2
$\begingroup$

Can the concept of twin paradox be applied to length contraction as well? meaning that the twin which is in spaceship will have its meter rod "actually" contracted while he will see his brother's meter rod contracted which is in fact will be an "apparent" effect.

$\endgroup$
  • 1
    $\begingroup$ related: “Reality” of length contraction in SR $\endgroup$ – AccidentalFourierTransform Apr 25 '16 at 11:32
  • $\begingroup$ relaed: This post in What is time dilation really?. $\endgroup$ – user36790 Apr 25 '16 at 12:04
  • $\begingroup$ All matter is constantly in motion within that 4D environment known as space-time, and does so with the same magnitude of motion as do photons have spatial motion, hence c motion. A change of direction of travel within the 4D environment, also leads to 4D rotation. Less motion across time, hence more motion across space, equals more rotation, and thus greater spatial length contraction. See entire goo.gl/fz4R0I video collection to get a better grasp. $\endgroup$ – Sean Apr 26 '16 at 17:21
2
$\begingroup$

Despite their superficial similarity Lorentz contraction and time dilation are different things and this is why there isn't a distance version of the twin paradox.

To see the difference you need to understand that a clock is a form of odometer. Suppose you start at the origin and travel 100 metres, then the odometer you carry will show the total distance you've moved in space i.e. 100m. Suppose you are also carrying a clock, which you set to zero at the moment you start walking, then after you've finished the walk your clock will show some time $t$. This time $t$ is the distance you've moved in time, just like the odometer shows the distance moved in space.

The twin paradox is that the two twins end up having moved different distances in time. That is, they both start at the same point on the time axis $t = 0$ but when they meet again they find have travelled different distances in time so their clocks show different times. (At this point we should issue the obligatory statement that the twin paradox isn't a paradox.)

If there were a distance version of the twin paradox it would be that on their return each twin observes the other twin to have moved a different total distance in space. Actually this is indeed the case, though since moving at different speeds in distance isn't as interesting as moving at different rates through time the distance version of the twin paradox has yet to catch on.

But when we talk about Lorentz contraction we normally aren't talking about the distance moved in space. Instead we consider some object, like a metre ruler, and calculate how the length of that object decreases with velocity. While length contraction is a real effect, what actually happens is that the metre ruler rotates in spacetime. As viewed by the static observer the proper length of the ruler remains constant, but the two ends of the ruler shift to slightly different times. In the case of our twins each twin would see the other twin's ruler rotated in spacetime by equal and opposite amounts, so actually the situation is perfectly symmetrical.

$\endgroup$
  • 1
    $\begingroup$ Sorry John, but a clock is not an odometer, and t is not the distance you've moved in time. A clock features a pendulum or a rocker or a vibrating crystal, or some other mechanism which moves in a regular cyclical fashion. This "local" motion is counted and/or accumulated and/or geared down, and is typically displayed as the motion of the big hand and the little hand - we don't call it a clock movement for nothing. Open up a clock and look at those cogs moving. That motion isn't through time, or spacetime, it's through space. $\endgroup$ – John Duffield Apr 26 '16 at 7:17
  • $\begingroup$ we can use the length contraction to calculate the traveling twin's path, and for example at speed 0.8c if the total travel distance was 10light years, it will come down to only 6 light years length contracted. So the clock of the traveling twin will show a time elapsed accordingly to that ratio that the earth twin's clock shows has elapsed. (I know this works vica-versa, but let's disregard that) Now is this result the same as if we would calculate with the Rindler metric and account for the acceleration at the turnaround point? $\endgroup$ – Árpád Szendrei Nov 11 '16 at 21:13
2
$\begingroup$

This might or might not be responsive to the question you intended to ask:

Suppose you've got a meter stick. Over a period of time, I apply identical forces to the front and back ends of that meter stick, causing them to accelerate in the same direction. Therefore the entire stick, being a rigid body, accelerates in that direction. After a while, the forces stop, so the stick is now moving at a fixed velocity.

In your frame (the frame the stick was in to begin with), the length of the stick can't change, because we applied identical forces to the front and back, so the distance between the front and the back can't change. On the other hand, the Lorentz contraction tells you that the (now-moving) stick is shorter in your frame than it is in its own frame. This means the moving stick, in its own frame, is now more than a meter long. It has stretched.

(In its own frame, the moving stick says that it has stretched because the front started accelerating before the back did.)

But there is a limit to how far you can stretch a stick. So if you get the velocity high enough (no matter how gently you accelerate it to that velocity) the stick must snap. And of course all observers must agree that it has snapped. I think we can count that as a real effect, which no observer can dismiss as merely "apparent".

I'm still a little unclear on whether we can give a completely non-relativistic explanation of why the meter stick snaps. I once asked a closely related question here.

$\endgroup$
  • $\begingroup$ Good stuff will. Think about the wave nature of matter and spin and spherical harmonics, then simplify an electron to a flat circular harmonic which looks like this from the side: |. When you move it, one point on the circumference traces a helical path, and all points map out a cylinder. The electron is then "smeared out". If that electron could see, it would see everything else looking length-contracted. The moot point is that when you step on the gas in your gedanken spaceship, you see a star looking flattened because you changed. The star didn't change one iota. $\endgroup$ – John Duffield Apr 26 '16 at 7:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.