info
Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss
Tom Robert's Tachyon Disproof
226 views
Skip to first unread message
Gary Harnagel
Dec 20, 2020, 9:17:26 PM12/20/20
to
> Consider inertial frames S and S' with coordinates (x,y,z,t) and
> (x',y',z',t'), having parallel x/x' axes with S' moving in the +x
> direction with speed v relative to S; ignore y,z,y',z'. Arrange the
> coordinates so x=0,t=0 coincides with x'=0,t'=0. Let g=1/sqrt(1-v^2) and
> L be a positive distance. Let w be the speed of the tachyonic signal
> relative to the inertial frame in which the transmitter and receiver are
> at rest. Use units with c=1, so 0<v<1 and w>1.
>
> At x=0,t=0 a transmitter at rest in S sends a tachyonic signal in the +x
> direction, which is received at x=L,t=L/w by a receiver at rest in S.
That is, at t = 0:
A --> w ______________ B
t = L/w:
A ______________ w --> B
> We have pre-arranged a transmitter at rest in S' to coincide with
> x=L,t=L/w, so the receiver in S can instantly send a normal signal to
> this transmitter in S' over zero distance. The transmitter in S'
> immediately sends a second tachyonic signal in the -x' direction.
That is, at t = L/w:
________________ w <-- D
A ____________________ B
> That signal was clearly sent at:
> x' = g(x - v t) = g L - g v L/w
> t' = g(t - v x) = g L/w - g v L
>
> In passing note the following:
> * x' > 0 (as expected)
> * if 1/w > v: t'>0 and there can be no causality violation.
> * if 1/w < v: t'<0 and the analysis continues; this
> condition always applies in the limit v -> 1.
> -----------------
> We have also pre-arranged a receiver at rest in S' to coincide with x=0
> when it receives that second tachyonic signal. It will immediately send
> a normal signal over zero distance to the transmitter in S, completing a
> causal loop -- the question is: does it arrive at t>0 (causality OK), or
> t<0 (causality violation)?
That is, at t = L/w + L/u:
C <-- u ________________ D
A ____________________ B
Is this arrangement actually possible? Tom used w =10c, v = 0.5c, but
if causality is NOT violated for w = ∞, it won't be violated for any slower
speed. So let's simplify the situation and let w = ∞:
What is tC' and tD' when the signal is launched from D?
tC' = 0, xC = 0, xD' = γL
tD' = γ( L/w - vL/c²) = -γvL/c²
There is some controversy about the speed of the return signal. Tom
assumed it was u' = - w, which would be u' = -∞. That would mean that
tC' = tD' = -γvL/c², but that occurred when t = -vL/c²:
t = -vL/c²:
C <-- u ________________ D
____A ____________________ B
At that time, C and A weren't adjacent, so A would have to wait until
xC = 0, at which time tC' = γ(0 - 0) = 0. Therefore, since ∆x' = -γL and
∆t' = 0 - (0 -γvL/c²) = γvL/c², u' = -c²/v. And THAT is exactly what
relativity of simultaneity requires.
So the time to send a signal signal message around a loop, even
infinitely-fast, is NOT less than zero, causality cannot be violated.
And since tachyon receivers can't really detect a signal with zero
energy, the speed must be a bit less than infinitely-fast, so ∆t > 0
for any possible speed.
> (x',y',z',t'), having parallel x/x' axes with S' moving in the +x
> direction with speed v relative to S; ignore y,z,y',z'. Arrange the
> coordinates so x=0,t=0 coincides with x'=0,t'=0. Let g=1/sqrt(1-v^2) and
> L be a positive distance. Let w be the speed of the tachyonic signal
> relative to the inertial frame in which the transmitter and receiver are
> at rest. Use units with c=1, so 0<v<1 and w>1.
>
> At x=0,t=0 a transmitter at rest in S sends a tachyonic signal in the +x
> direction, which is received at x=L,t=L/w by a receiver at rest in S.
That is, at t = 0:
A --> w ______________ B
t = L/w:
A ______________ w --> B
> We have pre-arranged a transmitter at rest in S' to coincide with
> x=L,t=L/w, so the receiver in S can instantly send a normal signal to
> this transmitter in S' over zero distance. The transmitter in S'
> immediately sends a second tachyonic signal in the -x' direction.
That is, at t = L/w:
________________ w <-- D
A ____________________ B
> That signal was clearly sent at:
> x' = g(x - v t) = g L - g v L/w
> t' = g(t - v x) = g L/w - g v L
>
> In passing note the following:
> * x' > 0 (as expected)
> * if 1/w > v: t'>0 and there can be no causality violation.
> * if 1/w < v: t'<0 and the analysis continues; this
> condition always applies in the limit v -> 1.
> -----------------
> We have also pre-arranged a receiver at rest in S' to coincide with x=0
> when it receives that second tachyonic signal. It will immediately send
> a normal signal over zero distance to the transmitter in S, completing a
> causal loop -- the question is: does it arrive at t>0 (causality OK), or
> t<0 (causality violation)?
That is, at t = L/w + L/u:
C <-- u ________________ D
A ____________________ B
Is this arrangement actually possible? Tom used w =10c, v = 0.5c, but
if causality is NOT violated for w = ∞, it won't be violated for any slower
speed. So let's simplify the situation and let w = ∞:
What is tC' and tD' when the signal is launched from D?
tC' = 0, xC = 0, xD' = γL
tD' = γ( L/w - vL/c²) = -γvL/c²
There is some controversy about the speed of the return signal. Tom
assumed it was u' = - w, which would be u' = -∞. That would mean that
tC' = tD' = -γvL/c², but that occurred when t = -vL/c²:
t = -vL/c²:
C <-- u ________________ D
____A ____________________ B
At that time, C and A weren't adjacent, so A would have to wait until
xC = 0, at which time tC' = γ(0 - 0) = 0. Therefore, since ∆x' = -γL and
∆t' = 0 - (0 -γvL/c²) = γvL/c², u' = -c²/v. And THAT is exactly what
relativity of simultaneity requires.
So the time to send a signal signal message around a loop, even
infinitely-fast, is NOT less than zero, causality cannot be violated.
And since tachyon receivers can't really detect a signal with zero
energy, the speed must be a bit less than infinitely-fast, so ∆t > 0
for any possible speed.
mitchr...@gmail.com
Dec 20, 2020, 9:49:25 PM12/20/20
to
What makes them go backward in time?
And what is their future source?
Where are the coming from in the
future?
And what is their future source?
Where are the coming from in the
future?
Gary Harnagel
Dec 21, 2020, 12:29:25 AM12/21/20
to
On Sunday, December 20, 2020 at 7:49:25 PM UTC-7, mitchr...@gmail.com wrote:
>
> What makes them go backward in time?
You don't GET it: My proof says they DON'T go back in time.
>
> What makes them go backward in time?
> And what is their future source?
> Where are the coming from in the future?
Yves Beegle
Dec 21, 2020, 8:54:50 AM12/21/20
to
mitchr...@gmail.com wrote:
> What makes them go backward in time? And what is their future source?
> Where are the coming from in the future?
Are you giving any money to the poor?
> What makes them go backward in time? And what is their future source?
> Where are the coming from in the future?
Yves Beegle
Dec 21, 2020, 9:03:55 AM12/21/20
to
Gary Harnagel wrote:
> On Sunday, December 20, 2020 at 7:49:25 PM UTC-7, mitchr...@gmail.com
> wrote:
>>
>> What makes them go backward in time?
>
> You don't GET it: My proof says they DON'T go back in time.
Wow, that "proof" is quite a stretch. Photons are not going back in time,
> On Sunday, December 20, 2020 at 7:49:25 PM UTC-7, mitchr...@gmail.com
> wrote:
>>
>> What makes them go backward in time?
>
> You don't GET it: My proof says they DON'T go back in time.
and things travelling bellow that speed. Higher than that _IS_ going into
the future, which is *_absurd_*. Back to your arduino and your
scintillator. I don't even know what a scintillator is, let alone the
kitchen table version of it.
Yves Beegle
Dec 21, 2020, 9:08:06 AM12/21/20
to
Gary Harnagel wrote:
>> And what is their future source?
>
> You don't GET it: Since they don't go back in time, the source is in
> the past.
>
>> Where are the coming from in the future?
>
> You don't GET it: they're not coming from the future.
Did you start giving any money to the poor? Eating and drinking with the
>> And what is their future source?
>
> You don't GET it: Since they don't go back in time, the source is in
> the past.
>
>> Where are the coming from in the future?
>
> You don't GET it: they're not coming from the future.
money changers and the tax collectors is not good enough. You are not
good enough, son. Not even self-sufficient.
Gary Harnagel
Dec 21, 2020, 2:21:30 PM12/21/20
to
violate causality regardless of whether it is assumed that the signal
is received by C at xC = 0 or C at xC = -v²L/c² (i.e., tC' = -γvL/c²), there
is another class which assumes that xC = 0 BEFORE the signal arrives
at C:
t = -vL/c²:
C __________________ D
A ____________________ B
Again, it is assumed that w = ∞.
t = 0:
__ C ________________ D
A --> w ______________ B
__ C <-- u _____________ D
A ____________________ B
So if it is assumed that u' = -∞, then tC' = tD' = -γvL/c², and the signal goes
into the past of the stationary lab of A and B, and causality is violated.
