Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Europan Ocean on 05/02/2015 11:15:21
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I am studying transmissions and reception, radio waves, microwaves... I find it difficult to understand that to make it simple, the wavelength and frequency of blue light can have a different amplitude. Wave height. How does this work? What difference would there be in the blue light with great amplitude compared to small amplitude?
Also some waves are not received under bridges, others are, how does that work?
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I am studying transmissions and reception, radio waves, microwaves... I find it difficult to understand that to make it simple, the wave length and frequency of blue light can have a different amplitude. Wave height. How does this work? What difference would there be in the blue light with great amplitude compared to small amplitude?
Also some waves are not received under bridges, others are, how does that work?
There are some different things here. Keeping it simple:
Frequency and wavelength of electromagnetic waves describes the same thing they are related by the speed of light.
Speed of light = wavelength x frequency. In light the wavelength is equivalent to colour, longer wavelengths towards red, shorter towards blue.
Amplitude. With water waves the amplitude is the height, with sound waves the equivalent is the peak pressure, Radio it's the intensity of the electric and magnetic components. We don't usually refer to amplitude of light, but intensity. That is brightness. You could think number of photons per second, but to really understand you'll have to get to the point of doing the maths of EM waves before this will click properly.
Wave can be affected by objects eg water waves by bridges, to create shadows. Water waves can also be refracted, reflected and diffracted just like radio or light waves, also affected by depth of water. This is a whole fascinating subject of its own. Look around for some surf science sites, or how waves on a beach are refracted as the water shallows.
So, in the case of EM radiation, The wavelength of the radio wave and size of object will affect how they respond to the object, creating either shadows or interference patterns. One example of this is higher frequency (shorter wavelength) radio waves which can't go beyond the horizon, but lower frequencies (longer wavelengths) can 'bend' over the horizon. Waves can also be reflected off surfaces or atmospheric layers.
Keep up the studies, you'll soon get there.
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In the case of sunlight and indoor lighting, the light consists of many different frequencies, with random phase. (Early radio transmitters also had many different frequencies, but this was banned soon after Titanic...)
Modern radio transmitters, microwaves and lasers are "coherent": the waves have a single frequency and are all in phase (well, pretty much). This can produce some interesting interference effects if you shine a laser pointer at a wall or a sheet of paper or a bubble - you see speckles and bands of light, where the reflection is stronger in some parts, and almost cancelled at other locations.
When you drive under a bridge, the radio waves reflect off the metal structure of the bridge, causing interference patterns which cancel out in some lanes (while getting stronger in others), at some positions under the bridge (but not others), and for some radio stations (but not others). This is why the effect appears so random.
So if you have a 1mW red laser (at around 430 THz), it has almost twice the wavelength of a 1mW blue laser (at around 790 THz), because the frequency has a ratio of around 2:1; this means that the interference rings are spaced much more closely with blue light than with red light. Now compare that to an AM radio station with a frequency of 1MHz - the wavelength is far longer, enough for you to hear distinct periods of loud and quiet as you drive towards and under a bridge. (It is less pronounced with FM radio at frequencies around 100MHz because this form of modulation does not convey information on the amplitude of the signal, and the interference bands are much closer.)
You will notice that wireless microphone receivers (and modern WiFi receivers) have 2 or 3 antennas, so that if one antenna is receiving no signal due to reflections off the building, the other antenna (just a few inches away) is likely to be receiving a good signal.
Extending the photon approach: Radio waves and light can be thought of as a stream of photons, the "particle" of light (which also has some wave-like properties). The photon Energy (http://en.wikipedia.org/wiki/Photon#Physical_properties) is proportional to frequency. The red laser puts out almost twice as many photons per second as the blue blue laser because of the frequency difference. However, they both produce the same power (amplitude). Then compare this to a wireless microphone producing 1mW of radio power at around 100MHz, which puts out far more photons than the red or blue laser, because the frequency ratio is quite extreme, but it also has the same power. But photon properties do not become obvious until you are talking about very low intensities or very high frequencies, where events occur 1 photon at a time. Otherwise, it is better to use the ocean wave analogy.
[Where I could use some help: In electrical circuits, power is proportional to the voltage squared; in a radio transmitter, power is proportional to the voltage squared; in ocean waves, power increases more than linearly with wave height. Is the power of a free-space electromagnetic wave proportional to the square of the electric field strength?]
