Four-vector

A four-vector, a term used in relativity theory , is a vector in a real, four-dimensional space with an indefinite square of length. For example, the time and location coordinates of an event in space-time are the components of a four-vector, as are the energy and momentum of a particle.

In two mutually moving inertial systems , the components of the two four-vectors can be converted into one another by a Lorentz transformation .

Notation

The abbreviations are used

• ${\displaystyle a^{\mu }=(a^{0},a^{1},a^{2},a^{3})}$for the contravariant
• ${\displaystyle a_{\mu }=(a_{0},a_{1},a_{2},a_{3})=(a^{0},-a^{1},-a^{2},-a^{3})}$for the covariant representation of a four-vector. (Details on contra- and covariant vectors ↓ )

Mostly, Greek indices are used when they run through the values ​​0, 1, 2, 3, while Latin indices only run through the values ​​1, 2, 3 of the spatial coordinates. In the theory of relativity , letters are preferred . ${\displaystyle \mu ,\nu }$

The metric of the Minkowski space of the special relativity theory was used and the associated metric tensor , in the general relativity theory the (location-dependent) metric tensor is to be selected. ${\displaystyle \eta _{\mu \nu }}$ ${\displaystyle g_{\mu \nu }}$

Position vector

The position vector or position four-vector of a particle contains both the time coordinate and the space coordinates of an event. In the theory of relativity, the time coordinate is multiplied by the speed of light , so that, like the space coordinates, it has the dimension of a length. ${\displaystyle t}$ ${\displaystyle \mathbf {x} =(x,y,z)}$ ${\displaystyle c}$

The contravariant representation of the location four-vector is

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

The fact that a contravariant four-vector is a contravariant follows from the fact that it is a coordinate vector to an orthonormal basis of the Minkowski space and accordingly changes contravariant by means of a Lorentz transformation when the basis is changed. ${\displaystyle x^{\mu }}$

In the flat space-time metric , the time coordinate has the opposite sign of the three space coordinates:

${\displaystyle \mathrm {d} s^{2}=c^{2}\mathrm {d} t^{2}-\mathrm {d} x^{2}-\mathrm {d} y^{2}-\mathrm {d} z^{2}}$

The metric therefore has the signature (+ - - -) or (- + + + ). Especially in texts on the special theory of relativity, the first signature is predominantly used, but this is only a convention and varies depending on the author.

Derived four-vectors

Further four-vectors can be derived and defined from the location four-vector.

Speed ​​of four

The four-vector of the velocity results from differentiation of the position vector according to the proper time : ${\displaystyle u^{\mu }}$ ${\displaystyle x^{\mu }}$ ${\displaystyle d\tau }$

${\displaystyle u^{\mu }={\frac {\mathrm {d} x^{\mu }}{\mathrm {d} \tau }}}$

with the proper time , which is linked to the coordinate time via the time dilation : ${\displaystyle \tau }$ ${\displaystyle t}$

${\displaystyle \mathrm {d} \tau ={\frac {1}{\gamma }}\,\mathrm {d} t,}$

with the Lorentz factor ${\displaystyle \gamma ={\frac {1}{\sqrt {1-\left({\dfrac {\mathbf {v} }{c}}\right)^{2}}}}.}$

From this it follows for the speed of four:

${\displaystyle u^{\mu }=\gamma \ {\frac {\mathrm {d} }{\mathrm {d} t}}(c\ t,\ x,\ y,\ z)=\gamma \ \left(c,\ {\dot {x}},\ {\dot {y}},\ {\dot {z}}\right)=\gamma \ (c,\ \mathbf {v} )}$

The standard of the four-speed results in both specific as well as in the general theory of relativity to

${\displaystyle |u^{\mu }|={\sqrt {u^{\mu }\ u^{\nu }\ \eta _{\mu \nu }}}={\sqrt {u_{\mu }\ u^{\mu }}}={\sqrt {\gamma ^{2}\ (c^{2}\ -\ \mathbf {v} ^{2})}}=c}$ .

