Four-force

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

F

=

d

P

d

τ

=

[

d
t

d
τ

d

d
t

m
γ

c

2

/

c

d
t

d
τ

d

d
t

m
γ

v

x

(
t
)

d
t

d
τ

d

d
t

m
γ

v

y

(
t
)

d
t

d
τ

d

d
t

m
γ

v

z

(
t
)

]

=

(

γ

f

⋅

v

c

,
γ

f

)

=
m

A

=
m

[

γ

4

a
v

c

2

c

γ

4

a
v

c

2

v

x

γ

2

a

x

(
t
)

γ

4

a
v

c

2

v

y

γ

2

a

y

(
t
)

γ

4

a
v

c

2

v

z

γ

2

a

z

(
t
)

]

{\displaystyle \mathbf {F} ={\mathrm {d} \mathbf {P} \over \mathrm {d} \tau }=\left[{\begin{array}{ll}{\frac {dt}{d\tau }}{\frac {d}{dt}}m\gamma c^{2}/c\\{\frac {dt}{d\tau }}{\frac {d}{dt}}m\gamma v\_{x}(t)\\{\frac {dt}{d\tau }}{\frac {d}{dt}}m\gamma v\_{y}(t)\\{\frac {dt}{d\tau }}{\frac {d}{dt}}m\gamma v\_{z}(t)\end{array}}\right]=\left(\gamma {\mathbf {f} \cdot \mathbf {v} \over c},\gamma {\mathbf {f} }\right)=m\mathbf {A} =m\left[{\begin{array}{ll}\gamma ^{4}{\frac {\mathbf {av} }{c^{2}}}c\\\gamma ^{4}{\frac {\mathbf {av} }{c^{2}}}v\_{x}+\gamma ^{2}a\_{x}(t)\\\gamma ^{4}{\frac {\mathbf {av} }{c^{2}}}v\_{y}+\gamma ^{2}a\_{y}(t)\\\gamma ^{4}{\frac {\mathbf {av} }{c^{2}}}v\_{z}+\gamma ^{2}a\_{z}(t)\\\end{array}}\right]}

.

## 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:

F

=

d

P

d

τ

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

.
For a particle of constant invariant mass

m
>
0

{\displaystyle m>0}

,

P

=
m

V

{\displaystyle \mathbf {P} =m\mathbf {V} }

 where

V

=
γ
(
c
,

v

)

{\displaystyle \mathbf {V} =\gamma (c,\mathbf {v} )}

 is the four-velocity, so we can relate the four-force with the four-acceleration

A

{\displaystyle \mathbf{A}}

 as in Newton's second law:

F

=
m

A

=

(

γ

f

⋅

v

c

,
γ

f

)

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

.
Here f is the rate of change of proper momentum (mass times proper velocity)

f

=

d

p

d

t

=

d

d

t

(

γ
m

v

)

=

v
c

γ

3

a
c

m
a

{\displaystyle {\mathbf {f} }={\mathrm {d} \mathbf {p} \over \mathrm {d} t}={\mathrm {d} \over \mathrm {d} t}\left(\gamma m{\mathbf {v} }\right)={\frac {v}{c}}\gamma ^{3}{\frac {a}{c}}ma}

and

f

⋅

v

=

d

d

t

(

γ
m

c

2

)

=

d

E

k
i
n
e
t
i
c

d

t

.

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

where

v

{\displaystyle \mathbf{v} }

,

p

{\displaystyle \mathbf{p}}

 and

f

{\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,

f

⋅

v

{\displaystyle \mathbf {f} \cdot \mathbf {v} }

, apart from relativistic corrections

γ

/

c

{\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

h

{\displaystyle h}

, besides the power

f

⋅

v

{\displaystyle \mathbf {f} \cdot \mathbf {v} }

. Note that work and heat cannot be meaningfully separated, though, as they both carry inertia. This fact extends also to contact forces, that is, to the  stress-energy-momentum tensor.

Therefore, in thermo-mechanical situations the time component of the four-force is *not* proportional to the power

f

⋅

v

{\displaystyle \mathbf {f} \cdot \mathbf {v} }

 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, and which in the Newtonian limit becomes

h

f

⋅

v

{\displaystyle h+\mathbf {f} \cdot \mathbf {v} }

.

## 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.

F

λ

:=

D

P

λ

d
τ

=

d

P

λ

d
τ

Γ

λ

μ
ν

V

μ

P

ν

{\displaystyle F^{\lambda }:={\frac {DP^{\lambda }}{d\tau }}={\frac {dP^{\lambda }}{d\tau }}+\Gamma ^{\lambda }{}\_{\mu \nu }V^{\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. 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

μ

=
(

F

0

,

F

)

{\displaystyle F^{\mu }=(F^{0},\mathbf {F} )}

 acting on a particle of mass

m

{\displaystyle m}

 which is momentarily at rest in a coordinate system. The relativistic force

f

μ

{\displaystyle f^{\mu }}

 in another coordinate system moving with constant velocity

v

{\displaystyle v}

, relative to the other one, is obtained using a Lorentz transformation:

f

=

F

(
γ
−
1
)

v

v

⋅

F

v

2

,

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

f

0

=
γ

β

⋅

F

=

β

⋅

f

.

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

where

β

=

v

/

c

{\displaystyle {\boldsymbol {\beta }}=\mathbf {v} /c}

.

In general relativity, the expression for force becomes

f

μ

=
m

D

V

μ

d
τ

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

with covariant derivative

D

/

d
τ

{\displaystyle D/d\tau }

. The equation of motion becomes

m

d

2

x

μ

d

τ

2

=

f

μ

−
m

Γ

ν
λ

μ

d

x

ν

d
τ

d

x

λ

d
τ

,

{\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

Γ

ν
λ

μ

{\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

f

f

α

{\displaystyle f\_{f}^{\alpha }}

 is the correct expression for force in a freely falling frame

ξ

α

{\displaystyle \xi ^{\alpha }}

, we can use then the equivalence principle to write the four-force in an arbitrary coordinate

x
μ

{\displaystyle x^\mu}

:

f

μ

=

∂

x

μ

∂

ξ

α

f

f

α

.

{\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:

F

μ

=
q

F

μ
ν

V

ν

{\displaystyle F\_{\mu }=qF\_{\mu \nu }V^{\nu }}

,
where

* F

μ
ν

{\displaystyle F\_{\mu\nu}}

 is the electromagnetic tensor,
* V

ν

{\displaystyle V^{\nu }}

 is the four-velocity, and
* q

{\displaystyle q}

 is the electric charge.

## See also

* four-vector
* four-velocity
* four-acceleration
* four-momentum
* four-gradient