In history there was an attempt to reach (+, +, +, +) by replacing "ct" with "ict", still employed today in form of the "Wick rotation". Wick rotation supposes that time is imaginary. I wonder if there is another way without need to have recourse to imaginary numbers.

Minkowski spacetime is based on the signature (-, +, +, +). In a Minkowski diagram we get the equation: $$ \delta t^2 - \delta x^2 = \tau^2 $$ Tau being the invariant spacetime interval or the proper time.

By replacing time with proper time on the y-axis of the Minkowski diagram, the equation would be $$ \delta x^2 + \tau^2 = \delta t^2$$ In my new diagram this equation would describe a right-angled triangle, and the signature of (proper time, space, space, space) would be (+, +, +, +).

enter image description here

I am aware of the fact that the signature (-, +, +, +) is necessary for the majority of physical calculations and applications (especially Lorentz transforms), and thus the (+, +, +, +) signature would absolutely not be practicable.( Edit: In contrast to some authors on the website about Euclidian spacetime mentioned in alemi’s comment below)

But I wonder if there are some few physical calculations/ applications where this signature is useful in physics (especially when studying the nature of time and of proper time).

Edit (drawing added): Both diagrams (time/space and proper time/space) are observer's views, even if, as it has been pointed out by John Rennie, dt is frame dependent and τ is not.

  • 4
    $\begingroup$ This probably does not count as "application", but Greg Egan, an extremely math-oriented science fiction author, has written a novel set in a space-time with signature $(+,+,+,+)$. Some physical consequences of this change of signature are explored in the book and become an essential part of the story. $\endgroup$ Commented Jun 4, 2014 at 10:18
  • 1
    $\begingroup$ This related question might give some further insight: physics.stackexchange.com/q/107443/23473 $\endgroup$
    – Jim
    Commented Jun 4, 2014 at 14:47
  • 1
    $\begingroup$ You may be interested in this website which has compiled a whole list of papers exploring this idea, some of which seem to lean in the crackpot direction. It seems it turns out not be be equivalent to special relativity as the velocity addition formula gets modified from the true form. $\endgroup$
    – alemi
    Commented Aug 8, 2014 at 4:54
  • $\begingroup$ @alemi, thank you for this important hint! Surely, it would be wrong to think this model could replace Minkowski spacetime. But it permits certain considerations for which Minkowski spacetime is not designed for, in particular an improved description of time (because any time derives from proper time). $\endgroup$
    – Moonraker
    Commented Aug 8, 2014 at 5:41

4 Answers 4


The significance of the metric:

$$ d\tau^2 = dt^2 - dx^2 $$

is that $d\tau^2$ is an invarient i.e. every observer in every frame, even accelerated frames, will agree on the value of $d\tau^2$. In contrast $dt$ and $dx$ are coordinate dependant and different observers will disagree about the relative values of $dt$ and $dx$.

So while it is certainly true that:

$$ dt^2 = d\tau^2 + dx^2 $$

this is not (usually) a useful equation because $dt^2$ is frame dependant.

  • $\begingroup$ So you agree that the application I presented (tau in the y-axis) is not entirely excluded from physical consideration? This would answer my question. $\endgroup$
    – Moonraker
    Commented Jun 4, 2014 at 8:27
  • 1
    $\begingroup$ No. I think the point that John is trying to make is - one has to have frame-dependent quantities on one side, and an invariant on the other side. That helps because in different frames, $dt$ and $dx$ shall vary, but in a way such that the interval defined as above stays invariant. Of course, you can take a quantity on the other side, but that does not improve anything here. Now you will have variants on both sides. $\endgroup$
    – 299792458
    Commented Jun 4, 2014 at 8:41
  • $\begingroup$ @Moonraker: I can't think of any useful application of having $\tau$ on the $y$ axis. Apart from anything else such a plot would exclude any spacelike connected regions of the spacetime (because $\tau$ would be imaginary). $\endgroup$ Commented Jun 4, 2014 at 9:06
  • 2
    $\begingroup$ @Moonraker, $\tau$ is not a coordinate. Plotting space coordinates against $\tau$ is plotting space coordinates against a parameter. The resulting plot will no longer be a spacetime diagram since you will have lost the time coordinate. $\endgroup$ Commented Jun 4, 2014 at 12:09

By defintion, Minkowski space $\mathbb{R}^{p,q}$ must have signature $(p,q)=(1,d-1)$, with metric,

$$ds^2 = -dt^2 +dx_1^2 + dx^2_2 + \dots$$

The signature $(+,+,\dots)$ corresponds to Euclidean space, which is obtained by a Wick rotation,

$$t\to -i\tau$$

to imaginary time $\tau$, and the metric is modified, in the case of Wick rotated Minkowski space, to $\delta_{\mu\nu}$. In many cases, it is convenient to do so, e.g. for the evaluation of the path integral. Specifically, as an example, in bosonic string theory, we rotate the Polyakov action to,

$$S=\frac{1}{4\pi \alpha'}\int d^2 \sigma \, \partial_\alpha X^\mu \partial_\beta X^\nu \delta_{\mu\nu}$$

