Minkowski spacetime: Is there a signature (+,+,+,+)?

Question

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 (+, +, +, +).

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $\vec x$

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

where

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

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

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

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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.