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| PritishKamat |
 Posted: Mon May 12, 2008 10:56 pm Post subject: Particle nature |
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I have a question: what differentiates photons which make up different wavelengths of light?
I know, the probable answer is the energy, E=hf (f=frequency), that the photons possess. Now answer this: where does this energy come from, considering that photons have no mass ( no mass energy and no kinetic energy)?
Also, since photons have zero mass, they have zero momentum and so, infinite De' Broglie wavelength. But this isn't the fact. why does de' broglie's equation fail here?
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| Dishmaster |
| Posted: Mon May 12, 2008 11:44 pm Post subject: Re: Particle nature |
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| PritishKamat wrote: |
I have a question: what differentiates photons which make up different wavelengths of light?
I know, the probable answer is the energy, E=hf (f=frequency), that the photons possess. Now answer this: where does this energy come from, considering that photons have no mass ( no mass energy and no kinetic energy)?
Also, since photons have zero mass, they have zero momentum and so, infinite De' Broglie wavelength. But this isn't the fact. why does de' broglie's equation fail here? |
De Broglie's equation does not fail. You just have to abandon the idea that momentum (impulse) is always related to a mass. In quantum mechanics it is not.
The momentum has been confirmed in many experiments and is an important ingredient in many astrophysical phenomena (radiation pressure). It is interesting that there is a bond between relativity and quantum mechanics here, because the relativistic expression for its energy is:
E = p*c
With de Broglie's equation you get: E = h*c/lambda = h*nu
The energy of the photons can have its origin in many different processes. But initially, they are emitted by atoms that reduce their energy level. You surely know that atoms consist of a nucleus and a "cloud" of electrons. Now, whenever one electron reduces its energy state to a lower orbit around the nucleus, the energy difference is emitted as a photon. This is a result of the fact that photons are the transmitter of the fundamental electro-magnetic force. In this example, it was used to keep the electron on a higher orbit around the positively charged nucleus and therefore "invisible". It was "liberated" as soon as it was not "needed" anymore. |
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| PritishKamat |
| Posted: Tue May 13, 2008 12:19 am Post subject: |
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isnt it 1/2p*c?
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| Dishmaster |
| Posted: Tue May 13, 2008 4:29 am Post subject: |
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| PritishKamat wrote: |
isnt it 1/2p*c? |
Nope. This is the total relativistic energy of a massless particle. Perhaps you are vaguely remembering the expression of the kinetic energy?
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| PritishKamat |
| Posted: Tue May 13, 2008 8:54 am Post subject: |
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how's it in relativity?
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| mitchellmckain |
| Posted: Tue May 13, 2008 1:37 pm Post subject: |
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| PritishKamat wrote: |
| how's it in relativity? |
Kinetic energy is the difference between the total energy of a moving object and the rest mass energy.
The total energy is given by E = sqrt((mc˛)˛+c˛p˛)
For a massless object this becomes E = pc
For a massive object this becomes E = γmc˛
(this is because for a massive particle p = γmv and when you plug this
into the first equation you get E = mc˛ sqrt(1+γ˛v˛/c˛) = γmc˛)
Thus in general, kinetic energy = sqrt((mc˛)˛+c˛p˛) - mc˛
For massless particles this becomes kinetic energy = total energy = c p
For a massive particle this becomes kinetic energy = γmc˛ - mc˛ = (γ-1)mc˛
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| PritishKamat |
| Posted: Tue May 13, 2008 11:19 pm Post subject: |
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But, isn't gamma infinity for a massive particle travelling at the speed of light? [ I hope the gamma you're referring to is the factor commecting rest mass and mass at speed of light]
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| Dishmaster |
| Posted: Wed May 14, 2008 2:04 am Post subject: |
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| PritishKamat wrote: |
| But, isn't gamma infinity for a massive particle travelling at the speed of light? [ I hope the gamma you're referring to is the factor commecting rest mass and mass at speed of light] |
This is gamma:
It is equal to one for no velocity and infinity for v=c. |
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| mitchellmckain |
| Posted: Wed May 14, 2008 7:43 am Post subject: |
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| PritishKamat wrote: |
| But, isn't gamma infinity for a massive particle travelling at the speed of light? [ I hope the gamma you're referring to is the factor commecting rest mass and mass at speed of light] |
Massive particles never travel at the speed of light and massless particles never travel at anything but the speed of light. There is never any reason to calculate gamma for a particle going at the speed of light, except to say that for a massive object to accelerate to the speed of light would require an infinite amount of energy, so it is not possible.
