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The LT, FTL Travel and Causality
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Gary Harnagel
Jan 13, 2015, 9:07:44 AM1/13/15
to
Consider movement of a ship, S, from point A to point B, where both points
are in the same inertial frame and usually spoken of as the "stationary"
frame. The Lorentz transform (LT) says that an observer aboard S will
observe that clocks at A and B appear to be running slower than his clock
and that observers at A and B will observe that the clock in S will appear
to be running slow. But what REALLY counts is when clocks are at the same
position and they can be compared directly. Under this regimen, the LT
says that the clock in S will have gained less time than the clock at B,
and since the B clock is synchronized with the clock at A, the S clock
will be behind the A clock, too.
So now let's consider that S is a faster-than-light (FTL) ship. Using the
same regimen as above, let's see what happens when S moves from A to B at
velocity n*c, where n is some number greater than unity. Let the distance
from A to B be d in the AB inertial frame and S leaves A at t = 0. Then
when S arrives at B, B's clock will read t1 = d/(n*c). If S immediately
turns around and goes back to A, A's clock will read t2 = 2*d/(n*c).
There is no problem here.
The problem arises when we include another ship, M, traveling at sub-light
velocity toward B. Let it pass A at the same time S leaves A (at t0 = 0).
According to the LT, M will say that the clock at B reads
tB' = -gamma*v*d/c^2 when tA = 0
(and tB = 0 also). And when S reaches B at t1 = d/(n*c), an observer
aboard M will read his clock as
(1) t1' = gamma*(t1 - v*x/c^2)
where x = d and gamma = 1/sqrt(1 - v^2/c^2). So
(2) t1' = gamma*(t1- v*d/c^2) = gamma*(d/c)*(1/n - v/c)
So the time it takes S to go from A to B is
(3) t1' - t0' = t1' = -gamma*(d/c)*(1/n - v/c)
according to M, which is a negative number for n > c/v. In other words,
S would arrive at B before it left A. This would seem to presnt a
causality problem.
But if we insist on the regimen that what counts is the situation when B
and M are coincident, which occurs when t3 = d/v, then
(4) t3' = gamma*(d/v - v*d/c^2) = gamma*(d/v)*(1 - v^2/c^2)
(5) t3' = (d/v)/gamma
M would find that S has been at B (if S had not returned to A) for
(6) t3' - t1' = (d/v)/gamma - gamma*(d/c)*(1/n - v/c)
(7) t3' - t1' = gamma*(d/v)*[1 - v/(n*c)]
But is this really a problem? Is this any more magical than B's clock
mysteriously found to be ahead of M's clock when B and M coincide?
So what about causality? Consider the case where S travels to B
instantaneously. According to M, S arrives at B at time
t1' = gamma*(0 - v*d/c^2) = -gamma*v*d/c^2
And now suppose S sends back a superluminal missile aimed to destroy M.
Can M be destroyed before S leaves A?
The missile moves at n*c in the AB frame. In that frame, the missile gets
back to A at time d/(n*c). So even if n is infinity, M cannot be destroyed
at any time less than zero, and causality is not violated. So if a method
of FTL travel is ever perfected (perhaps an Alcubierre or Natario warp
drive or some other technology), it does not appear that reality will be
shattered by destroying causality.
So it seems that the young lady from Bight whose speed was much faster
than light could never arrive the previous night :-)
Gary
are in the same inertial frame and usually spoken of as the "stationary"
frame. The Lorentz transform (LT) says that an observer aboard S will
observe that clocks at A and B appear to be running slower than his clock
and that observers at A and B will observe that the clock in S will appear
to be running slow. But what REALLY counts is when clocks are at the same
position and they can be compared directly. Under this regimen, the LT
says that the clock in S will have gained less time than the clock at B,
and since the B clock is synchronized with the clock at A, the S clock
will be behind the A clock, too.
So now let's consider that S is a faster-than-light (FTL) ship. Using the
same regimen as above, let's see what happens when S moves from A to B at
velocity n*c, where n is some number greater than unity. Let the distance
from A to B be d in the AB inertial frame and S leaves A at t = 0. Then
when S arrives at B, B's clock will read t1 = d/(n*c). If S immediately
turns around and goes back to A, A's clock will read t2 = 2*d/(n*c).
There is no problem here.
The problem arises when we include another ship, M, traveling at sub-light
velocity toward B. Let it pass A at the same time S leaves A (at t0 = 0).
According to the LT, M will say that the clock at B reads
tB' = -gamma*v*d/c^2 when tA = 0
(and tB = 0 also). And when S reaches B at t1 = d/(n*c), an observer
aboard M will read his clock as
(1) t1' = gamma*(t1 - v*x/c^2)
where x = d and gamma = 1/sqrt(1 - v^2/c^2). So
(2) t1' = gamma*(t1- v*d/c^2) = gamma*(d/c)*(1/n - v/c)
So the time it takes S to go from A to B is
(3) t1' - t0' = t1' = -gamma*(d/c)*(1/n - v/c)
according to M, which is a negative number for n > c/v. In other words,
S would arrive at B before it left A. This would seem to presnt a
causality problem.
But if we insist on the regimen that what counts is the situation when B
and M are coincident, which occurs when t3 = d/v, then
(4) t3' = gamma*(d/v - v*d/c^2) = gamma*(d/v)*(1 - v^2/c^2)
(5) t3' = (d/v)/gamma
M would find that S has been at B (if S had not returned to A) for
(6) t3' - t1' = (d/v)/gamma - gamma*(d/c)*(1/n - v/c)
(7) t3' - t1' = gamma*(d/v)*[1 - v/(n*c)]
But is this really a problem? Is this any more magical than B's clock
mysteriously found to be ahead of M's clock when B and M coincide?
So what about causality? Consider the case where S travels to B
instantaneously. According to M, S arrives at B at time
t1' = gamma*(0 - v*d/c^2) = -gamma*v*d/c^2
And now suppose S sends back a superluminal missile aimed to destroy M.
Can M be destroyed before S leaves A?
The missile moves at n*c in the AB frame. In that frame, the missile gets
back to A at time d/(n*c). So even if n is infinity, M cannot be destroyed
at any time less than zero, and causality is not violated. So if a method
of FTL travel is ever perfected (perhaps an Alcubierre or Natario warp
drive or some other technology), it does not appear that reality will be
shattered by destroying causality.
So it seems that the young lady from Bight whose speed was much faster
than light could never arrive the previous night :-)
Gary
Vitro
Jan 13, 2015, 11:19:59 AM1/13/15
to
shooting at each other FTL missiles traveling at Vm>1 (wrt the shooter).
A shoots at event E1(0,t1), missile hits B at event E2(x2,t2)=E2(0,t2')
and that's when B shoots back hitting A at E3(x3',t3')=E3(0,t3).
If I did the math right t3=t1*(1-Vs^2)*Vm^2/(Vm-Vs)^2 so t3 will be
less than t1 for (1-Vs^2)*Vm^2/(Vm-Vs)^2<1.
rotchm
Jan 13, 2015, 11:57:47 AM1/13/15
to
On Tuesday, January 13, 2015 at 9:07:44 AM UTC-5, Gary Harnagel wrote:
> Consider movement of a ship, S, from point A to point B, where both points
> are in the same inertial frame and usually spoken of as the "stationary"
> frame. The Lorentz transform (LT) says that an observer aboard S will
> observe that clocks at A and B appear to be running slower than his clock
> and that observers at A and B will observe that the clock in S will appear
> to be running slow.
Thats just a crude description of what the LT's says. What they say is that if you compare coincident clocks, they will have this and that value. (this too is very crude way to say it but reflects the essence of your post)
> Consider movement of a ship, S, from point A to point B, where both points
> are in the same inertial frame and usually spoken of as the "stationary"
> frame. The Lorentz transform (LT) says that an observer aboard S will
> observe that clocks at A and B appear to be running slower than his clock
> and that observers at A and B will observe that the clock in S will appear
> to be running slow.
> But what REALLY counts is when clocks are at the same
> position and they can be compared directly.
> Under this regimen, the LT
> says that the clock in S will have gained less time than the clock at B,
> and since the B clock is synchronized with the clock at A, the S clock
> will be behind the A clock, too.
> So now let's consider that S is a faster-than-light (FTL) ship...<snip>
ok...
> The problem arises when we include another ship, M, traveling at sub-light
> velocity toward B. Let it pass A at the same time S leaves A (at t0 = 0).
> According to the LT, M will say that the clock at B reads
>
> tB' = -gamma*v*d/c^2 when tA = 0
> (and tB = 0 also).
> And when S reaches B at t1 = d/(n*c), an observer
> aboard M will read his clock as
> (1) t1' = gamma*(t1 - v*x/c^2)
> where x = d and gamma = 1/sqrt(1 - v^2/c^2). So
> (2) t1' = gamma*(t1- v*d/c^2) = gamma*(d/c)*(1/n - v/c)
> So the time it takes S to go from A to B is
> (3) t1' - t0' = t1' = -gamma*(d/c)*(1/n - v/c)
> according to M, which is a negative number for n > c/v.
