Definition of Spacetime Interval

[greg_rack]

Hi guys,
I'll attach an excerpt from my textbook which isn't, in my opinion, very clear in explaining a spacetime interval(or I'm just missing the key to get the concept).

"How do we combine two different measurements such as time and space, to form an invariant variable? We can simply write 't=x/v 'and 'x=vt'; in other words, space and time are interchangeable using velocity as a unit [are they? Rearranged as such, isn't $t$ measured in $v^{-1}$ and $x$ in $v$?, ed.].
Let's introduce a conversion velocity 'c'. We can now measure time in meters, taking any time interval and multiplying it by the conversion velocity(t=ct) [how is that??, ed.]."

Now the book points out that, for calculating the squared hypotenuse $s^2$ of the triangle with legs $ct$(still didn't get the sense of writing a time as $ct$, nor the connection between this and the anticipation of rewriting $t$ and $x$ in terms of $v$) and $x$, we cannot use the basic Pythagorean theorem, but its hyperbolic version $s^2=(ct)^2-x^2$.
But my second question is: if the locus is a hyperbole, how could $s$ be invariant? Isn't it always different, depending on the frame of reference from which I measure $x$ and $t$(points in the curve haven't the same distance from the origin)?!
Shouldn't it be something like a semi-circle, for $s$ to be constant?

I know I'm a mess, but at least I hope I was somewhat clear in explaining my doubts... unfortunately I'm studying such topics on my own, and sometimes I even start doubting that 2+2=4 :)

 

[PeroK]

I think you are asking too many questions before the author gets to the point. You need to have the Lorentz transformation as well as the concept of hyperbolic geometry.

 

[greg_rack]

I think you are asking too many questions before the author gets to the point. You need to have the Lorentz transformation as well as the concept of hyperbolic geometry.

I have studied Lorentz transformation, but no hyperbolic geometry.
Would it be enough to at least grasp the invariant interval?

 

[PeroK]

I have studied Lorentz transformation, but no hyperbolic geometry.
Would it be enough to at least grasp the invariant interval?

The idea is that if you have: $$t' = \gamma(t - \frac{vx}{c^2}), \ \ x' = \gamma(x - vt)$$ then $$(s')^2 = (ct')^2 - (x')^2 = \dots = (ct)^2 - (x)^2 = s^2$$ and we have an invariant spacetime length using the hyperbolic formula.

I guess you could do it the other way round and assume $(s')^2 = s^2$ and show that the Lorentz Transformation holds.

 

[Sagittarius A-Star]

I have studied Lorentz transformation, but no hyperbolic geometry.

The Lorentz transformation satifies the invariance of the spacetime interval, see equations (8), (8a) and (11a):
https://en.wikisource.org/wiki/Rela...mple_Derivation_of_the_Lorentz_Transformation

 

[greg_rack]

The idea is that if you have: $$t' = \gamma(t - \frac{vx}{c^2}), \ \ x' = \gamma(x - vt)$$ then $$(s')^2 = (ct')^2 - (x')^2 = \dots = (ct)^2 - (x)^2 = s^2$$ and we have an invariant spacetime length using the hyperbolic formula.

I guess you could do it the other way round and assume $(s')^2 = s^2$ and show that the Lorentz Transformation holds.

So $s^2$ is $(ct)^2-x^2$ by mere definition?

It's simply a quantity that is proven to be "untouched" by Lorentz trans.? Or is there a more profound meaning/derivation?

 

[PeroK]

So $s^2$ is $(ct)^2-x^2$ by mere definition?

It's simply a quantity that is proven to be "untouched" by Lorentz trans.? Or is there a more profound meaning/derivation?

Yes, by definition; and, yes invariant under Lorentz transformations. That is quite profound!

 

[greg_rack]

Great!
I got the fact that a spacetime interval is invariant under LTs.
But above all: what is a spacetime interval? How do we define it, how do we "merge" two diverse quantities such as length and time?
How do we get to define it(distance between two points in a $x-t$ graph) as $\sqrt{(ct)^2-x^2}$?

 

[Nugatory]

It's simply a quantity that is proven to be "untouched" by Lorentz trans.? Or is there a more profound meaning/derivation?

It's also a quantity that has proven to be very useful, and it has physical significance that becomes apparent if you consider the two special cases $\Delta x=0$ and $\Delta t=0$. It's by far the cleanest way of explaining why the twin paradox isn't a paradox after all, and is essential to understanding General Relativity.

The invariance of the spacetime interval implies the Lorentz transformations and vice versa, so we can consider either one to be the more interesting/profound/fundamental thing from which the other is derived. Special relativity usually starts from the Lorentz transformations and then derives the spacetime interval formula, perhaps using it in an explanation of the twin paradox. When you're learning general relativity, you'll more likely start with the assumption that there is an invariant spacetime interval between any two events; the Lorentz transformations of special relativity appear when you apply the mathematical methods of GR to the special (that's why it's called "special" relativity) case of gravity-free flat spacetime.

