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Next: Proper time Up: Relativity and electromagnetism Previous: The physical significance of

Space-time

In special relativity we are only allowed to use inertial frames to assign coordinates to events. There are many different types of inertial frames. However, it is convenient to adhere to those with standard coordinates. That is, spatial coordinates which are right-handed rectilinear Cartesians based on a standard unit of length and time-scales based on a standard unit of time. We shall continue to assume that we are employing standard coordinates. However, from now on we shall make no assumptions, unless specifically stated, about the relative configuration of the two sets of spatial axes and the origins of time when dealing with two inertial frames. Thus, the most general transformation between two inertial frames consists of a Lorentz transformation in the standard configuration plus a translation (this includes a translation in time) and a rotation of the coordinate axes. The resulting transformation is called a general Lorentz transformation, as opposed to a Lorentz transformation in the standard configuration which will henceforth be termed a standard Lorentz transformation.

In Section 2.2 we proved quite generally that corresponding differentials in two inertial frames $S$ and $S'$ satisfy the relation

\begin{displaymath}
dx^2 + dy^2 + dz^2 - c^2dt^2 = d{x'}^2 + d{y'}^2+d{z'}^2 -c^2d{t'}^2.
\end{displaymath} (79)

Thus, we expect this relation to remain invariant under a general Lorentz transformation. Since such a transformation is linear it follows that
$\displaystyle (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2 -c^2(t_2-t_1)^2
=$      
$\displaystyle (x_2'-x_1')^2 + (y_2'-y_1')^2 + (z_2'-z_1')^2 -c^2(t_2'-t_1')^2,$     (80)

where $(x_1, y_1, z_1, t_1)$ and $(x_2, y_2, z_2, t_2)$ are the coordinates of any two events in $S$ and the primed symbols denote the corresponding coordinates in $S'$. It is convenient to write
\begin{displaymath}
-dx^2 -dy^2-dz^2 +c^2 dt^2 =ds^2,
\end{displaymath} (81)

and
\begin{displaymath}
-(x_2-x_1)^2- (y_2-y_1)^2 - (z_2-z_1)^2 +c^2(t_2-t_1)^2 = s^2.
\end{displaymath} (82)

The differential $ds$, or the finite number $s$, defined by these equations is called the interval between the corresponding events. Equations (2.51) and (2.52) express the fact that the interval between two events is invariant, in the sense that it has the same value in all inertial frames. In other words, the interval between two events is invariant under a general Lorentz transformation.

Let us consider entities defined in terms of four variables

\begin{displaymath}
x^1=x,~~x^2=y,~~x^3= z,~~x^4=ct,
\end{displaymath} (83)

and which transform as tensors (see Eqs. (2.30)-(2.32)) under a general Lorentz transformation. From now on such entities will be referred to as 4-tensors.

Tensor analysis cannot proceed very far without the introduction of a non-singular tensor $g_{ij}$, the so-called fundamental tensor, which is used to define the operations of raising and lowering suffixes (see Eqs. (2.42)-(2.44)). The fundamental tensor is usually introduced using a metric $ds^2 = g_{ij}\,dx^i dx^j$, where $ds^2$ is a differential invariant. We have already come across such an invariant, namely

$\displaystyle ds^2$ $\textstyle =$ $\displaystyle -dx^2 -dy^2 -dz^2 +c^2 dt^2$  
  $\textstyle =$ $\displaystyle -(dx^1)^2 -(dx^2)^2 -(dx^3)^2 + (dx^4)^2$  
  $\textstyle =$ $\displaystyle g_{\mu \nu} \,dx^{\mu} dx^{\nu},$ (84)

where $\mu, \nu$ run from 1 to 4. Note that the use of Greek suffixes is conventional in 4-tensor theory. Roman suffixes are reserved for tensors in three dimensional Euclidian space, so-called 3-tensors. The 4-tensor $g_{\mu \nu}$ has the components $g_{11}=g_{22}=g_{33}=-1, g_{44}=1,$ and $g_{\mu\nu}=0$ when $\mu\neq \nu$, in all permissible coordinate frames. From now on $g_{\mu \nu}$, as defined above, is adopted as the fundamental tensor for 4-tensors. $g_{\mu \nu}$ can be thought of as the metric tensor of the ``space'' whose points are the events $(x^1, x^2, x^3, x^4)$. This ``space'' is usually referred to as space-time, for obvious reasons. Note that space-time cannot be regarded as a straightforward generalization of Euclidian 3-space to four dimensions, with time as the fourth dimension. The distribution of signs in the metric ensures that the time coordinate $x^4$ is not on the same footing as the three space coordinates. Thus, space-time has a non-isotropic nature which is quite unlike Euclidian space with its positive definite metric. According to the relativity principle, all physical laws are expressible as interrelationships between 4-tensors in space-time.

