On Wednesday, May 26, 2021 at 5:38:34 PM UTC-6, Dono. wrote:
>
> On Wednesday, May 26, 2021 at 7:09:55 AM UTC-7, hit...@yahoo.com wrote:
> >
> > Again, this is completely general. D has the option of setting his signal velocity selector to any
> > value -∞ < u' < -c²/v.
>
> Look up, cretinoid.

Look down, your fly's open.

> Repeating the same imbecility doesn't make it right

Repeating truth is always right. Apparently you're too stupid to realize that -∞ < u' < -c²/v
valuess are what's required to drive a nail at A's feet, given the initial conditions. If you
understood even high school algebra, you'd know this. Or are you so stupid as to send
the signal AWAY from the receiver?

I have radioactive rocks in my deck box that have more brains than you do.

On Wednesday, May 26, 2021 at 6:38:53 PM UTC-7, hit...@yahoo.com wrote:
> that -∞ < u' < -c²/v
> valuess are what's required to drive a nail at

you are not allowed to tie the signal speeds to the speeds of the observers, dishonest imbecile

On Thursday, May 27, 2021 at 12:09:55 AM UTC+10, hit...@yahoo.com wrote:

> This post is more than long enough, so I'm stopping here and waiting for your response
> to the foregoing.

Thank you for stopping, because throughout the post I saying "But you're not answering the scenario !!! " 8->

Before getting to that, couple of things :

Firstly, I think we need to really nail down the "logistics" :

At the instant that one individual ("Peter") is colocated-with while being passed by another ("Paul") with v < c, then at that instant Peter can always successfully hand off to Paul any information (a "message") that Peter has available (ie that he received previously or at that instant). For example, we can imagine that Peter could attach that information as a text file to an email that is sent by radio, so always perfectly capable of being received by Paul. Naturally, how Peter originally got the information (eg mode of transmission, such as radio or tachyons) is totally irrelevant to this.

IOTTMCO, agreed ? :-)

Secondly: just looking at your closing paragraphs :

If we allow D to send the
> signal at u ' = -∞, C can plant the nail at A's feet, making time to reverse in S. and causality is
> violated
.....
> Again, D has no choice, He MUST send the signal at u' = -∞, and time in A goes backward.
> But time CANNOT go backward,

So you are saying that emitting a tachyon at (near-)∞ results in "time in <another frame> goes backwards", which is not allowed. Upshot is that tachyon cannot have been emitted at (near-)∞ - right ?

Thing is, that is an entirely general scenario. EVERY (near-)∞ tachyon emission would impact SOME "<another frame>" by having its "time goes backward".

That therefore rules out EVERY (near-)∞ tachyon emission - agreed ? >;->

OK, back to why I was saying "But you're not answering the scenario !!! " For any discussion to make progress, it is important to stay-on-topic, taking one thing at a time until that is nailed down and agreed. If we don't do that, we'll just end up going around in circles.

For this thread, that one thing we are currently discussing is the nailgun scenario under the heading of "OPTION 1 - D emits NEAR-infinite speed tachyons ". To reiterate, this scenario is :

At the instant that D is co-located with B, D sets his tachyon-speed-dial to (near-)∞ and so emits tachyons at that speed to C, who at the instant she receives them will fire a nail into the ground of the AB frame. For this single reality, where will the nail end up when viewed from the AB frame, and when viewed from the CD frame ?

That is THE SPECIFIC SCENARIO that we are currently discussing.

Instead of addressing that specific scenario, you said :
> From this view, it appears that D must send the signal at u' = -c²/v. This is true IF we require that
> C fire the gun at the moment she receives it. This is an artificial restriction. If D sends the signal
> at u' = -∞, it will arrive when tC' = tD' = -γvL/c². That is, when t = -vL/c²:

We'll be dealing with D setting his tachyon-speed dial to -c²/v and emitting tachyons at that speed when we get to "OPTION 2 - D emits tachyon at velocity -c²/v", but it is off-topic for now (and derailing from the specific scenario under discussion would, as said earlier, only lead to us going round in circles).

Now, in a previous post, you said :

> > CONTRADICTION. That single nail can only end up at one location - either at A's feet or not at A's feet.
> > So D sending that tachyon at near-infinite speed leads directly to a contradiction - Agreed ? >8->
> Sure, and this is because you have posited an impossible situation, a logistics conundrum.

