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Noise is a measurement

Science · Thermal & Statistical Physics · ages 22-24

Name ______________________ Date ____________

A resistor is in thermal equilibrium with no battery attached. What produces its Johnson-Nyquist voltage noise?
- A steady current supplied by a hidden battery
- Thermal motion of its charge carriers
- A changing resistance caused only by broken contacts
At fixed volume and particle number in the canonical ensemble, which expression gives energy variance?
- k_B T^2 C_V
- k_B T^2 / C_V
- k_B C_V / T^2
For an ordinary extensive short-range system away from critical regimes, root-mean-square energy fluctuations relative to a nonzero extensive mean usually shrink as N grows.
Circle one: True False
Two canonical samples at the same positive T have fixed V,N and different heat capacities. Which has larger root-mean-square energy fluctuations?
- The sample with the smaller heat capacity
- Neither; both have equal fluctuations
- The sample with the larger heat capacity
An ideal resistor at known positive T has constant R in the classical regime. An ideal nonloading measurement has unit gain within a band of known width delta f and zero gain outside. From its measured mean-square voltage noise, what can be inferred?
- R = mean-square voltage / (4 k_B T delta f)
- R = mean-square voltage times (4 k_B T delta f)
- R = mean-square voltage / (4 k_B T), regardless of bandwidth
What does the fluctuation-dissipation relation connect near thermal equilibrium in the linear-response regime?
- Equilibrium fluctuation spectra and dissipative response
- Equilibrium fluctuation spectra and only nondissipative response
- Arbitrary strong-drive response and a temperature-free noise value
Near a critical point, energy correlations extend across much of a sample. Can particle count alone justify relative energy fluctuations proportional to 1/sqrt(N)?
- Yes; every large sample has independent energy contributions
- Yes; critical correlations force energy variance to vanish
- No; the regular scaling assumptions need further checking
Rafi says, 'Energy variance and mean both grow as N, so relative root-mean-square fluctuations stay constant.' In the regular extensive regime, which calculation repairs this?
- N/N = 1, because variance is fluctuation size
- sqrt(N)/N = 1/sqrt(N), because fluctuation size is sqrt(variance)
- N/sqrt(N) = sqrt(N), because mean energy grows as sqrt(N)

Answer key

For grown-ups. Fold this page away before handing over the rest.

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Thermal motion of its charge carriers · Thermal charge motion produces equilibrium voltage fluctuations without a driven current.
k_B T^2 C_V · The canonical identity is Var(E)=k_B T^2 C_V. Since k_B and C_V each have units energy/kelvin, multiplying by T^2 gives the energy-squared units required for variance.
True · The fluctuation size grows as sqrt(N), while the mean grows as N. Their ratio therefore falls as 1/sqrt(N).
The sample with the larger heat capacity · At the same T, variance increases with C_V. Its square root also increases with C_V.
R = mean-square voltage / (4 k_B T delta f) · The ideal nonloading measurement gives mean-square voltage = 4 k_B T R delta f. Unit gain inside the band and zero gain outside make delta f its noise bandwidth. Divide by 4 k_B T delta f to isolate R.
Equilibrium fluctuation spectra and dissipative response · It links equilibrium fluctuation correlations or spectra to dissipative linear response, with temperature and frequency factors.
No; the regular scaling assumptions need further checking · Long correlations can change fluctuation scaling. The usual regular-regime argument cannot be assumed from size alone.
sqrt(N)/N = 1/sqrt(N), because fluctuation size is sqrt(variance) · Root-mean-square fluctuation is the square root of variance. Its ratio to the mean falls as 1/sqrt(N).