research31 min studyLesson 833 in Physics

Geodesic deviation and strain response

Relate TT curvature to relative acceleration and long-wavelength interferometer response without confusing coordinate motion with measured light phase.

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What you’ll understand

  • Calculate tidal acceleration from a weak TT waveform.
  • Distinguish a single-arm length change from differential detector strain.
  • Check the wavelength hierarchy behind a simple arm-response approximation.

A gravitational wave is observable because it changes the relation between nearby freely falling worldlines and the light exchanged between them. Geodesic deviation expresses that effect through curvature. In one useful gauge the test masses keep fixed coordinates while proper arm lengths vary; in a local freely falling frame their separation coordinates can move. Those descriptions agree on measured phase or distance when treated consistently. A strain formula must therefore specify its frame, its approximation, and whether it refers to one arm or a differential detector output.

Curvature drives relative acceleration

Retain the previous signature, x0=ct and Riemann sign convention. In the TT representation of a weak plane wave, the relevant tidal curvature is minus one half the second time derivative of the spatial strain tensor divided by c squared. Geodesic deviation then gives a relative acceleration equal to positive one half the strain's second derivative acting on the unperturbed separation. The sign of acceleration at a positive strain maximum is consequently restoring for a sinusoid.

This is an expansion for neighboring freely falling bodies and weak perturbations. A local detector-frame description neglects curvature variation across the apparatus when its size is much smaller than a wavelength. Integrating the acceleration gives a leading displacement equal to one half the strain tensor acting on the initial separation, after choosing the initial conditions that remove unrelated free drift. An arbitrary integration constant is not a new wave polarization.

R0i0j=−12c2h¨ijTT,ξ¨i=−c2Ri0j0ξj≃12h¨ijTTξ0j,δξi=12hijTTξ0jR_{0i0j}=-\frac{1}{2c^2}\ddot h^{\mathrm{TT}}_{ij},\qquad \ddot\xi^i=-c^2R^i{}_{0j0}\xi^j\simeq\frac12\ddot h^{\mathrm{TT}}_{ij}\xi_0^j,\qquad \delta\xi^i=\frac12h^{\mathrm{TT}}_{ij}\xi_0^j
R has inverse-length-squared units, ξ is separation in metres, dots denote seconds, and the acceleration is in m/s². The displacement uses the oscillatory particular solution with unrelated free motion removed. Weak amplitude, free test masses and a small detector region are assumed.↔ Scroll wide equations horizontally.

An interferometer contracts strain with arm directions

For an arm along a unit vector u, the fractional length perturbation in the long-wavelength limit is one half of u contracted twice with the TT tensor. Orthogonal arms along x and y respond oppositely to a plus wave arriving perpendicular to their plane. Their differential length change is therefore twice the single-arm change, while division by the common arm length gives the usual optimal differential strain. Missing this distinction creates a common factor-of-two error.

The actual optical instrument compares light phases accumulated along round trips and through its optical configuration. Treating that output as a calibrated strain invokes a response model. Suspensions, control loops and finite light travel time matter outside the ideal free-mass long-wavelength regime. The tensor contraction provides the geometric leading response, while a frequency-dependent transfer function handles the full instrument rather than redefining the incoming wave amplitude.

δLuL=12uiujhijTT,hdet=δLu−δLvL=DijhijTT,Dij=12(uiuj−vivj)\frac{\delta L_u}{L}=\frac12u^iu^jh^{\mathrm{TT}}_{ij},\qquad h_{\mathrm{det}}=\frac{\delta L_u-\delta L_v}{L}=D^{ij}h^{\mathrm{TT}}_{ij},\qquad D^{ij}=\frac12(u^iu^j-v^iv^j)
u and v are unit arm directions with equal unperturbed length L. hdet is a dimensionless differential response in the ideal long-wavelength, freely falling limit; it is not automatically either polarization component for arbitrary sky direction or orientation.↔ Scroll wide equations horizontally.

