I What accelerates the light back to its constant speed?

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Light does not actually accelerate or decelerate when transitioning between mediums; its speed remains constant at c in a vacuum. When light enters a medium like water, it appears to slow down due to absorption and re-emission processes at the atomic level, which is not visible to the observer. Upon exiting the medium, light resumes its speed in vacuum without requiring additional energy for acceleration. The energy of the light decreases when it enters the medium, and some of that energy is absorbed, which explains the difference in energy levels before and after the transition. Thus, the focus should be on where the energy goes rather than where it comes from.
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TL;DR
What accelerates the light back to its constant speed, when it changes medium?

what energy is used to accelerate the light to its constant speed in vacuum after it exits the drop of water?
For example, let's suppose that there is a drop of water in vacuum. The light travels with the constant speed of light, then enters through the drop of water, decelerates, since the density is different, then it exits back to vacuum, supposedly traveling with the constant speed of light in vacuum.

There is a difference between the speed of the light in the medium of the drop of water, and the vacuum afterwards.

But the question is: what energy is used to accelerate the light to its constant speed in vacuum after it exits the drop of water?
 
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sibir1us said:
TL;DR Summary: What accelerates the light back to its constant speed, when it changes medium?
The medium can change but light does not accelerate or decelerate, its speed is always c.
 
sibir1us said:
what energy is used to accelerate the light to its constant speed in vacuum after it exits the drop of water?
None. It doesn't take any energy, because the light is not "accelerated" after it exits the drop. See further comments below.

javisot20 said:
The medium can change but light does not accelerate or decelerate, its speed is always c.
While this statement is not false, it also leaves out the key missing piece that answers the OP's question.

The key missing piece is: the slower speed of light in a medium is only apparent. At a deeper level, what is happening is that light is being absorbed and re-emitted by the atoms in the medium, and that leads to an apparent slowdown in its speed, because we don't observe the individual absorptions and re-emissions. All we observe is their macroscopic effect: that it takes longer for light to cover a given distance through a medium than it does for it to cover the same distance through a vacuum, because although the light travels at ##c## between each emission and absorption, the emissions and absorptions take additional time. And since we can't see those individual microscopic events, we attribute that longer time it takes to the light traveling at a slower speed.
 
sibir1us said:
TL;DR Summary: What accelerates the light back to its constant speed, when it changes medium?

what energy is used to accelerate the light to its constant speed in vacuum after it exits the drop of water?

For example, let's suppose that there is a drop of water in vacuum. The light travels with the constant speed of light, then enters through the drop of water, decelerates, since the density is different, then it exits back to vacuum, supposedly traveling with the constant speed of light in vacuum.

There is a difference between the speed of the light in the medium of the drop of water, and the vacuum afterwards.

But the question is: what energy is used to accelerate the light to its constant speed in vacuum after it exits the drop of water?
Although you've posted this under quantum physics, you can use classical electromagnetism to explain refraction, reflection and transmission of light waves using Maxwell's equations.(Mentor note: this thread has since been moved out of the QM subforum) to If A light wave travelling through one medium (vacuum, say) reaches the boundary with another medium, then some of the wave is reflected and some continues through the medium. The speed of the wave in the medium is determined by its refractive index. The total energy of the wave does not increase.

The same process happens when the transmitted wave exits the medium. Some of the wave is reflected back into the medium and some is transmitted into the vacuum. The speed of the wave in the vacuum is determined by Maxwell's equations and must be ##c##. But, again, there is no increase in total energy.

In fact, the energy of the incoming wave will be greater than the total energy of the reflected wave plus the final transmitted wave. Some of the energy will have been absorbed by the medium.

The process of reflection and transmission at a boundary is covered in most undergraduate textbooks on classical eletromagentism. E.g. in Griffiths, section 9.3.2.
 
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sibir1us said:
what energy is used to accelerate the light to its constant speed in vacuum after it exits the drop of water?
First, how much energy is required to accelerate a massless particle to ##c##? Once you figure out how much is required then you can start looking for what is missing that amount of energy.
 
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sibir1us said:
But the question is: what energy is used to accelerate the light to its constant speed in vacuum after it exits the drop of water?
I recommend the video below to understand what the change in speed of light in a medium means.



As for energy: The light that leaves the water has less energy, than the light that enters the water. So you should be rather asking where the energy went to, instead of where it came from.
 
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I was using the Smith chart to determine the input impedance of a transmission line that has a reflection from the load. One can do this if one knows the characteristic impedance Zo, the degree of mismatch of the load ZL and the length of the transmission line in wavelengths. However, my question is: Consider the input impedance of a wave which appears back at the source after reflection from the load and has traveled for some fraction of a wavelength. The impedance of this wave as it...

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