So to reiterate, this specific scenario is what you have classed as "an impossible situation" :

====At the instant that D is co-located with B, D sets his tachyon-speed-dial to (near-)∞ and so emits tachyons at that speed to C, who at the instant she receives them will fire a nail into the ground of the AB frame. For this single reality, where will the nail end up when viewed from the AB frame, and when viewed from the CD frame ?
====
Sorry, but if you reckon this is "an impossible situation", then I find it difficult to think of any situation you would think IS possible ! ;-) After all, the impossibility is clearly not SR's PoR nor RoS, nor C's firing of the nailgun. That simply leaves those (near-)∞ tachyons as the impossibility - can you think of anything else ???

On Thursday, May 27, 2021 at 4:53:05 AM UTC-6, Rob Acraman wrote:
>
> =====
> At the instant that D is co-located with B, D sets his tachyon-speed-dial to (near-)∞ and so emits
> tachyons at that speed to C, who at the instant she receives them will fire a nail into the ground
> of the AB frame. For this single reality, where will the nail end up when viewed from the AB
> frame, and when viewed from the CD frame ?
> =====
>
> Sorry, but if you reckon this is "an impossible situation", then I find it difficult to think of any
> situation you would think IS possible ! ;-) After all, the impossibility is clearly not SR's PoR
> nor RoS, nor C's firing of the nailgun. That simply leaves those (near-)∞ tachyons as the
> impossibility - can you think of anything else ???

Hi Rob,

I deleted everything except the crux because I'm using my *&%@# laptop and it's old and
cranky.

Now, I think RoS is intimately involved in the solution. It's clearly involved in the
limitation of communication speed for Method I (sources and receivers in relative
motion}. And because of Morin's principle (I think it should be a principle rather than
a mere strategy), it's a crucial factor in Method II. So, we must be clear about the
particular configuration we're dealing with. For example, this one requires that D send
the signal at u = -∞ and u' = -c²/v:

t = 0 (S-frame perspective):
C --> v ____________ u' <---- D --> v
A _______________________ B

OTOH, this is the one that requires D to send it at u' = -∞:

t' = 0 (S'-frame perspective):
v <-- A ___________________ v <-- B
____C __________________ u' <---- D
Figure 1 Configuration 1

So let's look at it from S:

t = 0 (S-frame perspective):
C --> v _____________ D --> v
A _______________________ B
Figure 2a Configuration 2

t = vL/c² (S-frame perspective):
____ C --> v ________ u' <---- D --> v
A _______________________ B
Figure 2b

In order for u' = -∞, time in S must go from t = vL/c² to t = 0. We both agree that this
is impossible, yes?

The equation of motion of C is xC = vt and that of the signal is xS = L + u(t - vL/c²). The
point where they meet is x = xC = xS, and this will be at t = (1 - uv/c²)L/(v - u).

Mathematically, u is < 0 UNLESS u' < -c²/v, at which point it changes sign.

Now, it COULD be that this is impossible because tachyons don't exist. But Method I has
no such constraint, so it's also possible that something is wrong with the construction
of Method II scenarios.

Well, one thing that might be wrong is that u becomes infinite when u' = -c²/v. Infinities
in physics mean that something's wrong. At u = ∞, a tachyon is everywhere at once! It's
nonlocal. Pushing the velocity addition equation beyond this point by insisting that u'
can be less than -c²/v is certainly problematical.

c²/v IS RoS. I encourage you to look at Method I (Section 2 of my paper). The fact that
all the analyses of that method in print have been wrong should give everyone pause
about rash assertions that tachyons violate causality by Method II, particularly when
the mathematics has dubious validity.

On Thursday, May 27, 2021 at 7:02:07 AM UTC-7, hit...@yahoo.com wrote:
> Mathematically, u is < 0 UNLESS u' < -c²/v, at which point it changes sign.
>

Looks like the coin is starting to drop.

> Now, it COULD be that this is impossible because tachyons don't exist.

Looks like the coin is starting to drop.

