- #1
SteveinLondon
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The expansion of spacetime stretches and red shifts the wavelength of light. Is the amplitude of the wave stretched as well? So that very distant stars appear brighter, and therefor nearer?
So that very distant stars appear brighter, and therefor nearer?
Electromagnetic waves are not transverse vibrations of a medium. There is no actual sideways motion.
Electromagnetic radiation (often abbreviated E-M radiation or EMR) is a form of energy exhibiting wave-like behavior as it travels through space. EMR has both electric and magnetic field components, which oscillate in phase perpendicular to each other and perpendicular to the direction of energy propagation.
bcrowell said:...
Although this question can be answered purely classically, I think it's easier to analyze if you think in terms of photons. The photon's frequency is reduced by a factor z, so E=hf is also reduced by that factor. The volume occupied by the wave is increased by a factor of z^3, so the energy density is reduced by a factor of z^4. Since the energy density is reduced by z^4, the electric and magnetic fields are also reduced, by factors of z^2.
marcus said:The energy density of the photons is actually decreased by a factor of (1+z)4
Naty1 said:No medium, I get that..what does the second part mean..."actual sideways motion" ??
The field oscillates that way, right...?
Wiki has a (correct) concise explanation and associated diagram here:
http://en.wikipedia.org/wiki/Electromagnetic_wave
The energy density of the photons is actually decreased by a factor of (1+z)4
Please take another look at the final sentence of #2.Bill_K said:Sorry, but the original question was not about the energy density, was it. It says 'does the amplitude change'?
It's not a Lorentz transformation. This isn't a special-relativistic kinematic Doppler shift. There are no global Lorentz transformations in GR.Bill_K said:The amplitude of an electromagnetic wave is just |E|. The issue is simply a matter of how the Lorentz transformation acts on E and B. So I say it should be just one factor of 1+z.
marcus said:so the number of photons per unit volume is reduced by 1.53 (they are spread out in a larger volume).
So the whole effect on the energy density is to reduce it by a factor of 1.54 = a bit over 5.
Naty1 said:"I'm sure you meant 1.52 here for the inverse square law, didn't you?
That's for the unit area...
Marcus was addressing the unit volume...
V = 4/3(pi)r3
The amplitude of a stretched wave refers to the maximum displacement of the wave from its rest position. This can be measured by the height or depth of the wave's peaks or troughs. When a wave is stretched, its amplitude increases, meaning that the magnitude of its oscillations becomes greater.
The amplitude of a wave does not directly affect red shift. Red shift is a phenomenon that occurs when light or other electromagnetic radiation from a source is shifted towards the red end of the spectrum. This can happen due to several factors, such as the relative motion of the source and observer, or the expansion of the universe. The amplitude of a wave is not a factor in determining the amount of red shift.
No, the amplitude of a wave is a measure of the physical properties of the wave itself, while red shift is a measure of the change in frequency or wavelength of the wave. These are two different characteristics of a wave and cannot be directly measured in terms of each other.
The amplitude of a wave is directly proportional to its energy. This means that as the amplitude increases, so does the energy of the wave. This relationship is described by the equation E = hc/λ, where E is the energy of the wave, h is Planck's constant, c is the speed of light, and λ is the wavelength of the wave.
The amplitude of a wave is not affected by red shift. As mentioned earlier, red shift is a change in the frequency or wavelength of a wave, which does not directly impact the amplitude. However, red shift can affect the perceived amplitude of a wave, as the change in frequency may cause the wave to appear more or less intense to an observer.