> In other words, S would arrive at B before it left A.
> This would seem to present a causality problem.
Nope... as I said, this is a normal well known scenario discussed in many books and articles.
Enough now...
John Heath
Jan 13, 2015, 12:34:47 PM1/13/15
to
On Tuesday, January 13, 2015 at 9:07:44 AM UTC-5, Gary Harnagel wrote:
John H
Gary Harnagel
Jan 13, 2015, 7:23:35 PM1/13/15
to
On Tuesday, January 13, 2015 at 9:19:59 AM UTC-7, Vitro wrote:
>
> Let's have A and B travel away from each other at speed Vs<1 (c=1) and
> shooting at each other FTL missiles traveling at Vm>1 (wrt the shooter).
> A shoots at event E1(0,t1), missile hits B at event E2(x2,t2)=E2(0,t2')
> and that's when B shoots back hitting A at E3(x3',t3')=E3(0,t3).
>
> If I did the math right t3=t1*(1-Vs^2)*Vm^2/(Vm-Vs)^2 so t3 will be
> less than t1 for (1-Vs^2)*Vm^2/(Vm-Vs)^2<1.
Hi Vitro,
>
> Let's have A and B travel away from each other at speed Vs<1 (c=1) and
> shooting at each other FTL missiles traveling at Vm>1 (wrt the shooter).
> A shoots at event E1(0,t1), missile hits B at event E2(x2,t2)=E2(0,t2')
> and that's when B shoots back hitting A at E3(x3',t3')=E3(0,t3).
>
> If I did the math right t3=t1*(1-Vs^2)*Vm^2/(Vm-Vs)^2 so t3 will be
> less than t1 for (1-Vs^2)*Vm^2/(Vm-Vs)^2<1.
Well, there's certainly no causality problem when viewed from A's frame,
right? When viewing from the B frame, it doesn't matter because B is not
at A when the missile hits, applying the regimen in the "rule" I laid out.
Gary
Gary Harnagel
Jan 13, 2015, 7:28:14 PM1/13/15
to
On Tuesday, January 13, 2015 at 10:34:47 AM UTC-7, John Heath wrote:
>
> Nice post. Where in the paper on electrodynamics of moving bodies did you
> assume the Doppler effect is to be tossed in the garbage bin?
Hi John,
>
> Nice post. Where in the paper on electrodynamics of moving bodies did you
> assume the Doppler effect is to be tossed in the garbage bin?
The Doppler effect is external to the LT, as is time delay of light signals.
Those things have to be treated separately, in addition to the LT.
> Doppler effect is well and alive in SR.
Gary
John Heath
Jan 13, 2015, 9:36:21 PM1/13/15
to
"So it seems that the young lady from Bight whose speed was much faster
than light could never arrive the previous night :-)"
John H
Vitro
Jan 13, 2015, 11:31:00 PM1/13/15
to
On 15-01-13 07:23 PM, Gary Harnagel wrote:
> On Tuesday, January 13, 2015 at 9:19:59 AM UTC-7, Vitro wrote:
>>
>> Let's have A and B travel away from each other at speed Vs<1 (c=1)
>> and shooting at each other FTL missiles traveling at Vm>1 (wrt the
>> shooter). A shoots at event E1(0,t1), missile hits B at event
>> E2(x2,t2)=E2(0,t2') and that's when B shoots back hitting A at
>> E3(x3',t3')=E3(0,t3).
>>
>> If I did the math right t3=t1*(1-Vs^2)*Vm^2/(Vm-Vs)^2 so t3 will
>> be less than t1 for (1-Vs^2)*Vm^2/(Vm-Vs)^2<1.
>
> Hi Vitro,
>
> Well, there's certainly no causality problem when viewed from A's
> frame, right?
Actually there is. The causal chain is E1 causes E2 causes E3, where E3
> On Tuesday, January 13, 2015 at 9:19:59 AM UTC-7, Vitro wrote:
>>
>> Let's have A and B travel away from each other at speed Vs<1 (c=1)
>> and shooting at each other FTL missiles traveling at Vm>1 (wrt the
>> shooter). A shoots at event E1(0,t1), missile hits B at event
>> E2(x2,t2)=E2(0,t2') and that's when B shoots back hitting A at
>> E3(x3',t3')=E3(0,t3).
>>
>> If I did the math right t3=t1*(1-Vs^2)*Vm^2/(Vm-Vs)^2 so t3 will
>> be less than t1 for (1-Vs^2)*Vm^2/(Vm-Vs)^2<1.
>
> Hi Vitro,
>
> Well, there's certainly no causality problem when viewed from A's
> frame, right?
happens before E1 on A's worldline and prevents E1 from happening.
A fires first at E1, missile hits B at E2 but B survives and retaliates.
B's missile hits A at E3 and kills him so A is dead before E1 and cannot
fire the first round which started the fight in the first place.
Times t1 and t3 in the equation above are both measured by A at A.
Gary Harnagel
Jan 14, 2015, 9:05:10 AM1/14/15
to
On Tuesday, January 13, 2015 at 9:31:00 PM UTC-7, Vitro wrote:
>
> On 15-01-13 07:23 PM, Gary Harnagel wrote:
> >
> > On Tuesday, January 13, 2015 at 9:19:59 AM UTC-7, Vitro wrote:
> > >
> > > Let's have A and B travel away from each other at speed Vs<1 (c=1)
> > > and shooting at each other FTL missiles traveling at Vm>1 (wrt the
> > > shooter). A shoots at event E1(0,t1), missile hits B at event
> > > E2(x2,t2)=E2(0,t2') and that's when B shoots back hitting A at
> > > E3(x3',t3')=E3(0,t3).
> > >
> > > If I did the math right t3=t1*(1-Vs^2)*Vm^2/(Vm-Vs)^2 so t3 will
> > > be less than t1 for (1-Vs^2)*Vm^2/(Vm-Vs)^2<1.
> >
> > Hi Vitro,
> >
> > Well, there's certainly no causality problem when viewed from A's
> > frame, right?
>
> Actually there is. The causal chain is E1 causes E2 causes E3, where E3
> happens before E1 on A's worldline and prevents E1 from happening.
>
> A fires first at E1, missile hits B at E2 but B survives and retaliates.
> B's missile hits A at E3 and kills him so A is dead before E1 and cannot
> fire the first round which started the fight in the first place.
>
> Times t1 and t3 in the equation above are both measured by A at A.
If you are calculating from the frame of A, you must stay in the A frame.
>
> On 15-01-13 07:23 PM, Gary Harnagel wrote:
> >
> > On Tuesday, January 13, 2015 at 9:19:59 AM UTC-7, Vitro wrote:
> > >
> > > Let's have A and B travel away from each other at speed Vs<1 (c=1)
> > > and shooting at each other FTL missiles traveling at Vm>1 (wrt the
> > > shooter). A shoots at event E1(0,t1), missile hits B at event
> > > E2(x2,t2)=E2(0,t2') and that's when B shoots back hitting A at
> > > E3(x3',t3')=E3(0,t3).
> > >
> > > If I did the math right t3=t1*(1-Vs^2)*Vm^2/(Vm-Vs)^2 so t3 will
> > > be less than t1 for (1-Vs^2)*Vm^2/(Vm-Vs)^2<1.
> >
> > Hi Vitro,
> >
> > Well, there's certainly no causality problem when viewed from A's
> > frame, right?
>
> Actually there is. The causal chain is E1 causes E2 causes E3, where E3
> happens before E1 on A's worldline and prevents E1 from happening.
>
> A fires first at E1, missile hits B at E2 but B survives and retaliates.
> B's missile hits A at E3 and kills him so A is dead before E1 and cannot
> fire the first round which started the fight in the first place.
>
> Times t1 and t3 in the equation above are both measured by A at A.
Your t2' term indicates that you have not done so. The missile, traveling
at Vm, meets B when
Vm*t2 = d + Vs*t2
t2 = d/(Vm -Vs)
That is a positive number.
B is at:
x2 = Vm*t2 = Vm*d/(Vm - Vs)
B immediately fires back, and the second missile reaches A:
x3 = 0 = x2 + Vn*(t3 - t2)
Note that the missile does NOT come back at speed Vm because it was fired
from a moving platform (it doesn't really matter to prove that the missile
doesn't get back to A at t < 0, but just to be precise here :-):
Vn = (Vs - Vm)/(1 - Vs*Vm)
t3 = -x2/Vn + t2 = -Vm*d*(1 - Vs*Vm)/(Vm - Vs)^2 + d/(Vm - Vs)
So even if Vm is infinite, t3 = d*Vm^2/Vm^2 = d
which is, again, a positive number (dimensional irrationality bugs me,
which happened because some of the V terms are really V/c).