> Well, one thing that might be wrong is that u becomes infinite when u' = -c²/v. Infinities
> in physics mean that something's wrong. At u = ∞, a tachyon is everywhere at once! It's
> nonlocal. Pushing the velocity addition equation beyond this point by insisting that u'
> can be less than -c²/v is certainly problematical.
>

Looks like the coin dropped.

> c²/v IS RoS. I encourage you to look at Method I (Section 2 of my paper). The fact that
> all the analyses of that method in print have been wrong should give everyone pause
> about rash assertions that tachyons violate causality by Method II, particularly when
> the mathematics has dubious validity.

Still a crank, I see.

On Thursday, May 27, 2021 at 8:44:31 AM UTC-6, Dono. wrote:
>
> On Thursday, May 27, 2021 at 7:02:07 AM UTC-7, hit...@yahoo.com wrote:
> >
> > Mathematically, u is < 0 UNLESS u' < -c²/v, at which point it changes sign.
>
> Looks like the coin is starting to drop.

Nope, just being rigorously honest, something you don't understand.

> > Now, it COULD be that this is impossible because tachyons don't exist.
>
> Looks like the coin is starting to drop.

Nope, just rigorously covering all the possibilities, something you're too stupid to do.
..
> > Well, one thing that might be wrong is that u becomes infinite when u' = -c²/v. Infinities
> > in physics mean that something's wrong. At u = ∞, a tachyon is everywhere at once! It's
> > nonlocal. Pushing the velocity addition equation beyond this point by insisting that u'
> > can be less than -c²/v is certainly problematical.
>
> Looks like the coin dropped.

Looks like you have no clue about what this means.

> > c²/v IS RoS. I encourage you to look at Method I (Section 2 of my paper). The fact that
> > all the analyses of that method in print have been wrong should give everyone pause
> > about rash assertions that tachyons violate causality by Method II, particularly when
> > the mathematics has dubious validity.
>
> Still a crank, I see.

Still a mathematically-incompetent moron, blissfully unaware what it means to get infinity
for an answer. It usually means that your inference is NOT reality because:

"Only two things are infinite, the universe and human stupidity, and I'm not
sure about the former." -- Albert Einstein

Of course, you are a shining example of the latter.

On Friday, May 28, 2021 at 12:02:07 AM UTC+10, hit...@yahoo.com wrote:

> Hi Rob,
>
> I deleted everything except the crux because I'm using my *&%@# laptop and it's old and
> cranky.

Yep, and just think that laptop was WONDERFUL when you first got it ! :-D

I'll just skip to the crux as well :

>
> OTOH, this is the one that requires D to send it at u' = -∞:
>
> t' = 0 (S'-frame perspective):
> v <-- A ___________________ v <-- B
> ____C __________________ u' <---- D
> Figure 1 Configuration 1
>
> So let's look at it from S:
>
> t = 0 (S-frame perspective):
> C --> v _____________ D --> v
> A _______________________ B
> Figure 2a Configuration 2
>
> t = vL/c² (S-frame perspective):
> ____ C --> v ________ u' <---- D --> v
> A _______________________ B
> Figure 2b
>
> In order for u' = -∞, time in S must go from t = vL/c² to t = 0. We both agree that this
> is impossible, yes?

Yes - impossible. Let's also note that this impossibility does not constrain the value of "v" other than the sign (ie same direction as the tachyon) - so this impossibility is not specific to this one particular pair of frames, but is a GENERAL case.

In other words, you can NEVER give me a contradiction-free example of a (near-)∞ speed tachyon, since there will ALWAYS be other frames (like AB) that would be "hurled back in time" by that tachyon (which we agree must be impossible).

And since there are no contradiction-free examples of a (near-)∞ speed tachyon, that simply means (near-)∞ speed tachyons must be impossible themselves.

Agreed ?

On Thursday, May 27, 2021 at 5:01:38 PM UTC-7, hit...@yahoo.com wrote:
> On Thursday, May 27, 2021 at 8:44:31 AM UTC-6, Dono. wrote:
> >
> > On Thursday, May 27, 2021 at 7:02:07 AM UTC-7, hit...@yahoo.com wrote:
> > >
> > > Mathematically, u is < 0 UNLESS u' < -c²/v, at which point it changes sign.
> >
> > Looks like the coin is starting to drop.
> Nope, just being rigorously honest, something you don't understand.
> > > Now, it COULD be that this is impossible because tachyons don't exist..
> >
> > Looks like the coin is starting to drop.
> Nope, just rigorously covering all the possibilities,

Yes, I thought that you finally understood, I was sadly wrong, you are a hardened crank.

