As the universe is expanding light from further away it goes towards

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As the universe is expanding light from further away it goes towards the red end of the spectrum but will there ever become a point theoretically that it shifts so far it'll go off the red end of the spectrum and we wouldn't be able to see it
 

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  • #2
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Actually forget it the beers have clouded my head and I forgot about infra red etc. but what happens at the very end of the em spectrum
 
  • #3
phyzguy
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There is no "end of the EM spectrum". The wavelength just keeps getting longer and longer and the energy per photon lower and lower.
 
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There is no "end of the EM spectrum". The wavelength just keeps getting longer and longer and the energy per photon lower and lower.
But it DOES end.
On the lower side it "flatlines" so-to-speak, with the analog being DC.
On the high end we have cosmic rays(or is it "hard x-rays") To my knowledge there is no photon activity category above that.
 
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do you mean Dc as direct non reversal eletrical current
 
  • #6
JDoolin
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But it DOES end.
On the lower side it "flatlines" so-to-speak, with the analog being DC.
On the high end we have cosmic rays(or is it "hard x-rays") To my knowledge there is no photon activity category above that.
Don't forget gamma rays: http://en.wikipedia.org/wiki/Electromagnetic_spectrum


Those upper and lower limits both involve a relative velocity equal to the speed of light. There is a paradox associated with traveling along with a changing magnetic field, so that in your view, it is no longer changing. If there is no changing, there is no electric field, and hence, no photon.

On the other hand, we are familiar with sonic booms, when the plane has been traveling at the speed of sound for a while, the wavelengths all accumulate. In such a case, you have a wavelength that is, at least, momentarily zero.

I think there is no theoretical limit on how low or high a wavelength can be, except that it can't be zero wavelength, and it can't be zero frequency. I wonder though, hypothetically, if a zero wavelength photon would destroy a solar system if it hit anything.... or would it just drill a perfectly straight microscopic hole through everything?
 
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  • #7
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do you mean Dc as direct non reversal eletrical current
Yes. No sine wave at all.
 
  • #8
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I did forget gamma rays. Yes, they are above(higher frequency) than "hard" x-rays.
My bad, thanks for clarifying.
 
  • #9
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I don't think that there is any such thing as a "zero frequency" photon.
 
  • #10
Nabeshin
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I did forget gamma rays. Yes, they are above(higher frequency) than "hard" x-rays.
My bad, thanks for clarifying.
I sincerely hope you realize that just because we call everything at a certain energy and above gamma rays that this does not imply that there is a maximum photon energy... What we call them is absolutely irrelevant.

A zero-frequency photon is a non-existent photon, since it has zero energy and therefore cannot be said to exist at all. Similarly, a zero-wavelength photon cannot exist since its energy would be infinite. That said, within the context of normal special relativity and quantum theory, you can have a photon with arbitrarily small frequency or one with arbitrarily small wavelength. I.e. there can be no smallest frequency or wavelength.
 
  • #11
sophiecentaur
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This is what happens at the limit of the observable Universe. Space is expanding between us and the 'edge' at c so nothing can get here.

btw, why do photons have to come into every description of light? This is a Relativity thing not a QM thing.
 
  • #12
JDoolin
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This is what happens at the limit of the observable Universe. Space is expanding between us and the 'edge' at c so nothing can get here.

btw, why do photons have to come into every description of light? This is a Relativity thing not a QM thing.
Good question. The answer is that every detection of light above a certain frequency is found in descrete quanta of, if I recall correctly h bar times the frequency. I can't remember if that's the momentum or the energy.

At lower frequencies, we assume that this property holds, even though I'm not sure we have the equipment that detects single photons at the really low wavelengths. It probably takes lots of low energy photons, hitting all at once, to register on those big radio antenna systems, and that looks more like a wave than a bunch of photons.
 
  • #13
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Here is an intruder layman that want to expres his opinion about things that he don't understant, but want to understand. So let me explain my opinion:
1- About wave length i think that can't be more than 1.481818608*10^7cm because of the plank constant E = h / 1 and can't be shorter than 1.1885794* 10^-31 cm.because of Plank radius.
2- What if the red shift becomes from the scaterings of photons which colide with mass inpurities in space? Is reasonable that the number of collision with impurities will augmented with distance.When we have to do with parsec distances, let be number of impurities very small, the number of collisions will be very high. By the way moon light has any red shift? Sorry from my ignorance.
 