Vitro
Jan 14, 2015, 12:46:50 PM1/14/15
to
On 1/14/2015 9:05 AM, Gary Harnagel wrote:
>
> If you are calculating from the frame of A, you must stay in the A
> frame.
I disagree here since frame A is in no way special. More so both events
>
> If you are calculating from the frame of A, you must stay in the A
> frame.
E1 and E3 happen on A's worldline, which is timelike, and so any
observer will agree on the sequence.
<I'm removing the rest, it can be found in the previous post>
Here's my calculation, using the standard setup with common origin (0,0)
for the two frames. A is the un-primed frame, B is the primed frame.
Event E1(0, t1) A fires the first missile. Vm will be wrt A.
Missile hits B at event E2(x2, t2) with
x2 = Vs*t2 = Vm*(t2-t1)
t2 = t1 * Vm/(Vm-Vs)
We note that t2 > t1.
We now transform E2 to frame B => E2(0, t2') where
t2' = gamma*(t2 - Vs*x2)
This is when B fires his missile, at -Vm wrt B.
It hits A at event E3(x3', t3') where
x3' = -Vs*t3' = -Vm*(t3'-t2')
t3' = t2' * Vm/(Vm - Vs)
We transform E3 to frame A, E3(0, t3), to get A's time
t3 = gamma*(t3' + Vs*x3')
and start substituting, first x3'
t3 = gamma*(t3' - Vs^2*t3') = gamma * t3' * (1-Vs^2)
then t3'
t3 = gamma * t2' * Vm/(Vm-Vs) * (1-Vs^2)
then t2'
t3 = gamma^2 * (t2 - Vs*x2) * Vm/(Vm-Vs) * (1-Vs^2)
then x2
t3 = gamma^2 * (t2 - Vs^2*t2) * Vm/(Vm-Vs) * (1-Vs^2)
t3 = gamma^2 * t2 * (1 - Vs^2) * Vm/(Vm-Vs) * (1-Vs^2)
get rid of gamma
t3 = t2 * Vm/(Vm-Vs) * (1-Vs^2)
and finally substitute t2
t3 = t1 * Vm^2/(Vm-Vs)^2 * (1-Vs^2)
Now just pick Vs and Vm to make t3<t1.
Gary Harnagel
Jan 14, 2015, 1:38:44 PM1/14/15
to
On Wednesday, January 14, 2015 at 10:46:50 AM UTC-7, Vitro wrote:
>
> On 1/14/2015 9:05 AM, Gary Harnagel wrote:
> >
> > If you are calculating from the frame of A, you must stay in the A
> > frame.
>
> I disagree here since frame A is in no way special.
Of course it's "special": It's the frame from which the measurements are
>
> On 1/14/2015 9:05 AM, Gary Harnagel wrote:
> >
> > If you are calculating from the frame of A, you must stay in the A
> > frame.
>
> I disagree here since frame A is in no way special.
made, and where they REALLY count.
> More so both events E1 and E3 happen on A's worldline, which is timelike,
> and so any observer will agree on the sequence.
the sequence is positive in A's frame.
> <I'm removing the rest, it can be found in the previous post>
>
> Here's my calculation, using the standard setup with common origin (0,0)
> for the two frames. A is the un-primed frame, B is the primed frame.
>
> Event E1(0, t1) A fires the first missile. Vm will be wrt A.
>
> Missile hits B at event E2(x2, t2) with
>
> x2 = Vs*t2 = Vm*(t2-t1)
> t2 = t1 * Vm/(Vm-Vs)
>
at d, the distance between A and B at t1. The missile leaves A at t1 = 0 and
travels a distance x2 = d + Vs*t2 = Vm*t2.
> We note that t2 > t1.
> We now transform E2 to frame B => E2(0, t2')
than A's.
> where
>
> t2' = gamma*(t2 - Vs*x2)
>
> This is when B fires his missile, at -Vm wrt B.
> It hits A at event E3(x3', t3') where
>
> x3' = -Vs*t3' = -Vm*(t3'-t2')
> t3' = t2' * Vm/(Vm - Vs)
> We transform E3 to frame A, E3(0, t3), to get A's time
>
> t3 = gamma*(t3' + Vs*x3')
>
> and start substituting, first x3'
>
> t3 = gamma*(t3' - Vs^2*t3') = gamma * t3' * (1-Vs^2)
>
> then t3'
>
> t3 = gamma * t2' * Vm/(Vm-Vs) * (1-Vs^2)
>
> then t2'
>
> t3 = gamma^2 * (t2 - Vs*x2) * Vm/(Vm-Vs) * (1-Vs^2)
>
> then x2
>
> t3 = gamma^2 * (t2 - Vs^2*t2) * Vm/(Vm-Vs) * (1-Vs^2)
> t3 = gamma^2 * t2 * (1 - Vs^2) * Vm/(Vm-Vs) * (1-Vs^2)
>
> get rid of gamma
>
> t3 = t2 * Vm/(Vm-Vs) * (1-Vs^2)
>
> and finally substitute t2
>
> t3 = t1 * Vm^2/(Vm-Vs)^2 * (1-Vs^2)
>
> Now just pick Vs and Vm to make t3<t1.
Want to try that again using the correct procedures? (I don't need to as I
am confident that the sequence will still be the same as in A's frame).
Gary
Vitro
Jan 14, 2015, 3:32:18 PM1/14/15
to
Frames A and B have common origin O(0, 0), which means observers A and B
were together at time 0. Time t1>0 is some arbitrary later time in frame A.
All t, t', x and x' in my calculation represent coordinates, not
distances or elapsed times.
At time t1 observer B is at x1 = Vs*t1, I'm guessing this is your 'd'
but this is not needed anywhere.
At time t2 B is at x2 = Vs*t2.
The missile fired by A travels from event E1(0, t1) to event E2(x2, t2)
so it travels a distance of x2-0 in time t2-t1 at speed Vm, IOW
x2 = Vm * (t2-t1)
but x2 is also where B is at t2, x2 = Vs*t2 from above, which makes
t2 = t1 * Vm/(Vm-Vs)
You may want to try again remembering that I'm working only with
coordinates. If you still have doubts we can go over it together
step-by-step.
Gary Harnagel
Jan 14, 2015, 4:30:48 PM1/14/15
to
On Wednesday, January 14, 2015 at 1:32:18 PM UTC-7, Vitro wrote:
>
> On 1/14/2015 1:38 PM, Gary Harnagel wrote:
> >
> > > t3 = t1 * Vm^2/(Vm-Vs)^2 * (1-Vs^2)
> > >
> > > Now just pick Vs and Vm to make t3<t1.
> >
> > My, but you went through such a convoluted mess, and you did it wrong
> > anyway. Want to try that again using the correct procedures? (I
> > don't need to as I am confident that the sequence will still be the
> > same as in A's frame).
> >
> > Gary
>
> Let me try to explain it better, since it seems we're not on the same page.
>
> Frames A and B have common origin O(0, 0), which means observers A and B
> were together at time 0. Time t1>0 is some arbitrary later time in frame A.
Why make things more complicated than they have to be? Make t1 = t1' = 0
>
> On 1/14/2015 1:38 PM, Gary Harnagel wrote:
> >
> > > t3 = t1 * Vm^2/(Vm-Vs)^2 * (1-Vs^2)
> > >
> > > Now just pick Vs and Vm to make t3<t1.
> >
> > My, but you went through such a convoluted mess, and you did it wrong
> > anyway. Want to try that again using the correct procedures? (I
> > don't need to as I am confident that the sequence will still be the
> > same as in A's frame).
> >
> > Gary
>
> Let me try to explain it better, since it seems we're not on the same page.
>
> Frames A and B have common origin O(0, 0), which means observers A and B
> were together at time 0. Time t1>0 is some arbitrary later time in frame A.
and be done with it.
> All t, t', x and x' in my calculation represent coordinates, not
> distances or elapsed times.
>
> At time t1 observer B is at x1 = Vs*t1, I'm guessing this is your 'd'
> but this is not needed anywhere.
And I don't understand why you can't look at it from that way, too.
> At time t2 B is at x2 = Vs*t2.
>
> The missile fired by A travels from event E1(0, t1) to event E2(x2, t2)
> so it travels a distance of x2-0 in time t2-t1 at speed Vm, IOW
>
> x2 = Vm * (t2-t1)
>
> but x2 is also where B is at t2, x2 = Vs*t2 from above, which makes
>
> t2 = t1 * Vm/(Vm-Vs)
>
> and so on...
> You may want to try again remembering that I'm working only with
> coordinates. If you still have doubts we can go over it together
> step-by-step.
frame, the missile gets back AFTER B leaves A. I haven't worked through
that unholy mess of equations you've generated, but if you did them right,
they could not disagree with my direct approach.
