-5
$\begingroup$

It is well known that one can create an alternate form of special relativity where the speed of light is different in different directions and yet make all the same predictions of orthodox special relativity. This is because the one way speed of light is impossible to measure without making presuppositions. See this paper here, or wikipedia article here for reference. In the case of light traveling at different speeds in different directions, this also leads to different amounts of time dilation occurring in different directions.

So if one were to take a pair of synchronized clocks and have them force each other apart so that they travel the same speed in either direction, then have the clocks stop when they reach a premeasured equal amount in both directions, could you compare the two stopped clock readings and determine if time dilation was anisotropic or isotropic, and therefore whether the speed of light was also anisotropic or isotropic?

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ There are countless experiments that indicate that the laws of nature are isotropic. That's why all of our models assumes this symmetry. Your suggested experiment is just another one of those. But none of those constitute a direct measurement of the "one way speed of light" in the sense you are asking about, and your suggested experiment does not either. $\endgroup$ Commented Oct 15 at 15:34
  • 5
    $\begingroup$ This is a repost by the author of the previously closed question physics.stackexchange.com/q/860876 Please be aware that this is not the way to handle things here. You are supposed to fix the problems with the original and submit it to be reopened. $\endgroup$ Commented Oct 15 at 16:37
  • 1
    $\begingroup$ @DavidSantoPeitro your question is correctly closed (both times) as a duplicate of the above questions. The answers to the above questions show why there is no possible experiment which can measure the OWSOL. This is a common approach in physics where we solve a general case and then with that general solution we have answered all of the specific cases. The fact that your specific case is not individually addressed does not make the general answers inapplicable. This question is a duplicate, as are all future questions about any specific experiment intended to measure the OWSOL $\endgroup$ Commented Oct 15 at 16:44
  • 1
    $\begingroup$ @Dale It is not a duplicate. I see this all the time on this site with other peoples questions. Moderators are "duplicate happy" and try to mindlessly say one question is the same as another without really thinking about it. $\endgroup$ Commented Oct 15 at 18:21
  • 1
    $\begingroup$ Would you take it as an answer if, as you suggested, we just referred you to Misner/Thorne/Wheeler's Gravitation or to Einstein's Relativity? After a reading of either text, isotropy should be apparent. $\endgroup$ Commented Oct 15 at 19:41

2 Answers 2

Reset to default
0
$\begingroup$

So if one were to take a pair of synchronized clocks and have them force each other apart so that they travel the same speed in either direction, then have the clocks stop when they reach a premeasured equal amount in both directions, could you compare the two stopped clock readings and determine if time dilation was anisotropic or isotropic, and therefore whether the speed of light was also anisotropic or isotropic?

No. Look at the Lorentz factor for flat 1+1-dimensional spacetime (e.g. that from special relativity):

$$\gamma(v)=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.$$

Note that it is symmetric about the transformation $v\to -v$, so $\gamma(v)=\gamma(-v)$. Time dilation will necessarily be isotropic if special relativity is obeyed.

But wait, there might be a term in there linear in $v$ and which would break $v\to -v$ symmetry! Yes, you get this for example in the metric

$$ds^2=c^2dt^2-k^2dt^2-dx^2-2kdxdt.$$

In that case a term linear in $v$ does get added to the denominator in the Lorentz factor, arising directly from the $dxdt$ term in the line element. But of course, in these scenarios, you can coordinate-transform that term away. That line element is that of a coordinate system that has undergone a Galilean boost by $v_\text{boost}=k$, so of course anisotropy is introduced as a given, and you shouldn’t expect in this scenarios for ordinary SR’s isotropies to hold till you get back to a proper coordinate system.

There are situations where there is manifest curvature that produces off-diagonal terms above that can’t be whooshed away as easily. But in those cases, it only seems like the speed of light is anisotropic (or equivalently like two clocks would have different time dilation) if you try to check from a weird frame. Near a rotating black hole, off-diagonal terms appear as part of the Kerr metric, and won’t go away trivially, but the solution is to note that you are technically “physically rotating” around the black hole if you are not “coordinate-rotating” with it in the direction of rotation, insofar as for constant angular coordinates the metric near the BH looks exactly like one for a boosted coordinate system. So the proper thing to do is to boost into the frame where you are not physically rotating (ignoring the fact that you wouldn’t survive because in reality gravity would pull you in if you’re not orbiting), insodoing adding a rotation along with the black hole, and removing the terms that generate time dilation anisotropies.

All that is to say: the only scenario in GR where time dilation would be anisotropic is one where a simple Galilean boost (or Lorentzian if you prefer) removes the anisotropy.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Nothing about the original question required general relativity. This can all be discussed without curved spacetime, and introducing curvature simply obfuscates the core issue at hand. $\endgroup$ Commented Oct 15 at 18:27
  • $\begingroup$ @DavidSantoPietro Note that special relativity (e.g. no curved spacetime) is a limiting case of general relativity. I don't see how my first few paragraphs don't address your question. I actually specifically reference the Lorentz factors in special relativity here too, and identify metrics that seem to allow anisotropy but are in fact still special-relativistic and thus isotropic. If it wasn't obvious from the start, see the sentence I just bolded. $\endgroup$ Commented Oct 15 at 19:40
  • $\begingroup$ I obviously know that the Lorentz factor is symmetric in v for orthodox special relativity. And making a sentence in bold that effectively says "because special relativity says so" is not helpful at all. The time dilation formulas for anisotropic special relativity do depend on the direction of motion and do have linear terms in v. However, the velocities of objects in either direction for anisotropic special relativity also change and it all cancels out. I was able to figure that out after reading the aforementioned and referenced paper. $\endgroup$ Commented Oct 15 at 22:35
  • $\begingroup$ @DavidSantoPietro even if you already figured it out, still, see my answer: terms linear in $v$ are just due to an implicit choice of reference frame. Anisotropic special relativity is thus identical to regular special relativity. $\endgroup$ Commented Oct 15 at 23:02
-1
$\begingroup$

The answer is no. In the case of anisotropic speeds of light, the velocities of objects also get affected, and so the speeds of the moving clocks will not necessarily be moving the same speeds left and right, even if pushed apart by a seeming equal and opposite force. See pages 1 and 2 in this reference. The clock that travels faster will have a lesser time dilation and will end up ticking off the same amount of time as the clock that travels slower, which will have a greater time dilation, and the two clocks will end up ticking off the same amount of time total by the time they reach the predetermined equal distances to the left and right even though time dilation is different in each direction.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Self-answers to new questions are not an acceptable way to complain about the site. We do have Physics Meta if you want to post a complaint there. Also, on the note of your reference: I replied in the chat room generated on your last post as to why their description of OWSOL is not the same as what you are thinking. $\endgroup$ Commented Oct 15 at 14:45

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.