2. Physics 212 Lecture 23, Slide 1 Physics 212 Lecture 23

3. Physics 212 Lecture 27Music Who is the Artist? A)Soul Rebels Brass Band B)John Boutte C)New Orleans Nightcrawlers D)Paul Sanchez & Shammar Allen E)Alex McMurray and Matt Perrine Why? Threadhead Records Beats Shazam !! Fan-funded and volunteer run record company Wonderful music from New Orleans Hint: Thursday’s artists also did a great set at Lagniappe stage last Jazzfest

4. Physics 212 Lecture 23, Slide 3 Your Comments 05 “The first part of the pre-lecture was awesome. Then the energy was just so much material in so little time. Please go over energy but slower and with more examples please” “Not too bad, but clarify all the stuff on energy. Energy density and total energy are confusing.” We’ll do another example of power in e.m. waves “I thought the prelecture was confusing. It will present an equation and then declare that the rms variable should be used instead. I don’t completely understand the difference and what root mean square is used for. Thanks!” We’ll mention this again – pls chk back to Lec. 21 “i wish every lecture was this straightforward” “Is pointing vector really spelled ‘poynting’; vector???” “Why did the electromagnetic field give away all its energy? Because there was no poynt in keeping it. Why is the electromagnetic wave insecure about his looks? Because he’s a ‘plain’ wave. I hope you enjoyed this LIGHT humor...” “you don’t read lecture thoughts after midnight right? ;(”

5. Physics 212 Lecture 23, Slide 4 Plane Waves from Last Time E and B are perpendicular and in phase Oscillate in time and space Direction of propagation given by E X B E 0 = cB 0 Argument of sin/cos gives direction of propagation

6. Physics 212 Lecture 23, Slide 5 Understanding the speed and direction of the wave ExEx z t = 0 E x = E o sin(kz-  t) What has happened to the waveform in this time interval? It has MOVED TO THE RIGHT by /4 t =  /2  ExEx z

7. Physics 212 Lecture 23, Slide 6 Checkpoint 1a “since the electomagnetic wave is in the x direction it will be Ex” “e is in the y direction and since k and t are different signs, the wave it moving in the positive direction” “B field propagates in the positive z direction and lies in the yz-plane.”

8. Physics 212 Lecture 23, Slide 7 No – moving in the minus z direction No – has E y rather than E x Checkpoint 1a

9. Physics 212 Lecture 23, Slide 8 Wavelength is equal to the speed of light divided by the frequency. Checkpoint 2a c=3.0 x 10 8 m/s Check: Look at size of antenna on base unit

10. Physics 212 Lecture 23, Slide 9 Doppler Shift The Big Idea As source approaches: Wavelength decreases Frequency Increases

11. Physics 212 Lecture 23, Slide 10 Doppler Shift for e-m Waves What’s Different from Sound or Water Waves ? Sound /Water Waves : You can calculate (no relativity needed) BUT Result is somewhat complicated: is source or observer moving wrt medium? Electromagnetic Waves : You need relativity (time dilation) to calculate BUT Result is simple: only depends on relative motion of source & observer  > 0 if source & observer are approaching  < 0 if source & observer are separating  = v/c

12. Physics 212 Lecture 23, Slide 11 Doppler Shift for e-m Waves v or f’f vff’ The Doppler Shift is the SAME for both cases ! The Doppler Shift is the SAME for both cases ! f’/f ONLY DEPENDS ON THE RELATIVE VELOCITY

13. Physics 212 Lecture 23, Slide 12 Doppler Shift for e-m Waves A Note on Approximations WHY ??  << 1 Taylor Series: Expand around  = 0 Evaluate: NOTE:Remember  > 0 for approach  < 0 for separation

14. Physics 212 Lecture 23, Slide 13 Our Sun Star in a distant galaxy Red Shift (for Astronomical Objects) wavelength Wavelengths shifted higher Frequencies shifted lower Star separating from us (Expanding Universe)

15. Physics 212 Lecture 23, Slide 14 Red Shift “the dopper effect in relation to magnetic fields. Is this used in any practical applications? Is anything moving fast enough on earth that relative to the speed of light that there would be any significant shift? I understand that this makes sense in looking at the universe and motion of planets and stars and such but there is no way you can "run" fast enough to ever see an iclicker signal.” Star Clusters and Distant, Red Galaxies Near Edge of Galaxy NGC 3370 (Hubble Space Telescope)

16. Physics 212 Lecture 23, Slide 15 Police radars get twice the effect since the EM waves make a round trip: v  f’f’-f 30 m/s (67 mph) 1.000 x 10 -7 24,000,004,8004800 Hz 31 m/s (69 mph) 1.033 x 10 -7 24,000,004,9594959 Hz If f = 24,000,000,000 Hz (k-band radar gun) c = 300,000,000 m/s Example

17. Physics 212 Lecture 23, Slide 16 Checkpoint 2b f iclicker = 900 MHz A) B) C) “Running away from an iClicker will decrease the frequency which may let you see something.” “You need to increase the frequency at which the waves reach you, so you should run towards the iclicker” “Whether you move or the clicker moves, no visible wave will be availableency.”

