Notes

Jonathan's Convolved Data:

  • The convolution output data has a wavelength and a calibration shift.
  • Since several wavelengths are needed to perform the convolution, the output wavelength is the average.
  • A line is fit to the output data in order to find the calibration shift for the Keck data.

For Professor Griest
Derivation:
Before following along with the derivation, understand the meaning of these variables.

  • Measurements such as $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \lambda_{m}$ and $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega_{m}$ refer to the directly observed values from the Keck.
  • There are also the values of constants here on earth such as $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \alpha_0$ and $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \lambda_0$.
  • Another constant is the independently calculated $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} q$ value for each individual ion transition.
  • The value of $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \tilde{z}$ is calculated in the case that $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \alpha$ does not change.
  • Lastly there are the true values of $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} z$ (the redshift allowing for a variable $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \alpha$), $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \lambda_z$, $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega_z$, and $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \alpha_z$ (the values of in the QSO rest frame).

The variation in Energy levels due to a variable $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \alpha$

(1)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} E_z & = & E_0 + Q_1 [(\frac{\alpha_z}{\alpha_0})^2 - 1] + K_1(LS)(\frac{\alpha_z}{\alpha_0})^2 \\ & & + Q_2 [(\frac{\alpha_z}{\alpha_0})^4 - 1] + K_2(LS)^2(\frac{\alpha_z}{\alpha_0})^4 \end{eqnarray}

Energy is proportional to wavenumber ($\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega$)

(2)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} E & = & h\nu \\ & = & \frac{hc}{\lambda} \\ & = & hc\omega \\ \end{eqnarray}

The dependence of $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega$ on $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \alpha$

(3)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega_z & = & \omega_0 + \frac{Q_1}{hc} [(\frac{\alpha_z}{\alpha_0})^2 - 1] + \frac{K_1}{hc}(LS)(\frac{\alpha_z}{\alpha_0})^2 \\ & & + \frac{Q_2}{hc} [(\frac{\alpha_z}{\alpha_0})^4 - 1] + \frac{K_2}{hc}(LS)(\frac{\alpha_z}{\alpha_0})^4 \\ & = & \omega_0 +q_1[(\frac{\alpha_z}{\alpha_0})^2 - 1] + q_2 [(\frac{\alpha_z}{\alpha_0})^4 - 1] \\ \mbox{ where } & : & \omega_0 = \frac{1}{hc}[E_0 + K_1(LS) + K_2(LS)^2] \\ & & q_1 = \frac{1}{hc}[Q_1 + K_1(LS)] \\ & & q_2 = \frac{1}{hc}[Q_2 + K_2(LS)^2] \\ \end{eqnarray}

Dropping the $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} q_2$ term

(4)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \mbox{ since } & : & q_1 \gg q_2 \mbox{ , } q_1 \rightarrow q \\ \omega_z & = & \omega_0 + q [(\frac{\alpha_z}{\alpha_0})^2 - 1] \end{eqnarray}

Taylor expanding $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega_z$

(5)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega_z & = & \omega_z |_{\alpha=\alpha_0} + \frac{1}{1!} \frac{d\omega_z}{d\alpha} (\alpha_z-\alpha) |_{\alpha=\alpha_0} + \frac{1}{2!} \frac{d^2\omega_z}{d\alpha^2} (\alpha_z-\alpha)^2 |_{\alpha=\alpha_0} + \ldots \\ & = & \omega_z |_{\alpha=\alpha_0} + q \frac{2\alpha}{\alpha_0^2} (\alpha_z-\alpha) |_{\alpha=\alpha_0} + q \frac{1}{\alpha_0^2} (\alpha_z-\alpha)^2 |_{\alpha=\alpha_0} + 0 \\ & = & \omega_0 + q [(\frac{\alpha_0}{\alpha_0})^2 - 1] + q \frac{2\alpha_0}{\alpha_0^2} (\alpha_z-\alpha_0) + q \frac{1}{\alpha_0^2} (\alpha_z-\alpha_0)^2\\ & = & \omega_0 + 2q\frac{\alpha_z-\alpha_0}{\alpha_0} + q (\frac{\alpha_z-\alpha_0}{\alpha_0})^2 \\ & = & \omega_0 + 2q\frac{\Delta\alpha}{\alpha} + q (\frac{\Delta\alpha}{\alpha})^2 \\ \mbox{ where } & : & |\frac{\Delta\alpha}{\alpha_0}| = |\frac{\alpha_z-\alpha_0}{\alpha_0}| \ll 1 \end{eqnarray}

