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Back to Ontology of Space and Time

or

What does it all mean, again?

Ontology of Space and Time

John
D. Norton

Department of History and Philosophy of Science

University of Pittsburgh

This page uses mathematics typeset with mathjax, for which an internet connection is needed. Apologies for any presentation anomalies. I am not sufficiently expert in mathjax to know how to stop them.

The coordinate T coincides with the time read by clocks. We introduce a new time coordinate t(T) as a function of T. Then small increments of T, ΔT, transform according to

$$\Delta t=\frac{dt}{dT}\Delta T$$

or equivalently

$$\Delta T=\frac{dT}{dt}\Delta t$$

If t = log_{10}T so that T = 10^{t}, then

$$\frac{dT}{dt} = \frac{10^t}{dt} = \frac{d e^{t \thinspace \text{log}_e10}} {dt} = (\text{log}_e10) e^{t \thinspace \text{log}_e10} = (\text{log}_e10) 10^t = 2.3025 \thinspace 10^t = 2.3015 \thinspace T$$

Represent the spacetime interval by s. In a coordinate system (T, X, Y, Z) adapted to an inertial frame of reference in a Minkowski spacetime, the small difference of interval between two neighboring events ds is given by the expresssion for the line element:

ds^{2} = - dT^{2}
+ dX^{2} + dY^{2} + dZ^{2}

where units are chosen so that c=1.

For events that are spacelike separated, ds^{2}
corresponds to the ordinary Euclidean distance. For example, consider two
events that differ only in their X coordinate and do so by a small amount
dX, for example (0, X, 0, 0) and (0, X+dX, 0, 0). The interval-squared
separating them is ds^{2} = dX^{2}, which corresponds to a
measured distance of dX.

For events that are timelike separated, ds^{2}
corresponds to the time elapsed on clocks that move on a geodesic between
them. For example, consider two events that differ only in their T
coordinate and do so by a small amount dT, for example (T, 0, 0, 0) and
(T+dT, 0, 0, 0). The interval-squared separating them is ds^{2} =
- dT^{2}. To recover the proper time difference, we ignore the
minus sign and just use dT.

Events that are lightlike separated--that is can be connected by a light signal--are zero interval apart. Since we have set c=1, an example is a pair of events (T, X, 0, 0) and (T+1, X+1, 0, 0). More generally, setting ds = 0 picks out all the lightlike curves and thus specifies the lightcone structure of the spacetime.

To see this, we have

ds^{2} = - dT^{2}
+ dX^{2} + dY^{2} + dZ^{2} = 0

entails

dX^{2} + dY^{2}
+ dZ^{2} = dT^{2}

so that

$$1 = \frac{dX^2 + dY^2 + dZ^2}{dT^2} = \frac{dX^2}{dT^2} + \frac{dY^2} {dT^2} + \frac{dZ^2}{dT^2} = \left( \frac{dX}{dT} \right)^2 + \left( \frac{dY}{dT} \right)^2 + \left( \frac{dZ}{dT} \right)^2 = |\mathbf v|^2$$

where the velocity along the trajectory is **v**
with

$$ \mathbf v = \left ( \frac{dX}{dT} , \frac{dY}{dT} ,\frac{dZ}{dT} \right) $$

That is, if ds = 0 along a trajectory, that trajectory is one that is traced out by a point moving at unit speed. Since we have set c=1, unit speed is the speed of light.

Let us introduce a new spacetime coordinate system x^{μ}
= (x^{0}, x^{1}, x^{2}, x^{3}). Since
ds is a quantity ascertainable by direct measurement, independent of
the choice of the coordinate system, it must remain unchanged when
determined in the new coordinate system.

To ensure this sameness, we recover an expression for ds by replacing each of the small coordinate differences in the original coordinate system by corresponding differences in the new coordinate system. To do this we use expressions:

$$dT = \frac{\partial T}{\partial x^0}dx^0 + \frac{\partial T}{\partial x^1}dx^1 + \frac{\partial T}{\partial x^2}dx^2 + \frac{\partial T}{\partial x^3}dx^3$$

...

$$dZ = \frac{\partial Z}{\partial x^0}dx^0 + \frac{\partial Z}{\partial x^1}dx^1 + \frac{\partial Z}{\partial x^2}dx^2 + \frac{\partial Z}{\partial x^3}dx^3$$

After substituting for dT, ..., dZ, we recover an intimidatingly large expression for the line element:

$$ds^2 = \left ( \frac{\partial T}{\partial t} \right )^2
\left ( dx^0 \right )^2

+ \left [ \left ( \frac{\partial T}{\partial x^0} \right )\left (
\frac{\partial T}{\partial x^1} \right )

+ \left ( \frac{\partial X}{\partial x^0} \right )\left (
\frac{\partial X}{\partial x^1} \right )

+... \right ]dx^0dx^1

+ ...$$

This is formally rather unwieldy. We can simplify the notation if we write

T = X^{0}, X = X^{1}, Y = X^{2}, Z
= X^{3}.

