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a1call
2008-Oct-23, 05:56 AM
Hi,

*- Space station X is equipped with a perfectly monochromatic blue beacon

*- Space ship A departs at a constant speed such that the beacon light is red shifted by a single quantum step/frequency

*- Space ship C departs at a constant speed such that the beacon light is red shifted by 2 quantum steps/frequencies

*- Space ship B departs at a constant speed equal to the average of the speeds of space ships A & C

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**- What level of red shift would the photons from the beacon possess when observed from space ship B?

**- Could we conclude that speed is quantized and an average between 2 consecutive quantum speeds can not exist?

Thanks in advance.

StupendousMan
2008-Oct-23, 01:21 PM
[FONT=verdana]Hi,

*- Space station X is equipped with a perfectly monochromatic blue beacon

*- Space ship A departs at a constant speed such that the beacon light is red shifted by a single quantum step/frequency

*- Space ship C departs at a constant speed such that the beacon light is red shifted by 2 quantum steps/frequencies

[FONT]

I don't understand what you mean when you say "is red shifted by a single quantum step/frequency." Please explain further.

Until you can be more specific, we can't help to answer your question.

Hornblower
2008-Oct-23, 01:25 PM
To the best of my knowledge there is no such quantization. B's redshift would be halfway between the others.

a1call
2008-Oct-23, 01:42 PM
Hi StupendousMan,
Thank you for your reply.

My understanding is that energy levels of photons are not distributed linearly and in fact increase in quantum steps/leaps. In other words it is always an integer multiple of some value which is the lowest indivisible energy level possible.

So what I was trying to convey was that spaceship A would observe the lowest possible energy-level shift towards the longer wavelength and spaceship C the 2nd lowest possible.

As such common sense would imply that spaceship B would observe photons with energy levels in between those observed by the other two ships. However such intermediary value is not possible.

Hence I am asking for confirmation that speed is also quantized and changes in quantum leaps which is probably a given considering it is a related to mass and energy of the ship which are quantised values.

My problem is grasping this when speed is also a factor of coordinates in space which is not confirmed to be a quantum parameter.

BTW, I know I have poor expressability skills.

trinitree88
2008-Oct-23, 03:43 PM
[QUOTE=a1call;1349134] SNIPPET...

My understanding is that energy levels of photons are not distributed linearly and in fact increase in quantum steps/leaps. In other words it is always an integer multiple of some value which is the lowest indivisible energy level possible.

END SNIPPET

a1call. Your understanding of photon energy is incomplete. E=hv The energy can be any value as the frequency (v) can be any value. Sometimes texts give specific examples using frequencies reported to several significant figures, kind of implying that a change in the last figure makes them incremental. But, in principle, you could measure a frequency to 10, 15, 15,000 figures....and have any old energy as a result. The spectrum of an incandescent light bulb is a continuum, all the colors, frequencies are present. If you get down to your filament being made of two atoms and then one....then you get a line spectrum for that element when it ionizes. For macroscopic fliaments, it's continuous. pete

PraedSt
2008-Oct-23, 04:43 PM
Well, I sort of see where you're going with this, and I think it's a good question!
-------> Running to do some reading :)

Ken G
2008-Oct-23, 06:53 PM
Others have said this too-- there can be quantization of the emission process on the spaceship, but there is no similar quantization of redshift of the nature you imagine-- the redshift is continuous and does not know or care what mechanism quantifies the initial emission process.

PraedSt
2008-Oct-23, 07:04 PM
Others have said this too-- there can be quantization of the emission process on the spaceship, but there is no similar quantization of redshift of the nature you imagine-- the redshift is continuous and does not know or care what mechanism quantifies the initial emission process.

That's what I thought. But I was stuck with measurement. How do you measure the frequency, if the redshift just happens to shift it to one that doesn't correspond to an energy level difference? Somewhere, in some material.. Where am I going wrong with this? :confused:

a1call
2008-Oct-23, 07:17 PM
So this is one of those instances where light reminds us that it has a wave nature.

Still, it is significant that when individual photons' energy levels are measured aboard B you get an un-quantized value.

The significance is that it turns upside down everything I thought I knew, again.:(

StupendousMan
2008-Oct-23, 08:54 PM
That's what I thought. But I was stuck with measurement. How do you measure the frequency, if the redshift just happens to shift it to one that doesn't correspond to an energy level difference? Somewhere, in some material.. Where am I going wrong with this? :confused:

You are laboring under a misapprehension that the only type of photon-matter interaction is the type that occurs between an isolated, neutral atom and a photon. If an isolated, neutral hydrogen atom is excited to some energy below the ionization point, it will emit one (or more) photons which do come from a discrete set of energies (or frequencies, or wavelengths).

However, if the atom is part of a group -- in a dense gas, or in a liquid, or in a solid -- then the atom may emit photons of _any_ energy below the ionization energy. That's why dense materials produce continuous spectra when heated, whereas tenuous gases produce line spectra.

The same works in reverse: a photon which impinges on a solid surface _will_ be absorbed or reflected, no matter what wavelength it has (below the ionization energy).

Does that help?

PraedSt
2008-Oct-23, 09:05 PM
I was indeed labouring :)
So...it's a matter of numbers; that turn a discrete distribution into a continuous one, so to speak?
Thanks

a1call
2008-Oct-24, 01:44 AM
Thank you all for the replies and discussion.