  • #14
sophiecentaur
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Good question. The answer is that every detection of light above a certain frequency is found in descrete quanta of, if I recall correctly h bar times the frequency. I can't remember if that's the momentum or the energy.

At lower frequencies, we assume that this property holds, even though I'm not sure we have the equipment that detects single photons at the really low wavelengths. It probably takes lots of low energy photons, hitting all at once, to register on those big radio antenna systems, and that looks more like a wave than a bunch of photons.
ALL EM radiation consists of Photons. There's no sudden 'gear change' at some highish frequency. If you wouldn't bring Photons into the RF propagation theory then why should it be necessary, always, to get into Photons when dealing with light? It does not automatically enhance any argument to bring the little chaps into it. (And how big are they, anyway?)
When you get down to it, it's due to the fact that the particulate nature of light was introduced to explain phenomena like the Photoelectric Effect and it's still a 'craze' with casual (and some not-so-casual) students of Physics. What people fail to get is that the wave behaviour or at least the use of wave theory in predicting things about photons, is a very valid way of describing the behaviour of light. The wave approach also is a bit more demanding, mathematically, than an arm waving corpuscular description. Could that be why photons are so often the popular option?
To put it another way, would anyone automatically use de Broglie waves to describe what goes on on a snooker table? That is an equivalent situation because there is particle / wave duality for items with mass.
 
  • #15
JDoolin
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ALL EM radiation consists of Photons. There's no sudden 'gear change' at some highish frequency. If you wouldn't bring Photons into the RF propagation theory then why should it be necessary, always, to get into Photons when dealing with light? It does not automatically enhance any argument to bring the little chaps into it. (And how big are they, anyway?)
How big is a "photon?" In my arm-waving corpuscular description, they are roughly the size of the antenna that you're using to detect them. The lower energy photons are humongous, needing an Very Large Arry of dishes to pick up. The higher energy photons can be detected by the nucleus of an atom.

When you get down to it, it's due to the fact that the particulate nature of light was introduced to explain phenomena like the Photoelectric Effect and it's still a 'craze' with casual (and some not-so-casual) students of Physics. What people fail to get is that the wave behaviour or at least the use of wave theory in predicting things about photons, is a very valid way of describing the behaviour of light. The wave approach also is a bit more demanding, mathematically, than an arm waving corpuscular description. Could that be why photons are so often the popular option?
To put it another way, would anyone automatically use de Broglie waves to describe what goes on on a snooker table? That is an equivalent situation because there is particle / wave duality for items with mass.
The main reason I tend toward describing things in terms of photons is emission spectra. When an electron pops down from one shell to another, if I were to think in terms of waves, I would expect there to be a circle of energy propagating out from the atom. The further away from the atom, you would get a smaller energy from the wave. But that is not what happens. No matter how far away you are, you either see the event, or you don't. And if you do see the event, you see the whole energy of the event. You don't just see a proportion of the event. That energy that reached your eye has the same energy as the energy that was created by the event. There's no energy left over, and that means nobody else saw the event. Effectively, it was a direct interaction between the atom and your eye.

Well, I am arm-waving to a certain extent. I'm not entirely sure what you mean by a "bit more demanding." It's long been at the end of my to-do list to work out some level of comprehension of the wave equation. http://www.youtube.com/watch?v=YLlvGh6aEIs" makes it sound straightforward: "Start with maxwell's equations in a vacuum, and take the curl of Faraday's Law. Now the trick is to take the vector identity for the curl of curl, and then you'll see that everything simplifies, because Del dot E is zero, and
Ampere's Law simplifies the other side; put it all together and derive a wave equation for the E-Field. Take the curl of ampere's law and we'll get a similar equation for B, too." Then you have, a set of linear differential equations whose solution yields a complex transverse plane wave traveling at the speed of light.