Another example: In deriving the Bekenstein-Hawking entropy formula, we choose to approximate the partition function, normally given by a path integral, as

$$Z \sim \sum_{\text{classical solns.}} e^{-I_E}$$

where $I_E$ is the Euclidean Einstein-Hilbert action supplemented by necessary boundary terms. For the Schwarzschild metric, we'd Wick rotate to Euclidean space,

$$ds^2_E = \left( 1-\frac{2GM}{r}\right)d\tau^2 + \left( 1-\frac{2GM}{r}\right)^{-1}dr^2 + r^2 d\Omega^2_{\text{II}}$$

and impose periodicity on $\tau$ with period $\beta=1/T$. These are just a few instances of many where the signature $(+,+,+,+)$ is useful for computational purposes. As John Rennie pointed out correctly, just manipulating the invariant line element to,

$$dt^2 = ds^2 + dx^2$$

will achieve no effect, the metric is still technically $(1,1)$, and $dt^2$ is certainly frame dependent.

  • $\begingroup$ Thank you for your information about Wick rotation. Thank you also for your edit because it seems that there is a mistake in the formula of Wikipedia (I noted that you replaced proper time tau by time). I edited my question. $\endgroup$
    – Moonraker
    Commented Jun 4, 2014 at 8:36
  • 1
    $\begingroup$ Nice examples, +1. The only thing I really know about Wick rotation is its rather infamous involvement in the Black-Scholes equation used for option pricing: this is equivalent to a heat diffusion equation and a Wick rotation converts it to a Schrödinger-like equation which path integral techniques are then applied on. All very good in theory as long as stock prices move following white Gaussian processes, which insight Black and Scholes were awarded the Nobel economics prize for. However, they then thought they were hot shot fund managers and they ended up pushing things well ... $\endgroup$ Commented Jun 4, 2014 at 14:43
  • 1
    $\begingroup$ ..beyond the underlying assumptions - the history of Long Term Capital Management Inc makes interesting reading and a sobering lesson on the need to confront theory with experiment! $\endgroup$ Commented Jun 4, 2014 at 14:44
  • $\begingroup$ @WetSavannaAnimalakaRodVance very interesting, thanks. Also, you might be interested to read this book which also discusses why we can't use theory to predict the financial world. Jamal +1, really nice. Another important example would be instantons which are also very interesting (IMHO). $\endgroup$
    – Hunter
    Commented Jun 4, 2014 at 15:43
  • 1
    $\begingroup$ @Hunter: Yes, I absolutely share your enthusiasm for instantons! Unfortunately, I couldn't really include it without giving it justice by providing a full background in the answer first. $\endgroup$
    – JamalS
    Commented Jun 4, 2014 at 15:44

The pathological $i\,c\,t$ and / or a "trivial signature" seem very slick and simplying ideas on first glance, but the differences between Minkowsky and Euclidean space are actually rather deep and cannot simply be spirited away so readily.

Witness the following differences:

  1. The metric element (first fundamental form) in Euclidean space is a true metric: the distance between two elements in this space can only be nought if the two are the same points, and it is subadditive, i.e. fulfills the triangle inequality $d(x,z)\leq d(x,y) + d(y,z)$. This latter is very intuitve and asserts the everyday notion of "qualitative transitivity of nearness": roughly it means if $x$ is near to $y$ and $y$ near to $z$ then $z$ is "kind of" near to $x$.

  2. The metric element in Minkowski spacetime fulfils neither of these crucial properties: events separated by a null vector (which is different from the zero vector) have a distance of nought between them, and the metric element is NOT subadditive: the triangle inequality does not hold. So the Minkowsky "norm" is not even a seminorm in the mathematical sense.

With Euclidean spaces you're dealing with norms and inner products in the wonted, mathematician's sense. Their counterparts in Minkowsky spacetime are not of these kingdoms, even though they have some likenesses.

The Lorentz group is the set of all matrices which conserve the Minkowsky "norm": they conserve the quadratic form with the $+,\,-,\,-,\,-$ signature, and this can be shown to imply that the group members conserve the Minkowsky inner product as well. The introduction of complex numbers beclouds and messes up everything in this elegant description, because there is no concept of "signature" with complex matrix groups: in this case the notion of signature generalises to "matrices diagonisable to a matrix with terms of the form $e^{i\,\phi}$ along its leading diagonal". In such a group, one can follow paths that continuously deform the $e^{i\,\phi}$ terms into each other, so the notion of signature is lost.

