Relativity Metric: UNIZOR.COM - Relativity 4 All - Einstein View

Notes to a video lecture on UNIZOR.COM

Relativity Metric


Metric is one of the most important characteristic of space we live in, and it defines a distance between two points.

Prior to Theory of Relativity our space was considered three-dimensional Euclidian and, therefore, the square of a distance between two points
A(X_A,Y_A,Z_A) and B(X_B,Y_B,Z_B)
was equal to
d²(A,B) = (X_B−X_A)² +
+ (Y_B−Y_A)² + (Z_B−Z_A)²

We have analyzed how this distance transformed by Galilean transformation of coordinates from one inertial system into another, moving relative to the first, and came up with its invariance relative to this transformation.

Indeed, the Galilean transformations from inertial reference frame α{X,Y,Z} into inertial reference frame β{x,y,z} that moves relative to α with speed v along its X-axis are
t = T
x = X − v·T
y = Y
z = Z
where T=t is absolute time in both reference frames.

Then the distance between points A and B in β-frame is
d_β²(A,B) = (x_B−x_A)² +
+ (y_B−y_A)² + (z_B−z_A)² =
= (X_B−v·T−X_A+v·T)² +
+ (Y_B−Y_A)² + (Z_B−Z_A)² =
= d_α²(A,B)

With a progress of the Theory of Relativity we came up with Lorentz Transformation of coordinates that takes time into account.
This type of transformation seemed to better correspond to our theories and experimental data, but a concept of length in our traditional sense was no longer invariant (see Length Transformation lecture in this part of a course).

An interesting approach would be to find a different definition of metric in our space that takes time into account and is invariant relative to Lorentz Transformation analogously to classic definition of Euclidean metric being invariant relative to Galilean transformation (see Metric Invariance lecture in the Galilean View part of a course).

Let's attempt to apply Lorentz Transformation to classical definition of Euclidean metric, as mentioned above, and see where it leads us.

Assume, we have two reference frames α{T,X,Y,Z} and β{t,x,y,z}.
Further assume that β-frame moves relative to α-frame along X-axis with constant velocity vector V whose α-coordinates are {v,0,0}.

Then Lorentz transformation of coordinates from α to β is
t = γ·(T − v·X/c²)
x = γ·(X − v·T)
y = Y
z = Z
where γ = 1/√1−v²/c².

Applied this transformation to coordinates of points A and B, we obtain
x_A = γ·(X_A − v·T_A)
y_A = Y_A
z_A = Z_A
and
x_B = γ·(X_B − v·T_B)
y_B = Y_B
z_B = Z_B

Now we are ready to calculate the Euclidean distance between these two points in both reference frames α{T,X,Y,Z} and β{t,x,y,z}.

d_α²(A,B) = (X_B−X_A)² +
+ (Y_B−Y_A)² + (Z_B−Z_A)²

d_β²(A,B) = (x_B−x_A)² +
+ (y_B−y_A)² + (z_B−z_A)²

As seen from the coordinate transformation above,
x_B−x_A =
= γ·[(X_B−v·T_B) − (X_A−v·T_A)] =
= γ·(X_B−X_A)−γ·v·(T_B−T_A)
y_B−y_A = Y_B−Y_A
z_B−z_A = Z_B−Z_A

Obviously, unless v=0 and, consequently, γ=1,
x_B−x_A ≠ X_B−X_A
and, as a result,
d_α²(A,B) ≠ d_β²(A,B)
in a general case of reference frames moving relative to each other.
Lorentz Transformation does not preserve the Euclidean metric from one inertial reference frame to another, unless they are not moving relative to each other.

Let's proceed in our quest for a different definition of metric in our space that takes time into account and is invariant relative to Lorentz Transformation.

We change the definition of Euclidean metric d²(A,B) by adding a time-dependent component:
D²(A,B) = (X_B−X_A)² +
+ (Y_B−Y_A)² + (Z_B−Z_A)² +
+ f(T_B−T_A)
where f(T) is some function of time coordinate.

The first and the easiest attempt would be to add a component linearly dependent on a square of time differences similarly to dependency on space coordinates:
f(T) = k·T²
It's reasonable to try to find a factor k that would assure the invariance of Lorentz Transformation. If we succeed, great. If not, we'll try some other solution.

According to our newly proposed metric, the relativistic distance from A to B in α-frame is
D_α²(A,B) = (X_B−X_A)² +
+ (Y_B−Y_A)² + (Z_B−Z_A)² +
+ k·(T_B−T_A)²

To find unknown coefficient k (if possible, of course), we equate
D_α²(A,B) = D_β²(A,B)

Considering only X-coordinate and time change from α-frame to β-frame, our equation for k is
(X_B−X_A)² + k(T_B−T_A)² =
(x_B−x_A)² + k(t_B−t_A)²

If the solution for k is independent of the relative speed v of β-frame moving relative to α-frame and independent of coordinates of points A and B, our problem is solved.

Solving the above equation using Lorentz Transformation from α- to β-frame.
(X_B−X_A)² + k(T_B−T_A)² =
γ²·[(X_B−v·T_B) − (X_A−v·T_A)]² +
+ k·γ²·[(T_B−v·X_B/c²) −
− (T_A−v·X_A/c²)]²

Let's simplify the right side of this equation.

It's equal to

γ²·[(X_B−X_A) − v·(T_B−T_A)]² +
+ k·γ²·[(T_B−T_A) −
− (v/c²)·(X_B−X_A)]² =

= γ²·[(X_B−X_A)² −
− 2v·(X_B−X_A)·(T_B−T_A) +
+ v²·(T_B−T_A)² +
+ k·(T_B−T_A)² −
− 2(k·v/c²)·(X_B−X_A)·(T_B−T_A) +
+ k·(v/c²)²·(X_B−X_A)²] =

= γ²·[(1+k·(v/c²)²)(X_B−X_A)²
− 2(v+k·v/c²)·(X_B−X_A)·(T_B−T_A)
+ (v²+k)·(T_B−T_A)²] = D_α²(A,B)

Recall the metric in α-frame is
D_α²(A,B) =
(X_B−X_A)² + k·(T_B−T_A)²

To satisfy the equation
D_α²(A,B) = D_β²(A,B)
the corresponding coefficients in the increments along each coordinate must be equal: γ²·(1+k·(v/c²)²) = 1
2(v+k·v/c²) = 0
γ²·(v²+k) = k

If our approach is right, the three equations above must have one solution for k that satisfies all of them, and it must not depend on speed v.
Is it possible?

The second equation above can be used first, and the value of k from it is k=−c².
Let's check if this is a solution to the first and the third equations.

The first equation with k=−c² is
γ²·(1−c²·(v/c²)²) = 1
Considering γ² = 1/(1−v²/c²),
the left side is unconditionally equal to 1.

The third equation with k=−c² looks as follows
γ²·(v²−c²) = −c²
Obviously, this is an identity.

It looks like the value k=−c² fits all our criteria and the relativistic distance in α-frame
D_α²(A,B) =
= (X_B−X_A)² − c²·(T_B−T_A)²
is equal to relativistic distance in α-frame
D_β²(A,B) =
= (x_B−x_A)² − c²·(t_B−t_A)²
and, therefore, is relativistic invariant.