Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Mr Andrew on 06/09/2007 01:29:29
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What is the speed of light from the reference frame of a photon? Do other photons appear to travel at the speed of light with respect to them or do they appear stationary? What if they are aimed in opposite directions? If light travels at the speed of light with respect to light then how do photons interact because the one photon must look like a point to the first but how does interference work then (which can't be between a point and a wave, only two waves)?
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Would the photon exist at all or would it be energy? This here is question doesn't come up much around the bunkhouse but perhaps some of the other members more lucid on theoretical physics will turn up out of the woodwork soon. It is the beginning of the fall term in most place right now.
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I vaguely recollect something (will have to get my brain back into gear to verify it) but at the speed of light I believe time stops (one of the reasons why you cannot exceed the speed of light).
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What is the speed of light from the reference frame of a photon? Do other photons appear to travel at the speed of light with respect to them or do they appear stationary? What if they are aimed in opposite directions? If light travels at the speed of light with respect to light then how do photons interact because the one photon must look like a point to the first but how does interference work then (which can't be between a point and a wave, only two waves)?
It doesn't exist.
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you can find very good sources about it on the following links
http://www.whatusearch.net/Science/Physics/
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My reasoning behind this question stemmed from the idea that light waves appear to us as points (photons) and behave like ideal gases because they are moving at the speed of light (where time is zero), thus by the equation d = rt that everyone learns in elementary or middle school, the length should be zero (d = c*0 = 0). But since light has to move at the 'speed of light' in every reference frame one might ask if it should to another photon. However, reference frames can't exist at 'the speed of light' so that rule doesn't apply. Another way to think about it is that if a light wave moved at c relative to another light wave, then relative to it, that other wave would be a point. Interference can't happen between a point and a wave, only two waves, so light must appear as a wave to another light wave (so they can interact in an interference pattern).
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one photon fired all by itself will experience interference waves. and i believe from the frame of reference of a photon, if the photon was traveling in the same direction, it would appear stationary, and if it was in the opposite direction it would appear as traveling at the speed of light.
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In a vacuum one normal sort of photon is not "aware" of other photons because they do not interact. Also because they are massless and travel at the speed of light photons are not "aware" of time a photon that is created is only aware that it has been absorbed again whether it is a picosecond or a billion years.
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So, how does it 'become aware' of another photon when it interferes with it?
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So, how does it 'become aware' of another photon when it interferes with it?
Photons intended as point-like travelling particles in space are meaningless.
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My reasoning behind this question stemmed from the idea that light waves appear to us as points (photons) and behave like ideal gases because they are moving at the speed of light (where time is zero), thus by the equation d = rt that everyone learns in elementary or middle school, the length should be zero (d = c*0 = 0). But since light has to move at the 'speed of light' in every reference frame one might ask if it should to another photon. However, reference frames can't exist at 'the speed of light' so that rule doesn't apply. Another way to think about it is that if a light wave moved at c relative to another light wave, then relative to it, that other wave would be a point. Interference can't happen between a point and a wave, only two waves, so light must appear as a wave to another light wave (so they can interact in an interference pattern).
This statement imply you can find a ref. frame stationary with respect to a light wave, but it doesn't exist.
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Two photons interfering with each other are NOT interacting they are both there it is just that the energy at particular locations is zero because the two photons null each other out at that location. To have a null at one location it is essential that there is a peak at some other location.
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Isn't nulling each other out imply that the photons are interacting?
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Isn't nulling each other out imply that the photons are interacting?
All your reasonings are good for EM waves. Photons are something else.
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...aren't photons just EM waves? I mean...if they move at the speed of light relative to us, shouldn't they appear contracted to a point, in other words a photon?
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...aren't photons just EM waves? I mean...if they move at the speed of light relative to us, shouldn't they appear contracted to a point, in other words a photon?
No, photons are not EM waves. This sound strange, but photons are MUCH more strange, believe me. No one still knows what photons "really" are, apart from "quantums of electromagnetic field".
We can talk as we like about electromagnetic wave packets and discuss their properties and behaviour, but they are NOT photons.
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Can you please explain the difference between "quantums of electromagnetic field" and "electromagnetic wave packets?" They sound the same to me.
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There is no difference between photons and electromagnetic waves and there is nothing really mysterious about them the photon just defines the smallest quantity of energy that we can have in a wave at a particular frequency.
