Ciro Santilli @cirosantilli 34

 Incoming links: Twin paradox

Length contraction  Updated 2024-12-15  +Created 1970-01-01

Suppose that a rod has is length

L

measured on a rest frame

S

(or maybe even better: two identical rulers were manufactured, and one is taken on a spaceship, a bit like the twin paradox).

Question: what is the length

L^{'}

than an observer in frame

S^{'}

moving relative to

S

as speed

v

observe the rod to be?

The key idea is that there are two events to consider in each frame, which we call 1 and 2:

the left end of the rod is an observation event at a given position at a given time: $x_{1}$ and $t_{1}$ for $S$ or $x_{1}^{'}$ and $t_{1}^{'}$ for $S^{'}$
the right end of the rod is an observation event at a given position at a given time : $x_{2}$ and $t_{2}$ for $S$ or $x_{2}^{'}$ and $t_{2}^{'}$ for $S^{'}$

Note that what you visually observe on a photograph is a different measurement to the more precise/easy to calculate two event measurement. On a photograph, it seems you might not even see the contraction in some cases as mentioned at en.wikipedia.org/wiki/Terrell_rotation

Measuring a length means to measure the

x_{2} - x_{1}

difference for a single point in time in your frame (

t 2 = t 1

So what we want to obtain is

x_{2}^{'} - x_{1}^{'}

for any given time

t^{'} 2 = t^{'} 1

In summary, we have:

L L^{'} = x_{2} = x_{2}^{'} - x_{1} - x_{1}^{'} t_{2}^{'} = t_{1}^{'}

(1)

By plugging those values into the Lorentz transformation, we can eliminate

t_{2} a n d t_{1}

, and conclude that for any

t_{2}^{'} = t_{1}^{'}

, the length contraction relation holds:

L^{'} = \frac{L}{γ}

(2)

The key question that needs intuitive clarification then is: but how can this be symmetric? How can both observers see each other's rulers shrink?

And the key answer is: because to the second observer, the measurements made by the first observer are not simultaneous. Notably, the two measurement events are obviously spacelike-separated events by looking at the light cone, and therefore can be measured even in different orders by different observers.

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Twin paradox  Updated 2024-12-15  +Created 1970-01-01

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The key question is: why is this not symmetrical?

One answer is: because one of the twin accelerates, and therefore changes inertial frames.

But the better answer is: understand what happens when the stationary twin sends light signals at constant time intervals to each other. When does the travelling twin receives them?

By doing that, we see that "all the extra aging happens immediately when the twin turns around":

on the out trip, both twins receive signals at constant intervals
when the moving twin turns around and starts to accelerate through different inertial frames, shit happens:
- the moving twin suddenly notices that the rate of signals from the stationary twin increased. They are getting older faster than us!
- the stationary twin suddenly notices that the rate of signals from the moving twin decreased. They are getting older slower than us!
then when the moving twin reaches the return velocity, both see constant signal rates once again

Figure 1.
Twin paradox illustration with twins sending light signals at regular intervals
. Source.

Another way of understanding it is: you have to make all calculations on a single inertial frame for the entire trip.

Supposing the sibling quickly accelerates out (or magically starts moving at constant speed), travels at constant speed, and quickly accelerates back, and travels at constant speed setup, there are three frames that seem reasonable:

the frame of the non-accelerating sibling
the outgoing trip of the accelerating sibling
the return trip of the accelerating sibling

If you do that, all three calculations give the exact same result, which is reassuring.

Another way to understand it is to do explicit integrations of the acceleration: physics.stackexchange.com/questions/242043/what-is-the-proper-way-to-explain-the-twin-paradox/242044#242044 This is the least insightful however :-)

Bibliography:

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