# B Need help on a Lorentz Geometry Diagram

#### berlinspeed

Summary
As described.
Can someone tell me why the proper time between αβ is not t2+x2 but rather
t2-x2? Background: Last edited:
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#### Ibix

Because this is Minkowski geometry, not Euclidean.

Lorentz, by the way, not Lorenz. Lorenz was a different mathematician and physicist.

#### berlinspeed

Because this is Minkowski geometry, not Euclidean.

Lorentz, by the way, not Lorenz. Lorenz was a different mathematician and physicist.
Yes but why? Can you derive it?

#### Ibix

See any textbook on relativity. They all derive the invariant interval.

#### Pencilvester

Try constructing an inertial coordinate system $(x’ , t’)$ with the following criteria: events A and B both have $x’_A =x’_B =0$, and the slope of the line connecting P, Q, and the origin in the primed coordinates is equal to $- x_B / t_B$ (in the unprimed coordinates), and the light rays connecting P to B and B to Q keep their slopes of 1 and -1 respectively. When you do this, which of these quantities remains invariant despite using different coordinates: $t^2 + x^2$ and $t’^2 + x’^2$, or $t^2 - x^2$ and $t’^2 - x’^2$?

#### berlinspeed

Try constructing an inertial coordinate system $(x’ , t’)$ with the following criteria: events A and B both have $x’_A =x’_B =0$, and the slope of the line connecting P, Q, and the origin in the primed coordinates is equal to $- x_B / t_B$ (in the unprimed coordinates), and the light rays connecting P to B and B to Q keep their slopes of 1 and -1 respectively. When you do this, which of these quantities remains invariant despite using different coordinates: $t^2 + x^2$ and $t’^2 + x’^2$, or $t^2 - x^2$ and $t’^2 - x’^2$?
I failed to show it graphically..

#### Nugatory

Mentor
Yes but why? Can you derive it?
Yes. If you play with the Lorentz transformations some, you’ll be able to show that they preserve the quantity $\Delta{t}^2-\Delta{x}^2$ but not $\Delta{t}^2+\Delta{x}^2$. Transforming coordinates doesn’t change any of the real directly observable physics, so this is a pretty strong hint that the former and not the latter quantity is the physically significant one.
We can see that when $\Delta{t}$ is zero the two points we’re considering are the positions of the two ends of an object at the same time, so clearly that $\Delta{t}^2-\Delta{x}^2$ expression is giving us the length of that object when it is at rest in whatever frame we're using. Similarly when $\Delta{x}$ is zero we're considering the difference between two consecutive readings of a clock that is at rest in whatever frame we're using. Thus it makes sense to interpret this invariant quantity as the distance between the two events.
(Be aware that this is a somewhat oversimplified summary of what @Pencilvester said above and what @Ibix correctly says will be any intro textbook).

Unfortunately the surface of the sheet of paper on which we draw our spacetime diagrams is Euclidean, and that means that the drawing cannot accurately represent distances. These diagrams are enormously useful for showing how different events are related to one another: you can easily pick out the worldlines of objects that are moving inertially so at rest in some inertial frame; if you're careful with your axes you can see which events happen before, after, or at the same time as other events; these diagrams are by far the best way of visualizing the x and t coordinates; you can make sense of the pole-barn paradox, the twin paradox, and many of the other classic paradoxes of special relativity; and they are an essential starting point for general relativity.

But the surface of the paper is still Euclidean not Minkowski, so there's one thing that doesn't work. You cannot plop a ruler down on the surface of the paper and measure the distance between two points. That will give you the Euclidean distance, which isn’t right for spacetime.

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• berlinspeed

#### Nugatory

Mentor
I failed to show it graphically..
You can't. You're drawing your graph on a sheet of paper that obeys Euclidean geometry so the Pythagorean theorem works and the distance between points is given by $\Delta{s}^2=\Delta{t}^2+\Delta{x}^2$ and that's not the right relationship between distance and displacement in Minkowski geometry.

You can use the graph to show the relationship between the various intervals (that's what MTW are doing in the image you posted), but you can't make the distances come out right except in a few special cases (and it would be a good exercise to figure out which these are).

#### vanhees71

Gold Member
Have a look at the first chapter of

#### berlinspeed

You can't. You're drawing your graph on a sheet of paper that obeys Euclidean geometry so the Pythagorean theorem works and the distance between points is given by $\Delta{s}^2=\Delta{t}^2+\Delta{x}^2$ and that's not the right relationship between distance and displacement in Minkowski geometry.

You can use the graph to show the relationship between the various intervals (that's what MTW are doing in the image you posted), but you can't make the distances come out right except in a few special cases (and it would be a good exercise to figure out which these are).
Hi do you mind showing me the special cases where the distances come out right? I'm still wrapping my head around it..

#### Nugatory

Mentor
Hi do you mind showing me the special cases where the distances come out right? I'm still wrapping my head around it..
I don't mind, but I'm going to give you another day to think about it first. There is a very big hint in post #7.

#### berlinspeed

I don't mind, but I'm going to give you another day to think about it first. There is a very big hint in post #7.
Okay I see now with the help of Minkowski diagram, what I'm seeing is that when a particle is at rest in the primed frame, its $\Delta t$ is exactly the line segment in the unprimed frame given by the Pythagorean theorem $\Delta x^2 + \Delta t^2$ however scaled differently in the primed frame (its value equals to $\Delta t^2 - \Delta x^2$). So my question now is why and how is $\Delta t^2 - \Delta x^2$ invariant mathematically in the Minkowski space?

