# Invariance of the relativistic interval

Imagine we have two events, $$E_1, E_2$$ in the coordinate systems $$K, K'$$ (with coordinates $$(x,y,z,t),\ (x',y',z',t')$$), whilst $$K'$$ ist moving with the speed $$\vec v$$ in regard to $$K$$. From the constance of the speed of light, the following equation is supposed to be obvious: $$S^2= c^2(t_2-t_1)^2 - (x_2-x_1)^2-(y_2-y_1)^2- (z_2-z_1)^2=\\= c^2(t_2'-t_1')^2- (x_2'-x_1')^2- (y_2'-y_1')^2- (z_2'-z_1')^2$$ Which is called the relativistic interval, which (under assumption of this equation) obviously is invariant and can be visualized as a four-dimensional sphere. With the help of this invariance one can display the Lorentz-transformation as a four-dimensional rotation, but how does one get this invariance?

As far as I understood you could rewrite this like that: $$S^2=\Delta s_c^2-|\Delta \vec s|^2=\Delta s_c'^2-|\Delta\vec s'|^2$$ Whereas $$\Delta s_c$$ is the distance light would have travelled in the given amount of time and $$\Delta\vec s$$ is the distance between the two events. But why is this invariant? How can one conclude the given invariance from the constance of the speed of light? I'm feeling as if I missed something important here.

• One correction, the invariant interval is not a sphere in 4-dim. Look at the relative signs. – ggcg Dec 20 '18 at 22:08
• @ggcg you're right - now this makes even less sense to me – MetaColon Dec 20 '18 at 22:10
• That's funny. Are you confused about the invariance of this interval or the connection between this symmetry and the consistency of the speed of light? Or both? – ggcg Dec 20 '18 at 22:14
• Rigid rotations that do not affect time will leave the interval invariant too, the spatial part is a sphere. The transformations that mix time and a space parameter are called boosts (or sometimes a pseudo rotation). They can be expressed in the same manner as a rotation matrix but with an imaginary angle, using sinh(v/c) and cosh(v/c). – ggcg Dec 20 '18 at 22:16
• "$|\Delta \vec{s}|$ is the distance between the two events" - in what frame is this distance being measured? – probably_someone Dec 21 '18 at 15:48

Here's a geometric interpretation... (admittedly a top down approach).
(The essence of the idea is inspired by the Bondi k-calculus and
by the "product of times" formula seen in A.A. Robb's "Optical geometry of motion"

update (in case this is part of the question)
Why is the invariant of this form $$S^2=\Delta t^2-\Delta x^2$$?

• A good motivation is a radar measurement of an event $$P=(t_P,x_P)$$ not on your worldline.

Suppose you are an inertial observer.
To measure event $$P$$, imagine sending a light signal to $$P$$ and waiting for its echo, and noting the time on your wristwatch when you sent it $$t_{send}$$ and when you receive it $$t_{rec}$$. From those two times, you would assign event $$P$$ the following coordinates:

• time coordinate $$t_P=\frac{1}{2}(t_{rec}+t_{send})$$ [the midway time during the round trip]
• spatial coordinate $$x_P=\frac{1}{2}(t_{rec}-t_{send})$$ [half of the roundtrip time (multiplied by c)]

Note that $$t_{rec}=(t_P+x_P)$$ and $$t_{send}=(t_P-x_P)$$.

Consider another inertial observer who met you when your wristwatch read zero and they set their wristwatch to zero. They would make analogous measurements of event $$P$$.
Thus, note that $$t'_{rec}=(t'_P+x'_P)$$ and $$t'_{send}=(t'_P-x'_P)$$.

Taking an image from Bondi's "Relativity and Common Sense" It turns out for events joined by a future-directed light-signal
that $$t'_{send}=K t_{send}$$ (where $$K$$ is a proportionality constant
[which depends on the relative velocities of the observers]) and
that $$t_{rec}=K t'_{rec}$$ (the same proportionality constant).
(This is actually the Lorentz Boost transformation.)

So, it turns out that while $$t_P\neq t'_P$$ and $$x_P\neq x'_P$$, it turns out
that $$t_{rec}t_{send}=t'_{rec}t'_{send}.$$ (This is the product of times formula [seen in Robb and in Geroch].)