 

[PeroK]

Great!
I got the fact that a spacetime interval is invariant under LTs.
But above all: what is a spacetime interval? How do we define it, how do we "merge" two diverse quantities such as length and time?
How do we get to define it(distance between two points in a $x-t$ graph) as $\sqrt{(ct)^2-x^2}$?

If we have Euclidean space, then the distance between two points: $$(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$$ represents something physically meaningful. Also, it is invariant under rotations, which means that it doesn't depend on your choice of $x, y, z$ coordinates. These two things go together.

Spacetime is not Euclidean, but is hyperbolic. The equivalent invariant quantity is $$(c\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2$$ This is still invariant under spatial rotations, but also invariant under Lorentz Transformations, representing a constant relative velocity between two reference frames.

The geometric view of spacetime, therefore, views relative velocity as a (hyperbolic) rotation of spacetime. Essentially two reference frames are using the same spacetime, but each has a rotated view relative to the other.

 

[Ibix]

So $s^2$ is $(ct)^2-x^2$ by mere definition?

It's simply a quantity that is proven to be "untouched" by Lorentz trans.? Or is there a more profound meaning/derivation?

If you define the interval, then you can find transforms that leave it invariant. Then you can derive the kinematics of objects in such a space and show that they are consistent with experiment. So the interval and the fact that we assert that it is invariant are postulates of the theory, in this derivation. By definition they have no proof, except the post hoc discovery that they are useful.

These are not the same as the postulates Einstein used. He used his famous two postulates, from which he deduced the Lorentz transforms. Then he discovered that there are some quantities that are invariant under transform, including the interval. In that case, the postulates of the theory are Einstein's and we justify the importance of the interval because it turns out to be useful.

Essentially, the interval is important because it turns out to be useful. You can either assert it as the fundamental portion of the theory, or you can show that it ought to be significant because it turns out to be invariant if you start somewhere else. Historically, I believe that it was Minkowski who recognised that the Lorentz transforms look like "rotations" in a specific kind of four-dimensional space, and put relativity on to the geometric basis that we mostly use these days.

 

[etotheipi]

hey greg, another thing you might find cool is that if a particle has a (necessarily timelike) worldline $\mathscr{L}$ through the spacetime , and $\mathrm{d}\mathbf{\vec{x}}$ is an infinitesimal vector connecting two points on the curve, then the line element of the curve is given by$$||\mathrm{d}\mathbf{\vec{x}}||_{\mathbf{g}} = \sqrt{\pm \mathbf{g}(\mathrm{d}\mathbf{\vec{x}}, \mathrm{d}\mathbf{\vec{x}}) }$$don't worry about this $\mathbf{g}$ thing for now, just important to know that given an orthonormal basis and an origin [which defines a chart $p \mapsto (t,x,y,z)$] that thing above turns out to be the same as$$||\mathrm{d}\mathbf{\vec{x}}||_{\mathbf{g}} = \sqrt{(\mathrm{d}t)^2 - (\mathrm{d}x)^2 - (\mathrm{d}y)^2 - (\mathrm{d}z)^2}$$which is pretty much what you're talking about above, i.e. it's just a line element along a curve except now a curve through a space $\mathbf{R}^4$ equipped with a lorentzian metric. Anyway you can also parameterise the curve however you want but turns out the parameter $\tau$ defined by $||\mathrm{d}\mathbf{\vec{x}}||_{\mathbf{g}} := c\mathrm{d}\tau$ is a pretty nice choice, because it corresponds to the time the particle experiences as it goes along the curve.

The other neat thing is that $\mathbf{g}(\mathrm{d}\mathbf{\vec{x}}, \mathrm{d}\mathbf{\vec{x}})$ is coordinate independent (fancy terminology: a tensor that's eaten two vectors...), so if you decide choose a set of coordinates, parameterise $(t(\lambda), x(\lambda), y(\lambda), z(\lambda))$ and then bang some integral signs around that line element and do some mafs, the arc length won't depend on any arbitrary choice of chart.

 

[PeterDonis]

Shouldn't it be something like a semi-circle, for $s$ to be constant?

No, because of the minus sign in the formula. The local where $s$ is constant would be a circle if there were a plus sign in the formula, as in the usual Pythagorean theorem. But if there is a minus sign, the locus where $s$ is constant is a hyperbola.

 

[robphy]

In (1+1)-dim,
the square-interval between two events is a measure of the
area of the causal diamond between them.
For simplicity, let O and Q be timelike related, with O to the past of Q.
The causal diamond of OQ is intersection of
the future light cone of the past event O
and the past light cone of the future event Q.

