A static spacetime has a well-defined concept of simultaneity
Suppose that we have a static spacetime whose spatial metric is euclidean. Suppose that clocks run slower close to the origin of spatial coordinates, but far away, the space is the Minkowski space.
Since the spacetime is static, we have a well-defined concept of simultaneity in it. We can map the time of any spacetime event X to the time of a far-away static observer A by letting a light signal travel from A to X and back. A maps X to
(t_0 + t_1) / 2,
where t_0 is the signal departure time in A's clock and t_1 is the arrival time.
We can define A's time as the global time. In the Minkowski space area, it is also the clock time of all other static observers, but close to the origin, static observers think that the global time runs faster than their own clocks.
A fast moving perfectly rigid grid
Let us assume that we have a static spacetime. Let us assume that the spatial metric is euclidean throughout the spacetime, but time may run slower in the area T where -1 < x < 1 and -1 < y < 1. Outside that area, the spacetime is strictly Minkowski.
We define a global time coordinate by a clock of a Minkowski observer, and global spatial coordinates through the euclidean metric.
| |
D --------------
| |
b(D)-------------- ----> v
| |
Let us have a perfectly rigid grid moving to the direction of the x axis. We assume that its speed measured in the Minkowski area is a constant v, and that in the Minkowski area at each global time t, the grid is perfectly rectangular and it is aligned along the directions of the global x and y axes.
Let us look at the movement of the grid bar B whose y coordinate is 0 and compare it to the bar B_2 whose y coordinate is 2.
Let us paint a dot D in B and a brother dot b(D) in B_2, so that their x coordinates in the internal coordinated of the grid are the same.
Let us denote by A the area -1 < x < 1.
The dot D enters A at the same global time as its brother b(D), and exits at the same global time, because the grid is perfectly rigid and rectangular outside the special area T.
The average speed, measured in the global coordinates, of D and b(D), during the journey through the area A is the same. D can move at the most at the local speed of light, measured in the global coordinates.
If we let v approach the speed of light c in the Minkowski space, then we know that the average speed of light in the special area T must be at least as high.
This still leaves open that at some spots, the speed of light might be very slow.
If the speed of the grid in the area T only depends on the position, v(x), and not on the time, can we show that length contraction forces the speed of light to the same as the Minkowski speed, throughout the area?
Let us do a little perturbation calculation.
Suppose that it takes a time 2 - d for D to go from x = -1 to 0, and a time 2 + d to x =1.
The average speed is 0.5. Let the speed of light be a constant 1 in the area. The inverse length contraction is
1 / sqrt(1 - 1 / (2 - d)^2)
+ 1 / sqrt(1 - 1 / (2 + d)^2).
How does this compare to the constant speed inverse contraction
2 / sqrt(1 - 1/4)?
It depends on the second derivative of
1 / sqrt(1 - 1 / t^2) = f(t)^-1
at t = 2. The derivative of the denominator f(t),
f'(t) = (1 - 1 / t^2)^-0.5 * t^-3
is positive.
The second derivative is
f''(t) = (...)^-0.5 * -3 * t^-4
-0.5 * (...)^-1.5 * 2 * t^-3 * t^-3.
The second derivative is negative.
The derivative of f(t)^-1 is
- f(t)^-2 * f'(t).
The second derivative,
2 f(t)^-3 * f'(t)^2 - f(t)^-2 * f''(t)
is positive. We see that "perturbing" the constant speed tends to increase the inverse length contraction.
Finding the minimal inverse length contraction when the average speed through the area T is set, and the speed v(x) only depends on x, is a problem of variational calculus. If the speed of light is constant in T, then the optimum might be at a constant speed v(x) throughout the area.
The rigid grid forces the inverse length contraction to be the same in the bars B and B_2, in the area A. If the speed of light is the same throughout the area, v(x) = v gives the minimal inverse length contraction for B. It is the same as for B_2.
If the speed of light is lower at some spot of T, then the inverse length contraction grows for all v(x). Then the inverse length contraction of B is necessarily larger than B_2, which contradicts our assumption about rigidity.
