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\line{\hfill Doppler Shift Reference}
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\centerline{The Doppler Shift Formula is:}
$$\Delta \lambda / \lambda = v / c $$

$c = $ the speed of light = $3.0 \times 10^8$ m/s or 300,000 km/s.

$v = $ the radial velocity of the object (negative if toward the observer).

$\lambda = $ The rest wavelength of the spectral line at which you are looking.

$\Delta \lambda = \lambda_{\rm obs} - \lambda_{\rm rest}$, where

$\Delta \lambda$ is the (usually small) change in wavelength of the
spectral feature due to motion.

($\lambda_{\rm obs} = $ the observed (shifted) wavelength, and
$\lambda_{\rm rest} = $ the zero-velocity  laboratory wavelength.

1 nm = $10^{-9}$ m = 10 \AA .

1 \AA\ = $10^{-10}$ m.

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Example: If a star is coming toward you at 5 km/s, what is the
Doppler shift for a line in the visible (500 nm)?
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Solution: First, since the object is coming toward us, there will be a
blueshift, and we need a negative sign for the velocity: $v=-5$
km/s. Second, we plug into the formula.
$$ \Delta \lambda / (500\ {\rm nm}) = (-5 \ {\rm km/s}) / (300000
\ {\rm km/s})$$
The answer cranks out as $\Delta \lambda = -0.008$ nm. ($= -0.08$\AA .)

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Example: Suppose you observe a star through a spectroscope and see a
line due to iron at 527.05 nm. This line is normally (in the lab) seen
at 527.00 nm. How fast is this star receeding?
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Solution: First, $\Delta \lambda = 0.05 $ nm. Right? Right. So we plug
into the formula:
$$ (+0.05\ {\rm nm}) / (527\ {\rm nm}) = (v \ {\rm in\ km/s}) / (3.0 \times 10^5
\ {\rm km/s})$$
And we find $v=28.5$ km/s.

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Problem: A galaxy's H$\beta$ line (normally at 486 nm) is observed to
have shifted 1.62 nm redder. How fast is this galaxy receeding from
us?


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Problem: An expanding spherical supernova blast wave is expanding at
2000 km/s away from the explosion center. Since the shell is
transparent one can see both the side coming toward and the side
going away. What are the two wavelengths between which the Oxygen line
($\lambda$5007 \AA ) appears?

Answer: $\Delta \lambda = $ 33.4\AA\ from the formula, so the two
bracketing wavelengths are plus and minus this amount from 5007\AA :
4974\AA\ and 5040\AA .


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