In arelativistic theory ofphysics, aLorentz scalar is ascalar expression whose value isinvariant under anyLorentz transformation. A Lorentz scalar may be generated from, e.g., the scalar product of vectors, or by contracting tensors. While the components of the contracted quantities may change under Lorentz transformations, the Lorentz scalars remain unchanged.

A simple Lorentz scalar inMinkowski spacetime is thespacetime distance ("length" of their difference) of two fixed events in spacetime. While the "position"-4-vectors of the events change between different inertial frames, their spacetime distance remains invariant under the corresponding Lorentz transformation. Other examples of Lorentz scalars are the "length" of 4-velocities (see below), or theRicci curvature in a point in spacetime fromgeneral relativity, which is a contraction of theRiemann curvature tensor there.

Inspecial relativity the location of a particle in 4-dimensionalspacetime is given by$${\displaystyle x^{\mu }=(ct,\mathbf{x})}$$where${\displaystyle \mathbf{x}=\mathbf{v}t}$ is the position in 3-dimensional space of the particle,${\displaystyle \mathbf{v}}$ is the velocity in 3-dimensional space and${\displaystyle c}$ is thespeed of light.

The "length" of the vector is a Lorentz scalar and is given by$${\displaystyle x_{\mu }{x}^{\mu }={\eta }_{\mu \nu }{x}^{\mu }{x}^{\nu }=(ct{)}^2-\mathbf{x}\cdot \mathbf{x}\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}(c\tau {)}^2}$$where${\displaystyle \tau }$ is the proper time as measured by a clock in the rest frame of the particle and theMinkowski metric is given byThis is a time-like metric.

Often the alternate signature of theMinkowski metric is used in which the signs of the ones are reversed.This is a space-like metric.

In the Minkowski metric the space-like interval${\displaystyle s}$ is defined as$${\displaystyle x_{\mu }{x}^{\mu }={\eta }_{\mu \nu }{x}^{\mu }{x}^{\nu }=\mathbf{x}\cdot \mathbf{x}-(ct{)}^2\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}{s}^2.}$$

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

The velocity in spacetime is defined as$${\displaystyle v^{\mu }\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\frac{d{x}^{\mu }}{d\tau }=\left(c\frac{dt}{d\tau },\frac{dt}{d\tau }\frac{d\mathbf{x}}{dt}\right)=\left(\gamma c,\gamma \mathbf{v}\right)=\gamma \left(c,\mathbf{v}\right)}$$where$${\displaystyle \gamma \stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\frac{1}{\sqrt{1-\frac{\mathbf{v}\cdot \mathbf{v}}{c^2}}}.}$$

The magnitude of the 4-velocity is a Lorentz scalar,$${\displaystyle v_{\mu }{v}^{\mu }=-c^2.}$$

Hence,⁠${\displaystyle c}$⁠ is a Lorentz scalar.

The 4-acceleration is given by$${\displaystyle a^{\mu }\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\frac{d{v}^{\mu }}{d\tau }.}$$

The 4-acceleration is always perpendicular to the 4-velocity$${\displaystyle 0=\frac{1}{2}\frac{d}{d\tau }\left(v_{\mu }{v}^{\mu }\right)=\frac{d{v}_{\mu }}{d\tau }{v}^{\mu }=a_{\mu }{v}^{\mu }.}$$

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:$${\displaystyle \frac{dE}{d\tau }=\mathbf{F}\cdot \mathbf{v}}$$where${\displaystyle E}$ is the energy of a particle and${\displaystyle \mathbf{F}}$ is the 3-force on the particle.

The 4-momentum of a particle is$${\displaystyle p^{\mu }=m{v}^{\mu }=\left(\gamma mc,\gamma m\mathbf{v}\right)=\left(\gamma mc,\mathbf{p}\right)=\left(\frac{E}{c},\mathbf{p}\right)}$$where${\displaystyle m}$ is the particle rest mass,${\displaystyle \mathbf{p}}$ is the momentum in 3-space, and$${\displaystyle E=\gamma m{c}^2}$$ is the energy of the particle.

Consider a second particle with 4-velocity${\displaystyle u}$ and a 3-velocity${\displaystyle {\mathbf{u}}_2}$. In the rest frame of the second particle the inner product of${\displaystyle u}$ with${\displaystyle p}$ is proportional to the energy of the first particle$${\displaystyle p_{\mu }{u}^{\mu }=-E_1}$$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.${\displaystyle E_1}$, the energy of the first particle in the frame of the second particle, is a Lorentz scalar. Therefore,$${\displaystyle E_1={\gamma }_1{\gamma }_2{m}_1{c}^2-{\gamma }_2{\mathbf{p}}_1\cdot {\mathbf{u}}_2}$$in any inertial reference frame, where${\displaystyle E_1}$ is still the energy of the first particle in the frame of the second particle.

In the rest frame of the particle the inner product of the momentum is$${\displaystyle p_{\mu }{p}^{\mu }=-(mc{)}^2.}$$

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${\displaystyle m_0}$ to avoid confusion with the relativistic mass, which is${\displaystyle \gamma m_0}$.

Note that$${\displaystyle {\left(\frac{p_{\mu }{u}^{\mu }}{c}\right)}^2+p_{\mu }{p}^{\mu }=\frac{E_1^2}{c^2}-(mc{)}^2=\left({\gamma }_1^2-1\right)(mc{)}^2={\gamma }_1^2{\mathbf{v}}_1\cdot {\mathbf{v}}_1{m}^2={\mathbf{p}}_1\cdot {\mathbf{p}}_1.}$$

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.

The 3-speed, in the frame of the second particle, can be constructed from two Lorentz scalars$${\displaystyle v_1^2={\mathbf{v}}_1\cdot {\mathbf{v}}_1=\frac{{\mathbf{p}}_1\cdot {\mathbf{p}}_1}{E_1^2}{c}^4.}$$

Scalars may also be constructed from the tensors and vectors, from the contraction of tensors (such as${\displaystyle F_{\mu \nu }{F}^{\mu \nu }}$), or combinations of contractions of tensors and vectors (such as${\displaystyle g_{\mu \nu }{x}^{\mu }{x}^{\nu }}$).

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

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