9
$\begingroup$

Is there a relativistic effect at play that explains the following thought experiment, that I am unaware of?

The apparatus depicted in the image has two straight bars in the middle emitting light from both ends. In the reference frame that is moving with the apparatus the following measurements are made: A = B, and α = β.

At an equal distance apart horizontally there are three vertical bars. The vertical bar in the middle is emitting two beams of light towards the vertical bars at each end. The angle between the light beams emitted to the left hand side is α, and the distance between the points of impact on the left hand side vertical bar is A. On the right side the angle is β and the distance is B.

Now lets consider what the apparatus would look like from a reference frame from which the apparatus is observed as moving to the right at velocity v. In my understanding A = B would be also observed from this reference frame. X would be observed to be shorter due to length contraction, but it would still be equal when comparing the two sides of the apparatus with each other.

Light would need to travel a longer distance to reach B than to reach A due to the observed movement of the apparatus. For A = B to hold, it must follow that α > β. Would the outside observer observe the bars in the middle to bend producing the inequality between α and β? I assume that I have misunderstood something, as this feels somewhat paradoxical.

Edit: Here are some calculations I have attempted in order to understand what would be observed by the external observer. My intent is to understand what is observed as opposed to what is seen. Hopefully this helps clarify what exactly it is that I am struggling with.

Due to length contraction the observed horizontal length of each side of the apparatus X' could be calculated: $X' = X * \sqrt{1−\frac{v^2}{c^2}}$ Based on the observed movement of the apparatus, any light reaching the right side wall at time $t$ would have been emitted at time $t - t_{B'}$. The following relationship between the horizontal and vertical component of the light path, and $t_{B'}$ should exist:

$t_{B'} = \frac{\sqrt{(\frac{B'}{2})^2 + (X' + t_{B'}*v)^2}}{c}$

The angle $\beta'$ then is:

$\beta' = 2\arctan (\frac{\frac{B’}{2}}{X’ + t_{B'}*v})$

Same for the left side:

$t_{A'} = \frac{\sqrt{(\frac{A'}{2})^2 + (X' - t_{A'}*v)^2}}{c}$

$\alpha' = 2\arctan (\frac{\frac{A’}{2}}{X’ - t_{A'}*v})$

As an example let's examine a case where X = 1, A = 1, v = 0.5, and c = 1 As assumed before, A = A' = B = B'.

$X' = X * \sqrt{1−\frac{0.5^2}{1^2}} \approx 0.866$

The system of equations can then be simplified and solved:

$ \begin{equation} \left\{ \begin{array}{lcr} t_{B'} &=& \frac{\sqrt{(\frac{1}{2})^2 + (0.866 + t_{B'}*0.5)^2}} \\ \\ \beta' &=& 2\arctan (\frac{\frac{1}{2}}{0.866 + t_{B'}*0.5}) \\ \\ t_{A'} &=& \frac{\sqrt{(\frac{1}{2})^2 + (0.866 - t_{A'}*0.5)^2}} \\ \\ \alpha' &=& 2\arctan (\frac{\frac{1}{2}}{0.866 - t_{A'}*0.5}) \end{array} \right. \end{equation} $

WolframAlpha gives me the following solution:

$t_{B'} \approx 1.87$, $\beta' \approx 0.54$, $t_{A'} \approx 0.71$, and $\alpha' \approx 1.56$

Is there some other equation that I would need to properly calculate what the outside observer would observe?

