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twin paradox with distant mirror

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  twin paradox with distant mirror gertjacobusse@gmail.com - 2017-02-06 11:55 -0800
    Re: twin paradox with distant mirror Ned Latham <nedlatham@woden.valhalla.oz> - 2017-02-06 22:33 +0000
    Re: twin paradox with distant mirror "Dono," <sa_ge@comcast.net> - 2017-02-06 15:11 -0800
      Re: twin paradox with distant mirror gertjacobusse@gmail.com - 2017-02-06 23:44 -0800
        Re: twin paradox with distant mirror gertjacobusse@gmail.com - 2017-02-07 02:46 -0800
        Re: twin paradox with distant mirror "Paul B. Andersen" <relativity@paulba.no> - 2017-02-07 16:29 +0100
    Re: twin paradox with distant mirror rotchm <rotchm@gmail.com> - 2017-02-06 21:40 -0800
      Re: twin paradox with distant mirror "Dono," <sa_ge@comcast.net> - 2017-02-06 22:27 -0800
    Re: twin paradox with distant mirror "Paul B. Andersen" <relativity@paulba.no> - 2017-02-07 12:55 +0100
      Re: twin paradox with distant mirror gertjacobusse@gmail.com - 2017-02-07 04:51 -0800
        Re: twin paradox with distant mirror gertjacobusse@gmail.com - 2017-02-07 05:33 -0800
          Re: twin paradox with distant mirror "Dono," <sa_ge@comcast.net> - 2017-02-07 05:45 -0800
            Re: twin paradox with distant mirror gertjacobusse@gmail.com - 2017-02-07 06:00 -0800
              Re: twin paradox with distant mirror "Dono," <sa_ge@comcast.net> - 2017-02-07 06:17 -0800
              Re: twin paradox with distant mirror Odd Bodkin <bodkinodd@gmail.com> - 2017-02-07 08:21 -0600
                Re: twin paradox with distant mirror gertjacobusse@gmail.com - 2017-02-07 06:51 -0800
                  Re: twin paradox with distant mirror "Dono," <sa_ge@comcast.net> - 2017-02-07 07:09 -0800
                Re: twin paradox with distant mirror gertjacobusse@gmail.com - 2017-02-08 06:23 -0800
                  Re: twin paradox with distant mirror mlwozniak@wp.pl - 2017-02-08 06:51 -0800
                    Re: twin paradox with distant mirror gertjacobusse@gmail.com - 2017-02-08 13:30 -0800
                      Re: twin paradox with distant mirror mlwozniak@wp.pl - 2017-02-08 22:49 -0800

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#408139 — twin paradox with distant mirror

I am trying to understand the twin paradox, and I would like an explanation of the following. I hope someone can help me, thanks for your patience!

A stays on earth, B moves at 0.99 times the speed of light towards point X at a distance of one light year (from A's reference frame), and instantly returns at the same speed (leaving, returning and arriving back with incredible acceleration that will totally alter his reference frame).

At point X there is a mirror. One day after B leaves, A sends an extremely powerful flash of light towards X. From the perspective of A, the flash of light will be back exactly two years later (because X is one light year away).

But where will B be, relative to the flash of light?

If I try to reason from the fact that the speed of light is constant from every reference frame:

- from reference frame B, B will have traveled 0.99 light day, so the flash will catch up in less then one day
- from reference frame A, B is almost a light day away, but 0.99 light day later he will almost be two light days away (and not caught up by the flash yet)

I guess my reasoning is too simple because the speed and/or location of B may not be the same from the reference frames of A and B. But what will happen, from both perspectives?

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#408152

gertjacobusse wrote:
>
> I am trying to understand the twin paradox, and I would like
> an explanation of the following. I hope someone can help me,
> thanks for your patience!
>
> A stays on earth, B moves at 0.99 times the speed of light
> towards point X at a distance of one light year (from A's
> reference frame), and instantly returns at the same speed
> (leaving, returning and arriving back with incredible
> acceleration that will totally alter his reference frame).
>
> At point X there is a mirror. One day after B leaves, A
> sends an extremely powerful flash of light towards X.
> From the perspective of A, the flash of light will be back
> exactly two years later (because X is one light year away).
>
> But where will B be, relative to the flash of light?

By setting the distance relative to A, you have chosen A's
reference frame for whe whole experinebt, Sending the
flash of light from point A does the same. A's reference
frame is the meaningdul one.

