When I learned about the Minkowski Space and it's coordinates, it was explained such that the metric turns out to be $$ ds^{2} = -(c^{2}dx^{0})^{2} +(dx^{1})^{2} + (dx^{2})^{2} + (dx^{3})^{2} $$ where $ x^0,x^1,x^2,x^3 $ come from $ x^{\mu} : \mu = 0,1,2,3 $, and $ c $ is the speed of light. The first resource I had access to--I have to do a bit of digging for the exact paper--of course addressed that this invariant takes the tensor form: $$ ds^2 = g_{\mu \nu}dx^{\mu}dx^{\nu} $$ as well. This two things I've seen in all of my texts and online resources regarding resources. The element in question that varies between authors, is the time coordinate, $ x^0 $. When it was first explained to me, it was using the standard Cartesian representation for the spatial portion of the coordinates, and the time coordinate was labeled as $ x^0 = ict $. Squaring gives $ (x^0)^2 = -c^2t^2 $, and applying differential calculus, we get $ (dx^0)^2 = -c^2dt^2 $. Sensible and expected to have the tensor formula spit out the Minkowski Metric. The author then later explicitly states the coordinates are $ x^0 = ict, x^1=x,x^2=y,$ and $x^3 = z$.

My questions is then, why is it most authors on the subject omit the imaginary unit on the time coordinate? For example, here.

The only reason I can fathom omission is if the author is using metric signature $ [+,-,-,-] $, where I started off learning the theory with signature $ [-,+,+,+] $ which may be the reason seeing the time coord with no imaginary unit seems dissonant to me. All help appreciated!

Edit: After reading the other answers, my questions is now one of why and how (mathematically) do we obtain the Minkowski Metric Signature. More specifically the one element with a different sign.

  • $\begingroup$ Sort of related: en.wikipedia.org/wiki/Wick_rotation $\endgroup$
    – BMS
    Apr 8, 2014 at 18:08
  • $\begingroup$ Very helpful, but still not getting to the root of why :C $\endgroup$ Apr 8, 2014 at 18:10
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    $\begingroup$ Really, the root answer is that you don't get much by adding all of the machinery of complex numbers -- what is the conjugate of a spacetime point, and what does it mean? Should inner products be $v^{*}\cdot v$? Why is the time coordinate the only one that takes a complex number? If you say "there's just a minus sign" in the signature, you avoid these problems. $\endgroup$ Apr 8, 2014 at 18:47
  • $\begingroup$ Hmmm. So from what I understand, the assumptions and "guesses" that we make in mathematical physics are literally just that. The means to an end? $\endgroup$ Apr 8, 2014 at 18:56
  • $\begingroup$ It's one thing to ask why (+,-,-,-) versus (-,+,+,+), but is an entirely separate matter to question the signature, itself: to ask why 3 go together and 1 goes the other way. At that point, one might actually question your question with: "as opposed to what?". With (+,+,+,+) then you don't have a finite non-zero invariant speed, so the speed of waves propagating in a vacuum is now relative (and actually "speed" would now be "slope" since time would be a spatial dimension, if (+,+,+,+)). A more constructive question would be: "what plays the role of signature for non-relativistic physics?". $\endgroup$
    – NinjaDarth
    Jan 21, 2023 at 20:52

9 Answers 9


as you wrote, the spacetime invariant can be expressed as: $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ and from that we normally get: $$ds^2=-c^2dt^2+dx^2+dy^2+dz^2$$ This is not because of some arbitrary imaginary time unit, this is because the metric ($g_{\mu\nu}$) is a diagonal matrix with the coefficients of each term of the $ds^2$ equation: $$g_{\mu\nu}=\left(\begin{array}{l}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)$$ and the coordinates are listed as you would assume: $$dx^{\mu}=\left(\begin{array}{l}cdt\\dx\\dy\\dz\end{array}\right)$$ Then, you should note that $$g_{\mu\nu}dx^{\mu}=dx_{\nu}=\left(\begin{array}{l}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)\left(\begin{array}{l}cdt\\dx\\dy\\dz\end{array}\right)=\left(-cdt~~dx~~dy~~dz\right)$$ Also, $v_{\mu}v^{\mu}$ is the inner product, meaning: $$dx_{\nu}dx^{\nu}=\left(-cdt~~dx~~dy~~dz\right)\left(\begin{array}{l}cdt\\dx\\dy\\dz\end{array}\right)=-c^2dt^2+dx^2+dy^2+dz^2$$ This is the equation you want without any imaginary unit omission. The reason for the $-1$ in the $g_{\mu\nu}$ is that it makes the system Lorentz invariant; it maintains $ds^2$ as a spacetime invariant quantity.

