Speed of a photon go to zero when it reflects of a mirror

AI Thread Summary
The speed of a photon does not reach zero when it reflects off a mirror at a 90-degree angle; instead, the photon is effectively destroyed and recreated in the reflection process. This interaction highlights the wave-particle duality of photons, where their behavior is influenced by boundary conditions. In particle accelerators, larger circular paths are necessary to manage synchrotron radiation emitted by electrons, which increases with energy and smaller radii. Protons, on the other hand, require larger radii to reduce the strength of magnetic fields needed to maintain their circular trajectory. Overall, the design of particle accelerators balances energy efficiency and the physical properties of the particles being accelerated.
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Light

Does the speed of a photon go to zero when it reflects of a mirror at a 90 degree angle? Just to be sure, I mean if the photons’ comes straight down on a mirror lying flat and reflects straight up, does the photon’s velocity reach zero at any point in time?

also

Particle accelerators:
If the particles travel in a circle at 99.99% of light why do they have to be built so big? Couldn’t they just travel in circles until the distance needed is reached?
 
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You have to remember that light ha a wave-particle duality. So a photon act as a wave. When it reflect off of a mirror it does not change speed.
As for the particle accelerator: Imagine a large circle and a small circle. If you take a sector of the same length of each circle the sector of the larger circle will be flatter/ straighter than the sector of the small circle. Since that such speeds particles will have enormous momentums (mass will increase by the theory of relativity), it will be extremely hard to force the particle to deviate its trajectory form a straight line. In a large circular accelerator this deviation per "sector" is smaller thus less energy is spent. It is thus more efficient to have a large circular accelerator.
 


TeeTeeKnow said:
Light

Does the speed of a photon go to zero when it reflects of a mirror at a 90 degree angle? Just to be sure, I mean if the photons’ comes straight down on a mirror lying flat and reflects straight up, does the photon’s velocity reach zero at any point in time?

That's because you think of a photon as a bullet, which must have some "continuous trajectory". Photons are defined as increments in the excitation state of electromagnetic modes. You can choose whether you pick the free field, or whether you include the boundary conditions (such as those implied by a mirror).

If you take the free field modes, then the mirror introduces an interaction between different free modes. An incoming photon is destroyed, and an outgoing photon is created in a reflection process.

If you include the mirror in the boundary conditions, then the mirror is now part of the EM field description, and the photon corresponds to an electromagnetic mode which already includes reflection. So the photon is a step in this mode.

also

Particle accelerators:
If the particles travel in a circle at 99.99% of light why do they have to be built so big? Couldn’t they just travel in circles until the distance needed is reached?

There are two different reasons to have large radii. If the particles are electrons, then the reason to have a large radius is that electrons emit synchrotron radiation, and the higher the energy, and the smaller the radius, the more synchrotron radiation they emit. That's energy loss which needs to be compensated by RF cavities. For a synchrotron radiation facility, that's perfect, but if you want to reach high energies, it puts a burden on the RF power you need to supply: only a part goes into accelerating the particles, most of it goes into compensating the synchrotron radiation losses. There's a trade off.

If the particles are protons, then at high energies their masses make them want to go on a straight line, and it takes a strong magnetic field to curb them into a circle. Technologically, one is limited to the magnetic field strength one can apply, which then gives you the lowest curvature radius that you can obtain at a given energy.

Electrons are curved easily with magnetic fields, but emit synchrotron radiation, which must be compensated. That's why for electrons, one wants large radius, to limit the power wasted.

Protons almost don't emit synchrotron radiation, but need strong magnetic fields to be curbed. That's why for protons, one wants a large radius, to limit the magnetic field.
 
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