You might like to check out my exposition of $SO^+(1,3)$ here for further details.

Any other "device" that "smooths over" the signature is thus likely to have limited application.

  • $\begingroup$ FYI so you can improve your awesome blog: In your exposition, you have a LaTeX error in equation 57, and just before equation 67 you've LaTeX'ified a normal sentence. $\endgroup$
    – JamalS
    Commented Jun 4, 2014 at 11:07
  • $\begingroup$ @JamalS Thanks muchly. At the moment it is all still very much a "brain dump and first draught stage" - it's grown from a set of notes I have written over the years to teach myself and also teach (with wildly variable success) colleagues who might use Lie theory for optical system design. One "mistake" I know for a fact is that I have written up a big body of stuff with the Adjoint representation around the wrong way: I've written $Ad(\gamma)\,X=\gamma^{-1}\,X\,\gamma$ whereas the wonted convention is the other way around - I dread fixing this one up and keep putting it off! It works of ... $\endgroup$ Commented Jun 4, 2014 at 12:14
  • $\begingroup$ ... course if one is consistent (which is why it took so long to pull myself up on this one), but it's simply confusing to break a widely used convention. I think it was because different group theorists use different orderings of conjugation - Emil Artin's Galois Theory book for example writes conjugation as $\gamma^{-1}\, \zeta\,\gamma$. $\endgroup$ Commented Jun 4, 2014 at 12:16
  • $\begingroup$ If you need to just replace a single expression, copy and paste the entire code in word, using the 'find' tool to automatically replaced all instances of a particular expression, and copy and paste back. $\endgroup$
    – JamalS
    Commented Jun 4, 2014 at 12:28

Consider a 2-D Euclidean vector $\mathbf v$. The length squared is

$$r^2 = \mathbf v \cdot \mathbf v = x^2 + y^2$$

where $x$ and $y$ are the components of the vector on some basis.

$$x = \mathbf v \cdot \hat e_x $$

$$y = \mathbf v \cdot \hat e_y $$

Now, we could write the following equation

$$y^2 = x^2 - r^2$$

but this would not imply that $r$ is a component of any vector because it isn't - $r$ is not a coordinate.

Nor could we interpret this as changing the Euclidean inner product to a Minkowski inner product. The right hand side is not an inner product since, in fact, the above equation is just

$$y^2 = x^2 - \mathbf v \cdot \mathbf v$$

Similarly, $\tau$ is not a coordinate and is not a component of a four-vector. We write, for a time-like displacement four-vector $\vec x$

$$\tau^2 = \vec x \cdot \vec x = x^{\mu}x_{\mu} = t^2 - r^2$$


$$t = \vec x \cdot \hat e_0 $$

Thus, though we could certainly write the equation

$$t^2 = \tau^2 + r^2$$

we do not interpret the right hand side as an inner product since the above equation is just

$$t^2 = \vec x \cdot \vec x + r^2$$

By replacing time with proper time on the y-axis of the Minkowski diagram

First of all, and most importantly, the resulting diagram would not be a spacetime diagram at all since the time coordinate would be suppressed; $\tau$ is not a coordinate.

Whereas a directed line segment between two events in a spacetime diagram is a four-vector, such a line segment between two points in your diagram would not be a four-vector.

A line or curve in your diagram could be interpreted as a plot of a family of world lines; a plot of the spatial coordinates of the events making up the world lines against the proper time along the world line.

However, from this diagram, we cannot identify the actual events along the world line since, on your diagram, the time coordinate is suppressed.

  • $\begingroup$ I think you did not think about the consequences of such a diagram: Why should tau (proper time, invariable spacetime interval) not be a time coordinate (replacing the Minkowski time coordinate) ? Why should Minkowski time disappear? As I pointed out, Minkowski (observer's) time is the diagonal of each rectangular triangle of worldliness. Why would there be no 4-vector between two events? In summary, the difference of such a diagram is that the universal proper time replaces the relative observer's time of the Minkowski diagram (which is becoming the abovementioned diagonal) $\endgroup$
    – Moonraker
    Commented Jun 4, 2014 at 14:29
  • 1
    $\begingroup$ @Moonraker, because $\tau$ is not a coordinate just as the length of a vector is not a coordinate. Either you see this distinction or you don't. If you don't, you've got some thinking to do. $\endgroup$ Commented Jun 4, 2014 at 14:48
  • $\begingroup$ Sorry, in fact I did not see that, this is an additional answer to my question (I did well to upgrade your answer)! In order to go on I would like to ask if proper time can be considered as a vector in my alternative spacetime diagram, because proper time is represented exclusively in upward direction (y-axis), thus the scalar of the spacetime interval (proper time) has received a (unique) direction. $\endgroup$
    – Moonraker
    Commented Jun 4, 2014 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.