I am wondering whether or not I fully understood what the variables represent in the Lorentz time and distance transformations.

What I understand is:

t′ is the 'proper time', the time taken to move by another object measured by a distant 'stationary' observer.

t is the time taken to move from one point to another, measured by the 'moving' observer.

x′ is the 'proper distance/length' of the 'moving' observer, measured by the 'stationary' observer.

x the distance/length from one point to another, measured by the 'moving' observer.

u is the speed of the 'moving' object relative to the stationary observer.

Can someone confirm this to be correct? I am very confused, because since observed movement is 'relative', how can I be certain which object in a question is the 'moving' one, and which one is the 'stationary' object?


3 Answers 3


The bad news is that your understanding of the coordinates x, t, x', t' is badly jumbled . The good news is that your last sentence asks a good question, so let's start with it:

Q: "how can I be certain which object in a question is the 'moving' one, and which one is the 'stationary' object?"

A: You just "name" them that way! You can even switch the "names" around, but once you do, you need to stick to your chosen denomination. This is what the principle of relativity is about: there really isn't a truly "stationary" or a truly "moving" frame/object/observer. To avoid the confusion altogether, let's just call one A and the other B. Now we can deal with the main question.

Q: "what the variables represent in the Lorentz time and distance transformations"?

A: Say both A and B observe an event E and measure separately where and when E takes place. The Lorentz transformations relate the position and time of E as measured in frame A (or by observer A) to its position and time as measured in frame B (or by observer B):

Say in frame A, E is observed at position $x_A$ and time $t_A$.

In frame B the same E is seen at position $x_B$ and time $t_B$.

If B moves at velocity $u$ relative to A, as seen from A, then the Lorentz transformations tell us that

$$ x_B = \gamma(x_A - u t_A) \\ t_B = \gamma(t_A - \frac{u x_A}{c^2}) $$

where $\gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$. Conversely, as seen from B, A moves relative to B at velocity -u, and the Lorentz transformation from B to A reads

$$ x_A = \gamma(x_B + u t_B) \\ t_A = \gamma(t_B + \frac{u x_B}{c^2}) $$

To get to your original notation and definitions, let's identify

$x_A \rightarrow x$, or the position where E occurs in the "stationary" frame

$t_A \rightarrow t$, or the time at which E occurs in the "stationary" frame

$x_B \rightarrow x'$, or the position where E occurs in the "moving" frame

$t_B \rightarrow t'$, or the time at which E occurs in the "moving" frame

Notice that the terms "proper time" and "proper length" do not appear at all up to this point. We can say that time $t_A$ ($t_B$) is the proper time of frame A (B), but we cannot attach a proper time to event E unless we know that it concerns an object C that "co-moves with"/"is stationary with respect to" frame A (B). If C is co-moving with A (B), we can also call the length of C as seen in A (B) the "proper length" of C. If however C is moving with respect to both A and B, we cannot identify the coordinates of event E in frame A or frame B as "proper" coordinates.

For example, say A is the frame of Earth and B is the frame of a spaceship moving away from Earth at relative velocity u (as seen from Earth). Then $x_A$ and $t_A$ are the position and time of event E as measured from Earth, while $x_B$ and $t_B$ are the position and time of E as measured on the spaceship. If E occurs on Earth we can say that its proper-time is $t_A$, if it occurs on the spaceship, its proper time is $t_B$. But if E occurs on a space probe moving away from both the spaceship and Earth neither $t_A$ nor $t_B$ are its proper-times. Same goes for object lengths.


Statement than one reference frame is stationary means that we choose its point of view for the problem at hand. It doesn't mean that it is special in any way - only that it is convenient for problem at hand. So "stationary", " static" and other statements like that are always some sort of shortcut. It can be misleading at first. Remember that we could have chosen any other reference frame to be "static". Note that we know that Earth is moving (and accelerating!) relative to Sun which is moving relative to something too, but for problems in classical mechanics it is convenient and usually precise enough to treat surface of Earth as fixed, static background.

In special relativity it is taken for granted that every inertial observer can coordinatize whole spacetime with four numbers, say $t,x,y,z$. Other observer can do it too obtaining equally valid description of spacetime, but in terms of different four numbers, say $t',x',y',z'$. Lorentz transformation allows you to translate those descriptions into each other. That is given coordinates of some event in one reference frame you can calculate its coordinates in different reference frame. To do this you need relative velocity of the two observers.


No, your understanding of the variables $t',t,x',x$ is incorrect. Lorentz transformation relates two coordinates that are in relative motion. One way to state the principle of relativity which serves as an axiom of SR is that there's no absolute state of rest, so given two observers who are in relative motion and whose coordinates are given by $(x,y,z,t)$ and $(x',y',z',t')$ respectively, one cannot say who is "truly" moving and who is "truly" stationary. So the way you interpret the variables is wrong. All you can say about them is that they're the coordinates(the rulers and the clocks) that are used by two observers in relative motion.

$(x,y,z,t)$ and $(x',y',z',t')$ are just different coordinates used by two observers to denote the same events* in the spacetime.

*events are something that happen at a given location in space at a given moment in time. If you're familiar with spacetime diagrams, they're represented geometrically as points.


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.