0
$\begingroup$

I was doing a problem on relativity and I was thinking about the problem in terms of the postulates of Special Relativity. Special Relativity only considers motion in inertial reference frames and I was doing this problem over here;

Essentially, if I have a rod which is tilted at an angle (theta)0, with respect to the x-axis, and moving along the x-axis at speed v, and if that rod has a length L0 observed from its frame of reference, what is the length of the rod as observed by an observer in a stationary frame with respect to the frame of the rod. Also, what is the angle, as observed by that stationary observer, of the rod from the x-axis?

I got the answer correct and it's presented below. nenter image description here

The problem that I with this is that the angle itself changes, which makes me think that rotation has occurred while observing the rod from one frame to the next. And if the rod itself is the second reference frame, in that case, would it not be a rotating reference frame and therefore, not within the realm of special relativity? I mean, if it is a rotating reference frame as observed from other reference frames, then it's going to have an acceleration no?

Sorry if this is a stupid question, I'm just really confused and, yet, excited about this cos it's so interesting.

$\endgroup$
  • $\begingroup$ The angle does not "change". This was always the angle measured by the stationary observer, and this remains constant for him at all times. $\endgroup$ – Deep Mar 11 '17 at 11:34
  • 1
    $\begingroup$ We must be careful to distinguish between the calculated length contraction and what's actually observed. The observer moving relative to the angled rod doesn't exactly see the angle you calculate because light that left the different parts of the rod simultaneously doesn't arrive at the observer's eye simultaneously. See Terrell rotation. $\endgroup$ – PM 2Ring Mar 11 '17 at 11:56
  • $\begingroup$ @PM2Ring The Terrell effect animation has me rethinking my understanding of so-called constituent-counting rules in exclusive processes (e.g. $d(\gamma, p)n$). $\endgroup$ – JEB Jan 24 '18 at 4:17
  • 1
    $\begingroup$ Please mark up your math using mathjax rather than posting a photo of a piece of paper. $\endgroup$ – Ben Crowell Apr 25 '18 at 5:20
0
$\begingroup$

interpretation of your result

You should not interpret this is a rotation of the angle. It is merely a result of there only being length contraction along the motion of the observer.

The y length of the rod changes but it's x length will change due to length contraction (as you correctly derived.) which changes the angle.

extra "riddle"

You mentioned that you are really interested in this effect so let me give you a related problem that leads directly to general relativity:

Imagine a really fast merry go round with some observer riding the attraction.

  • The length along his motion (circumference of the attraction) will get contracted
  • The length perpendicular to his motion (radius of the attraction) will remain unchanged

The result is that this observer will measure a circle with $Circumference \neq 2\pi radius$!

This is not possible in flat spacetime such that the observer must be observing curved spacetime!

$\endgroup$
  • $\begingroup$ The merry go round can be done in flat space. It does not lead to general relativity. GR is about gravity. $\endgroup$ – mmesser314 Mar 11 '17 at 13:21
  • $\begingroup$ For an external observer the merry go round is indeed a normal flat spacetime, and indeed and external observer will not feel a gravitational field. An observer riding the merry go round will not observe flat spacetime as I explained above. And indeed, such an observer will experience gravity in the form of "the centrifugal force". Remember the equivalence principle: "pseudoforces due to acceleration are equivalent to gravity" this is exactly what this example tells us. $\endgroup$ – gertian Mar 11 '17 at 14:11
  • $\begingroup$ I'm sorry, GR doesn't necessarily have to be about gravity, yes? GR deals with accelerations and, by the Equivalence Principle, accelerations are all the same, no matter the cause. Please correct me if I'm wrong? $\endgroup$ – Abhijeet Vats Mar 11 '17 at 14:12
  • $\begingroup$ The equivalence principle tells us that we cannot distinguish acceleration and gravity (c.f.r. the elevator experiment). So you could interpret GR as describing accelerated motion or gravity depending on your point of view... But hey, all I wanted to do was to give OP some extra thinking material. Didn't mean to go off topic this far... $\endgroup$ – gertian Mar 11 '17 at 14:15
  • $\begingroup$ It's fine, chill, I get more and more excited when the discussion goes more in depth. Don't worry about it $\endgroup$ – Abhijeet Vats Mar 11 '17 at 14:16

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.