Does this make sense? What about the fact that A and B KNOW that C is
at xC = +v²L/c², not at xC = 0? In Tom's scenario, it didn't matter whether
the signal went to a previous position of C at xC = -v²L/c² or at xC = 0 in
order to complete the message loop. In this scenario, if the signal goes
to xC = +v²L/c², the loop cannot be completed. If it goes to xC = 0 the
loop is completed but causality is violated.
But there are problems with the signal getting past C at xC = +v²L/c², which
is where A and B OBSERVE it to be. Can we discount the observation of A
and B? Doesn't that represent the physical presence of C?
When A sends the first signal to B, C and D have the same problem: they
observe B between A and x' = γL, not at x' = γL (where B was BEFORE A sent
the signal. So again, if the OBSERVATION of C and D is correct, then the
message loop cannot be formed, but if we discount their observation, the
signal goes into the past of C and D.
My preference is to accept the observations of all constituents. That's REALITY.
Thus, in REALITY, tachyons do NOT violate causality. Those who choose the
alternative allow themselves to be infused with fantasies that pretend to disprove
the existence of faster-than-light phenomena, but the the argument is specious.
Yves Beegle
Dec 21, 2020, 3:49:16 PM12/21/20
to
Gary Harnagel wrote:
>> So the time to send a signal signal message around a loop, even
>> infinitely-fast, is NOT less than zero, causality cannot be violated.
>> And since tachyon receivers can't really detect a signal with zero
>> energy, the speed must be a bit less than infinitely-fast, so ∆t > 0
>> for any possible speed.
>
> Although the particular scenario presented by Tom Roberts doesn't
> violate causality regardless of whether it is assumed that the signal is
> received by C at xC = 0 or C at xC = -v²L/c² (i.e., tC' = -γvL/c²),
> there is another class which assumes that xC = 0 BEFORE the signal
> arrives at C:
Moeny to the poor, son, without which you cannot enter.
>> So the time to send a signal signal message around a loop, even
>> infinitely-fast, is NOT less than zero, causality cannot be violated.
>> And since tachyon receivers can't really detect a signal with zero
>> energy, the speed must be a bit less than infinitely-fast, so ∆t > 0
>> for any possible speed.
>
> Although the particular scenario presented by Tom Roberts doesn't
> violate causality regardless of whether it is assumed that the signal is
> received by C at xC = 0 or C at xC = -v²L/c² (i.e., tC' = -γvL/c²),
> there is another class which assumes that xC = 0 BEFORE the signal
> arrives at C:
Revelation of John 12:9:
And the great dragon was cast out, that old serpent, called the Devil,
and Satan, which deceiveth the whole world: he was cast out into the
earth, and his angels were cast out with him.
Yves Beegle
Dec 21, 2020, 3:56:39 PM12/21/20
to
Gary Harnagel wrote:
> So if it is assumed that u' = -∞, then tC' = tD' = -γvL/c², and the
> signal goes into the past of the stationary lab of A and B, and
> causality is violated.
>
> Does this make sense? What about the fact that A and B KNOW that C is
> at xC = +v²L/c², not at xC = 0? In Tom's scenario, it didn't matter
> whether the signal went to a previous position of C at xC = -v²L/c² or
> at xC = 0 in order to complete the message loop. In this scenario, if
> the signal goes to xC = +v²L/c², the loop cannot be completed. If it
> goes to xC = 0 the loop is completed but causality is violated.
James 2:3:
> So if it is assumed that u' = -∞, then tC' = tD' = -γvL/c², and the
> signal goes into the past of the stationary lab of A and B, and
> causality is violated.
>
> Does this make sense? What about the fact that A and B KNOW that C is
> at xC = +v²L/c², not at xC = 0? In Tom's scenario, it didn't matter
> whether the signal went to a previous position of C at xC = -v²L/c² or
> at xC = 0 in order to complete the message loop. In this scenario, if
> the signal goes to xC = +v²L/c², the loop cannot be completed. If it
> goes to xC = 0 the loop is completed but causality is violated.
And ye have respect to him that weareth the gay clothing, and say unto
him, Sit thou here in a good place; and say to the poor, Stand thou
there, or sit here under my footstool:
James 2:5:
Hearken, my beloved brethren, Hath not God chosen the poor of this world
rich in faith, and heirs of the kingdom which he hath promised to them
that love him?
Revelation of John 3:17:
Because thou sayest, I am rich, and increased with goods, and have need
of nothing; and knowest not that thou art wretched, and miserable, and
poor, and blind, and naked:
James 2:6:
But ye have despised the poor. Do not rich men oppress you, and draw you
before the judgment seats?
Yves Beegle
Dec 21, 2020, 4:09:35 PM12/21/20
to
Gary Harnagel wrote:
> That is, at t = L/w + L/u:
> C <-- u ________________ D A ____________________ B
>
> Is this arrangement actually possible? Tom used w =10c, v = 0.5c, but
> if causality is NOT violated for w = ∞, it won't be violated for any
> slower speed. So let's simplify the situation and let w = ∞:
>
> What is tC' and tD' when the signal is launched from D?
James 2:2:
> That is, at t = L/w + L/u:
> C <-- u ________________ D A ____________________ B
>
> Is this arrangement actually possible? Tom used w =10c, v = 0.5c, but
> if causality is NOT violated for w = ∞, it won't be violated for any
> slower speed. So let's simplify the situation and let w = ∞:
>
> What is tC' and tD' when the signal is launched from D?
For if there come unto your assembly a man with a gold ring, in goodly
apparel, and there come in also a poor man in vile raiment;
mitchr...@gmail.com
Dec 21, 2020, 6:09:18 PM12/21/20
to
On Sunday, December 20, 2020 at 9:29:25 PM UTC-8, hit...@yahoo.com wrote:
> On Sunday, December 20, 2020 at 7:49:25 PM UTC-7, mitchr...@gmail.com wrote:
> >
> > What makes them go backward in time?
> You don't GET it: My proof says they DON'T go back in time.
So where do they come from?
> On Sunday, December 20, 2020 at 7:49:25 PM UTC-7, mitchr...@gmail.com wrote:
> >
> > What makes them go backward in time?
> You don't GET it: My proof says they DON'T go back in time.
Can a negative HIggs give them their energy?
> > And what is their future source?
> You don't GET it: Since they don't go back in time, the source is in the past.
> > Where are the coming from in the future?
> You don't GET it: they're not coming from the future.
That is what tachyons do...
Mitchell Raemsch
Gary Harnagel
Dec 21, 2020, 7:09:03 PM12/21/20
to
On Monday, December 21, 2020 at 4:09:18 PM UTC-7, mitchr...@gmail.com wrote:
> On Sunday, December 20, 2020 at 9:29:25 PM UTC-8, hit...@yahoo.com wrote:
> > On Sunday, December 20, 2020 at 7:49:25 PM UTC-7, mitchr...@gmail.com wrote:
> > >
> > > What makes them go backward in time?
> >
> > You don't GET it: My proof says they DON'T go back in time.
>
> So where do they come from?
Tachyons are HYPOTHETICAL particles that obey special relativity:
> On Sunday, December 20, 2020 at 9:29:25 PM UTC-8, hit...@yahoo.com wrote:
> > On Sunday, December 20, 2020 at 7:49:25 PM UTC-7, mitchr...@gmail.com wrote:
> > >
> > > What makes them go backward in time?
> >
> > You don't GET it: My proof says they DON'T go back in time.
>
> So where do they come from?
> Can a negative HIggs give them their energy?
> > > And what is their future source?
> >
> > You don't GET it: Since they don't go back in time, the source is in the past.
> >
> > > Where are the coming from in the future?
> >
> > You don't GET it: they're not coming from the future.
> >
> If they are in the positive lab contained they will annihilate with it...
> That is what tachyons do...
>
> Mitchell Raemsch
mitchr...@gmail.com
Dec 21, 2020, 8:04:51 PM12/21/20
to
Michael Moroney
Dec 21, 2020, 9:47:28 PM12/21/20
to
still not antimatter.
mitchr...@gmail.com
Dec 21, 2020, 9:59:28 PM12/21/20
to
Mitchell Raemsch
Michael Moroney
Dec 21, 2020, 10:23:01 PM12/21/20
to
positive energy which can be released as photons; for example an electron (511
keV mass) annihilates with an antimatter positron (also 511 keV mass) to
release two 511 keV photons.
mitchr...@gmail.com
Dec 21, 2020, 10:25:38 PM12/21/20
to
That would annihilate with negative energy
tachyons you moron...
Michael Moroney
Dec 21, 2020, 10:52:20 PM12/21/20
to
with positive energy to release photons with positive energy.
Tachyons aren't antimatter, if they exist at all.
>That would annihilate with negative energy
>tachyons you moron...
number, and their mass would also be an imaginary number if they had a real
valued energy (positive or negative)... You were, after all, looking for a
reason for the existence of imaginary numbers, there you are, tachyons with an
imaginary gamma by exceeding c... don't believe me? Do the math, calculate
gamma for v > c....go ahead...
🤯 <--- Mitch's mind after reading this...
Yves Beegle
Dec 22, 2020, 9:06:22 AM12/22/20
to
Michael Moroney wrote:
>>That would annihilate with negative energy tachyons you moron...
