# Euclidean space to Minkowski spacetime

Can you continuously deform (i.e., shrink, twist, stretch, etc. in any way without tearing) four-dimensional Euclidean space to make it four-dimensional Minkowski spacetime?

What do u mean by continuity here? In Mathematics, both Euclidean space (3+1) and Minkowski space are homeomorphic to $$\mathbb{R}^4$$ as a topological manifold. Then the identity map between them is the homeomorphism(also a continuous map)

Both 4D-Euclidean space and (3+1)D-Minkowski spacetime are 4D-vector spaces.
Indeed, $$\vec R=\vec A+\vec B$$ is the same operation in both spaces.

What differs is the assignments of square-magnitudes to the vectors and the assignments of "angles" between the vectors, which are both provided by a metric structure added to the vector space structure.

To continuously transform from one to the other, leave the vector space structure alone,
and change the signature of the metric tensor field.

Write $$g_{ab}=\left( \begin{array}{cccc} 1 & 0 & 0 &0 \\ 0 & -E & 0 &0 \\ 0 & 0 & -E &0 \\ 0 & 0 & 0 &-E \end{array}\right)$$ and let $$E$$ vary from $$-1$$ (Euclidean space) to $$+1$$ (Minkowski spacetime), with $$0$$ as the degenerate time-metric of the Galilean spacetime.

Try my Desmos visualization:

robphy v8e spacetime diagrammer for relativity (v8e-2021) - t-UP https://www.desmos.com/calculator/emqe6uyzha
and play with the E-slider. Of course!

It is called Wick rotation. You have to change the time coordinate $$t$$ to $$-it$$, which changes the sign of the action in the partition function of a physical system. It is a basic tool is statistical physics. The partition function then changes from $$\exp(-iS)$$ to $$\exp(S)$$. As I'm writing its rather backwards, typically the action of a physical system is given in Lorentzian signature and the rotation changes it to Euclidean. Then you perform calculations/simulations in Euclidean signature and perform an analytic continuation and rotate the solution back to Lorentzian.

There are some caveats in particular cases, but this is how it works.

• Not sure if $t \rightarrow -it$ is continuous. It looks discontinuous. But perhaps you can say $t \rightarrow \exp(i \theta) t$ and let $\theta$ vary smoothly from $0$ to $-\pi/2$. Questioner might be looking for an answer restricted to real numbers? Oct 27, 2021 at 20:26
• @AndrewSteane You maybe right. If continuity is important, then one has to find a function that interpolates between 1 and $-i$ continously and should multiply with such a term, as you are proposing. Oct 27, 2021 at 20:40

It sounds like you looking for a way to highlight the similarities between the two spaces. There are deep differences between the two. This is a truly counter-intuitive subject, every bit as bad as quantum mechanics. Long story short - time is very different from what everyday experience leads you to think it is. It mostly gets glossed over, because all you really need to know follows from "the speed of light is c".

In math, a vector space is a set of objects that can be added together and multiplied by a number. The space often has a metric that says how long the vectors are. In physics, there is an additional requirement that the space behaves as expected under coordinate transformations. E.G. Lengths and angles must be preserved under rotations.

In classical mechanics, if you have a stick at rest, you can measure its length with a ruler. You hold the ruler against the stick and note which marks line up with the ends.

You could do the same measurement while running by the stick. If you are moving, it is important to match up marks with the two ends at the same time. But you expect to get the same length. Length is a property of the stick. Your velocity (or equivalently, the stick's velocity) does not change it.

You could use vectors to do the same measurement. You set up a coordinate system (essentially a whole lot of rulers). You note the coordinates of the ends, and subtract.

You could define a coordinate system that follows you as you run by. Again you expect length is length.

Finally, you can rotate the stick. This changes the x, y, and z components of the length. But you still get the same total length. You are not confused. You know that x or y or z alone is not the complete length. You need all of them.

The important thing is some quantities (length) are invariant. They are in some sense real. They exist, independently of how you measure them. You get the same answer no matter how you measure them. Others (coordinates) are not complete invariant quantities. They do change, but work together to produce an invariant.

When people explored electromagnetism, they discovered something very unexpected. The speed of light is always $$c$$, no matter how you measure it.

Classically, this is impossible. If it is true, something is very, very wrong with our everyday understanding of the world around us. And it is true. So using everyday intuition to understand relativity often points in the wrong direction. Right answers are often "impossible."

One impossible result can be used to derive others. And so we have the failure of simultaneity, length contraction, and time dilation. We get twin and ladder paradoxes - classically impossible behavior that illustrates how the universe really works.

The first step to making sense out of it comes from constructing spacetime.

An interval of space is not invariant, as it was classically. A moving observer measures a contracted length. Time intervals are not invariant. A moving observer measures a dilated time.

But can construct an invariant interval of spacetime. This puts physics back on the footing of invariants that exist and can be measured how you like.

So the question becomes why is time so much like space that you can put them together and expect $$\Delta t^2$$ to work just like another coordinate together with $$\Delta x^2$$? The answer is that time is not at all what you think it is.

Classically, space is all there is. Space keeps changing state. Time describes the progression from past to present to future. But space in the present instant is all that exists. The past is over and gone. The future hasn't happened yet.

If you wanted, you could stack all the different versions of space on top of each other into a 4-D block. But it isn't needed. You can do physics without it. If you did this, time would uniquely label each slice. Given any 4-D point, everyone would agree on which other points have that same t.

It isn't a space the same way 3-D space is. Left/right, front/back, and up/down are all the same kind of thing. Given two points, you can use coordinates to calculate the distance between them.

