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

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

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 the speed of light.

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

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

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

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

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.

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

This is a space-like metric.

In the Minkowski metric the space-like interval ${\displaystyle s}$ is defined as

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

We use the space-like 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

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

where

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

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

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

Hence, c is a Lorentz scalar.

The inner product of acceleration and velocity

The 4-acceleration is given by

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

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

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

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:

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

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

Measurement of the energy of a 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.

Measurement of the rest mass of the 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 ${\displaystyle \gamma m_{0}}$.

Measurement of the 3-momentum of the particle

Note that

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

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

$${\displaystyle v_{1}^{2}=\mathbf {v} _{1}\cdot \mathbf {v} _{1}={\frac {\mathbf {p} _{1}\cdot \mathbf {p} _{1}}{E_{1}^{2}}}c^{4}.}$$