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[Where I could use some help: In electrical circuits, power is proportional to the voltage squared; in a radio transmitter, power is proportional to the voltage squared; in ocean waves, power increases more than linearly with wave height. Is the power of a free-space electromagnetic wave proportional to the square of the electric field strength?]
Spot on.
Although the electric and magnetic fields are perpendicular vectors they can be resolved with rate of energy transfer perpendicular to both, in the direction of the wave propagation and proportional to the product of the two. Using E= B x c you can substitute B and end up with E squared.
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the wave length and frequency of blue light can have a different amplitude.
Amplitude of light in classic physics, is replaced by quantity of photons in quantum physics..
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I spoke with an amateur radio enthusiast about interference effects on FM radio. He described it as a "picket fence" effect.
Imagine a fence whose fenceposts are separated by gaps of equal width. If you drive past this fence in your car (with the window down), you will hear a rapidly changing echo of your car.
Driving under a bridge, or near the limits of an FM transmitter's coverage, you will hear this rapid variation of the signal, which sounds quite different from the slow fading of AM radio (at the same speed).
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Frequency is the inverse of wavelength, and describes the "quality" of electromagnetic radiation. At very high frequencies (very short wavelengths), from infrared to cosmic gamma radiation, where it is useful to consider individual photons, frequency is directly related to photon energy.
Amplitude describes the "intensity" of radiation: the strength of a radio signal or the brightness of a light.
Propagation of radio waves through tunnels is a bigger subject than can be accommodated in a chatroom! At some wavelengths, commensurate with the tunnel dimenskions, it may act as a waveguide, producing a fairly unattenuated signal at any point inside. At very long wavelengths the electric field may simply oscillate between the ends of the tunnel, again producing a strong signal throughout. Diffraction may propagate short wave (e.g. mobile phone) signals some way into the tunnel but many long railway and road tunnels use "leaky feeders" or repeater stations to actively generate FM (usually public service stations) and phone signals inside the tunnel.
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Frequency is the inverse of wavelength,
Not exactly.
f = c/wavelength, when we're talking about light.
Inverse of frequency is period.
T=1/f
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True. Frequency is (a constant multiplied by) the inverse of wavelength where the constant is the propagation speed of whatever in whichever. I admit to being outpedanted.
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I think the point of confusion for Europan is that an EM wave involves changes to the field itself as opposed to changes in the location of some object within the field, which is how we detect the wave.
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Spoton Colin
Quote "Although the electric and magnetic fields are perpendicular vectors they can be resolved with rate of energy transfer perpendicular to both". That makes the EM or light energy transferred 3D with electric current at right angles to the magnetic flux and the energy transfer voltage at right angles to both. Thus we have a volume of energy to vibrate in an enclosure which may vary in shape and cancel or amplify the resultant vibration.
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Susskind sums it up nicely here: Europan's question comes up at about 28:30. He addresses Evan's question just before that. The energy of a light wave is proportional to the square of its amplitude. More specifically, A2 is proportional to nhf, where n is the number of photons and f is the frequency. Power is how fast energy changes over time so there is another factor of f involved there. More on photon wave amplitude here: https://physics.stackexchange.com/questions/47105/amplitude-of-an-electromagnetic-wave-containing-a-single-photon.
Note that waves in electric circuits are different because the wave medium has inertial mass, which means current lags voltage. Water waves are even more different because the water molecules move back and forth as well as up and down.
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The stackexchange guys make a good point, which is that Heisenberg allows the amplitude of any given photon wave to be arbitrarily large or small. It is only the average that satisfies E=A2.
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Once you invoke Heisenberg you are fishing in dangerous intellectual waters.
The problem with EM radiation is that classical wave equations work very well for radio and microwaves, but gamma radiation is best modelled by particle equations, and the bit in between, visible light to soft x-rays, needs a touch of both.
Classically, the brightness of a monochromatic light can be modelled by the amplitude of a wave, and this describes most of what we do with lasers, for instance, where we can use the interference of coherent waves to make all sorts of measurements. Likewise diffraction and refraction are nicely modelled by wave mechanics.
Problems arise when we look at the intensity of a photon source. Each photon has a defined energy E = hf so the intensity of a source (energy emitted per second) depends on the number of photons emitted per second (flux) and their energy. However the relationship c = fl means that we can also associate a wavelength with a single photon, and hence predict and interpret x-ray diffraction patterns, but the "amplitude" of a single photon is clearly meaningless, and it is not helpful to assign an effective amplitude to a beam or a bunch of photons when we can talk about flux, spectrum and intensity.
