2 Methods

To find $\avg{x^2}$, the raw photodiode voltages $V_p(t)$ are converted to distances $x(t)$ using the photodiode sensitivity $\sigma_p$ (the slope of the voltage vs. distance curve of data taken while the tip is in contact with the surface) via

$$x(t) = \frac{V_p(t)}{\sigma_p}$$ (5)

Rather than computing the variance of $x(t)$ directly, we attempt to filter out noise by fitting the spectral power density ( $\operatorname{PSD}$) of $x(t)$ to the theoretically predicted $\operatorname{PSD}$ for a damped harmonic oscillator (eq. 53)

$$\ddt{x} + \beta\dt{x} + \omega_0^2 x = \frac{F_\text{thermal}}{m}$$ (6)

$$\ensuremath{\operatorname{PSD}}(x, \omega) = \frac{G_1}{(\omega_0^2-\omega^2)^2 + \beta^2\omega^2},$$ (7)

where $G_1\equiv G_0/m^2$, $\omega_0$, and $\beta$ are used as the fitting parameters (see eqn.s 53). The variance of $x(t)$ is then given by eq. 59

$$\avg{x(t)^2} = \frac{\pi G_1}{2\beta\omega_0^2},$$ (8)

which we can plug into the equipartition theorem (eqn. 1) yielding

$$k = \frac{2 \beta \omega_0^2 k_BT}{\pi G_1}.$$ (9)

From eqn. 61, we find the expected value of $G_1$ to be

$$G_1 \equiv G_0/m^2 = \frac{2}{\pi m} k_BT \beta.$$ (10)

2.1 Fitting deflection voltage directly

In order to keep our errors in measuring $\sigma_p$ seperate from other errors in measuring $\avg{x(t)^2}$, we can fit the voltage spectrum before converting to distance.

$$\ddt{V_p}/\sigma_p + \beta\dt{V_p}/\sigma_p + \omega_0^2 V_p/\sigma_p = F_$$ (11)

$$\ddt{V_p} + \beta\dt{V_p} + \omega_0^2 V_p = \sigma_p\frac{F_\text{thermal}}{m}$$ (12)

$$\ddt{V_p} + \beta\dt{V_p} + \omega_0^2 V_p = \frac{F_\text{thermal}}{m_p}$$ (13)

$$\ensuremath{\operatorname{PSD}}(V_p, \omega) = \frac{G_{1p}} { (\omega_0^2-\omega^2)^2 + \beta^2\omega^2 }$$ (14)

$$\avg{V_p(t)^2} = \frac{\pi G_{1p}}{2\beta\omega_0^2} = \frac{\pi \sigma_p^2 G_{1}}{2\beta\omega_0^2} = \sigma_p^2 \avg{x(t)^2},$$ (15)

where $m_p\equiv m/\sigma_p$, $G_{1p}\equiv G_0/m_p^2=\sigma_p^2 G_1$. Plugging into the equipartition theorem yeilds

$$k = \frac{\sigma_p^2 k_BT}{\avg{V_p(t)^2}} = \frac{2 \beta\omega_0^2 \sigma_p^2 k_BT}{\pi G_{1p}}.$$ (16)

From eqn. 10, we find the expected value of $G_{1p}$ to be

$$G_{1p} \equiv \sigma_p^2 G_1 = \frac{2}{\pi m} \sigma_p^2 k_BT \beta.$$ (17)

2.2 Fitting deflection voltage in frequency space

Note: the math in this section depends on some definitions from section 3.

As yet another alternative, you could fit in frequency $f\equiv\omega/2\pi$ instead of angular frequency $\omega$. But we must be careful with normalization. Comparing the angular frequency and normal frequency unitary Fourier transforms

$$\ensuremath{{\mathcal F}\left\{ {x(t)} \right\}}(\omega) \equiv \frac{1}{\sqrt{2\pi}} \ensuremath{\int_{-\infty}^{\infty} {x(t) e^{-i \omega t}} \dd{t}}$$ (18)

$$\ensuremath{{\mathcal F}_f\left\{ {x(t)} \right\}}(f) \equiv \ensuremath{\int_{-\infty}^{\infty} {x(t) e^{-2\pi i f t}}... ...sqrt{2\pi}\cdot\ensuremath{{\mathcal F}\left\{ {x(t)} \right\}}(\omega=2\pi f),$$ (19)

from which we can translate the $\operatorname{PSD}$

$$\ensuremath{\operatorname{PSD}}(x, \omega) \equiv \ensuremath{\lim_{{t_T}\rightarrow \infty}}\frac{1}{t_T} ... ...\vert{ \ensuremath{{\mathcal F}\left\{ {x(t)} \right\}}(\omega) }\right\vert^2}$$ (20)

$$\ensuremath{\operatorname{PSD}}_f(x, f) \equiv \ensuremath{\lim_{{t_T}\rightarrow \infty}}\frac{1}{t_T} ... ...pi f) }\right\vert^2} = 2\pi \ensuremath{\operatorname{PSD}}(x, \omega=2\pi f).$$ (21)

The variance of the function $x(t)$ is then given by plugging into eqn. 34 (our corollary to Parseval's theorem)

$$\avg{x(t)^2} = \ensuremath{\int_{0}^{\infty} {\ensuremath{\operatorname{PSD}}(... ...\ensuremath{\int_{0}^{\infty} {\ensuremath{\operatorname{PSD}}_f(x,f)} \dd{f}}.$$ (22)

Therefore

$$\ensuremath{\operatorname{PSD}}_f(V_p, f) = 2\pi\ensuremath{\operatorname{PSD}}(V_p,\omega) = \frac{2\pi G_... ...2 4\pi^2f^2} = \frac{G_{1p}/8\pi^3}{(f_0^2-f^2)^2 + \frac{\beta^2 f^2}{4\pi^2}}$$ (23)

$$= \frac{G_{1f}}{(f_0^2-f^2)^2 + \beta_f^2 f^2}$$ (24)

$$\avg{V_p(t)^2} = \frac{\pi G_{1f}}{2\beta_f f_0^2}.$$ (25)

where $f_0\equiv\omega_0/2\pi$, $\beta_f\equiv\beta/2\pi$, and $G_{1f}\equiv G_{1p}/8\pi^3$. Finally

$$k = \frac{\sigma_p^2 k_BT}{\avg{V_p(t)^2}} = \frac{2 \beta_f f_0^2 \sigma_p^2 k_BT}{\pi G_{1f}}.$$ (26)

From eqn. 10, we expect $G_{1f}$ to be

$$G_{1f} = \frac{G_{1p}}{8\pi^3} = \frac{\sigma_p^2 G_1}{8\pi^3} = ... ...\pi m} \sigma_p^2 k_BT \beta}{8\pi^3} = \frac{\sigma_p^2 k_BT \beta}{4\pi^4 m}.$$ (27)

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2 Methods
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Drexel Physics