On Thursday, May 27, 2021 at 6:23:58 PM UTC-6, Rob Acraman wrote:
>
> On Friday, May 28, 2021 at 12:02:07 AM UTC+10, hit...@yahoo.com wrote:
> >
> > Hi Rob,
> >
> > I deleted everything except the crux because I'm using my *&%@# laptop and it's old and
> > cranky.
>
> Yep, and just think that laptop was WONDERFUL when you first got it ! :-D

Actually, I bought it used to run a gamma spectrometer. It works fine for that but is a pain
to use when I'm away from home.

> I'll just skip to the crux as well :
>
> > OTOH, this is the one that requires D to send it at u' = -∞:
> >
> > t' = 0 (S'-frame perspective):
> > v <-- A ___________________ v <-- B
> > ____C __________________ u' <---- D
> > Figure 1 Configuration 1
> >
> > So let's look at it from S:
> >
> > t = 0 (S-frame perspective):
> > C --> v _____________ D --> v
> > A _______________________ B
> > Figure 2a Configuration 2
> >
> > t = vL/c² (S-frame perspective):
> > ____ C --> v ________ u' <---- D --> v
> > A _______________________ B
> > Figure 2b
> >
> > In order for u' = -∞, time in S must go from t = vL/c² to t = 0. We both agree that this
> > is impossible, yes?
>
> Yes - impossible. Let's also note that this impossibility does not constrain the value of "v"
> other than the sign (ie same direction as the tachyon) - so this impossibility is not specific
> to this one particular pair of frames, but is a GENERAL case.
>
> In other words, you can NEVER give me a contradiction-free example of a (near-)∞ speed tachyon,
> since there will ALWAYS be other frames (like AB) that would be "hurled back in time" by that
> tachyon (which we agree must be impossible).

Not so, Rob, I have shown you such an example:

t = 0:
C --> v ___________ D --> v
____ A ______________ B

t = vL/c²:
____ C --> v _____ u' <---- D --> v
____ A ________________ B
tC' = γvL/c², tD' = 0; xC' = γ(0 - v²L/c²) = -γv²L/c², xD' = γ(L - v²L/c²) = L/γ

Since tA = vL/c² and tB = vL/c², u = -∞ and since u' = (u - v)/(1 - uv/c²), u' = -c²/v.

Now for the S' perspective:

t' = 0:
____ v <-- A ________ v <-- B
C _______________ u' <---- D
xD' - xC' = L/γ + γv²L/c² = (L/γ)/(1 - v²/c²) = γL

t' = γvL/c²:
v <-- A ________ v <-- B
____ C u' <--- ___________ D
u' = -γL/(tC' - tD') = -c²/v

> And since there are no contradiction-free examples of a (near-)∞ speed tachyon,
> that simply means (near-)∞ speed tachyons must be impossible themselves.
>
> Agreed ?

Obviously not, since I just demonstrated that one exists. Furthermore. it agrees with
Method I: D cannot send the signal directly to A faster than u' = -c²/v, otherwise it won't
have energy to be detected. However, it will appear to A and B that it traveled at u = -∞.

So, first of all, it's not true that tachyon speeds are limited to c²/v since they're infinitely-
fast in the other frame.

This should be sufficient proof to the analytical mind that the other configurations
which fail are not proof against the existence of tachyons, but are proof that RoS (and
logistics) must be rigorously obeyed, which is Morin's strategy. Raising it to a Principle:
there must be two analyses, one analyzed completely from the S frame and the other
analyzed completely from the S' frame,and both analyses must agree. Otherwise, the
configuration is invalid and cannot be used to conclude anything about tachyons.

