Doppler Effect - Astronomy: The Human Quest for Understanding

$$\frac{v}{c} = \frac{\Delta\lambda}{\lambda_\text{emit}} = \frac{\lambda_\text{obs}-\lambda_\text{emit}}{\lambda_\text{emit}}\quad[\text{Eq. (8.2)}]$$

Note: If you feel uncomfortable with working with variables, fractions, or percentages, you should review the following tutorials before working through this tutorial:

Introduction

The Doppler effect equation provides a straightforward and convenient way to measure the velocity of an object coming toward or away from the observer, but it does not provide any information about the object’s motion perpendicular to the line-of-sight to that object. (Astronomers refer to the line-of-sight velocity as “radial velocity” and the perpendicular motion as “proper motion.”)

You are certainly familiar with the related Doppler effect for sound: when a car is coming toward you that is producing a sound (such as a car horn), the sound has a higher pitch than when the object is moving away from you, and as the car passes you, you can hear the pitch slide down. When the car is coming toward you, the wavelength is compressed (shortened) relative to when the car is not moving, and when the car is moving away from you, the wavelength is stretched (lengthened). A shorter wavelength corresponds to a higher pitch and a longer wavelength corresponds to a lower pitch.

The Doppler effect for light is similar in that an approaching light source has its wavelength shortened, resulting in a higher frequency for the light, and for a light source that is moving away (receding), its wavelength is lengthened, resulting in a lower frequency. Light with a shorter wavelength is bluer than light with a longer wavelength. As a result, an approaching light source is referred to as having its light “blueshifted” and the light from a receding source is “redshifted.”

Equation(8.2) calculates the speed of the source relative to the observer ($v$) if the shift in wavelength is known ($\lambda_\text{obs} – \lambda_\text{emit}$), where $\lambda_\text{obs}$ is then observed wavelength and $\lambda_\text{emit}$ is the wavelength emitted by the source). Caveat: Equation (8.2) only applies if the relative motion of the source and the observer is much less than the speed of light ($c$). If the relative speed is greater than about 10% of the speed of light ($0.1c$), then Einstein’s special theory of relativity must be used, and the resulting equation gets more complicated; something we won’t worry about in this textbook.

If you look closely at Equation (8.2) you will see that there are two general possibilities; either $\Delta\lambda = \lambda_\text{obs} – \lambda_\text{emit}$ is greater than zero, meaning that $\lambda_\text{obs}$ is greater than $\lambda_\text{emit}$ ($\lambda_\text{obs} > \lambda_\text{emit}$) or $\lambda_\text{obs} – \lambda_\text{emit}$ is less than zero, meaning that $\lambda_\text{obs}$ is less than $\lambda_\text{emit}$ ($\lambda_\text{obs} < \lambda_\text{emit}$). In the first case the observed wavelength is stretched relative to the emitted wavelength ($v>0$), and the source is therefore moving away from the observer, and in the second case the opposite is true ($v<0$).

Examples

Astronomers studying a star find that one of the spectral lines of carbon has an observed wavelength of 578 nm, but the wavelength in a laboratory is determined to be 589 nm. Is the light redshifted or blueshifted? Is the star moving toward the telescope or away from it? How fast is the star moving toward or away from the telescope in terms of the speed of light?

Since the observed wavelength is shorter than the wavelength emitted in a laboratory (or by the star), the light is blueshifted and the star is moving toward the telescope. To calculate the star’s speed relative to the telescope, $$\Delta\lambda = 578\text{ nm} – 589\text{ nm} = -11\text{ nm},$$ and $$\frac{v}{c} = \frac{\Delta\lambda}{\lambda_\text{emit}} = \frac{-11\text{ nm}}{589\text{ nm}} = 0.0187.$$ This means that the star’s speed is 1.87% of the speed of light toward the telescope.
Helium has a spectral line with a wavelength of 444 nm as measured in a laboratory. The spectral line is detected in a star at a wavelength of 448 nm. What is the star’s speed relative to the telescope in units of km/s? Is the star moving toward or away from the telescope?

The change in wavelength of the spectral line is 4 nm, implying that the speed of the star relative to the speed of light is $$\frac{v}{c} = \frac{4\text{ nm}}{444\text{ nm}} = 0.00909.$$ Given that the speed of light is nearly $c = \text{300,000 km/s}$, the star’s speed relative to the telescope is $$v = 0.00909c = 0.00909\times \text{300,000 km/s} = 2700\text{ km/s}.$$ Since the difference between the observed wavelength and the emitted wavelength is positive, the wavelength is stretched out because the star is moving away from the telescope.
In an undergraduate research project, a student observes a star that has a hydrogen spectral line of 657 nm. Referring to Table 7.2 on page 252, what is the laboratory wavelength that is closest to the observed value? Is the star moving toward or away from the telescope, and at what speed?

The closest wavelength listed in Table 7.2 is 656 nm (this is hydrogen’s $\text{H}\alpha$ Balmer line). Since the observed wavelength is longer than the laboratory wavelength, the star is moving away from the telescope at a speed of $$\frac{v}{c} = \frac{1\text{ nm}}{656\text{ nm}} = 0.00152\qquad\text{or}\qquad0.152\%\text{ of the speed of light.}$$

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where $10$ raised to an integer (either positive or negative) replaces $\times 10^\text{exponent}$ with ${\rm E(exponent)}$ or ${\rm e(exponent)}$; in other words, $3.14159 \times 10^9$ is represented by $3.14159\text{E}9$. If your calculator has the $\Large{\hat{ }}$ symbol, then you would enter the number as $3.14159 \times 10{\Large{\hat{ }}}9$. If your calculator has a $10^x$ button then you would enter the number as $3.14159\times9\,\,10^x$. If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

(Answers are available below.)

Hydrogen has a very important radio-wavelength spectral line of 21.1 cm that is useful for measuring the motions of hydrogen clouds in galaxies. If a galaxy is observed to have a hydrogen line at a wavelength of 21.3 cm, what is the speed of the galaxy relative to the telescope? Is the galaxy approaching the telescope or is it moving away? Is the wavelength blueshifted or redshifted?
Sodium has a spectral line of 498 nm. That specific spectral line is observed in a star at a wavelength of 496 nm. Is the star approaching or receding from the telescope and at what speed? Is the wavelength blueshifted or redshifted?
Nitrogen has a spectral line with a laboratory wavelength of 505 nm. That line is seen in a star at a wavelength of 520 nm. Is the line redshifted or blueshifted? Is the star moving toward or away from the telescope, and at what speed.
The $\text{H}\gamma$ line of hydrogen has a laboratory wavelength of 434 nm, but it is observed in a star at a wavelength 438 nm. The star also displays a sodium line that has a laboratory wavelength of 515 nm. What would the observed wavelength of the sodium line be? (Note: This problem requires a bit of algebra in order to solve for $\lambda_\text{emit}$ once $v/c$ is known from the hydrogen data.)

Answers

$0.00948c$. The observed galaxy is moving away. The wavelength is redshifted.
$0.00402c$. The star is approaching and the wavelength is blueshifted.
The line is redshifted. The star is moving away from the telescope at a speed of $0.0297c$, or $8910\text{ km/s}$.
519.7 nm, or approximately 520 nm.