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So here's a scenario I made up and I want to know if it is correct or not.

I'm was standing on the earth, l$_0$ meters away from a stationary rocket in space. The rocket then starts travelling towards me at a constant speed v. From my frame of reference the distance l$_0$ will be contracted and hence I will observe the total distance between me and the rocket to be $\frac{l_0}{\gamma}$. From the frame of reference of the rocket the distance observed will also be $\frac{l_0}{\gamma}$.

Is this correct? If so then what if the rocket was travelling away from me at the same speed, how would length be contracted in such scenario?

Also, If the the rocket was travelling to the moon and the distance between the rocket and the moon initially (i.e. when me, the rocket, and the moon are stationary) is x$_0$. What would I observe the distance x$_0$ to be?

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  • $\begingroup$ Are you suggesting that just because the front of the rocket starts to move, you suddenly change your mind about where it's located? If so, you need to stop and think for a minute. $\endgroup$
    – WillO
    Sep 7, 2016 at 16:55
  • $\begingroup$ once the rocket starts moving the distance between me and the rocket is contracted, right? $\endgroup$
    – BarCouSeH
    Sep 7, 2016 at 17:00
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    $\begingroup$ @BarCouSeH: no. $\endgroup$ Sep 7, 2016 at 17:06

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In relativity we measure distance using a reference frame, and generally this is our rest frame i.e. the frame in which we are at rest at the origin. So in my rest frame here on Earth I measure the distance to you in your rocket as $\ell_0$.

Since you and the rocket start off at rest relative to me we share the same reference frame, except that we may disagree where the origin is, so you also measure the distance between us to be $\ell_0$.

Lorentz contraction happens because the reference frame used by a object moving relative to me doesn't line up with my reference frame. Specifically we disagree about where the time and distance axes are so what looks to me like a distance will look to you like a combination of distance and a time, and vice versa.

Now back to your question about the rocket. When you and the rocket start accelerating my reference frame doesn't change, and none of the distances measured by me in that frame change. So the distance from me to you is still $\ell_0$, apart of course from the fact it is now decreasing with time because you're moving towards me.

But your reference frame has changed. You started in a frame at rest with respect to me, and now you're in a frame moving at speed $v$ with respect to me. So the distances you measure will have changed, and your measurement of the distance between us will have decreased to $\ell_0/\gamma$.

But how do we know that you are the one who changed frames and not me? Well that's because you are the one who accelerated i.e. you felt the g-forces as your rocket motor cut in. In special relativity it is unambiguous who accelerates and who doesn't, because all you have to do is measure the g-forces you feel. Sitting here on Earth I felt no g-forces so I know that I haven't accelerated and my reference frame hasn't changed.

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  • $\begingroup$ Okay now suppose that the rocket didn't accelerate and has always been travelling at a constant speed. How can we tell which one changed frames in this case? $\endgroup$
    – BarCouSeH
    Sep 7, 2016 at 19:38
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The length $l_0$ exists in your reference frame, so you may measure it without any length contraction. The rocket does not jump towards you due to length contraction when it starts moving quickly (it will move toward you, though). The length that exists in the rocket's moving frame is the length of the rocket itself. The rocket will appear to get shorter in your frame when it starts moving quickly.

From the rocket's perspective, its own length does not change since that length exists in its frame, in which it is at rest. However, space outside of the rocket would warp when the rocket suddenly started moving at a great speed, and you would appear to get closer due to length contraction in addition to the rocket's new velocity closing the gap over time.

In conclusion, $l_0$ does not contract in the Earth's reference frame, but it does in the rocket's frame.

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  • $\begingroup$ Omg the first time I read this it made no sense but now I read it and it makes total sense. The main thing I wasn't getting is that there's a difference between space contraction and "object" contraction. From my frame of reference I see the rocket itself contracted, not the space between it and me. However from the rocket's frame of reference, it sees the whole space contracted. $\endgroup$
    – BarCouSeH
    Sep 11, 2016 at 6:31
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    $\begingroup$ Spaces and objects are both just lengths. The difference is that one length (the space between the earth and the rocket) is in the reference frame of the observer on Earth, and the other length (the rocket itself) is in the reference frame of the rocket. You could put space buoys (with magic station-keeping rockets or something) at even intervals between you and the rocket. If they are spaced every $x$, then the rocket will think they're spaced every $x/\gamma$ once it gets going (since they'll be moving relative to the rocket), yet the earthling will think they're still spaced every $x$. $\endgroup$
    – EL_DON
    Mar 27, 2019 at 3:23

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