Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: jeffreyH on 04/09/2015 23:02:30
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As gravity tends to redshift wavelengths would the wavelength tend to infinite at infinity. Would this imply that the wave is dependent upon the presence of gravity.
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Would this imply that the wave is dependent upon the presence of gravity?
No, electromagnetic waves propagate quite well in the absence of a strong gravitational field (eg between galaxy clusters - that's how astronomers can see quasars).*
would the wavelength tend to infinite at infinity
A photon emitted from the surface of the Earth is slightly red-shifted as it rises through Earth's gravitational field, and slightly more as it escapes the Sun's gravitational field, and slightly more as it escapes the galaxy's gravitational field. But the Earth has a very low mass, and these effects are very small; the photon will arrive "at infinity" with almost the same wavelength that it left, provided you measure the wavelength with a spectrometer moving in the same frame of reference as the Earth.
If, however, you measure the wavelength with a spectrometer moving in the frame of reference of a local galaxy (say) 12 billion light years away, the light will be severely red-shifted, due to the general expansion of the universe.
For a more dramatic example, if you take a photon emitted near a significant mass (like a black hole), it will be severely red-shifted before it travels 1000km from the black hole. And the closer you release the photon to the black hole, the more red-shifted it will be by the time it reaches 1000km altitude. In the limit, a photon released right at the event horizon will be red-shifted to infinity by the time it reaches 1000km altitude; no need to wait until it reaches infinite distance!
*As for how light would propagate in a universe containing zero mass - well, we don't have any empty lab universes where we can test this hypothetical scenario...
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In the limit, a photon released right at the event horizon will be red-shifted to infinity by the time it reaches 1000km altitude….
So what does being “red-shifted to infinity” actually mean? Red-shifting involves an increase in the distance between wave crests. What does the wave profile look like when it has red-shifted to infinity?
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As gravity tends to redshift wavelengths would the wavelength tend to infinite at infinity. Would this imply that the wave is dependent upon the presence of gravity.
If you're asking whether the frequency or wavelength is a function of the gravitational potential where the photon is relative to an observer then yes. This means that frequency, f, is a function of gravitational potential, phi, i.e. f, = f(phi) where f(0) = f0 is the frequency in the absence of the gravitational potential.
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Thanks Pete and Evan your input helps to clear some things up.
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what does being “red-shifted to infinity” actually mean?
One measure of red-shift is the wavelength emitted vs the wavelength detected. (Other measures compare frequency or photon energy, but these all bear a simple relationship to the wavelength in a vacuum.)
Red shift can occur due to Doppler shift (or its relativistic equivalent), Cosmological expansion of the universe, or Gravitational time dilation (which has no classical equivalent).
Astronomers calculate the measure "z" (https://en.wikipedia.org/wiki/Redshift#Measurement.2C_characterization.2C_and_interpretation) as a ratio of wavelengths to define the redshift of objects in space.
- The highest reported redshift (https://en.wikipedia.org/wiki/Redshift#Highest_redshifts) for a galaxy had z=8.6
- The redshift of the CMBR is currently around z=1089
As red shift "approaches infinity", the detected wavelength "approaches infinity" (and the equivalent frequency & photon energy approach zero). As the photon energy approaches zero, it becomes harder & harder to detect the photons, let alone measure their wavelength, so the source becomes almost undetectable.
So I guess you could say we have not yet succeeded in measuring a redshift "approaching infinity" - but it is still a useful concept if we ever find a tame black hole to experiment on....
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So I guess you could say we have not yet succeeded in measuring a redshift "approaching infinity"
If by "approaching infinity" you mean approaching a point where we might as well consider it to be infinite, for all practical purposes; I’m OK with that.
On the other hand if you are talking about actually "approaching infinity", there is the problem that however much approaching you do, you will still be infinitely far away.
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If a wave function flatlines then it is infinite.
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If a wave function flatlines then it is infinite.
How does that differ from not being a wave?
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You have to consider the time component. If a disc is spinning at 10 rpm and we move it horizontally and trace the path we get a wave. However if we slow its spin rate to 1 nano metre a year and move it at the same rate horizontally we will trace out what we perceive to be a straight line.
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If a wave function flatlines then it is infinite.
How does that differ from not being a wave?
Strictly speaking, it doesn't. If it's flat, there's no wave. So that's why electromagnetic waves don't escape the BH event horizon; it's as if space is 'falling' or 'stretching' into the BH faster than c.
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However if we slow its spin rate to 1 nano metre a year and move it at the same rate horizontally we will trace out what we perceive to be a straight line.
This is a good example of what I mean by "approaching infinity" used to mean "approaching a point where we might as well consider it to be infinite, for all practical purposes". Given enough time, your wave will emerge, but only in the far future, so why bother with it?
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If it's flat, there's no wave
Mathematically, you can treat a constant value as a wave of 0Hz; for example COS(2π0t) is a constant.
Using the Fourier transform, the average value of a periodic waveform appears as the 0Hz component of that wave.
Where it impacts physics is that a changing electric (or magnetic or gravitational) field will release some energy which propagates away "to infinity". However, a constant electric (or magnetic or gravitational) field does not radiate.
In the limit, as the frequency approaches zero, the radiated energy approaches zero, and the ability to detect any such radiation also approaches zero. So in practice, radiation "red-shifted to infinity" cannot be differentiated from "no radiation" by the receiver.
However, there is a difference at the transmitter end - in one case the transmitter is losing power as radiation, while in the constant case, the transmitter is not radiating any power.
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However, there is a difference at the transmitter end - in one case the transmitter is losing power as radiation, while in the constant case, the transmitter is not radiating any power.
Fair enough, but at the transmitter end, has the wave flat-lined?
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Fair enough, but at the transmitter end, has the wave flat-lined?
That depends.
If the cause is redshift then no, it is still transmitting, the effect is due to motion.
If as described by JefferyH, in the disc example, the transmitter slows down to infinite slowness then it 'flatlines' at the transmitter. Most of us would say the transmitter has stopped transmitting.
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Most of us would say the transmitter has stopped transmitting.
Would we not be correct in saying that? If it has really flatlined (unlike Jeffrey's disc example), what would it be transmitting?
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As gravity tends to redshift wavelengths would the wavelength tend to infinite at infinity. Would this imply that the wave is dependent upon the presence of gravity.
Yes, but only of mechanical waves...
Electromagnetic wave propagation is unaffected by any known force fields...
Redshift of light is a result of refraction and not gravitational field...