$\endgroup$
5
  • $\begingroup$ Write primes for values in the frame where the apparatus is moving horizontally. Then $A=B=A'=B'$ while $x'<x$ so $\alpha'=\beta'>\alpha=\beta$. You seem to mostly understand this. Then you write " For A = B to hold, it must follow that α > β." This is wrong, but I can't pinpoint your error because I'm not sure what led you to this. It does not follow from your observation about light travel times and I'm not sure why you think it does. $\endgroup$
    – WillO
    Commented Jan 27 at 14:03
  • $\begingroup$ I concluded α′ > β′ should be the case if A′=B′, based on the observed distance traveled by the light before it hits each of the walls. The right side wall is moving away from the point from which the light was originally emitted while the light makes it's way from the center to the wall. Similarly the light traveling in the left direction needs to travel a shorter distance to reach the wall. As the angle is a result of the horizontal and vertical components of the lights' paths, should the angle not be different when the vertical component is equal, but the horizontal component is not? $\endgroup$
    – Joona
    Commented Jan 27 at 14:15
  • 1
    $\begingroup$ Check out the "Velocity Raptor" game starting at level 25+. In these levels the world is shown as the raptor would perceive it. And yes, because of time-of-flight delays from different parts of objects they appear bent. testtubegames.com/velocityraptor.html $\endgroup$
    – Jagerber48
    Commented Jan 27 at 15:08
  • 1
    $\begingroup$ Peripheral but (just possibly) may be of interest. NOT the same but ... : Cameras with roller blind shutters (or electronic equivalents) produce a image by moving a sampling window across the image. When the observer is travelling at speed (typically photos taken from a car) if the direction of travel is orthogonal to the axis/direction as the shutter window travel a series of vertical objects at right angles to the direction of travel appear bent (eg bars on a motorway roadside fence). BUT if the camera is rotated at 90 degrees so that window travel matches direction of travel ... $\endgroup$ Commented Jan 29 at 2:20
  • 1
    $\begingroup$ ... bars are not bent. They will now be either compressed in x axis dimension or xpanded in x axis direction depending on whether window travel is with or against vehicle motion - but this effect is usually less visually perceptible. $\endgroup$ Commented Jan 29 at 2:21

5 Answers 5

15
$\begingroup$

Your calculations are basically correct, for calculating the angles between light rays going forwards or backwards.

Angle $\beta'$ between the forward going light rays should indeed be smaller than $\alpha'$ between the rear going light rays. The forward focused rays are emitted at a tighter angle in a beam. You have basically rediscovered "the relativistic beaming effect".

There also appears to be a bit of confusion between measuring the angle between the bars and the angle between the light paths. What you have calculated, is the angle between the light paths and they are different fore and aft. The individual bars (e.g green) do not bend, but the angle between then green and blue bars does change with relative velocity, but it is the same fore and aft.

The angle between the bars themselves is given by:

$\theta' = \arctan \left(\frac{B}{X \sqrt{1-v^2/c^2}}\right) \approx 0.8570 $ front and back

where B and X are as depicted in the OP diagram.

Check out this Geogebra animation I created. (Right click the 't' slider and select animate from the drop down menu).

enter image description here

You will be able to see how the paths of the photons create a different angle from the tubes, while remaining within the tubes. Relativity is clever :-)

The angle between the green tube and the blue tube increases as the velocity increases, and the angle between them does indeed seem to bend in your reference frame but the angle between them is the same on either side, fore and aft. On the other hand, the angle between the light paths on either side is different. All these angles are what you would actually measure in your reference frame and are not due to light travel times to your eye.

Is there some other equation that I would need to properly calculate what the outside observer would observe?

You can cross check your results with the formulas for relativistic beaming. You can also use the Lorentz transforms. Calculate all the x and t measurements in the reference frame where the tubes are not moving and then use the Lorentz transforms to calculate the expected x' and t' measurements.

$\endgroup$
1
  • 2
    $\begingroup$ Thank you for the well put together and insightful answer @KDP! The Geogebra animation is very helpful. Suddenly it is made intuitive how the light is observed traveling at different angles on either side, while the angle of the tubes is observed to be the same on both sides. Thanks for naming the relativistic beaming effect as well for further reading. This is exactly what I needed. $\endgroup$
    – Joona
    Commented Jan 28 at 9:58
7
$\begingroup$

You have to take into account the difference between what is seen and what is observed. They are two different things in special relativity.

What your eye (or a hypothetical high-speed camera) would see in a particular scene depends on what photons are reaching it at a particular instant. However, not all of these photons would have left the objects in that scene at the same time; the photons travelling from farther objects had to leave earlier. This can, as you suspect, lead to rigid objects appearing to bend as you travel past them.

The video below (quite lo-res but still worth watching) illustrates this very well. The camera accelerates down a road in a universe where $c$ is 1 m/s. In the first rendering, the travel time of the light is ignored, and everything looks as you'd expect. In the second rendering, the travel time of the light is included, and objects appear to bend. For example, the bottom of the road sign appears bend towards the camera as the camera passes it; this is because light from the bottom of the pole generally has less distance to travel to the camera than light from the top of the pole.

https://www.youtube.com/watch?v=IgAII_crHHc

However, when we say "observe" in special relativity, we assume that we have effectively taken all of these travel-time effects into account. In other words, an "observation" of the behaviour of the road sign would be couched in terms of events: "At $t = 0$, the top of the sign was at $(x,y,z) = (1,0,1)$ and the bottom was at $(1,0,0)$. At $t = 1$, the top of the sign was at $(2,0,1)$ and the bottom was at $(2,0,0)$" And so forth.

