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Proper Acceleration and Accelerometer Readings

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An ideal accelerometer measures the spatial components of a worldline’s four-acceleration in the instrument’s instantaneous orthonormal frame. Its scalar reading is their Euclidean magnitude. This is proper acceleration: it belongs to the worldline, not to a coordinate description of that worldline.

Required background. Special Relativity provides Minkowski causal geometry, four-vectors, and mass-shell kinematics; Relativistic Observers and Measurement provides observer rest spaces and invariant decompositions of measured quantities. Helpful background. Calculus and Linear Algebra provides multivariable differentiation, linear maps, inner products, and basis changes; Ordinary Differential Equations provides initial-value problems, linearization, and asymptotic solution methods.

Proper acceleration as an instrument reading

Section titled “Proper acceleration as an instrument reading”

Let xμ(τ)x^\mu(\tau) be a timelike worldline parametrized by proper time. Its four-velocity and four-acceleration are

uμ=dxμdτ,aμ=Duμdτ=uν∇νuμ.u^\mu=\frac{dx^\mu}{d\tau}, \qquad a^\mu=\frac{D u^\mu}{d\tau} =u^\nu\nabla_\nu u^\mu.

In inertial Cartesian components the connection vanishes, so the second expression reduces to duμ/dτdu^\mu/d\tau. The covariant form remains valid in the accelerated and curvilinear coordinates used later. Because the connection is metric-compatible and uμuμ=−1u^\mu u_\mu=-1,

0=Ddτ(uμuμ)=2uμaμ.0=\frac{D}{d\tau}(u^\mu u_\mu)=2u_\mu a^\mu.

Thus aμa^\mu is orthogonal to the timelike vector uμu^\mu and is therefore spacelike or zero. The invariant

α=aμaμ≥0\boxed{\alpha=\sqrt{a^\mu a_\mu}\geq0}

is the proper-acceleration magnitude. The instantaneous-rest-frame interpretation follows directly in Hughes 2024, Lecture 13, §13.2, printed pp. 104–105, Eqs. (13.1)–(13.5).

Carry an orthonormal tetrad ea^μe_{\hat a}{}^\mu with e0^μ=uμe_{\hat0}{}^\mu=u^\mu. The accelerometer reports

ai^=ei^μaμ,a0^=−uμaμ=0,α2=δi^j^ai^aj^.a^{\hat i}=e^{\hat i}{}_{\mu}a^\mu, \qquad a^{\hat0}=-u_\mu a^\mu=0, \qquad \alpha^2=\delta_{\hat i\hat j}a^{\hat i}a^{\hat j}.

The decomposition of a carried orthonormal frame into Fermi–Walker and spatial-rotation parts is shown in Gourgoulhon 2010, PDF slides 17–18.

Operationally, the proof mass tends to free drift relative to the driven casing, but a working sensor usually suspends or servos it. The calibrated relative displacement—or the feedback force required to hold the mass—measures ai^a^{\hat i}; optical or capacitive readout only reports that response. NIST 2021, “A Better Way to Measure Acceleration” describes both a beam-suspended proof mass and an interferometric displacement readout. In the flat-spacetime scope of this chapter, this is the device’s specific-force reading. A freely drifting instrument reads zero even in curvilinear coordinates; an instrument held at a fixed accelerated-coordinate position can read nonzero even though its coordinate acceleration vanishes.

The direction components depend on how the spatial axes are carried. The magnitude α\alpha does not. Defining those axes without hidden rotation is the purpose of Fermi–Walker transport on the next page.

In an inertial chart, write uμ=γ(1,v)u^\mu=\gamma(1,\mathbf v), where γ=(1−v2)−1/2\gamma=(1-v^2)^{-1/2} and A=dv/dt\mathbf A=d\mathbf v/dt. For v≠0v\ne0, define v^=v/v\hat{\mathbf v}=\mathbf v/v, A∥=(A⋅v^)v^\mathbf A_\parallel=(\mathbf A\mathbin{\cdot}\hat{\mathbf v})\hat{\mathbf v}, and A⊥=A−A∥\mathbf A_\perp=\mathbf A-\mathbf A_\parallel. Then

α2=γ6A∥2+γ4A⊥2.\boxed{\alpha^2=\gamma^6 A_{\parallel}^2+\gamma^4 A_{\perp}^2}.