Quadruple pulse

The four-pulse is defined as analogous to the classic pulse

${\displaystyle p^{\mu }\ =\ m\ u^{\mu }=(\gamma \ m\ c,\ \gamma \ m\ \mathbf {v} ),}$

where is the mass of the body. In comparison with Newtonian mechanics , the combination is sometimes interpreted as “dynamically increasing mass” and referred to as “rest mass”, which, however, can easily lead to incorrect conclusions due to an inadequate classical approach. In the consequent four-part calculus without reference to non-relativistic physics, only the coordinate-independent mass is of practical importance. ${\displaystyle m}$ ${\displaystyle \gamma m}$ ${\displaystyle m}$ ${\displaystyle m}$

With the equivalence of mass and energy , the quad pulse can be written as ${\displaystyle E\ =\ \gamma \ m\ c^{2}}$

${\displaystyle p^{\mu }\ =\ \left(E/c,\ \mathbf {p} \right)}$

with the relativistic spatial momentum , which differs from the classical momentum vector by the Lorentz factor . ${\displaystyle \mathbf {p} \ =\ \gamma \ m\ \mathbf {v} }$ ${\displaystyle \gamma }$

Since the four-momentum unites the energy and the spatial momentum, it is also known as the energy-momentum vector .

The energy-momentum relationship results from the square of the norm of the four -pulse ${\displaystyle p_{\mu }\ p^{\mu }}$

${\displaystyle E^{2}\ -\ \mathbf {p} ^{2}\ c^{2}\ =\ m^{2}\ c^{4},}$

from which a time- and location-independent Hamilton function for free , relativistic particles can be derived.

Four acceleration

By re-deriving the four-velocity after obtaining the four acceleration. ${\displaystyle u^{\mu }\ =\ {\frac {\mathrm {d} x^{\mu }}{\mathrm {d} \tau }}}$ ${\displaystyle \tau }$

The 0th component of the quad acceleration is determined to

${\displaystyle {\frac {\mathrm {d} u^{0}}{\mathrm {d} \tau }}\ =\ c\ {\frac {\mathrm {d} }{\mathrm {d} \tau }}\ \gamma \ =\ c\ {\frac {\mathrm {d} t}{\mathrm {d} \tau }}\ {\frac {\mathrm {d} }{\mathrm {d} t}}\gamma \ =\ c\ \gamma \ \cdot \ {\frac {\gamma ^{3}}{c^{2}}}\ \left(\mathbf {v} \ \cdot \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right)\ =\ {\frac {\gamma ^{4}}{c}}\ \left(\mathbf {v} \ \cdot \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right).}$

The spatial components of the quad acceleration are

${\displaystyle {\frac {\mathrm {d} u^{j}}{\mathrm {d} \tau }}\ =\ {\frac {\mathrm {d} t}{\mathrm {d} \tau }}\ {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\gamma \ \mathbf {v} \right)\ =\ \gamma \ {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\gamma \ \mathbf {v} \right)\ =\ {\frac {\gamma ^{4}}{c^{2}}}\ \left(\mathbf {v} \ \cdot \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right)\ \cdot \ \mathbf {v} \ +\ \gamma ^{2}\ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}.}$

Overall, you get the result for the four-speed acceleration

${\displaystyle {\frac {\mathrm {d} u^{\mu }}{\mathrm {d} \tau }}\ =\ {\frac {\gamma ^{4}}{c^{2}}}\ \mathbf {v} \ \cdot \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\ (c,\ \mathbf {v} )\ +\ \gamma ^{2}\ \left(0,\ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right).}$

Four acceleration consists of a part with a factor and a part with . Thus, for accelerations, one obtains parallel and orthogonal to different accelerations of four. With the Graßmann identity ${\displaystyle {\frac {\gamma ^{4}}{c^{2}}}}$ ${\displaystyle \gamma ^{2}}$ ${\displaystyle \mathbf {v} }$

${\displaystyle \mathbf {v} \ \times \ \left(\mathbf {v} \ \times \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right)\ =\ \mathbf {v} \ \left(\mathbf {v} \ \cdot \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right)\ -\ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\ \left(\mathbf {v} \ \cdot \ \mathbf {v} \right)}$

one can transform the expression for the spatial part of the four-vector. One finds that

