generaly we consider that E=hv (v=frequency of light)
what I mean is there any loss in energy of light when it travel via glass or other material having heigher optical density .if it does then how?
When light travels through air, then through glass, and then reemerges into air, the final frequency is the same as before it entered the glass (and the color is unchanged). Hence the photon energy has not changed. I’m not sure what the proper description is for the photon when inside the glass, but clearly there is no energy change.
if there is no variation in energy
then
there will be defect in Einstine
E=mc2
because,we have so far consider that speed of light does not hold constant for different medium of different optical density,so by above equation there will be variation in energy E
in other case,speed of light will be same for every medium.
Please read my blog article again. Its subject is that the speed of light is always c = 3e8 m/s. Also, note that in E = hf, the frequency f is the same in the air and in the glass. As for E = mc^2, that famous equation is not a general one: it just describes the energy content of an object of nonzero mass m that is at rest, and a photon is never at rest. For objects with mass, the correct relationship between mass and energy is E = gamma*mc^2, where gamma is 1/sqrt(1-v^2/c^2), and even this more general equation is irrelevant for massless particles (m=0) such as the photon, for which the most one can say is that E = pc, where p is the momentum. For particles with nonzero mass, one can say E^2 β (pc)^2 = (mc^2)^2.
thank u sir for your kind information.if you don’t mind,can you please discribe me about the phenomina change and about permitivity of a particular substance wrt light speed
The latest issue (March 2013) of the American Journal of Physics has a very good article that has tangential bearing on this topic. “There are no particles, there are only fields”, is about as accessible a description I’ve seen that outlines the lessons of quantum field theory on the question of field/particle duality. The title almost says it all, but in the above discussion it fits with the field descriptors of the situation implying that thinking of this propagation in terms of photons may be a bit of a red herring.
It puzzles me that you don’t mention the group velocity of light when discussing such a fundamental topic. It can be equal to the phase velocity if the medium is not dispersive. So it is c/n , in this case, which means the energy of the light pulse is delayed compared to the free space propagation case. And, clearly, the measurement of the time delay in the experiment you proposed must prove that. So it seems βthe speed of light in a medium with refractive index n is 3e8/n m/sβ is true or pretty close to that. Correct me if I’m wrong.
If you suddenly turn on a single-frequency sine wave, downstream of the material you will first see light arrive at the speed of c = 3e8 m/s. It is only after many cycles that a steady state is established in which there is a phase shift at the observation location that is consistent with what would happen if light moved through the material at a speed of c/n. The speed of light is everywhere and always 3e8 m/s, but the superposition of all the fields produced by all the accelerated charges, the original ones and those in the material, makes in the steady state a situation in which the phase advances through the material with a speed of c/n. The first energy to arrive downstream, before the steady state is established, arrives at the speed of 3e8 m/s.
Well I agree with you on the question of phase speed. But it is still unclear what you mean by the the first energy to arrive downstream. If we have a short pulse which is convinient to use for measuring time delay, then it has a certain shape, an envelope of the wave which propagates through space at a group velocity. And this velocity can easily be equal to c/n or close to that, which means the slab of glass actually delays the light pulse comparing with its propagation in free space. Sorry, I still cannot see where I’m wrong.
Consider a very detailed mechanistic view. You want to make a short pulse, which we can do by accelerating a point charge upwards from rest and then decelerate it to rest. This pulse will propagate downstream where it encounters a slab of glass, placed so that a normal to the glass passes through the source charge location. The electric field accelerates charges in the glass, mainly the electrons, because the nuclei are very massive. These accelerated electrons (re)radiate. At an observation location downstream (along a line to the source charge that is perpendicular to the slab) we observe an electric field that is the superposition of the radiation from the original source charge and the radiation from the electrons in the glass. With the exception of the small number of electrons lying exactly on the line connecting source and observation locations, all re-radiation from electrons in the glass is retarded compared to the original radiation, because the distance from source to off-axis electron in the glass to the observation location is longer than the direct path. This means that the first nonzero electric field is observed at a time determined by 3e8 m/s.
