Let's say for example that I'm travelling through space in a spaceship at a speed of 0.9C and my shutters are closed but my internal light is on. Things should look pretty normal, since I'm travelling in the same frame of reference as the light source. Then I switch that internal light off and open the shutters and suddenly everything shrinks (due to me being in a different frame of reference to the external light source, assumed to be a nearby star). But if along with the length contraction time has dilated in equal proportion, would I even notice a difference?

I'm assuming the answer is No, but that's assuming I understand SR properly (particularly my assumption of length contraction countering the effects of time dilation or vice versa).


Nothing changes for anything moving with the same velocity (speed and direction) as you. Everything else is contracted in length and dilated in time.

  • $\begingroup$ So what you're saying is that what I see is independent of the frame of reference of the source of the light that allows me to see it? $\endgroup$ – Alan Gee Aug 24 '16 at 12:26
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    $\begingroup$ The shape of things, yeah. You see the ship because it's absorbing and re-emitting light. The light will be red- or blue-shifted depending on its outside source, so the colours might be a lil funky, but the shape of the ship won't change so long as you're moving with the same velocity. $\endgroup$ – hebetudinous Aug 24 '16 at 12:32
  • $\begingroup$ Actually, the more I think about it, the more I'm convincing myself that nothing would look different inside the space ship. The material of the ship would absorb and re-emit the same wavelengths - the source of those wavelengths being what would normally be either infrared blue-shifted to visible or ultraviolet red-shifted to visible. $\endgroup$ – hebetudinous Aug 24 '16 at 12:57
  • $\begingroup$ but if the wavelength remains unchanged when it's re-emitted then the length contraction (which is due to the wavelength) must be the same for the spaceship as for the external world. $\endgroup$ – Alan Gee Aug 24 '16 at 13:44
  • $\begingroup$ Not necessarily. $\endgroup$ – hebetudinous Aug 24 '16 at 13:44

For understanding length contraction and the relation between length contraction and time dilation, I recommend not to refer to some shrinking of lengthes (a phenomenon which perhaps we never will be able to observe) but to an example of distance contraction which seems of much more relevance, even if we are not yet traveling at speed of light:

A spaceship is traveling near light speed from Earth to a far-away star. For the crew, time passes slower than for people on Earth (due to time dilation). They are aging less. But it is important to note that the velocity of the spaceship is the same from the point of view of people on Earth and from the point of view of the spaceship. Accordingly, for the crew in the spaceship, the distance must be contracted (because $v = s/t$).

This example demonstrates two things: 1. time dilation and Lorentz contraction are not compensating each other, they are playing together, they are working in the same sense. 2. Lorentz contraction can be derived from time dilation.

  • $\begingroup$ What I'm trying to establish is, does the combination of time dilation and length contraction make things appear pretty much the same as they would if neither was occurring? $\endgroup$ – Alan Gee Aug 24 '16 at 13:09
  • $\begingroup$ Such a statement would be correct for velocity (see my answer). $\endgroup$ – Moonraker Aug 24 '16 at 13:19
  • $\begingroup$ But do you agree with me that the way I see the spaceship is dependent on the reference frame of the source of light shining on it, or do you agree with hebetudinous that it is dependent on the frame of reference of the spaceship itself? $\endgroup$ – Alan Gee Aug 24 '16 at 17:17
  • $\begingroup$ You always must look at the reference frame of the observer you are considering, in this case the reference frame of the spaceship. $\endgroup$ – Moonraker Aug 24 '16 at 18:55
  • $\begingroup$ Yes the reference frame of the observer is always one of the reference frames to consider, the other, which was the one I was querying, is the reference frame of the observed. $\endgroup$ – Alan Gee Aug 24 '16 at 21:23

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