3

Consider the following situation:

You are moving with high speed on a road on earth and are looking at the night sky, in particular at two stars.

In the rest frame of earth, you and your eye are length contracted. This implies that the proper distance between the images of the stars on your retina is greater than the distance between the images on the retina of an observer at rest w.r.t. the earth (in my calculations the angles also change, but this does not influence the distance, because both angles change).

But because the position of the dots on your retina (or more specifically, the cells that are excited by starlight) are independent of the frame of reference, you also observe a distance between the two stars that is greater than the other observer. Now that is a problem, because special relativity says you should observe a smaller distance (the sky is length contracted).

So the question is did I make a mistake? Is this actually what happens?

Qmechanic
  • 220,844
Phlipp
  • 31
  • 2

3 Answers3

1

You are not making a mistake, except in thinking that this is a problem. As always, the key to solving this dilemma is the relativity of simultaneity.

For simplicity consider the following geometry: in the earth frame the distant stars are at rest and are at equal and opposite angles from the vertical along the direction of travel for the car. At that moment the rider in the car receives a momentary flicker in the light from each star.

In the earth frame those flickers occurred simultaneously, and so the optical separation of the light sources is equal to the distance between the stars.

In the car frame, however, those two flickers occurred at different times. This means that the optical separation is not equal to the distance in this frame. Between the times of the two flickers the stars have moved such that the optical separation is greater than the distance between the stars in the car frame.

Dale
  • 117,350
0

This might not directly answer your question, but I want to remind you that "length contraction" might not mean that "in your view, the length is contracted" since the lights from the objects need time to propagate to your eyes.

Here is a video trying to simulate the visual effect of special relativity: https://www.youtube.com/watch?v=JQnHTKZBTI4. You can see that the object might not be actually "contracted" in your view. Here is another video explaining the reason behind the optical effect: https://www.youtube.com/watch?v=l44JSW-1RvE.

Hope this answer clarifies some part of your question.

creaple
  • 19
0

You are moving with high speed on a road on earth and are looking at the night sky, in particular at two stars.

From a somewhat practical point of view, even the speed of any (section of) road itself, on Earth's surface, with respect to a non-rotating quasi-inertial frame which is momentarily comoving with Earth's center) is comparatively high compared to typical speeds of cars "going really fast" wrt. the road surface (except for road section in "high polar regions" of Earth); and even more so the speed, namely roughly 30 km/s of any (section of) road on Earth's surface wrt. the non-rotating quasi-inertial frame which is comoving wrt. the center of the Sun.

[...] But because the position of the dots on your retina (or more specifically, the cells that are excited by starlight)

... it is certainly correct and useful to be just as specific about the concrete identifiable constituents (a.k.a. participants) of the setup under consideration ...

are independent of the frame of reference,

That's (bordering on being) incorrect, and consequential misleading. Recall W. Rindler's dictum that

"An inertial frame is simply an infinite set of point particles sitting still in space relative to each other".

In this sense, the set of starlight-collecting cells in (or rather, making up) the retina of the driver you described constitutes itself a (small, or at least partial) particular frame of reference.

you also observe a distance between the two stars that is greater than the other observer.

That's (bordering on being) incorrect. No: distances or not ("plainly") observed, but distances, or more specifically: ratios of distances, are measured (in ideal cases, including thought experiments); or, in more practical circumstances, distance ratios are estimated (usually accompanied by a quantitative stateent, or at least a qualitative discussion, about the range of confidence of the estimate).

This is especially relevant when understanding "length contraction" as referring to the measured distance ratio between certain pairs of participants who belong to different (inertial) reference frames which move (uniformly) wrt. each other; in Einstein's popular example specifically the distance ratio between

  • the distance (A'B') of two specific identifiably constituents, A' and B', of a train (which moves with some speed v on a railway embankment) and

  • the distance (AB) of two specific identifiably constituents, A and B, of the railway embankment, which are identified (in rlation to train constituents A' and B') by the requirement that the (clock) indication of A at the passage of A' is simultaneous to the (clock) indication of B at the passage of B'.

It can be shown and has been shown that this specific distance ratio, (A'B') / (AB), has the value $\frac{1}{\sqrt{1 - (v/c)^2}}$; cmp. for instance eq. $(1\rm d)$ in this derivation.

user12262
  • 4,360