7
$\begingroup$

Here is a thought experiment.

Consider a frame S in which an observer is at rest and she looks ahead to find a right angle triangle of sides 3 (x-axis),4(y-axis) and hypotenuse 5. Now consider another frame S' in which is moving at a speed 'v' moving along x-axis, with respect to frame S, that has another observer.

Will the observer in the frame S' be able to accurately deduce the Pythagoras theorem or will length contraction along the x-axis lead to some other x-axis length for the triangle?

$\endgroup$
5
  • $\begingroup$ Note that the hypotenuse has a component in the $x$-direction. $\endgroup$
    – jwimberley
    Jul 20 at 1:47
  • $\begingroup$ And although it's not directly relevant to this question, it's crucial to understand the ladder paradox (en.wikipedia.org/wiki/Ladder_paradox) if you're thinking about geometric objects' shapes from a relativistic perspective. $\endgroup$
    – jwimberley
    Jul 20 at 1:48
  • $\begingroup$ Yes. I did think of the x-direction of the hypotenuse. I should work out the math and chek if it works. $\endgroup$ Jul 20 at 1:51
  • $\begingroup$ It seems to me that the length contraction might present a faint but marginally discernible difference in the apparent thickness of the x-axis, which would prevent the exact formulation of the theorem. $\endgroup$
    – Edouard
    Jul 20 at 2:25
  • 2
    $\begingroup$ I think there are two distinct parts to your question that are perhaps not clearly delineated: (1) are the two observers independently able to deduce the Pythagorean theorem (by studying what to each of them, in their own frame, appear to be right triangles), and (2) under what conditions is a right triangle in one frame also a right triangle in the other frame (albeit not necessarily a similar one - e.g. maybe it's squashed somehow, but the right angle is preserved). $\endgroup$ Jul 20 at 18:37
15
$\begingroup$

If you draw a right triangle on a sheet of rubber, and you uniformly stretch the rubber in a direction aligned with one of the non-hypotenuse sides...you still have a right triangle. However, the angle has changed. Do you suppose there could be any parallels with your thought experiment?

$\endgroup$
3
  • 4
    $\begingroup$ I don't see a lot of value added in this kind of "what-if" type of response, looking down on the asker. $\endgroup$
    – Olorin
    Jul 21 at 14:29
  • 5
    $\begingroup$ @Olorin - actually, this is not looking down on the asker, this nudges the asker (who is clearly studying SR and trying to understand this stuff more deeply) in the right direction, and is providing them with an opportunity to have an small Aha! moment, which, if successful, is going to be far more valuable to them then a straight up answer. Now, while this is a Q&A site, sometimes what the asker is asking is just a superficial facet of what they really want to know. So this in fact shows empathy, and rather then addressing a superficial need, it tries to address the deeper question. $\endgroup$ Jul 21 at 16:27
  • 2
    $\begingroup$ The only issue with this answer, is that it isn‘t future proof definitive understanding for the next person with a similar or exact question, rather is only helpful directly to the OP if they solve the riddle or to those who already know the answer. $\endgroup$
    – morbo
    Jul 21 at 18:35
9
$\begingroup$

I have made a diagram to clarify the situation

black is for the $S$ coordinate system

red is for the $S'$ coordinate system with motion $v'$ relative to $S$

The black triangle is what $S$ sees. The red triangle with smaller reduced base $b'$ is what $S'$ sees. $S$ and $S'$ will both see right triangle and both will obey the Pythagorean Theorem. $$ a^2+b^2=c^2 $$ and also $$ a'^2+b'^2=c'^2 $$

In fact they will also both know what each other sees by using the Length Contraction formulas.

I hope that helps clarify your question.

enter image description here

$\endgroup$
5
  • $\begingroup$ I'm not a physicist, but doesn't relativistic length contraction also contract and expand angles as part of the Lorentz transformation? I'm not so sure that a relativistic right triangle will still be a right triangle. $\endgroup$
    – nick012000
    Jul 21 at 13:49
  • $\begingroup$ It does but it depends on which direction the length contraction occurs, note that an angle in the triangle does change! It’s just the one with 90 degrees! $\endgroup$
    – Fredrik Sy
    Jul 25 at 19:41
  • $\begingroup$ I mean to say that "It's just NOT the one with 90 degrees $\endgroup$
    – Fredrik Sy
    Jul 30 at 3:09
  • $\begingroup$ @FredricSy Tilt the triangle so that the length contraction occurs perpendicular to the hypotenuse, and I'm pretty sure that the ninety degree angle would no longer be ninety degrees. $\endgroup$
    – nick012000
    Jul 30 at 4:28
  • $\begingroup$ You are correct, as I had mentioned in the above comment, however the length contraction happens along the x axis according to the question. $\endgroup$
    – Fredrik Sy
    Jul 30 at 4:55
6
$\begingroup$

In all inertial frames the metric is $ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2$. For any moment in time we can set $dt=0$ and recover the Pythagorean theorem.

So the Pythagorean theorem holds for space in all inertial frames in special relativity. So, if one leg of a right triangle is length contracted, then you can use the Pythagorean theorem to determine the contraction of the hypotenuse.

$\endgroup$
4
$\begingroup$

As an alternative to the other answers, you could just use the principle of relativity rather than any explicit form of the Lorentz transformation (which is deduced from it): this says that the laws of physics can be formulated in a way that is the same in all inertial frames.

This implies that there is no experiment that would tell you your state of (uniform) motion, in particular if the Pythagorean theorem holds in one frame, it holds in all frames.

$\endgroup$
3
$\begingroup$

Your thought experiment needs to start with a more basic questions: "What is a straight line?" and "What is a right angle?" and "How to measure distance?"

If we are doing real physics experiment, and we say use a beam of laser as a "straight line" and an atomic clock to measure the time of transit between points thus measuring distance, and a mirror to reflect laser in a right angle. And you put the whole setup in a spacecraft traveling at near light speed.

From the observer inside the spacecraft, obviously Pythagoras Theorem holds. For a stationary observer outside the spacecraft, with the Lorenz transform, I believe Pythagoras Theorem still holds.

However if the spacecraft fly next to a black hole, the space is not Cartesian, thus Pythagoras Theorem does not hold. (Same applies to accelerating frame.)

$\endgroup$

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.