From my Insight
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
and my article in AJP ( 84, 344 (2016); https://doi.org/10.1119/1.4943251 )
[early draft: https://arxiv.org/abs/1111.7254 ]

The square-interval is 16 (count it).
There are 4 (=sqrt(16)) ticks ("clock diamonds"="unit causal diamonds") along the diagonal OQ.
(Divide OQ into 4 congruent pieces. Then sketch the corners... to produce 4 diamonds [similar to OQ].)
Note that OQ has velocity 3/5... count 5 "ticks" up from O, and count 3 "sticks" to the right to get to Q.

Note the area of the diamonds (parallelograms with sides parallel to the asymptotes [the lightcone]) with one corner at the center and the opposite corner on the hyperbola
is independent of the point on the hyperbola.

Observe that all of these parallelograms have the same area. See the hyperbola?

Here are slides from a talk at the 2018 AAPT meeting. (The affiliation and contact information there is not current.)
https://www.aapt.org/docdirectory/m...dGraphPaper-CalculatingWithCausalDiamonds.pdfBy the way, "hyperbolic trigonometry" [in flat Minkowski spacetime] is what is needed,
not [curved] "hyperbolic geometry"... unless you want to study the mass shell or the space of velocities.

Analogously, "circular trigonometry" [in flat Euclidean space]
is different from [curved] "elliptic [spherical] geometry".

 

[greg_rack]

Thanks a lot guys, I really appreciate your time, now it is much much clearer!
You are wonderful

By the way, I even managed to find a cool and pretty straightforward derivation of the spacetime interval formula, which doesn't require the notion of hyperbolic space nor any advanced math, which makes it suitable even for a rookie like me :)

Define two events separated by a $\Delta x$ distance on space and $\Delta t$ on time;
In order for them to be timelike related:
$$-c\leqslant \frac{\Delta x}{\Delta t} \leqslant c$$
Which can be rewritten as:
$$\Delta x^2 \leqslant (c\Delta t)^2 \rightarrow (c\Delta t)^2 -\Delta x^2 \geqslant 0$$
Now we define "spacetime interval" the quantity
$$\Delta s^2=(c\Delta t)^2-\Delta x^2$$

 

[Sagittarius A-Star]

Shouldn't it be something like a semi-circle, for $s$ to be constant?

By the way, I even managed to find a cool and pretty straightforward derivation of the spacetime interval formula, which doesn't require the notion of hyperbolic space

I even managed to find a cool and pretty straightforward derivation of the spacetime interval formula, which does require the notion of hyperbolic space:

$\cosh^2 \zeta - \sinh^2 \zeta = 1$
...
$\beta = \tanh \zeta$, $\ \ \ \ \ \gamma = \cosh \zeta$, $\ \ \ \ \ \gamma\beta = \sinh \zeta$
...
$\begin{pmatrix} a'^0 \\ a'^1 \\ a'^2 \\ a'^3 \end{pmatrix} = \ \ \ ... \ \ \ = \begin{pmatrix} a^0 \cosh \zeta - a^1 \sinh \zeta\\ -a^0 \sinh \zeta + a^1 \cosh \zeta\\ a^2 \\ a^3 \end{pmatrix}$
...
We see that the above Lorentz transformation is similar (but notidentical) to the expression for the 3-D Euclidean geometry spatial rotation!

Source (see PDF page 6):
http://web.hep.uiuc.edu/home/serrede/P436/Lecture_Notes/P436_Lect_16.pdf

 

[robphy]

I even managed to find a cool and pretty straightforward derivation of the spacetime interval formula, which does require the notion of hyperbolic space:

$\cosh^2 \zeta - \sinh^2 \zeta = 1$

Source (see PDF page 6):
http://web.hep.uiuc.edu/home/serrede/P436/Lecture_Notes/P436_Lect_16.pdf

This is a common misconception...

That's not hyperbolic space [a curved non-euclidean geometry]...
it's just hyperbolic-trigonometry on a flat Minkowskian (non-Euclidean) geometry...
...
like circular-trigonometry on a flat Euclidean geometry...
(not spherical or elliptic space [a curved non-euclidean geometry]).

(You can say "hyperbolic rotation", just as one can say "circular rotation".
But this isn't hyperbolic or elliptic space.)

In Euclidean, Minkowski, and Galilean geometries (which are flat),
given a line and a point not on that line, there is exactly one parallel line through that point. (Playfair's version of the Parallel Postulate.)
In curved geometries, it's not exactly one.
(In hyperbolic space, it's more than one.
In spherical/elliptic space, it's zero.)UPDATE:
In the posted notes, I see the statement on page 18.
It should say "hyperbolic trigonometry" and "circular trigonometry".

By the way, to study the hyperboloids [staying on that surface] at the bottom of page 18, one could use hyperbolic geometry... just as to study a sphere (surface of the earth) one would use spherical/elliptic geometry.

At glance, the rest of those notes look useful.