When the flash of light overtakes B both have covered the
same distance and B has taken one day, or 86400 seconds,
longer than the flash. You can use Newtin's formula relating
distance, speed and time to work out that the flash overtakes
B after 99 days (at which time B has been travelling for 100
days).

And I see no paradox.

----snup----

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#408159

On Monday, February 6, 2017 at 11:55:56 AM UTC-8, gertja...@gmail.com wrote:
> I am trying to understand the twin paradox, and I would like an explanation of the following. I hope someone can help me, thanks for your patience!
> 
> A stays on earth, B moves at 0.99 times the speed of light towards point X at a distance of one light year (from A's reference frame), and instantly returns at the same speed (leaving, returning and arriving back with incredible acceleration that will totally alter his reference frame).
> 
> At point X there is a mirror. One day after B leaves, A sends an extremely powerful flash of light towards X. From the perspective of A, the flash of light will be back exactly two years later (because X is one light year away).
> 
> But where will B be, relative to the flash of light?
> 
> If I try to reason from the fact that the speed of light is constant from every reference frame:
> 
> - from reference frame B, B will have traveled 0.99 light day, so the flash will catch up in less then one day
> - from reference frame A, B is almost a light day away, but 0.99 light day later he will almost be two light days away (and not caught up by the flash yet)
> 
> I guess my reasoning is too simple because the speed and/or location of B may not be the same from the reference frames of A and B. But what will happen, from both perspectives?

When the light came back from the mirror, the total proper elapsed time is t_a=2yrs.
From B's perspective, the total elapsed time is t_b=t_a \sqrt{1-\frac{v_b^2}{c^2}}. much less than 2 yrs. How much space did B cover in t_b years traveling at v_b speed? The answer may require knowing the acceleration at the turning point :-)

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#408195

Thanks for your helpful responses.

> From B's perspective, the total elapsed time is t_b=t_a \sqrt{1-\frac{v_b^2}{c^2}}. much less than 2 yrs. How much space did B cover in t_b years traveling at v_b speed? The answer may require knowing the acceleration at the turning point :-)

This is what still puzzles me and why I introduced the mirror. We know the acceleration: the speed change of B is instant, like a flash of light that hits a mirror. And in my opinion I defined how much space B did cover: two light years in the reference frame of A. So, if much less than 2 yrs passed for B, then B actually traveled much faster than the speed of light... or is the amount of space different for B (length contraction?)

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#408211

> We know the acceleration: the speed change of B is instant, like a flash of light that hits a mirror.

Hm, reasoning from the intertial reference frame before B returns, that speed change is impossible because it is 2*0.99*c. Is that where I went wrong?

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#408239

On 07.02.2017 08:44, gertjacobusse@gmail.com wrote:
>
> This is what still puzzles me and why I introduced the mirror.
> We know the acceleration: the speed change of B is instant,
> like a flash of light that hits a mirror.
> And in my opinion I defined how much space B did cover:
> two light years in the reference frame of A.

https://paulba.no/pdf/TwinsByMetric.pdf
See 2.2.

v = 0.99c and distance 2 light year [ly], TA = 2/0.99 year ~= 2.02 year

> So, if much less than 2 yrs passed for B,

TB = T sqrt(1-v^2/c^2) = 2.02 * 0.141 year = 0.285 year
(Same as Dono's equation)

> then B actuallytraveled much faster than the speed of light...

Quite.
v' = 2 ly/0.285y ~= 7.02c!

Note also that v' = v/sqrt(1-v^2/c^2) = 0.99c/0.141 = 7.02c

See below.

> or is the amount of space different for B (length contraction?)

When B is at the mirror when her clock shows (T/2)sqrt(1-v^2/c^2),
then A has moved away with the speed v, and will be at a distance
v (T/2)sqrt(1-v^2/c^2) as measured in B's rest frame.
The speed of A is obviously v in the B's rest frame.
And even more obvious, the speed of B in B's rest frame is zero.

So what is moving with the speed 7.02c?

It is dx/dt', that is the distance measured in A's rest frame
divided by B's proper time.
There is no upper limit to this speed.
I have a book where this speed is called 'proper speed'.
However, this book is from 1965 and I think this name
may be obsolete.

If we look at the four velocity, we have the components:
u = [c dt/dt',dx/dt', dy/dt',dz/dt']
In our case:
u = [c/sqrt(1-v^2/c^2),v/sqrt(1-v^2/c^2),0,0]

So v/sqrt(1-v^2/c^2) ~= 7c is the spatial component
of the four-velocity.

