# Why Minkowski spacetime and not Euclidean spacetime?

1. Aug 6, 2005

### cefarix

Why does everyone use +---/-+++ Minkowski spacetime over ++++ Euclidean spacetime? Minkowski spacetime preserves spacetime intervals under Lorentz transformations but so does Euclidean spacetime under equivalent rotational transformations from which SR can also be deduced.
(someone show me how to write eqs.)
If for example we take theta = arcsin v/c, then the rotational transformation equations would be (assuming the moving frame moves parallel to the x-axis)
x' = x * cos theta + ct sin theta = x * cos theta + vt
y' = y
z' = z
ct' = ct * cos theta - x sin theta = ct * cos theta - vx/c
Note that in this framework the Lorentz factor becomes gamma = sec theta.
The distance between any two points would still be an invariant among any frames. Taking an example, v = 0.5c, x = 10, and ct = 2, gives sqrt (104) as the distance in the original frame. The y and z coordinates will be ignored here for simplicity (they are zero) but you should be able to see that a transform will work with them as well. theta = 30 deg, so the Lorentz factor is sec 30 deg = 1.155. This agrees with 1/sqrt(1-v^2/c^2). x' is then 10 * 0.866 + 2 * 0.5 = 9.66 and ct' is -3.27. You can see from this that the distance is the same as the original frame, sqrt(104). So this preserves intervals across frame rotations due to relative velocity as well.
So why is this approach not used?

2. Aug 7, 2005

### Mortimer

You have stumbled over Euclidean relativity like a number of others did before you, see http://www.rfjvanlinden171.freeler.nl/#morelinks for references.
I guess the main reasons why Minkoswki is preferred:
- It has been succesfully used since 1907 in virtually all textbooks and lecture notes, so why change a winning team?
- There are some inconsistencies between Minkowski space-time and Euclidean space-time in situations involving 3 or more reference frames, like e.g. the velocity addition or certain particle collisions. Depending on the specific Euclidean approach (there is not yet general consensus), calculations may give deviating results.

Last edited by a moderator: Apr 21, 2017
3. Aug 7, 2005

### robphy

In Special Relativity (in Minkowski spacetime), the metric signature +--- or -+++ tells us that there are THREE types of vectors, timelike, spacelike, or null (sometimes called lightlike). Any Lorentz transformation preserves the type of the vector. That is, a spacelike vector cannot be transformed into a null or timelike vector... and likewise for the other two. A Euclidean metric does not have this structure.

Let me emphasize an important point included in the preceding paragraph. That any Lorentz Transformation preserves the type of a null (lightlike) vector encodes the statement that "the speed of light is the same for all observers". A vector (ct,x,y,z)=(k,k,0,0), where k is any real number, is a null vector under Lorentz transformations. It transforms to (k',k',0,0). What happens in your Euclidean approach?

Expanding further on the first paragraph, the null and timelike vectors are each divided into two classes, future and past. Spacelike vectors have no invariant distinction like this (... this encodes the relativity of simultaneity). Thses classes are presrved by orthochronous Lorentz transformations. That is, a future pointing vector cannot become a past pointing vector.

Here is another point comparing the Lorentz Transformations and the Euclidean Transformations. The parameter for the Lorentz boosts, called the rapidity which I will denote here as $$\theta_M$$, ranges from $$-\infty\mbox{ to }\infty$$. The Euclidean rotations have a parameter, called the angle and denoted by $$\theta_E$$, with a finite range, say, $$-\pi \mbox{ to }\pi$$, which, if extended to a larger range, shows a periodicity.

The rapidity is related to the spatial velocity of an observer (with a necessarily future timelike unit 4-velocity) by $$v=\tanh(\theta_M)$$, and it is an additive parameter in the following sense. For clarity, since spatial velocities are "relative", write (dropping the M subscript) $$v_{10}=\tanh(\theta_1-\theta_0)$$, which measures the relative velocity of observer 1 with respect to observer 0. One can write $$v_{21}$$ similarly. So, what is $$v_{20}$$ in terms of $$v_{21}$$ and $$v_{10}$$? Write $$v_{20}=\tanh(\theta_2-\theta_0)=\tanh( (\theta_2 - \theta_1)-(\theta_1 - \theta_0))$$, then use the trig identities for the hyperbolic tangent. You'll get the standard velocity composition formula. Exercise(?): the method I laid out here is straightforward... carry this out using your Euclidean approach... what do you get...and what does it explicitly look like? I'd be curious to see.

4. Aug 7, 2005

### Mortimer

The velocity addition formula indeed doesn't hold in the Euclidean approach and this could be Achilles' heel of Euclidean relativity. I've seen a derivation from Montanus that gives correct results but requires some sort of absolute reference frame.
I like to remark however that there seems to be not much consensus on the correct way the velocity addition formula should be derived. I believe it was pmbphy or robphy who recently showed me yet another type of derivation that I didn't know of besides 5 or 6 different ways (really different in terms of the used reasoning) that I had already encountered before. I therefor cautiously tend to doubt the experimentally verifiable correctness of this equation. I did an extensive search for experimental verifications of this specific equation in the area of velocities where the Euclidean form shows significant deviations (at velocities above let's say 1/2 c for both 2nd and 3rd frame) and found none. I'd still be very interested in references though.

