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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.

Simple scalars in special relativity

The length of a position vector

World lines for two particles at different speeds.

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

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

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

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

where $\tau$ is c times the proper time as measured by a clock in the rest frame of the particle and the 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 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 Minkowski metric in the rest of this article.

The length of a velocity vector

The velocity vectors in spacetime for a particle at two different speeds. In relativity an acceleration is equivalent to a rotation in spacetime

The velocity in spacetime is defined as

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

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 and is minus one,

$v_{\mu }v^{\mu }=-1\,$.

The 4-velocity is therefore, not only a representation of the velocity in spacetime, is also a unit vector in the direction of the position of the particle in spacetime.

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

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

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:

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

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

$p^{\mu }=mv^{\mu }=\left(\gamma m,\gamma {m\mathbf {v} \over c}\right)=\left(\gamma m,{\mathbf {p} \over c}\right)=\left({E \over c^{2}},{\mathbf {p} \over c}\right)$

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

$E=\gamma mc^{2}\,$

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

$p_{\mu }u^{\mu }=-{E_{1} \over c^{2}}$

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

${E_{1} \over c^{2}}=\gamma _{1}\gamma _{2}m_{1}-\gamma _{2}\mathbf {p} _{1}\cdot \mathbf {u} _{2}$

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

$p_{\mu }p^{\mu }=-m^{2}\,$.

Therefore $m^{2}$ is a Lorentz scalar. The relationship remains true independent of the frame in which the inner product is calculated.

Measurement of the 3-momentum of the particle

Note that

$\left(p_{\mu }u^{\mu }\right)^{2}+p_{\mu }p^{\mu }={E_{1}^{2} \over c^{4}}-m^{2}=\left(\gamma _{1}^{2}-1\right)m^{2}=\gamma _{1}^{2}{{\mathbf {v} _{1}\cdot \mathbf {v} _{1}} \over c^{2}}m^{2}=\mathbf {p} _{1}\cdot \mathbf {p} _{1}$.

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

$v_{1}^{2}=\mathbf {v} _{1}\cdot \mathbf {v} _{1}={{\mathbf {p} _{1}\cdot \mathbf {p} _{1}c^{6}} \over {E_{1}^{2}}}$.

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

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• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534