Ask a physicist anything. (7)

[Upisoft]

I don't know. I was thinking more in terms of refraction. Naively, it seems likely to me that reflection can be traced to a single event, but it doesn't always happen at the exact same depth in the material.

Well, it could be not a single event. That would not surprise me at all.

Well, this is just basic wave mechanics. It is fundamentally impossible to physically localize a wave which only consists of one wavelength. This is, in fact, the origin of the Heisenberg Uncertainty Principle: the more localized a wave is in space, the less localized it is in frequency (which is related to momentum).

If you have continuous beam of light between the source and the detector that does not hold. If you start pulsing the source, then I could agree. In fact continuous beam of light is spread across all the possible points a photon could be. Only the effect of detection (which is absorption) makes the beam to stop. You don't think that putting a detector in a beam of light would have effect on the light that have not reached the detector, do you? Thus putting a limit of how far the photons can travel should not change their energy/wavelength before they hit the target. Yes it changes the wavelength from X to 0 when absorption effect takes place at the target.

I don't get what you're saying here.

I'm saying that photons should behave in the same way between points A and B, regardless if they are going to travel until the end of time or are going to be observed at point C, that is after points A and B.

 

[Chalnoth]

If you have continuous beam of light between the source and the detector that does not hold.

I don't see why you'd think that. Localizing a photon, in any direction, requires summing up different wavelengths.

You don't think that putting a detector in a beam of light would have effect on the light that have not reached the detector, do you? Thus putting a limit of how far the photons can travel should not change their energy/wavelength before they hit the target.

It depends, but typically no, this won't happen, because the individual photons are far more localized than the length of the beam.

 

[Upisoft]

I don't see why you'd think that. Localizing a photon, in any direction, requires summing up different wavelengths.

That is equivalent to say you can change the frequency of a photon by putting a detector on the path of a photon, while the fact of existence of a photon and its disappearance is pretty good explanation for the form of the wave. If the photons existed forever, then you'd need your Fourier transform, but fortunately they can be absorbed.

It depends, but typically no, this won't happen, because the individual photons are far more localized than the length of the beam.

Yeah, just what I said above. Thus you can have photons with the same energy in packets. And as they are bosons it is more probable they will be with the same energy. Since it is probability based and there is an effect of amplification, LASERs usually have very little number of photons with different energy than the main energy of emission, but there is nothing that forbids existence of photon packets with the same energy and no other energy. You may not know this energy exactly, but you may know it is the same energy for all photons in the packet. There is nothing preventing that knowledge.

 

[Chalnoth]

That is equivalent to say you can change the frequency of a photon by putting a detector on the path of a photon, while the fact of existence of a photon and its disappearance is pretty good explanation for the form of the wave. If the photons existed forever, then you'd need your Fourier transform, but fortunately they can be absorbed.

Why do you think those two things are equivalent?

Yeah, just what I said above. Thus you can have photons with the same energy in packets. And as they are bosons it is more probable they will be with the same energy. Since it is probability based and there is an effect of amplification, LASERs usually have very little number of photons with different energy than the main energy of emission, but there is nothing that forbids existence of photon packets with the same energy and no other energy. You may not know this energy exactly, but you may know it is the same energy for all photons in the packet. There is nothing preventing that knowledge.

Except that the beam is localized, so it can't be perfectly-constrained in energy. This can probably be understood as due to the energy-time uncertainty relationship, which means that unstable states do not have an absolute energy: they have a width in energy that is related to the time it takes that state to decay. Since lasers are emitted from unstable states returning to stable ones, they aren't going to be perfect delta functions in energy.

 

[Upisoft]

Why do you think those two things are equivalent?

Because localizing a beam of light would, in your opinion will change the energy of the photons. And I guess you believe shorter the beam, more dispersed the energy of the photons will be.... But you are not going to have gamma rays just by making the length of the beam very short.

Except that the beam is localized, so it can't be perfectly-constrained in energy.

You mean it can't be perfectly measured...

This can probably be understood as due to the energy-time uncertainty relationship, which means that unstable states do not have an absolute energy: they have a width in energy that is related to the time it takes that state to decay. Since lasers are emitted from unstable states returning to stable ones, they aren't going to be perfect delta functions in energy.

Huh? Why do you think that something that decays spontaneously does not have absolute energy? How conservation of energy is going to work otherwise?

 

[Chalnoth]

Because localizing a beam of light would, in your opinion will change the energy of the photons. And I guess you believe shorter the beam, more dispersed the energy of the photons will be.... But you are not going to have gamma rays just by making the length of the beam very short.

The photons are localized long before they strike the detector. They are localized by the physics of how they are emitted.

But by the way, yes, if you put a detector within a wavelength of the emitter, that detector is going to have a significant effect upon how the photons are emitted.

Huh? Why do you think that something that decays spontaneously does not have absolute energy? How conservation of energy is going to work otherwise?

