Re: A discussion of 'Tachyons, the 4-momentum ...'

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Sujet : Re: A discussion of 'Tachyons, the 4-momentum ...'
De : hitlong (at) *nospam* yahoo.com (gharnagel)
Groupes : sci.physics.relativity
Date : 05. Oct 2024, 20:35:12
Autres entêtes
Organisation : novaBBS
Message-ID : <428875ea108f76fa154169b3e097aa39@www.novabbs.com>
References : 1
User-Agent : Rocksolid Light
On Thu, 3 Oct 2024 10:10:06 +0000, gharnagel wrote:
>
In DOI: 10.13189/ujpa.2023.170101, Figure 4 depicts a thought
experiment involving FTL communication.  This figure is
attached to this post.  We, the viewers, are at rest with
observers A and B.
>
At event E1, observer A, at x = 0, t = vL/c^2, sends an
infinitely-fast tachyon signal to observer B, who receives
it at x = L, t = vL/c^2.  B immediately passes it to D, who
is moving at velocity v.  D's clock reads t' = 0.
>
D is now going to send it to C.  The question is, where is
C?  Is C at x = 0 or is C at x = Lv^2/c^2?  According to
us (the viewers), C is at x = Lv^2/c^2, but if we (the
viewers) switched frames so we're at rest with C and D,
(see Figure 5 in DOI: 10.13189/ujpa.2023.170101).  Now,
it appears that D can send a signal to c at t' = t = 0
(infinitely-fast).
>
But according to our (the viewers) present position in
Figure 5, A (at t = 0) could only send the original signal
to B, who would receive it at t = vL/c^2, or the loop
would not be completed (because of the relativity of
simultaneity).
>
Let's move back to Figure 4.  So in order to complete
the loop, D must send the signal to where C is not (at
t = 0, x = 0), which flouts RoS.  But at t = vL/c^2, C
is not adjacent to A, so no closed loop can be completed
in this scenario.
This scenario is described as Method II, which demonstrates
that RoS limits the ability to complete a closed message
loop that violates causality.  There are some that still
can't understand the logic, so let's look at Method I.
Method I shows that E = mc^2/sqrt(u^2/c^ - 1) for tachyons,
where the tachyon mass is im and u is the tachyon velocity
in frame S.  The valid range of u is split: -\infty < u < -c
and c < u < \infty.  It is VERY important to note that
E NEVER becomes negative for ANY valid value of u.
By the Principle of Relativity, the tachyon velocity in frame
S' is E' = mc^2/sqrt(u'^2/c^2 - 1), where u' has the same
range as in S.  E' may be derived from the E equation by the
relativistic velocity composition equation (RVCE):
u' = (u - v)/(1 - uv/c^2)
This is just as valid as the Lorentz tranformation equations.
Applying this,
E = mc^2 sqrt[(1 - uv/c^2)^2]/sqrt(u^2/c^2 - 1)
Conventional physicists have made an error at this point which
isn't wrong unless 1 - uv/c^2 < 0.  When they do this, they
proclaim that E becomes negative, which E NEVER does throughout
the whole valid range for u, as demonstrated above!
This should alert any thoughtful physicist that the RCVE has
a limited domain of applicability.  This must mean that the
LT also has the same limitation.
Those that assert that Method II can violate causality do so
by working in the region where 1 - uv/c^2 is negative, which
is outside of the applicability range of the LT.  They have
no foundation for their assertions.

Date Sujet#  Auteur
3 Oct12:10 * A discussion of 'Tachyons, the 4-momentum ...'11gharnagel
3 Oct13:25 +- Re: A discussion of 'Tachyons, the 4-momentum ...'1Maciej Wozniak
3 Oct14:36 +* Re: A discussion of 'Tachyons, the 4-momentum ...'7Richard Hachel
3 Oct19:30 i`* Re: A discussion of 'Tachyons, the 4-momentum ...'6gharnagel
3 Oct20:09 i `* Re: A discussion of 'Tachyons, the 4-momentum ...'5Richard Hachel
4 Oct13:28 i  `* Re: A discussion of 'Tachyons, the 4-momentum ...'4gharnagel
4 Oct13:55 i   `* Re: A discussion of 'Tachyons, the 4-momentum ...'3Richard Hachel
4 Oct21:36 i    `* Re: A discussion of 'Tachyons, the 4-momentum ...'2gharnagel
4 Oct22:01 i     `- Re: A discussion of 'Tachyons, the 4-momentum ...'1Richard Hachel
5 Oct20:35 `* Re: A discussion of 'Tachyons, the 4-momentum ...'2gharnagel
5 Oct20:57  `- Re: A discussion of 'Tachyons, the 4-momentum ...'1Richard Hachel

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