Supposed we have the standard boosting configuration, with stationary reference frame $O$ and moving reference frame $O'$ traveling at velocity $\vec{v} = |v|\hat{x}$. The origins of the two frames are equal at $t=0$, where a flash of light is emitted. The transformation equations are standard: $$ x' = \gamma (x-vt)\\ ct' = \gamma (ct - \beta x) $$ where $\beta = \frac{v}{c}$. So along the $+x$ direction, this makes sense, as the quantity $x-vt$ represents the distance from $O'$ to the crest of the wave in the $O$ coordinate system, which must be multiplied by $\gamma $ to get the equivalent distance that light travels in the $O'$ frame. Now, suppose we are interested in the other end of the emitted wave that travels along the $-x$ direction. In the $O'$ frame, the light must take a circular path, therefore the length $x'$ should be equal to $-x'$. But when drawing out the problem, we see that in the lab frame, the trajectory of light after time $t$ is $-X$, while an observer from the $O'$ system would see light take a path equivalent to $\lambda (-x - vt)$ for some constant $\lambda$. In this scenario, would $\lambda$ need to be $\frac{1}{\gamma}$ to shrink this larger distance into the correct path length in the $O'$ reference frame? How are we supposed to interpret the lack of symmetry between transformations along the direction of motion and anti parallel to the direction of motion?

  • 1
    $\begingroup$ Light travels along straight lines, not along circles. The Lorentz transform is linear, so a path that is straight in one frame is straight in all. $\endgroup$
    – WillO
    Commented Jan 11, 2022 at 0:49
  • $\begingroup$ If you think of the problem in 3D, light generates a sphere of radius ct. After time t, the sphere is centered in the S coordinate system, but off-center of the S' system. In the moving frame s', looking towards the -x direction, you see light travel ct', but this length transformation appears different when looking in the -x direction vs the x direction due to the direction of v. $\endgroup$
    – jakerz
    Commented Jan 11, 2022 at 2:19
  • $\begingroup$ @jakerz you said "the sphere is centered in the S coordinate system, but off-center of the S' system" this is incorrect. The sphere is centered in both coordinate systems. $\endgroup$
    – Dale
    Commented Jan 11, 2022 at 13:53
  • $\begingroup$ My mistake. I mean, an observer in S will believe that the sphere is off center in the S' system. $\endgroup$
    – jakerz
    Commented Jan 11, 2022 at 20:21

5 Answers 5


You are forgetting about the relativity of simultaneity. The planes of constant time in O and O' are tilted relative to each other. The tilts are such that in either frame a plane of constant time equates to a sloping slice through time in the other frame, the slope rising in the direction of travel, so clocks in the other frame seem to be getting progressively more ahead of time in the forward direction and progressively behind time in the opposite direction.

It is the fact that the planes of constant time in the two frames are tilted that allows the speed of light to be constant relative to both frames.

Suppose the origins of O and O' coincide at t=0 when a light is flashed from the common origin to the left and right, and that O' is moving to the right at around 0.5c. After a second has passed in O, the light will be about 300,000km away in either direction. At the same times that light is at those positions in O, it is about 150,000km ahead of O' to the right and about 450,000km away to the left. Since the speed of light in O' is constant in both directions, that means the time in the O' frame where the light is 150,000km away must be around 0.5s, while the time where the light is 450,000km away to the left must be around 1.5s.

So you should be able to see that from the O'perspective, the time 150,000km ahead in O seems to be running 0.5s ahead of the local time in O', while the time 450,000km behind seems to be lagging 0.5s behind the local time in O'.

  • $\begingroup$ So in the case of a charged particle in motion with velocity v (in motion viewed from O), we start to record the field moving radially outward as it passes the origin in O. In the coordinate system of the particle (O'), the field radiates in all directions equally, but in the O system, radiation appears to start being emitted in the direction of v before the reverse direction? $\endgroup$
    – jakerz
    Commented Jan 11, 2022 at 22:38
  • $\begingroup$ Not quite. In each frame the radiation is emitted at the same moment. At a fixed time in each frame the radiation is at an equal distance from the origin of the frame. The point is that a fixed time in one frame is two different times in the other and vice versa. Keep trying to draw some diagrams and figure it out. $\endgroup$ Commented Jan 11, 2022 at 22:44

I think you are mistakenly applying the time-dilation/length-contraction factor $\gamma$ to wavelengths. I think it should be the Doppler factor.

Length contraction applies to (say) the length of a ruler, involving the separation between two timelike lines.

Wavelengths, however, involve the separation between two lightlike-lines. This involves the Doppler factor.

My answer to this Deriving Relativistic Doppler Effect through length contraction might be helpful.


For an interpretation...

The lack of symmetry in your two transformations is a result of maintaining the invariance of the interval, which is geometrically realized as the invariance of the area of this "diamond" (since the determinant of a boost is 1).

Have a look at my visualization https://www.geogebra.org/m/Jq4jDMRW
where velocity=3/5 corresponds to a Doppler factor $k=\sqrt\frac{1+v}{1-v}=2$..
a stretch of the forward light segment by 2 and shrinkage of the backward light segment by 2. robphy-RRGP-clockDiamond-geogebra

It might be enlightening to think in terms of rapidity $\theta$,
where $v=\tanh\theta$ and $\gamma=\cosh\theta$ and $k=\exp\theta$. (Note that $e^{-\theta}=\frac{1}{e^\theta}$.)


A light flashes. According to Alice, there are stationary screens in all directions at distance $x$ light-years from the light source. The light hits all of those screens at time $t=x$.

Also according to Alice, Bob is traveling east and Carol is traveling west, both at speed $v$.

According to Bob, the light hits the eastmost screen at a location $x_1=\sqrt{1-v}/\sqrt{1+v}$ light-years east of the orginal flash and the westmost screen at a location $x_2=\sqrt{1+v}/\sqrt{1-v}$ light-years west of the original flash. It hits these screens at times $x_1$ and $x_2$.

Carol says something similar, with $\pm$ signs reversed.

Of course there is no symmetry saying that Bob and Carol have to agree on which screen gets hit first, because Bob and Carol are traveling in different directions.

What's troubling you about all that?


The equation for a photon traveling in the -x direction in the O' reference frame is: $$ x' = -ct' $$ When we transform that equation to the O reference frame, using the Lorentz transformation, we get: $$ \gamma (x - vt) = -\gamma (ct - \frac{v}{c} x) \\ x - vt = \frac{v}{c} x - ct \\ x (1 - \frac{v}{c}) = - (c - v) t \\ x = -ct $$ That is, a photon traveling with speed c in the -x' direction in the O' frame will be traveling in the -x direction with speed c when viewed from the O frame. There is no "lack of symmetry"


From 'reality', if a sound wave is emitted at a point P (shock wave), do we hear the explosion once or twice, even if we are moving with a speed v, change the direction of propagation of the wave, it is like being interested in the light emitted by the sun in the other direction of the axis (earth-sun), it is up to the extraterrestrials to make their transformations, in our reference frames the event (emission of photons in this direction) never existed :-).

Reversing the direction of wave propagation has an indirect link with the absorber theory of Wheeler and Feynman (also called the Wheeler–Feynman time-symmetric theory)

On this diagram to the right of the screen, we take $ \alpha = 0 $, the light goes from M to O and O ', if the light goes from M in the opposite direction, how can we conceive of its encounter with the frames of reference $\mathcal{R}_{(O)}$ and $\mathcal{R}'_{(O')}$? and how to synchronize the clocks using signals that do not exist in the two frames of reference (for observers) and what time are we going to talk about?


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