As I said, it's due to the energy-time uncertainty relationship, described here:
Uncertainty principle - Wikipedia, the free encyclopedia

Nevertheless, Einstein and Bohr understood the heuristic meaning of the principle. A state that only exists for a short time cannot have a definite energy. To have a definite energy, the frequency of the state must accurately be defined, and this requires the state to hang around for many cycles, the reciprocal of the required accuracy.

For example, in spectroscopy, excited states have a finite lifetime. By the time-energy uncertainty principle, they do not have a definite energy, and each time they decay the energy they release is slightly different. The average energy of the outgoing photon has a peak at the theoretical energy of the state, but the distribution has a finite width called the natural linewidth. Fast-decaying states have a broad linewidth, while slow decaying states have a narrow linewidth.

As you can't get photon emission from a stable state (by definition), no photon can be emitted in a precise eigenstate of energy.

 

[Upisoft]

The photons are localized long before they strike the detector. They are localized by the physics of how they are emitted.

But by the way, yes, if you put a detector within a wavelength of the emitter, that detector is going to have a significant effect upon how the photons are emitted.

Why only "within a wavelength"? Don't you think the effect, smaller or even negligible, must exist in other cases?

As I said, it's due to the energy-time uncertainty relationship, described here:
Uncertainty principle - Wikipedia, the free encyclopedia

What makes you think then that so called "stable" states do have definite energy? After all they are limited in time from the big bang...

As you can't get photon emission from a stable state (by definition), no photon can be emitted in a precise eigenstate of energy.

You can, it is called virtual photon.

 

[Chalnoth]

Why only "within a wavelength"? Don't you think the effect, smaller or even negligible, must exist in other cases?

Well, obviously the effect is going to drop off extremely rapidly once you go beyond the localization size of the photons, which is likely to be a few wavelengths.

What makes you think then that so called "stable" states do have definite energy? After all they are limited in time from the big bang...

This is why the energy-time uncertainty principle is an inequality, not an equality. Unstable states must have some width in energy. Stable states don't necessarily need to. The energy-time uncertainty principle only sets a lower bound on the uncertainty. But in practice, the more stable a state is, the sharper-peaked its energy.

You can, it is called virtual photon.

Virtual photons don't escape. So they can't be said to be truly emitted.

 

[Upisoft]

Well, obviously the effect is going to drop off extremely rapidly once you go beyond the localization size of the photons, which is likely to be a few wavelengths.

But still, even if the effect drops off very rapidly it is still present. What changes is our ability to measure it. Right?

This is why the energy-time uncertainty principle is an inequality, not an equality. Unstable states must have some width in energy. Stable states don't necessarily need to. The energy-time uncertainty principle only sets a lower bound on the uncertainty. But in practice, the more stable a state is, the sharper-peaked its energy.

Thus the life time of the universe makes all states we consider stable slightly dispersed. With time passing stable states will be more definite. Damn... that means at the time=0+epsilon, the energy state of the universe must have been very undefined. LOL. It have to be called the Big What? theory....

Virtual photons don't escape. So they can't be said to be truly emitted.

How do you explain shielding then? You can shield something from surrounding magnetic field. Thus you can make virtual photons to be absorbed in different place. But without emission there is no absorption.

 

[Chalnoth]

But still, even if the effect drops off very rapidly it is still present. What changes is our ability to measure it. Right?

To a degree. But by the time the stuff that happens one meter further away changes the behavior of the photons by less than one part in 10^100, we really don't care any longer, do we? The falloff of these effects tends to drop off exponentially-fast with distance.

Though it is actually true that the behavior of everything in the universe must effect everything else, and that is through gravity. My bet is that is going to vastly dominate the effect of the extent of the wave functions, though we would need a theory of quantum gravity to say for sure.

Thus the life time of the universe makes all states we consider stable slightly dispersed. With time passing stable states will be more definite. Damn... that means at the time=0+epsilon, the energy state of the universe must have been very undefined. LOL. It have to be called the Big What? theory....

I don't think that's correct. I don't think the energy width changes due to the age of the system. Not that it matters in practice: once you reach durations of even one second, the energy width is positively minuscule (5e-35 J, or 3e-16 eV).

How do you explain shielding then? You can shield something from surrounding magnetic field. Thus you can make virtual photons to be absorbed in different place. But without emission there is no absorption.

I don't understand what you're trying to say here. As I said, virtual photons are not emitted: an emitted photon is, by definition, a real photon. Virtual photons (and other particles) instead act as sort of intermediaries between other particles. Some even think they are nothing more than artifacts of a particular representation of the mathematics, and have no real existence whatsoever. I tend to lean on them being real in a sense, but the point remains: you can't produce a real photon from a stable state. Stable states cannot emit anything. Otherwise they would be losing energy, which would violate their definition as stable states.

 

[Tuddrussell]

Wikipedia describes Vibranium, a fictional marvelverse metal, as having "The ability to absorb all vibrations in the vicinity as well as kinetic energy directed at it.http://en.wikipedia.org/wiki/Vibranium#cite_note-8 The energy absorbed is stored within the bonds between the molecules that make up the substance."