On Friday, May 28, 2021 at 12:29:14 PM UTC+10, hit...@yahoo.com wrote:
> On Thursday, May 27, 2021 at 6:23:58 PM UTC-6, Rob Acraman wrote:
> >
> > > OTOH, this is the one that requires D to send it at u' = -∞:
> > >
> > > t' = 0 (S'-frame perspective):
> > > v <-- A ___________________ v <-- B
> > > ____C __________________ u' <---- D
> > > Figure 1 Configuration 1
> > >
> > > So let's look at it from S:
> > >
> > > t = 0 (S-frame perspective):
> > > C --> v _____________ D --> v
> > > A _______________________ B
> > > Figure 2a Configuration 2
> > >
> > > t = vL/c² (S-frame perspective):
> > > ____ C --> v ________ u' <---- D --> v
> > > A _______________________ B
> > > Figure 2b
> > >
> > > In order for u' = -∞, time in S must go from t = vL/c² to t = 0. We both agree that this
> > > is impossible, yes?
> >
> > Yes - impossible. Let's also note that this impossibility does not constrain the value of "v"
> > other than the sign (ie same direction as the tachyon) - so this impossibility is not specific
> > to this one particular pair of frames, but is a GENERAL case.
> >
> > In other words, you can NEVER give me a contradiction-free example of a (near-)∞ speed tachyon,
> > since there will ALWAYS be other frames (like AB) that would be "hurled back in time" by that
> > tachyon (which we agree must be impossible).
> Not so, Rob, I have shown you such an example:

Well, the above points were clearly relating to the example shown in the figures above, which were of the tachyon being emitted by D at u' = -(near-)∞, whereas your counter-example is the tachyon being emitted by D at -c²/v.

But OK, I should have made explicit what was implicit in the claim, so let's fix that :-)

======
> In order for u' = -∞, time in S must go from t = vL/c² to t = 0. We both agree that this
> is impossible, yes?

Yes - impossible. Let's also note that this impossibility does not constrain the value of "v" other than the sign (ie same direction as the tachyon) - so this impossibility is not specific to this one particular pair of frames, but is a GENERAL case.

In other words, you can NEVER give me a contradiction-free example of a tachyon being emitted at (near-)∞ speed (such as by D), since there will ALWAYS be other frames (like AB) that would be "hurled back in time" by that tachyon (which we agree must be impossible).

And since there are no contradiction-free examples of tachyons being emitted at (near-)∞ speed , that simply means that tachyons being emitted at (near-)∞ speed must be impossible.

Agreed ? :-)

This post replaces an earlier one that I have just deleted. While the earlier one was valid, this post is more general.

> > > OTOH, this is the one that requires D to send it at u' = -∞:
> > >
> > > t' = 0 (S'-frame perspective):
> > > v <-- A ___________________ v <-- B
> > > ____C __________________ u' <---- D
> > > Figure 1 Configuration 1
> > >
> > > So let's look at it from S:
> > >
> > > t = 0 (S-frame perspective):
> > > C --> v _____________ D --> v
> > > A _______________________ B
> > > Figure 2a Configuration 2
> > >
> > > t = vL/c² (S-frame perspective):
> > > ____ C --> v ________ u' <---- D --> v
> > > A _______________________ B
> > > Figure 2b
> > >
> > > In order for u' = -∞, time in S must go from t = vL/c² to t = 0. We both agree that this
> > > is impossible, yes?
> >
> > Yes - impossible. Let's also note that this impossibility does not constrain the value of "v"
> > other than the sign (ie same direction as the tachyon) - so this impossibility is not specific
> > to this one particular pair of frames, but is a GENERAL case.
> >
> > In other words, you can NEVER give me a contradiction-free example of a (near-)∞ speed tachyon,
> > since there will ALWAYS be other frames (like AB) that would be "hurled back in time" by that
> > tachyon (which we agree must be impossible).
> Not so, Rob, I have shown you such an example:
>
> t = 0:
> C --> v ___________ D --> v
> ____ A ______________ B
>
> t = vL/c²:
> ____ C --> v _____ u' <---- D --> v
> ____ A ________________ B
> tC' = γvL/c², tD' = 0; xC' = γ(0 - v²L/c²) = -γv²L/c², xD' = γ(L - v²L/c²) = L/γ
>
> Since tA = vL/c² and tB = vL/c², u = -∞ and since u' = (u - v)/(1 - uv/c²), u' = -c²/v.
>
> Now for the S' perspective:
>
> t' = 0:
> ____ v <-- A ________ v <-- B
> C _______________ u' <---- D
> xD' - xC' = L/γ + γv²L/c² = (L/γ)/(1 - v²/c²) = γL
>
> t' = γvL/c²:
> v <-- A ________ v <-- B
> ____ C u' <--- ___________ D
> u' = -γL/(tC' - tD') = -c²/v

OK, so let's say it is B that emits the tachyon at u = -∞ - is that contradiction-free ?

