I have a problem transforming from one system to another when the direction of motion is changed. To demonstrate the problem I'll set up an easy example with intuitive numbers:

Problem (Animation) enlargeleft: external observer (correct), right: moving observer (obviously wrong); v=c/2

I have a square with side length $S = 1 \text{ Ls}$ (lightsecond), and therefore the circumference $U = 4 \text{ Ls}$. In its upper right I place a runner, which runs around the square counterclockwise with the velocity $v=\frac{c}{2}$. He emits one red photon to the front, and a blue one in the clockwise direction.

If I ask for the time and the location where the first photon outpaces the runner, this is easy from the view of an external observer who is at rest relative to the route: The blue photon passes the runner at $t = \frac{U}{c+v} = \frac{8}{3} \text{ sec}$.

The place where the first impact happens is $t\cdot v = \frac{4}{3} \text{ Ls}$, the upper third on the left side of the square.

We note that the blue photon meets the red photon in the lower right corner; after $2 \text{ sec}$ both have travelled $2 \text{ Ls}$ and therefore exactly 2 side lengths.

So far so good. But now I'm starting to struggle:

If I try to transform the scene into the system of the runner with $v=\frac{c}{2}$, I first transform the $\{1\}\times \{1\}$ square into a rectangle with the side lengths $\{1\} \times \left\{ \sqrt{1-\frac{v^2}{c^2}} \right\}$ = $\{1\}\times\{0.866\}$ - because of the Lorentz transformation the leghts in direction of movement shall contract.

The red photon has $c$ relative to the runner, while the route is moving towards him with $v$. So from the view of the runner the photon is moving with $c+v = 1.5 c$ relative to the route.

After the point where the photon is turning to its left and therefore changes its direction from horizontal to vertical, its vertical velocity must be $\sqrt{c^2-v^2}$, so the total velocity relative to the runner can be $c$ (Pythagoras).

Now I calculate:

The Lorentz-contracted side length $S' = \frac{S}{\gamma} = \frac{\sqrt{3}}{2} = 0.866 \text{ Ls}$. I divide those through $c+v$ to get the time until the red photon makes its first turn:

$$\tau_1 = \frac{S/ \gamma}{c+v} = \frac{1\cdot \sqrt{1-(\frac{1}{2})^2}}{1+\frac{1}{2}}$$

$$\tau_1 = \frac{1}{\sqrt{3}} \text{ sec}$$

Because at this time the runner is still moving horizontal, but the photon vertical, the length of the vertical side is uncontracted relative to the runner, thus $S = 1 \text{ Ls}$. The time until the red photon reaches the lower left corner is therefore $S/(c^2-v^2)$:

$$\tau_2=\frac{S}{\sqrt{c^2-v^2}} = \frac{1}{\sqrt{1^2-(\frac{1}{2})^2}}$$

$$\tau_2= \frac{2}{\sqrt{3}} \text{ sec}$$

The total time until the red photon reaches the lower left corner is then

$$\tau_1+\tau_2 = \sqrt{3} \text{ sec}$$

This is also exactly the time the runner needs to travel his upper contracted side:

$$\frac{S}{\gamma \cdot v} = \sqrt{3} \text{ sec}$$

Here the problem becomes obvious:

From the view of the external observer (left image) we know that the red photon meets the blue photon in the lower left corner.

Runner and photon start moving towards each other while both start from opposite directions on the same path; the runner moving straight down, and the photon straight up. Because the runner's velocity relative to the route $v=\frac{c}{2}$, the blue photons velocity relative to the route is $c-v$ which is also $\frac{c}{2}$. Runner and photon now have the same speed in opposite directions (so the photon has $c$ relative to the runner).

Because of this they now must meet on the half way; but from the perspective of the external observer we know that they meet not half way, but in the upper third.

What did I do wrong? Can anyone find my mistake?