What would the practical applications of such a metal be? Does this metal break any "rules" physics-wise? Would such a metal do anything interesting?

 

[Chalnoth]

Wikipedia describes Vibranium, a fictional marvelverse metal, as having "The ability to absorb all vibrations in the vicinity as well as kinetic energy directed at it.http://en.wikipedia.org/wiki/Vibranium#cite_note-8 The energy absorbed is stored within the bonds between the molecules that make up the substance."

What would the practical applications of such a metal be? Does this metal break any "rules" physics-wise? Would such a metal do anything interesting?

Well, I could think of two ways that a metal might absorb vibrations:
1. The metal could vibrate itself.
2. The metal could dampen the vibrations to produce heat.

The first option is pretty simple and straightforward, because it just means that the metal transmits sound, which metals tend to do very well. It doesn't really "store" the energy, however. The wave just passes right on through. So I don't think this is useful.

The second option could work. You'd need a material that was very good at dissipating incoming pressure waves of a variety of frequencies (it's probably impossible for a material to dissipate waves of *all* frequencies). By dissipating the waves, their energy is turned into heat. This sort of material seems very plausible to me, but I'm not really sure what it could be used for, except the obvious use of dampening vibrations. Sound waves typically don't contain enough energy to really do a whole lot. But at least the material would make for a rather good sound insulator.

 

[Upisoft]

I don't think that's correct. I don't think the energy width changes due to the age of the system. Not that it matters in practice: once you reach durations of even one second, the energy width is positively minuscule (5e-35 J, or 3e-16 eV).

We are constantly finding out that some things we considered stable are not stable. Bismuth-209 was found to be radioactive in 2003 and considered stable before that. Its half life time is over billion times larger than the current number we put for the age of the universe. There are isotopes with even higher half-life than that. Thus the energy width should be unimaginable for such events. If you had only one atom of Bismuth-209 you may never know it was unstable.

Anyway how you can be certain that the universe had definite energy at the big bang event? The time it existed was so short that deltaE should had extremely high value. So, if you know what energy in the universe is, i.e. deltaE, you will know what is the smallest deltaT possible.

Otherwise they would be losing energy, which would violate their definition as stable states.

Why bidirectional transfer of energy to be considered "losing energy"? There is some energy in the field, it does not disappear.

 

[Chalnoth]

We are constantly finding out that some things we considered stable are not stable.

That's just a change in our knowledge, not a change in physical reality. But like I said, the line width for something that has a mean lifetime of even one second is absurdly small. Let alone a process which takes billions of years.

Anyway how you can be certain that the universe had definite energy at the big bang event? The time it existed was so short that deltaE should had extremely high value. So, if you know what energy in the universe is, i.e. deltaE, you will know what is the smallest deltaT possible.

I didn't say it did. What I did say is that the age has nothing to do with it. It's the mean transition time that is important, not the age.

Why bidirectional transfer of energy to be considered "losing energy"? There is some energy in the field, it does not disappear.

That isn't emission.

 

[Upisoft]

That's just a change in our knowledge, not a change in physical reality. But like I said, the line width for something that has a mean lifetime of even one second is absurdly small. Let alone a process which takes billions of years.

I didn't say it did. What I did say is that the age has nothing to do with it. It's the mean transition time that is important, not the age.

So you take some arbitrary time and say that everything that happens faster than that is not stable. In the case of radioactive decay what do you consider "transition time"?

That isn't emission.

Then it is another process that transfers energy from a particle to the field. But still it is a process that transfers energy.

 

[Chalnoth]

So you take some arbitrary time and say that everything that happens faster than that is not stable.

Huh?

In the case of radioactive decay what do you consider "transition time"?

It's the mean lifetime. Related to the half-life by a factor of ln(2).

Then it is another process that transfers energy from a particle to the field. But still it is a process that transfers energy.

No, it really isn't. A stable system generates a constant field. There is no transfer. The energy of the field is part of the energy of the system.

 

[pgp_protector]

Q. Big Bang or Big Splat?

 

[Tuddrussell]

If you were given the chance to explore a fictional world in person, like say the Star Wars universe or Narnia or whathave you, which genre would be most interesting from a physicist's perspective and why?

Sword and sorcery? Sci-fi? Steampunk? Religious?

 

[Wiccan_Child]

If you were given the chance to explore a fictional world in person, like say the Star Wars universe or Narnia or whathave you, which genre would be most interesting from a physicist's perspective and why?

Sword and sorcery? Sci-fi? Steampunk? Religious?

From a physical point of view, I'd like to experiment with the gun from Portal -have one portal above the other in a tube, pour water down the tube, attach to a generator, and bam, eternal free energy. Perhaps.

But I'd love to experiment with magic, be it in Dungeons & Dragons or The Elder Scrolls or even Pokémon... ah, we can but dream

 

[Lillen]

How do a soccerplayer spin the ball over 102 cm from its original course when kicking the ball. To me it seems like a impossibility to screw a ball, no matter it's baseball or soccer!