In the special theory of relativity, four-force is a four-vector that replaces the classical 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,:

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

For a particle of constant invariant 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 the four-velocity. In analogy to Newton's second law, we can also relate the four-force to the four-acceleration, ${\displaystyle \mathbf {A} }$, by equation:

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

Here

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

and

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

where ${\displaystyle \mathbf {u} }$, ${\displaystyle \mathbf {p} }$ and ${\displaystyle \mathbf {f} }$ are 3-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 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 ${\displaystyle h}$, 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 the stress–energy–momentum tensor.^[3][2]

Therefore, in thermo-mechanical situations the time component of the four-force is not 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 ${\displaystyle h+\mathbf {f} \cdot \mathbf {u} }$.

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.

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

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.^[5] 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 ${\displaystyle F^{\mu }=(F^{0},\mathbf {F} )}$ 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:

$${\displaystyle {\begin{aligned}\mathbf {f} &=\mathbf {F} +(\gamma -1)\mathbf {v} {\mathbf {v} \cdot \mathbf {F} \over v^{2}},\\f^{0}&=\gamma {\boldsymbol {\beta }}\cdot \mathbf {F} ={\boldsymbol {\beta }}\cdot \mathbf {f} .\end{aligned}}}$$

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

In general relativity, the expression for force becomes

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

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

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

where ${\displaystyle \Gamma _{\nu \lambda }^{\mu }}$ 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 ${\displaystyle f_{f}^{\alpha }}$ is the correct expression for force in a freely falling frame ${\displaystyle \xi ^{\alpha }}$, we can use then the equivalence principle to write the four-force in an arbitrary coordinate ${\displaystyle x^{\mu }}$:

$${\displaystyle f^{\mu }={\partial x^{\mu } \over \partial \xi ^{\alpha }}f_{f}^{\alpha }.}$$

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

where

• ${\displaystyle F_{\mu \nu }}$ is the electromagnetic tensor,
• ${\displaystyle U^{\nu }}$ is the four-velocity, and
• ${\displaystyle q}$ is the electric 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.