Nope.

D's emission of a (near-)∞ tachyon was impossible since it caused another frame (the AB frame) to "be hurled back in time".
Likewise, B's emission of a (near-)∞ tachyon is equally impossible since it causes another frame (let's introduce the EF frame) to "be hurled back in time" as well.

All we do is choose the EF frame such that the relationship that EF has to AB is the same as the relationship that AB has to CD.

So : AB is moving past CD at velocity v, so EF is moving past AB at velocity v as well
The proper length C-to-D is L * gamma, and A-to-B is L, so E-to-F is L / gamma

Same scale, so same result :-)

B emitting u = (near-)∞ tachyons causes the EF frame to "be hurled back in time" and must therefore be equally impossible

That (near-)∞ tachyons remain equally impossible.

Agreed ? ;-)

On Friday, May 28, 2021 at 12:43:51 AM UTC-6, Rob Acraman wrote:
> This post replaces an earlier one that I have just deleted. While the earlier one was valid, this post is more general.
> > > > OTOH, this is the one that requires D to send it at u' = -∞:
> > > >
> > > > t' = 0 (S'-frame perspective):
> > > > v <-- A ___________________ v <-- B
> > > > ____C __________________ u' <---- D
> > > > Figure 1 Configuration 1
> > > >
> > > > So let's look at it from S:
> > > >
> > > > t = 0 (S-frame perspective):
> > > > C --> v _____________ D --> v
> > > > A _______________________ B
> > > > Figure 2a Configuration 2
> > > >
> > > > t = vL/c² (S-frame perspective):
> > > > ____ C --> v ________ u' <---- D --> v
> > > > A _______________________ B
> > > > Figure 2b
> > > >
> > > > In order for u' = -∞, time in S must go from t = vL/c² to t = 0. We both agree that this
> > > > is impossible, yes?
> > >
> > > Yes - impossible. Let's also note that this impossibility does not constrain the value of "v"
> > > other than the sign (ie same direction as the tachyon) - so this impossibility is not specific
> > > to this one particular pair of frames, but is a GENERAL case.
> > >
> > > In other words, you can NEVER give me a contradiction-free example of a (near-)∞ speed tachyon,
> > > since there will ALWAYS be other frames (like AB) that would be "hurled back in time" by that
> > > tachyon (which we agree must be impossible).
> > Not so, Rob, I have shown you such an example:
> >
> > t = 0:
> > C --> v __________ D --> v
> > ____ A ______________ B
> >
> > t = vL/c²:
> > ____ C --> v _____ u' <---- D --> v
> > ____ A ________________ B
Figure 3b]

> > tC' = γvL/c², tD' = 0; xC' = γ(0 - v²L/c²) = -γv²L/c², xD' = γ(L - v²L/c²) = L/γ
> >
> > Since tA = vL/c² and tB = vL/c², u = -∞ and since u' = (u - v)/(1 - uv/c²), u' = -c²/v.
> >
> > Now for the S' perspective:
> >
> > t' = 0:
> > ____ v <-- A ________ v <-- B
> > C _______________ u' <---- D
> > xD' - xC' = L/γ + γv²L/c² = (L/γ)/(1 - v²/c²) = γL
> >
> > t' = γvL/c²:
> > v <-- A ________ v <-- B
> > ____ C u' <--- ___________ D
> > u' = -γL/(tC' - tD') = -c²/v
>
> OK, so let's say it is B that emits the tachyon at u = -∞ - is that contradiction-free ?
>
> Nope.

Yes, it IS contradiction-free. Let's look at the details instead of just making an assertion.
Look at Figure 3b above (which I just labeled). tC' = γvL/c², tD' = 0, time doesn't go backward.
Furthermore, u' = (u - v)/(1 - uv/c²). Since u = -∞, u' = -c²/v, exactly what we found u' must be
when D sent the signal.