18. A) B) C) Physics 212 Lecture 23, Slide 17 Checkpoint 2b Need to shift frequency UP Need to approach i>clicker (  > 0) How fast would you need to run to see the i>clicker radiation? f iclicker = 900 MHz Approximation Exercise:

19. Physics 212 Lecture 23, Slide 18 Waves Carry Energy

20. Physics 212 Lecture 23, Slide 19 Sunlight on Earth: I ~ 1000J/s/m 2 I ~ 1000J/s/m 2 ~ 1 kW/m 2 ~ 1 kW/m 2 Intensity Intensity = Average energy delivered per unit time, per unit area Area = A Length = c dt

21. Physics 212 Lecture 23, Slide 20 Waves Carry Energy

22. Physics 212 Lecture 23, Slide 21 Comment on Poynting Vector Just another way to keep track of all this - Its magnitude is equal to I – Its direction is the direction of propagation of the wave

23. Power in EM Waves: Example A cell phone tower has a transmitter with a power of 100 W. What is the magnitude of the peak electric field a distance 1500 m (~ 1 mile) from the tower? Assume the transmitter is a point source. Physics 212 Lecture 23, Slide 22 What is the intensity of the wave 1500 m from the tower? A) 1.5 nW/m 2 B) 3.5  W/m 2 C) 6 mW/m 2

24. Physics 212 Lecture 23, Slide 23 Momentum of light is E/c  light can exert pressure!! Analogy from mechanics: For E-M waves: pressure Radiation pressure Light has Momentum!

25. Physics 212 Lecture 23, Slide 24 Momentum of light is p = U/c  light can exert pressure!! Recall force from mechanics: Light has Momentum! Approximate  p by p: But intensity of light is: Therefore:

26. Physics 212 Lecture 23, Slide 25 Momentum of light is p = U/c  light can exert pressure!! Light has Momentum!

27. Physics 212 Lecture 23, Slide 26 Checkpoint 1b Which of the following actions will increase the energy carried by an electromagnetic wave? A. Increase E keeping  constantB. Increase  keeping E constant C. Both of the above will increase the energyD. Neither of the above will increase the energy “increasing E will increase amplitude, therefore engergy” “By increasing the frequency, we will increase the energy.” “both dont effect because energy isnt dependant” “Increasing the frequency or the field would increase the energy carried”

28. Physics 212 Lecture 23, Slide 27 Checkpoint 1b But then again, what are we keeping constant here? WHAT ABOUT PHOTONS? Which of the following actions will increase the energy carried by an electromagnetic wave? A. Increase E keeping  constantB. Increase  keeping E constant C. Both of the above will increase the energyD. Neither of the above will increase the energy The energy of one photon is E photon = hf = h  /2  U wave = N photons E photon = 1/2  0 E 0 2

29. Physics 212 Lecture 23, Slide 28 PHOTONS We believe the energy in an e-m wave is carried by photons Question: What are Photons? Answer: Photons are Photons. Photons possess both wave and particle properties Particle: Energy and Momentum localized Wave: They have definite frequency & wavelength (f = c) Question: How can something be both a particle and a wave? Answer: ~It can’t (when we observe it) What we see depends on how we choose to measure it ! The mystery of quantum mechanics: More on this in PHYS 214 h = 6.63e -34 J-s Planck’s constant Connections seen in equations: E = hf p = h/ p = h/

30. Physics 212 Lecture 23, Slide 29 Calculation 1 An electromagnetic wave is described by: where is the unit vector in the +y direction. Wave moves in +z direction (A) (B) (C) (D) Which of the following graphs represents the z-dependence of B x at t = 0? X E and B are “in phase” (or 180 o out of phase) X x y z E B x y z

31. Physics 212 Lecture 23, Slide 30 Calculation 2 An electromagnetic wave is described by: Wave moves in –z direction +z points out of screen -z points into screen x y What is the form of B for this wave? (A) (B) (C) (D) B E x y z

32. Physics 212 Lecture 23, Slide 31 Calculation 3 An electromagnetic wave is described by: Wave moves in negative z-direction +z points out of screen -z points into screen x y B E Which of the following plots represents B x (z) at time t =  /2  ? (A) (B) (C) (D) at  t =  /2:

33. Physics 212 Lecture 23, Slide 32 Calculation 4 A certain unnamed physics professor was arrested for running a stoplight. He said the light was green. A pedestian said it was red. The professor then said: “We are both being truthful; you just need to account for the Doppler effect !” As professor approaches stoplight, the frequency of its emitted light will be shifted UP The speed of light does not change Therefore, the wavelength (c/f) would be shifted to a smaller value If he goes fast enough, he could observe a green light ! Is it possible that the professor’s argument is correct? ( green = 500 nm, red = 600 nm) (A) YES (B) NO

34. Physics 212 Lecture 23, Slide 33Follow-Up A certain unnamed physics professor was arrested for running a stoplight. He said the light was green. A pedestian said it was red. The professor then said: “We are both being truthful; you just need to account for the Doppler effect !” How fast would the professor have to go to see the light as green? ( green = 500 nm, red = 600 nm) (A) 540 m/s (B) 5.4 X10 4 m/s (C) 5.4 X 10 7 m/s (D) 5.4 X 10 8 m/s Relativistic Doppler effect: Note approximation for small  is not bad: c = 3 X 10 8 m/s  v = 5.4 X 10 7 m/s Change the charge to SPEEDING!