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A nonzero four-force cannot have all components constant in an inertial frame along an entire constant-mass trajectory. If $U=U_0+F\tau /m$, preserving $U^2=c^2$ for all proper times requires both $U_0\cdot F=0$ and $F^2=0$. The orthogonal complement of timelike $U_0$ is negative definite, forcing $F=0$. A constant laboratory three-force or a four-force of constant invariant magnitude is a different, consistent condition.

The four-force is the proper time derivative of four-momentum. For constant rest mass and coordinate three-force $\mathbf{F}=d\mathbf{p}/dt$, its components are $K^{\mu }=\gamma (\mathbf{F}\cdot \mathbf{v}/c,\mathbf{F})$ in the convention $p^{\mu }=(E/c,\mathbf{p})$. Thus it packages the power transfer and three-force into one Lorentz-covariant vector.

Covered by the official notes: definition of a pure force via constant rest mass (PDF p. 13; handwritten p. 8).^[ref]

The page labels $d{p}^{\mu }/d\tau$ as a proper force; that is the four-force. The requested proper force is the three-force measured in the instantaneous rest frame of the object. For a pure force its magnitude is $m_0$ times the proper acceleration, and in that frame the four-force has components $(0,{\mathbf{f}}_0)$.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

At any event on a massive observer's worldline, choose an orthonormal tetrad with timelike basis vector $e_{(0)}=U/c$. Its components obey $g(e_{(a)},e_{(b)})={\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, so locally $d{s}^2={\eta }_{ab}d{X}^{(a)}d{X}^{(b)}$: the Minkowski metric. A local inertial frame can also remove the Christoffel symbols at that event. Curvature cannot be removed on a finite region, and a frame following an accelerated observer need not remain inertial in its neighbourhood.

In general coordinates, $X^{\mu }(\tau )$ labels the trajectory, while the four-velocity and four-acceleration areCoordinate values $X^{\mu }$ are not themselves a vector under arbitrary nonlinear coordinate changes; their tangent $U$ is. For fixed rest mass $m$, the four-momentum is $P^{\mu }=m{U}^{\mu }$ and the four-force is $f^{\mu }=D{P}^{\mu }/d\tau =m{A}^{\mu }$. With the $+---$ metric signature,where $a$ is the proper acceleration. The middle identity follows by differentiating the constant velocity norm; in the instantaneous rest frame, $U=(c,0)$ and $A=(0,\mathbf{a})$, proving the last identity.

The proper acceleration is $\alpha =\sqrt{-A^2}$, the magnitude of the instantaneous rest-frame three-acceleration. A pure force is a four-force preserving rest mass, or equivalently satisfying $F\cdot U=0$. Indeed $F=d(mU)/d\tau$ gives $F\cdot U=c^{2}dm/d\tau$.

For the right-hand branch of the hyperbola, use $c=1$, $x>0$ and $x=\sqrt{L^2+t^2}$, taking $L>0$. Direct differentiation givesThe sign of $v$ follows the sign of $t$, whereas the coordinate acceleration on this branch is positive. ConsequentlyThe proper-time parameterization $t=L\sinh (\tau /L)$, $x=L\cosh (\tau /L)$ independently gives constant proper acceleration. Restoring dimensions, a hyperbola $x^2-c^2{t}^2=L^2$ has $\alpha =c^2/L$. Constant proper acceleration does not mean that all inertial-frame components of the four-force are constant.

For a massive particle on a timelike world line, the four-velocity is $U^{\mu }=d{X}^{\mu }/d\tau =\gamma (c,\mathbf{v})$, with $U^2=c^2$. Its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$ and its four-force is $F^{\mu }=d{P}^{\mu }/d\tau$. Here $\tau$ is proper time. A massless trajectory does not admit this massive-particle proper-time parametrization.

The asserted classification of arbitrary orthogonal vectors is false. With the chosen Minkowski metric, $D=(0,1,0,0)$ and $G=(0,0,1,0)$ are nonzero, orthogonal and both spacelike vectors. Also $D=(1,1,0,0)$ and $G=2D$ are orthogonal null vectors. The valid statement is that a nonzero vector orthogonal to a timelike vector is spacelike. To prove it, choose the rest frame of the timelike vector, $D=(d,0,0,0)$; then $D\cdot G=d{G}^0=0$ forces $G^0=0$, so $G^2=-|\mathbf{G}{|}^2<0$.

The relativistic energy-momentum relation givesTherefore $\boxed{P^2=0⟹m=0}$. A photon is an example: $E=c|\mathbf{p}|$. A collection of photons need not itself have zero invariant mass.