>
> I hate to tell you, but your beloved gamma for tachyons would be an
> imaginary number, and their mass would also be an imaginary number if
> they had a real valued energy (positive or negative)... You were, after
> all, looking for a reason for the existence of imaginary numbers, there
> you are, tachyons with an imaginary gamma by exceeding c... don't
> believe me?
> Do the math, calculate gamma for v > c....go ahead...
>
> 🤯 <--- Mitch's mind after reading this...
bull gates real name is billy gatinsky and Brad Pitt real name is Brandy
>>That would annihilate with negative energy tachyons you moron...
>
> I hate to tell you, but your beloved gamma for tachyons would be an
> imaginary number, and their mass would also be an imaginary number if
> they had a real valued energy (positive or negative)... You were, after
> all, looking for a reason for the existence of imaginary numbers, there
> you are, tachyons with an imaginary gamma by exceeding c... don't
> believe me?
> Do the math, calculate gamma for v > c....go ahead...
>
> 🤯 <--- Mitch's mind after reading this...
Pittinsky. Bam! Proofs:
ACTOR BRAD PITT DARKEST SECRET NOW EXPOSED
https://www.bitchute.com/video/PIIQE0EIXKHD/
Tom Roberts
Dec 26, 2020, 4:38:26 PM12/26/20
to
You ignored the fact that relative to S the second is
going BACKWARDS IN TIME (which cannot be captured in
such diagrams). This is shown below (for w=∞): relative
to S the second tachyon propagates from x=L,t=0 to
x=0,t=-vL. (My original calculation holds for all w>1).
Note that A and B are locations at rest in S, and C and D are locations
at rest in S'. In S, C is adjacent to A BEFORE D is adjacent to B.
Unlike your mis-drawings, my original gedanken is possible (within the
usual limits of a gedanken).
Your drawings hold for 0<w<1, because in that case the
second signal propagates forward in time relative to
all frames, including S. But that's not a tachyon.
> Tom used w =10c, v = 0.5c, but
> if causality is NOT violated for w = ∞, it won't be violated for any slower
> speed. So let's simplify the situation and let w = ∞:
my original calculations. To explicitly show you your error, I'll re-do
the calculations below.
OK, use w=∞ (infinity). Ignore the fact that such a tachyon is in
principle undetectable because it has zero energy.
Causality is still violated when one calculates correctly (you didn't).
See below.
[Note I use c=1, you don't; I use g where you use γ.]
> What is tC' and tD' when the signal is launched from D?
>
> tC' = 0, xC = 0, xD' = γL
> tD' = γ( L/w - vL/c²) = -γvL/c²
signal is launched from D" is too ambiguous -- you MUST specify in which
frame "when" applies -- so I ignore that.
The signal launched by D happens when the first tachyon arrives at
x=L,t=0. Remember D is at rest in S'. So the second tachyon is launched at:
x' = g(x - v t) = gL
t' = g(t - v x) = -g v L
I presume these are what you mean by xD' and tD'.
D launches its tachyon signal at x'=gL,t'=-gvL. Since its speed relative
to S' is ∞, it arrives at C:
t' = -g v L
By construction, C is at rest in S', and has been pre-arranged to be
co-located with x=0 when the tachyon arrives at C, so C is located at [#]
x' = g v^2 L
I presume these are what you intended to mean by tC' and xC', but you
calculated incorrectly.
[#] Remember the location x=0 is moving with speed -v
relative to S'. So at t'=-gvL, the location x=0 is at
x' = g v^2 L -- that is where C is pre-positioned.
Check: x = g(x' + v t') = g(g v^2 L - g v^2 L) = 0.
Note in passing that C is at rest in S', and since 0<v<1, C is located
between x'=0 and D (also at rest in S', located at x'=gL). So the second
tachyon did indeed propagate in the -x' direction, as constructed.
> There is some controversy about the speed of the return signal.
it propagates with speed w'=∞ (infinity) relative to frame S', in the
-x' direction.
> Tom
> assumed it was u' = - w, which would be u' = -∞.
the Principle of Relativity (PoR), which is part and parcel of SR.
The PoR states (essentially) that the laws of physics
are the same relative to all inertial frames. So if you
can launch a tachyon with speed w relative to frame S
from a source at rest in S, then you can launch one with
speed w relative to frame S' from a source at rest in S'.
Indeed this holds for all w, not just w>1 (except for
|w|<1 we would not apply the term "tachyon").
> That would mean that
> tC' = tD' = -γvL/c², but that occurred when t = -vL/c²:
from simultaneity in S' -- these two events are simultaneous in S'
(tC' = tD'), so they are NOT simultaneous in S:
- the second tachyon is launched at xD'=gL,tD'=-gvL,
which corresponds to x=L,t=0.
- the second tachyon is received at xC'=gv^2L,tC'=-gvL,
which corresponds to x=0,t=-vL
Check:
x = g(x' + v t') = g(g v^2 L - g v^2 L) = 0
t = g(t' + v x') = -g^2 v L + g^2 v^3 L = - g^2 v L (1-v^2)
= - v L
This is nothing new and agrees algebraically with my initial post: the
second tachyon arrives at A (x=0) BEFORE the first tachyon was launched
from A (x=0). This is a causality-violating loop.
> [... further errors based on the above mistake]
In a later post you say:
> So if it is assumed that u' = -∞, then tC' = tD' = -γvL/c², and the signal goes
> into the past of the stationary lab of A and B, and causality is violated.
Yes. But u' = -∞ is part of my original gedanken, and is required by the
> into the past of the stationary lab of A and B, and causality is violated.
PoR. Yes, causality is violated.
The PoR states (essentially) that the laws of physics
are the same relative to all inertial frames. So if you
can launch a tachyon with speed w relative to frame S
from a source at rest in S, then you can launch one with
speed w relative to frame S' from a source at rest in S'.
Indeed this is so for all w, not just w>1 (except for
|w|<1 we would not apply the term "tachyon").
> Does this make sense? What about the fact that A and B KNOW that C is
> at xC = +v²L/c², not at xC = 0?
position in S (as you claim here) -- C has a given value of x ONLY at a
single value of t.
Your mistakes make your attempt to avoid my disproof invalid. This has
been well known for about a century: in SR, tachyon signaling implies
causality violations.
Tom Roberts
Jim Fox
Dec 26, 2020, 5:02:39 PM12/26/20
to
Tom Roberts wrote:
> Your mistakes make your attempt to avoid my disproof invalid. This has
> been well known for about a century: in SR, tachyon signaling implies
> causality violations.
Not true. His only mistake in tachyons is that he doesn't understand
> Your mistakes make your attempt to avoid my disproof invalid. This has
> been well known for about a century: in SR, tachyon signaling implies
> causality violations.
relativity, not even remotely introductory level. Somebody should write a
book entitled *Introduction to Relativity*. He will not read it, feeling
insulted pushed to read something having *Introduction* in it.
Gary Harnagel
Dec 26, 2020, 7:02:24 PM12/26/20
to
series of diagrams. But "backward-in-time" doesn't make
sense. Certainly, A can send a signal to B, and some moving
observers C and D can observe that something happened
at B before something else happened at A, but they can't
observe tachyons going backward in time (they have no
energy at u = c²/v, which is the demarcation point).
> This is shown below (for w=∞): relative
> to S the second tachyon propagates from x=L,t=0 to
> x=0,t=-vL.
actually OBSERVES in his laboratory? He OBSERVES C at
xC = 0, tC' = γ(L/w - vL/c²)) when the tachyon is launched by
D, right?
How can a signal traveling between D and C (in S') have any
effect whatsoever on a clock in S? It can't, of course.
> (My original calculation holds for all w>1).
>
> Note that A and B are locations at rest in S, and C and D are locations
> at rest in S'. In S, C is adjacent to A BEFORE D is adjacent to B.
be impossible to complete the loop.
> Unlike your mis-drawings, my original gedanken is possible (within the
> usual limits of a gedanken).
(1) D didn't have the message to transmit the signal to C, so the message
loop fails and causality isn't violated.
(2) When D gets the message (being adjacent to B), C wasn't at x = 0, so
so the message loop fails and causality isn't violated.
Both (1) and (2) are what the physicist in the lab OBSERVES.
> Your drawings hold for 0<w<1, because in that case the
> second signal propagates forward in time relative to
> all frames, including S. But that's not a tachyon.
going between moving transceivers:
A -- > w __________ D --> v
A cannot send the signal faster than w = c²/v, otherwise the
signal will have no energy relative to D. For the same reason,
D cannot send a signal faster that u = -c²/v to A.
So those signals ARE tachyons. This method, referred to as
Method I, is easily understood and allows w approaching
infinity as v approaches zero, and w approaches c as v approaches
c.
I think the first question that should be answered is: How can
adding two additional observers, as in the four-observer hand-off
method, referred to as Method II, as conventionally touted, be in
disagreement with the direct Method I?
It seems to me that there's a lot of mystique and speculation
surrounding Method II that leads to fallacious conclusions.
Ignoring the physicist in the lab and promoting the mystique of
time as a physical spatial dimension is responsible for some of
that.
> OK, use w=∞ (infinity). Ignore the fact that such a tachyon is in
> principle undetectable because it has zero energy.
> Causality is still violated when one calculates correctly (you didn't).
> See below.
