# 5.7 Doppler effect for light  (Page 2/2)

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${f}_{\text{obs}}={f}_{s}\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}.$

Notice that the signs are different from those of the wavelength equation.

## Calculating a doppler shift

Suppose a galaxy is moving away from Earth at a speed 0.825 c . It emits radio waves with a wavelength of
0.525 m. What wavelength would we detect on Earth?

## Strategy

Because the galaxy is moving at a relativistic speed, we must determine the Doppler shift of the radio waves using the relativistic Doppler shift instead of the classical Doppler shift.

## Solution

1. Identify the knowns: $u=0.825c;{\lambda }_{s}=0.525\phantom{\rule{0.2em}{0ex}}\text{m.}$
2. Identify the unknown: ${\lambda }_{\text{obs}}.$
3. Express the answer as an equation:
${\lambda }_{\text{obs}}={\lambda }_{s}\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}.$
4. Do the calculation:
$\begin{array}{}\\ \\ \\ \\ \\ \hfill {\lambda }_{\text{obs}}& ={\lambda }_{s}\sqrt[]{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}\hfill \\ & =\left(0.525\phantom{\rule{0.2em}{0ex}}\text{m)}\sqrt{\frac{1+\frac{0.825c}{c}}{1-\frac{0.825c}{c}}}\hfill \\ & =1.70\phantom{\rule{0.2em}{0ex}}\text{m.}\hfill \end{array}$

## Significance

Because the galaxy is moving away from Earth, we expect the wavelengths of radiation it emits to be redshifted. The wavelength we calculated is 1.70 m, which is redshifted from the original wavelength of 0.525 m. You will see in Particle Physics and Cosmology that detecting redshifted radiation led to present-day understanding of the origin and evolution of the universe.

Check Your Understanding Suppose a space probe moves away from Earth at a speed 0.350 c . It sends a radio-wave message back to Earth at a frequency of 1.50 GHz. At what frequency is the message received on Earth?

We can substitute the data directly into the equation for relativistic Doppler frequency:
${f}_{\text{obs}}={f}_{s}\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}=\left(1.50\phantom{\rule{0.2em}{0ex}}\text{GHz)}\sqrt{\frac{1-\frac{0.350c}{c}}{1+\frac{0.350c}{c}}}=1.04\phantom{\rule{0.2em}{0ex}}\text{GHz.}$

The relativistic Doppler effect has applications ranging from Doppler radar storm monitoring to providing information on the motion and distance of stars. We describe some of these applications in the exercises.

## Summary

• An observer of electromagnetic radiation sees relativistic Doppler effects if the source of the radiation is moving relative to the observer. The wavelength of the radiation is longer (called a red shift) than that emitted by the source when the source moves away from the observer and shorter (called a blue shift) when the source moves toward the observer. The shifted wavelength is described by the equation:
${\lambda }_{\text{obs}}={\lambda }_{s}\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}.$

where ${\lambda }_{\text{obs}}$ is the observed wavelength, ${\lambda }_{s}$ is the source wavelength, and v is the relative velocity of the source to the observer.

## Conceptual questions

Explain the meaning of the terms “red shift” and “blue shift” as they relate to the relativistic Doppler effect.

What happens to the relativistic Doppler effect when relative velocity is zero? Is this the expected result?

There is no measured change in wavelength or frequency in this case. The relativistic Doppler effect depends only on the relative velocity of the source and the observer, not any speed relative to a medium for the light waves.

Is the relativistic Doppler effect consistent with the classical Doppler effect in the respect that ${\lambda }_{\text{obs}}$ is larger for motion away?

All galaxies farther away than about $50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{ly}$ exhibit a red shift in their emitted light that is proportional to distance, with those farther and farther away having progressively greater red shifts. What does this imply, assuming that the only source of red shift is relative motion?

It shows that the stars are getting more distant from Earth, that the universe is expanding, and doing so at an accelerating rate, with greater velocity for more distant stars.]

## Problems

A highway patrol officer uses a device that measures the speed of vehicles by bouncing radar off them and measuring the Doppler shift. The outgoing radar has a frequency of 100 GHz and the returning echo has a frequency 15.0 kHz higher. What is the velocity of the vehicle? Note that there are two Doppler shifts in echoes. Be certain not to round off until the end of the problem, because the effect is small.

can someone explain normalization condition
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Gitanjali
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Omsai
particle nature
Omsai
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Anil
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Neptune
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Amalesh
excuse me yash what negative
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coz, electron contained minus ion
Manish
negative sign rika shrestha ji
YASH
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yes yash ji
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yes prema
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yes
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how can you confirm?
Amalesh
sry
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If they are same then why they named differently?
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because beta particles give the information that the electron is ejected from the nucleus with very high energy
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atomic n.o
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no of atoms present in nucleus
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miss sanjana your question is wrong...question should be no of nucleus present in an atom...and answer is 1...
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Owaise is right
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