I'll work it out (eventually) by switching frames, but it seems an utter
waste of time. It seems unlikely that starting M after S leaves A could
cause the problem, but what do I know?
Gary
Gary Harnagel
Jan 14, 2015, 4:52:19 PM1/14/15
to
the missile back. D's clock is synchronized to A's clock and the missile
fires at t2, which is a positive number. The missile cannot get back to A
in less than zero time. Q.E.D.
Gary
Vitro
Jan 14, 2015, 6:58:18 PM1/14/15
to
analysing the following:
A and B travel away from each other at Vs = 0.5. A fires a missile
toward B at speed Vm = 4 (wrt A).
What's the speed of the missile in frame B? What's odd about it's
trajectory as observed by B?
It helps to draw a diagram.
Vitro
Jan 14, 2015, 7:01:24 PM1/14/15
to
On 15-01-14 06:57 PM, Vitro wrote:
>
> Wrong, and that's actually the whole point. You can see that by
> analysing the following:
>
> A and B travel away from each other at Vs = 0.5. A fires a missile
> toward B at speed Vm = 4 (wrt A).
>
> What's the speed of the missile in frame B? What's odd about it's
> trajectory as observed by B?
>
> It helps to draw a diagram.
>
**its trajectory
>
> Wrong, and that's actually the whole point. You can see that by
> analysing the following:
>
> A and B travel away from each other at Vs = 0.5. A fires a missile
> toward B at speed Vm = 4 (wrt A).
>
> What's the speed of the missile in frame B? What's odd about it's
> trajectory as observed by B?
>
> It helps to draw a diagram.
>
Gary Harnagel
Jan 14, 2015, 11:01:47 PM1/14/15
to
relativistic velocity addition (subtraction, actually) formula:
Vn = (Vm - Vs)/(1 - Vm*Vs)
Yes? You have realized that at certain values of Vm and Vs the
denominator becomes zero, yes? That stopped me for quite a while, too,
until I decided to look deeper and see what is really going on there.
And what is really going on there does not produce the anomaly even
though the formula does.
THAT is why I insist on staying in the AD frame. Doing so does not steer
you wrong like the formula does.
You must admit that if staying in A does not produce the anomaly, staying
in B must not do so either. Can't you see that?
Besides, how can it be wrong to stay in A's frame? It is irrational to
deny that.
Gary
Vitro
Jan 15, 2015, 9:57:03 AM1/15/15
to
On 1/14/2015 11:01 PM, Gary Harnagel wrote:
> On Wednesday, January 14, 2015 at 4:58:18 PM UTC-7, Vitro wrote:
>>
>> On 15-01-14 04:52 PM, Gary Harnagel wrote:
>>>
>>> Okay, I thought. Put an observer D at x2, the point and time
>>> that B launches the missile back. D's clock is synchronized to
>>> A's clock and the missile fires at t2, which is a positive
>>> number. The missile cannot get back to A in less than zero time.
>>> Q.E.D.
>>>
>>> Gary
>>>
>>
>> Wrong, and that's actually the whole point. You can see that by
>> analysing the following:
>>
>> A and B travel away from each other at Vs = 0.5. A fires a missile
>> toward B at speed Vm = 4 (wrt A).
>>
>> What's the speed of the missile in frame B? What's odd about it's
>> trajectory as observed by B?
>>
>> It helps to draw a diagram.
>
> What helps is to think PROPERLY. I assume you are referring to the
> relativistic velocity addition (subtraction, actually) formula:
>
> Vn = (Vm - Vs)/(1 - Vm*Vs)
>
> Yes?
No, that's not meant for FTL. What you do is take a series of events
> On Wednesday, January 14, 2015 at 4:58:18 PM UTC-7, Vitro wrote:
>>
>> On 15-01-14 04:52 PM, Gary Harnagel wrote:
>>>
>>> Okay, I thought. Put an observer D at x2, the point and time
>>> that B launches the missile back. D's clock is synchronized to
>>> A's clock and the missile fires at t2, which is a positive
>>> number. The missile cannot get back to A in less than zero time.
>>> Q.E.D.
>>>
>>> Gary
>>>
>>
>> Wrong, and that's actually the whole point. You can see that by
>> analysing the following:
>>
>> A and B travel away from each other at Vs = 0.5. A fires a missile
>> toward B at speed Vm = 4 (wrt A).
>>
>> What's the speed of the missile in frame B? What's odd about it's
>> trajectory as observed by B?
>>
>> It helps to draw a diagram.
>
> What helps is to think PROPERLY. I assume you are referring to the
> relativistic velocity addition (subtraction, actually) formula:
>
> Vn = (Vm - Vs)/(1 - Vm*Vs)
>
> Yes?
E(x, t) along the missile path in frame A then LT them to frame B.
So I said the missile travels at Vm=4 wrt A. Let the firing event be
E1(0, 0), the next E2(4, 1), then E3(8, 2), E4(12, 3), etc.
Now let's transform them to frame B:
x' = gamma*(x - Vs*t)
t' = gamma*(t - Vs*x)
E1(0, 0), E2(4.041452, -1.1547), E3(8.0827, -2.3094),
E4(12.12435, -3.4641), ...
What do you notice about the successive event times as measured by B?
The point is if something travels FTL in one frame then you can find
another frame where that thing travels backwards in time. So the missile
fired by A actually travels back in time in frame B, and the missile
fired by B also travels back in time in frame A.
So, in the scenario I proposed we had A fire at t1, hits B at t2>t1,
return missile hits A at t3<t1, meaning the return missile from B
travels to A in negative time!
> You have realized that at certain values of Vm and Vs the denominator
> becomes zero, yes? That stopped me for quite a while, too, until I
> decided to look deeper and see what is really going on there. And
> what is really going on there does not produce the anomaly even
> though the formula does.
>
> THAT is why I insist on staying in the AD frame. Doing so does not
> steer you wrong like the formula does.
causality is violated in one frame it is so in every frame.
What steers you wrong is "thinking PROPERLY" instead of just applying
the theory without letting intuition bias creep in. Golden rule: always
translate your scenarios in terms of events, then use and transform them
correctly. It will not mater which frames you use, how many and how
often you transform between them, the result should be the same regardless.
>
> You must admit that if staying in A does not produce the anomaly,
B to A in negative time.
> staying in B must not do so either. Can't you see that?
>
> Besides, how can it be wrong to stay in A's frame? It is irrational
> to deny that.
>
> Gary
>
Gary Harnagel
Jan 15, 2015, 10:44:32 AM1/15/15
to
On Thursday, January 15, 2015 at 7:57:03 AM UTC-7, Vitro wrote:
>
> On 1/14/2015 11:01 PM, Gary Harnagel wrote:
> >
> > What helps is to think PROPERLY. I assume you are referring to the
> > relativistic velocity addition (subtraction, actually) formula:
> >
> > Vn = (Vm - Vs)/(1 - Vm*Vs)
> >
> > Yes?
>
> No, that's not meant for FTL.
Well, I'm glad that you recognize that. It took me quite a while to be
>
> On 1/14/2015 11:01 PM, Gary Harnagel wrote:
> >
> > What helps is to think PROPERLY. I assume you are referring to the
> > relativistic velocity addition (subtraction, actually) formula:
> >
> > Vn = (Vm - Vs)/(1 - Vm*Vs)
> >
> > Yes?
>
> No, that's not meant for FTL.
emboldened to go beyond it. It does, however, appear to work for
(Vm + Vs)/(1 + Vm*Vs), at least it doesn't make any ridiculous claims
like the resultant velocity has reversed.
> What you do is take a series of events E(x, t) along the missile path in
> frame A then LT them to frame B.
with the RVA formula.
> So I said the missile travels at Vm=4 wrt A. Let the firing event be
> E1(0, 0), the next E2(4, 1), then E3(8, 2), E4(12, 3), etc.
that you have done so. You are merely considering the missile going on
and on forever. What's the point?
> Now let's transform them to frame B:
>
> x' = gamma*(x - Vs*t)
> t' = gamma*(t - Vs*x)
>
> E1(0, 0), E2(4.041452, -1.1547), E3(8.0827, -2.3094),
> E4(12.12435, -3.4641), ...
>
> What do you notice about the successive event times as measured by B?
without specifying a moving frame other than the missile itself, which
is beyond calculation by the LT?
> The point is if something travels FTL in one frame then you can find
> another frame where that thing travels backwards in time.
a frame coincides with the missile. B would have to travel back to a to
verify that A has been destroyed before it fired its missile. That
did NOT happen in A's frame.
> So the missile fired by A actually travels back in time in frame B, and
> the missile fired by B also travels back in time in frame A.
mathematical shenanigan.
> So, in the scenario I proposed we had A fire at t1, hits B at t2>t1,
> return missile hits A at t3<t1, meaning the return missile from B
> travels to A in negative time!