>
> [Note I use c=1, you don't; I use g where you use γ.]
> > What is tC' and tD' when the signal is launched from D?
> >
> > tC' = 0, xC = 0, xD' = γL
> > tD' = γ( L/w - vL/c²) = -γvL/c²
> These values for C are wrong; I calculate them below. Also: "when the
> signal is launched from D" is too ambiguous -- you MUST specify in which
> frame "when" applies -- so I ignore that.
Obviously because t = 0.
> The signal launched by D happens when the first tachyon arrives at
> x=L,t=0. Remember D is at rest in S'. So the second tachyon is launched at:
> x' = g(x - v t) = gL
> t' = g(t - v x) = -g v L
> I presume these are what you mean by xD' and tD'.
> D launches its tachyon signal at x'=gL,t'=-gvL. Since its speed relative
> to S' is ∞, it arrives at C:
> t' = -g v L
(A) t = -vL/c²:
C ____________________ D
A _______________________ B
Which doesn't make any sense because D hasn't received the message
from B. Or we start at any convenient time such that C is adjacent to A
and D is adjacent to B:
(B) t = 0:
C ____________________ D
A ____________________ B
Thus D CAN receive the message from B and send the signal to C,
BUT ... C isn't adjacent to A when C gets the presumed signal in the
past of the S frame, so the loop fails. Or, just admit that the physicist
in his lab isn't myopic and KNOWS that C is at x = 0, t = 0 and THAT'S
where the signal is received.
> By construction, C is at rest in S', and has been pre-arranged to be
> co-located with x=0 when the tachyon arrives at C, so C is located at [#]
> x' = g v^2 L
from D (in S') to C (in S') can cause time in S to go backward, then the
signal can be received at x = 0. Why not assume that C at x = v²L/c² be
the location where the signal is received? Unfortunately, no loop :-) BUT
set up a situation where there is no C at x > 0 when the signal arrives:
(C) t = 0:
C _____________________ D --> v
____ A _____________________ B
A launches w = c²/v, it arrives at B at t = vL/c² and is passed to D at tD' = 0.
D sends a signal to C at u' = -c²/v and it arrives at C and C passes it to A at
t = vL/c².
No problem with missing connections, no problem with dragging the lab along
with A and B kicking and screaming into the past, no backward-in-time in either
S or S', no causality violation and no disagreement with the straight-forward
Method I.
> The PoR states (essentially) that the laws of physics
> are the same relative to all inertial frames. So if you
> can launch a tachyon with speed w relative to frame S
> from a source at rest in S, then you can launch one with
> speed w relative to frame S' from a source at rest in S'.
can launch tachyons at any speed c < w < ∞, but that is no
guarantee that a message loop can be completed. You have
to force together puzzle pieces that don't fit to make it look
like it can happen.
There has also been some misunderstanding about willy-
nilly varying v and proclaiming "AHA! Causality is violated!"
This is a canard. The initial positions of C and D (at t = 0)
are xC = -v²L/c² and xD = L(1 - v²/c²), so changing v without
changing xC and xD is verboten.
> Indeed this holds for all w, not just w>1 (except for
> |w|<1 we would not apply the term "tachyon").
>
> > That would mean that
> > tC' = tD' = -γvL/c², but that occurred when t = -vL/c²:
>
> THIS IS COMPLETELY WRONG.
tD' = -γvL/c², and C is at x = 0, so tC' -γvL/c² (as you assert),
and since t' = γ(t - vx/c²), so what else can time in S be?
I think this post is already too long and there's plenty to discuss.
Jim Fox
Dec 26, 2020, 7:59:07 PM12/26/20
to
Gary Harnagel wrote:
>> You ignored the fact that relative to S the second is going BACKWARDS
>> IN TIME (which cannot be captured in such diagrams).
>
> I disagree with you on that, Tom. It CAN be captured in a series of
> diagrams. But "backward-in-time" doesn't make sense. Certainly, A can
> send a signal to B, and some moving observers C and D can observe that
> something happened at B before something else happened at A, but they
> can't observe tachyons going backward in time (they have no energy at u
> = c²/v, which is the demarcation point).
Leave me alone. You are like people from *modeRNA*, 6 died from the
>> You ignored the fact that relative to S the second is going BACKWARDS
>> IN TIME (which cannot be captured in such diagrams).
>
> I disagree with you on that, Tom. It CAN be captured in a series of
> diagrams. But "backward-in-time" doesn't make sense. Certainly, A can
> send a signal to B, and some moving observers C and D can observe that
> something happened at B before something else happened at A, but they
> can't observe tachyons going backward in time (they have no energy at u
> = c²/v, which is the demarcation point).
vaccine in the control group, 7 died in the placebo group. Therefore the
vaccine is more than 100% safe, because less than zero died because the
covid vaccines. Amazing what you can do with statistics. You just kill
people on the other side, similar electronics, canceling an pole by null
point overlapped same place. Your theory apparently works from A to B to
C to D, but not from only A to B.
Brad Gassaway
Dec 30, 2020, 9:20:57 AM12/30/20
to
Gary Harnagel wrote:
> (B) t = 0: C ____________________ D A ____________________ B
> Thus D CAN receive the message from B and send the signal to C,
> BUT ... C isn't adjacent to A when C gets the presumed signal in the
> past of the S frame, so the loop fails. Or, just admit that the
> physicist in his lab isn't myopic and KNOWS that C is at x = 0, t = 0
> and THAT'S where the signal is received.
Your "big" theory is no more than, a begin to move, making an opposite
> (B) t = 0: C ____________________ D A ____________________ B
> Thus D CAN receive the message from B and send the signal to C,
> BUT ... C isn't adjacent to A when C gets the presumed signal in the
> past of the S frame, so the loop fails. Or, just admit that the
> physicist in his lab isn't myopic and KNOWS that C is at x = 0, t = 0
> and THAT'S where the signal is received.
appearance to appear.
Tom Roberts
Jan 3, 2021, 12:51:49 AM1/3/21
to
> But "backward-in-time" doesn't make
> sense.
> Certainly, A can send a signal to B, and some moving
> observers C and D can observe that something happened
> at B before something else happened at A, but they can't
> observe tachyons going backward in time (they have no
> energy at u = c²/v, which is the demarcation point).
has any receiver or observer trying to observe any tachyon that was not
emitted by a transmitter at rest in the same frame as the
receiver/observer. So your objection here simply does not apply.
If A can send a tachyon signal to B, with both at rest in inertial frame
S, then if SR is valid, D can send a similar tachyon signal to C, with
both of them at rest in inertial frame S'. In my previous post I
computed the coordinates of all transmission and reception events in
both S and S'. It is INESCAPABLE that in frame S the final reception
event at A happens BEFORE the initial transmission event at A (this
final reception is NOT a tachyon signal, it is an ordinary signal over
zero distance with zero delay from C to A, which are co-located at that
event).
It does not matter what the dynamics of tachyon signaling are, or
whether you think they have no energy -- if the speed of a tachyon
is >c relative to the frame in which its transmitter is at rest, and a
receiver at rest in the same frame can receive it, then my conclusion
follows. NOBODY ever has to "observe a tachyon going backwards in time",
but yet the (ordinary) signal from C to A closing the loop comes to A
BEFORE the first tachyon was launched by A.
********
* Here are diagrams drawn from the standpoint of frame S [@],
* with w=∞, v=0.5 and L=16 characters. Note that S' is "Lorentz
* contracted" here, so in these diagrams 0-to-D in S' is the
* same number of characters as A-to-B in S; 0-to-C in S' is 1/4
* of those characters (= v^2). As before, ignore the vertical
* separation on the page.
*
* [@] I.e. the positions of everything are drawn at the
* value of t given for the diagram, as measured in S.
*
* Here is the situation at t=-vL, when A receives the signal from C:
* S': 0---C-----------D (Moving to the right at v (= 0.5)
* S: ----A---------------B (note A is at x=0)
*
* Here is the situation at t=0, when A launches the first tachyon,
* B receives the first tachyon, B sends a zero-length signal with
* zero delay to D, and D launches the second tachyon:
* S': 0---C-----------D
* S: ----A---------------B
********
> How is it possible to dispute what the physicist
> actually OBSERVES in his laboratory? He OBSERVES C at
> xC = 0, tC' = γ(L/w - vL/c²)) when the tachyon is launched by
> D, right?
Moreover, no possible physicist can observe both C and D. So this
paragraph is impossible and I ignore it.
Note that we are analyzing this from a global perspective not limited to
either frame or to any location. As this is a gedanken, we can COMPUTE
when and where each event occurs, relative to either frame, and I did
just that in my previous post.
The first tachyon propagates instantaneously relative to S, but the
second tachyon propagates instantaneously relative to S' -- that is NOT
simultaneous in S as you seem to claim here. Indeed, as I calculated in
the previous post, that second tachyon induces C to send a zero-distance
zero-delay signal that arrives at A at x=0,t=-vL, so it comes to A
BEFORE the first tachyon was launched by A at t=0.
> How can a signal traveling between D and C (in S') have any
> effect whatsoever on a clock in S? It can't, of course.
the second tachyon, C is adjacent to A and can send an ordinary signal
to A over zero distance with zero delay. THAT signal closes the loop,
and comes to A before A launched the first tachyon.
