Lorentz scalar

In physics, a Lorentz scalar is a scalar which is invariant under a Lorentz transformation. A Lorentz scalar may be generated from multiplication of vectors or tensors. While the components of vectors and tensors are in general altered by Lorentz transformations, scalars remain unchanged.

A Lorentz scalar is not necessarily a scalar in the strict sense of being a (0,0)-tensor, that is, invariant under any base transformation. For example, the determinant of the matrix of base vectors is a number that is invariant under Lorentz transformations, but it is not invariant under any base transformation.

Simple scalars in special relativity

The length of a position vector

In special relativity the location of a particle in 4-dimensional spacetime is given by

$x^{\mu} = (ct, \mathbf{x} )$

where $\mathbf{x} = \mathbf{v} t$ is the position in 3-dimensional space of the particle, $\mathbf{v}$ is the velocity in 3-dimensional space and $c$ is the speed of light.

The "length" of the vector is a Lorentz scalar and is given by

$x_{\mu} x^{\mu} = \eta_{\mu \nu} x^{\mu} x^{\nu} = (ct)^2 - \mathbf{x} \cdot \mathbf{x} \ \stackrel{\mathrm{def}}{=}\ (c\tau)^2$

where $\tau$ is the proper time as measured by a clock in the rest frame of the particle and the Minkowski metric is given by

$\eta^{\mu\nu} =\eta_{\mu\nu} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}$.

This is a time-like metric.

Often the alternate signature of the Minkowski metric is used in which the signs of the ones are reversed.

$\eta^{\mu\nu} =\eta_{\mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$.

This is a space-like metric.

In the Minkowski metric the space-like interval $s$ is defined as

$x_{\mu} x^{\mu} = \eta_{\mu \nu} x^{\mu} x^{\nu} = \mathbf{x} \cdot \mathbf{x} - (ct)^2 \ \stackrel{\mathrm{def}}{=}\ s^2$.

We use the space-like Minkowski metric in the rest of this article.

The length of a velocity vector

The velocity in spacetime is defined as

$v^{\mu} \ \stackrel{\mathrm{def}}{=}\ {dx^{\mu} \over d\tau} = \left (c {dt \over d\tau}, { dt \over d\tau}{d\mathbf{x} \over dt} \right ) = \left ( \gamma c, \gamma { \mathbf{v} } \right ) = \gamma \left ( c, { \mathbf{v} } \right )$

where

$\gamma \ \stackrel{\mathrm{def}}{=}\ { 1 \over {\sqrt {1 - {{\mathbf{v} \cdot \mathbf{v} } \over c^2} } } }$.

The magnitude of the 4-velocity is a Lorentz scalar,

$v_{\mu} v^{\mu} = -c^2\,$.

Hence, c is a Lorentz scalar.

The inner product of acceleration and velocity

The 4-acceleration is given by

$a^{\mu} \ \stackrel{\mathrm{def}}{=}\ {dv^{\mu} \over d\tau}$.

The 4-acceleration is always perpendicular to the 4-velocity

$0 = {1 \over 2} {d \over d\tau} \left ( v_{\mu}v^{\mu} \right ) = {d v_{\mu} \over d\tau} v^{\mu} = a_{\mu} v^{\mu}$.

Therefore, we can regard acceleration in spacetime as simply a rotation of the 4-velocity. The inner product of the acceleration and the velocity is a Lorentz scalar and is zero. This rotation is simply an expression of energy conservation:

${d E \over d\tau} = \mathbf{F} \cdot { \mathbf{v}}$

where $E$ is the energy of a particle and $\mathbf{F}$ is the 3-force on the particle.

Energy, rest mass, 3-momentum, and 3-speed from 4-momentum

The 4-momentum of a particle is

$p^{\mu} = m v^{\mu} = \left ( \gamma m c, \gamma { m \mathbf{v} } \right ) = \left ( \gamma m c, { \mathbf{p} } \right ) = \left ( {E \over c } , { \mathbf{p} } \right )$

where $m$ is the particle rest mass, $\mathbf{p}$ is the momentum in 3-space, and

$E = \gamma m c^2 \,$

is the energy of the particle.

Measurement of the energy of a particle

Consider a second particle with 4-velocity $u$ and a 3-velocity $\mathbf{u}_2$. In the rest frame of the second particle the inner product of $u$ with $p$ is proportional to the energy of the first particle

$p_{\mu} u^{\mu} = - { E_1}$

where the subscript 1 indicates the first particle.

Since the relationship is true in the rest frame of the second particle, it is true in any reference frame. $E_1$, the energy of the first particle in the frame of the second particle, is a Lorentz scalar. Therefore

${ E_1} = \gamma_1 \gamma_2 m_1 c^2 - \gamma_2 \mathbf{p}_1 \cdot \mathbf{u}_2$

in any inertial reference frame, where $E_1$ is still the energy of the first particle in the frame of the second particle .

Measurement of the rest mass of the particle

In the rest frame of the particle the inner product of the momentum is

$p_{\mu} p^{\mu} = - (mc)^2 \,$.

Therefore the rest mass (m) is a Lorentz scalar. The relationship remains true independent of the frame in which the inner product is calculated. In many cases the rest mass is written as $m_{0}$ to avoid confusion with the relativistic mass, which is $\gamma m_{0}$

Measurement of the 3-momentum of the particle

Note that

$\left ( p_{\mu} u^{\mu} /c \right ) ^2 + p_{\mu} p^{\mu} = { E_1^2 \over c^2 } -(mc)^2 = \left ( \gamma_1^2 -1 \right ) (mc)^2 = \gamma_1^2 { {\mathbf{v}_1 \cdot \mathbf{v}_1 } }m^2 = \mathbf{p}_1 \cdot \mathbf{p}_1$.

The square of the magnitude of the 3-momentum of the particle as measured in the frame of the second particle is a Lorentz scalar.

Measurement of the 3-speed of the particle

The 3-speed, in the frame of the second particle, can be constructed from two Lorentz scalars

$v_1^2 = \mathbf{v}_1 \cdot \mathbf{v}_1 = { { \mathbf{p}_1 \cdot \mathbf{p}_1 c^4 } \over { E_1^2 } }$.

More complicated scalars

Scalars may also be constructed from the tensors and vectors, from the contraction of tensors, or combinations of contractions of tensors and vectors.

References

• Misner, Charles; Thorne, Kip S. & Wheeler, John Archibald (1973). Gravitation. San Francisco: W. H. Freeman. ISBN 0-7167-0344-0.
• Landau, L. D. and Lifshitz, E. M. (1975). Classical Theory of Fields (Fourth Revised English Edition). Oxford: Pergamon. ISBN 0-08-018176-7.