> D's emission of a (near-)∞ tachyon was impossible since it caused another frame (the AB frame)
> to "be hurled back in time". Likewise, B's emission of a (near-)∞ tachyon is equally impossible
> since it causes another frame (let's introduce the EF frame) to "be hurled back in time" as well.
>
> All we do is choose the EF frame such that the relationship that EF has to AB is the same as the
> relationship that AB has to CD.

No, it's NOT the "same relationship" you think it is. Note that the position of D at t = 0 is xD = L/γ².
γ contains v, so if you posit another frame noving at v1 instead of v, γ will be different. So your EF
requires different conditions at t = 0, and with those different conditions, there are no contradictions
in the above figures.

> So : AB is moving past CD at velocity v, so EF is moving past AB at velocity v as well
> The proper length C-to-D is L * gamma, and A-to-B is L, so E-to-F is L / gamma
>
> Same scale, so same result :-)
>
> B emitting u = (near-)∞ tachyons causes the EF frame to "be hurled back in time" and must therefore
> be equally impossible
>
> That (near-)∞ tachyons remain equally impossible.
>
> Agreed ? ;-)

Absolutely not, Rob. Perhaps you're thinking of the cases where an EF configuration doesn't meet the
requirements I've specified. Those cases can have E2 (a signal is received) happening before E1 (a signal
is sent). This is normal LT behavior, but those frames aren't configured to interact as CD is: that is, C can
drive a nail at A's feet, but E can't do that unless at t = 0, xE = -v²L/c², where γ = 1/sqrt(1 - v1²/c²). That only
happens if v1 = v or xE = -v1²L/c² and xF = L(1 - v1²/c²).

On Thursday, May 27, 2021 at 11:43:51 PM UTC-7, Rob Acraman wrote:
> This post replaces an earlier one that I have just deleted. While the earlier one was valid, this post is more general.
> > > > OTOH, this is the one that requires D to send it at u' = -∞:
> > > >
> > > > t' = 0 (S'-frame perspective):
> > > > v <-- A ___________________ v <-- B
> > > > ____C __________________ u' <---- D
> > > > Figure 1 Configuration 1
> > > >
> > > > So let's look at it from S:
> > > >
> > > > t = 0 (S-frame perspective):
> > > > C --> v _____________ D --> v
> > > > A _______________________ B
> > > > Figure 2a Configuration 2
> > > >
> > > > t = vL/c² (S-frame perspective):
> > > > ____ C --> v ________ u' <---- D --> v
> > > > A _______________________ B
> > > > Figure 2b
> > > >
> > > > In order for u' = -∞, time in S must go from t = vL/c² to t = 0. We both agree that this
> > > > is impossible, yes?
> > >
> > > Yes - impossible. Let's also note that this impossibility does not constrain the value of "v"
> > > other than the sign (ie same direction as the tachyon) - so this impossibility is not specific
> > > to this one particular pair of frames, but is a GENERAL case.
> > >
> > > In other words, you can NEVER give me a contradiction-free example of a (near-)∞ speed tachyon,
> > > since there will ALWAYS be other frames (like AB) that would be "hurled back in time" by that
> > > tachyon (which we agree must be impossible).
> > Not so, Rob, I have shown you such an example:
> >
> > t = 0:
> > C --> v ___________ D --> v
> > ____ A ______________ B
> >
> > t = vL/c²:
> > ____ C --> v _____ u' <---- D --> v
> > ____ A ________________ B
> > tC' = γvL/c², tD' = 0; xC' = γ(0 - v²L/c²) = -γv²L/c², xD' = γ(L - v²L/c²) = L/γ
> >
> > Since tA = vL/c² and tB = vL/c², u = -∞ and since u' = (u - v)/(1 - uv/c²), u' = -c²/v.
> >
> > Now for the S' perspective:
> >
> > t' = 0:
> > ____ v <-- A ________ v <-- B
> > C _______________ u' <---- D
> > xD' - xC' = L/γ + γv²L/c² = (L/γ)/(1 - v²/c²) = γL
> >
> > t' = γvL/c²:
> > v <-- A ________ v <-- B
> > ____ C u' <--- ___________ D
> > u' = -γL/(tC' - tD') = -c²/v
> OK, so let's say it is B that emits the tachyon at u = -∞ - is that contradiction-free ?
>
> Nope.
>
> D's emission of a (near-)∞ tachyon was impossible since it caused another frame (the AB frame) to "be hurled back in time".
> Likewise, B's emission of a (near-)∞ tachyon is equally impossible since it causes another frame (let's introduce the EF frame) to "be hurled back in time" as well.
>
> All we do is choose the EF frame such that the relationship that EF has to AB is the same as the relationship that AB has to CD.
>
> So : AB is moving past CD at velocity v, so EF is moving past AB at velocity v as well
> The proper length C-to-D is L * gamma, and A-to-B is L, so E-to-F is L / gamma
>
> Same scale, so same result :-)
>
> B emitting u = (near-)∞ tachyons causes the EF frame to "be hurled back in time" and must therefore be equally impossible
>
> That (near-)∞ tachyons remain equally impossible.
>
> Agreed ? ;-)
You are trying to have a rational discussion with a profoundly dishonest crank.