Note also |u| = sqrt(-(c/sqrt(1-v^2/c^2)^2+(v/sqrt(1-v^2/c^2)^2) = -c

-- 
Paul

https://paulba.no/

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#408187

On Monday, February 6, 2017 at 2:55:56 PM UTC-5, gertja...@gmail.com wrote:

> A stays on earth, B moves at 0.99 times the speed of light 
> towards point X at a distance of one light year (from A's 
> reference frame), and instantly returns at the same speed 
> (leaving, returning and arriving back with incredible 
> acceleration that will totally alter his reference frame).

Well, B's accelerations does alter his "reference frame". In fact, 
B DOES NOT have an associated inertial reference frame; B is not inertial.
You may however consider B as two frames; the outgoing B and another incoming BB. These two frames have different 'simultaneities' and is often a source of errors in those tackling the twin paradox (TP). So, take great care in introducing a "turning point".
 
> At point X there is a mirror. One day after B leaves, A sends 
> an extremely powerful flash of light towards X. From the 
> perspective of A, the flash of light will be back exactly 
> two years later (because X is one light year away).
> 
> But where will B be, relative to the flash of light?

Analysing from the inertial frame A: This is done by classical kinematics; using units of Years & LY, thus c = 1 & v = 0.99.

B leaves at time 0, reaches X at time 1/0.99, instantly turns around and comes back to A (x=0) at time  T1 = 2/v = 2/0.99.

The pulse is sent out at time 1/365, arrives at X at time 1/365+1 and arrives back at A at time T2 = 1/365 + 2.

Note that T1 > T2, so the pulse arrives back before B. That is, B is 
T1 - T2 units of time away as the pulse reaches home. Since B is traveling at speed v, B is thus a distance of v*(T1-T2) units away from A.

----Here, we are referring to two events: E1 being the pulse
----coinciding back with A at time T2,  and E2 being traveler
----B coinciding with the position v*(T1-T2) at time T2. This is 
----the simultaneity of these two events in A's frame and
----how you judge the distance in this scenario. 
----Now, since these two events are simul in frame A, they will
----not be simul in frame(s) B. Since they are not simul, then
----what do *you* mean by the distance (location of the pulse 
----wrt frame B or traveler B)? 
----Here, I will assume you mean the difference in positions, 
----the Δx' of those two events *in the incoming inertial frame*. 
----By the interval form of the Lorentz contraction factor, Δx' =
---- = v*(T1-T2)*gamma. 

> - from reference frame B, 

But there is no "reference frame of B". You need to be very careful in what you mean by that. 

> B will have traveled 0.99 light day,

No. Wrt the "person" B, he never travels wrt himself; he always remains at his own position (x' = 0, say). Needless to say, I stop here, since from this point, your description (words) no longer make sense. 

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#408192

On Monday, February 6, 2017 at 9:40:26 PM UTC-8, rotchm wrote:
> 
> Analysing from the inertial frame A: This is done by classical kinematics; using units of Years & LY, thus c = 1 & v = 0.99.
> 
Turkey

Since v=0.99c, only a turkey would use classical kinematics. 

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#408213

On 06.02.2017 20:55, gertjacobusse@gmail.com wrote:
> I am trying to understand the twin paradox, and I would like an explanation of the following.
> I hope someone can help me, thanks for your patience!
>
> A stays on earth, B moves at 0.99 times the speed of light
> towards point X at a distance of one light year (from A's reference frame),
> and instantly returns at the same speed (leaving, returning and arriving
> back with incredible acceleration that will totally alter his reference frame).
>
> At point X there is a mirror. One day after B leaves,
> A sends an extremely powerful flash of light towards X.
> From the perspective of A, the flash of light will be
> back exactly two years later (because X is one light year away).
>
> But where will B be, relative to the flash of light?

I suppose the question is:
Were is B when the flash hits her and when will the flash hit?

I will make it a bit more general:
Let the speed of B in A's rest frame be v,
and let the flash be emitted at the time T in A's rest frame.