Post Script:

The only explanation I might think of is that the photons make a jump when the runner changes direction (or at least what seems like a jump for the runner when his deceleration and acceleration times are infinitesimaly short). I'm not sure if this is physically correct, nor do I have any idea how to calculate the jumped distances correctly without tricking around... What I did here was a kind of a cheat; I solved for the distance where the photon would have to start so it can meet the runner at the right spot (but I have calculated the right spot in the outer observers system, see left image).

Possible Soultion enlarge

  • 1
    $\begingroup$ Here's the notation we use, LaTeX. $\endgroup$
    – HDE 226868
    Jan 27, 2015 at 21:36
  • $\begingroup$ if your runner changes direction, then he gets accelerated. If he gets accelerated, then the speed of light is no longer c (radar distance is not even symmetric for accelerated observer), and you need general relativity to compute exactly the path of a photon from an accelerated observer. $\endgroup$
    – sure
    Feb 3, 2015 at 16:05
  • 2
    $\begingroup$ @sure: You do not need general relativity to handle acceleration. In the absence of gravity special relativity should do. $\endgroup$
    – Yukterez
    Feb 3, 2015 at 16:16
  • $\begingroup$ The right image shall not be a transformation, it just shows the system of the runner. The only reason why the runner is moving and the route is fixed is for better display on the monitor. If the runner (black point) were fixed in the animation, the route would move in the same way just in opposite directions. (It seems that the answear I'm refering to has been deleted while I'm typing my reply to it, so nevermind) $\endgroup$
    – Yukterez
    Feb 3, 2015 at 17:40
  • $\begingroup$ well, you don't necessarily do that's true, but i'm not sure that special relativity (with clock hypothesis) allows you to integrate the motion here (moreover its on smooth so) $\endgroup$
    – sure
    Feb 3, 2015 at 18:47

2 Answers 2


This is a twin paradox in disguise. Here the external observer is the inertial one, and the moving one (black point) is a travelling one. You were right when in postscript you supposed that the photons "jump" to another place. This is because, when the traveller accelerates, it "scans" the external observer's time at an increased speed, and in the limit it just jumps to another point in external observer's time. This is better illustrated with wikipedia's image:

enter image description here

See how the simultaneity planes' intersections with $ct$ axis discontinuously change from blue set to red set.

Let's now calculate the coordinates where red and blue photons will appear after acceleration.

First, switch to the initial proper frame of black point moving along $x$ and express the world line of the point after acceleration, as seen in this frame, in terms of external observer's time. From now on, I'll denote world line coordinates via $\{\tau,x,y\}$, taking $c=1$. So world line after acceleration, as seen in initial frame, is

$$W_0=\left\{\frac{2t-1}{\sqrt3},\frac{t-2}{\sqrt3},1-\frac t2\right\}.$$

Now move the acceleration time to the origin, so that the proper time $\tau=0$ at the moment of acceleration. We achieve this by subtracting $W_0(t=2)=\{\sqrt3,0,0\}$ from the world line coordinates above, and get

$$W_1=\left\{\frac{2t-4}{\sqrt3},\frac{t-2}{\sqrt3},1-\frac t2\right\}.$$

Now we transform to the new frame by applying two Lorentz transforms: first the one getting us away from leftward moving frame, and then the one accelerating us downwards:

$$W_2=L_y(v)\cdot L_x(-v)\cdot W_1,$$

and this gives


which, of course, is to be expected — the runner rests at the positional origin in this frame.

Now we can transform our blue and red particles' world lines to this frame. Their world lines in original leftward moving frame are:

$$W_{b0}=\left\{\frac{2t-1}{\sqrt3},\frac{t-2}{\sqrt3},t-3\right\},$$ $$W_{r0}=\left\{\sqrt3(t-1),\sqrt3(t-2),-1\right\}.$$

After transformation they'll be:

$$W_{b2}=\left\{\frac{3t-7}{\sqrt3},0,\frac{3t-8}{\sqrt3}\right\},$$ $$W_{r2}=\left\{\frac{2t-5}{\sqrt3},t-2,\frac{t-4}{\sqrt3}\right\}.$$

Solving the equations $\tau=0$ for $t$ for both particles gives us

$$t_b=\frac73,$$ $$t_r=\frac52.$$

Substituting them into $W_{?2}$ gives us finally the points where these particles start motion after acceleration, as seen from the new frame:

$$W_{b2}=\left\{0,0,-\frac1{\sqrt3}\right\},$$ $$W_{r2}=\left\{0,\frac12,-\frac{\sqrt3}2\right\},$$

which is what we wanted to find.