Continuing to observe after first noting a nonzero electric field, the shape of the pulse will be different from the original pulse because there is a transient behavior of the bound electrons in the glass which is often modeled by an electron bound to the atom by a spring-like force, with damping. Of course the usual term for this is dispersion, that the phase velocity depends on the frequency.
I’ve deliberately avoided using the terms “phase velocity” or “group velocity” to make the point as clearly as I know how. This doesn’t mean that these concepts aren’t useful, but the fundamental physics of the situation is that the speed of light is 3e8 m/s.
Thanks for your extensive reply. I think I begin to understand your point. Theses are just two different ways to look at the same situation. One way is to say that a light pulse travels at a group velocity an is therefore slowed down by a slab of glass, because this velocity is less than c in the glass. This is the conception I had in mind initially. But your point is that light always travels at 3e8 m/s and the reason why we observe the delay after it passed through a slab of glass is that at the observation location we have a superposition of the original wave and the wave from the accelerated charges in the glass. These two waves interfere destructively or constructively at different moments so that it appears to us that the light pulse was delayed.
In order for these two approaches to give the same result we must suppose that the original and the secondary wave are in antiphase at the first moment when they arrive at the observation location. And they stop canceling out each other after a delay which corresponds to the delay we can obtain from the group velocity conception. Makes sense to me. Is this what you meant?
No, that’s definitely not what I meant. I say again that in the case of the momentarily accelerated/decelerated charge, the first appearance of a nonzero field at the observation location takes place at a time that is d/3e8 seconds after the start of the acceleration, where d is the distance between source charge and observation location, for the reasons I gave. There is NO delay in the first appearance of a nonzero field. With or without the glass slab, the time when you first notice a field is d/3e8 seconds. What the glass slab does is to change E(t) at the observation location (the pulse shape), due to reradiation, to be different from what it would be without the slab.
In the case of turning on pure sinusoidal acceleration of the source charge, after a transient, in the steady state, the timing of maxima of E at the observation location is the same that you would get if light traveled at a speed c/n through the glass, and this effect is due to the superposition of the field you would get without the glass and the field due to reradiation.
generaly we consider that E=hv (v=frequency of light)
what I mean is there any loss in energy of light when it travel via glass or other material having heigher optical density .if it does then how?
When light travels through air, then through glass, and then reemerges into air, the final frequency is the same as before it entered the glass (and the color is unchanged). Hence the photon energy has not changed. I’m not sure what the proper description is for the photon when inside the glass, but clearly there is no energy change.
if there is no variation in energy
then
there will be defect in Einstine
E=mc2
because,we have so far consider that speed of light does not hold constant for different medium of different optical density,so by above equation there will be variation in energy E
in other case,speed of light will be same for every medium.
Please read my blog article again. Its subject is that the speed of light is always c = 3e8 m/s. Also, note that in E = hf, the frequency f is the same in the air and in the glass. As for E = mc^2, that famous equation is not a general one: it just describes the energy content of an object of nonzero mass m that is at rest, and a photon is never at rest. For objects with mass, the correct relationship between mass and energy is E = gamma*mc^2, where gamma is 1/sqrt(1-v^2/c^2), and even this more general equation is irrelevant for massless particles (m=0) such as the photon, for which the most one can say is that E = pc, where p is the momentum. For particles with nonzero mass, one can say E^2 β (pc)^2 = (mc^2)^2.
thank u sir for your kind information.if you don’t mind,can you please discribe me about the phenomina change and about permitivity of a particular substance wrt light speed
No, sorry; I haven’t the time to reply, as this is a very complex subject. See an advanced E&M textbook for such discussions.
The latest issue (March 2013) of the American Journal of Physics has a very good article that has tangential bearing on this topic. “There are no particles, there are only fields”, is about as accessible a description I’ve seen that outlines the lessons of quantum field theory on the question of field/particle duality. The title almost says it all, but in the above discussion it fits with the field descriptors of the situation implying that thinking of this propagation in terms of photons may be a bit of a red herring.
Thanks much for your comment, Joel. I found the AJP article to be very interesting indeed.
It puzzles me that you don’t mention the group velocity of light when discussing such a fundamental topic. It can be equal to the phase velocity if the medium is not dispersive. So it is c/n , in this case, which means the energy of the light pulse is delayed compared to the free space propagation case. And, clearly, the measurement of the time delay in the experiment you proposed must prove that. So it seems βthe speed of light in a medium with refractive index n is 3e8/n m/sβ is true or pretty close to that. Correct me if I’m wrong.