5. Aug 7, 2005

### learningphysics

These transformation equations seem incorrect to me. Suppose we take a light ray with equation x = ct... transform this into the S' frame we get:

$$x' = \frac{c*cos\theta+v}{c*cos\theta - v} ct'$$

So the speed of light is no longer c in the S' frame unless we take v=0 (remain in the same frame), or costheta = 0.

6. Aug 7, 2005

### robphy

With all of the symmetries enjoyed by Minkowski spacetime, one should expect (and possibly be pleasantly be surprised) that there are many ways to get the same result. Is there ONE proof of the Pythagorean theorem in Euclidean space?

Experimentally speaking (with our current technology), I would think that one would look at collision experiments for verification. In fact, one could use the velocity composition law to derive the form of the 4-momentum. (I have to look up the reference. Before I found that reference, I had been unhappy with the way the form of 4-momentum gets motivated.)

Last edited: Aug 7, 2005
7. Aug 8, 2005

### Mortimer

I checked this and more or less came to the conclusion that in all collision experiments the lab frame is used as the single frame of reference for all measurements. There are no moving frames from within measurements take place (at least I haven't found such experiments). This is hardly surprising of course. It would be difficult to have measuring devices move at the particle's speeds.

8. Aug 8, 2005

### Mortimer

In what I've seen, most Euclidean approaches keep the postulate of the constancy of lightspeed simply upright against all odds (like Einstein did). I think this essentially means that for a moving observer the velocity vector of photons must also rotate at the same angle in SO(4).

9. Aug 8, 2005

### pervect

Staff Emeritus

If the Lorentz interval of a lightbeam is zero, any transformation which preserves the Lorentz interval must preserve the speed of light.

This leads me to wonder whether the proposed transformation actually does leave the Lorentz interval invariant. Unless I've missed something, it does not

x1 := x * cos (theta) + c*t* sin (theta);
t1 := t * cos (theta) - x * sin(theta)/c;

which was the posted transformation, doesn't seem to work (using maple's simplify command, it doesn't simplify out).

Replacing the first line with

x1 := x * cos (theta) - c*t* sin (theta);

comes closer to working

x1^2 - c^2*t1^2 = (2*cos(theta)^2-1)*(x^2-t^2*c^2)

which doesn't leave the Lorentz interval invariant, but multiplies it by a scale factor depending on theta.

I suspect that this may be the actual proposal, a transform that leaves 'c' invariant while not preserving the Lorentz interval. However, that was not the original claim as I read it.

10. Aug 8, 2005

### neopolitan

Orthogonality

This may not be the standard reason given, but for me, the reason why spacetime is a "curved pseudo Riemannian manifold with a metric of signature of (-+++)" is because the temporal dimension is universally orthogonal to the spatial dimensions.

The temporal dimension is also markedly different to the spatial dimensions. I can choose whatever spatial axes I like and I can move in any direction I like along those axes. The same cannot be said for the temporal dimension.

Note also that I can move perpendicularly to any spatial axis I choose to select, but I cannot move perpendicularly to the temporal axis - to do so would mean instantaneous translation from one point to another.

If you use ++++ Euclidean spacetime, you are basically implying that you could choose axes which combine space and time. It'd be an interesting exercise to see what problems you'd run into trying to get that to work.

11. Aug 8, 2005

### cefarix

Ok, it seems to me now that my initial take on this was mistaken.
In Euclidean spacetime, when being used for position, t = 0 always. t is non-zero when dealing with 4-momentum. The t component of 4-momentum is the mass. When a particle is accelerated, from a "lab" frame, 4-momentum is added to the particle's 4-momentum which is perpendicular to the time-axis. So the new 4-momentum is tilted, and you can define an angle theta between the time axis and the 4-momentum vector. The sine of theta multiplied by c would give the particle's new velocity in the lab frame. With respect to itself, of course, the particle is at rest and its 4-momentum is parallel to its time axis. But since the 4-momentum increased in magnitude, for the particle itself, its mass has increased. The mapping of coordinates can be defined as: x = x' cos theta, where x' is in the frame of the particle, and x is in the lab frame. Also, if the particle were initially moving, and we fired a photon at it in the same direction it was moving, so that the photon would collide with the particle and transfer its momentum to it, then it can be shown that whether we look at it from the lab frame or the particle frame, it gives us the same result. In the lab frame, the photon's momentum is p, the particle's time-momentum is m, and the particle's space-momentum is q, then we get r = sqrt(m^2 + (p+q)^2), the 4-momentum of the particle after the collision. The new angle theta' would be theta' = arctan ((p+q)/m), so the new velocity, in the lab frame, becomes v' = c sin ( arctan ((p+q)/m) ). If we look at it from the particle's frame, then the photon's momentum is no longer p, but p' = p cos theta. Since in the particle's own frame its at rest, it has no space-momentum, but only some time-momentum, m'. The resulting momentum r' = sqrt(m'^2 + p'^2). From here we can calculate the change in angle dtheta from within the particle's frame, and it would be: dtheta = arctan (p'/m'). dtheta back in the lab frame would be equal to dtheta' = dtheta cos theta, and the new resultant theta in the lab frame would be theta_r = dtheta' + theta. The new velocity in the lab frame is then v' = c sin theta_r