Coordinate descriptions and propagation limits must agree

In TT coordinates initially resting freely falling masses remain at fixed spatial coordinates, while the spatial metric changes their proper separation. In a local inertial description the leading tidal displacement appears as motion. Neither picture should be used halfway through the calculation with the other's light-propagation rules. The final observable is a comparison performed by a physical clock and light path, which is why changing gauge cannot remove a genuine measured tidal effect.

A laboratory-sized detector can often use a locally planar incoming wave even when the source is cosmological. Propagation then determines the observer-frame frequency and amplitude before the local response is applied. If the arm length is no longer small compared with wavelength, the wave phase varies along the light path and the simple instantaneous length formula fails. This limitation is especially relevant for long space-based baselines; a correct response includes time delays rather than assuming a uniform strain across every arm.

Single-arm motion and differential response

A plus wave arrives along +z with h+=A cos(2πft), h×=0, A=10⁻²¹ and f=100 Hz. Ideal orthogonal x/y arms have L=4.0 km and freely falling endpoints. Find the single-arm length amplitude, differential length amplitude, and x-arm tidal-acceleration amplitude. Use π²≈9.870.

  1. One arm has displacement amplitude AL/2=(10⁻²¹)(4000)/2=2.0×10⁻¹⁸ m. The other arm has the same magnitude with opposite sign under this polarization and orientation.

  2. Subtracting gives differential amplitude AL=4.0×10⁻¹⁸ m. Dividing by L yields detector strain amplitude 10⁻²¹, equal to h+ only because this example uses the optimal perpendicular orientation.

  3. Twice differentiating the sinusoid gives single-arm acceleration amplitude (AL/2)(2πf)2≈7.90×10−13 m/s2(AL/2)(2\pi f)^2\approx7.90\times10^{-13}\,\mathrm{m/s^2}. The wavelength is about 3000 km, comfortably larger than the 4 km arm.

The result

The ideal single-arm and differential amplitudes are 2.0×10⁻¹⁸ m and 4.0×10⁻¹⁸ m; the single-arm acceleration amplitude is 7.90×10⁻¹³ m/s². These are leading free-mass response quantities, not literal TT-coordinate mirror motion.

Explore the model

Interactive lab

Interactive lab · vectors

Arm orientation enters quadratically

For a pure plus tensor, the arm contraction is proportional to ux²−uy². Reset and set both controls to Angle 0° and Magnitude 1; read (ux,uy)=(1,0), giving contraction factor 1. For a fresh comparison, Reset and set both controls to Angle 45° and Magnitude 1; read equal components about 0.707, giving factor 0.

Arm orientation enters quadraticallyPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forUnit-vector projections illustrate directional sensitivity. The vector lab neither moves freely falling mirrors nor propagates light, and it has no spacetime strain, curvature or interferometer transfer function.

Check your understanding

For an optimal plus wave and equal orthogonal arms, how does differential length amplitude compare with one arm's amplitude?

Try it yourself

An idealized instrument has arm length 500 km and receives a 1000 Hz wave. Using c=3.0×10⁸ m/s, calculate L/λ and the one-way wave-phase change 2πfL/c. Is the instantaneous uniform-arm approximation justified?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute λ=c/f before comparing the two length scales. The approximation requires a small phase change across an arm.

Compare with the solution

The wavelength is 300 km, so L/λ=5/3≈1.67. The phase change is 2π(5/3)=10π/3≈10.47 rad. Neither quantity is small, so the simple uniform-strain arm formula is not justified. A time-delayed optical response is required. A small wave amplitude does not rescue this approximation: amplitude smallness and the wavelength hierarchy are distinct conditions.

Take it with you

  • Geodesic deviation relates measurable relative acceleration to tidal curvature.
  • Single-arm displacement and differential strain have different normalization factors.
  • Weak amplitude does not guarantee a long-wavelength detector response or eliminate optical transfer functions.

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