Epimetheus,
While the rotating hypercube even exists as a well-known screensaver (under Linux => xscreensaver), the 4d space of relativistic physics is different in an essential aspect:
It's the metric! The 4d space is called a
Minkowski space while that hypercube lives in a so-called 4d
Euclidean space.
Without getting too mathematical, the metric determines e.g. how the length^2 of a vector is calculated in terms of it's components.
Let X=(x0,x1,x2,x3)
be a general 4-vector, then in
Euclidean space we calculate the length^2 as
X.X = X^2 = x0^2 + x1^2 + x2^2 + x3^2,
i.e as the straightforward generalization of what we know and love in 3 dimensions. For the
Minkowski space of relativity, there are some crucial '-' signs involved, such that the "length^2" of a vector is NOT necessarily positive!!!
X.X = X^2 = x0^2 - x1^2 - x2^2 - x3^2
(Note: sometimes you see another convention, which is /physically/ equivalent: X.X = - x0^2 + x1^2 + x2^2 + x3^2)
So from what I understand the extra element is time element, letting t represent time, so, in 4D vectors we have (x,y,z, t).
Almost...Time and space don't have the same physical dimensions, hence cannot be the components of one vector
. What is lacking is c, the speed of light: c*t = x0 is the first, extra component (remember [velocity] = [length/time]), and x1=x, x2=y, x3=z.
So a 4d positional vector has the components
X= (c*t, x,y,z)
The amazing fact is, that with the Minkowski metric, the
length^2 of any such vector is the SAME in EVERY relativistic reference frame, it's an
invariant under Lorentz frame transformations as we say.
So we can evaluate the length^2 of X in ANY frame that is CONVENIENT, without changing the underlying physics!!Consider the components of the vector X in 2 such frames, this means
X.X=X^2 = (c*t)^2 - x^2 -y^2 -z^2 = (c*t')^2 - x'^2 -y'^2 - z'^2
Solving that equation properly, will immediately give you the famous
length contraction and time dilatation of special relativity!
The
proper time tau that parametrizes the world lines is nothing but the time in a special frame: t'=tau, x'=y'=z'=0, such that
X^2 = (c*t)^2 - x^2 - y^2 -z^2 = (c*tau)^2
We can immediately parametrize the components of the vector X in some frame as function of the proper time such that this equation is automatically satisfied. Would you know how this is done??
(Hint use the hyperbolic sinus/cosinus, sinh(tau), cosh(tau), satisfying sinh^2 - cosh^2 =1 )
The task is the complete analog of the well-known parametrization of a 3d unit vector in terms of cos and sin with cos^2 + sin^2 =1, namely
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X_3d = (x1,x2,x3) = (sin(tau), sin(phi)*cos(tau), cos(phi)*cos(tau))
then
X_3d^2 = sin^2(tau) + (cos^2(phi) + sin^2(phi))* cos^2(tau) = sin^2(tau) + cos^2(tau) = 1
[You see, while a general 3d-vector depends on 3 components x1,x2,x3, the constraint X_3d^2=1 eliminates one. The above trigonometric parametrization makes this manifest in terms of only tau and phi!]
Since a speed vector is generally the derivative of the position vector, we now define a 4d-velocity vector, the /tangent vector/, by forming the derivative wrto the /proper time tau/
V = dX/dtau = (c*dt/dtau, dx/dtau,dy/dtau,dz/dtau)
Since the 4d velocity is once more a proper 4-vector, again, it's length^2
V.V = V^2
is frame independent!
Moreover one easily finds that the vector product
V.X
is also an invariant under Lorentz frame transformations.
+++++++++++++++++++
Hence, after identifying all invariants under Lorentz frame transformations, the world line orbits can be parametrized easily in terms of invariants and covariants such that one can read off what part of the orbit is frame independent and what is frame dependent. The straight analog of this proven procedure is what
I am proposing since long for making the frame dependence of the non-relativistic orbits of Celestia manifest!+++++++++++++++++++
F.
