Fun With Photons!

[Xamonas Chegwé]

There is actually a corollary to the more commonly known uncertainty principle which links that the uncertainty in the energy of an event to the uncertainty of the time of the event. The more precisely you know when something happens, the less accurately you can know how much energy was involved - and vice versa.

[MiM]

There is actually a corollary to the more commonly known uncertainty principle which links that the uncertainty in the energy of an event to the uncertainty of the time of the event. The more precisely you know when something happens, the less accurately you can know how much energy was involved - and vice versa.

Yes, that was in the link I gave .The minimum uncertainty of the energy of a state is defined by its longevity dE*dT>=hslash/2. Short lived states have bigger width than long lived.

So to have a very precise (very, still not exact) gamma you need to keep a nucleus in a very precise position for a very long time, and just wait for it to decay (a millenia or so...). If yo use a big amount of long lived nuclei, you will not know which of them decayed, and your information on the origin of the position is lost, unless you measure it, and then we are back to the question of the interference of the measurement... Stating that you know the exact wavelength of a photon, is in my mind very close to stating that you know the exact velocity of a massive particle.

I am far from convinced, that the difficulties in assigning a position to a photon (until observed), can be boiled down to your simple calculations around the uncertainty principle.

[Xamonas Chegwé]

There is actually a corollary to the more commonly known uncertainty principle which links that the uncertainty in the energy of an event to the uncertainty of the time of the event. The more precisely you know when something happens, the less accurately you can know how much energy was involved - and vice versa.

Yes, that was in the link I gave .The minimum uncertainty of the energy of a state is defined by its longevity dE*dT>=hslash/2. Short lived states have bigger width than long lived.

So to have a very precise (very, still not exact) gamma you need to keep a nucleus in a very precise position for a very long time, and just wait for it to decay (a millenia or so...). If yo use a big amount of long lived nuclei, you will not know which of them decayed, and your information on the origin of the position is lost, unless you measure it, and then we are back to the question of the interference of the measurement... Stating that you know the exact wavelength of a photon, is in my mind very close to stating that you know the exact velocity of a massive particle.

I am far from convinced, that the difficulties in assigning a position to a photon (until observed), can be boiled down to your simple calculations around the uncertainty principle.

I never said it was right!

It was just some thinking I was doing that seemed to make sense and, importantly, it agrees with the Feynmann sum-over-histories approach to photon paths - ie. that a particle takes every possible path between its origin and point of interaction. How I got there appeared logical but, given the nature of quantum physics, is almost certainly far too simplistic!

∆s∆a ≥ h (The uncertainty in simplicity, times the uncertainty in accuracy, can never be less than the reduced Planck constant!)

[MiM]

∆s∆a ≥ h (The uncertainty in simplicity, times the uncertainty in accuracy, can never be less than the reduced Planck constant!)

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[mistermack]

On the subject of E = mc2 only applying to particles at rest - this is usually the case. However, it does apply to photons in a vacuum. I skipped any attempt at justification as I was trying to present a simplified, easy to follow argument.

A photon has no rest mass, so E = mc2 can't apply directly to a photon, whether in a vacuum or not.
If a photon had a rest mass, it would take infinite energy to accelerate it to c.
E = mc2 can apply indirectly, in that if a body emits a photon, it will lose a mass of E/c2, where E is the energy of the photon. And conversely, if a body absorbs a photon, it will gain that amount of mass.

I can't see E = mc2 applying to anything travelling at c, as nothing with a rest mass CAN travel at c.

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[Xamonas Chegwé]

On the subject of E = mc2 only applying to particles at rest - this is usually the case. However, it does apply to photons in a vacuum. I skipped any attempt at justification as I was trying to present a simplified, easy to follow argument.

A photon has no rest mass, so E = mc2 can't apply directly to a photon, whether in a vacuum or not.
If a photon had a rest mass, it would take infinite energy to accelerate it to c.
E = mc2 can apply indirectly, in that if a body emits a photon, it will lose a mass of E/c2, where E is the energy of the photon. And conversely, if a body absorbs a photon, it will gain that amount of mass.

I can't see E = mc2 applying to anything travelling at c, as nothing with a rest mass CAN travel at c.

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You missed my point. It applies to photons in a vacuum travelling at c because they have no rest mass.

reduces to E=mc2 for photons because rest mass (m0) = 0

They only possess relativistic mass while moving.

[mistermack]

You missed my point. It applies to photons in a vacuum travelling at c because they have no rest mass.

reduces to E=mc2 for photons because rest mass (m0) = 0

They only possess relativistic mass while moving.

OK. That actually reduces to Ek = MC2, where Ek is the kinetic energy of the photon.
The traditional understanding of E = MC2 is where E is the rest energy of a body, ie, energy not including kinetic energy.

[Xamonas Chegwé]

You missed my point. It applies to photons in a vacuum travelling at c because they have no rest mass.

reduces to E=mc2 for photons because rest mass (m0) = 0

They only possess relativistic mass while moving.

OK. That actually reduces to Ek = MC2, where Ek is the kinetic energy of the photon.
The traditional understanding of E = MC2 is where E is the rest energy of a body, ie, energy not including kinetic energy.

"E=mc2 either means E=m0c2 for an object at rest, or E=mrelc2 when the object is moving."

http://en.wikipedia.org/wiki/Mass-energ ... istic_mass

[mistermack]

"E=mc2 either means E=m0c2 for an object at rest, or E=mrelc2 when the object is moving."

http://en.wikipedia.org/wiki/Mass-energ ... istic_mass

Yes, I can see where the cross purposes came from. If they stuck to one definition of mass, it would make things clearer.
Just below that line, they write :

When modern physicists say "mass", they are usually talking about rest mass, since if they meant "relativistic mass", they would just say "energy".

It's amazing that a supposedly precise science should use a term like mass so loosely.

[Xamonas Chegwé]

"E=mc2 either means E=m0c2 for an object at rest, or E=mrelc2 when the object is moving."

http://en.wikipedia.org/wiki/Mass-energ ... istic_mass

Yes, I can see where the cross purposes came from. If they stuck to one definition of mass, it would make things clearer.
Just below that line, they write :

When modern physicists say "mass", they are usually talking about rest mass, since if they meant "relativistic mass", they would just say "energy".

It's amazing that a supposedly precise science should use a term like mass so loosely.

Not really, it reflects the evolution of science. In Newtonian physics, mass is simply mass (and was assumed invariant) - it is only when you overlay Einsteinian relativity that imprecision creeps in. But by then, the term was well established, so subscripts are used where the meaning is ambiguous. The same is true in a lot of maths, where the same symbols and names have come to mean very different things and must be used contextually for precision.

[Gawdzilla Sama]

Spoiler that shit!