# Homework Help: Doppler effect spectrum of a star

1. Jul 31, 2007

### mpswee2

1. The problem statement, all variables and given/known data

An astronomer observes a hydrogen line in the spectrum of a star. The wavelength of hydrogen in the laboratory is 6.563 x 10-7m, but the wavelength in the star’s light is measured at 6.56186 x 10-7m. Which of the following explains this discrepancy?

A) The star is moving away from Earth.

B) The wavelength of light that the star is emitting changes constantly.

C) The frequency of light that the star is emitting changes constantly.

D) The star is approaching Earth.

I chose A, the star is moving away from the observer. D is correct. My reasoning: Since the photon's wavelength observed in the laboratory (lambda') is greater than the actual wavelength (lambda) issued from the star, I thought this meant the star was moving AWAY. The wavelength is "stretched out" as the star recedes from the observer, and the observed frequency is smaller.

Am I switching around the the Doppler effect on wavelength and frequency?
Could someone clarify the whole concept here, which I must have misunderstood.
(The effect above is called a blue shift-- but I'm not sure why.)

2) This question makes me wonder if, for photons with wavelength in the visible range, can the Doppler effect have an effect on the color perceived?? In other words, if an object emitting yellow light were receding from the observer, might the light actually be perceived as green or blue (ie shorter wavelength)?

Thanks a lot.

Last edited: Jul 31, 2007
2. Jul 31, 2007

### meopemuk

The correct answed is D. Think about it this way: the wavelength emitted by hydrogen in the reference frame of the star is 6.563 x 10-7m, i.e., the same as for hydrogen in the laboratory frame. However, the light arrives to Earth with a lower wavelength. This means that the frequency of the arriving light is higher than normal (frequency is inversely proportional to the wavelength). Increased frequency means that the source is moving toward us. Remember ambulance siren? It has a higher pitch (a higher frequency) when the ambulance moves toward us.

3. Aug 1, 2007

### Matthaeus_

The star is approaching the earth with a speed $$v \approx 52\ \mathrm{km/s}$$, which you can calculate by solving:

$$\displaystyle \lambda^{\prime} = \lambda \left( \frac{1+\beta}{1-\beta}\right)^{1/2} \quad \text{ where } \quad \beta = \frac{v}{c}$$

You probably overlooked that $$6.561$$ is smaller than $$6.563$$.
Anyway, if the wavelength is shorter than it should, it means that the line is more on the blue side of the visible spectrum, because blue light has a shorter wavelength than red light, and that's why we call this phenomenon blueshift.

Yes, of course. But it would be quite the contrary: if an object were receding from you, you would see it redshifted. A yellow object would become orange or red.

4. Aug 1, 2007

### mpswee2

Doppler in depth...

thanks, folks. can we think of the approaching emitter as (sort of) "hurrying" the photons along-- as if pushing-- which compresses the wavelength, increases the frequency, and thus adds energy (E=hf)? Which is the reason we call the effect a blueshift, ie increasing photon E?

In this case, is it true that photons from an approaching emitter arrive with more energy (presumably the emitter's kinetic energy) than photons from a stationary emitter, which arrive with more energy than a receding emitter??

The trouble with this conclusion is that the speed, v, of higher energy particles (blue) would be greater than the speed of lower energy particles (red) (E=1/2mv^2); yet the speed of light in a vacuum, c, is constant, independent of the photon's energy. How can we say red (less energy) and UV light (higher energy) travel at the same speed, c, assuming both photons have the same mass? Shouldn't UV light travel faster by virtue of its greater energy (E= hf= 1/2mv^2)??

Do the two equations for energy-- E=hf and KE=1/2mv^2-- refer to different forms of energy? Or are applied under different circumstances?

A little jumbled. Thanks for your all's help.

Last edited: Aug 1, 2007
5. Aug 1, 2007

### Matthaeus_

Gee! You're using the wrong equations for energy! A photon carries a momentum
$$p = h/\lambda$$.

According to special relativity:

$$\displaystyle E^2 = p^2 c^2 + m_0^2 c^4$$. But photons have no mass!! Therefore, no $$\frac 1 2 mv^2$$!

6. Aug 1, 2007

### mpswee2

Ah! I see. Thanks.

But what about the interpretation of Doppler...

In this case, is it true that photons from an approaching emitter arrive with more energy (presumably the emitter's kinetic energy) than photons from a stationary emitter, which arrive with more energy than a receding emitter??

7. Aug 1, 2007

### Matthaeus_

Sure. That immediately follows from $$~E = hf$$.