Let me be historical. In Euclidean 3-D coordinates, you find the interval between positions as $$\Delta d_{Eucl}^2=(X_2-X_1)^2+(Y_2-Y_1)^2+(Z_2-Z_1)^2$$ When incorporating relativity and time, the interval becomes a spacetime quantity. Because relativity sets the maximum speed of information as $c$, we make the interval $$\Delta s^2=\Delta d_{Eucl}^2-c^2(t_2-t_1)^2$$ This represents the original interval - the distance between the two events - minus the maximum distance the information could travel in the time between the two events. That difference lets us determine if the events happened in a definite chronological order ($\Delta s^2<0$) or if they occurred in two distinctly separate positions ($\Delta s^2>0$), since in relativity we can't always be sure. It is from this that the $-1$ in the metric originates. Space and time coordinates are given opposite signs here. We keep the metric in terms of $s^2$ because we simply can't be sure if $s$ is positive or negative. There was no original imaginary time coordinate, that was simply someone's poor interpretation and it has been (thankfully) dropped for the most part.

I should probably also point out that the imaginary time coordinate can not come out of Euclidean 4-D either. If one ignores relativity, then there is no maximum velocity. If there is no maximum velocity, there is no natural way of equating spatial and temporal coordinates. Therefore, not only would it not be right to use $c$ in the $ict$ coordinate, it also would not make sense to add time to space because there would be no agreeable conversion between them. However, if you don't ignore relativity, then you must subtract the time term from the 3-D interval in order to comply with the notion of a maximum velocity. So the Euclidean signature, $(1,1,1,1)$ can not be used to describe 4-D spacetime! So you never define the time coordinate as imaginary.

  • $\begingroup$ I was under the impression that the metric has that negative 1 because of the imaginary unit, but this is clearly not the case. Is there a more rigorous derivation of the choice of metric signature? I hate to believe it was arbitrary and just worked. $\endgroup$ Apr 8, 2014 at 18:06
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    $\begingroup$ Sadly, it was arbitrary and just worked. Well, no, it wasn't arbitrary. When Einstein wrote his original paper on special relativity he had to write time with an opposite sign to to space, but he didn't write the metric in the way we do these days. He was more driven by the physics. When Minkowski reformulated SR we got the metric with the signature we use today. $\endgroup$ Apr 8, 2014 at 18:12
  • $\begingroup$ Have links or names to any Minkowski papers off the top? ( I'm searching now ) $\endgroup$ Apr 8, 2014 at 18:13
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    $\begingroup$ It was not arbitrary at all, it is from the spacetime interval. In Euclidean space, an interval is the sum of the squares of the difference between the three coordinates of two positions.... $\endgroup$
    – Jim
    Apr 8, 2014 at 18:14
  • $\begingroup$ when you introduce relativity, you take that interval and find the difference between it and the distance light could travel in that amount of time, which is $s^2=d_{Eucl}^2-c^2t^2$. That effectively finds the spacetime separation between two events. This is done because relativity takes $c$ as the max speed, so the true interval when incorporating time would have to be the measured interval minus the maximum interval in the given time $\endgroup$
    – Jim
    Apr 8, 2014 at 18:17

If you define $x^0=ict$, then I assume one takes $x_\mu=x^\mu$ so that the metric is actually $\eta_{\mu\nu}=\text{diag}(1,1,1,1)=\delta_{\mu\nu}$, i.e. you're dealing with a Euclidean metric. Then $$ds^2=\delta_{\mu\nu}dx^\mu dx^\nu$$ gives the usual outcome : $$ds^2=-c^2dt^2+d\vec{x}^2$$ The usual conventions are as follows:

Option one: One defines $x^\mu=(ct,\vec{x})$ and $x_\mu=\eta_{\mu\nu}x^\nu=(-ct,\vec{x})$ where $\eta_{\mu\nu}=\text{diag}(-1,1,1,1)$. This results in $$ds^2=-c^2dt^2+\vec{x}^2$$This convention is usually taken in treatments that focus on (general) relativity and/or spacetime structure.

