# Where does the negative signature case come from in the Pythagorean derivation of distances in spacetime?

I am reading Why does $$E=mc^2$$ (and why should we care?) by Brian Cox and Jeff Forshaw. I want to understand these three sentences (from page 76/77):

Once we follow Occam and make these two simplifying assumptions, we are left with only two possible choices as to how to calculate distances in spacetime. The length of the hypotenuse must be either $$s^2 = (ct)^2 + x^2$$ or $$s^2 = (ct)^2—x^2$$. There is no other option.

The two simplifying assumptions are that

1. "at least the space part of our spacetime should be Euclidean (in the mathematical sense). In other words, flat." and

2. "spacetime is unchanging and the same everywhere."

Where does $$s^2=(ct)^2 - x^2$$ come from as an equation for the length of $$s$$? I thought $$s^2=(ct)^2 + x^2$$ would be the only (single) option.

• Possible duplicates: physics.stackexchange.com/q/304799/2451 and links therein. Commented Sep 5, 2023 at 11:51
• The factor can (mathematically) be -1, 0 or 1. 1 is just 4D Euclidean spacetime, 0 is the "Newtonian" case, and -1 is the "other option". As @ConnorBehan has pointed out, -1 agrees with observations and experiments. Commented Sep 5, 2023 at 12:09
• Another way of looking at this (which I find superior to the argument presented here) is that the Einstein postulate that $c$ is fixed in all reference frames leads to the concept of the Lorentz factor and the Lorentz boost. The only way for the interval to remain invariant under the boost transforms is for it to have the form $s^2 = c^2t^2 - x^2$, which is one of your two "choices." Commented Sep 5, 2023 at 16:30
• The $+$ would make time just another spatial dimension. Everyday experience shows it isn’t. Commented Sep 5, 2023 at 22:12
• The book doesn't say that "spacetime should be Euclidean", it says that at least the space part of spacetime should be Euclidean, which then allows for your other option. Commented Sep 5, 2023 at 23:19

(copied from my answer to Minkowski Metric Signature [with some modifications])

Here's an argument essentially due to Bondi.
It is physically motivated by radar measurements.

First, an introduction to Bondi's k-calculus.

(This is based on a diagram from Bondi's "E=mc2: An Introduction to Relativity" (http://www.worldcat.org/title/emc2-an-introduction-to-relativity/oclc/156217827), which accompanied Bondi's series of lectures "E=mc2: Thinking Relativity Through", a series of ten lectures on BBC TV running from Oct 5 to Dec 7, 1963. It had a typo that I corrected.)

Two inertial observers (Bondi will call) Alfred and Brian meet at event O.

Alfred performs a radar measurement to assign coordinates to event P on Brian's worldline.

After a time $$T$$ on Alfred's wristwatch, he sends a light signal to Brian. Brian receives the signal at a time $$kT$$ on Brian's watch (event P), where $$k$$ is a proportionality constant (independent of $$T$$). [This $$k$$ turns out to be the Doppler factor].

When this light-signal is reflected by Brian's worldline (at event P), the reflected signal back arrives at Alfred's worldline when Alfred's watch reads $$k(kT)$$, where the same factor of $$k$$ is used because of the Principle of Relativity. (We've also used that the speed of light is the same for these observers.)
[Side note: These two triangles, with two timelike legs and one lightlike leg, are similar in Minkowski spacetime.]

So, Alfred can assign a time-coordinate and a space-coordinate to the distant event P (displacements from event O): $$\Delta t_{P}=(\mbox{half of the elapsed time})=\frac{t_{rec}+t_{send}}{2}=\frac{k^2T+T}{2}$$ $$\Delta x_{P}=(\mbox{half of the roundtrip distance})=c\frac{t_{rec}-t_{send}}{2}=c\frac{k^2T-T}{2}.$$

By division, one can get $$\quad v_{BA}=\displaystyle\frac{\Delta x_P}{\Delta t_P}=\frac{k^2-1}{k^2+1}\quad$$ (independent of $$T$$),
which can be solved for $$k$$ to get the Doppler formula.

Note that
by addition: $$\quad \Delta t_{P}+(1/c)\Delta x_{P}=t_{rec}\quad$$, and
by subtraction: $$\Delta t_{P}-(1/c)\Delta x_{P}=t_{send}$$.

Now consider two inertial observers making radar measurements, assigning coordinates to a distant event (call it Q).

Each observer sends a light-signal and waits for its echo to be received, noting his wristwatch reading at these two events on his worldline. (Geometrically, we have the light-cone of Q intersecting the two inertial worldlines that met at event O.)

[Side note: Although not necessary, event Q could be on the worldline of a third observer (call her Carol). Then these radar measurements would involve $$k_{CB}$$ and $$k_{CA}$$, relating Carol and Brian and Carol and Alfred.
The "$$k$$" used above in the first part and in the part below could be called $$k_{BA}$$ to relate Brian and Alfred.]

(The diagram is from Bondi's "Relativity and Common Sense".)