Therefore the value of $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega$ in the QSO rest frame is

(6)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega_z & = & \omega_0 + 2q\frac{\Delta\alpha}{\alpha_0} \\ \Delta\omega & = & 2q\frac{\Delta\alpha}{\alpha_0} \end{eqnarray}

Defining the true redshift ($\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} z$)

(7)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} z & = & \frac{\lambda_{observed}}{\lambda_{emitted}} - 1 \\ & = & \frac{\lambda_{m}}{\lambda_{z}} - 1 \\ & = & \frac{\omega_{z}}{\omega_{m}} - 1 \end{eqnarray}

Defining the apparent redshift ($\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \tilde{z}$)

(8)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \tilde{z} & = & \frac{\lambda_{m}}{\lambda_0} - 1 \\ & = & \frac{\omega_0}{\omega_{m}} - 1 \end{eqnarray}

Solving in terms of $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega$

(9)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \omega_z & = & \omega_{m}(1 + z) \\ \omega_0 & = & \omega_{m}(1 + \tilde{z}) \\ \omega_z - \omega_0 & = & \omega_{m}[(1+z)-(1+\tilde{z})] \\ & = & \omega_{m}( z - \tilde{z}) \\ & = & \frac{\omega_z}{1 + z} (z - \tilde{z}) \\ {\Delta\omega} & = & \frac{-\Delta z}{\lambda_z(1+ z)} \\ \mbox{where} & : & \Delta\omega = \omega_z - \omega_0 \\ & & \Delta z = \tilde{z} - z \end{eqnarray}

From Eq (6) & (9) the change in redshift is

(10)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \frac{-\Delta z}{\lambda_z(1+ z)} & = & 2q\frac{\Delta\alpha}{\alpha_0} \\ \Delta z & = & -2q\frac{\Delta\alpha}{\alpha_0}\lambda_z(1+ z) \\ \end{eqnarray}

Variable $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \alpha$ leads to a variable $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} v$

(11)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} c & = & \lambda_0 \nu_0 \\ v_z & = & \lambda_z \nu_0 \\ \frac{\Delta v}{c} & = & \frac{v_z - c}{c} \\ & = & \frac{\lambda_z \nu_0 - \lambda_0 \nu_0}{\lambda_0 \nu_0} \\ & = & \frac{\lambda_z - \lambda_0}{\lambda_0} \\ \end{eqnarray}

Evaluating $\definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \Delta\omega$ in terms of velocity

(12)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} \Delta\omega & = & \frac{1}{\lambda_z} - \frac{1}{\lambda_0} \\ & = & \frac{\lambda_0 - \lambda_z}{\lambda_z \lambda_0} \\ & = & -\frac{\lambda_z - \lambda_0}{\lambda_0} \frac{1}{\lambda_z} \\ & = & -\frac{\Delta v}{c} \frac{1}{\lambda_z} \\ \end{eqnarray}

From Eq (6) & (12) the change is velocity is

(13)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} -\frac{\Delta v}{c} \frac{1}{\lambda_z} & = & 2q \frac{\Delta\alpha}{\alpha_0} \\ \Delta v & = & -2q \lambda_z c \frac{\Delta\alpha}{\alpha_0} \\ & = & -2q \lambda_0 c \frac{\Delta\alpha}{\alpha_0} \\ \mbox{where} & : & \lambda_z \approx \lambda_0 \end{eqnarray}

Finding the velocities for FeII 1608 and 1611

(14)
\begin{eqnarray} \definecolor{blue}{rgb}{.525,.631,.922}\pagecolor{blue} v & = & -2q\lambda_0 c \frac{\Delta\alpha}{\alpha} \\ v (1608\AA) & = & -2[-1.03 \times10^5 m^{-1}][1608\times10^{-10} m][3.00\times10^8 m/s][(-5.43 \pm 1.16)\times10^{-6}] \\ & = & (-54.0 \pm 11.5) m/s \\ v (1611\AA) & = & -2[1.56\times10^5 m^{-1}][1611\times10^{-10} m][3.00\times10^8 m/s][(-5.43 \pm 1.16)\times10^{-6}] \\ & = & (+81.9 \pm 17.5) m/s \\ \Delta v & = & (+135.9 \pm 20.9) m/s \end{eqnarray}
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