Option two: One defines $x^\mu=(ct,\vec{x})$ and $x_\mu=\eta_{\mu\nu}x^\nu=(ct,-\vec{x})$. where $\eta_{\mu\nu}=\text{diag}(1,-1,-1,-1)$. This results in $$d\tilde{s}^2=-ds^2=c^2dt^2-\vec{x}^2$$ This approach is usually taken when the focus is on particle physics results. My personal theory is that this is because it results in the equation $p_\mu p^\mu=m^2$ (as opposed to $p_\mu p^\mu=-m^2$), which is one of the main equations in particle physics and arguably looks slightly more pleasing than the alternative. As was pointed out in a comment, it also makes time-like intervals positive, which might be preferred when dealing with particles (which, of course, always travel along time-like or light-like trajectories).

Of course, the two approaches are completely equivalent to each other. The convention with the imaginary time coordinate has fallen out of use a little bit. I can see why this would happen: It's not a particularly helpful convention for intuition, nor does it seem to carry over well to general relativity.

  • $\begingroup$ I love it when the answer is convention! Thank you very much for clarifying that! $\endgroup$ Apr 8, 2014 at 18:02
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    $\begingroup$ I think particle physicist choose that signature because it makes timelike intervals positive, which is nice if you are dealing with mostly particle trajectories. $\endgroup$
    – George G
    Apr 8, 2014 at 18:09

In general, the expression for the metric and the expression for the coordinates need to work together to give you the correct line element. So the following combinations all have the same line element $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$:

$g_{\mu\nu}=diag(1,1,1,1)$ with $x^{\mu}=(ict,x,y,z)$


$g_{\mu\nu}=diag(-1,1,1,1)$ with $x^{\mu}=(ct,x,y,z)$


$g_{\mu\nu}=diag(-c^2,1,1,1)$ with $x^{\mu}=(t,x,y,z)$

The primary reason to abandon the "ict" notation and use one of the last two is because we eventually want to go beyond special relativity and do general relativity. That is best done using (pseudo-) Riemannian geometry, and Riemannian geometry requires real-valued coordinates. Furthermore, when using Riemannian geometry you are not even guaranteed to have a time coordinate, so you cannot easily be sure where to put the "i".


The only real reason to introduce $ict$ coordinates is to stress the similarity (for didactic purposes I guess) between Lorentz transformation and orthogonal rotations in more used-to Euclidian space.

Note that Minkowski pseudo-Euclidian space obtains exactly "normal" Euclidian form if complex time is introduced,namely: metric signature becomes $++++$: exactly like if it was regular Euclidian 4-space. Also, more vivid: matrix of Lorentz transformation obtains exactly form of real orthogonal matrix due to $\cos(ix) = \cosh x$ (similar for $\sin x$). Thus, you rotate through complex angle, but matrix looks like regular orthogonal, for example boost in $x$-direction, $$ \left( \begin{matrix} \cos z & \sin z & 0 & 0\\ -\sin z & \cos z & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix} \right) $$

where $z$ is now strictly imaginary.

  • $\begingroup$ I've been wondering if that would be the "origin" of spinors in QM? $\endgroup$
    – DWin
    Dec 4, 2023 at 1:16

Someone gave me the following nice picture that always stuck afterwards:

A light source in vacuum turned on at the point $(x_0,y_0,z_0)$ at time $t_0$ will form a sphere growing at the speed of light with radius $r = c (t-t_0)$. The equation for the sphere for $x = x(t)$, $y = y(t)$, $z = z(t)$ is

$$ r^2 = c^2(t-t_0)^2 = (x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2. $$

In the infinitesimal limits $ t \to t_0$, $x \to x_0$, etc...:

$$ d(ct)^2 = dx^2 + dy^2 + dz^2. $$

Rewriting as

$$ 0 = - d(ct)^2 + dx^2 + dy^2 + dz^2 \qquad \textrm{ or } \qquad 0 = d(ct)^2 - dx^2 - dy^2 - dz^2 $$

shows where the relative sign comes from, and also illustrates why either $\eta^{\mu \nu }=\textrm{diag}(-1,1,1,1)$ (first case) or $\eta^{\mu \nu} = \textrm{diag}(1,-1,-1,-1)$ (second case) is fine.

Since this is for light-like separated events (i.e. the wave front of an actual light wave), the zero is actually the interval. This usually only gets me into trouble for remembering which is the right sign for the spacetime interval. Turns out it's

$$ ds^2 = - d(ct)^2 + dx^2 + dy^2 + dz^2 $$

which I guess you can get by requiring proper distance is a real number:

$$ d\sigma = \sqrt{ds^2} = \sqrt{- d(ct)^2 + dx^2 + dy^2 + dz^2} . $$


"... my questions is now one of why and how (mathematically) do we obtain the Minkowski Metric Signature. More specifically the one element with a different sign."