To see where the powers of γ\gamma enter, use d/dτ=γ d/dtd/d\tau=\gamma\,d/dt and

dγdt=γ3v⋅A.\frac{d\gamma}{dt}=\gamma^3\mathbf v\mathbin{\cdot}\mathbf A.

The temporal component is a0=γ4v⋅Aa^0=\gamma^4\mathbf v\mathbin{\cdot}\mathbf A, while

a=γ2A+γ4(v⋅A)v.\mathbf a=\gamma^2\mathbf A +\gamma^4(\mathbf v\mathbin{\cdot}\mathbf A)\mathbf v.

Contracting −(a0)2+a2-(a^0)^2+\mathbf a^2 first gives

aμaμ=γ4A2+γ6(v⋅A)2.a^\mu a_\mu =\gamma^4\mathbf A^2 +\gamma^6(\mathbf v\mathbin{\cdot}\mathbf A)^2.

Resolving A\mathbf A relative to v\mathbf v produces the boxed invariant. At v=0v=0 the parallel direction is undefined but the unsplit expression is regular and gives α=∣A∣\alpha=|\mathbf A|. This calculation also shows why simply multiplying every laboratory-acceleration component by one common γ\gamma factor fails.

For one-dimensional motion, α=γ3∣dv/dt∣\alpha=\gamma^3|dv/dt|. For uniform circular motion at laboratory radius RR and angular speed Ω\Omega, require ∣Ω∣R<1|\Omega|R\lt1. The speed is constant, A⊥=Ω2RA_\perp=\Omega^2R, and

α=γ2v2R=γ2Ω2R,γ=(1−Ω2R2)−1/2.\alpha=\gamma^2\frac{v^2}{R} =\gamma^2\Omega^2R, \qquad \gamma=(1-\Omega^2R^2)^{-1/2}.

The calibrated proper-acceleration vector points inward, even though a suspended proof mass may be displaced outward relative to the casing. The coordinate components of aμa^\mu vary around the circle, but its magnitude is constant.

  • In the instantaneous rest frame, ua^=(1,0,0,0)u^{\hat a}=(1,0,0,0) and aa^=(0,arest)a^{\hat a}=(0,\mathbf a_{\rm rest}), so α=∣arest∣\alpha=|\mathbf a_{\rm rest}|. This fixes the sign and interpretation.
  • In the low-speed limit, γ→1\gamma\to1, so proper and laboratory acceleration agree.
  • A second coordinate derivative d2xi/dt2d^2x^i/dt^2 is neither a four-vector nor generally an accelerometer reading. It also changes under a reparametrization of time.
  • “Momentarily at rest” does not mean unaccelerated. It says v=0\mathbf v=0 at one event, not aμ=0a^\mu=0 there.
  • Proper acceleration is local. It does not, by itself, specify a reference frame or a network of synchronized clocks.

Transfer task: a constant accelerometer reading

Section titled “Transfer task: a constant accelerometer reading”

A rocket moves along the xx axis, starts from rest, and keeps its thrust and proper-acceleration vector in the fixed +x+x direction with magnitude α=a0>0\alpha=a_0\gt0. Find its laboratory velocity v(t)v(t).

Solution

The one-dimensional relation gives

a0=γ3dvdt=1(1−v2)3/2dvdt.a_0=\gamma^3\frac{dv}{dt} =\frac{1}{(1-v^2)^{3/2}}\frac{dv}{dt}.

Integrating from v=0v=0 at t=0t=0,

a0t=v1−v2,v(t)=a0t1+a02t2.a_0t=\frac{v}{\sqrt{1-v^2}}, \qquad \boxed{v(t)=\frac{a_0t}{\sqrt{1+a_0^2t^2}}}.

Hence v≃a0tv\simeq a_0t for a0t≪1a_0t\ll1, while v→1v\to1 without exceeding the speed of light. The associated trajectory is derived on the Rindler page.

Continue to Fermi–Walker Transport and Nonrotating Frames to determine how an accelerating observer can carry spatial directions without rotating them.