If you suddenly turn on a single-frequency sine wave, downstream of the material you will first see light arrive at the speed of c = 3e8 m/s. It is only after many cycles that a steady state is established in which there is a phase shift at the observation location that is consistent with what would happen if light moved through the material at a speed of c/n. The speed of light is everywhere and always 3e8 m/s, but the superposition of all the fields produced by all the accelerated charges, the original ones and those in the material, makes in the steady state a situation in which the phase advances through the material with a speed of c/n. The first energy to arrive downstream, before the steady state is established, arrives at the speed of 3e8 m/s.
Well I agree with you on the question of phase speed. But it is still unclear what you mean by the the first energy to arrive downstream. If we have a short pulse which is convinient to use for measuring time delay, then it has a certain shape, an envelope of the wave which propagates through space at a group velocity. And this velocity can easily be equal to c/n or close to that, which means the slab of glass actually delays the light pulse comparing with its propagation in free space. Sorry, I still cannot see where I’m wrong.
Consider a very detailed mechanistic view. You want to make a short pulse, which we can do by accelerating a point charge upwards from rest and then decelerate it to rest. This pulse will propagate downstream where it encounters a slab of glass, placed so that a normal to the glass passes through the source charge location. The electric field accelerates charges in the glass, mainly the electrons, because the nuclei are very massive. These accelerated electrons (re)radiate. At an observation location downstream (along a line to the source charge that is perpendicular to the slab) we observe an electric field that is the superposition of the radiation from the original source charge and the radiation from the electrons in the glass. With the exception of the small number of electrons lying exactly on the line connecting source and observation locations, all re-radiation from electrons in the glass is retarded compared to the original radiation, because the distance from source to off-axis electron in the glass to the observation location is longer than the direct path. This means that the first nonzero electric field is observed at a time determined by 3e8 m/s.
Continuing to observe after first noting a nonzero electric field, the shape of the pulse will be different from the original pulse because there is a transient behavior of the bound electrons in the glass which is often modeled by an electron bound to the atom by a spring-like force, with damping. Of course the usual term for this is dispersion, that the phase velocity depends on the frequency.
I’ve deliberately avoided using the terms “phase velocity” or “group velocity” to make the point as clearly as I know how. This doesn’t mean that these concepts aren’t useful, but the fundamental physics of the situation is that the speed of light is 3e8 m/s.
Thanks for your extensive reply. I think I begin to understand your point. Theses are just two different ways to look at the same situation. One way is to say that a light pulse travels at a group velocity an is therefore slowed down by a slab of glass, because this velocity is less than c in the glass. This is the conception I had in mind initially. But your point is that light always travels at 3e8 m/s and the reason why we observe the delay after it passed through a slab of glass is that at the observation location we have a superposition of the original wave and the wave from the accelerated charges in the glass. These two waves interfere destructively or constructively at different moments so that it appears to us that the light pulse was delayed.
In order for these two approaches to give the same result we must suppose that the original and the secondary wave are in antiphase at the first moment when they arrive at the observation location. And they stop canceling out each other after a delay which corresponds to the delay we can obtain from the group velocity conception. Makes sense to me. Is this what you meant?
No, that’s definitely not what I meant. I say again that in the case of the momentarily accelerated/decelerated charge, the first appearance of a nonzero field at the observation location takes place at a time that is d/3e8 seconds after the start of the acceleration, where d is the distance between source charge and observation location, for the reasons I gave. There is NO delay in the first appearance of a nonzero field. With or without the glass slab, the time when you first notice a field is d/3e8 seconds. What the glass slab does is to change E(t) at the observation location (the pulse shape), due to reradiation, to be different from what it would be without the slab.
In the case of turning on pure sinusoidal acceleration of the source charge, after a transient, in the steady state, the timing of maxima of E at the observation location is the same that you would get if light traveled at a speed c/n through the glass, and this effect is due to the superposition of the field you would get without the glass and the field due to reradiation.
All right. I see your point. Many thanks for finding time to explain! π