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As others has said, wavelength and frequency describe the character of a wave, and are related to the speed of light. Since a wave has both crests and troughs, the amplitude; height, of the wave has both positive; crest, and negative; trough, aspects, which always add to zero, no matter what the amplitude. Wavelength and frequency are always finite and not zero.
For example the speed of sound in air is about 1100 feet/sec. Waves moving in air will sound different, based on the wavelength and/or frequency. The amplitude is a measure of how loud the sound is. Since the crests and troughs are always equal and opposite, and each wave have one crest and one trough, that always adds to zero. The result is the pitch remains the same, but only appears louder.
What is interesting is, the red shift only impacts the wavelength and frequency, but not the amplitude. A red shift may change the color of the light, but it does not impact the intensity of the new color. This leads to some interesting questions. Since the amplitude is a part of the wave, but does not red shift, what would happen if the amplitude changed suddenly? Would that appear like it is a Doppler shift?
The analogy is watching the TV with very low volume. You need to move you head forward and strain to hear it. If I suddenly crank the volume, to beyond comfortable, you will move your head away from the sound, creating relative motion.
As an experiment, play a single note on a keyboard, that is amplified to be very loud. While maintaining the same note, lower the volume to very low. Does the pitch appear to change like the note is moving away, even though the note has not changed?
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As an experiment, play a single note on a keyboard, that is amplified to be very loud. While maintaining the same note, lower the volume to very low. Does the pitch appear to change like the note is moving away, even though the note has not changed?
Pitch change with loudness is a common acoustic illusion which has been well researched since the early part of the last century. The actual pitch does not change, only our perception of it.
It works best with pure tones and can often be heard in the dying reverberation of a concert organ.
What is interesting is, the red shift only impacts the wavelength and frequency, but not the amplitude. A red shift may change the color of the light, but it does not impact the intensity of the new color. This leads to some interesting questions. Since the amplitude is a part of the wave, but does not red shift, what would happen if the amplitude changed suddenly? Would that appear like it is a Doppler shift?
You probably have missed Alan's post. Worth reading if you want to understand light intensity.
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Amplitude. With water waves the amplitude is the height, with sound waves the equivalent is the peak pressure, Radio it's the intensity of the electric and magnetic components. We don't usually refer to amplitude of light, but intensity. That is brightness. You could think number of photons per second, but to really understand you'll have to get to the point of doing the maths of EM waves before this will click properly.
Wave can be affected by objects eg water waves by bridges, to create shadows. Water waves can also be refracted, reflected and diffracted just like radio or light waves, also affected by depth of water. This is a whole fascinating subject of its own. Look around for some surf science sites, or how waves on a beach are refracted as the water shallows.
So, in the case of EM radiation, The wavelength of the radio wave and size of object will affect how they respond to the object, creating either shadows or interference patterns. One example of this is higher frequency (shorter wavelength) radio waves which can't go beyond the horizon, but lower frequencies (longer wavelengths) can 'bend' over the horizon. Waves can also be reflected off surfaces or atmospheric layers.
I'd like to make one point very clear. There is absolutely nothing wrong with speaking of the amplitude of visible light since light is electromagnetic radiation. Light just at a much higher frequency. Its just not used much since in the lab one measures either energy of photons or the intensity of a beam of light.
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What is interesting is, the red shift only impacts the wavelength and frequency, but not the amplitude.
The amplitude of a coherent laser beam is easier to define and measure than the amplitude of white light from a galaxy. The amplitude white light can be considered to be related to its power or intensity (where power is proportional to the amplitude squared).
Red shift reduces the frequency of spectral features in the white light, which also reduces the power delivered (and amplitude), for two reasons:
- redder photons carry less energy
- because the galaxy is farther away this second than it was last second, we receive fewer photons in 1 second than were emitted in 1 second.
This resolves a paradox generally associated with Olbers (although others had raised the question before him):
- If you look out into space, everywhere you look should end up on a star (this was before they discovered galaxies)
- So the whole sky should be as bright as the surface of the Sun
- And we should be deep-fried
- The resolution is that the expansion of the universe causes red-shift, which reduces the amplitude of the light from these stars, so we haven't been burnt to a crisp.
See: https://en.wikipedia.org/wiki/Olbers%27s_paradox