On Saturday, May 29, 2021 at 1:50:03 AM UTC+10, hit...@yahoo.com wrote:
> On Friday, May 28, 2021 at 12:43:51 AM UTC-6, Rob Acraman wrote:
> > This post replaces an earlier one that I have just deleted. While the earlier one was valid, this post is more general.
> > > > > OTOH, this is the one that requires D to send it at u' = -∞:
> > > > >
> > > > > t' = 0 (S'-frame perspective):
> > > > > v <-- A ___________________ v <-- B
> > > > > ____C __________________ u' <---- D
> > > > > Figure 1 Configuration 1
> > > > >
> > > > > So let's look at it from S:
> > > > >
> > > > > t = 0 (S-frame perspective):
> > > > > C --> v _____________ D --> v
> > > > > A _______________________ B
> > > > > Figure 2a Configuration 2
> > > > >
> > > > > t = vL/c² (S-frame perspective):
> > > > > ____ C --> v ________ u' <---- D --> v
> > > > > A _______________________ B
> > > > > Figure 2b
> > > > >
> > > > > In order for u' = -∞, time in S must go from t = vL/c² to t = 0. We both agree that this
> > > > > is impossible, yes?
> > > >
> > > > Yes - impossible. Let's also note that this impossibility does not constrain the value of "v"
> > > > other than the sign (ie same direction as the tachyon) - so this impossibility is not specific
> > > > to this one particular pair of frames, but is a GENERAL case.
> > > >
> > > > In other words, you can NEVER give me a contradiction-free example of a (near-)∞ speed tachyon,
> > > > since there will ALWAYS be other frames (like AB) that would be "hurled back in time" by that
> > > > tachyon (which we agree must be impossible).
> > > Not so, Rob, I have shown you such an example:
> > >
> > > t = 0:
> > > C --> v __________ D --> v
> > > ____ A ______________ B
> > >
> > > t = vL/c²:
> > > ____ C --> v _____ u' <---- D --> v
> > > ____ A ________________ B
> Figure 3b]
> > > tC' = γvL/c², tD' = 0; xC' = γ(0 - v²L/c²) = -γv²L/c², xD' = γ(L - v²L/c²) = L/γ
> > >
> > > Since tA = vL/c² and tB = vL/c², u = -∞ and since u' = (u - v)/(1 - uv/c²), u' = -c²/v.
> > >
> > > Now for the S' perspective:
> > >
> > > t' = 0:
> > > ____ v <-- A ________ v <-- B
> > > C _______________ u' <---- D
> > > xD' - xC' = L/γ + γv²L/c² = (L/γ)/(1 - v²/c²) = γL
> > >
> > > t' = γvL/c²:
> > > v <-- A ________ v <-- B
> > > ____ C u' <--- ___________ D
> > > u' = -γL/(tC' - tD') = -c²/v
> >
> > OK, so let's say it is B that emits the tachyon at u = -∞ - is that contradiction-free ?
> >
> > Nope.
> Yes, it IS contradiction-free. Let's look at the details instead of just making an assertion.
> Look at Figure 3b above (which I just labeled). tC' = γvL/c², tD' = 0, time doesn't go backward.
> Furthermore, u' = (u - v)/(1 - uv/c²). Since u = -∞, u' = -c²/v, exactly what we found u' must be
> when D sent the signal.