${\displaystyle {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\ +\ {\frac {1}{c^{2}}}\ \left(\mathbf {v} \ \times \ \left(\mathbf {v} \ \times \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right)\right)\ =\ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\ \left(1\ -\ {\frac {\mathbf {v} ^{2}}{c^{2}}}\right)\ +\ {\frac {1}{c^{2}}}\ \mathbf {v} \ \left(\mathbf {v} \ \cdot \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right)\ =\ {\frac {1}{\gamma ^{2}}}\ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\ +\ {\frac {1}{c^{2}}}\ \mathbf {v} \ \left(\mathbf {v} \ \cdot \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right)}$

is. It follows

${\displaystyle {\frac {\mathrm {d} u^{j}}{\mathrm {d} \tau }}\ =\ \gamma ^{4}\ \left({\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\ +\ {\frac {1}{c^{2}}}\ \left(\mathbf {v} \ \times \ \left(\mathbf {v} \ \times \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right)\right)\right).}$

and thus overall

${\displaystyle {\frac {\mathrm {d} u^{\mu }}{\mathrm {d} \tau }}\ =\ \gamma ^{4}\ \left({\frac {1}{c}}\ \mathbf {v} \ \cdot \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}},\ \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\ +\ {\frac {1}{c^{2}}}\ \left(\mathbf {v} \ \times \ \left(\mathbf {v} \ \times \ {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}\right)\right)\right)}$

Quadruple force and equation of motion

As with the four-momentum, a four- force, also called a Minkowski force, can be defined analogously to the corresponding Newtonian force :

${\displaystyle K^{\mu }={\frac {\mathrm {d} p^{\mu }}{\mathrm {d} \tau }}=\gamma {\frac {\mathrm {d} p^{\mu }}{\mathrm {d} t}}}$

This is the equation of motion of the special theory of relativity. It describes accelerated movements in an inertial system .

Furthermore, the four force can be related to the Newtonian force : In the inertial system in which the mass is approximately at rest (it is at rest at the point in time , then for sufficiently small due to the limited acceleration:) , the classical Newtonian equation must apply: ${\displaystyle \mathbf {F} }$ ${\displaystyle t=0}$ ${\displaystyle t}$ ${\displaystyle v\ll c}$

${\displaystyle {\begin{pmatrix}0\\\mathbf {F} \end{pmatrix}}={\frac {\mathrm {d} p^{\mu }}{\mathrm {d} t}}={\frac {1}{\gamma }}{\frac {\mathrm {d} p^{\mu }}{\mathrm {d} \tau }}\Rightarrow \mathbf {F} ={\frac {1}{\gamma }}K^{i}\Leftrightarrow K^{i}=\gamma \,\mathbf {F} }$

with the spatial part of the quadruple. ${\displaystyle K^{i}=(K^{1},K^{2},K^{3})}$

In any inertial system we have

${\displaystyle {\begin{pmatrix}K^{0}\\K^{1}\\K^{2}\\K^{3}\end{pmatrix}}={\begin{pmatrix}{\frac {1}{c}}\mathbf {u} \mathbf {F} \\\mathbf {F} _{\perp \mathbf {u} }+\gamma \mathbf {F} _{\|\mathbf {u} }\end{pmatrix}}={\begin{pmatrix}{\frac {1}{c}}\mathbf {u} \mathbf {F} \\\left(\mathbf {F} -{\frac {\mathbf {u} \mathbf {F} }{u}}{\frac {\mathbf {u} }{u}}\right)+\gamma {\frac {\mathbf {u} \mathbf {F} }{u}}{\frac {\mathbf {u} }{u}}\end{pmatrix}}}$,

where the spatial component is the quad speed. That means, the space part of the Minkowski force is the Newtonian force, whereby the part parallel to the velocity is multiplied by. ${\displaystyle \mathbf {u} =\gamma \mathbf {v} }$ ${\displaystyle \gamma }$

The power transmitted by the acceleration is . ${\displaystyle K^{\mu }}$ ${\displaystyle cK^{0}}$

In the special case that a Newtonian force acts only parallel to the speed, the relationship between Newtonian force and spatial acceleration follows from the equation of motion for four-vectors: ${\displaystyle \mathbf {F} }$

${\displaystyle \mathbf {F} =\gamma ^{3}m\mathbf {a} }$

For spatial forces perpendicular to the direction of movement , however, follows

${\displaystyle \mathbf {F} =\gamma m\mathbf {a} }$.

