Quadruple pulse

In relativistic physics , the four -momentum or energy-momentum vector of a particle or system is summarized as its energy and its momentum in the form of a four-vector , i.e. H. a vector with four components (energy + 3 spatial directions of the momentum). The quadruple momentum is a conserved quantity , i.e. i.e. it remains constant as long as the particle or system does not experience any external influences.

Energy-momentum relation

In units of measurement in which the speed of light has the dimensionless value , the relationship between energy and momentum of a particle of mass and speed results : ${\displaystyle c=1}$ ${\displaystyle E}$ ${\displaystyle {\boldsymbol {p}}}$ ${\displaystyle m}$ ${\displaystyle {\boldsymbol {v}}}$

${\displaystyle {\begin{pmatrix}E\\{\boldsymbol {p}}\end{pmatrix}}={\begin{pmatrix}{\frac {m}{\sqrt {1-{\boldsymbol {v}}^{2}}}}\\{\frac {m\,{\boldsymbol {v}}}{\sqrt {1-{\boldsymbol {v}}^{2}}}}\end{pmatrix}}}$

Invariant mass

The square of the length of the four-momentum is - regardless of the speed - always equal to the square of the mass (and therefore - like every scalar or every scalar product of four-vectors - invariant under Lorentz transformation ):

${\displaystyle {\begin{pmatrix}E\\{\boldsymbol {p}}\end{pmatrix}}\cdot {\begin{pmatrix}E\\{\boldsymbol {p}}\end{pmatrix}}=E^{2}-{\boldsymbol {p}}^{2}=m^{2}}$

Mass bowl

This energy-momentum relation or energy-momentum relation , which is fundamental for relativistic kinematics, means geometrically that the possible four -momentum of particles of mass in the four-dimensional momentum space is all based on the equation ${\displaystyle {\boldsymbol {p}}}$ ${\displaystyle m}$ ${\displaystyle {\begin{pmatrix}E\\{\boldsymbol {p}}\end{pmatrix}}}$

${\displaystyle E^{2}-{\boldsymbol {p}}^{2}=m^{2}}$

described three-dimensional hypersurface (a two-shell hyperboloid ) whose asymptotes form the light cone of the momentum space. Because a four-pulse is always forward-looking (i.e. lies inside the forward light cone), only one of the two shells of the hyperboloid comes into question, namely the one given by the equation

${\displaystyle E={\sqrt {{\boldsymbol {p}}^{2}+m^{2}}}}$

described mass shell .

For virtual particles, the following applies if the mass of the same particle is in the real state. In technical jargon one says: They are “not on the mass shell” or: They are not “on-shell” but “off-shell”. ${\textstyle m^{2}\neq E^{2}-{\boldsymbol {p}}^{2}}$ ${\textstyle m}$

Derivation of the velocity dependence of energy and momentum

How the energy and the momentum of a particle of mass depend on its velocity , results in the relativity theory from the fact that energy and momentum are additive conserved quantities for every observer. We refer to them collectively as . If one particle has an additive conserved quantity and another particle has the conserved quantity , then the system of both particles has the conserved quantity . ${\displaystyle m}$ ${\displaystyle \mathbf {v} \,(|\mathbf {v} |<c)}$ ${\displaystyle p}$ ${\displaystyle p_{1}}$ ${\displaystyle p_{2}}$ ${\displaystyle p=p_{1}+p_{2}}$

Even a moving observer determines the conservation quantities and for both particles , but they do not necessarily have the same, but transformed values. However, it must be the case that the sum of these values ​​is the transform of the sum: ${\displaystyle p_{1}^{\prime }}$ ${\displaystyle p_{2}^{\prime }}$

${\displaystyle p_{1}^{\prime }+p_{2}^{\prime }=(p_{1}+p_{2})^{\prime }}$

In the same way (for all numbers ) there is a multiplied system with a conservation magnitude for the moving observer the multiplied conservation magnitude ${\displaystyle a}$ ${\displaystyle a\,p}$

${\displaystyle (a\,p)^{\prime }=a\,p^{\prime }}$

to. Mathematically, this means that the conserved quantities that a moving observer measures by means of a linear transformation ${\displaystyle L}$

${\displaystyle p^{\prime }=Lp}$

are related to the conserved quantities of the observer at rest.

The linear transformation is limited by the fact that such an equation has to hold for every pair of observers, the reference systems of the observers being derived from each other by Lorentz transformations and shifts. If the reference systems of the first and second observer are connected through and from the second to a third through , then the reference system is connected from the first to the third through . The corresponding transformations of the conserved quantities must be exactly the same ${\displaystyle \Lambda }$ ${\displaystyle \Lambda _{1}}$ ${\displaystyle \Lambda _{2}}$ ${\displaystyle \Lambda _{2}\circ \Lambda _{1}}$

${\displaystyle L(\Lambda _{2})\circ L(\Lambda _{1})=L(\Lambda _{2}\circ \Lambda _{1})}$

fulfill.

In the simplest case is . Since Lorentz transformations - matrices are so concerns the simplest, non-trivial transformation law, and not just a true, four conserved quantities , like the space-time coordinates transform as a four-vector: ${\displaystyle L(\Lambda )=\Lambda }$ ${\displaystyle 4\times 4}$ ${\displaystyle p^{\prime }=p}$ ${\displaystyle p}$

${\displaystyle p^{\prime }=\Lambda p}$

In anticipation of the result of our consideration, we call this four-vector the four-momentum.

In particular, a particle at rest does not change when it rotates. Therefore, those components of its quadruple momentum that, like a three-dimensional position vector, change into a rotated vector when rotated , do not change either . But the only such vector is the zero vector . So the quadruple momentum of a particle at rest has a value ${\displaystyle p}$ ${\displaystyle p}$

${\displaystyle p_{\text{Ruhe}}={\begin{pmatrix}m\\0\\0\\0\end{pmatrix}}}$

The name is chosen in anticipation of the later result, but initially stands for some value. ${\displaystyle m}$

For along the moving axis observer the particle has a speed and a Lorentz transformed four pulse (we expect the sake of simplicity in measurement systems with ): ${\displaystyle x}$ ${\displaystyle v}$ ${\displaystyle c=1}$

${\displaystyle {\begin{pmatrix}{\frac {1}{\sqrt {1-v^{2}}}}&{\frac {v}{\sqrt {1-v^{2}}}}&&\\{\frac {v}{\sqrt {1-v^{2}}}}&{\frac {1}{\sqrt {1-v^{2}}}}&&\\&&1&\\&&&1\\\end{pmatrix}}{\begin{pmatrix}m\\0\\0\\0\end{pmatrix}}={\begin{pmatrix}{\frac {m}{\sqrt {1-v^{2}}}}\\{\frac {m\,v}{\sqrt {1-v^{2}}}}\\0\\0\end{pmatrix}}}$

If one develops the four conserved quantities according to the speed:

${\displaystyle {\begin{pmatrix}{\frac {m}{\sqrt {1-v^{2}}}}\\{\frac {m\,v}{\sqrt {1-v^{2}}}}\\0\\0\end{pmatrix}}={\begin{pmatrix}m+{\frac {1}{2}}\,mv^{2}+\dots \\m\,v+\dots \\0\\0\end{pmatrix}}}$

and if one compares with Newton's mechanics, the physical meaning of the components of the four-momentum is revealed: the first component is the energy and the three components, which change with rotations like a position vector, are the momentum:

${\displaystyle p={\begin{pmatrix}{\text{Energie}}\\{\text{Impuls}}\end{pmatrix}}}$

As in Newton's mechanics, the speed-independent parameter in the relation that indicates the momentum of a particle as a function of its speed is called mass. It must be positive according to all observations. ${\displaystyle m}$

Consideration in SI units

The equation given in the first section for the quadruple momentum only applies if the speed of light has the dimensionless value . In other systems of measurement the factor is to be inserted as follows: ${\displaystyle c=1}$ ${\displaystyle c}$

${\displaystyle {\begin{pmatrix}{\frac {E}{c}}\\{\boldsymbol {p}}\end{pmatrix}}={\begin{pmatrix}{\frac {mc}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}\\{\frac {m\,{\boldsymbol {v}}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}\end{pmatrix}}}$

Hence the energy is:

${\displaystyle E({\boldsymbol {v}})={\frac {m\,c^{2}}{\sqrt {1-{\frac {{\boldsymbol {v}}^{2}}{c^{2}}}}}}=\gamma (v)\cdot m\,{c^{2}}}$

with the Lorentz factor ${\displaystyle \gamma \geq 1.}$

It is limited downwards by the rest energy :

${\displaystyle E({\boldsymbol {v}})\geq E_{\text{Ruhe}}=E({\boldsymbol {v}}=0)=m\,c^{2}}$

The relativistic impulse is:

${\displaystyle {\boldsymbol {p}}({\boldsymbol {v}})={\frac {m\,{\boldsymbol {v}}}{\sqrt {1-{\frac {{\boldsymbol {v}}^{2}}{c^{2}}}}}}=\gamma (v)\cdot m{\boldsymbol {v}}}$

The relativistic energy-momentum relation results from the square of the energy:

${\displaystyle {\begin{aligned}E^{2}&=\gamma ^{2}\cdot (mc^{2})^{2}\\&=(1-{\boldsymbol {\beta }}^{2}+{\boldsymbol {\beta }}^{2})\cdot \gamma ^{2}\cdot (mc^{2})^{2}\\&=\underbrace {(1-{\boldsymbol {\beta }}^{2})\cdot \gamma ^{2}} _{=\,1}\cdot (mc^{2})^{2}+\gamma ^{2}\cdot m^{2}\cdot \underbrace {{\boldsymbol {\beta }}^{2}c^{4}} _{=\,{\boldsymbol {v}}^{2}c^{2}}\\&=(mc^{2})^{2}+{\boldsymbol {p}}^{2}c^{2}\end{aligned}}}$

with the assignment

${\displaystyle {\boldsymbol {\beta }}={\boldsymbol {v}}/c.}$

If you split off the mass from the quadruple momentum, the quadruple speed remains : ${\displaystyle u}$

${\displaystyle {\boldsymbol {p}}=m\,{\boldsymbol {u}}}$

It is the derivation of the world line that the particle traverses according to its proper time : ${\displaystyle {\tilde {\boldsymbol {x}}}(\tau )=(ct(\tau ),x(\tau ),y(\tau ),z(\tau ))}$ ${\displaystyle \tau }$

${\displaystyle {\boldsymbol {u}}={\frac {\mathrm {d} {\tilde {\boldsymbol {x}}}}{\mathrm {d} \tau }}={\begin{pmatrix}{\frac {c}{\sqrt {1-{\frac {{\boldsymbol {v}}^{2}}{c^{2}}}}}}\\{\frac {\boldsymbol {v}}{\sqrt {1-{\frac {{\boldsymbol {v}}^{2}}{c^{2}}}}}}\end{pmatrix}}}$

d. That is, the speed of four is the normalized tangential vector to the world line:

${\displaystyle (u^{0})^{2}-{\boldsymbol {u}}^{2}=c^{2}}$

The differential of the proper time is - in contrast to - a scalar quantity and gives the denominator . ${\displaystyle \mathrm {d} \tau }$ ${\displaystyle \mathrm {d} t}$ ${\displaystyle {\sqrt {1-{{\boldsymbol {v}}^{2}/c^{2}}}}}$

Application: equation of motion and the force / power quad vector

In the moving system there is and remains zero as long as no force acts. However, if a force is exerted over a period of time and an external power is applied at the same time , both the speed and the energy of the particle increase (in the same frame of reference as before!). Due to the impulse of force and the power supply , the equation of motion then applies : ${\displaystyle {\boldsymbol {v}}=0}$ ${\displaystyle \delta \tau }$ ${\displaystyle {\boldsymbol {K}}}$  ${\displaystyle L}$

${\displaystyle \delta \,{\boldsymbol {p}}=\delta (m\,{\boldsymbol {u}})={\begin{pmatrix}{\frac {L}{c}}\\{\boldsymbol {K}}\end{pmatrix}}\delta \tau }$

The right-hand side of this equation defines the force-power quad vector. So it will u. a. the rest energy of the system increases from to , d. that is, the mass is increased slightly; see. Equivalence of mass and energy . At the same time, the impulse increases the speed - and thus the kinetic energy . It is assumed that the speed starting from zero after the increase still remains small compared to the speed of light, so that Newtonian physics is valid in the moving system . ${\displaystyle mc^{2}}$ ${\displaystyle mc^{2}+L\delta \tau }$

See also

• Energy-momentum tensor

Individual evidence

1. ↑ See e.g. B. Volume 2 of the textbook series by Landau / Lifschitz , Harri Deutsch V., Frankfurt / Main