# Does the Lorentz transformation only apply to corresponding observers?

Let $$S$$ and $$S'$$ be inertial frames moving at a relative velocity $$v$$ in the $$x$$-direction. Imagine sending observers to all points in each reference frame. The following rules hold:

(1) In a given frame, all observers agree to measure the spatial coordinates $$(x, y, z)$$ of an event $$P$$ with respect to the origin of their reference frame.

(2) In a given frame, all observers' clocks are synchronized using some method (such as Einstein synchronization).

Consider two observers in the $$S$$-frame. Let observer $$A$$ be located at $$(x, y, z) = (x_A,y_A,z_A)$$. Let observer $$B$$ be located at $$(x, y, z) = (x_B,y_B,z_B)$$. Assume an event $$P$$ occurs at some other point in the $$S$$-frame. By rule (1), both observers $$A$$ and $$B$$ will agree on the spatial position of $$P$$, say $$(x, y, z) =(x_P, y_P, z_P)$$. However, since the speed of light is a finite constant in the inertial frame $$S$$, observers $$A$$ and $$B$$ will disagree on the time of the event P, as $$A$$ and $$B$$ are located at different positions and their clocks are synchronized by rule (2). Hence, we have that $$A$$ measures the spacetime coordinate of $$P$$ to be $$(t, x, y, z) = (t_A, x_P,y_P, z_P)$$, while observer $$B$$ measures the spacetime coordinate of $$P$$ to be $$(t, x, y, z) = (t_B, x_P,y_P, z_P)$$. Here, $$t_A \neq t_B$$.

Now we consider the $$S'$$ frame. Let observer $$A'$$ be located at $$(x',y',z') = (x_A',y_A',z_A')$$. Here, $$x_A = x_A', \; y_A = y_A', \; z_A = z_A'$$. Let observer $$B'$$ be located at $$(x',y',z') = (x_B',y_B',z_B')$$. Here, $$x_B = x_B', \; y_B = y_B', \; z_B = z_B'$$. In other words, $$A$$ and $$A'$$ and $$B$$ and $$B'$$ are located at the same position relative to their origins, (i.e., they "correspond").

Now surely, the Lorentz transformations would relate the measurements of the spacetime coordinates of the event $$P$$ made by $$A$$ and $$A'$$. They would also relate the measurements made $$B$$ and $$B'$$. However, those same transformations could not relate the measurements made by $$A$$ and $$B'$$ measurements or the measurements made by $$B$$ and $$A'$$ right? This is what I mean when I say that the Lorentz transformation only applies to corresponding observers.

• Your phrase "corresponding observers" is not one that has a commonly understood definition. That's why I didn't use it in the title I substituted for your original title.
– user4552
Nov 29 '19 at 15:33
• @ Ben Crowell I have edited my question. Nov 29 '19 at 16:53
• (1) In a given frame, all observers agree to measure the spatial coordinates (x,y,z) of an event P with respect to the origin of their reference frame. [...] By rule (1), both observers A and B will agree on the spatial position of P, say (x,y,z)=(xP,yP,zP). By this, do you mean that $x_A=x_B$, etc.? If so, then this is wrong. Rule 1 doesn't imply this.
– user4552
Nov 29 '19 at 17:57
• @ Ben Crowell I agree that rule (1) does not imply that $x_a = x_b, y_a = y_b, z_a = z_b$ etc. However, I am confused as to why rule (1) does not imply that both observers will measure the same spatial position of $P$. Nov 29 '19 at 18:35
• I think the fundamental issue is a conceptual confusion. You use the words "same spatial position," and insist that observers should agree on that. SR does not have an absolute notion of "same place," and in fact Galilean relativity lacks this as well. Secondarily, you're overcomplicating things by having separate notions of observers and Minkowski coordinate bases. These can be treated as being the same thing. Defining a Minkowski coordinate basis defines an observer, and vice versa.
– user4552
Nov 29 '19 at 19:33

I see two errors in your reasoning. First, on S, even if the light signal reaches A and B at different times, the times measured are the same, the observed time of arrival of the signal is corrected by the observer because he know this delay is d/c, where d is the distance to the event. Second, it is false that $$x_A = x_A', \; y_A = y_A', \; z_A = z_A'$$ in general. See the lorentz transformation at https://en.wikipedia.org/wiki/Lorentz_transformation.

So the notion of corresponding observers is ill defined. Using the Lorentz transformations you can relate the observations between any observers, just plug the corresponding coordinates.

• @ Wolphram jonny "Second, it is false that $𝑥_A=𝑥′_A, 𝑦_A=𝑦′_A ,𝑧_A=𝑧′_A$ in general. See the lorentz transformation..." You're absolutely right. There is no stipulation in the Lorentz transformations that $𝑥_A=𝑥′_A, 𝑦_A=𝑦′_A, 𝑧_A=𝑧′_A$. However, you say that " the times measured are the same, the observed time of arrival of the signal is corrected by this delay." How would this delay correct the observed time? Nov 29 '19 at 19:54
• This answer is correct. The observers A and B agree on the time of event P since they are all using synchronized clocks.
– Dale
Nov 29 '19 at 20:26
• @ Dale If the clocks are synchronized, then they read the same time in general right? So if the light signal reaches $A$ and $B$ at different times, how could they agree on the time of the event $P$? Nov 29 '19 at 20:34
• The observer knows the distance d to the event and so he know the event happened at t-d/c. Of course, the observers infers the time based on the time of detection, the clock itself does not correct anything
– user65081
Nov 29 '19 at 20:39
• @ Wolphram jonny Ah I see! Thank you so much! Nov 29 '19 at 20:47