These four coordinates can then be written simply as X^{μ}
where μ takes the values 0, 1, 2, 3.

Then, following the Einstein summation convention
explained below, the expression for ds^{2} is written as

$$ds^2 = \eta _{\mu\nu} dX^\mu dX^\nu$$

where μ and ν take values 0, 1, 2, 3 and the matrix
$$\eta _{\mu\nu}=\begin{bmatrix}

-1 & 0 & 0 & 0\\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0\\

0 & 0 & 0 & 1

\end{bmatrix}$$

is the representation of the metric tensor of Minowski spacetime in this coordinate system.

In Einstein's convention, when an index is repeated
"upstairs" and "downstairs" we sum over that index. So the expression for
ds^{2} is really:

$$ds^2 =\sum_{\mu =0}^{3} \sum_{\nu =0}^{3} \eta _{\mu\nu} dX^\mu dX^\nu$$

We will continue to use this convention below.

We transform the expression for ds^{2} to the new
coordinate system as before, by substituting the new coordinate
differences dx^{α} for the old coordinate differences dX^{μ}:

$$ds^2 = \eta _{\mu\nu}

\left (\frac{\partial X^\mu}{\partial x^\alpha} \right )

\left (\frac{\partial X^\nu}{\partial x^\beta} \right ) dx^\alpha
dx^\beta$$

This expression within the new coordinate system can be rewritten as

$$ds^2 = g _{\alpha\beta} dx^\alpha dx^\beta$$

where the Minkowski metric tensor in this new coordinate system is given by

$$g_{\alpha\beta} = \eta _{\mu\nu}

\left (\frac{\partial X^\mu}{\partial x^\alpha} \right )

\left (\frac{\partial X^\nu}{\partial x^\beta} \right )$$

This last equation shows that g_{μν} is a tensor.
For the defining property of a tensor is the rule that is used to
transform it to a new coordinate system, when we have it given in an old
coordinate system.

The new components are linear functions of the old components, where the coefficients in the linear transformations are coordinate derivatives like $$\left (\frac{\partial X^\mu}{\partial x^\alpha} \right )$$

The way these coordinate derivatives enter into the transformation equations determine whether the tensor is known as a "covariant tensor" or a "contravariant tensor."

The metric tensor g_{μν} is "covariant" since the
coordinate derivatives in the transformation equations have the form

$$\left (\frac{\partial X^\mu}{\partial x^\alpha} \right ) = \left (\frac{\text{old coordinate}}{\text{new coordinate}} \right )$$

The coordinate differences $$dx^\alpha$$ transform contravariantly:

$$dx^\mu = \left ( \frac{\partial x^\mu}{\partial X^\nu} \right ) dX^\nu$$

What makes this an example of a contravariant transformation is that the coordinate derivatives are reversed and have the form

$$\left (\frac{\text{new coordinate}}{\text{old coordinate}} \right )$$

An illustration of these transformations is provided by the transformation from a Minkowski spacetime in an inertial coordinate system to one in a uniformly accelerated coordinate system.

In the inertial coordinate system (T, X, Y, Z) = (X^{0},
X^{1}, X^{2}, X^{3}), we have the line element

ds^{2} = - dT^{2}
+ dX^{2} + dY^{2} + dZ^{2}

and the metric matrix

$$\eta _{\mu\nu}= \begin{bmatrix}

-1 & 0 & 0 & 0\\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0\\

0 & 0 & 0 & 1

\end{bmatrix}$$

We transform to a uniformly accelerated coordinate system
(t, x, y, z) = (x^{0}, x^{1}, x^{2}, x^{3}).
Following A. Einstein and N. Rosen, "The Particle Problem in the General
Theory of Relativity," *Physical Review*, 48 (1935), pp. 73 - 77,
on p. 74, we have for the transformation equations that:

T = x sinh(at) X = x cosh(at) Y = y Z = z

We seek the form of the Minkowski metric, g_{μν},
in this new coordinate system, using the rule for transforming the
covariant tensor η_{μν},

$$g_{\alpha\beta} = \eta _{\mu\nu}

\left (\frac{\partial X^\mu}{\partial x^\alpha} \right )