A tensor of rank one is called a 4-vector. We shall also have occasion to use ordinary vectors in three dimensional Euclidian space. Such vectors are called 3-vectors and are conventionally represented by boldface symbols. We shall use the Latin suffixes $i,j,k,$ etc. to denote the components of a 3-vector; these suffixes are understood to range from 1 to 3. Thus, ${\bfm u} = u^i = dx^i/dt$ denotes a velocity vector. For 3-vectors we shall use the notation $u^i=u_i$ interchangeably; i.e., the level of the suffix has no physical significance.

When tensor transformations from one frame to another actually have to be computed, we shall usually find it possible to choose coordinates in the standard configuration, so that the standard Lorentz transform applies. Under it, any contravariant 4-vector $T^\mu$ transforms according to the same scheme as the difference in coordinates $x_2^\mu-x_1^\mu$ between two points in space-time. It follows that

$\displaystyle T^{1'}$ $\textstyle =$ $\displaystyle \gamma(T^1-\beta \,T^4),$ (85)
$\displaystyle T^{2'}$ $\textstyle =$ $\displaystyle T^2,$ (86)
$\displaystyle T^{3'}$ $\textstyle =$ $\displaystyle T^3,$ (87)
$\displaystyle T^{4'}$ $\textstyle =$ $\displaystyle \gamma(T^4 - \beta \,T^1),$ (88)

where $\beta = v/c$. Higher rank 4-tensors transform according to the rules (2.30)-(2.32). The transformation coefficients take the form
$\displaystyle p_{\mu}^{\mu'}$ $\textstyle =$ $\displaystyle \left\lgroup
\begin{array}{cccc}
\gamma & 0 & 0 & -\gamma\beta \\...
... 0 \\
0 & 0 & 1 & 0 \\
-\gamma\beta & 0 & 0 & \gamma
\end{array}\right\rgroup$ (89)
$\displaystyle p_{\mu'}^{\mu}$ $\textstyle =$ $\displaystyle \left\lgroup
\begin{array}{cccc}
\gamma & 0 & 0 & \gamma\beta \\ ...
...& 0 \\
0 & 0 & 1 & 0 \\
\gamma\beta & 0 & 0 & \gamma
\end{array}\right\rgroup$ (90)

Often the first three components of a 4-vector coincide with the components of a 3-vector. For example, the $x^1$, $x^2$, $x^3$ in $R^\mu= (x^1, x^2, x^3, x^4)$ are the components of ${\bfm r}$, the position 3-vector of the point at which the event occurs. In such cases we adopt the notation exemplified by $R^\mu = ({\bfm r}, ct)$. The covariant form of such a vector is simply $R_\mu = (-{\bfm r}, ct)$. The squared magnitude of the vector is $(R)^2 = R_\mu R^\mu = -r^2 + c^2 t^2$. The inner product $g_{\mu\nu}R^\mu Q^\nu = R_\mu Q^\mu$ of $R^\mu$ with a similar vector $Q^\mu = ({\bfm q}, k)$ is given by $ R_\mu Q^\mu = - {\bfm r}\!\cdot\!{\bfm q} + ct\,k$. The vectors $R^\mu$ and $Q^\mu$ are said to be orthogonal if $R_\mu Q^\mu = 0$.

Since a general Lorentz transformation is a linear transformation, the partial derivative of a 4-tensor is also a 4-tensor;

\begin{displaymath}
\frac{\partial A^{\nu\sigma}}{\partial x^{\mu}} = {A^{\nu\sigma}}_{,\mu}.
\end{displaymath} (91)

Clearly, a general 4-tensor acquires an extra covariant index after partial differentiation with respect to the contravariant coordinate $x^\mu$. It is helpful to define a covariant derivative operator
\begin{displaymath}
\partial_\mu \equiv \frac{\partial}{\partial x^\mu} = \left(\nabla,
\frac{1}{c}\frac{\partial}{\partial t}\right),
\end{displaymath} (92)

where
\begin{displaymath}
\partial_\mu A^{\nu\sigma} \equiv {A^{\nu\sigma}}_{,\mu}.
\end{displaymath} (93)