>
> > You have realized that at certain values of Vm and Vs the denominator
> > becomes zero, yes? That stopped me for quite a while, too, until I
> > decided to look deeper and see what is really going on there. And
> > what is really going on there does not produce the anomaly even
> > though the formula does.
> >
> > THAT is why I insist on staying in the AD frame. Doing so does not
> > steer you wrong like the formula does.
>
> Irrelevant.
> Physical processes don't care about frames and observers. If causality
> is violated in one frame it is so in every frame.
> What steers you wrong is "thinking PROPERLY" instead of just applying
> the theory without letting intuition bias creep in.
> Golden rule: always translate your scenarios in terms of events, then
> use and transform them correctly. It will not mater which frames you use,
> how many and how often you transform between them, the result should be
> the same regardless.
>
> > You must admit that if staying in A does not produce the anomaly,
>
> It wrongfully makes you think that the return missile cannot travel from
> B to A in negative time.
> > staying in B must not do so either. Can't you see that?
> >
> > Besides, how can it be wrong to stay in A's frame? It is irrational
> > to deny that.
>
> Not wrong, if done correctly. Wrong in your case.
a valid argument.
Gary
Tom Roberts
Jan 15, 2015, 12:57:47 PM1/15/15
to
to derive the Doppler effect. Indeed the derivation for the Doppler effect for
light parallels the derivation of the Doppler effect for sound using Newtonian
mechanics; the difference is using the Lorentz transform and its composition of
velocities.
IOW: the Doppler effect in SR is required for consistency.
I see nothing "ad hoc" about that.
Tom Roberts
Tom Roberts
Jan 15, 2015, 1:57:28 PM1/15/15
to
On 1/13/15 1/13/15 - 8:07 AM, Gary Harnagel wrote:
> [...]
violate causality in the rather narrow sense of arriving before leaving. The
limerick abandons accuracy for humor.
There is a large literature on this, and a search for "tachyon" will find much
of it.
(Tachyons are hypothetical particles that travel faster than c
(never slower). They have many curious properties (e.g. the
slower they go the higher their kinetic energy). There have
been many experimental searches for them, but none have ever
been found.)
There are, however, serious theoretical problems with FTL travel. In particular,
the time order of "leaving A" and "arriving at B" is not well defined; this
causes problems with causality as normally defined. If A launches a FTL missile
that hits B, in some other frame of reference the time ordering is different and
the missile travels from B to A -- that's quite strange, because presumably the
effort to launch a missile is very different from what happens when it hits B.
[This does not affect QFTs like QED, in which FTL (virtual)
photons are common; the emission and absorption of a photon
are time reversals of each other, and the theory retains its
Lorentz invariance.]
You can also find frames in which that FTL missile is going backward in time.
That is, for such a missile the future is NOT WELL DEFINED. This does not
necessarily lead to self-contradictions (e.g. as in QED), but it is certainly
counter to our experience!
This is what Vitro has been trying to tell you.
You have been adamantly attempting to "stay in frame A", but
that's invalid -- anybody can observe the world and describe
what happens; you cannot restrict them to certain frames. For
an FTL missile those descriptions can be WILDLY different, and
inconsistent with our experience that the future is always in
the future and is the same for everyone.
Another more serious problem arises in GR, where the existence of FTL effects
destroys global hyperbolicity. That basically means that the field equation
cannot be solved as a boundary-value problem, and the entire theory collapses.
Note that "warp metrics" like that of Alcubierre do NOT
involve anything FTL. They are merely rather curious manifolds
in which between certain trajectories, some paths for timelike
objects arrive before other paths for lightlike objects. They
are also HIGHLY nonphysical (unobtanium with negative mass
density is essential)!
Of course to date no FTL signal (or travel) has ever been observed, so GR
survives, and the whole discussion is purely academic.
Tom Roberts
> [...]
> So it seems that the young lady from Bight whose speed was much faster
> than light could never arrive the previous night :-)
Yes. It has been known for a long time (many decades) that FTL travel need not
> than light could never arrive the previous night :-)
violate causality in the rather narrow sense of arriving before leaving. The
limerick abandons accuracy for humor.
There is a large literature on this, and a search for "tachyon" will find much
of it.
(Tachyons are hypothetical particles that travel faster than c
(never slower). They have many curious properties (e.g. the
slower they go the higher their kinetic energy). There have
been many experimental searches for them, but none have ever
been found.)
There are, however, serious theoretical problems with FTL travel. In particular,
the time order of "leaving A" and "arriving at B" is not well defined; this
causes problems with causality as normally defined. If A launches a FTL missile
that hits B, in some other frame of reference the time ordering is different and
the missile travels from B to A -- that's quite strange, because presumably the
effort to launch a missile is very different from what happens when it hits B.
[This does not affect QFTs like QED, in which FTL (virtual)
photons are common; the emission and absorption of a photon
are time reversals of each other, and the theory retains its
Lorentz invariance.]
You can also find frames in which that FTL missile is going backward in time.
That is, for such a missile the future is NOT WELL DEFINED. This does not
necessarily lead to self-contradictions (e.g. as in QED), but it is certainly
counter to our experience!
This is what Vitro has been trying to tell you.
You have been adamantly attempting to "stay in frame A", but
that's invalid -- anybody can observe the world and describe
what happens; you cannot restrict them to certain frames. For
an FTL missile those descriptions can be WILDLY different, and
inconsistent with our experience that the future is always in
the future and is the same for everyone.
Another more serious problem arises in GR, where the existence of FTL effects
destroys global hyperbolicity. That basically means that the field equation
cannot be solved as a boundary-value problem, and the entire theory collapses.
Note that "warp metrics" like that of Alcubierre do NOT
involve anything FTL. They are merely rather curious manifolds
in which between certain trajectories, some paths for timelike
objects arrive before other paths for lightlike objects. They
are also HIGHLY nonphysical (unobtanium with negative mass
density is essential)!
Of course to date no FTL signal (or travel) has ever been observed, so GR
survives, and the whole discussion is purely academic.
Tom Roberts
nick
Jan 15, 2015, 2:34:19 PM1/15/15
to
Tom Roberts wrote:
> On 1/13/15 1/13/15 - 6:28 PM, Gary Harnagel wrote:
>> On Tuesday, January 13, 2015 at 10:34:47 AM UTC-7, John Heath wrote:
>>> Doppler effect is well and alive in SR.
>>
>> Only in an ad hoc manner.
>
> I'm not sure what you mean by this. Certainly no new postulates are
Nothing, he is just stupid. Can't even spell ad-hock. He just likes
> On 1/13/15 1/13/15 - 6:28 PM, Gary Harnagel wrote:
>> On Tuesday, January 13, 2015 at 10:34:47 AM UTC-7, John Heath wrote:
>>> Doppler effect is well and alive in SR.
>>
>> Only in an ad hoc manner.
>
> I'm not sure what you mean by this. Certainly no new postulates are
talking to scientists.
> needed in SR to derive the Doppler effect. Indeed the derivation for the
> Doppler effect for light parallels the derivation of the Doppler effect
> for sound using Newtonian mechanics; the difference is using the Lorentz
> transform and its composition of velocities.
>
> IOW: the Doppler effect in SR is required for consistency.
>
> I see nothing "ad hoc" about that.
>
>
> Tom Roberts
pnal...@gmail.com
Jan 15, 2015, 3:43:33 PM1/15/15
to
On Thursday, January 15, 2015 at 11:34:19 AM UTC-8, nick wrote:
> ... Can't even spell ad-hock.
Ad-hock? Really? Just who is the stupid one here?
http://itknowledgeexchange.techtarget.com/writing-for-business/ad-hoc-ad-hock-add-hoc-add-hock/
> ... Can't even spell ad-hock.
Ad-hock? Really? Just who is the stupid one here?
http://itknowledgeexchange.techtarget.com/writing-for-business/ad-hoc-ad-hock-add-hoc-add-hock/
Gary Harnagel
Jan 15, 2015, 4:15:51 PM1/15/15
to
On Thursday, January 15, 2015 at 11:57:28 AM UTC-7, tjrob137 wrote:
>
> On 1/13/15 1/13/15 - 8:07 AM, Gary Harnagel wrote:
> > [...]
> > So it seems that the young lady from Bight whose speed was much faster
> > than light could never arrive the previous night :-)
>
> Yes. It has been known for a long time (many decades) that FTL travel
> need not violate causality in the rather narrow sense of arriving before
> leaving. The limerick abandons accuracy for humor.
>
> There is a large literature on this, and a search for "tachyon" will
> find much of it.
Yes, I started digging further into SR on the tachyon subject a decade
>
> On 1/13/15 1/13/15 - 8:07 AM, Gary Harnagel wrote:
> > [...]