>> Note that A and B are locations at rest in S, and C and D are locations
>> at rest in S'. In S, C is adjacent to A BEFORE D is adjacent to B.
>
> In which case, D couldn't have received the message from B, so it would
> be impossible to complete the loop.
to B when B receives the first tachyon, and C is adjacent to A when C
receives the second tachyon [#]. Note that C and D are at rest in S',
and C is located between x'=0 and D (x'=gL), so it is QUITE CLEAR that C
is adjacent to A before t=0 -- see the above diagrams. Yes, the scenario
must start before t=0 (and also before t'=0).
[#] As I said before, we must PRE-COMPUTE locations
for C and D and place them there, so they are indeed
where the scenario needs them to be. Since this is a
gedanken, that is easy to do (in the real world it
would be a challenge). My previous post gives the
values and the above diagrams use them.
When observers in frame S observe the coordinates of the second tachyon,
via direct observation at B and via a zero-delay signal from C at A,
they conclude it propagated backwards in time, but they never actually
had to observe the second tachyon. I did that because detecting such
tachyons is problematical -- in this scenario tachyons are received by
receivers at rest in the same frame as the transmitter.
> (1) D didn't have the message to transmit the signal to C,
first tachyon signal, and B passed an ordinary signal to D over zero
distance with zero delay. THAT'S WHAT THE SCENARIO IS.
> (2) When D gets the message (being adjacent to B), C wasn't at x = 0, so
> so the message loop fails and causality isn't violated.
B at that event), it simply does not matter where C is located. C's
location only matters when C receives the second tachyon signal from D,
and at that event C is adjacent to A (i.e. x=0). THAT'S WHAT THE
SCENARIO IS.
You seem to have confused simultaneity between the
two frames -- they are DIFFERENT.
The second tachyon propagates with infinite speed RELATIVE TO S', not S.
In S the second tachyon is effectively going BACKWARDS IN TIME, which
cannot be expressed by any speed -- this tachyon is observed only in S'
at C; in S the ordinary zero-delay signal C->A permits S to INFER when
the second tachyon arrived at C, and thus at A since they are co-located
at this event. See the above diagrams.
Bottom line:
The scenario and calculations I presented demonstrate the impossibility
of preserving all of these:
A) the validity of SR
B) tachyon signaling at speed > c relative to an inertial frame
C) ordinary concepts of causal order (transmission of a signal
must occur before its reception, both referred to the same
inertial frame) [for w=∞, before => simultaneous]
You keep making mistakes, such as using "when" without specifying a
frame, and so remain confused. But you have presented NOTHING that
disputes the scenario I presented, in which causal ordering is violated.
Tom Roberts
Prokaryotic Capase Homolog
Jan 3, 2021, 5:38:04 AM1/3/21
to
On Saturday, January 2, 2021 at 11:51:49 PM UTC-6, tjrob137 wrote:
> * Here is the situation at t=0, when A launches the first tachyon,
> * B receives the first tachyon, B sends a zero-length signal with
> * zero delay to D, and D launches the second tachyon:
> * S': 0---C-----------D
> * S: ----A---------------B
The spacing between the colon and 0 seems to have been collapsed in the
> * Here is the situation at t=0, when A launches the first tachyon,
> * B receives the first tachyon, B sends a zero-length signal with
> * zero delay to D, and D launches the second tachyon:
> * S': 0---C-----------D
> * S: ----A---------------B
S' diagram. I have to add three spaces. Is this what other people see, or is
this unique to "new and improved" Google Groups?
Barb Ungaro
Jan 3, 2021, 8:33:16 AM1/3/21
to
usury oriented regime, planetary scale. When you arrive on Earth with
your spaceship, from distant galaxies, you arrive to a feudal usury
oriented oppressive regime planet. In the middle of nowhere. You will
start asking yourself, how on earth did I not miss this shithole here.
Tom Roberts
Jan 3, 2021, 2:44:13 PM1/3/21
to
in your quote three spaces are missing between "S':" and "0". In the
original diagram for t=0, S' 0 is directly above A, and D is directly
above B. These diagrams render correctly for me, both in the original
edit and when viewing the post in the newsgroup (via Thunderbird).
It is likely that Google Groups collapses multiple space to a single
space, as it surely renders via HTML, which automatically does that. You
could look at the page source, and perhaps it will be correct.
HTML has to resolve this, but it surely won't
work in a newsgroup posting, or in non-HTML newsreaders.
This implies that in future diagrams we should not rely on multiple
spaces to render faithfully.
Also, I often indent paragraphs by 4 spaces or a tab, such as the
paragraph above beginning "HTML has...". Please let me know if that
paragraph is indented 1/2 to 3/4 inch, or not (it is in Thunderbird).
Tom Roberts
Prokaryotic Capase Homolog
Jan 4, 2021, 1:59:56 PM1/4/21
to
On Sunday, January 3, 2021 at 1:44:13 PM UTC-6, tjrob137 wrote:
> Also, I often indent paragraphs by 4 spaces or a tab, such as the
> paragraph above beginning "HTML has...". Please let me know if that
> paragraph is indented 1/2 to 3/4 inch, or not (it is in Thunderbird).
The tab character \t seems to be ignored. "HTML" is flush with all of the other lines
> Also, I often indent paragraphs by 4 spaces or a tab, such as the
> paragraph above beginning "HTML has...". Please let me know if that
> paragraph is indented 1/2 to 3/4 inch, or not (it is in Thunderbird).
in your post. Below is the source:
\r\u003cbr\u003e\r\u003cbr\u003e\tHTML has \u0026amp;nbsp; to resolve this,
mitchr...@gmail.com
Jan 4, 2021, 2:11:33 PM1/4/21
to
You only need Gamma math to disprove a tachyon. That Gamma math does not
exist..
Gary Harnagel
Jan 6, 2021, 8:08:39 AM1/6/21
to
On Saturday, January 2, 2021 at 10:51:49 PM UTC-7, tjrob137 wrote:
>
> On 12/26/20 6:02 PM, Gary Harnagel wrote:
> >
> > On Saturday, December 26, 2020 at 2:38:26 PM UTC-7, tjrob137 wrote:
> > >
> > > On 12/20/20 8:17 PM, Gary Harnagel wrote:
> > > >
> > > > That is, at t = 0:
> > > > A --> w ______________ B
> > > >
> > > > t = L/w:
> > > > A ______________ w --> B
> > > >
> > > >
>
> On 12/26/20 6:02 PM, Gary Harnagel wrote:
> >
> > On Saturday, December 26, 2020 at 2:38:26 PM UTC-7, tjrob137 wrote:
> > >
> > > On 12/20/20 8:17 PM, Gary Harnagel wrote:
> > > >
> > > > That is, at t = 0:
> > > > A --> w ______________ B
> > > >
> > > > t = L/w:
> > > > A ______________ w --> B
> > > >
> > > >
> > > > That is, at t = L/w:
> > > > ________________ w <-- D
> > > > A ____________________ B
> > > >
> > > > ________________ w <-- D
> > > > A ____________________ B
> > > >
> > > > That is, at t = L/w + L/u:
> > > > C <-- u ________________ D
> > > > A ____________________ B
> > > >
> > > > Is this arrangement actually possible?
> > >
> > > NO! You drew it WRONG.
> > >
> > > You ignored the fact that relative to S the second is
> > > going BACKWARDS IN TIME (which cannot be captured in
> > > such diagrams).
> >
> > I disagree with you on that, Tom.
>
> You are wrong.
Well, It's mortifying to be called wrong, but I must remember
> > > > C <-- u ________________ D
> > > > A ____________________ B
> > > >
> > > > Is this arrangement actually possible?
> > >
> > > NO! You drew it WRONG.
> > >
> > > You ignored the fact that relative to S the second is
> > > going BACKWARDS IN TIME (which cannot be captured in
> > > such diagrams).
> >
> > I disagree with you on that, Tom.
>
> You are wrong.
that WHO'S right is not as important as WHAT'S right:
“Discussion is always better than argument, because argument
is to find WHO is right and discussion is to find WHAT is
right.” – Anon.
So what IS right?
> > But "backward-in-time" doesn't make sense.
>
> Right! Which is why your claims also make no sense.
"D launches a signal at (L,0) in S, or [γL,-γvL/c²] in S'
and E4 is "C receives information at [0,-γvL/c²] in S' or
(0,-vL/c²) in S."
This presumes that C existed from the viewpoint of an observer
at rest in S at that time and was adjacent to A and passed a
message to A:
t = -vL/c²
C ____________ D
A ________________ B
This, of course, allows all kinds of paradoxical situations if
we try to complete a message loop.
(1) We presume that A intiated the message at (0,0), but received
it before he wrote it, which violates causality.
(2) He received it at t = -vL/c², but refused to send it to B,
thus producing a "Heisenberg failure":
https://en.wikipedia.org/wiki/Thiotimoline
(3) He saw D launch it at t = 0 and then time for A went backward.
(4) He didn't receive it from C at all.
We can both agree that (4) is the only reasonable conclusion. Your
position is (4a) FTL signals are impossible and that's the end of
the story.
