Help   |   Contact Us

 

Lecture 15: Doppler Effect, Sound, EM Radiation

OCW Scholar

« Previous | Next »

Lecture Topics

Someone holds a cleear balloon with black dots on it.
  • Doppler Effect for Sound and for EM Radiation
  • Binary Stars
  • Black Holes
  • Expansion of the Universe

Learning Objectives

By the end of this lecture, you should:

  • understand the Doppler effect in a medium used as a reference (sound).
  • understand the Doppler effect for relative motion (EM waves).
  • describe blue and red shifts of stellar light.
  • describe use of radial velocity in determining properties of binary star systems.
  • describe black holes and evidence for them in binary systems.
  • describe use of Doppler-like redshifts in determining the expansion of the universe.

Lecture Activities

Check Yourself

  • Most car horns actually have two horns of a slightly different frequency, which makes beats to make the sound more noticeable. We will ignore this fact and assume that you have a very close brush with a car going by at 50 km/hour, sounding a horn of frequency 500 Hz. What frequencies do you hear as the car approaches and recedes?

› View/hide answer

In this case the observer is stationary so vr=0 and we will take the speed of sound vs as 340 m/s. The speed of the car vt is 50 km/hour = 50000m/3600s = 13.9 m/s, but the sign will change as it approaches and recedes. So \[f' = f\left( {\frac{{{v_s} - {v_r}}}{{{v_s} - {v_t}}}} \right) = 500\left( {\frac{{340}}{{340 \pm 13.8}}} \right)\]and when the car is approaching, the frequency will sound like 521 Hz and when receding, 480 Hz. Since, in the demo, a much smaller speed variation produced a detectable frequency change, this change would be very apparent.

 

  • The Doppler formula for EM radiation was not derived but simply given (in terms of wavelength rather than frequency, since that is often what astronomers measure) as \[\lambda ' = \lambda \left( {\frac{{1 - \beta \cos \theta }}{{\sqrt {1 - {\beta ^2}} }}} \right)\] where \(\beta = \frac{v}{c}\) and θ is the angle between the object’s velocity and the vector from it to the observer. For motion perpendicular to the line of sight, normal intuition would not lead one to expect any effect on the wavelength. In this relativistic formula, there is in fact a change in wavelength for perpendicular motion. For the Ca absorption line mentioned in the lecture, but observed in the Sun, calculate the effect on the wavelength of Earth’s orbital speed of 30 km/s perpendicular to the Earth-Sun line.

› View/hide answer

Here θ=90º and \[\beta = \frac{{30 \times {{10}^3}}}{{3 \times {{10}^8}}} = {10^{ - 4}}\] Then \[\lambda ' = 3933.664\left( {\frac{1}{{\sqrt {1 - {{10}^{ - 8}}} }}} \right) = 3933.66402\] Even with the high precision that astronomers can attain, this relativistic effect would likely be too small to measure in this case.

 

  • Older police radars work at about 10.5 GHz, a very similar frequency to that used in some of the demonstrations. Radar works by shooting out a beam which makes electrons move in the target at the frequency they receive, and then detects that reflected beam with a Doppler shift again. Assume this just means you get two times the Doppler shift and figure out what frequency change comes from a car going 110 km/h. If you were a radar designer, how would you cheaply detect this change?

› View/hide answer

It is likely most straightforward to use the wavelength form supplied. Since \(f\lambda = c\) we convert to a wavelength of \[\lambda = \frac{c}{f} = \frac{{3 \times {{10}^8}}}{{10.5 \times {{10}^9}}} = {\text{0}}{\text{.028571428}},\] where we have kept a lot of decimal places since we know the effect will be small and we assume that the original frequency is known very accurately (see below). To easily catch a speeder you likely have to be in front of them, so θ=0 and its cosine is 1. In the previous question we found that the denominator had little effect even at orbital speeds, so for cars it can be ignored. At 110 km/h we have \[\beta = \frac{{110 \times 1000/3600}}{{3 \times {{10}^8}}} = {\text{1}}{\text{.01851852}} \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}.\] Keeping in mind the factor of 2 (the radar beam is shifted as viewed from the speeding car and the reflected beam is shifted again as viewed from the police car), we then get \[\lambda ' = \lambda \left( {1 - 2\beta } \right){\text{ = 0}}{\text{.028571428(1 - 2}}{\text{.03703704e - 7}}) = {\text{0}}{\text{.028571422}}.\] Converting back to frequency we get \[f = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}}}{{{\text{0}}{\text{.028571425}}}} = {\text{10500002348}}Hz\] or 10.5 GHz + 2348 Hz. This may seem impossible to detect, but look what has happened. A microwave signal has been reflected at a very slightly different frequency. This is ideal for making beats if part of the outgoing signal is combined with the received signal, and the beat frequency is in the easily detected audio range. In fact, the beats approach even means that the outgoing signal frequency need not be very precisely maintained. Police radar is more evidence that physics works, no matter how painful!

 

« Previous | Next »