> > So it seems that the young lady from Bight whose speed was much faster
> > than light could never arrive the previous night :-)
>
> Yes. It has been known for a long time (many decades) that FTL travel
> need not violate causality in the rather narrow sense of arriving before
> leaving. The limerick abandons accuracy for humor.
>
> There is a large literature on this, and a search for "tachyon" will
> find much of it.
or so ago. The first thing I found was that they were claimed to
"violate causality." I'm re-thinking that :-)
> There are, however, serious theoretical problems with FTL travel.
particles beating on your ship :-)
> In particular, the time order of "leaving A" and "arriving at B" is not
> well defined; this causes problems with causality as normally defined.
> If A launches a FTL missile that hits B, in some other frame of reference
> the time ordering is different and the missile travels from B to A --
> that's quite strange, because presumably the effort to launch a missile
> is very different from what happens when it hits B.
supposedly strikes or you are at A where it supposedly strikes. Anyone
else in some other place is experiencing phenomena similar to mutual
time dilation. Their cries of alarm mean nothing unless they are right
where the action is.
> [This does not affect QFTs like QED, in which FTL (virtual)
> photons are common; the emission and absorption of a photon
> are time reversals of each other, and the theory retains its
> Lorentz invariance.]
>
> You can also find frames in which that FTL missile is going backward in
> time.
> That is, for such a missile the future is NOT WELL DEFINED. This does not
> necessarily lead to self-contradictions (e.g. as in QED), but it is
> certainly counter to our experience!
But I think there is a better argument for it: looking at the RVA equation:
Vn = (Vs + Vm)/(1 + Vs*Vm/c^2)
for the limit as Vm goes to infinity, the result is
Vn = c^2/Vs
which seems to me to mean that if infinite velocity is observed in one
frame, it is less-than-infinite in another. So a moving ship cannot fire
an infinitely-fast missile in a "stationary" target's frame.
> This is what Vitro has been trying to tell you.
>
> You have been adamantly attempting to "stay in frame A", but
> that's invalid -- anybody can observe the world and describe
> what happens; you cannot restrict them to certain frames.
the sequence is backwards in one frame, it's backwards in all frames :-)
> For an FTL missile those descriptions can be WILDLY different, and
> inconsistent with our experience that the future is always in
> the future and is the same for everyone.
so he has one at A who observes that the missile actually hits A and after
t = 0.
> Another more serious problem arises in GR, where the existence of FTL
> effects destroys global hyperbolicity. That basically means that the
> field equation cannot be solved as a boundary-value problem, and the
> entire theory collapses.
> Note that "warp metrics" like that of Alcubierre do NOT
> involve anything FTL. They are merely rather curious manifolds
> in which between certain trajectories, some paths for timelike
> objects arrive before other paths for lightlike objects. They
> are also HIGHLY nonphysical (unobtanium with negative mass
> density is essential)!
>
> Of course to date no FTL signal (or travel) has ever been observed, so GR
> survives, and the whole discussion is purely academic.
>
> Tom Roberts
FTL travel is discovered (Alcubierre-type, wormhole-type, or just vanish
here and appear way over there). In our relatively-flat region of
space, before and after the warp-field is on, the LT should apply. It
bothers me if it can't handle the situation.
Sorry, that's just the way I am.
Gary
nick
Jan 15, 2015, 4:29:51 PM1/15/15
to
Gary Harnagel
Jan 15, 2015, 7:14:00 PM1/15/15
to
On Thursday, January 15, 2015 at 1:43:33 PM UTC-7, pnal...@gmail.com wrote:
>
> On Thursday, January 15, 2015 at 11:34:19 AM UTC-8, nick the nimrod
your watch :-)
>
> On Thursday, January 15, 2015 at 11:34:19 AM UTC-8, nick the nimrod
'nym-shifting nutjob wrote:
> >
> > ... Can't even spell ad-hock.
>
> Ad-hock? Really? Just who is the stupid one here?
>
> http://itknowledgeexchange.techtarget.com/writing-for-business/ad-hoc-ad-hock-add-hoc-add-hock/
ad hock: What happens when you try to tell the time after you hocked
> >
> > ... Can't even spell ad-hock.
>
> Ad-hock? Really? Just who is the stupid one here?
>
> http://itknowledgeexchange.techtarget.com/writing-for-business/ad-hoc-ad-hock-add-hoc-add-hock/
your watch :-)
Bobby Wagner
Jan 16, 2015, 2:13:59 PM1/16/15
to
Tom Roberts wrote:
> Another more serious problem arises in GR, where the existence of FTL
> effects destroys global hyperbolicity. That basically means that the
> field equation cannot be solved as a boundary-value problem, and the
> entire theory collapses.
Due to the nature of EM, as part of matter (now separated by a distance,
> Another more serious problem arises in GR, where the existence of FTL
> effects destroys global hyperbolicity. That basically means that the
> field equation cannot be solved as a boundary-value problem, and the
> entire theory collapses.
space) a Faster than Light speed or travel makes no any sense. For a
photon (a result of shift in energy level) it is the same matter,
disregard distances between a sender atom and a receiver atom.
This clearly settles the discussion. Einstein made a correct assumption
for whatever completely different reason. Nevertheless, he guessed right.
There CANNOT be Faster than Light travel. (unless one fakes the distances)
Gary Harnagel
Jan 18, 2015, 11:45:22 AM1/18/15
to
On Thursday, January 15, 2015 at 11:57:28 AM UTC-7, tjrob137 wrote:
>
>
> You have been adamantly attempting to "stay in frame A", but
> that's invalid -- anybody can observe the world and describe
> what happens; you cannot restrict them to certain frames.
Hi Tom,
> that's invalid -- anybody can observe the world and describe
> what happens; you cannot restrict them to certain frames.
But I don't STAY in one frame. I calculate in one frame and then use the
LT to shift to B's frame. Thus, the time S arrives at D is
t2 = d/Vs
and for B that time is
t2' = gamma*(t2 - Vm*d/c^2) = gamma*d*(1- Vm*Vs/c^2)/Vs
x2' = gamma*(d - Vm*d/Vs) = gamma*d*(1 - Vm/Vs)
If S starts back immediately upon reaching D at the same speed, he arrives
at A at time
t3 = 2*d/Vs
and for B, that's
t3' = gamma*(2*d/Vs - Vm*0) = 2*gamma*d/Vs
which is certainly greater than zero; hence, S cannot shoot himself from
D before he leaves A in any frame, thus removing that particular causality
violation scenario.
The problem with Vitro's gedanken is that it's completely different from
mine and he makes some ... um ... ill-founded assumptions. I believe my
gedanken is much more reasonable: 1) In our neck of the galactic woods,
all potential targets are moving at about the same speed (the galactic
flat velocity profile), so having S land on a relativistic rocket is not
something to consider at this early stage of interstellar exploration,
and 2) What would be the momentum of a warp ship when it turns off its
bubble? Would momentum be conserved? If so, then S would have to match
velocities with B in normal space, so S would arrive at D first, and my
gedanken would apply.
> For an FTL missile those descriptions can be WILDLY different, and
> inconsistent with our experience that the future is always in
> the future and is the same for everyone.
to me that claims like this are wrong:
"If such particles did exist, they could be used to build a tachyonic
antitelephone and send signals faster than light, which (according to special
relativity) would lead to violations of causality."
http://en.wikipedia.org/wiki/Tachyon
> Another more serious problem arises in GR, where the existence of FTL effects
> destroys global hyperbolicity. That basically means that the field equation
> cannot be solved as a boundary-value problem, and the entire theory collapses.
>
> Note that "warp metrics" like that of Alcubierre do NOT
> involve anything FTL. They are merely rather curious manifolds
> in which between certain trajectories, some paths for timelike
> objects arrive before other paths for lightlike objects.
when a warp bubble is formed at A and is turned off at D? Now you have
flat space-time again and this embarrassing ship which disappeared from A
and reappeared at D in less time than it took light to do the trip. So
whether no "FTL" was technically involved, the LT now has to deal with it,
no?
> They are also HIGHLY nonphysical (unobtanium with negative mass density is
> essential)!
much faith in that, but there are other possibilities. And I understand
they have huge supplies of unobtainium on the planet Pandora :-)
Gary
fuller...@hotmail.com
Jan 18, 2015, 11:57:26 AM1/18/15
to
> Well, there are options. Vacuum energy has been proposed. I don't place
> much faith in that, but there are other possibilities. And I understand
> they have huge supplies of unobtainium on the planet Pandora :-)
>
> Gary
If a Bathysphere could be placed at the center of the earth,
While the gravitational acceleration would be Zero,
Time would still be Dilated, yes ??