Expressing this back in terms of the $$t_P$$s and $$x_P$$s, this says
that $$({t_P}^2-{x_P}^2)=({t'_P}^2-{x'_P}^2).$$ (This is the invariance of the square-interval.)

Let's consider the (1+1)-dim case and write $$S^2=\Delta t^2-\Delta x^2=(\Delta t+\Delta x)(\Delta t-\Delta x)=\Delta u \Delta v.$$ I'm going to define quantities $$u=t+x$$ and $$v=t-x$$
called (up to sign and scaling conventions) "light-cone coordinates" since these are coordinates with axes along the light-cone.

Written in this way $$S^2=\Delta u \Delta v$$ looks like the "area of a diamond" ( a parallelogram whose sides are parallel to the light cone). One corner could be taken to be the origin and the opposite corner traces out a hyperbola (the Minkowski-circle, the curve of constant interval from the origin) as you do a Lorentz boost-transform. (Recall that for a point $$(x,y)$$ along the hyperbola $$xy=1$$, a rectangle with corners at $$(0,0)$$ and at $$(x,y)$$, with sides parallel to the x- and y-axes, has area 1.)

Some important features of the Lorentz boost-transformation in (1+1)-Minkowski spacetime.

• It has a determinant of 1. So, it preserves areas.
• It has eigenvectors along the light-cone (this is the invariance of the speed of light).
• The eigvenvalues are $$k$$ and $$1/k$$ (since the product is equal to the determinant).
(It turns out that the eigenvalues are the Doppler factors.)

So, under a boost-transformation
these diamonds transform (are reshaped) into other diamonds with the same area.

Check out my visualization: https://www.geogebra.org/m/Jq4jDMRW I exploited this "area of a diamond" to visualize proper time along an inertial worldline and developed a method to do calculations on "rotated graph paper".

Check out my visualization: https://www.geogebra.org/m/HYD7hB9v The idea is written up here:
"Relativity on rotated graph paper"
American Journal of Physics 84, 344 (2016); https://doi.org/10.1119/1.4943251

• I'm sure you explained the topic greatly, but sadly I didn't understand that much. Maybe I'll gather some more knowledge in the paper you suggested and get back to your answer later. – MetaColon Dec 21 '18 at 13:18
• The paper is an application of the idea that the area is equal to the square interval. Can you see that t^2-x^2=(t+x)(t-x)=uv? I will interpret u as the width and v and the height of a parallelogram. Thus uv=area (as in the diagram/simulation). – robphy Dec 21 '18 at 14:21
• Maybe I misunderstood the your main concern... are you asking why that quantity t^2-x^2 is invariant? – robphy Dec 21 '18 at 14:29
• I just updated the answer with a proof inspired by H. Bondi's "Relativity and Common Sense". – robphy Dec 21 '18 at 15:17
• You're right - this is what I was asking for now. Now this actually makse sense. – MetaColon Dec 22 '18 at 18:55

You are asking a question right at the foundations of relativity.

There are two ways to develop special relativity. In the first approach one talks about inertial frames, the principle of relativity, and the invariance of the maximum speed (called speed of light). In the second approach one merely announces that spacetime is a kind of space (mathematically an example of a 'differentiable manifold') and that invariant distances in the space are given by the interval (as in the equations you are asking about).

In the first approach one can derive the Lorentz transformation from the two main principles (relativity and speed of light), and then one can prove that the interval is invariant. I wouldn't call it obvious, but it does come out reasonably quickly. In the second approach one defines the Lorentz transformation as that kind of coordinate transformation which preserves the interval. One then derives other results, such as that something traveling at the maximum speed relative to one coordinate frame will also have that same speed relative to others. It must do, because it moves between events separated by a null interval. Such an interval is null no matter who is measuring it, since it is invariant after all!

You get if from the postulate of the constancy of the speed of light. If we take the speed of light to be the same in all inertial frames, then it follows that

$$c^2= \left(\frac{d\mathbf{r}}{dt}\right)^2 = \left(\frac{d\mathbf{r'}}{dt'}\right)^2$$

That is,

$$dr^2 = c^2dt^2 \implies dr^2 - c^2dt^2 =0$$

and

$$dr'^2 = c^2dt'^2 \implies dr'^2 - c^2dt'^2 =0$$

Setting the two equal we get

$$dr^2 - c^2dt^2 = dr'^2 - c^2dt'^2.$$

From this derivation, you can easily see that the choice of the Minkowski metric signature being 2 or -2 is arbitrary.