The fact that photons are nulling each other at one point and reinforcing each other at another does not imply that the photons themselves are in any way changed by this process and they are not changed by this process so therefore they are not interacting in the nuclear physics sense. For the photons to interact there must be some form of non linear material present that causes them to be absorbed, change frequency or create other particles like electron positron pairs
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There is no difference between photons and electromagnetic waves
I wish it was! If what you say were true, then a photon should have:
1) Position in space
2) Dimensions
A wave packet does have position and dimensions; a photon doesn't have.
3) Wave properties only. But a photon has particle properties also.
A photon can be associated to an EM wave, but it's not it.
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So, in addition to generating an electromagnetic wave, an electron dropping energy levels emits a photon. Light then consists of both an electromagnetic wave and a photon (which itself has wave and particle properties?). Am I interpreting your post correctly? So, by your reasoning, is it possible to have an electromagnetic wave or a photon alone, without the other?
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No it is not they are one and the same thing, that is what is meant by the wave particle duality. To measure a photon of electromagnetic radiation you always have to destroy it. wherther you measure it as a wave or a particle depends on the type of measuring technique you use.
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Ok, that's what I thought. Can you clarify a point here for me?...if an EM wave is moving at the speed of light, then shouldn't it appear as a point to us? So is it possible then for photons just to be the particle representation (due to our observations from an inertial refrence frame) of EM waves? For all practical purposes this is the same as what you have stated, but it is interesting, no?
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Ok, that's what I thought. Can you clarify a point here for me?...if an EM wave is moving at the speed of light, then shouldn't it appear as a point to us?
You switch on a lamp for 1.5 seconds then you switch off it. The EM wave packet is long more than Earth-Moon distance. Why should it be a point? If, instead, the lamp is moving towards you, than that distance is reduced; easy to compute how much. If the lamp would move towards you at (almost) light speed, then the wave packet lenght would become (almost) zero. You don't have a finite lenght object which then is accelerated to light speed: you have a finite lenght object already travelling at light speed.So is it possible then for photons just to be the particle representation (due to our observations from an inertial refrence frame) of EM waves?
What do you mean exactly?
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Whoa!, hold on...I thought that light from the lamp consisted of more than one wave-packet. Each atom in the tungsten filament (assuming of course that it's an incandescent bulb) emitted a seperate wave packet. Each one of those is due to an electron losing energy (which then gets converted into a wave-packet). Those electrons don't take 1.5 seconds to lose that energy so how can one wave-packet be that long? I can see the stream of wave-packets being that long but each individual wave-packet (which is moving at light speed) should be contracted to a point then as a consequence of SR.
To clarify your question lightarrow, I mean, is a photon simply a wave-packet length-contracted to a point as a result of SR? This makes sense to me but I do not know all there is to know about E&M and as I'm only in highschool, you certainly know more about it than me. I want to understand the modern viewpoint, which I'm assuming you hold as a modern scientist, and reconcile it with what I see in the equations and explanations I am given in school. Once I understand physics as it is now, I can begin to understand it in a new way, thus making the jump from scientist to Research Scientist.
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Whoa!, hold on...I thought that light from the lamp consisted of more than one wave-packet.
You can define a wave packet as any finite waveform. Each single jump of an electron within one of the light's atoms will create a tiny wavepacket. Since there's many atoms all being excited at the same time, and this goes on for 1.5 seconds, you can add all these tiny wavepackets together to get one big wavepacket that lasts for 1.5 seconds. (Waves in quantum mechanics and in E&M can be added like this.) So you can think of it as one long wavepacket consisting of the addition of a lot of small wavepackets.
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Whoa!, hold on...I thought that light from the lamp consisted of more than one wave-packet. Each atom in the tungsten filament (assuming of course that it's an incandescent bulb) emitted a seperate wave packet. Each one of those is due to an electron losing energy (which then gets converted into a wave-packet). Those electrons don't take 1.5 seconds to lose that energy so how can one wave-packet be that long? I can see the stream of wave-packets being that long but each individual wave-packet (which is moving at light speed) should be contracted to a point then as a consequence of SR.
No. The electron generates an EM wave that already travels at the speed of light. Read again my previous post: it's not an object of finite lenght which is then accelerated to light's speed, so you cannot talk about Lorentz contraction, here.To clarify your question lightarrow, I mean, is a photon simply a wave-packet length-contracted to a point as a result of SR? This makes sense to me but I do not know all there is to know about E&M and as I'm only in highschool, you certainly know more about it than me. I want to understand the modern viewpoint, which I'm assuming you hold as a modern scientist, and reconcile it with what I see in the equations and explanations I am given in school. Once I understand physics as it is now, I can begin to understand it in a new way, thus making the jump from scientist to Research Scientist.