Well, If you check Einstein's "Relativity: The Special and General Theory", you will find in the Appendix I (just before equation (10)) that Einstein simply started off with the Pythagorean Theorem in 4D, which he put like this:

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

He then squared both sides and moved $c^2t^2$ to the left changing the sign. This gave him the signature [-, +, +, +] Obviously, you are equally free to move the $x^2 + y^2 + z^2$ to the right, in which case you will get the signature reversed [+, -, -, -] (apparently, that's what Minkowski did).

Curiously enough, from the fact that Einstein used the Pythagorean Theorem we can see that he is talking about lengths/distances here, and not coordinates (which are points) as he later keeps calling them (and everybody after him). On the other hand, it is quite obvious even without looking at the derivation, if you come to think about it. Simply, you cannot shrink points (coordinates) into infinitesimals like dx or dy, since they cannot get any smaller then they already are - their extension is exactly null. You cannot also square points to make them $x^2$, $y^2$, $z^2$. Shrunk point and squared point is still the same point. At the same time, ds is called the "line-element" (Einstein called it also "linear element"). Line suggest distance or length, and not a point (or coordinate for that matter), doesn't it?

And by the way, this equation implies that time "travels" orthogonally to x, y, z. Sure, we do draw t orthogonally to x for instance, in order to better visualize various functions, like acceleration. But when we observe physical things, like a car, accelerate without changing direction, they do not "draw" a hiperbole in space, do they?


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

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

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

Their wristwatch readings are related by $$\left( \Delta t_Q' - \frac{\Delta x_Q'}{c}\right) = k\left( \Delta t_Q - \frac{\Delta x_Q}{c} \right)$$ and $$\left( \Delta t_Q' + \frac{\Delta x_Q'}{c}\right) = \frac{1}{k}\left( \Delta t_Q + \frac{\Delta x_Q}{c} \right)$$

By multiplication, we get the following equation: $$\left({\bf \mbox{invariant square interval}}\right)=\left( \Delta t_Q'^2 - \frac{\Delta x_Q'^2}{c^2}\right) =\left( \Delta t_Q^2 - \frac{\Delta x_Q^2}{c^2}\right),$$ 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


Since $g_{μυ}=\vec{e_{μ}} \cdot \vec{e_{υ}}$ , $g_{00}=\vec{e_{0}} \cdot \vec{e_{0}}$ , And $ \begin{pmatrix} ds\end{pmatrix}^{2}=g_{μυ}x_{μ}x_{υ}=-(ct)^{2}+x^{2}+...$;
If: $g_{μυ}=\begin{pmatrix} ict\\x\\y\\z\end{pmatrix} $.is Cartesian Coordinate system then $\vec{r}=ict \cdot \vec{ict} +x \cdot \vec{x} ... $,$g_{00}=\vec{e_{0}} \cdot \vec{e_{0}}=\vec{ict} \cdot \vec{ict}=1$;

So in $g_{μυ}=\begin{pmatrix} ct\\x\\y\\z\end{pmatrix} $.$ \frac{∂r}{∂\begin{pmatrix} ct\end{pmatrix}}=i\cdot \vec{ict}=\vec{e_{0}}$,so its Curvilinear Coordinates Systemand then $g_{00}=-1$;

Also same in $g_{μυ}=\begin{pmatrix} t\\x\\y\\z\end{pmatrix} $. $ \frac{∂r}{∂t}=ic\cdot \vec{ict}=\vec{e_{0}}$(its Curvilinear Coordinates Systemand too),then $g_{00}=-c^{2}$.


Wikipedia gives sufficient coverage on this topic https://en.wikipedia.org/wiki/Minkowski_space#Metric_signature

I cite: "In general, but with several exceptions, mathematicians and general relativists prefer spacelike vectors to yield a positive sign, (− + + +), while particle physicists tend to prefer timelike vectors to yield a positive sign, (+ − − −)." "Arguments for" (− + + +) "include "continuity" from the Euclidean case corresponding to the non-relativistic limit $c \rightarrow \infty$". There are many reasons to prefer to use (− + + +), for example four-gradient to become $(−c\frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}) $. Using one signature via another is just a matter of choice and convinience.


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