I want to know how an ordinary angle $\theta$ transforms under a Lorentz boost. For that purpose I consider a 4-vector given by $$ a ^ \mu = ( t , \cos \theta , \sin \theta , 0 ) .$$ The angle I will analyze is the one between this 4-vector and the $a$ axis, $\theta$. I consider a boost $$ \Lambda ^\mu _{\ \ \ \nu} = \left( \begin{matrix}\gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{matrix} \right) .$$ Thus the transformed 4-vector is $$ a ^{\prime \mu} = ( \gamma (t - \beta \cos \theta), \gamma (\cos \theta - t \beta), \sin \theta, 0 ) $$ Now the angle subtended between $a^\prime$ and the $x^\prime$ axis should be the transformed $\theta^\prime$, i.e. $$ \theta ^ \prime = \tan ^-1 \left( \frac{\sin \theta}{\gamma (\cos \theta - t \beta)} \right) .$$

Now I have some doubts:

  1. Which should be the correct value of $t$, the time component of the 4-vector $a$ used to define the angle $\theta$? I would say $t=0$ in order to have the angle $\theta$ defined by a space-like vector, the same idea as when you define the proper distance between two points. In this case I would think of a sort of proper angle.

  2. In the case $t=0$ I find that $ \theta^\prime \leq \theta $... This is not consistent with the length contraction along the $x$ axis... In fact, the $x$ component of $a^\prime$ is greater than that of $a$...

I was expecting to obtain something like this:

enter image description here

but I didn't although I have transformed the vector using the transformation law... What is wrong?


Length measurement in the moving frame should be done at simultaneous times in the moving frame, so you need to do a Lorentz transformation fixing $t'= 0$.

This gives

$$ t = \beta \cos \theta $$


$$ a'^\mu = ( 0 , \frac {\cos \theta} \gamma , \sin \theta , 0 ) $$

So $$ \theta ' = tan^{-1} ( \gamma \tan \theta ) $$

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  • $\begingroup$ Wow, much simpler! I was trying to adapt the example from Wikipedia to learn myself. $\endgroup$ – HiddenBabel Oct 4 '18 at 5:50
  • $\begingroup$ If I replace $t = \beta \cos \theta$ in my expression for $a^\prime$ I do not obtain what you are saying. I keep on thinking about this problem and found the solution. Measurements must be taken at simultaneous times in both frames, not only in the moving frame. I will post the solution I found. $\endgroup$ – user171780 Oct 4 '18 at 9:41
  • $\begingroup$ Now I have understood what you said. Please let me post the same answer but with more detail about what I was doing wrong. $\endgroup$ – user171780 Oct 4 '18 at 11:11
  • $\begingroup$ Good - yes please do - that is the key calculation for why "time dialates" but "space contracts" even though it look symmetric is x and t at first sight. Beautiful! $\endgroup$ – Bruce Greetham Oct 4 '18 at 12:57

I answer my own question after realizing what I was doing wrong. It is very similar to Bruce Greetham's answer but with more detail.

What I was doing is trying to use a unique 4-vector $a$ to measure the distance in both frames, and this is wrong because distance measurements must be simultaneous in BOTH frames. Thus, what I was trying to do can be depicted in spacetime diagrams as:


As can be seen, if $a$ is simultaneous in one frame, lets say the lab frame, then it is not simultaneous in the boost frame, and vice versa. So trying to use the projection of only one 4-vector is not enough.

After thinking about this I realized that I had to consider the world line of an object used to measure the distance. This allows to define two 4-vectors $a$ and $b$ as follows:

This is nice!

So in the lab frame the distance must be measured using $a$ (that coincides to be the space projection of $b$ just because it is at rest, but may not be the case) and in the boost frame the distance must be measured using $b$. The important requirement is that $a$ is a purely space like 4-vector in the lab frame and the same for $b$ in the boost frame.

Now lets move to the math. The coordinates of each 4-vector in each frame are:


So I have defined $a$ and $b$ in the lab frame and then I have applied a boost to obtain the coordinates in the boost frame. Now I will impose the requirement that $b$ must be a space like 4-vector in the boost frame. This means that


where the notation $ \left\lceil b_t \right\rfloor _{\text{Lab frame}} $ reads "the time component of the 4-vector $b$ as seen in the lab frame". Replacing $t = - \beta \cos \theta $ everywhere and using that $ \gamma = \frac{1}{\sqrt{1-\beta^2}} $ we get

enter image description here

Now we can calculate $ \theta $ as seen in both frames. First, in the lab frame we get what we expected:

enter image description here

In the boost frame this reads:

enter image description here

If we plot the transformation for some values of $\beta$ we have:

enter image description here

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If you're moving, the angle is going to change. What you want to do is to define the origins of your reference frames $S$ and $S'$ at $t_0 = t_0' = 0, x_0 = x_0' = 0$. That is, $t_0 = t_0'$ is when observers in each frame perform their angle measurements. But how do they make a measurement? There was some event at $(x,y), t < 0$ which created light. In order to see the event, the light must have traveled some distance to enter their eyes. This is why you can't use $t = 0$ as the event time, because if the event happens anywhere away from the origin, there is no time for light to travel to the observers. But since $S'$ is traveling w.r.t. $S$, the distance the light had to travel is not the same.

So let's say the observers are measuring the angle from their position to the Sun. For simplicity's sake, say that at $t_0$, the Sun is located at $(x,y) = (1,1)$. This means the observer at $S$ measures the Sun at a $45^\circ$ angle, but we get another piece of info: the light that observer $S$ measures must have travelled a distance equal to $\sqrt{1^2 + 1^2} = \sqrt{2}$, so we can find out the time that light was emitted,

$$-ct_e = \sqrt{2} \Leftrightarrow t_e = -\frac{\sqrt{2}}{c}$$

Now we can boost $x$ to $x'$ to get,

$$x' = \gamma(-c\beta t_e + x) = \gamma(\sqrt{2}\beta + 1)$$

which is always larger than $x$ (for positive velocity).

Since $y' = y$, the angle $S'$ measures is

$$\tan^{-1}\left(\frac{1}{\gamma(\sqrt{2}\beta + 1)}\right) $$

which is always smaller than $45^\circ$.

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