• I think this proves that null intervals are invariant, but it doesn't apply to a general pair of events $E_1$, $E_2$. In other words, your proof works even if $\gamma$ (the Lorentz factor) is arbitrary. – MannyC Dec 21 '18 at 0:08

A proof can be made with the following thought experiment. (In relativity thought experiments are ubiquitous). I'm sorry if I can't make drawings, but if you search "light clock experiment" you will see what I'm talking about with drawings and possibly a more detailed explanation.

Take a box with two reflective walls at distance $$d/2$$ and imagine that a beam of light is bouncing back and forth, and imagine that a detector at one of the two walls counts the light bounces and uses it to track time.

If you look at the box in its rest frame you'll see the light beam bouncing every $$\Delta t = d/c$$ seconds. Let's say that there is a built in mechanism that makes the box, i don't know, ring an alarm, after $$n$$ bounces. That is, the box alarm will go off after a time $$n \Delta t$$.

Now let's go into a reference frame moving at constant velocity $$v$$ with respect to the original frame, along the direction orthogonal to the light beam. The time interval for a full bounce is $$\Delta t'$$ and the distance travelled is $$\mathrm{dist} = 2\sqrt{(\Delta x')^2 + (\Delta y')^2}\,,\qquad\Delta x = \frac{v\Delta t'}{2}\,,$$ where $$\Delta y'$$ is the distance travelled in the vertical direction in the new frame. It must equal $$d/2$$ because intervals orthogonal to $$v$$ don't get transformed under a boost (see later for an explanation of that). From the fact that $$c$$ is the same in both frames we conclude $$c^2 (\Delta t')^2 = v^2 (\Delta t')^2 + d^2\;\;\Rightarrow\;\;(\Delta t')^2 = \frac{c^2}{c^2-v^2}\Delta t^2\,.$$ We need to show that $$\Delta y' = \Delta y = d/2$$. There's another funny little thought experiment to prove that 1. Say I have two halves of a pipe painted differently and kept separated. They are running towards each other along the main axis of the pipe. If the transverse directions would shrink after a Lorentz transformation then in one frame I would see one half pipe enclosing the other, and in one other frame the opposite. Clearly the color seen by an observer after the pipes meet must be a frame invariant thing, so the solution is that the pipes don't shrink/enlarge along the orthogonal directions.

Ok, now let's come to the punch line. The events $$E_1$$ and $$E_2$$ are, respectively, the light clock starts and the light clock alarm rings. Let us compute the distance $$S^2=c^2(t_1 - t_1)^2 - (\vec{x}_1 - \vec{x}_2)^2\,,$$ in the two cases. For the original frame $$S^2 = (nc\Delta t )^2$$. For the other frame $$S^2 = (n c \Delta t')^2 - (n v \Delta t')^2 = (c^2 - v^2) \frac{(nc\Delta t )^2}{c^2 - v^2}\,.$$ So the two factors simplify and it works out!

But we are not done yet. This proves that the distance is invariant under arbitrary boosts for positive intervals (usually called time-like). Another answer given previously proved it for null intervals (called light-like) in full generality. We are left to prove it for negative intervals (space-like) and also for pure rotations. Now the pure rotation case is just trivial because if we have $$\Delta t = \Delta t'$$ that's just the statement that the Euclidean norm is invariant under rotation. Whereas the proof for space-like intervals can be done with another thought experiment. Take two detectors along the $$x$$ direction and shine a laser from the middle point. In the rest frame $$\Delta t = 0$$ while in a frame boosted along $$x$$ the events won't be simultaneous. The math will work out similarly so you can do it as an exercise.

• But why is the distance (as you called S^2) the distance the light has travelled minus the distance the frame has moved? Shouldn't it be a summation? – MetaColon Dec 21 '18 at 13:14
• That's the point. You have to define it with a minus to obtain an invariant. The version with a plus is frame dependent. We are using the thought experiment to come up with a definition that has the propery of being frame invariant. – MannyC Dec 21 '18 at 14:10
• If the fact that I call "distance" a non positive definite quantity puzzles you, that's just a matter of getting used with terminology. – MannyC Dec 21 '18 at 14:12