I am studying special relativity but it isn't making any sense. I know that if I have space time co-ordinates in a stationary reference frame I need to use the Lorentz transformations to get the co-ordinates in a moving reference frame but I don't understand what this actually means physically and how to decide which clocks to apply the transformations to.

For example, two clocks are distant and synchronized. Clock A is stationary and clock B is moving fast towards clock A. The moment they pass an observer who travelled with clock B can see both clocks and sees clock A is slow and clock B is fast but an observer who stayed with clock A sees the opposite even though they are in the same place looking at the same pair of clocks at the same time? The clocks really physically read different values depending on which observer is looking?

Edit: I know acceleration changes things, so please assume clock B was always moving at a steady speed and the clocks synchronized with a flashing light source that was exactly half way between the clocks at the start of the experiment.

  • $\begingroup$ I'm sorry but I don't understand the difference, I thought the LT was what decided the reading on the clock. Are they different ideas? $\endgroup$ Feb 2, 2021 at 17:03
  • $\begingroup$ "two clocks are distant and synchronized". Synchronized in what frame? $\endgroup$
    – WillO
    Feb 3, 2021 at 5:01
  • $\begingroup$ Does this help? physics.stackexchange.com/a/383363/123208 There are lots of other answers by robphy using rotated graph paper. $\endgroup$
    – PM 2Ring
    Feb 3, 2021 at 5:45
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    $\begingroup$ BTW, to draw spacetime diagrams for your scenario you also need to specify the velocity of the flashing light source. $\endgroup$
    – PM 2Ring
    Feb 3, 2021 at 5:51
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    $\begingroup$ how to decide which clocks to apply the transformations to. Don't apply them to clocks (whatever you might mean by that). Apply them to the coordinates of an event. $\endgroup$
    – WillO
    Feb 5, 2021 at 0:52

3 Answers 3


Reference frames are just coordinate systems, nothing more.

Spacetime is not very much different from Euclidean space. The main difference is that there is a flipped sign in the spacetime version of the Pythagorean theorem.

Just as you can put Cartesian coordinates on Euclidean space, you can put inertial coordinates on spacetime.

The Lorentz transformation is just a rotation between different coordinate systems. It can be written

$$\begin{pmatrix}ct'\\x'\end{pmatrix} = \begin{pmatrix}γ&γβ\\γβ&γ\end{pmatrix} \begin{pmatrix}ct\\x\end{pmatrix}$$

where $β=v/c$ and $γ=1/\sqrt{1-β^2}$. A rotation (of less than 90°) in Euclidean space can be written

$$\begin{pmatrix}x'\\y'\end{pmatrix} = \begin{pmatrix}γ&γβ\\-γβ&γ\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}$$

where $β=\tan θ$ is the slope of each coordinate system relative to the other, and $γ=1/\sqrt{1+β^2}=\cos θ$. They're identical aside from a couple of sign flips and the factor of $c$ (which in advanced courses is usually set to $1$). In the special-relativistic case you can also write $β=\tanh α$ and $γ=\cosh α$, where $α$, called the rapidity, is an angle very much like $θ$.

All of the funny "effects" of special relativity are just geometric relationships between different coordinate systems. For example, in Euclidean geometry, if you have a vertical strip (parallel to the $y$ axis) whose width (measured along the $x$ axis) is $w$, and you measure its width along the $x'$ axis, which is inclined by a slope of $β$ to the $x$ axis, you'll get a width of $\sqrt{1+β^2}$. That's length contraction (with a sign flip). If you have a line parallel to the $y$ axis with tick marks on it at 1-unit intervals, and you measure the $y'$ coordinates of the same tick marks, you'll find that they're $1/\sqrt{1+β^2}$ units apart. That's time dilation (with a sign flip).

The reason you're confused is that special relativity is taught in a gratuitously confusing way. They say that "observers" "observe" events when they're really talking about assigning inertial (Cartesian) coordinates to events. Nobody actually sees what they claim is "observed". It's just coordinates.

  • $\begingroup$ Very nice answer. You may also want to mention that $\beta=\frac{v}{c}=\tanh(\psi)$, where $\psi$ is the "boost angle" or whatever it's called (to complement your next line regarding $\tan\theta$; thus emphasizing that "hyperbolic trig is to Lorentzian signature as circular trig is to Riemannian signature"). I have to say, your one paragraph about time dilation and length contraction was much more insightful than hordes of formulae and other "intuitive observer-type" of explanations. $\endgroup$
    – peek-a-boo
    Feb 4, 2021 at 12:43
  • $\begingroup$ The usual explanations about observers observing events and just the very language used to describe these concepts is confusing. I only recently started studying special relativity, and after two or three days of thinking about it, I finally came to the same conclusion as you wrote in that paragraph. So while I did learn quite a bit by struggling, I do wish I had seen such an explanation earlier, because this is really just a simple geometric problem (with a minor twist due to Lorentz signature... but I'm more math oriented anyway so that wasn't an issue for me). $\endgroup$
    – peek-a-boo
    Feb 4, 2021 at 12:49
  • $\begingroup$ @peek-a-boo I added a bit about boost angle (rapidity). $\endgroup$
    – benrg
    Feb 5, 2021 at 0:35

To simply answear your question. Yes. (Nice name btw)

For more details it seems that you want to find some absolute reference. Which is precisely what special relativity is breaking. Everything is relative.

In your scenario for instance the first sentence is wrong "Clock A is stationary and clock B is moving fast" Nothing is stationary nor moving. Or equivalently everything is stationary or moving in a certain frame.

Indeed observer that travels with clock A sees clock B slow and the one traveling with clock B sees clock A slow. And it is not a problem because neither time nor space are absolute.

Something that helps me (beware it is not exact at all) is this picture :

Instead of thinking space and time constant let's imagine a world where some speed is constant (it turns out to be speed of light but whatever).

Now speed writes like $V = \frac{L}{T}$ so in order to keep speed constant $L$ and $T$ may vary. Again it is a picture. Lorentz transform formalise this concept in the right way

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    $\begingroup$ Everything is relative. Not true! There are Lorentz-invariant quantities that all observers agree on. $\endgroup$
    – G. Smith
    Feb 2, 2021 at 18:33
  • $\begingroup$ Absolutely ;) and relativity is not relative $\endgroup$
    – kyril
    Feb 2, 2021 at 19:49

I won't try to answer your question in detail. I will only encourage you to give up the mental picture you have of observers 'looking at each other's clocks' as they pass each other. I did this when I was starting out in SR, and it tends to be a confusant. Unfortunately it's easy to fall into this picture because many textbooks do use this language. E.g. "observer A sees observer B's clock running more slowly".

Try to think about it more abstractly. Don't think about time dilation in terms of what the observers 'see' when they 'look at' each other's clocks. Instead, think about it in terms of how much time each observer measures as having elapsed ON THEIR OWN CLOCK between two events. If they are moving wrt each other, then one clock will have made more ticks than the other, and the difference will be bigger for larger relative velocity. It's as simple as that.

Rather than trying to picture this in your mind like a movie, it will be more clarifying for you to learn how to draw and analyze Space-Time Diagrams. Draw out the STD for the situation you're describing and it should be clear.


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