If you looked at all of the coordinates of the events that you observed, you'd find that your frame was still composed of straight lines at any given $t$. It wouldn't have looked like it when you flew past it; but since you're a smart person, you'd know that you had to take the light travel time into account to figure out where a particular point "really" was at a particular time. And once you did that, the bendiness would disappear.

$\endgroup$
4
  • 1
    $\begingroup$ Thank you for the comment! Perhaps I am indeed confusing what would be observed and what would be seen by the external observer. Unfortunately I am still unable to find where I make the mistake. I added some calculations to the question to clarify my thought process. $\endgroup$
    – Joona
    Commented Jan 27 at 19:45
  • $\begingroup$ @Joona: I don't think you're necessarily making a mistake (though it's certainly possible — I haven't checked your math.) From what I can tell, you're trying to figure out the positions of various parts of the apparatus for which the light gets to a particular point at a given time. In my language above, that is what's "seen" at that point. But it's not what an observer moving relative to the apparatus would "observe." ... $\endgroup$ Commented Jan 27 at 21:37
  • $\begingroup$ ... To find what this observer would "observe" you would figure out where the various points on the apparatus are at a single moment of time (in the observer's frame). I am fairly sure that if you do that you'll find that the angles end up being the same. $\endgroup$ Commented Jan 27 at 21:38
  • $\begingroup$ Saying seen<>observed contradicts your own definitions since you defined them to be the same thing. (Read your definitions.) Moreover using "seen" begs the question since "seen" has no technical meaning other what you would give it; it doesn't bring in any useful distinction from "observed" from the everyday meaning. And it is a confusing choice of word since in everyday sense it is a synonym for "observed". The distinction you are trying to make--per "you'd know that"--is observed vs deduced & the deduced here involves starting from further information about the setup than what is observed. $\endgroup$
    – philipxy
    Commented Jan 28 at 16:25
2
$\begingroup$

You can not draw a picture where A=B and $\alpha$ and $\beta$ are different, or can you?

So, therefore moving blobs of light can not draw lines inside a foggy spaceship cabin so that A=B and $\alpha$ and $\beta$ are different.

The aforementioned blobs of light drawing lines in the direction of motion of a very fast moving spaceship must move in a very small angle to each other, as observed in the frame in which the ship moves very fast.

And that above thing is the thing that you are thinking, when you are thinking that some angle must be small.

https://en.wikipedia.org/wiki/Relativistic_aberration

$\endgroup$
3
  • $\begingroup$ Actually, it's easy to draw a picture where $A=B$ and $\alpha\neq\beta$ if you believe the two $x$'s are no longer equal. I guess this must be what the OP is thinking, but it's very hard to tell. $\endgroup$
    – WillO
    Commented Jan 27 at 16:41
  • $\begingroup$ My confusion comes into play when the outside observer starts taking into consideration the time it takes for light to reach each of the sides of the apparatus once it has been emitted from the middle. I edited the question to add some details on how I would expect this to change the situation. My logic is surely flawed, but hopefully this helps understand how I came to the conclusion that $\alpha' > \beta'$ if A = B = A' = B'. $\endgroup$
    – Joona
    Commented Jan 27 at 19:53
  • $\begingroup$ @Joona There was no need to clarify your confusion. I got it. Now try to understand my answer. en.wikipedia.org/wiki/Relativistic_aberration $\endgroup$
    – stuffu
    Commented Jan 27 at 23:19
0
$\begingroup$

In the lab frame, light takes longer to go from $A$ to $B$ than it takes to get from $B$ to $A$. There are (at least) two possible reasons for this: Either $\alpha>\beta$ or the apparatus is moving. You've already stipulated that the apparatus is moving, so I don't understand why you think you can conclude that $\alpha>\beta$.

$\endgroup$
0
$\begingroup$

Other answers have given more details, but I think that it's important to note that as Lorentz transformations are linear, any curve that is straight in one reference frame will be straight in another (obviously this doesn't hold in general relativity, just special).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.