# Space-Time Diagram in Special Relativity (for S and S')

So, this is how it is typically done. We have a reference frame $$S$$ and the coordinates $$(x,t)$$ of an event, $$E$$ (or collection of events which we call world-lines) are mapped onto a 2-dimensional physical paper. (Typically $$x$$ is horizontal axis and $$t$$ is vertical axis and the two axis are orthogonal)

Now we analyse the same "event-space" (i.e. event $$E$$ and the world-lines etc) with respect to another reference frame $$S'$$ which is moving relative to $$S$$. We need new set of co-ordinates for $$S'$$ to "measure" the event-space (i.e to mark an event on the piece of paper with respect to $$S'$$). We get the new co-ordinates for $$x'$$ and $$t'$$ which are just "rotation" (in opp directions) of original axis (rotation depends on relative speed between $$S$$ and $$S'$$) and $$x'$$ and $$t'$$ are not orthogonal.

My question: We in the above construction implicitly assume that we do not need to "re-draw" the events and world-lines on the paper. The events and world-lines we marked in the $$S$$ co-ordinate system will continue to be on the same "physical" point (or line if world-line) on the paper in the new co-ordinate system of $$S'$$.

Why is above assumption acceptable? Why are we guaranteed that by mere rotation of the co-ordinate axis we will get the co-ordinate system for $$S'$$ in which the physical layout on the sheet of paper of the event-space remains same as it was in $$S$$ co-ordinate system? Is this unique to special-relativity?

• Note, that the new coordinates, S', are not a rotation but a Lorentz transformation. The will kinda squeeze both x' and t' closer to 45 degrees, or stretch them both away from 45 degrees (in units where c=1). Commented May 25, 2020 at 14:38
• yes true - I edit my Q Commented May 25, 2020 at 14:42

The construction of the S'-axes for Minkowski spacetime is analogous to the S'-axes for Euclidean geometry. So, it is not "unique to special relativity".

In other words, first ask your questions about Euclidean geometry [and how you might prove it] then see if the answers can be applied to Minkowskian geometry by analogy (possibly with some generalizations of some concepts).

Euclidean geometry uses a unit-circle.
The (solid red) tangent-line to the circle is orthogonal to the (dotted) radius (along the t'-axis).
Through the origin, draw the x'-axis as the parallel to this tangent line.
The x'-axis is orthogonal to the t'-axis.

Minkowskian geometry [for special relativity] uses a unit-hyperbola as its unit-"circle".
The (solid red) tangent-line to the "circle" is [Minkowski-]orthogonal to the (dotted) radius (along the t'-axis).
Through the origin, draw the x'-axis as the parallel to this tangent line.
Note that, under the implied Lorentz transformation, this x'-axis looks-from-a-Euclidean-viewpoint "oppositely rotated" to the t'-axis.
The x'-axis is [Minkowski-]orthogonal to the t'-axis.

Galilean geometry [for Galilean relativity] uses a horizontal line as its unit-"circle".
The (solid red) tangent-line to the "circle" is [Galilean-]orthogonal to the (dotted) radius (along the t'-axis).
Through the origin, draw the x'-axis as the parallel to this tangent line.
Note that, under the implied Galilean transformation, this x'-axis looks-from-a-Euclidean-viewpoint "not rotated".
The x'-axis is [Galilean-]orthogonal to the t'-axis.

Visit my visualization:
https://www.desmos.com/calculator/r4eij6f9vw
try the E-slider to see Euclidean (E=-1), Minkowski (E=+1), and Galilean (E=0).

• Thanks. I guess I'm not well trained to understand these geometries well. Hence unable to fully understand your answer. I guess I need to read up a little bit about these different geometries. Thanks Commented May 25, 2020 at 14:23
• If I can ask a question please - When you say X-geometry uses Y as it's "unit circle"m what do you mean by "unit circle" in this context? Thanks Commented May 25, 2020 at 14:36
• @aman_cc "Circle" means "curve of constant distance [interval]" from a point O. A "unit circle" is for "odometer reading = 1" for the Euclidean displacement along a radial line from O and "wristwatch reading = 1" for the spacetime displacement along an inertial worldline from O. Commented May 25, 2020 at 14:41
• Thanks. Any reference on net or book I can read a bit more about this. I'm not really well trained in these topics, Thanks Commented May 25, 2020 at 14:45
• @aman_cc Search for "Geometry of Special Relativity" by Tevian Dray. Commented May 25, 2020 at 15:02

"Why ...?" By construction. The way we plot the axis of the second coordinate frame (wiki), i.e. the way we map $$S'$$ onto the piece of paper, is chosen in such a way as to allow for every point allready plotted on paper, as being in $$S$$, to also be the corresponding event in $$S'$$.

You may ask wether such construction is non-contradictory, i.e. whether you can in fact build such a construction. From experience we know that it is. I am not a mathematician, but I would assume that this is because Lorentz transformations are bijective (one-to-one and onto).

You could think of other maps that do not have this property. For example, in 1D a map $$x'=x^2$$ for $$x\in\mathbb{R}$$ will map all real numbers, positive and negative, into non-negative numbers. In this example, your $$S'$$ would have no 'presence' in $$x'<0$$.

• @Cyro - Yes my question ultimately is as you say - whether such a construction is possible? If yes what are the properties of transformation that are necessary etc. yes as you likely say- this is more of maths now. I was trying to understand how those mathematical properties link to postulates of special relativity, Commented May 25, 2020 at 14:20

The reason we construct a mathematic representation where we describe the same physics in two different coordinate systems is because we want a mathematical description that obeys the Equivalence Principle. Essentially, the fundamental idea of relativity is that physics works the same way for all intertial frames of reference. It would be very strange if you and I observed an entirely different particle world line from different reference frames (much stranger even than strange things like time dilation). We may describe that world line differently (by using different coordinates) but the actual physical behavior is whatever it is, independent of the observer's frame.

We want a way to mathematically describe both what different observers will see differently, and what will be invariant. A very tidy and concise representation of this is to take word lines as physically real and notice how their components change under a particular coordinate transformation. We can only know that this formalism describes the world through observation.

• Hi dont fully agree. To draw world-lines having a co-ordinate system is pre-requisite. In-fact that is the basis of my question - How are we sure that when we change the co-ordinate system the same points on paper will represent the same world-line in the new co-ordinate system (of S'). To me it is clearly no-coincidence and hence must follow from postulates of special relativity and hence its mathematics. Commented May 25, 2020 at 14:53
• Yes, to draw the world line in the first place, you use some coordinate system. Then to represent what an observer who is moving relative to the first coordinate system, we discover that we can do this by applying a Lorentz transformation to the coordinate system. It didn't have to be the case that the universe worked in this way. But it is observed to be so. The postulate from SR that is represented by this is the Equivalence Principle. Commented May 26, 2020 at 14:15