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Why photon energy is finite?


DimaMazin

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Are you hinting towards the problem of the vaccum energy in QFT?

 

Well, you can only measure energy differences and these are finite. In particular, the energy of the ground state maybe infinite, but by redefining the Hamiltonian by subtracting infinite energy we can get something finite.

 

(You can also see this infinity as a problem with the ordering of the fields)

Edited by ajb
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Are you hinting towards the problem of the vaccum energy in QFT?

 

Well, you can only measure energy differences and these are finite. In particular, the energy of the ground state maybe infinite, but by redefining the Hamiltonian by subtracting infinite energy we can get something finite.

 

(You can also see this infinity as a problem with the ordering of the fields)

I don't understand language of QM. Photon energy is finite relative to field of any body.If photon is gamma photon for us and it travels to very very fast electron then can field of the electron destroy the photon before collision(because energy of the photon is higher than finite energy relative to the electron)?

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I don't understand language of QM. Photon energy is finite relative to field of any body.If photon is gamma photon for us and it travels to very very fast electron then can field of the electron destroy the photon before collision(because energy of the photon is higher than finite energy relative to the electron)?

An electron colliding woth a photon will not destroy the photon. It will scatter, and another photon will emerge. In the rest frame of the photon electron, the photon will have a lower energy.

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An electron colliding woth a photon will not destroy the photon. It will scatter, and another photon will emerge. In the rest frame of the photon electron, the photon will have a lower energy.

And so photon can have infinite energy relative to gravitational field of electron before collision?Does only strong gravitational field turns photon with high energy into mass or into few photons?

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Does only strong gravitational field turns photon with high energy into mass or into few photons?

 

The usual situation is that a gamma photon produces a positron-electron pair near a nucleus. The strong electric field (not gravitation) is necessary, and this is much more efficient with a neavy atom than a light one.

 

The energy of a photon depends on the observer, that's the Döppler effect (the gravitation potential has an influence too). Though, the photon's energy won't get zero nor infinite whatever the other observer's speed. The photon keeps its speed versus any observer and a finite energy, consistent with zero rest mass.

 

I have supposed very vaguely that photons of high energy arriving at a horizon event might help the creation of new particles, a process more efficient than Hawking radiation, but this is uneducated guess. Translate: not mainstream physics. Or shorter: probably crank.

http://www.scienceforums.net/topic/84942-seeded-hawking-radiation/

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It will scatter, and another photon will emerge. In the rest frame of the photon electron, the photon will have a lower energy.

 

Whether it will have more or less energy depends on velocity vector direction.

 

Particle traveling in direction of incoming photon has blueshift Relativistic Doppler effect,

[math]f=f_0*\sqrt{\frac{1+v}{1-v}}=f_0*(1+v)*\gamma[/math]

 

while particle travel in the same direction as photon has redshift Relativistic Doppler effect.

[math]f=f_0*\sqrt{\frac{1-v}{1+v}}=f_0*(1-v)*\gamma[/math]

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Whether it will have more or less energy depends on velocity vector direction.

 

Particle traveling in direction of incoming photon has blueshift Relativistic Doppler effect,

[math]f=f_0*\sqrt{\frac{1+v}{1-v}}=f_0*(1+v)*\gamma[/math]

 

while particle travel in the same direction as photon has redshift Relativistic Doppler effect.

[math]f=f_0*\sqrt{\frac{1-v}{1+v}}=f_0*(1-v)*\gamma[/math]

 

 

I specified the rest frame of the electron. There is no velocity vector to worry about.

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I specified the rest frame of the electron. There is no velocity vector to worry about.

 

??

Without velocity vector you don't know whether do redshift or blueshift..

 

Imagine situation:

the Sun is emitting highly accelerated electron, or other elementary particle, say it has v=0.5c relativistic velocity,

And then a while later the Sun is emitting photon, once photon reaches particle, we have to use redshift calculation to be able predict how particle behaved.

 

Now imagine two binary star system, particle from 1st star is emitted toward 2nd star in system,

and approaching emitted by this star photon, then we have to use blueshift calculation to be able predict how particle behaved.

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"In the rest frame of the electron"

Electron energy can be infinite relative to us . Isn't it? If so then blue shift of photon with finite energy (relative to us) can be infinite in electron frame . Otherwise photon can exist in our frame without existance in the electron frame . Physics is science about mathematical connections of its laws .

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Electron energy can be infinite relative to us . Isn't it? If so then blue shift of photon with finite energy (relative to us) can be infinite in electron frame . Otherwise photon can exist in our frame without existance in the electron frame . Physics is science about mathematical connections of its laws .

 

 

A basic tenet of relativity is that you can analyze a problem in any inertial frame. I chose a convenient one, that's all. But even if you choose another frame, the photon energy is finite, because a massive particle can't travel at c.

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A basic tenet of relativity is that you can analyze a problem in any inertial frame. I chose a convenient one, that's all. But even if you choose another frame, the photon energy is finite, because a massive particle can't travel at c.

 

So theoretically any photon can approach infinite energy by simply picking an inappropriate/useless enough frame…a frame where probably no mass exists at rest, and the excessive energy would have no significance

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So theoretically any photon can approach infinite energy by simply picking an inappropriate/useless enough frame…a frame where probably no mass exists at rest, and the excessive energy would have no significance

Yes. Approach infinity. But then, some photons will be moving in the opposite direction and be redshifted.

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A basic tenet of relativity is that you can analyze a problem in any inertial frame. I chose a convenient one, that's all. But even if you choose another frame, the photon energy is finite, because a massive particle can't travel at c.

Yes,photon energy isn't limited for object with smal mass . May gamma photon be destroyed by our gravitation and long time of existance in it?

Edited by DimaMazin
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Yes,photon energy isn't limited for object with smal mass .

 

It is limited. The photon energy is always finite.

 

May gamma photon be destroyed by our gravitation and long time of existance in it?

Why would a gamma exist for a long time in our gravitation? But no, gravitation is not going to destroy a photon. It will be destroyed by being absorbed by something.

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May gamma photon be destroyed by our gravitation and long time of existance in it?

 

Gamma photon is absorbed by particle, and f.e. two new photons with lower energies are emitted.

Or gamma photon is absorbed by particle, particle accelerated, and new lower energy photon is emitted.

It's cascade like effect. 1 MeV gamma photon could be after 20+ such interactions be split to 1 million photons with 1 eV energy each.

See Compton scattering f.e.

https://en.wikipedia.org/wiki/Compton_scattering

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