My position is (4b) Something's wrong with the reasoning involved
in the sequence.
t = 0:
____C ____________ D
A --> w __________ B
A launches the signal to B at w = ∞, D receives it from B at
tD' = -γvL/c² and sends it to C at u' = -∞.
t = 0:
____C _____ <-- u' D
A ________________ B
-------------------------------------------------------
Question: Does the act of sending the signal infinitely fast guarantee
that it will be received by C at tC' = tD' = -γvL/c²?
-------------------------------------------------------
From the perspective of C and D:
t' = -γvL/c²:
______C u'<--_______ D
v <-- A ____________ B
t' = 0:
______C ____________ D
A --> w ______ B
B has already passed D when A launches the signal.
-------------------------------------------------------
Question: Can the act of sending a signal infinitely fast in S cause
cause time to go backward in S'?
-------------------------------------------------------
Of course not. Note that the position of B at t' = 0 is not
adjacent to D, just as C wasn't adjacent to A at t = 0 (in S).
These phenomena are due to the fact that velocity is relative.
From the perspective of A and B (at rest in S), at t = 0, C is at
xC = v²L/c² and tC' = γ(v²/c²)vL/c². xC' = γv²L/c² and xD' = γL.
This is represented in the diagram above.
Sending a signal from D to C at u' = -∞ means either the signal arrives
at C at tC' = tD' = -γvL/c² and the act of sending causes time in S to
go backwards, which is absurd since an action by observers stationary
in S' cannot have any effect on S, OR, the signal is received by C at
xC = v²L/c².
If the former, then FTL signals are impossible and that's the end of the
story.
If the latter, then FTL signals do not violate causality but there's an
operational problem: Why can't D send an infinitely-fast signal to C?
Perhaps that's the wrong question. D CAN send an infinitely-fast signal
to C, but C may not be able to pass it to A.
Science is based upon experimental evidence, but FTL isn't a domain where
there is such for or against it. All we have at present are thought
experiments, and thinking has led to interesting insights, but it has
also led us astray.
I'm always skeptical of claims that something is impossible. I believe that
FTL phenomena (real phenomena, not just optical illusions) are a part of
nature and will be discovered. Consequently, the "impossibility" arguments
are flawed.
The fact that direct communication between a pair of moving transceivers
does NOT violate causality:
A --> w ________________ D --> v
A ________________ u' <-- D --> v
should be a touchstone for any FTL argument. It's straightforward and simple,
involving only two participants. It's illogical that adding more participants
could destroy such a rational conclusion.
BTW, the fact that Mermin, Norton, Tipler and Llewellyn used direct communication
incorrectly to "disprove" FTL. They all forgot that w and u' cannot be infinitely-fast.
Don't you think, Tom, that this should be noted in the journals, regardless of whether
your four-participant scenario is right or wrong?
Emr Tupanes
Jan 6, 2021, 9:52:05 AM1/6/21
to
Gary Harnagel wrote:
>> Right! Which is why your claims also make no sense.
>
> Your position is that, given w = ∞, u' = -∞, Event E3 is "D launches a
> signal at (L,0) in S, or [γL,-γvL/c²] in S'
> and E4 is "C receives information at [0,-γvL/c²] in S' or (0,-vL/c²) in
> S."
You always talk about vaccines. Here we go.
>> Right! Which is why your claims also make no sense.
>
> Your position is that, given w = ∞, u' = -∞, Event E3 is "D launches a
> signal at (L,0) in S, or [γL,-γvL/c²] in S'
> and E4 is "C receives information at [0,-γvL/c²] in S' or (0,-vL/c²) in
> S."
I’m a nurse in a nursing home and I’m on board with the contracting
something through the test kits. They test positive and after nearly a
year of isolation they are kept in their rooms, mostly in bed 24/7 and it
doesn’t take long before they develop pneumonia or they aspirate. More
and more I am convinced that the test kits are the key to people getting
the flu like symptoms in the first place. Of coarse the tests are testing
nothing but they know who received the tainted tests and will be tagged
as positive.
https://153news.net/watch_video.php?v=4UMX6N649BK3
Emr Tupanes
Jan 6, 2021, 9:57:53 AM1/6/21
to
Gary Harnagel wrote:
> Your position is that, given w = ∞, u' = -∞,
Ohh boy, -inf again, so you must be thinking your position lies in the
> Your position is that, given w = ∞, u' = -∞,
middle, are you?
Emr Tupanes
Jan 6, 2021, 10:17:30 AM1/6/21
to
Gary Harnagel wrote:
> I'm always skeptical of claims that something is impossible. I believe
> that FTL phenomena (real phenomena, not just optical illusions) are a
> part of nature and will be discovered. Consequently, the
> "impossibility" arguments are flawed.
> The fact that direct communication between a pair of moving transceivers
> does NOT violate causality:
> A --> w ________________ D --> v A ________________ u' <-- D --> v
> should be a touchstone for any FTL argument. It's straightforward and
> simple, involving only two participants. It's illogical that adding
> more participants could destroy such a rational conclusion.
You are a prolific liar. Here is the definition of a prolific liar:
> I'm always skeptical of claims that something is impossible. I believe
> that FTL phenomena (real phenomena, not just optical illusions) are a
> part of nature and will be discovered. Consequently, the
> "impossibility" arguments are flawed.
> The fact that direct communication between a pair of moving transceivers
> does NOT violate causality:
> A --> w ________________ D --> v A ________________ u' <-- D --> v
> should be a touchstone for any FTL argument. It's straightforward and
> simple, involving only two participants. It's illogical that adding
> more participants could destroy such a rational conclusion.
Prolific liars are those who report that they tell five or more lies per
day. They are more likely than the average person to believe that lying
is acceptable in some circumstances.
Dono.
Jan 6, 2021, 11:01:32 AM1/6/21
to
On Wednesday, January 6, 2021 at 5:08:39 AM UTC-8, hit...@yahoo.com wrote:
> BTW, the fact that Mermin, Norton, Tipler and Llewellyn used direct communication
> incorrectly to "disprove" FTL. They all forgot that w and u' cannot be infinitely-fast.
>
I see that you already have a shovel, do you want a pickaxe?
> BTW, the fact that Mermin, Norton, Tipler and Llewellyn used direct communication
> incorrectly to "disprove" FTL. They all forgot that w and u' cannot be infinitely-fast.
>
mitchr...@gmail.com
Jan 7, 2021, 3:40:41 PM1/7/21
to
Light above... atom below...
Gary Harnagel
Jan 9, 2021, 7:00:52 PM1/9/21
to
On Thursday, January 7, 2021 at 1:40:41 PM UTC-7, mitchr...@gmail.com wrote:
>
> Gamma law won the tachyon. There is a speed limit.
> Light above... atom below...
Well, Mitch, your "gamma law" doesn't work for photons, so why would anyone
>
> Gamma law won the tachyon. There is a speed limit.
> Light above... atom below...
in his right mind believe it'd work for tachyons?
Tom Roberts
Jan 11, 2021, 3:09:08 PM1/11/21
to
On 1/6/21 7:08 AM, Gary Harnagel wrote:
> [to me]
C is NOT at x'=0 as you keep claiming. C's position is determined by the
requirement that A and C be co-located at the instant that C receives
the second tachyon -- that requires C to be located at x'=γv²L, between
x'=0 and D (at x'=γL).
Repeating, with w=∞ and c=1:
A launches the first tachyon at (x=0,t=0) in S,
which is (x'=0,t'=0) in S'
B receives the first tachyon at (x=L,t=0) in S,
which is (x'=γL,t'=-γvL) in S'
D launches the second tachyon at (x'=γL,t'=-γvL) in S',
which is (x=L,t=0) in S
C receives the second tachyon at (x'=γv²L,t='=-γvL) in S',
which is (x=0,t=-vL) in S
As I said before, C is NOT at x'=0. C's position is determined by the
requirement that A and C be co-located at the instant that C receives
the second tachyon -- that requires C to be located at x'=γv²L, between
x'=0 and D (at x'=γL). Since S' is moving to the right relative to S
(i.e. increasing x), and C is located at a position with x'>0, it is
CRYSTAL CLEAR that C and A coincide BEFORE t=0.
> BTW, the fact that Mermin, Norton, Tipler and Llewellyn used direct communication
> incorrectly to "disprove" FTL. They all forgot that w and u' cannot be infinitely-fast.
In my original post in this thread, I calculated all the relevant
coordinates for any w>1, not just w=∞. Conclusion: in this scenario, for
any w>1 and any 0<v<1, the looped message arrives at A BEFORE A launches
the first tachyon.
Tom Roberts
> [to me]
> Your position is that, given w = ∞, u' = -∞, Event E3 is
> "D launches a signal at (L,0) in S, or [γL,-γvL/c²] in S'
> and E4 is "C receives information at [0,-γvL/c²] in S' or
> (0,-vL/c²) in S."
NO! Can't you read???? You keep MISREPRESENTING the location of C.
> "D launches a signal at (L,0) in S, or [γL,-γvL/c²] in S'
> and E4 is "C receives information at [0,-γvL/c²] in S' or
> (0,-vL/c²) in S."
C is NOT at x'=0 as you keep claiming. C's position is determined by the
requirement that A and C be co-located at the instant that C receives
the second tachyon -- that requires C to be located at x'=γv²L, between
x'=0 and D (at x'=γL).
