Learn Special Relativity: Understanding Graphs for X-T System

[jaumzaum]

Hello guys!
I just started to learn Special Relativity though a Stanford youtube channel, and I had some problems already in the first class
The teacher drew a graph with one spatial dimension (x-axis) and one temporal dimension (t-axis). Where X is the horizontal axis, T is the vertical, both ortogonal. We measure t in seconds and x in light-seconds, so that the line x=t is the light path. Then we introduce a new reference frame. The frame consistis of a person moving to the right with speed v. We prove that the x'and t'axis have the forms written below and we plot them in the graph. Until then, everything fine!

Now we consider the following experiment: I (the one who is moving to the right) have a clock and see in my clock that it has passed 1 second. So, for me, t'=1. When I use the below equations to calculate the time measured by the observer in the X-T system, I find that t=Lt', where L is the Lorentz factor. So t>t'. When I look to the graph, it clearly shows that the "size" of t is smaller than the "size" of t'. The only explanation I have for that is that maybe the graphs don't have the same scale. Is that right? If so, does it really worth it to do graphs like these, if we cannot trust in the size of things? Also, is there a graph that can be constructed in scale between both observers? And lastly, if the graph is not in scale, is there a geometry method to compare the sizes (i.e. find out if t is greater than t' looking to the graph)?

 

[PeterDonis]

When I look to the graph, it clearly shows that the "size" of t is smaller than the "size" of t'

The fact that you put "size" in quotes here suggests that you realize it is not the real, physical magnitude of what is being represented. You're right, it isn't. The "size" you are referring to is the Euclidean length of the line segments in question. But the geometry that is being represented in the diagram is not Euclidean geometry. So it is not possible for all of the Euclidean lengths in the diagram to match the real, physical magnitudes of the things the diagram is representing. That's why you have to calculate those magnitudes using the Lorentz transformations.

 

[PeterDonis]

is there a geometry method to compare the sizes (i.e. find out if t is greater than t' looking to the graph)?

Yes, but to do this you have to draw more curves on the graph. The geometry that the graph is representing is hyperbolic geometry; that means that the set of points that have a real, physical magnitude of $1$ for their proper time from the origin (i.e., an observer whose worldline was a straight line from the origin to any such point would measure 1 unit of time by his clock between the origin and that point) is a hyperbola, whose asymptotes are the lightlike lines $x = t$ and $x = -t$. If you draw such a hyperbola passing through the point you labeled A, with $t' = 1$, its intersection with the $t$ axis will be the point on that axis with $t = 1$.

 

[PeroK]

Yes, good question. The geometry of the these spacetime diagrams is not Euclidean. They are pseudo-Euclidean. The spacetime distance from the origin, say, the a point $(t, x)$ is given by $s = \sqrt{t^2 - x^2}$ and not $s = \sqrt{t^2 + x^2}$ (as would be normal in Euclidean geometry). This is all in the original unprimed frame.

If we look at some spacetime distances (from the origin):

First, to the point $(t, 0)$, we have $s = t$.

Next, to the point $(t, vt)$, we have $s = \sqrt{t^2 - v^2t^2} = \frac{t}{\gamma}$. Note that the diagonal line represents a shorter spacetime distance!

Now, to the point $(t, t)$ on the light path: $s = 0$. This path, characteristic of all light paths, has zero spacetime distance (it's called a "null" path).

Finally, to the point $(1, 2)$, which is outside the light path: $s = \sqrt{-3}$. This is called a "spacelike" path. Normally we redefine things here for spacelike paths to say that $s = -\sqrt{3}$.

I agree with you, there are subtleties in spacetime geometry.

 

[David Lewis]

The unit length on a primed axis = √(1+β²)/(1-β²)

 

[PeterDonis]

The unit length

For this to be useful, you have to explain what you mean by "the unit length", since the whole point of the discussion is that this term does not have a unique meaning in this context.

 

[pervect]

I'm not sure I understand your question, but there is an interesting geometric technique that says that the "area" of a light clock, when plotted on a space-time diagram, is constant when the diagram is scaled so that the slope of a light ray is 45 degrees.

The light clock represents both a unit of proper time, and a unit of proper distance, in a convenient graphical form. Light clocks are commonly discussed, but I can't be positive you're aware of them.

It's possible that this interesting result about light clocks may not be the question you're trying to ask. I can't really tell from the way your question is worded, unfortunately.

For more detail on the light clock, assuming you know what they are, see "Relativity on Rotating Graph Paper". You can do a search for physics forums, the arxiv version is at https://arxiv.org/abs/1111.7254, the version published in the American Jouranal of Physics is slightly different than the arxiv version, and is paywalled.

If you're not familiar with light clocks, this may not help, unfortunately.

 

[David Lewis]

...you have to explain what you mean by "the unit length"...

The distance on the paper between tick marks would have to be greater on the primed axes than on the unprimed axes in order to keep the graph to scale.

 

[PeterDonis]

The distance on the paper between tick marks

Meaning, the Euclidean distance, I take it?