Let's first do the calculations in A's rest frame.
B is starting at the time t = 0
At the time T, when the flash is emitted, B will be at position vT.
A time t later, when the flash hits B, B will be at position (T+t)v.
    ct = v(T+t) => t = vT/(c-v)
So the flash hits at the time and position in A's rest frame:
   t_1 = T + vT/(c-v) = cT/(c-v)
   x_1 = vcT/(c-v)

When the flash hits B, her clock will show:
  t_1' = (t_1 - v.x_1/c^2)/sqrt(1-v^2-c^2) = sqrt((c+v)/(c-v))T

A will move away with the speed v, and will be at the position
in B's rest frame:
   x_A' = - sqrt((c+v)/(c-v))vT

In both frames:
The faster B goes, the longer time will it take for the light
to catch up with B, and the farther away will the other
twin be when the light catches up.
But the time and distance will be less as measured in B's rest frame
than as measured in A's rest frame.

> If I try to reason from the fact that the speed of light is constant from every reference frame:
>
> - from reference frame B, B will have traveled 0.99 light day, so the flash will catch up in less then one day

T = 1 day and v = 0.99c yields: t_1' = 14.1, days x_A' = 13.6 light days

> - from reference frame A, B is almost a light day away,
> but 0.99 light day later he will almost be two light days away (and not caught up by the flash yet)

T = 1 day and v = 0.99c yields: t_1 = 100 days, x_1 = 99 light days
>
> I guess my reasoning is too simple because the speed and/or location of B may not be the same from
> the reference frames of A and B. But what will happen, from both perspectives?

This has nothing to do with the twin paradox, so what was your point?

https://paulba.no/pdf/TwinsByMetric.pdf
https://paulba.no/pdf/TwinsByDoppler.pdf
https://paulba.no/twins.html

-- 
Paul

https://paulba.no/

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#408215

> This has nothing to do with the twin paradox, so what was your point?

Thank you for explaining the calculations, this answers my question for the moment when the light catches up.

I think the twin paradox emerges when B arrives back on earth - and less time has passed for B than for A - while B has been back and forth to a location that is one light year away from A.

Wen I analyze that from the reference frame of A, B has traveled a distance of 2 light years in less than two years (faster than light?) - but has been caught up by a flash of light traveling at the speed of light... that's the part of the paradox I don't get.

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#408222

I do realize that, in the reference frame of A, it has not taken less than two years. But in the reference frame(s) of B it has; now will B agree on the distance of 2 light years - I guess not, but why?

@Paul thanks for your links, truly interesting!

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#408227

On Tuesday, February 7, 2017 at 5:33:26 AM UTC-8, gertja...@gmail.com wrote:
> I do realize that, in the reference frame of A, it has not taken less than two years. But in the reference frame(s) of B it has; now will B agree on the distance of 2 light years - I guess not, but why?
> 
> @Paul thanks for your links, truly interesting!

In B's frame of reference the distance is length contracted, this is why. 

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#408229

> In B's frame of reference the distance is length contracted, this is why.

Thank you, the fast and competent responses are really helpful, I think I understand a bit more now (although it is not parsimonious for me and my naive space/time perception to alter both space and time to maintain a constant c). I think I need to dive into the equations of Pauls links to understand more details.

For me at least one question remains: is this twin paradox thought experiment even theoretically possible? B needs to change from his outgoing frame (speed 0.99c relative to A) to an incoming frame (speed -0.99c relative to A). Relative to his own inertial frame, B would need to change his speed 2*0.99*c (faster than light) in the direction of A. Isn't that impossible?

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#408230

On Tuesday, February 7, 2017 at 6:01:01 AM UTC-8, gertja...@gmail.com wrote:
> > In B's frame of reference the distance is length contracted, this is why.
> 
>
> For me at least one question remains: is this twin paradox thought experiment even theoretically possible? 

Yes, it has been executed in practice (not at v=0.99!):

Vessot et al., “A Test of the Equivalence Principle Using a Space-borne Clock”, Gel. Rel. Grav., 10, (1979) 181–204.
“Test of Relativistic Gravitation with a Space borne Hydrogen Maser”, Phys. Rev. Lett. 45 2081–2084.

They flew a hydrogen maser in a Scout rocket up into space and back (not recovered). Gravitational effects are important, as are the velocity effects of SR. This experiment is also known as “Gravity Probe A”.
C. Alley, “Proper Time Experiments in Gravitational Fields with Atomic Clocks, Aircraft, and Laser Light Pulses,” in Quantum Optics, Experimental Gravity, and Measurement Theory, eds. Pierre Meystre and Marlan O. Scully, Proceedings Conf. Bad Windsheim 1981, 1983 Plenum Press New York, ISBN 0-306-41354-X, pg 363–427.