There is a corresponding contravariant derivative operator
\begin{displaymath}
\partial^\mu \equiv \frac{\partial}{\partial x_\mu} = \left(-\nabla,
\frac{1}{c}\frac{\partial}{\partial t}\right),
\end{displaymath} (94)

where
\begin{displaymath}
\partial^\mu A^{\nu\sigma} \equiv g^{\mu\tau} {A^{\nu\sigma}}_{,\tau}.
\end{displaymath} (95)

The 4-divergence of a 4-vector $A^\mu= ({\bfm A}, A^0)$ is the invariant
\begin{displaymath}
\partial^\mu A_\mu = \partial_\mu A^\mu =
\nabla\!\cdot \! {\bfm A} + \frac{1}{c}\frac{\partial A^0}
{\partial t}.
\end{displaymath} (96)

The four dimensional Laplacian operator, or d'Alembertian, is equivalent to the invariant contraction
\begin{displaymath}
\Box \equiv \partial_\mu \partial^\mu = -\nabla^2 + \frac{1}{c^2} \frac
{\partial^2}{\partial t^2}.
\end{displaymath} (97)

Recall that we still need to prove (from Section 2.2) that the invariance of the differential metric,

\begin{displaymath}
ds^2 = {dx'}^2 + {dy'}^2 + {dz'}^2 - c^2 {dt'}^2 =
{dx}^2 + {dy}^2 + {dz}^2 - c^2 {dt}^2,
\end{displaymath} (98)

between two general inertial frames implies that the coordinate transformation between such frames is necessarily linear. To put it another way, we need to demonstrate that a transformation which transforms a metric $g_{\mu\nu} \,dx^\mu dx^\nu$ with constant coefficients into a metric $g_{\mu'\nu'}\, dx^{\mu'} dx^{\nu'}$ with constant coefficients must be linear. Now
\begin{displaymath}
g_{\mu\nu} = g_{\mu'\nu'}\,p_{\mu}^{\mu'} p_{\nu}^{\nu'}.
\end{displaymath} (99)

Differentiating with respect to $x^\sigma$ we get
\begin{displaymath}
g_{\mu'\nu'}\,p_{\mu\sigma}^{\mu'} p_{\nu}^{\nu'} + g_{\mu'\nu'}\,
p_{\mu}^{\mu'}p_{\nu\sigma}^{\nu'} = 0,
\end{displaymath} (100)

where
\begin{displaymath}
p_{\mu\sigma}^{\mu'} = \frac{\partial p_\mu^{\mu'} }{\partia...
...\mu'}}{\partial x^{\mu}\partial
x^\sigma}=p_{\sigma\mu}^{\mu'}
\end{displaymath} (101)

etc. Interchanging the indices $\mu$ and $\sigma$ yields
\begin{displaymath}
g_{\mu'\nu'}\,p_{\mu\sigma}^{\mu'}
p_{\nu}^{\nu'} + g_{\mu'\nu'}\,p_{\sigma}^{\mu'}
p_{\nu\mu}^{\nu'} = 0.
\end{displaymath} (102)

Interchanging the indices $\nu$ and $\sigma$ gives
\begin{displaymath}
g_{\mu'\nu'}\,p_{\sigma}^{\mu'}
p_{\nu\mu}^{\nu'} + g_{\mu'\nu'}\,
p_{\mu}^{\mu'}p_{\nu\sigma}^{\nu'} = 0,
\end{displaymath} (103)

where the indices $\mu'$ and $\nu'$ have been interchanged in the first term. It follows from Eqs. (2.68), (2.70), and (2.71) that
\begin{displaymath}
g_{\mu'\nu'} \,p_{\mu\sigma}^{\mu'} p_{\nu}^{\nu'} =0.
\end{displaymath} (104)

Multiplication by $p_{\sigma'}^{\nu}$ yields
\begin{displaymath}
g_{\mu'\nu'} \,p_{\mu\sigma}^{\mu'} p_{\nu}^{\nu'}p_{\sigma'}^{\nu}
= g_{\mu'\sigma'}\,p_{\mu\sigma}^{\mu'} =0.
\end{displaymath} (105)

Finally, multiplication by $g^{\nu'\sigma'}$ gives
\begin{displaymath}
g_{\mu'\sigma'}g^{\nu'\sigma'}\,p_{\mu\sigma}^{\mu'} = p_{\mu\sigma}^{\nu'}
=0.
\end{displaymath} (106)

This proves that the coefficients $p_{\mu}^{\nu'}$ are constants and, hence, that the transformation is linear.


next up previous
Next: Proper time Up: Relativity and electromagnetism Previous: The physical significance of
Richard Fitzpatrick 2002-05-18