Their wristwatch readings are related by $$t'_{send}=k t_{send}$$ $$\left( \Delta t_Q' - \frac{\Delta x_Q'}{c}\right) = k\left( \Delta t_Q - \frac{\Delta x_Q}{c} \right)$$ and $$t'_{rec}=\frac{1}{k} t_{rec}$$ $$\left( \Delta t_Q' + \frac{\Delta x_Q'}{c}\right) = \frac{1}{k}\left( \Delta t_Q + \frac{\Delta x_Q}{c} \right)$$

By multiplication, we get the following equation, which is independent of $$k$$: $$t'_{send}t'_{rec}=t_{send} t_{rec}\mbox{... in terms of radar coordinates}$$ which can be written in terms of rectangular components as $$\left({\bf \mbox{invariant square interval}}\right)=\left( \Delta t_Q'^2 - \frac{\Delta x_Q'^2}{c^2}\right) =\left( \Delta t_Q^2 - \frac{\Delta x_Q^2}{c^2}\right),$$ with its minus-sign in front of the spatial coordinate.
(Calling this "the invariant square interval" and not "minus the invariant square interval" is the choice of sign convention.)

(Side note: By addition and subtraction, one gets the Lorentz transformations.)

The reason why this method works is that we are working in the eigenbasis of the Lorentz Transformation, where the the lightlike directions are the eigenvectors and the Doppler factor and its reciprocal are the eigenvalues.

This is based on a blog entry that I contributed here
https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/

UPDATE to address the question in the comment by @sleep.

visit my Desmos visualization robphy v8e spacetime diagrammer for relativity (v8e-2021) - t-UP

• The "unit circle" (the "curve of constant square-interval 1") has the form $$t^2-Ey^2=(1)^2$$ in the Euclidean ($$E=-1$$), Galilean ($$E=0$$), and Minkowskian ($$E=+1$$) case.
• "Perpendicular to a radius vector" (i.e. a "right angle" in that geometry) is defined using the tangent to the circle.
• "Hypotenuse of a right-triangle" is defined as the side opposite the right-angle.

The relative-slope (with respect to the vertical) in these cases is $$v=3/5$$.

• For $$E=-1$$, this triangle with adjacent side of 1 has a hypotenuse (here, 1.162) that is longer than the adjacent side. Indeed, $$(hyp)=\frac{(adj)}{\cos\phi}\geq (adj) \mbox{ since \cos\phi \leq 1, where \tan\phi=v}.$$

• For $$E=0$$, this triangle with adjacent side of 1 has a hypotenuse (here, 1) that is equal to the adjacent side.

• For $$E=+1$$, this triangle with adjacent side of 1 has a hypotenuse (here, 0.8) that is shorter than the adjacent side. Indeed, $$(hyp)=\frac{(adj)}{\cosh\theta}\leq (adj) \mbox{ since \gamma=\cosh\theta \geq 1, where \tanh\theta=v}.$$

If you want to see a Pythagorean triple in the Minkowski case, you can use my "Relativity on Rotated Graph Paper" method. You can construct other triples on the diagram of the elastic-collision at https://www.physicsforums.com/insights/relativity-rotated-graph-paper/

• The general policy is to not post duplicate answers. OTOH, the focus of this new question is rather different to that of Doryan Miller's question, so I don't think it would be appropriate to close this question as a dupe of that one. Commented Sep 5, 2023 at 17:35
• Great answer, this is how SR should be taught IMO. Notice people, no mention of "time dilation" or "relativity of simultaneity", just geometry & algebra. Commented Sep 5, 2023 at 19:25
• "We've also used that the speed of light is the same for these observers." We have also used that the speed of light is the same both ways (i.e. that physics is isotropic). Which is something that must be assumed and cannot be measured or confirmed in any way. Commented Sep 6, 2023 at 12:27
• @m4r35n357 Thanks. Bondi’s formulation emphasizes measurement first, with relatively easy to follow steps in logic and in algebra. Those “effects” and the various formulae can be developed next. Commented Sep 6, 2023 at 12:28
• @Arthur Yes, but for an introduction, this is probably sufficient. Commented Sep 6, 2023 at 12:30

It comes from the most fundamental observation of physics: past is qualitatively different from future. A Euclidean model of spacetime cannot accommodate this: time would be just another dimension. You could, for example, rotate yourself freely on the $$xt$$ plane, allowing you to live your life in the $$-t$$ direction, opposite the rest of us.

The Minkowskian model makes "boosts" (rotations that exchange space with time) different from ordinary rotations in a way that preserves the distinction between past and future for events that can either influence you or be influenced by you.

One way to see the emergence of the negative sign is by rearranging the following equation. Suppose a light beam leaves from the origin at time $$t = 0$$ and reaches some distance $$r = \sqrt{x^2 + y^2 + z^2}$$ at time $$t$$. Then

$$\sqrt{x^2 + y^2 + z^2} = ct$$

or $$c^2 t^2 - x^2 + y^2 + z^2 = 0 \,.$$

In particular, if the distance $$r > ct$$, then even light would not be able to reach it in time $$t$$. However, if $$r < ct$$, then something slower than light can also reach the point $$r$$ in time t.

For any general value of $$r$$ (i.e., not necessarily the one that satisfies the first equation, $$r=ct$$), the distance $$c^2 t^2 - r^2$$ can be called $$s^2$$. Therefore

$$s^2 = c^2 t^2 - x^2 + y^2 + z^2\,,$$

where for your case, we can set $$y=z=0$$. Consequently, cumulatively the set of points $$(r, t)$$ is called spacelike (not even reachable by light in time $$t$$), lightlike (only reachable by light in time $$t$$), or timelike (a massive particle can also reach it given its moving at a fast enough velocity) depending on whether $$s^2<0, s^2=0$$ or $$s^2>0$$.