Probably closer to 94 m/s/s
299792458 x ((1.6 x 10^8) / (5.101 x 10^14)) = 94.0340977847
Yeah probably closer to 100 ish than 900
((9.8 m/s/s) / 2) ^3 = 117.649
cubed for one each of the XYZ axis'
http://i59.tinypic.com/2rdvrz7.jpg
John Heath
Jan 18, 2015, 1:21:22 PM1/18/15
to
On Tuesday, January 13, 2015 at 9:07:44 AM UTC-5, Gary Harnagel wrote:
> Consider movement of a ship, S, from point A to point B, where both points
> are in the same inertial frame and usually spoken of as the "stationary"
> frame. The Lorentz transform (LT) says that an observer aboard S will
> observe that clocks at A and B appear to be running slower than his clock
> and that observers at A and B will observe that the clock in S will appear
> to be running slow. But what REALLY counts is when clocks are at the same
> position and they can be compared directly. Under this regimen, the LT
> says that the clock in S will have gained less time than the clock at B,
> and since the B clock is synchronized with the clock at A, the S clock
> will be behind the A clock, too.
>
> So now let's consider that S is a faster-than-light (FTL) ship. Using the
> same regimen as above, let's see what happens when S moves from A to B at
> velocity n*c, where n is some number greater than unity. Let the distance
> from A to B be d in the AB inertial frame and S leaves A at t = 0. Then
> when S arrives at B, B's clock will read t1 = d/(n*c). If S immediately
> turns around and goes back to A, A's clock will read t2 = 2*d/(n*c).
> There is no problem here.
>
> The problem arises when we include another ship, M, traveling at sub-light
> velocity toward B. Let it pass A at the same time S leaves A (at t0 = 0).
> According to the LT, M will say that the clock at B reads
>
> tB' = -gamma*v*d/c^2 when tA = 0
>
> (and tB = 0 also). And when S reaches B at t1 = d/(n*c), an observer
> aboard M will read his clock as
>
> (1) t1' = gamma*(t1 - v*x/c^2)
>
> where x = d and gamma = 1/sqrt(1 - v^2/c^2). So
>
> (2) t1' = gamma*(t1- v*d/c^2) = gamma*(d/c)*(1/n - v/c)
>
> So the time it takes S to go from A to B is
>
> (3) t1' - t0' = t1' = -gamma*(d/c)*(1/n - v/c)
>
> according to M, which is a negative number for n > c/v. In other words,
> S would arrive at B before it left A. This would seem to presnt a
> causality problem.
>
> But if we insist on the regimen that what counts is the situation when B
> and M are coincident, which occurs when t3 = d/v, then
>
> (4) t3' = gamma*(d/v - v*d/c^2) = gamma*(d/v)*(1 - v^2/c^2)
>
> (5) t3' = (d/v)/gamma
>
> M would find that S has been at B (if S had not returned to A) for
>
> (6) t3' - t1' = (d/v)/gamma - gamma*(d/c)*(1/n - v/c)
>
> (7) t3' - t1' = gamma*(d/v)*[1 - v/(n*c)]
>
> But is this really a problem? Is this any more magical than B's clock
> mysteriously found to be ahead of M's clock when B and M coincide?
>
> So what about causality? Consider the case where S travels to B
> instantaneously. According to M, S arrives at B at time
>
> t1' = gamma*(0 - v*d/c^2) = -gamma*v*d/c^2
>
> And now suppose S sends back a superluminal missile aimed to destroy M.
> Can M be destroyed before S leaves A?
>
> The missile moves at n*c in the AB frame. In that frame, the missile gets
> back to A at time d/(n*c). So even if n is infinity, M cannot be destroyed
> at any time less than zero, and causality is not violated. So if a method
> of FTL travel is ever perfected (perhaps an Alcubierre or Natario warp
> drive or some other technology), it does not appear that reality will be
> shattered by destroying causality.
Had a second thought on this. S goes from A to B. In this case I would like to use the speed of sound instead of light. S moves 2 times the speed of sound from A to B. There is a bat that lives on B . Mr. Bat sees in sound not light. What would Mr. Bat see when S passes at 2 times the speed of sound? His little ears will not detect S but they will detect the sonic boom. What would this sonic boom look like to our bat? The sonic boom is length contracted as sound can never travel faster than sound itself. The very notion is logically impossible. The energy density of the sonic boom is also greater due to being length contracted. The chronological order of the random noise made by S moving through air is preserved in the sonic boom. That is to say that the front of the sonic boom will have more resent sounds given off by S and the rear will have sounds that were made in the past. This would be confusing for the bat as his ears would tell him S is moving backwards in time from B to A. The fussy point I am taking longer than expected to make is information going backwards in time does not mean that S is going backwards in time. The notion of FTL travel violating causality by going backwards in time is false if the bat analogy holds true. Got it off my chest. Now I can have lunch :<).
> Consider movement of a ship, S, from point A to point B, where both points
> are in the same inertial frame and usually spoken of as the "stationary"
> frame. The Lorentz transform (LT) says that an observer aboard S will
> observe that clocks at A and B appear to be running slower than his clock
> and that observers at A and B will observe that the clock in S will appear
> to be running slow. But what REALLY counts is when clocks are at the same
> position and they can be compared directly. Under this regimen, the LT
> says that the clock in S will have gained less time than the clock at B,
> and since the B clock is synchronized with the clock at A, the S clock
> will be behind the A clock, too.
>
> So now let's consider that S is a faster-than-light (FTL) ship. Using the
> same regimen as above, let's see what happens when S moves from A to B at
> velocity n*c, where n is some number greater than unity. Let the distance
> from A to B be d in the AB inertial frame and S leaves A at t = 0. Then
> when S arrives at B, B's clock will read t1 = d/(n*c). If S immediately
> turns around and goes back to A, A's clock will read t2 = 2*d/(n*c).
> There is no problem here.
>
> The problem arises when we include another ship, M, traveling at sub-light
> velocity toward B. Let it pass A at the same time S leaves A (at t0 = 0).
> According to the LT, M will say that the clock at B reads
>
> tB' = -gamma*v*d/c^2 when tA = 0
>
> (and tB = 0 also). And when S reaches B at t1 = d/(n*c), an observer
> aboard M will read his clock as
>
> (1) t1' = gamma*(t1 - v*x/c^2)
>
> where x = d and gamma = 1/sqrt(1 - v^2/c^2). So
>
> (2) t1' = gamma*(t1- v*d/c^2) = gamma*(d/c)*(1/n - v/c)
>
> So the time it takes S to go from A to B is
>
> (3) t1' - t0' = t1' = -gamma*(d/c)*(1/n - v/c)
>
> according to M, which is a negative number for n > c/v. In other words,
> S would arrive at B before it left A. This would seem to presnt a
> causality problem.
>
> But if we insist on the regimen that what counts is the situation when B
> and M are coincident, which occurs when t3 = d/v, then
>
> (4) t3' = gamma*(d/v - v*d/c^2) = gamma*(d/v)*(1 - v^2/c^2)
>
> (5) t3' = (d/v)/gamma
>
> M would find that S has been at B (if S had not returned to A) for
>
> (6) t3' - t1' = (d/v)/gamma - gamma*(d/c)*(1/n - v/c)
>
> (7) t3' - t1' = gamma*(d/v)*[1 - v/(n*c)]
>
> But is this really a problem? Is this any more magical than B's clock
> mysteriously found to be ahead of M's clock when B and M coincide?
>
> So what about causality? Consider the case where S travels to B
> instantaneously. According to M, S arrives at B at time
>
> t1' = gamma*(0 - v*d/c^2) = -gamma*v*d/c^2
>
> And now suppose S sends back a superluminal missile aimed to destroy M.
> Can M be destroyed before S leaves A?
>
> The missile moves at n*c in the AB frame. In that frame, the missile gets
> back to A at time d/(n*c). So even if n is infinity, M cannot be destroyed
> at any time less than zero, and causality is not violated. So if a method
> of FTL travel is ever perfected (perhaps an Alcubierre or Natario warp
> drive or some other technology), it does not appear that reality will be
> shattered by destroying causality.
>
> So it seems that the young lady from Bight whose speed was much faster
> than light could never arrive the previous night :-)
>
> Gary
> So it seems that the young lady from Bight whose speed was much faster
> than light could never arrive the previous night :-)
>
Had a second thought on this. S goes from A to B. In this case I would like to use the speed of sound instead of light. S moves 2 times the speed of sound from A to B. There is a bat that lives on B . Mr. Bat sees in sound not light. What would Mr. Bat see when S passes at 2 times the speed of sound? His little ears will not detect S but they will detect the sonic boom. What would this sonic boom look like to our bat? The sonic boom is length contracted as sound can never travel faster than sound itself. The very notion is logically impossible. The energy density of the sonic boom is also greater due to being length contracted. The chronological order of the random noise made by S moving through air is preserved in the sonic boom. That is to say that the front of the sonic boom will have more resent sounds given off by S and the rear will have sounds that were made in the past. This would be confusing for the bat as his ears would tell him S is moving backwards in time from B to A. The fussy point I am taking longer than expected to make is information going backwards in time does not mean that S is going backwards in time. The notion of FTL travel violating causality by going backwards in time is false if the bat analogy holds true. Got it off my chest. Now I can have lunch :<).