To be completely honest, I've never sat down with paper and pencil and become familiar enough with it to explain it or felt I had a very good grasp of it. Maybe with a better grasp of it, I could better reconcile it with emission spectra. More likely, not, since the wave-particle duality of matter is a phenomenon some literature describes as beyond our capability of understanding.

This may be kind of speculative, but let's ask what happens at both ends? In the RF case, what we have is a generation of current/voltage in an antenna. In the emission/absorption case we have the movement of an electron into a higher or lower orbital. In either case there is an acceleration of charges. In the emission/abosrption case, obviously there are two charges involved in this acceleration, the protons in the nucleus and the electron that changes orbital. In the RF case, it is not so obvious, but I suspect there are also two charges involved, accelerating toward or away, or whatever.

So a photon, whichever way you look at it, consists of four events. At one end, two charges moving in such a way to convert potential energy into radiant energy, and at the other end, two charges moving in such a way to convert radiant energy into potential energy. In the corpuscular theory, this radiant energy travels in such a way that it affects these four, and only these four particles.

However, with 6.02 X 10^23 atoms per mole, there is little sense in concerning yourself with the corpuscular theory. In the wave theory, you would just take the superposition of all of the interactions present and think in terms of voltage and current and all of the random directional statistical stuff adds up to your inverse square laws, etc.
 
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  • #16
sophiecentaur
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This is rather simplistic in parts and there are several points here which need clearing up.

If photons of low frequency radiation need a "large array of dishes" to detect them, how is it that a portable LF receiver can be held in the hand (/15000 of the wavelength).

Electrons "popping" from one orbital to another only occurs in gases. In condensed matter, the energy levels occur in bands and a photon doesn't just interact with one electron; it interacts with many atoms during an energy transition. I think this proves my point about the photon thing being a short-cut to an unfounded belief that one has understanding.
You might be interested to know that classical EM theory accounts for 'light pressure' which is always explained, these days, by photons hitting an object and transferring momentum. I was looking at a treatment of this in an ancient textbook of mine and no photons are used. I should put your pencil and paper exercise a bit higher up your list and try to get to grips with some of the basics of EM theory. I think you might not be so wedded to the idea of explaining everything using photons - after all, you would find it very hard to describe the operation of a waveguide by considering photons alone.
I am not sure that a photon is always the way in which you describe it because photons are created in nuclear changes - and we don't have electrons in there, except when they are created during the process.

In your final para you are just saying that wave theory and QM tend to give the same answer. If course that's true but I could guarantee to solve a simple EM problem a lot quicker by not using Photons - in the same way that I could predict line spectra from an excited gas molecule by using them. If I were to insist on using wave theory then I would be unreasonable and wasting my time - I am just saying that the inverse is also true and that photons are a bit more popular than is appropriate. Because they are a simpler concept and can be visualised without loads of sweaty Maths.
 
  • #17
JDoolin
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I should put your pencil and paper exercise a bit higher up your list and try to get to grips with some of the basics of EM theory.
Indeed I should.
 
  • #18
JDoolin
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Here is an intruder layman that want to expres his opinion about things that he don't understant, but want to understand. So let me explain my opinion:
1- About wave length i think that can't be more than 1.481818608*10^7cm because of the plank constant E = h / 1 and can't be shorter than 1.1885794* 10^-31 cm.because of Plank radius.
2- What if the red shift becomes from the scaterings of photons which colide with mass inpurities in space? Is reasonable that the number of collision with impurities will augmented with distance.When we have to do with parsec distances, let be number of impurities very small, the number of collisions will be very high. By the way moon light has any red shift? Sorry from my ignorance.
Hi intruder Layman. I sometimes feel the same way myself. Don't think of the planck's constant as a hard limit. It is just a ratio between the frequency of a photon and an energy of a photon.

The Planck radius, on the other hand is a specific length. It would take a lot of energy to constrain a particle into a region with radius around the Planck Length, but an photon, or electromagnetic wave with a wavelength around the Planck Length is just a fairly high energy photon.