We proved that the speed of light everywhere in T must be the same as in the Minkowski space, but our proof hangs on proving the variational calculus result above.
If we allow v(t, x) depend on the time t, too, then the variational problem is harder. Some strange, fractal-like function might beat the constant speed in optimality. We need to check literature about special relativity and if there are any variational calculus results.
In newtonian mechanics, there is no length contraction, and it is easier to prove that a perfectly rigid grid stays rectangular at each global time moment t.
We define a global time coordinate by a clock of a Minkowski observer, and global spatial coordinates through the euclidean metric.
| |
D --------------
| |
b(D)-------------- ----> v
| |
Let us have a perfectly rigid grid moving to the direction of the x axis. We assume that its speed measured in the Minkowski area is a constant v, and that in the Minkowski area at each global time t, the grid is perfectly rectangular and it is aligned along the directions of the global x and y axes.
Let us look at the movement of the grid bar B whose y coordinate is 0 and compare it to the bar B_2 whose y coordinate is 2.
Let us paint a dot D in B and a brother dot b(D) in B_2, so that their x coordinates in the internal coordinated of the grid are the same.
Let us denote by A the area -1 < x < 1.
The dot D enters A at the same global time as its brother b(D), and exits at the same global time, because the grid is perfectly rigid and rectangular outside the special area T.
The average speed, measured in the global coordinates, of D and b(D), during the journey through the area A is the same. D can move at the most at the local speed of light, measured in the global coordinates.
If we let v approach the speed of light c in the Minkowski space, then we know that the average speed of light in the special area T must be at least as high.
This still leaves open that at some spots, the speed of light might be very slow.
If the speed of the grid is time-independent
If the speed of the grid in the area T only depends on the position, v(x), and not on the time, can we show that length contraction forces the speed of light to the same as the Minkowski speed, throughout the area?
Let us do a little perturbation calculation.
Suppose that it takes a time 2 - d for D to go from x = -1 to 0, and a time 2 + d to x =1.
The average speed is 0.5. Let the speed of light be a constant 1 in the area. The inverse length contraction is
1 / sqrt(1 - 1 / (2 - d)^2)
+ 1 / sqrt(1 - 1 / (2 + d)^2).
How does this compare to the constant speed inverse contraction
2 / sqrt(1 - 1/4)?
It depends on the second derivative of
1 / sqrt(1 - 1 / t^2) = f(t)^-1
at t = 2. The derivative of the denominator f(t),
f'(t) = (1 - 1 / t^2)^-0.5 * t^-3
is positive.
The second derivative is
f''(t) = (...)^-0.5 * -3 * t^-4
-0.5 * (...)^-1.5 * 2 * t^-3 * t^-3.
The second derivative is negative.
The derivative of f(t)^-1 is
- f(t)^-2 * f'(t).
The second derivative,
2 f(t)^-3 * f'(t)^2 - f(t)^-2 * f''(t)
is positive. We see that "perturbing" the constant speed tends to increase the inverse length contraction.
Finding the minimal inverse length contraction when the average speed through the area T is set, and the speed v(x) only depends on x, is a problem of variational calculus. If the speed of light is constant in T, then the optimum might be at a constant speed v(x) throughout the area.
The rigid grid forces the inverse length contraction to be the same in the bars B and B_2, in the area A. If the speed of light is the same throughout the area, v(x) = v gives the minimal inverse length contraction for B. It is the same as for B_2.
If the speed of light is lower at some spot of T, then the inverse length contraction grows for all v(x). Then the inverse length contraction of B is necessarily larger than B_2, which contradicts our assumption about rigidity.
We proved that the speed of light everywhere in T must be the same as in the Minkowski space, but our proof hangs on proving the variational calculus result above.
If we allow v(t, x) depend on the time t, too, then the variational problem is harder. Some strange, fractal-like function might beat the constant speed in optimality. We need to check literature about special relativity and if there are any variational calculus results.
In newtonian mechanics, there is no length contraction, and it is easier to prove that a perfectly rigid grid stays rectangular at each global time moment t.
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