They flew atomic clocks in airplanes that remained localized over Chesapeake Bay, and also which flew to Greenland and back.
Bailey et al., “Measurements of relativistic time dilation for positive and negative muons in a circular orbit,” Nature 268 (July 28, 1977) pg 301.
Bailey et al., Nuclear Physics B 150 pg 1–79 (1979).

They stored muons in a storage ring and measured their lifetime. When combined with measurements of the muon lifetime at rest this becomes a highly relativistic twin scenario (v ~0.9994 c), for which the stored muons are the traveling twin and return to a given point in the lab every few microseconds. Muon lifetime at rest: Meyer et al., Physical Review 132, pg 2693; Balandin et al., JETP 40, pg 811 (1974); Bardin et al., Physics Letters 137B, pg 135 (1984). Also a test of the clock hypotheses (below).

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#408232

On 2/7/2017 8:00 AM, gertjacobusse@gmail.com wrote:
>> In B's frame of reference the distance is length contracted, this is why.
>
> Thank you, the fast and competent responses are really helpful, I think I understand a bit more
> now (although it is not parsimonious for me and my naive space/time perception to alter both
> space and time to maintain a constant c). I think I need to dive into the equations of Pauls
> links to understand more details.
>
> For me at least one question remains: is this twin paradox thought experiment even theoretically
> possible? B needs to change from his outgoing frame (speed 0.99c relative to A) to an incoming
> frame (speed -0.99c relative to A). Relative to his own inertial frame, B would need to change
> his speed 2*0.99*c (faster than light) in the direction of A. Isn't that impossible?
>

The change in VELOCITY is 2*0.99*c, not the change in speed. Speed and 
velocity are distinct in physics. Speed is the magnitude of the 
velocity, sign removed. The change in velocity accounts also for change 
in direction (here sign).

I assure you, there's nothing forbidden about a change in velocity of 
2*0.99*c.

-- 
Odd Bodkin -- maker of fine toys, tools, tables

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#408235

Great insights, thank you for helping a curious physics noob!

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#408238

On Tuesday, February 7, 2017 at 6:51:36 AM UTC-8, gertja...@gmail.com wrote:
> Great insights, thank you for helping a curious physics noob!

you are welcomed

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#408376

> > For me at least one question remains: is this twin paradox thought experiment even theoretically
> > possible? B needs to change from his outgoing frame (speed 0.99c relative to A) to an incoming
> > frame (speed -0.99c relative to A). Relative to his own inertial frame, B would need to change
> > his speed 2*0.99*c (faster than light) in the direction of A. Isn't that impossible?
> >
> 
> The change in VELOCITY is 2*0.99*c, not the change in speed. Speed and 
> velocity are distinct in physics. Speed is the magnitude of the 
> velocity, sign removed. The change in velocity accounts also for change 
> in direction (here sign).
> 
> I assure you, there's nothing forbidden about a change in velocity of 
> 2*0.99*c.
> 
> -- 
> Odd Bodkin -- maker of fine toys, tools, tables

I'm having second thoughts about this explanation, let me ask the question in a different way:

When B is at the mirror before returning (one light year away from A in A's reference frame), then seen from B's own reference frame, A has been moving with 0.99*c for 0.142 years.

B perceives his own speed as 0, and A is still moving away at 0.99c. Then, from B's reference frame, what will be the minimal time required to reach A?

From B's reference frame before he returns, it seems impossible to reach A in 0.142 years - he would need to exceed the speed of light (because A is still moving!).

Is that possible? Wouldn't it imply that you can reach any destination in any time - by just changing your reference frame (accelerating) again and again.

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#408388

W dniu środa, 8 lutego 2017 15:24:02 UTC+1 użytkownik gertja...@gmail.com napisał:

> When B is at the mirror before returning (one light year away from A in A's reference frame), then seen from B's own reference frame, A has been moving with 0.99*c for 0.142 years.

And when you walk a street, from your point of
view trees and buildings running around you are
seen.
Relativistic moron said!!!
It must be true.

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#408453

Op woensdag 8 februari 2017 15:51:42 UTC+1 schreef mlwo...@wp.pl:

> And when you walk a street, from your point of
> view trees and buildings running around you are
> seen.
> Relativistic moron said!!!
> It must be true.

yes, true; but make sure not to get distracted, you might fall off the earth when you reach the edge

(if your post was meant to explain something: I would need more details...)

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