\left (\frac{\partial X^\nu}{\partial x^\beta} \right )$$

To apply the formula, we need first to calculate the partial derivatives in it. Most of them are zero valued. The only non-zero ones are

$$ \frac{\partial T}{\partial t} = \frac{\partial (x \sinh(at))}{\partial t} = ax \cosh(at) $$

$$ \frac{\partial T}{\partial x} = \frac{\partial (x \sinh(at))}{\partial x} = \sinh(at) $$

$$ \frac{\partial X}{\partial t} = \frac{\partial (x \cosh(at))}{\partial t} = ax \sinh(at) $$

$$ \frac{\partial X}{\partial x} = \frac{\partial (x \cosh(at))}{\partial x} = \cosh(at) $$

$$ \frac{\partial Y}{\partial y} = \frac{\partial Z}{\partial z} = 1 $$

Most of the terms in the summation for g_{μν}
vanish, which greatly simplifies the calculation. Here are the non-zero
terms:

$$

\begin{multline}

g_{00} =

\eta _{00} \left (\frac{\partial T}{\partial t} \right )^2

+

\eta _{11} \left (\frac{\partial X}{\partial t} \right )^2 \\

=

(-1) (ax \cosh(at))^2 + (+1) (ax \sinh(at))^2

= -a^2x^2

\end{multline}

$$

since cosh^{2}(at) - sinh^{2}(at) =1.

$$

\begin{multline}

g_{11} =

\eta _{00} \left (\frac{\partial T}{\partial x} \right )^2

+

\eta _{11} \left (\frac{\partial X}{\partial x} \right )^2 \\

=

(-1) (\sinh(at))^2 + (+1) (\cosh(at))^2

= 1

\end{multline}

$$

$$ g_{22} =

\eta _{22} \left (\frac{\partial Y}{\partial y} \right )^2

= (+1) (1)^2 = 1

$$

$$

g_{33} =

\eta _{33} \left (\frac{\partial Z}{\partial z} \right )^2

= (+1) (1)^2 = 1 $$

Combining we find that

$$ g _{\mu\nu}= \begin{bmatrix}

-a^2x^2 & 0 & 0 & 0\\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0\\

0 & 0 & 0 & 1

\end{bmatrix}$$

and that the Minkowski line element becomes

ds^{2} = - a^{2}x^{2}dt^{2}
+ dx^{2} + dy^{2} + dz^{2}

Finally, we might ask why the transformation

T = x sinh(at) X = x cosh(at) Y = y Z = z

is to a coordinate system adapted to uniformly accelerating frame of reference (or briefly "uniformly accelerated coordinates").

The Newtonian transformation for constant acceleration A is:

T = t and X = x + (1/2)At^{2}

A point at rest in the accelerating frame will have a constant x coordinate, so that its trajectory in the inertial coordinate system is

X = constant + (1/2)At^{2}
= constant + (1/2)AT^{2}

Its velocity is V = dX/dT = AT, which grows without limit with time and will eventually exceed the speed of light.

The formula using sinh and cosh functions assures that points at rest in the uniformly accelerated frame of reference approach the speed of light, c=1, asymptotically. Instead of the Newtonian parabolic motion, they have a hyperbolic motion.

A point at rest in the accelerating frame has x = constant = k, so that its trajectory is

T = k sinh(at) X = k cosh(at)

We can treat t as a path parameter and form

$$ \frac {dT} {dt} = ak \cosh(at) $$

$$ \frac {dX} {dt} = ak \sinh(at) $$

It follows that the speed of the point at rest in the accelerating frame of reference is

$$ \left( \frac {dX} {dT} \right ) = \left( \frac {dX} {dt} \right ) \left( \frac {dt} {dT} \right ) = \frac {ak \sinh(at)} {ak \cosh(at)} $$

$$ \lim_{t \to \infty} \left( \frac {dX} {dT} \right ) = \lim_{t \to \infty} \frac {ak \sinh(at)} {ak \cosh(at)} = 1 $$

For small t, however, the transformation is approximately the Newtonian. Then we have

sinh(at) ≈ at
and
cosh(at) ≈ 1 + (1/2)a^{2}t^{2}

Once again, we track a point at rest in the accelerating frame at x = constant = k, but now for small t. Then we have

T ≈ k at X
≈ k(1 + (1/2)a^{2}t^{2})

Substituting t = T/ka in the expression for X, we recover that the trajectory of the point is

X ≈ k(1 + (1/2)a^{2}t^{2})
= k(1 + (1/2)a^{2}(T/ka)^{2})= k(1 + (1/2) (1/k)^{2}
T^{2})

Hence we recover the Newtonian parabolic motion in a small
region around x=k, where the acceleration corresponding to the Newtonian
acceleration is A = (1/k)^{2}. This means that the corresponding
Newtonian acceleration differs according to the value of the
coordinate x=k of the point whose motion is tracked.

In another
chapter, you will find diagrams
of the parabolic motions of uniform acceleration and corresponding
hyperbolic motions that display these effects graphically.

Copyright John D. Norton. November 12, 2019.