Repeating, with w=∞ and c=1:
A launches the first tachyon at (x=0,t=0) in S,
which is (x'=0,t'=0) in S'
B receives the first tachyon at (x=L,t=0) in S,
which is (x'=γL,t'=-γvL) in S'
D launches the second tachyon at (x'=γL,t'=-γvL) in S',
which is (x=L,t=0) in S
C receives the second tachyon at (x'=γv²L,t='=-γvL) in S',
which is (x=0,t=-vL) in S
As I said before, C is NOT at x'=0. C's position is determined by the
requirement that A and C be co-located at the instant that C receives
the second tachyon -- that requires C to be located at x'=γv²L, between
x'=0 and D (at x'=γL). Since S' is moving to the right relative to S
(i.e. increasing x), and C is located at a position with x'>0, it is
CRYSTAL CLEAR that C and A coincide BEFORE t=0.
> BTW, the fact that Mermin, Norton, Tipler and Llewellyn used direct communication
> incorrectly to "disprove" FTL. They all forgot that w and u' cannot be infinitely-fast.
coordinates for any w>1, not just w=∞. Conclusion: in this scenario, for
any w>1 and any 0<v<1, the looped message arrives at A BEFORE A launches
the first tachyon.
Tom Roberts
Stone B Fusco
Jan 11, 2021, 3:56:12 PM1/11/21
to
Tom Roberts wrote:
> Repeating, with w=∞ and c=1:
> A launches the first tachyon at (x=0,t=0) in S,
> which is (x'=0,t'=0) in S'
> B receives the first tachyon at (x=L,t=0) in S,
> which is (x'=γL,t'=-γvL) in S'
> D launches the second tachyon at (x'=γL,t'=-γvL) in S',
> which is (x=L,t=0) in S
> C receives the second tachyon at (x'=γv²L,t='=-γvL) in S',
> which is (x=0,t=-vL) in S
> As I said before, C is NOT at x'=0. C's position is determined by the
> requirement that A and C be co-located at the instant that C receives
> the second tachyon -- that requires C to be located at x'=γv²L, between
> x'=0 and D (at x'=γL). Since S' is moving to the right relative to S
> (i.e. increasing x), and C is located at a position with x'>0, it is
> CRYSTAL CLEAR that C and A coincide BEFORE t=0.
> Repeating, with w=∞ and c=1:
> A launches the first tachyon at (x=0,t=0) in S,
> which is (x'=0,t'=0) in S'
> B receives the first tachyon at (x=L,t=0) in S,
> which is (x'=γL,t'=-γvL) in S'
> D launches the second tachyon at (x'=γL,t'=-γvL) in S',
> which is (x=L,t=0) in S
> C receives the second tachyon at (x'=γv²L,t='=-γvL) in S',
> which is (x=0,t=-vL) in S
> As I said before, C is NOT at x'=0. C's position is determined by the
> requirement that A and C be co-located at the instant that C receives
> the second tachyon -- that requires C to be located at x'=γv²L, between
> x'=0 and D (at x'=γL). Since S' is moving to the right relative to S
> (i.e. increasing x), and C is located at a position with x'>0, it is
> CRYSTAL CLEAR that C and A coincide BEFORE t=0.
> In my original post in this thread, I calculated all the relevant
> coordinates for any w>1, not just w=∞. Conclusion: in this scenario, for
> any w>1 and any 0<v<1, the looped message arrives at A BEFORE A launches
> the first tachyon.
Which proves his tachyon wrong, as it would arrive to A from its future.
> coordinates for any w>1, not just w=∞. Conclusion: in this scenario, for
> any w>1 and any 0<v<1, the looped message arrives at A BEFORE A launches
> the first tachyon.
Gary Harnagel
Jan 11, 2021, 10:27:25 PM1/11/21
to
On Monday, January 11, 2021 at 1:09:08 PM UTC-7, tjrob137 wrote:
>
> On 1/6/21 7:08 AM, Gary Harnagel wrote:
> > [to me]
> > Your position is that, given w = ∞, u' = -∞, Event E3 is
> > "D launches a signal at (L,0) in S, or [γL,-γvL/c²] in S'
> > and E4 is "C receives information at [0,-γvL/c²] in S' or
> > (0,-vL/c²) in S."
>
> NO! Can't you read????
> You keep MISREPRESENTING the location of C.
> C is NOT at x'=0 as you keep claiming.
C is ALWAYS at the same point in S'. The USUAL initial condition is
>
> On 1/6/21 7:08 AM, Gary Harnagel wrote:
> > [to me]
> > Your position is that, given w = ∞, u' = -∞, Event E3 is
> > "D launches a signal at (L,0) in S, or [γL,-γvL/c²] in S'
> > and E4 is "C receives information at [0,-γvL/c²] in S' or
> > (0,-vL/c²) in S."
>
> NO! Can't you read????
> You keep MISREPRESENTING the location of C.
> C is NOT at x'=0 as you keep claiming.
t = t' = 0 = xA = xC = xC'. So perhaps I made a MISTAKE. These things
happen, you know. No need to go high-order on this.
> C's position is determined by the requirement that A and C be co-located at
> the instant that C receives the second tachyon -- that requires C to be located
> at x'=γv²L, between x'=0 and D (at x'=γL).
See, Tom, I made a simple MISTAKE (I got xC = 0 at t = -vL/c² correct so it was
obviously a mistake, and irrelevant anyway since the problem is observed from
S, not S'.
> Repeating, with w=∞ and c=1:
> A launches the first tachyon at (x=0,t=0) in S,
> which is (x'=0,t'=0) in S'
>
> B receives the first tachyon at (x=L,t=0) in S,
> which is (x'=γL,t'=-γvL) in S'
>
> D launches the second tachyon at (x'=γL,t'=-γvL) in S',
> which is (x=L,t=0) in S
>
> C receives the second tachyon at (x'=γv²L,t='=-γvL) in S',
> which is (x=0,t=-vL) in S
CAUSED time to go backward in S. Completely ridiculous, yes?
> As I said before, C is NOT at x'=0. C's position is determined by the
> requirement that A and C be co-located at the instant that C receives
> the second tachyon -- that requires C to be located at x'=γv²L, between
> x'=0 and D (at x'=γL).
the fact that I had xC = 0 in S at t = -vL/c².
> Since S' is moving to the right relative to S
> (i.e. increasing x), and C is located at a position with x'>0, it is
> CRYSTAL CLEAR that C and A coincide BEFORE t=0.
t = -vL/c²
C ____________ D
A ________________ B
of two moving objects is as you say. This approach CAUSES time to
go backward in the observers' frame (A and B in S), which is absurd
since the mere observation of a tachyon signal can have no effect on
the observers. You take this as a proof that tachyons are impossible.
But what if tachyons are CONFIRMED!
Yes, I know, you'd prefer to cross that bridge if (when) we come to it.
I prefer to cross it now :-) because I believe we WILL come to it.
IF (WHEN) tachyons are confirmed, then the conventional analysis cannot
be correct. The position of C, as observed by A and B, IS the position of
C. It is VALID. Since the tachyons cannot cause time to reverse in S, the
position of C (at xC = v²L/c²) is where the signal will be received, and C
will be unable to pass it to A. From the viewpoint of C and D, they will see
no problem with C receiving the signal at tC' = -γvL/c², but they WILL have
a problem with A sending the first tachyon signal to B since it will drag t'
time backward. Thus they will claim B must receive the signal at t' = 0
where B isn't adjacent to D and thus the loop cannot be completed.
What CAN be done is what I proposed in Figure 7 of vixra 2011.0076. The
adjacency disconnect happens when u > c²/v, so THAT's the limit where
the hand-off method fails. For u <= c²/v, the signals don't cause time to be
reversed and causality isn't violated.
Furthermore, that limit is the SAME limit imposed by the direct communication
scenario. You might consider that a coincidence, I don't.
> > BTW, the fact that Mermin, Norton, Tipler and Llewellyn used direct communication
> > incorrectly to "disprove" FTL. They all forgot that w and u' cannot be infinitely-fast.
>
> In my original post in this thread, I calculated all the relevant
> coordinates for any w>1, not just w=∞. Conclusion: in this scenario, for
> any w>1 and any 0<v<1, the looped message arrives at A BEFORE A launches
> the first tachyon.
>
> Tom Roberts
DIRECT method with receivers moving relative to transmitters (which I dealt with
in vixra 2012.0108). Whether your method is right or wrong, their method is
incorrect and does NOT disprove tachyons. Do you believe that error should be
propagated in the literature?
Stone B Fusco
Jan 12, 2021, 8:36:15 AM1/12/21
to
Gary Harnagel wrote:
>> C's position is determined by the requirement that A and C be
>> co-located at the instant that C receives the second tachyon -- that
>> requires C to be located at x'=γv²L, between x'=0 and D (at x'=γL).
>
> Well, at t = -vL/c², xC = 0, therefore, xC' = γ(0 + vL/c²) = γvL/c²
> See, Tom, I made a simple MISTAKE (I got xC = 0 at t = -vL/c² correct so
> it was obviously a mistake, and irrelevant anyway since the problem is
> observed from S, not S'.
Amazing good your theory. It permits committing mistakes.
>> C's position is determined by the requirement that A and C be
>> co-located at the instant that C receives the second tachyon -- that
>> requires C to be located at x'=γv²L, between x'=0 and D (at x'=γL).