 

[jaumzaum]

@PeterDonis @PeroK @pervect @David Lewis Thanks you all for the responses!
I will try to learn more about non Euclidean spaces. I thought I had enough mathematical knowledge to understand the Physics in Special Relativity but I never had a subject covering non Euclidean geometry, and I think that is the key to my question.

@PeroK The equations you provided are valid only for the primed axis or for both axis?

Also, is this type of diagram the best (I.e. the most didactic) to represent a 2 frames in Special Relativity? I think @PeterDonis already answered this when he said it’s impossible to have all lengths to be physical and mathematical “equally sized” in a non Euclidean space, but I am wondering if there is another type of diagram or interpretation that could express the same information, with a different geometry so that it would be easier to comprehend the relation between the axis (even though they are not Euclidean). I think that @pervect was trying to answer that, but I am not really sure.

Also,I am sorry for my English :/
Lastly, are there some good materials that you would suggest me to read to learn more about Special Relativity?

 

[robphy]

So t>t'. When I look to the graph, it clearly shows that the "size" of t is smaller than the "size" of t'. The only explanation I have for that is that maybe the graphs don't have the same scale. Is that right? If so, does it really worth it to do graphs like these, if we cannot trust in the size of things? Also, is there a graph that can be constructed in scale between both observers? And lastly, if the graph is not in scale, is there a geometry method to compare the sizes (i.e. find out if t is greater than t' looking to the graph)?

Note that you have the analogous problem in ordinary [Galilean] position-vs-time diagrams,
where the t- and t'-axis are along the worldlines of inertial objects.
In Galilean case, you have t=t'
even though the "[Euclidean] size" of t (OB) is smaller than the "[Euclidean] size" of t' (OA).

As others have mentioned, the Euclidean size specified by a circle is
not appropriate for spacetime diagrams or Galilean position-vs-time diagrams.
For Spacetime diagrams in special relativity, you have hyperbolas as "the curves of constant separation".

You could use "hyperbolic graph paper" analogous to a polar-coordinate graph paper.
But that possibly unfairly distinguishes a point, especially if you want to turn (as in the clock effect/twin paradox).

You could use a "hyperbolic" analogue of a compass (used to draw circles in Euclidean geometry).Here is a visualization of three methods to show that
OB, the apparent elapsed time between O and A according to the lab frame
(since the lab frame along OB says that B is simultaneous with A)
is longer than
OA, the elapsed time recorded on the wristwatch of the inertial astronaut traveling from events O to A.

1. Draw concentric hyperbolas centered at the separation event O.
   The hyperbola corresponding to OB (the rest frame) is larger than the hyperbola for OA (traveler frame).
   So, OB > OA, where BA is simultaneous according to OB since the tangent line at event B meets event A.
   
   
   
   (At the intersection of the line along OA and the fifth hyperbola,
   if you draw the tangent line, it will intersect the line along OB at its intersection with the fourth hyperbola.
   This is a demonstration that time-dilation is symmetric between inertial observers that met briefly at O.)
2. This refers to the method described by @pervect above.
   Using rotated graph paper to help determine the area of "causal diamonds", observe that the causal diamond with timelike diagonal OB (rest frame) has an area larger than that of the diamond with OA.
   So, OB > OA, where BA is simultaneous according to OB since the spacelike diagonal of OB's diamond is parallel to BA. (The rotated grid represents "light-clock diamonds" which are traced out by light-signals in an observer's light -clock. In those units, area of the causal diamond of OB is equal to the square of the number of light-clock diamonds along one diagonal. Here OB^2=(5)^2 and OA^2=(4)^2. )
   
   
   
   The causal diamond encodes a lot the geometry of Minkowski spacetime.
   The area is measure of the square-interval of one of its diagonals.
   The aspect-ratio (width/height,
   with width along the future-frontward-lightlike vector
   and height along the future-backward-lightlike vector)
   is equal to the square of the Doppler factor (an eigenvalue of the the Lorentz transformation)
   encodes the velocity along the diamond's timelike diagonal.
   
   By varying the aspect-ratio while keeping event O fixed and the area constant (i.e. a Lorentz boost transformation), the opposite corner traces out a hyperbola.
   Thus, the family of causal diamonds with one corner at O and the other corner of the fifth hyperbola all have area equal to (5)^2 grid diamonds.
3. If you construct the ticks using Light-Clock Diamonds along the axes (with the help of rotated graph paper, where all ticks have the same area), you'll see
   that OB >OA.
   
   
   
   
   The construction is symmetric for the traveling observer.
   
   

For more information, you can consult my PhysicsForums Insights:
https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/

 

[PeroK]

@PeroK The equations you provided are valid only for the primed axis or for both axis?

Lastly, are there some good materials that you would suggest me to read to learn more about Special Relativity?

Those calculations were done for coordinates in the unprimed frame. You don't need a separate course in non-Euclidean geometry. You can learn the geometry as you learn SR and spacetime diagrams.

The first chapter of Morin's textbook is online here:

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

 

[vanhees71]

Perhaps, this helps too:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[eltodesukane]

see this video: