Four-force

The four-force is defined as the rate of change in the four-momentum of a particle with respect to the particle's proper time. Hence,:

$$\backslash mathbf\{F\}\; =\; \{\backslash mathrm\{d\}\backslash mathbf\{P\}\; \backslash over\; \backslash mathrm\{d\}\backslash tau\}.$$

For a particle of constant invariant mass $m\; >\; 0$, the four-momentum is given by the relation $\backslash mathbf\{P\}\; =\; m\backslash mathbf\{U\}$, where $\backslash mathbf\{U\}=\backslash gamma(c,\backslash mathbf\{u\})$ is the four-velocity. In analogy to Newton's second law, we can also relate the four-force to the four-acceleration, $\backslash mathbf\{A\}$, by equation:

$$\backslash mathbf\{F\}\; =\; m\backslash mathbf\{A\}\; =\; \backslash left(\backslash gamma\; \{\backslash mathbf\{f\}\backslash cdot\backslash mathbf\{u\}\; \backslash over\; c\},\backslash gamma\{\backslash mathbf\; f\}\backslash right).$$

Here

$$\{\backslash mathbf\; f\}=\{\backslash mathrm\{d\}\; \backslash over\; \backslash mathrm\{d\}t\}\; \backslash left(\backslash gamma\; m\; \{\backslash mathbf\; u\}\; \backslash right)=\{\backslash mathrm\{d\}\backslash mathbf\{p\}\; \backslash over\; \backslash mathrm\{d\}t\}$$

and

$$\{\backslash mathbf\{f\}\backslash cdot\backslash mathbf\{u\}\}=\{\backslash mathrm\{d\}\; \backslash over\; \backslash mathrm\{d\}t\}\; \backslash left(\backslash gamma\; mc^2\; \backslash right)=\{\backslash mathrm\{d\}E\; \backslash over\; \backslash mathrm\{d\}t\}\; .$$

where $\backslash mathbf\{u\}$, $\backslash mathbf\{p\}$ and $\backslash mathbf\{f\}$ are 3-space vectors describing the velocity, the momentum of the particle and the force acting on it respectively; and $E$ is the total energy of the particle.

From the formulae of the previous section it appears that the time component of the four-force is the power expended, $\backslash mathbf\{f\}\backslash cdot\backslash mathbf\{u\}$, apart from relativistic corrections $\backslash gamma/c$. This is only true in purely mechanical situations, when heat exchanges vanish or can be neglected.

In the full thermo-mechanical case, not only work, but also heat contributes to the change in energy, which is the time component of the energy–momentum covector. The time component of the four-force includes in this case a heating rate $h$, besides the power $\backslash mathbf\{f\}\backslash cdot\backslash mathbf\{u\}$.{{cite journal|last1=Grot|first1=Richard A.|last2=Eringen|first2=A. Cemal| title=Relativistic continuum mechanics: Part I – Mechanics and thermodynamics|date=1966|journal=Int. J. Engng Sci.| volume=4|issue=6|pages=611–638, 664|doi=10.1016/0020-7225(66)90008-5}} Note that work and heat cannot be meaningfully separated, though, as they both carry inertia.{{cite journal| last1=Eckart|first1=Carl| title=The Thermodynamics of Irreversible Processes. III. Relativistic Theory of the Simple Fluid|date=1940|journal=Phys. Rev.| volume=58| issue=10| pages=919–924| doi=10.1103/PhysRev.58.919| bibcode=1940PhRv...58..919E}} This fact extends also to contact forces, that is, to the stress–energy–momentum tensor.C. A. Truesdell, R. A. Toupin: The Classical Field Theories (in S. Flügge (ed.): Encyclopedia of Physics, Vol. III-1, Springer 1960). §§152–154 and 288–289.

Therefore, in thermo-mechanical situations the time component of the four-force is not proportional to the power $\backslash mathbf\{f\}\backslash cdot\backslash mathbf\{u\}$ but has a more generic expression, to be given case by case, which represents the supply of internal energy from the combination of work and heat,{{cite journal| last1=Maugin|first1=Gérard A.|title=On the covariant equations of the relativistic electrodynamics of continua. I. General equations|date=1978|journal=J. Math. Phys.| volume=19| issue=5| pages=1198–1205| doi=10.1063/1.523785| bibcode=1978JMP....19.1198M}} and which in the Newtonian limit becomes $h\; +\; \backslash mathbf\{f\}\; \backslash cdot\; \backslash mathbf\{u\}$.

