Question: How do you explain the "Twin experiment" from the point of view of the moving observer
In order to answer this question:
- Let us first define what the Twin Experiment is.
- Let us see what the answer is from the point of view of the Observer at rest.
Twin experiment
In the twin experiment we have two Observers A and B. Each Observer has a clock
Observer A stays at home.
Observer B makes a trip which consists of two parts. In the first part he has a constant speed v to a turnpoint P. In the second part, after reaching turnpoint P, he returns back home (to Observer A) which the same speed v.
After returning home the two Observers compare there clocks. The clock of Observer B runs behind. You can also say: The clock of Observer A (at rest) runs the fastest and or the moving clock B runs the slowest.
Two experiments
In order to study the results in more detail we perform the experiment as follows:
- When the clock of A reaches 1 time unit (t=1), A issues a light signal to B.
- When B receives this signal, this defines the turnpoint P, B returns back home
- The distance from A to point P is x.
- The time for the signal to go from A to B is t
- x = ct = (1+t)*v
- ct - vt = v
- t = v/(c-v) time units
Twin experiment example 1: v = c/2
Gamma = sqr(1-vē/cē)=sqr(1-1/4)= 1/2*sqr(3)
A4=4 B3=2*sq(3) A4=4 B3
|\ /|
| / |
| \ / |
| / |
| \ / |
A3=3 A3=3 |
| . \ / . |
| . / . |
| . \ C2/. . |
| . / . . |
| .\ / . .|
A2=2-----------B2=sq(3) A2=2------.----B2
| ./ \ . .|
C2=sq(3). . / \ .. |
| . . / \ .. |
| .. / \ . .|
c1|-.----B1=1 C1\-.----B1=1
A1=1 / A1=1\ |
| / \ |
| / \ |
| / \ |
| / \ |
|/ \|
-----A0=0-B0=0------------ --------A0=0-B0=0-------
- The time for the signal from Observer A at t=1 (this is point A1) to reach observer B (this is point B2) is t
- t = v/(c-v) = 0.5*c/(c-0.5*c) = 0.5/(1-0.5) = 0.5/0.5 = 1 time unit
- That means when the clock of A hits 2 time units the signal reaches observer B
- The distance between A2 - B2 = 1
- The events A2 and B2 are simultaneous for Observer A.
- When Observer B receives the signal and sends it back, then Observer A will receive it 1 time unit later. This will be at A3 (t=3)
- 1 Time unit later Observer B will return back home. This defines the point A4 (t=4)
- At that moment the clock of Observer B will show 4 * gamma = 4 * 1/2*sq(3) = 2*sq(3). This defines point B3.
- At the return point P Observer's B clock will show sq(3). This defines point B2.
- The point B1 defines the position where Observer's B clock hits 1 time units.
- At that same moment the clock of Observer A reaches 1/gamma = 2/sq(3). This defines point C1.
- The distance between C1 and B1 = 2/sq(3)*v = 1/sq(3)
- The signal issues at B1 reaches Observer A at C2. The time is than 2/sq(3)+1/sq(3)=2/sq(3)=sq(3)
Twin experiment example 2: v = 0.5*sq(3)*c
Gamma = sqr(1-vē/cē) = sqr(1-3/4) = sqr(1/4)= 1/2
A4=4sq(3)+8 B3=2*sq(3)+4
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A3=4sq(3)+7 \
| . \
| . \
| . \
| . \
| .\
A2=2sq(3)+4---------------------------------------------------B2=sq(3)+2
| ./
| . /
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c2=2+sq(3) . /
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c1=2-----------.-------/ B1=1
| . /
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A1=1 /
| /
| /
| /
| /
| /
-----A0=0-B0=0-------------
- The time for the signal from Observer A at t=1 (this is point A1) to reach observer B (this is point B2) is t
- t = v/(c-v) = 0.5*sq(3)*c/(c-0.5*sq(3)*c) = sq(3)/(2-sq(3)) = sq(3)*(2+sq(3))/(2-sq(3))*(2+sq(3)) = 2sq(3)+3 time unit
- That means when the clock of A hits 2sq(3)+4 time units the signal reaches observer B
- The distance between A2 - B2 = 2sq(3)+3
- The events A2 and B2 are simultaneous for Observer A.
- When Observer B receives the signal and sends it back, then Observer A will receive it 2sq(3)+3 time unit later. This will be at A3 (t=4sq(3)+7)
- 1 Time unit later Observer B will return back home. This defines the point A4 (t=4sq(3)+8)
- At that moment the clock of Observer B will show (4sq(3)+8) * gamma = (4sq(3)+8) * 1/2 = 2sq(3)+4. This defines point B3.
- At the return point P Observer's B clock will show sq(3)+2. This defines point B2.
- The point B1 defines the position where Observer's B clock hits 1 time units.
- At that same moment the clock of Observer A reaches 1/gamma = 2. This defines point C1.
- The distance between C1 and B1 = 2 * v = 2 * 1/2*sq(3) = sq(3)
- The signal issues at B1 reaches Observer A at C2. The time is than 2+sq(3)
Question: How do you explain the "Twin experiment" from the point of view of the moving observer
- In the above we have explain the Twin Experiment from the point of view of the Observer At rest by using Lorentz Transformation.
- Generally speaking when the two Observers meet we know what the reading is of the Observer A at rest. We apply Lorentz Transformation to calculate the reading of the Moving Observer B.
- By diving both times by two you know what the clock reading is at the return point of the moving Observer and the clock reading of the simultaneous event of the Observer at rest.
- How ever I doubt if it is possible to explain the same from the point of view of the moving Observer
- The right part of the first sketch tries to describe this but I doubt if this is correct.
IMO it is impossible to explain the behavior from the point of the moving Observer B.
The consequence is that Observer B has to accept what Observer A tells him.