  • $\begingroup$ Best answear so far, but how exactly do I compute that limit of √(3)/2 when acceleration time goes to zero without having to switch to the outer observers system? $\endgroup$
    – Yukterez
    Feb 3, 2015 at 17:42
  • $\begingroup$ @СимонТыран I just found the proper time by definition: $$\tau=\frac1c\sqrt{c^2t^2-\Delta x^2-\Delta y^2}=t\sqrt{c^2-v^2}=t\frac{\sqrt{3}}2.$$ I'm going to provide some more details tomorrow. $\endgroup$
    – Ruslan
    Feb 3, 2015 at 19:49
  • $\begingroup$ Ah, that's just the time through the gamma factor; the information I also used in the second image yukterez.ist.org/sagnac.ansatz.1.gif to transform from the unmoved clock's system to the runner's system and have the photon displaced just the right amount that it meets the runner at the right place, which also happens to be the right (contracted) time (what I called "cheating" in my original post because it doesn't actually involve taking the limit of the acceleration phase going to zero and getting the displacement of both photons after the runner's salient point turn). $\endgroup$
    – Yukterez
    Feb 3, 2015 at 22:10
  • $\begingroup$ If I just take the situation as viewed by an observer at rest to the route and transform into the moving frame I need the worldlines of the runner and the photon to meet at one point from where I can run the situation backwards to get the "jumping position", otherwise I could not get the new position of the blue photon in the moving frame; not to mention the position of the red photon, which at this moment is many salient points away from meeting the runner. $\endgroup$
    – Yukterez
    Feb 3, 2015 at 22:40
  • $\begingroup$ @СимонТыран I've updated the answer to give actual results. $\endgroup$
    – Ruslan
    Feb 4, 2015 at 12:07

Every time the runner hits a corner, the photons should suddenly move to a different part of the track. Due relativity of simultaneity, "now" changes whenever the point of reference of the runner changes. This is the basis of the Rietdijk–Putnam argument. Also, if the second picture is supposed to be from the point of reference of the runner, the runner should be staying still and the track should be moving.

It would make more sense for the photon in front to be blue due to the doppler effect. Although if you carry that logic too far it will get pointlessly complicated. We only care about which photon is which, not how much energy they have.


Right now, you have the photon teleport with the track. You are using a Lorentz transformation to find the new position of the photon. The problem is that you're still using the old time. You need to find the time from the new point of reference as well, and then trace the path of the photon in order to find out where it is now.

  • $\begingroup$ >> "Also, if the second picture is supposed to be from the point of reference of the runner, the runner should be staying still and the track should be moving." << I already answeared this with "The only reason why the runner is moving and the route is fixed is for better display on the monitor. If the runner (black point) were fixed in the animation, the route would move in the same way just in opposite directions." The relative velocity from the runner to both photons is always 1c. It does not matter if I move the route or the runner, everything stays the same. $\endgroup$
    – Yukterez
    Feb 3, 2015 at 23:38
  • $\begingroup$ As you can clearly see in the right part of the image the lenght of the route is contracted even if not moving. That is NOT because it shows the reference frame of an unmoving observer. It shows exactly the moving runner's system, but optically centered at the route just for viewing comfort. $\endgroup$
    – Yukterez
    Feb 3, 2015 at 23:52
  • $\begingroup$ >> "It would make more sense for the photon in front to be blue due to the doppler effect."<< This is not about doppler effects. The colors do not represent redshift, they are only to identify the photons. Calculating the redshift is not a problem (at least not during the unaccelerated phases). $\endgroup$
    – Yukterez
    Feb 4, 2015 at 4:51

Your Answer

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

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