In general, anything inside our galaxy (such as the moon) is going to have a fairly low redshift, because the relative velocity between objects inside the galaxy is relatively small. The Hubble flow is not a phenomenon that exists inside a galaxy, but between galaxies. The stars and planets within the galaxies, are orbiting around a common center; not drifting apart over time.
 
  • #19
JDoolin
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If photons of low frequency radiation need a "large array of dishes" to detect them, how is it that a portable LF receiver can be held in the hand (/15000 of the wavelength).

Electrons "popping" from one orbital to another only occurs in gases. In condensed matter, the energy levels occur in bands and a photon doesn't just interact with one electron; it interacts with many atoms during an energy transition. I think this proves my point about the photon thing being a short-cut to an unfounded belief that one has understanding.

I was told this by one of my professors a long time ago that the wavelength of the wave had to be on the order of magnitude of the antenna built to receive, and I believed it at the time, and it stuck in my head. I heard it and then said it, so that was hearsay.

The idea of a photon involving four particles accelerating toward or away from each other, though, does not rely on any kind of distance between them. Where condensed matter is concerned, that's certainly not my area of expertise, but I can't see how what you're saying conflicts with my idea. Maybe I don't know what you mean by "energy levels occur in bands" but in a conductor such as an antenna, I would think you'd have a nice fat conduction band, and in the phenomenon we know as radio, ther could be bajillions of interactions between individual electrons.

I found an article to help get some hard data:

http://www.eham.net/articles/9982

We have a 40.6 micro-Watt Antenna producing a 3.5 MHz signal.

Using the formula E = n h / f, where h=6.63 *10^-34 J-s, I calculated there are
2.14 * 10^41 photons per second coming out of that antenna.

The receiver 546.8 miles away is at the end of a sphere of radius 879,801 meters. The surface area is then, 4 Pi r^2 = 8.56 *10^18 m^2

So at that distance, you will be receiving how many photons per meter^2 per second?

2.14 * 10^41 / 8.56*10^18 = 2.2*10^28 photons per meter^2 per second.

22,000,000,000,000,000,000,000,000,000 photons per square meter per second and that is some kind of world record for the faintest signal detected in 2005.

I'm kind of curious about the size and shape of antenna used. The wavelength on the 3.5 MHz signal is a whopping 87 meters! And all this comes back to your point that knowing that "it's photons" hardly explains how you can fit 22 bajillion of 87 meter photons in a square meter. :) Or why it takes a "1000 foot Beverage antenna" (about 3 and a half wavelengths) to receive.

Meanwhile AM radio stations operate at 250 watts to 50 thousand watts. I'm not sure, but can orders of magnitude greater power allow orders of magnitude smaller antennas to pick up the signal? I'm still speculating a bit, but isn't it possible that the small hand-held LF receiver is in fact, only receiving part of each photon. That somewhere, for each electron accelerating in the antenna, there is another distant electron accelerating in the opposite direction. Maybe that's why when I stand at a certain position in my room, the radio comes in better, because I'm placing a set of protons/electrons in a position where they can better interact with the protons/electrons in the antenna.
 
  • #20
sophiecentaur
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Whaaat???
A "part of a photon"??
QM says that the photon is the lowest amount of energy associated with any particular frequency. It's a QUANTUM.

If you think that you 'know' that a photon has a size of a wavelength of the radiation then google "size of a photon". It's much much more complicated than you think and there is no reason to think that a photon actually occupies any volume at all. The wavelength of the radiation is an entirely separate issue and refers to the wave treatment of the phenomenon.
How would you ever receive a photon of radio waves with a length of 1500m through a small hole in a metal box if the photon were 1500m long / wide?
It is, in fact, very hard to construct a well screened room. The holes need to be minute fractions of the radiation that you want to screen against. The occasional photon still gets through. Sometimes, a problem makes more sense when you treat it as a wave phenomenon and sometimes it's best to treat it in QM terms. That's the thing about duality. Reconciling the two approaches is hard.

