In thespecial theory of relativity,four-force is afour-vector that replaces the classicalforce.

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

$${\displaystyle \mathbf{F}=\frac{\mathrm{d}\mathbf{P}}{\mathrm{d}\tau }.}$$

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

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

Here

$${\displaystyle \mathbf{f}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\gamma m\mathbf{u}\right)=\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t}}$$

and

$${\displaystyle \mathbf{f}\cdot \mathbf{u}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\gamma m{c}^2\right)=\frac{\mathrm{d}E}{\mathrm{d}t}.}$$

where${\displaystyle \mathbf{u}}$,${\displaystyle \mathbf{p}}$ and${\displaystyle \mathbf{f}}$ are3-space vectors describing the velocity, the momentum of the particle and the force acting on it respectively; and${\displaystyle E}$ 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,${\displaystyle \mathbf{f}\cdot \mathbf{u}}$, apart from relativistic corrections${\displaystyle \gamma /c}$. This is only true in purely mechanical situations, when heat exchanges vanish or can be neglected.

In the full thermo-mechanical case, not onlywork, but alsoheat contributes to the change in energy, which is the time component of theenergy–momentum covector. The time component of the four-force includes in this case a heating rate${\displaystyle h}$, besides the power${\displaystyle \mathbf{f}\cdot \mathbf{u}}$.^[1] Note that work and heat cannot be meaningfully separated, though, as they both carry inertia.^[2] This fact extends also to contact forces, that is, to thestress–energy–momentum tensor.^[3][2]

Therefore, in thermo-mechanical situations the time component of the four-force isnot proportional to the power${\displaystyle \mathbf{f}\cdot \mathbf{u}}$ 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,^[2][1][4][3] and which in the Newtonian limit becomes${\displaystyle h+\mathbf{f}\cdot \mathbf{u}}$.

Ingeneral relativity the relation between four-force, andfour-acceleration remains the same, but the elements of the four-force are related to the elements of thefour-momentum through acovariant derivative with respect to proper time.

$${\displaystyle F^{\lambda }:=\frac{D{P}^{\lambda }}{d\tau }=\frac{d{P}^{\lambda }}{d\tau }+{\mathrm{\Gamma }}^{\lambda }{}_{\mu \nu }{U}^{\mu }{P}^{\nu }}$$

In addition, we can formulate force using the concept ofcoordinate 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.^[5] Inspecial relativity the transformation will be a Lorentz transformation between coordinate systems moving with a relative constant velocity whereas ingeneral relativity it will be a general coordinate transformation.

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

where${\displaystyle \mathit{\beta }=\mathbf{v}/c}$.

Ingeneral relativity, the expression for force becomes

$${\displaystyle f^{\mu }=m\frac{D{U}^{\mu }}{d\tau }}$$

withcovariant derivative${\displaystyle D/d\tau }$. The equation of motion becomes

$${\displaystyle m\frac{d^2{x}^{\mu }}{d{\tau }^2}=f^{\mu }-m{\mathrm{\Gamma }}_{\nu \lambda }^{\mu }\frac{d{x}^{\nu }}{d\tau }\frac{d{x}^{\lambda }}{d\tau },}$$

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

$${\displaystyle f^{\mu }=\frac{\mathrm{\partial }{x}^{\mu }}{\mathrm{\partial }{\xi }^{\alpha }}{f}_f^{\alpha }.}$$

In special relativity,Lorentz four-force (four-force acting on a charged particle situated in an electromagnetic field) can be expressed as:$${\displaystyle f_{\mu }=q{F}_{\mu \nu }{U}^{\nu },}$$

where

• ${\displaystyle F_{\mu \nu }}$ is theelectromagnetic tensor,
• ${\displaystyle U^{\nu }}$ is thefour-velocity, and
• ${\displaystyle q}$ is theelectric charge.

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

• Rindler, Wolfgang (1991).Introduction to Special Relativity (2nd ed.). Oxford: Oxford University Press.ISBN 0-19-853953-3.

• Four-vectors
• Force

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• Short description is different from Wikidata