# Lorentz transformations in Minkowski space including invariants

Let's assume the reference frame of a still observer with the 2 axes: the x-axis and time axis t. Let's say there's another observer within that frame moving with a constant velocity v with respect to the first observer (who's still). The two transformed coordinates for this moving observer are given by x' and t' in a way (by Lorentz transformation):

$$x' = \frac{x-vt}{\sqrt{1-v^2}}$$ ... (1.1)

$$t' = \frac{t-vx}{\sqrt{1-v^2}}$$ ... (1.2)

assuming relativistic units.

Now, we know that $$\tau^2 = t^2-x^2 = t'^2 - x'^2$$ is an invariant called proper time. Let's say, for the moving observer, we move along his world line i.e. $$x' = 0$$. If that's true, we get $$\tau^2 = t'^2$$ or $$\tau = t'$$.

Similarly, if we move along $$x=0$$ i.e. the world line of the still observer in the same RF, we get $$\tau^2 = t^2$$ or $$\tau = t$$.

Since $$\tau$$ is an invariant, it doesn't change in the same reference frame. Clearly, from the above 2 calculations we get that $$\tau = t = t'$$ which is obviously not true since t $$\neq$$ t' (shown in equation 1.2).

I feel like I'm missing something rather obvious, something fundamental to the idea of Lorentz transformations, could someone please point that out?

• Are you sure the world line of still observer in the same RF is x=0? In the frame of moving person, wouldn't the stationary one be moving? Aug 12, 2022 at 7:33
• Yes, as x=0 is the definition of the 'still' frame that I've assumed. Aug 12, 2022 at 7:34
• Okay, I think I see where your confusion is Aug 12, 2022 at 7:36

## 1 Answer

Here is how the set up looks like:

We have the real world and three charts (osbervers viewing) it. The observer is denoted by it's reference frame (I am using Schutz's definition). In SR, we are interested in the chart transition maps, the one which relate the coordinate in one frame to another.

Now, there is a bit of trickery going on here conventional SR presentation use a bit of abuse of notation of these variables. They say use that $$(x,t)$$ is the coordinates of the trajectory of a particle in space time and also that it is coordinate of the whole grid. I prefer to be more careful and if I am to talk about the trajectory of a certain particle, I will suffix my variables with a letter.

For the case here, we study the motion of two particle $$A$$ and $$B$$ in the real world (2D space time manifold) in these three frames.

You have written:

Similarly, if we move along $$x=0$$ i.e. the world line of the still observer in the same RF, we get $$\tau^2 = t^2$$ or $$\tau = t$$.

To understand what went wrong we begin at the equation $$\tau^2 = t^2 - x^2= t'^2 - x'^2$$

In the above equation, the space time interval is invariant, if you plug in the points from the $$t-x$$ plane corresponding to the points in the $$t'-x'$$ plane by the Lorentz transformation.

The points on the line $$x=0$$ and the line $$x'=0$$ are not necessarily the same space time points so it doesn't make sense to compare their time.

The reason it looks the same is that if you are in a path where the $$x$$ coordinate doesn't change then proper time = normal time.

• I tried my best to explain it :/ I suck at explaning so Aug 12, 2022 at 8:13
• Ah, got it. Thanks a ton, and no, the explanation made it clear! Aug 12, 2022 at 8:58
• I can recommend you one good. Check out Rindler's Relativity, Special, Astro and General book. @UtkarshJetly Aug 12, 2022 at 14:50