Julian Lanham
Jan 18, 2015, 4:46:23 PM1/18/15
to
Gary Harnagel wrote:
> On Thursday, January 15, 2015 at 11:57:28 AM UTC-7, tjrob137 wrote:
>>
>> You have been adamantly attempting to "stay in frame A", but that's
>> invalid -- anybody can observe the world and describe what happens;
>> you cannot restrict them to certain frames.
>
> Hi Tom,
>
> But I don't STAY in one frame. I calculate in one frame and then use
> the LT to shift to B's frame. Thus, the time S arrives at D is
>
> t2 = d/Vs
>
> and for B that time is
>
> t2' = gamma*(t2 - Vm*d/c^2) = gamma*d*(1- Vm*Vs/c^2)/Vs
t2' = gamma*(t2 - Vm*d/c^2) = gamma*d*(1/Vs - Vm/c^2)
> On Thursday, January 15, 2015 at 11:57:28 AM UTC-7, tjrob137 wrote:
>>
>> You have been adamantly attempting to "stay in frame A", but that's
>> invalid -- anybody can observe the world and describe what happens;
>> you cannot restrict them to certain frames.
>
> Hi Tom,
>
> But I don't STAY in one frame. I calculate in one frame and then use
> the LT to shift to B's frame. Thus, the time S arrives at D is
>
> t2 = d/Vs
>
> and for B that time is
>
> t2' = gamma*(t2 - Vm*d/c^2) = gamma*d*(1- Vm*Vs/c^2)/Vs
LOL.
> The problem with Vitro's gedanken is that it's completely different from
> mine and he makes some ... um ... ill-founded assumptions. I believe my
> gedanken is much more reasonable:
> 1) In our neck of the galactic woods,
> all potential targets are moving at about the same speed (the galactic
> flat velocity profile), so having S land on a relativistic rocket is not
> something to consider at this early stage of interstellar exploration,
> and
> 2) What would be the momentum of a warp ship when it turns off its
> bubble? Would momentum be conserved? If so, then S would have to match
> velocities with B in normal space, so S would arrive at D first, and my
> gedanken would apply.
Tom that you cannot just make yourself arbiter arbitrating from a
privileged proffered frame. Are you really that stupid?
> Well, there are options. Vacuum energy has been proposed. I don't
> place much faith in that, but there are other possibilities. And I
> understand they have huge supplies of unobtainium on the planet Pandora
> :-)
Gary Harnagel
Jan 18, 2015, 7:04:10 PM1/18/15
to
On Sunday, January 18, 2015 at 2:46:23 PM UTC-7, Julian Lanham wrote:
>
> As observed by whom? You always refuse to tell. You've been told by Mr.
> Tom that you cannot just make yourself arbiter arbitrating from a
> privileged proffered frame. Are you really that stupid?
My, but the persistent pest resurfaces with a new pseudonym and is recognized
>
> As observed by whom? You always refuse to tell. You've been told by Mr.
> Tom that you cannot just make yourself arbiter arbitrating from a
> privileged proffered frame. Are you really that stupid?
immediately. Why does he do it? Is he insane? Is he so starved for
attention that he will make a fool of himself? Doesn't he know this truth?
"stop trying to fill your head with science--for to fill your heart with love
is enough." -- Richard P. Feynman
http://new.damn.com/took-just-four-minutes-bring-entire-middle-school-tears-wow/
Julian Lanham
Jan 18, 2015, 7:29:56 PM1/18/15
to
Gary Harnagel wrote:
> On Sunday, January 18, 2015 at 2:46:23 PM UTC-7, Julian Lanham wrote:
>>
>> As observed by whom? You always refuse to tell. You've been told by Mr.
>> Tom that you cannot just make yourself arbiter arbitrating from a
>> privileged proffered frame. Are you really that stupid?
>
> My, but the persistent pest resurfaces with a new pseudonym and is
> recognized immediately. Why does he do it? Is he insane? Is he so
> starved for attention that he will make a fool of himself? Doesn't he
> know this truth?
Apropos names, yours probably is an anagram, which indicates a rather
> On Sunday, January 18, 2015 at 2:46:23 PM UTC-7, Julian Lanham wrote:
>>
>> As observed by whom? You always refuse to tell. You've been told by Mr.
>> Tom that you cannot just make yourself arbiter arbitrating from a
>> privileged proffered frame. Are you really that stupid?
>
> My, but the persistent pest resurfaces with a new pseudonym and is
> recognized immediately. Why does he do it? Is he insane? Is he so
> starved for attention that he will make a fool of himself? Doesn't he
> know this truth?
Gabriel Archangel or similar. Your reaction and refusal to relate to the
posed issue, strictly suggests that Gabriel Archangel is evading as being
caught deep in stupidity. Have a nice day.
*plonk*
Tom Roberts
Jan 21, 2015, 1:15:23 PM1/21/15
to
On 1/15/15 1/15/15 - 3:15 PM, Gary Harnagel wrote:
> On Thursday, January 15, 2015 at 11:57:28 AM UTC-7, tjrob137 wrote:
>> In particular, the time order of "leaving A" and "arriving at B" is not
>> well defined; this causes problems with causality as normally defined.
>> If A launches a FTL missile that hits B, in some other frame of reference
>> the time ordering is different and the missile travels from B to A --
>> that's quite strange, because presumably the effort to launch a missile
>> is very different from what happens when it hits B.
>
> But that doesn't happen if you are at B when the purported missile
> supposedly strikes or you are at A where it supposedly strikes. Anyone
> else in some other place is experiencing phenomena similar to mutual
> time dilation. Their cries of alarm mean nothing unless they are right
> where the action is.
Their "cries of alarm" mean that causality is not well defined, as "arrival" can
> On Thursday, January 15, 2015 at 11:57:28 AM UTC-7, tjrob137 wrote:
>> In particular, the time order of "leaving A" and "arriving at B" is not
>> well defined; this causes problems with causality as normally defined.
>> If A launches a FTL missile that hits B, in some other frame of reference
>> the time ordering is different and the missile travels from B to A --
>> that's quite strange, because presumably the effort to launch a missile
>> is very different from what happens when it hits B.
>
> But that doesn't happen if you are at B when the purported missile
> supposedly strikes or you are at A where it supposedly strikes. Anyone
> else in some other place is experiencing phenomena similar to mutual
> time dilation. Their cries of alarm mean nothing unless they are right
> where the action is.
be seen to happen before "departure". This applies not only to your example of a
missile over many kilometers, but also to a tabletop experiment in a lab.
Accepting tachyon interactions with normal matter requires a MAJOR conceptual
shift, and a COMPLETE revision of our ideas about the world.
>> You can also find frames in which that FTL missile is going backward in
>> time.
>> That is, for such a missile the future is NOT WELL DEFINED. This does not
>> necessarily lead to self-contradictions (e.g. as in QED), but it is
>> certainly counter to our experience!
>
> Yes, but so is mutual time dilation.
ambiguity in causality for an FTL missile goes MUCH deeper.
> But I think there is a better argument for it: looking at the RVA equation:
> Vn = (Vs + Vm)/(1 + Vs*Vm/c^2)
> for the limit as Vm goes to infinity, the result is
> Vn = c^2/Vs
> which seems to me to mean that if infinite velocity is observed in one
> frame, it is less-than-infinite in another. So a moving ship cannot fire
> an infinitely-fast missile in a "stationary" target's frame.
applied to them. Remember: the kinetic energy of a tachyon increases as it SLOWS
DOWN. That is so counter-intuitive and unusual that most everything breaks down.
>> Of course to date no FTL signal (or travel) has ever been observed, so GR
>> survives, and the whole discussion is purely academic.
>
> What is academic at one time is not at another. Suppose some method of
> FTL travel is discovered (Alcubierre-type, wormhole-type, or just vanish
> here and appear way over there). In our relatively-flat region of
> space, before and after the warp-field is on, the LT should apply. It
> bothers me if it can't handle the situation.
But the LT cannot "handle" any situation in which gravitation is important, much
> FTL travel is discovered (Alcubierre-type, wormhole-type, or just vanish
> here and appear way over there). In our relatively-flat region of
> space, before and after the warp-field is on, the LT should apply. It
> bothers me if it can't handle the situation.
less such speculative things as warp drives and wormholes!
Tom Roberts
Gary Harnagel
Jan 21, 2015, 1:45:07 PM1/21/15
to
seems reasonable that SR should apply in those segments of time.
But you're right. It is speculation and we don't know the conditions
needed to make and break the bubble. Same for the unknown technology
that causes someone to vanish here and appear over there.
Gary
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