Classical time is different from space. Everyone will agree on the value of a time interval. But it is different from a space interval.

It doesn't make sense to ask about the separation between points in this 4-D construct. How far away is the point $$1$$ m to the left and $$2$$ sec ago?

For these reasons, classical space and time are described as 3+1. There are 3 dynamical variables, x, y, and z (or similar variables for other coordinate systems), and 1 parameter, t. The dynamical variables are coordinates of the space. Time is not.

Given that c is constant, you can show the universe doesn't work classically.

If you stand still and measure the length of the stick, you identify space and time coordinates for both ends of the stick. You make sure the times are equal and subtract the spatial coordinates.

If you run by and try this, you see that the times of those two points are no longer equal. You have to choose different $$4$$-D points to get a simultaneous measurement.

This is huge. This should should totally confuse you for a long time. It is the single most counter intuitive thing about special relativity.

If an event can be simultaneous with an event from my present in one frame, and a different event from my past in another, then this idea that the present is all that exists must be wrong. This leads to questions like What is time, does it flow, and if so what defines its direction?.

Most of the answers to that question invoke the block universe. This is the idea time does not flow. Instead, the past, present, and future all statically exist. It is much like all the points in 3-D space exist, even though you are only using one of them.

Another way to think of this is that time does flow. Objects trace trajectories in spacetime, always moving toward the future. But there is no universal way to match up the present instant of different trajectories. See my answer.

Another very important point is that time now has properties that you only think of space as having. This is best seen on a space time diagram. I stand by the road as a car passes by.

• As I move forward in time, my position is always $$O$$, or $$x = 0$$. At the instant the car passes me, its position is also $$O$$. I say whatever happens on the road at $$P_1$$ will happen someplace else.
• The driver says his position the same for all times and the world is passing by. At the instant I pass him, his position is $$O$$, or $$x' = 0$$. I momentarily have the same position as he does, but I move on. He says whatever will happen at $$P_1$$ will happen right here.
• It is OK for him to think of $$O$$ and $$P_1$$ happening in the same place, but for me in a different frame to think they happen in different places.
• So long as a point can be reached from $$O$$ by traveling slower than c, it doesn't matter how far away it is. A frame of reference (car speed below c) can be arranged where it happens at the same place as $$O$$.

Time has all these properties too.

• The entire road moves forward in time. At the instant the car passes me, the entire road is at $$t = 0$$. I say whatever happens on the road at $$P_2$$ will happen in the future.
• The driver says the entire road moves forward in time. At the instant I pass him, the entire road is at $$t' = 0$$. I momentarily have the same time as he does, but I move on. He says whatever happens on the road at $$P_2$$ is happening right now.
• It is OK for him to think of $$O$$ and $$P_2$$ happening at the same time, but for me in a different frame to think they happen at different times.
• So long as you would have to travel faster than light to reach a point from $$O$$, it doesn't matter how far in the past or future it is. A frame of reference (car speed below c) can be arranged where it happens at the same time as $$O$$.

Finally, $$ct$$ has units of distance.

It works out that time is so much like space that you can make a $$4$$-D space with $$4$$ dynamical variables $$t$$, $$x$$, $$y$$, and $$z$$. The space has a physically meaningful notion of length that can either be a distance or a time. It behaves like a space should under coordinate transformations. The name of this space is spacetime.

If you do find a transformation that continuously maps $$3$$+$$1$$-D space into $$4$$-D spacetime, it doesn't show how similar they are. Instead, it highlights their differences.

In $$3$$+$$1$$-D space, length and time intervals are separately invariants. In $$4$$-D spacetime, they are not. The mapping does not preserve the metric.

What is Minkowski space-time? It simply means that the allowed transformations which preserve the laws of physics are invariant under O(3,1) transformations.

What is Euclidean space? It simply means that the allowed transformations which preserve the geometric objects are invariant under O(4) transformations. i.e. a geometric object described by coordinates must be equivalent to the same geometric object after an O(4) transformation.

Both use the same $$\mathbb{R}^4$$ space.

On the other hand one can draw small light cones at every point in Minkowski space-time and hence impose a partial ordering on the points. O(3,1) transformations preserve this partial ordering. Thus one might define Minkowski space-time as $$\mathbb{R}^4$$ plus this partial ordering. In which case the two spaces might be seen as fundamentally different.

The point here is that it is not the spaces that are different but the equivalence classes of geometric objects defined in the spaces that are different.

Is it possible to find real functions $$f(x,y)$$ and $$g(x,y)$$, such that, for any four real numbers $$a,b,c,d$$ such that $$(a-c)^2-(b-d)^2\geq 0$$, we have: $$(a-c)^2-(b-d)^2=(f(a,b)-f(c,d))^2+(g(a,b)-g(c,d))^2$$

The context is that I want to map events on spacetime to transformed co-ordinate axes, such that the space-time interval between any two events becomes equal to the euclidean distance between the two events.

Assume $$f,g$$ are functions which solve your problem. Then, for $$a-c=d-b$$, ie $$a+b= c+d$$, we get : $$0 = f(a,b) - f(c,d) = g(a,b) - g(c,d)$$
this means that there must be functions $$F(x),F(y)$$ such that : $$f(x,y) = F(x+y)\qquad \text{and}\qquad g(x,y) = G(x+y)$$
But then, for $$a=b$$ and $$c=d=0$$, we get : $$0 = F(2a) - F(0) = G(2a) - G(0)$$ so $$F,G$$ are constant, which is absurd.
Remark : the solution I mentionned originally $$f(x,y) = x$$ and $$g(x,y) = iy$$ is used in physics. It is called Wick rotation