#### Nugatory

Mentor
Okay I see now with the help of Minkowski diagram, what I'm seeing is that when a particle is at rest in the primed frame, its $\Delta t$ is exactly the line segment
Yes, in that case and it also works when $\Delta{t}$ is zero.
So my question now is why and how is $\Delta t^2 - \Delta x^2$ invariant mathematically in the Minkowski space?
Because the x and t coordinates in one frame are related to the x and t coordinates in another frame by the Lorentz transformations. It's just an exercise in algebra (one that you'll find in any relativity textbook) to show that these transformations change x to x' and t to t' in such a way that $\Delta{t}^2-\Delta{x}^2=\Delta{t'}^2-\Delta{x'}^2$. Thus your question comes down to asking why the Lorentz transformations describe how the universe we live in works (if they didn't we wouldn't care about them or Minkowski space).

There are two mathematically consistent possibilities for how our universe works:
1) There is some speed that is the same in all frames and the x and t coordinates of different frames are related by the Lorentz transformations (where the constant $c$ is that invariant speed).
2) There is no speed that is the same in all frames and and the x and t coordinates of different frames are related by the Galilean transformations.

Experiments have proven far beyond any reasonable doubt that #1, with light moving at that invariant speed (which is why we call $c$ the speed of light) is how the universe works. But WHY it is #1 instead of #2? Physics, as an experimental science, cannot answer that question.

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• rrogers and vanhees71

#### berlinspeed

Yes, in that case and it also works when $\Delta{t}$ is zero.Because if the x and t coordinates in one frame are related to the x and t coordinates in another frame by the Lorentz transformations. It's just an exercise in algebra (one that you'll find in any relativity textbook) to show that these transformations change x to x' and t to t' in such a way that $\Delta{t}^2-\Delta{x}^2=\Delta{t'}^2-\Delta{x'}^2$. Thus your question comes down to asking why the Lorentz transformations describe how the universe we live in works (if they didn't we wouldn't care about them or Minkowski space).

There are two mathematically consistent possibilities for how our universe works:
1) There is some speed that is the same in all frames and the x and t coordinates of different frames are related by the Lorentz transformations (where the constant $c$ is that invariant speed). There is no speed that is the same in all frames and and the x and t coordinates of different frames are related by the Galilean transformations.

Experiments have proven far beyond any reasonable doubt that #1, with light moving at that invariant speed (which is why we call $c$ the speed of light) is how the universe works. But WHY it is #1 instead of #2? Physics, as an experimental science, cannot answer that question.
But why is light speed the same for all observer? I would think about Maxwell's equations give speed of light as the speed that the EM fields are propagating, but nowhere it gives satisfactory answer there..

#### Ibix

But why is light speed the same for all observer?
Depends on what you mean. The invariance of the speed of light was a postulate of relativity. Modern "one postulate" approaches show that an invariant speed - either finite or infinite - is a consequence of the principle of relativity, and experiment shows we live in a universe that has a finite invariant speed (the infinite invariant speed one is a Newtonian universe). In either case there isn't a "why", really. We just explore the consequences of it being true and find that they are an accurate description of our universe.

The other thing you could mean is why does light travel at the invariant speed. That's because that's what massless things do in a relativistic universe. Which isn't a terribly profound "why" but, is there at least.

• Nugatory and vanhees71

#### Nugatory

Mentor
But why is light speed the same for all observer? I would think about Maxwell's equations give speed of light as the speed that the EM fields are propagating, but nowhere it gives satisfactory answer there..
The historical flow of discovery was more or less:
We can calculate the speed of light directly from Maxwell's equations, and we get the answer $\frac{1}{\sqrt{\epsilon_0\mu_0}}$ where $\epsilon_0$ and $\mu_0$ are physical constants. The speed of the observer does not appear in this result so should not affect it; if everyone is using the same laws of electricity and magnetism they should find the same speed of light even if they are moving relative to one another. Thus, it is reasonable to adopt the postulate that the speed of light is the same for all observers and see where that postulate takes us. Einstein did this in his 1905 paper and it took him to the Lorentz transformations and the rest of relativity; the experimental evidence since then proves that he got it right.

However, that doesn't really answer your "why?" question. Yes, if we accept Maxwell's equations we are naturally led to the the invariant speed of light, Lorentz transformations, and the rest of relativity - but that just begs the question of why electromagnetism obeys Maxwell's equations and not some other equations that would take us somewhere else. There's no way that empirical science can answer that question: experiments will tell us what the laws of physics are, but not why we have these laws and not some other laws. Eventually we're left with no better answer than "Because that's the rules in the only universe we have".

[Edit: As @Ibix suggests above, there is a more modern route to the same conclusion, some discussion starting at post #5 of https://www.physicsforums.com/threads/derivation-of-the-lorentz-transformations.974098/. However, it doesn't really help with your "why?" question, it just leads to the unanswerable question of why we have an isotropic universe instead of some other]

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• vanhees71 and m4r35n357

"Need help on a Lorentz Geometry Diagram"

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