# Four-force

(Redirected from 4-force)

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

## In special relativity

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:

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

For a particle of constant invariant mass ${\displaystyle m>0}$, ${\displaystyle \mathbf {P} =m\mathbf {U} }$ where ${\displaystyle \mathbf {U} =\gamma (c,\mathbf {u} )}$ is the four-velocity, so we can relate the four-force with the four-acceleration ${\displaystyle \mathbf {A} }$ as in Newton's second law:

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

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.

## Including thermodynamic interactions

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.

If 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

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:

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

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

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 ${\displaystyle x^{\mu }}$:

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

## Examples

In special relativity, Lorentz four-force (four-force acting to charged particle situated in 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.