There is no nonzero, literally constant four-force that is a pure force throughout this motion. Initially $U=(c,0,0,0)$, so purity forces $F^0=0$. If the full four-vector were constant, $U(\tau )=U(0)+F\tau /M_0$; its square would be $c^2+F^2{\tau }^2/M_0^2$, which cannot remain $c^2$ for a nonzero initially spatial $F$. This is a genuine ambiguity in the source wording.

The usual consistent interpretation is constant laboratory relativistic force $f=d{p}_x/dt$, equivalently constant one-dimensional proper acceleration $f/M_0$. Take $f>0$ and let $s=ft/(M_{0}c)$. Integrating momentum conservation with the applied impulse gives $p_x=ft$, soThe corresponding four-force has components $F^{\mu }=\gamma f(\beta ,1,0,0)$, which are not constant, although its invariant magnitude is $F^2=-f^2$. Its temporal component is indeed $F^0=(\gamma /c)dE/dt$. The speed rises from zero and approaches $c$ asymptotically; $\gamma$ starts at one, initially rises quadratically and becomes linear in $t$ at late times.

If instead only the spatial four-force component $F^x=f_x$ is held constant, the distinct consistent answer is most simply parametrized by $u=f_x\tau /(M_{0}c)$:This follows by integrating $d{U}^x/d\tau =f_x/M_0$ and $dt/d\tau =U^0/c$. The temporal component must still vary. These two interpretations must not be conflated with a constant four-vector.

With constant rest mass, $\mathbf{p}=m\gamma \mathbf{v}$ and the chain rule gives $\dot{\gamma }={\gamma }^3(\mathbf{v}\cdot \mathbf{a})/c^2$. Hence the laboratory relativistic force isUse the vector triple product identity $\mathbf{v}\times (\mathbf{v}\times \mathbf{a})=\mathbf{v}(\mathbf{v}\cdot \mathbf{a})-v^2\mathbf{a}$, and ${\gamma }^3(1-v^2/c^2)=\gamma$, to obtainThe relativistic energy is $E=m\gamma c^2$, whose time derivative is $m{\gamma }^3\mathbf{v}\cdot \mathbf{a}$. Taking the dot product of the first expression for $\mathbf{F}$ with $\mathbf{v}$ gives the same quantity, because $\gamma +{\gamma }^3{v}^2/c^2={\gamma }^3$. Thus $dE/dt=\mathbf{F}\cdot \mathbf{v}$. Since $d/d\tau =\gamma d/dt$, the four-force isIts tensor contraction with the four-velocity vanishes: $F^{\alpha }{U}_{\alpha }=-{\gamma }^2\mathbf{F}\cdot \mathbf{v}+{\gamma }^2\mathbf{F}\cdot \mathbf{v}=0$. This orthogonality also follows by differentiating $p^{\alpha }{p}_{\alpha }=-m^2{c}^2$. Resolving the laboratory acceleration relative to $\mathbf{v}$ gives ${\mathbf{F}}_{\parallel }=m{\gamma }^3{\mathbf{a}}_{\parallel }$ and ${\mathbf{F}}_{\perp }=m\gamma {\mathbf{a}}_{\perp }$.

The three-force measured in a particle's instantaneous rest frame. For a pure force, ${\mathbf{f}}_0=m_0{\mathbf{a}}_0$, where ${\mathbf{a}}_0$ is the rest-frame acceleration. In that frame the four-force has zero temporal component and spatial part ${\mathbf{f}}_0$; the four-force itself is a different object.

A four-force preserving the particle's rest mass. For a massive particle, $P=mU$ and $U^2=c^2$ in the mostly-minus convention imply $F\cdot U=c^{2}dm/d\tau$. Thus constant mass is equivalent to orthogonality of the force and four-velocity. Energy transfer in a moving frame is allowed even though rest mass does not change.

For three-momentum $p=\gamma mv$, force and acceleration satisfy

The associated four-force uses proper time instead of coordinate time and is a different, four-dimensional object.

For a particle instantaneously at rest in a primed frame moving at $v\hat{\mathbf{x}}$ relative to the laboratory, the ordinary three-force transforms as $F_{\parallel }=F_{\parallel }^{\prime }$ and $F_{\perp }=F_{\perp }^{\prime }/\gamma$. The spatial components of the four-force are ${\gamma }_u\mathbf{F}$, whereas its time component is ${\gamma }_u\mathbf{F}\cdot \mathbf{u}/c$. For a co-moving material segment, length contraction also changes its laboratory length, so transforming total force is not the same as transforming force per unit length.