the form ξ = β (x - vt).

    But you have made no reasoning, you are using your knowledge
    of what the form of the transform should be to derive the transform.

    But never mind, let it pass.
    It is true that the transform is of the form ξ = β (x - vt)


    and

    x = β (ξ + v τ)

    then

    x = β² (x - vt) + β v τ

    β v τ = x (1 - β²) + β² v t

    τ = x (1 - β²)/βv + β t

    τ = β [(1/β² - 1) x/v + t]

    Replacing ξ and τ in ξ² = c² τ², it follows that

    β² (x - vt)² = c² β² [x (1/β² - 1) x/v + t]²

    Factoring the above equation with a little help of algebra, it's
    obtained

    [β² - c²β²/v² (1/β² - 1)] x² - 2 β² [v + c²/v (1/β² - 1)] xt = β² (c² -
    v²) t²

    To verify

    x² + y² + z² = c² t²

    It's required that, in the previous equation,

    [β² - c²β²/v² (1/β² - 1)] = 1

    [v + c²/v (1/β² - 1)] = 0

    β² (c² - v²) = c²

    From the last equation,

    β² = c²/(c² - v²)

    or

    β = 1/√(1 - v²/c²) , which is the modern factor Gamma (γ).

    With this result of β (γ), the transforms are

    ξ = β (x - vt) = (x - vt)/√(1 - v²/c²)

    τ = β [(1/β² - 1) x/v + t] = β (t - vx/c²) = (t - vx/c²)/√(1 - v²/c²)



    What is all dat stuff up there? i don't see 'any' numbers! It cannot be math if it ain't got any numbers...


    Any why so many equal signs???? and so many half moons )))))(((((()))))



    come on, give me at least ONE number! wait, is dat a ONE i see??? o thought it was another letter L....




    --
    The Starmaker -- To question the unquestionable, ask the unaskable,
    to think the unthinkable, mention the unmentionable, say the unsayable,
    and challenge the unchallengeable.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)