The concept of a “dynamic” relativistic mass for the term, which is sometimes introduced when considering momentum, is therefore misleading in comparison with Newton's equation of motion . Because for any spatial directions, the relationship between the spatial variables is and although linear , but no simple proportionality. ${\displaystyle \gamma m}$ ${\displaystyle \mathbf {F} }$ ${\displaystyle \mathbf {a} }$

Co- and contravariant vectors

The components of a contravariant four-vector change in Lorentz transformations into: ${\displaystyle a}$ ${\displaystyle \Lambda }$

${\displaystyle a^{\prime }=\Lambda \,a}$

You write your components with the numbers above: ${\displaystyle a=(a^{0},a^{1},a^{2},a^{3})}$

The components of a covariant four-vector follow the contra-redient (opposite) transformation law:

${\displaystyle b^{\prime }=\Lambda ^{-1\,{\text{T}}}\,b}$

You write your components with the numbers below: ${\displaystyle b=(b_{0},b_{1},b_{2},b_{3})}$

The two transformation laws are not the same, but equivalent , because by definition they satisfy:

${\displaystyle \Lambda ^{-1\,{\text{T}}}=\eta \,\Lambda \,\eta ^{-1}}$

with the usual Minkowski metric of the SRT:

${\displaystyle \eta _{\mu \nu }=\mathrm {diag} (1,-1,-1,-1)=\eta ^{\mu \nu }}$

Hence results

${\displaystyle a_{\mu }=(a_{0},a_{1},a_{2},a_{3})=\eta _{\mu \nu }a^{\nu }=\eta \,a=(a^{0},-a^{1},-a^{2},-a^{3})}$

the components of the covariant vector associated with the contravariant vector . ${\displaystyle a}$

Einstein's sum convention is used for the four-vector indices . The inner product of two four-vectors in the Minkowski space is given by:

${\displaystyle a_{\mu }b^{\mu }=\eta _{\mu \nu }a^{\nu }b^{\mu }=a_{0}b_{0}-a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3}}$

For example, the partial derivatives of a function are the components of a covariant vector. ${\displaystyle f(x)}$

Lorentz transformations map to: ${\displaystyle x}$

${\displaystyle x^{\prime }=\Lambda \,x}$

and define the transformed function by stating that it has the same value at the transformed location as the original function at the original location: ${\displaystyle f^{\prime }=f\circ \Lambda ^{-1}}$

${\displaystyle f^{\prime }(x^{\prime })=f(x)}$

With

${\displaystyle f^{\prime }(x)=f(\Lambda ^{-1}\,x)}$

Because of the chain rule, the partial derivatives transform contrarily:

${\displaystyle {\frac {\partial f^{\prime }}{\partial x^{m}}}(x)={\frac {\partial (\Lambda ^{-1}x)^{n}}{\partial x^{m}}}\,{\frac {\partial f}{\partial y^{n}}}_{|_{y=\Lambda ^{-1}\,x}}=\Lambda ^{-1\,n}{}_{m}\,{\frac {\partial f}{\partial y^{n}}}_{|_{y=\Lambda ^{-1}\,x}}=\Lambda ^{-1\,{\text{T}}}{}_{m}{}^{n}\,{\frac {\partial f}{\partial y^{n}}}_{|_{y=\Lambda ^{-1}\,x}}}$

See also

• Quadruple tensor

literature

• L. D. Landau : Textbook of theoretical physics. Volume 2: L. D. Landau, E. M. Lifschitz : Classical field theory. 12th revised edition. Verlag Harri Deutsch, Thun et al. 1997, ISBN 3-8171-1327-7 .
• Torsten Fliessbach : General Theory of Relativity. BI-Wissenschafts-Verlag, Mannheim et al. 1990, ISBN 3-411-14331-2 (with a chapter on special relativity theory).
• Walter Greiner : Theoretical Physics. Volume 3a: Walter Greiner, Johann Rafelski : Special Theory of Relativity. 2nd revised and expanded edition. Verlag Harri Deutsch, Thun et al. 1989, ISBN 3-8171-1063-4 .
• Reinhard Meinel : Special and general theory of relativity for Bachelor students. 2nd Edition. Springer-Verlag, Berlin 2019, ISBN 978-3-662-58966-3 .

Web links

• Norbert Dragon Geometry of the Theory of Relativity (PDF; 2.5 MB).