Apart your mistake of lenght contraction that I've showed you, there is another, more subtle, mistake: you still want to see a photon as a wave packet. A photon IS NOT a wave packet, nor a particle travelling from source to detector. This is common knowledge among physicists.
Now I add something more, but this is only my idea of a photon; I tell you it, to make you understand why photons are not obvious objects as we usually think:
my idea of a photon is "the quantized interaction between electromagnetic field and detector". Where do you see a moving particle (or wave packet) here?
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Ok, I must have been taught incorrectly in school because I always thought a photon represented the particle nature of light. So, a photon isn't really a particle, or a wave, it's just the symbolic name for the interaction between light and an electron (the detector) which can only occur if the light has an energy equal to any number of quantized values (energy levels). Understandable. And now the big question...Why does light behave both as a particle and a wave? That's what I've been trying to answer by applying the lorentz contraction to light (which I now understand is inapplicable to light because it would involve changing the wavelength to zero and thus making the energy infinite).
Ok, on to another question...since an oscillating electric charge creates an electric wave, shouldn't an oscillating mass create a gravitic wave? How does this wave interact with mass? Is it's absorbtion quantized, similar to a photon's.
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Ok, I must have been taught incorrectly in school because I always thought a photon represented the particle nature of light.
But this is indeed true, it's only that "the particle nature of light" has this meaning: when you detect light, its energy comes in packets, as if there was a travelling particle with that discrete energy. What we are sure is this discretization of energy, but it's not correct, in my opinion, to associate it to a real travelling particle; this hasn't been proved yet. So, a photon isn't really a particle, or a wave, it's just the symbolic name for the interaction between light and an electron (the detector) which can only occur if the light has an energy equal to any number of quantized values (energy levels). Understandable. And now the big question...Why does light behave both as a particle and a wave? That's what I've been trying to answer by applying the lorentz contraction to light (which I now understand is inapplicable to light because it would involve changing the wavelength to zero and thus making the energy infinite).
Light behaves both as particle and wave depending on the kind of the apparatus you use to measure its properties: if you look at its particle-like properties, it behaves as particle; if you look at its wave-like properties, it behaves as wave. It seems curious, but it's true! Also, shouldn't this fact make you think that light properties are not really objective, but depend on how do you look at them?Ok, on to another question...since an oscillating electric charge creates an electric wave, shouldn't an oscillating mass create a gravitic wave?
Yes, only that quadrupole oscillations are needed in that case.How does this wave interact with mass? Is it's absorbtion quantized, similar to a photon's.
It should be (according to logic and intuition), but no one can answer clearly to this question yet, since a quantistic theory of gravitation doesn't exist yet.
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Ah, ok. Now, suppose that atoms did not absorb light in quantized amounts (had continuous energy bands) then would the light hitting these atoms be percieved as having particle-like properties?
What do you mean by 'quadrupole oscillations?'
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Ah, ok. Now, suppose that atoms did not absorb light in quantized amounts (had continuous energy bands) then would the light hitting these atoms be percieved as having particle-like properties?
Yes, if the light's frequency is high enough to produce Compton scattering. In that effect, light interacts with electrons bumping them off the atom and computations shows that everything can be explained as if light were made of particles. The key-point however is still in the interaction between light and electrons (free or bound in atoms or inside a metal etc.)What do you mean by 'quadrupole oscillations?'
Let's make the example with electrostatics, which is more simple. You have many point charges, of both polarities, in a specific region of space. If you want to compute the electric field generated by this group of charges away from it, you can add these terms:
1) the total charge of the group (algebric sum of all the charges)
2) the total dipole moment; it's the fact that a sub-total + charge can be slightly displaced from a sub-total - charge (example: only two charges, one + and one -)
3) the total quadrupole moment; it's a more complex unbalancing of charges
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etc.
A system of charges can have total charge = 0 but total dipole moment ≠ 0; or charge = dipole moment = 0 but quadrupole moment ≠ 0, and so on.
When we compute the electromagnetic radiation generated by a system of moving charges, we see that we can sum the dipole moment variation, then the quadrupole moment variation...and so on.
The vantage of this procedure is that every term we add to the sum is less than the preceding one, so we can stop our sum up to the desired precision of the result.
General Relativity is much more complex than electrodynamics; it turns out that gravitational radiation (gravitational waves) cannot be produced by dipole moment variations (oscillations) only, as it is in electrodynamics, but they need at least quadrupole oscillations. Don't ask me why, however.