>
> Well, at t = -vL/c², xC = 0, therefore, xC' = γ(0 + vL/c²) = γvL/c²
> See, Tom, I made a simple MISTAKE (I got xC = 0 at t = -vL/c² correct so
> it was obviously a mistake, and irrelevant anyway since the problem is
> observed from S, not S'.
mitchr...@gmail.com
Jan 12, 2021, 2:58:40 PM1/12/21
to
The lab tachyon has never been measured.
They don't belong in science as Gamma math
disproves them... No reaching light speed
in any way for the atom... by gravity
or any propulsion for it.
Mitchell Raemsch
rotchm
Jan 12, 2021, 4:30:23 PM1/12/21
to
On Monday, December 21, 2020 at 7:09:03 PM UTC-5, hit...@yahoo.com wrote:
> On Monday, December 21, 2020 at 4:09:18 PM UTC-7, mitchr...@gmail.com wrote:
<snip>
> On Monday, December 21, 2020 at 4:09:18 PM UTC-7, mitchr...@gmail.com wrote:
You are seriously replying to that idiot? Everyone here knows that it's a troll.
As any other troll, its not serious in a discussion. It just wants to make you type. And you fell for it.
Stop feeding the troll ! That would be the smartest thing you can do. In fact, what you
should do is to report him. Report spam. That is what I do for that troll. The more of us that do
it the quicker they will get booted off. Don't fall for it again.
rotchm
Jan 12, 2021, 4:32:18 PM1/12/21
to
On Monday, December 21, 2020 at 10:23:01 PM UTC-5, Michael Moroney wrote:
> "mitchr...@gmail.com" <mitchr...@gmail.com> writes:
You got caught, falling for the troll. Stop feeding the troll.
Report it instead. The more of us who do report it,
the quicker they'll get booted off. Do your part.
> "mitchr...@gmail.com" <mitchr...@gmail.com> writes:
You got caught, falling for the troll. Stop feeding the troll.
Report it instead. The more of us who do report it,
the quicker they'll get booted off. Do your part.
Stone B Fusco
Jan 12, 2021, 4:46:29 PM1/12/21
to
astronomy. You despicable piece of sub-human. You are infesting this
forum with hate speech.
Complaints-To: groups...@google.com
Injection-Info: google-groups.googlegroups.com;
posting-host=184.160.32.227;
posting-account=BHsbrQoAAAANJj6HqXJ987nOEDAC1EsJ
NNTP-Posting-Host: 184.160.32.227
References: <127d3a22-5445-4125...@googlegroups.com>
rotchm
Jan 12, 2021, 6:51:26 PM1/12/21
to
On Tuesday, January 12, 2021 at 4:46:29 PM UTC-5, Stone B Fusco wrote:
<spam & whining snipped>
Spam reported.
I urge others to do the same.
<spam & whining snipped>
Spam reported.
I urge others to do the same.
Gary Harnagel
Jan 20, 2021, 5:10:41 PM1/20/21
to
he sometimes sounds more cogent than others who post here :-)
I just want to reiterate the diagrams defining Tom's scenario:
t = -vL/c²
C ____________ D
A ________________ B
____C ____________ D
A ________________ B
At t = 0, A launches signal traveling at w = ∞ to B.
B receives it at t = 0+, passes it to D, which is adjacent.
tD' = -vL/c², and D transmits it to C.
All this is from the viewpoint of an observer stationary (with A and B) in S.
The question is, where is C when C receives it?
According to A and B, the calculate where C meets the signal by solving
the pair of equations:
xC = v(t + vL/c²) and xS = L + ut, where u is u = (u' + v)/(1 + u'v/c²)
and u' is presumed to b u'= -w. Therefore, u = (-w + v)/(1 - wv/c²).
So for w < c²/v, u < 0, at w = c²/v, u = ∞. Actually u goes from -∞ to +∞ as
w goes from just under c²/v to just over c²/v. Just over c²/v, A and B would
see the signal meeting C at xC < v²L/c², so the signal would CAUSE time to
go backward in S. This, of course, cannot happen.
The conventional answer is to say that FTL signals are impossible. But this
reasoning is somewhat circular. Perhaps it's more reasonable to consider
that the solution of xC = v(t + vL/c²) = xS = L + ut becomes invalid for w > c²/v,
that is, for u' < - c²/v (i.e., u' > -∞).
After all, A OBSERVES that C is at xC = v²L/c² at t = 0. At u = -∞, the signal
travels to c at xC = v²L/c² instantaneously (in S), and C is NOT adjacent to A
so C cannot pass the message contained in the signal to A, so no message
loop is formed.
From the perspective of C and D, B isn't adjacent to D when A sends the signal
to B, so B can't pass the message to D. Either way, a message loop cannot be
completed.
Bubba Slaten
Jan 20, 2021, 5:21:32 PM1/20/21
to
extreme crank Gary Harnagel regurgitated:
> On Tuesday, January 12, 2021 at 2:30:23 PM UTC-7, rotchm wrote:
>> You are seriously replying to that idiot? Everyone here knows that it's
>> a troll.
>
> I just want to reiterate the diagrams defining Tom's scenario:
> t = -vL/c² C ____________ D A ________________ B
Ohh, this is nothing.
http://wiki.naturalphilosophy.org/index.php?title=Gary_Harnagel
David de Hilster
Born November 13, 1959
Residence Boca Raton, FL, United States
Nationality USA
Known for Particle Model, Light, Neomechanics, Expansion Tectonics,
Autodynamics
Scientific career
Fields Supercomputers, Artist, Filmmaker
David de Hilster is an American scientist, *artist*, filmmaker, bullshit
artist, musician and *dissident scientist* the later of which, started
with his meeting and working with Argentinian physicist Dr. Ricardo
Carezani in 1992 who showed Einstein's
*Special theory of Relativity wrong* in the early 1940s. David is
currently president and co-founder of the John Chappell Natural
Philosophy Society and is co-authoring a book with his father Bob de
Hilster on their Particle Model entitled "Principia Mathematica 2" which
claims to be the first complete physical model that is based on Newtonian
laws and having no math.
> On Tuesday, January 12, 2021 at 2:30:23 PM UTC-7, rotchm wrote:
>> You are seriously replying to that idiot? Everyone here knows that it's
>> a troll.
>
> He doesn't sound like the nym-shifting troll, although he he MAY be as
> you say. OTOH, he sometimes sounds more cogent than others who post
> here
Trolls arguing loads of 🐂💩 from one another.
> you say. OTOH, he sometimes sounds more cogent than others who post
> here
> I just want to reiterate the diagrams defining Tom's scenario:
> t = -vL/c² C ____________ D A ________________ B
http://wiki.naturalphilosophy.org/index.php?title=Gary_Harnagel
David de Hilster
Born November 13, 1959
Residence Boca Raton, FL, United States
Nationality USA
Known for Particle Model, Light, Neomechanics, Expansion Tectonics,
Autodynamics
Scientific career
Fields Supercomputers, Artist, Filmmaker
David de Hilster is an American scientist, *artist*, filmmaker, bullshit
artist, musician and *dissident scientist* the later of which, started
with his meeting and working with Argentinian physicist Dr. Ricardo
Carezani in 1992 who showed Einstein's
*Special theory of Relativity wrong* in the early 1940s. David is
currently president and co-founder of the John Chappell Natural
Philosophy Society and is co-authoring a book with his father Bob de
Hilster on their Particle Model entitled "Principia Mathematica 2" which
claims to be the first complete physical model that is based on Newtonian
laws and having no math.
Yohi Fitzgibbons
Jan 21, 2021, 5:58:34 PM1/21/21
to
Bubba Slaten wrote:
>> I just want to reiterate the diagrams defining Tom's scenario:
>> t = -vL/c² C ____________ D A ________________ B
>
> Ohh, this is nothing.
>
> http://wiki.naturalphilosophy.org/index.php?title=Gary_Harnagel
>
> David de Hilster Born November 13, 1959 Residence Boca Raton, FL,
> United States Nationality USA Known for Particle Model, Light,
> Neomechanics, Expansion Tectonics, Autodynamics Scientific career Fields
> Supercomputers,
> Artist, Filmmaker David de Hilster is an American scientist, *artist*,
> filmmaker, bullshit artist, musician and *dissident scientist* the later
> of which, started with his meeting and working with Argentinian
> physicist Dr. Ricardo Carezani in 1992 who showed Einstein's
All, but what the feck is *autodynamics* and *neomechanics*?? Something
>> I just want to reiterate the diagrams defining Tom's scenario:
>> t = -vL/c² C ____________ D A ________________ B
>
> Ohh, this is nothing.
>
> http://wiki.naturalphilosophy.org/index.php?title=Gary_Harnagel
>
> David de Hilster Born November 13, 1959 Residence Boca Raton, FL,
> United States Nationality USA Known for Particle Model, Light,
> Neomechanics, Expansion Tectonics, Autodynamics Scientific career Fields
> Supercomputers,
> Artist, Filmmaker David de Hilster is an American scientist, *artist*,
> filmmaker, bullshit artist, musician and *dissident scientist* the later
> of which, started with his meeting and working with Argentinian
> physicist Dr. Ricardo Carezani in 1992 who showed Einstein's
you get served in the moon landing america?? What was the name of the
first (female) dog sent into space, from america? I am waiting ..zZZzzz..
0 new messages