ANTENNAE.
An efficient antenna will often be a whole number of half-wavelengths in size but that doesn't mean that you can't transmit or receive on much smaller antennae. It is true that, if you look at the antenna on someone's chimney, you will usually be able to estimate the wavelength they are receiving by looking at the lengths of the elements. But there's more to it than that.
When talking about 'weak and strong' radio signals, it is also important to consider the amount of information they need to carry. The bandwidth of the signal is just as important as the signal power when deciding whether or not you can receive it satisfactorily. You could never use that signal in your example to transmit 625 line Colour TV.

BANDS.
There's a thing called the Pauli Exclusion Principle. This means that no two Fermions (the particles like electrons, protons etc)) can have the same Quantum Numbers (i.e. the same state) So an isolated atom can have discrete energy states but if you squash a lot of atoms up close, they can't all have the same states and what was a single energy value becomes spread out into a 'band' of energy values, around the original single value.
 
  • #21
JDoolin
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Whaaat???
A "part of a photon"??
QM says that the photon is the lowest amount of energy associated with any particular frequency. It's a QUANTUM.

If you think that you 'know' that a photon has a size of a wavelength of the radiation then google "size of a photon". It's much much more complicated than you think and there is no reason to think that a photon actually occupies any volume at all. The wavelength of the radiation is an entirely separate issue and refers to the wave treatment of the phenomenon.
How would you ever receive a photon of radio waves with a length of 1500m through a small hole in a metal box if the photon were 1500m long / wide?
It is, in fact, very hard to construct a well screened room. The holes need to be minute fractions of the radiation that you want to screen against. The occasional photon still gets through. Sometimes, a problem makes more sense when you treat it as a wave phenomenon and sometimes it's best to treat it in QM terms. That's the thing about duality. Reconciling the two approaches is hard.

ANTENNAE.
An efficient antenna will often be a whole number of half-wavelengths in size but that doesn't mean that you can't transmit or receive on much smaller antennae. It is true that, if you look at the antenna on someone's chimney, you will usually be able to estimate the wavelength they are receiving by looking at the lengths of the elements. But there's more to it than that.
When talking about 'weak and strong' radio signals, it is also important to consider the amount of information they need to carry. The bandwidth of the signal is just as important as the signal power when deciding whether or not you can receive it satisfactorily. You could never use that signal in your example to transmit 625 line Colour TV.

BANDS.
There's a thing called the Pauli Exclusion Principle. This means that no two Fermions (the particles like electrons, protons etc)) can have the same Quantum Numbers (i.e. the same state) So an isolated atom can have discrete energy states but if you squash a lot of atoms up close, they can't all have the same states and what was a single energy value becomes spread out into a 'band' of energy values, around the original single value.
Okay. I see at least part of my error. Let me explain what I was thinking and why it is wrong. I was thinking because of conservation of momentum, there needed to be two particles involved in the receiving of a photon, both accelerated in opposite directions perpendicular to the path of the photon (in the direction of the electric field component of the EM wave).

But really, it only needs one particle at the source and destination of the wave. Actually, that might help me clarify in my own head, a few things I was confused about.

But now that I've made this clarification, I think you said earlier that an individual photon could interact with lots and lots of charged particles, all at once, in a solid. I said, no, just two. Now you're saying "part of a photon" is nonsense, so that would mean it can only interact with one charged particle, I think. Can you clarify?
 
  • #22
sophiecentaur
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As far as I see it, you need to think of interaction with SYSTEMS rather than particles. Even with a Hydrogen atom, the Proton is part of the internal Energy arrangement. It happens to be 2000times more massive, so we sort of ignore it but it is still involved. Also, you can have molecules that have rotational energy, which involves more than one atom and yet which interacts with em radiation.
Wanting "part of a photon" must be a non starter because the quantum size is a property of the electromagnetic radiation but that doesn't mean that a system, involving several particles, can't interact with just one photon's worth of energy. You see, I have no problem with reconciling the idea of a Quantum of Energy not actually needing to be a Particle, as such. I don't know of any phenomenon that needs to be explained by the Photon being an actual particle with a defined extent.

Back to the "part of a photon" thing. If you have two masses held apart by a spring and you give them a bash (a quantum of energy) they will both be vibrating. But where is the energy? It will be in the spring, when compressed and in the motion of the masses, too. It's just 'in the system' and the share of energy keeps changing within the system.
 

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