In general relativity the relation between four-force, and four-acceleration remains the same, but the elements of the four-force are related to the elements of the four-momentum through a covariant derivative with respect to proper time.

$$F^\backslash lambda\; :=\; \backslash frac\{DP^\backslash lambda\; \}\{d\backslash tau\}\; =\; \backslash frac\{dP^\backslash lambda\; \}\{d\backslash tau\; \}\; +\; \backslash Gamma^\backslash lambda\; \{\}\_\{\backslash mu\; \backslash nu\}U^\backslash mu\; P^\backslash nu$$

In addition, we can formulate force using the concept of coordinate transformations between different coordinate systems. Assume that we know the correct expression for force in a coordinate system at which the particle is momentarily at rest. Then we can perform a transformation to another system to get the corresponding expression of force.{{cite book|last1=Steven|first1=Weinberg|title=Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity|date=1972|publisher=John Wiley & Sons, Inc.|isbn=0-471-92567-5|url-access=registration|url=https://archive.org/details/gravitationcosmo00stev_0}} In special relativity the transformation will be a Lorentz transformation between coordinate systems moving with a relative constant velocity whereas in general relativity it will be a general coordinate transformation.

Consider the four-force $F^\backslash mu=(F^0,\; \backslash mathbf\{F\})$ acting on a particle of mass $m$ which is momentarily at rest in a coordinate system. The relativistic force $f^\backslash mu$ in another coordinate system moving with constant velocity $v$, relative to the other one, is obtained using a Lorentz transformation:

$$\backslash begin\{align\}$$

\mathbf{f} &= \mathbf{F} + (\gamma - 1) \mathbf{v} {\mathbf{v}\cdot\mathbf{F} \over v^2}, \\

f^0 &= \gamma \boldsymbol{\beta}\cdot\mathbf{F} = \boldsymbol{\beta}\cdot\mathbf{f}.

\end{align}

where $\backslash boldsymbol\{\backslash beta\}\; =\; \backslash mathbf\{v\}/c$.

In general relativity, the expression for force becomes

$$f^\backslash mu\; =\; m\; \{DU^\backslash mu\backslash over\; d\backslash tau\}$$

with covariant derivative $D/d\backslash tau$. The equation of motion becomes

$$m\; \{d^2\; x^\backslash mu\backslash over\; d\backslash tau^2\}\; =\; f^\backslash mu\; -\; m\; \backslash Gamma^\backslash mu\_\{\backslash nu\backslash lambda\}\; \{dx^\backslash nu\; \backslash over\; d\backslash tau\}\; \{dx^\backslash lambda\; \backslash over\; d\backslash tau\},$$

where $\backslash Gamma^\backslash mu\_\{\backslash nu\backslash lambda\}$ is the Christoffel symbol. If there is no external force, this becomes the equation for geodesics in the curved space-time. The second term in the above equation, plays the role of a gravitational force. If $f^\backslash alpha\_f$ is the correct expression for force in a freely falling frame $\backslash xi^\backslash alpha$, we can use then the equivalence principle to write the four-force in an arbitrary coordinate $x^\backslash mu$:

$$f^\backslash mu\; =\; \{\backslash partial\; x^\backslash mu\; \backslash over\; \backslash partial\backslash xi^\backslash alpha\}\; f^\backslash alpha\_f.$$

In special relativity, Lorentz four-force (four-force acting on a charged particle situated in an electromagnetic field) can be expressed as:

$$f\_\backslash mu\; =\; q\; F\_\{\backslash mu\backslash nu\}\; U^\backslash nu\; ,$$

where

• $F\_\{\backslash mu\backslash nu\}$ is the electromagnetic tensor,
• $U^\backslash nu$ is the four-velocity, and
• $q$ is the electric charge.

• four-vector
• four-velocity
• four-acceleration
• four-momentum
• four-gradient

{{Reflist}}

• {{cite book | author = Rindler, Wolfgang | title=Introduction to Special Relativity | url = https://archive.org/details/introductiontosp0000rind | url-access = registration | edition=2nd | location= Oxford | publisher=Oxford University Press | year=1991 | isbn=0-19-853953-3}}

Category:Four-vectors

Category:Force