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| PritishKamat |
| Posted: Wed May 14, 2008 9:29 am Post subject: |
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got it. But one more question about relativity.
Time dilation in frames having some relative is often explained using the light and mirror experiment with two observers, one in the train with the apparatus and the other on the platform, both recording time intervals between two events. Well, now say they are A & B respectively.
Now, say both have similar, brand new , watches. the time difference between the two events can now be measured by their watches instead of applying any mathematics.
Wont these readings be the same? They should be, shouldn't they? considering they both see photons striking the mirrors at the same time.
The funny thing is, if the readings are same, on mathematical calculations, we see that speed of light depends on reference frame. Now how did this happen?! 
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| mitchellmckain |
| Posted: Wed May 14, 2008 10:06 am Post subject: |
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| PritishKamat wrote: |
got it. But one more question about relativity.
Time dilation in frames having some relative is often explained using the light and mirror experiment with two observers, one in the train with the apparatus and the other on the platform, both recording time intervals between two events. Well, now say they are A & B respectively.
Now, say both have similar, brand new , watches. the time difference between the two events can now be measured by their watches instead of applying any mathematics.
Wont these readings be the same? They should be, shouldn't they? considering they both see photons striking the mirrors at the same time.
The funny thing is, if the readings are same, on mathematical calculations, we see that speed of light depends on reference frame. Now how did this happen?! |
Well there are a lot of details left out in your description so I don't know exactly what you are describing, but the fact is that watches at rest in two different inertial frames are always going to measure different time intervals between the same two events if those events are separated by a spatial distance in the same direction as the relative velocity between those two intertial frames.
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| PritishKamat |
| Posted: Wed May 14, 2008 10:22 am Post subject: |
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well, if you haven't heard, the experiment goes like this :
2 observers, one in the train (A) at constant non zero velocity and one on the platform (B) try to measure the time difference between two same events.
A has with him, two mirrors and a light beam is oscillating (reflecting from each mirror) between them. Now, as both, A and B look at the light beam, A sees the light beam in a straight line, but B sees it moving in a V shape.
Now, the time difference between the two events is measured by; t=dist/C
since the dist is more for B, t is more and so, time dilation is observed.
Now, my argument goes like this:
Consider the two observers wearing identical watches and they calculate time difference according to their watches. now, my question is, wont the readings be the same? considering that both see photons striking the mirror simultaneously.
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| mitchellmckain |
| Posted: Wed May 14, 2008 6:27 pm Post subject: |
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| PritishKamat wrote: |
well, if you haven't heard, the experiment goes like this :
2 observers, one in the train (A) at constant non zero velocity and one on the platform (B) try to measure the time difference between two same events.
A has with him, two mirrors and a light beam is oscillating (reflecting from each mirror) between them. Now, as both, A and B look at the light beam, A sees the light beam in a straight line, but B sees it moving in a V shape.
Now, the time difference between the two events is measured by; t=dist/C
since the dist is more for B, t is more and so, time dilation is observed.
Now, my argument goes like this:
Consider the two observers wearing identical watches and they calculate time difference according to their watches. now, my question is, wont the readings be the same? considering that both see photons striking the mirror simultaneously. |
But they dont see the photons striking the mirror simultaneously. What is simultaneous in one inertial frame is not simultaneous in another inertial frame if the events are separated by a distance in the direction of the relative velocity between the two frames.
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| PritishKamat |
| Posted: Wed May 14, 2008 11:25 pm Post subject: |
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How come? I still can't digest this fact. By the way, how is time-keeping done in physics experiments on relativity?
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| mitchellmckain |
| Posted: Thu May 15, 2008 9:53 am Post subject: |
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| PritishKamat wrote: |
| How come? I still can't digest this fact. By the way, how is time-keeping done in physics experiments on relativity? |
That is too bad because this relativity of simultaneity is the key to everything!
Without that, time dilation doesn't make any sense at all, BECAUSE IT IS RELATIVE!!!
If a ship travels away from the earth at 86.6% of the speed of light, the people on the earth will think that the clocks on the ship are going 1/2 as fast as their own clocks, BUT..., and this is the KICKER, people on the ship will think that it is the clocks on the EARTH which are going 1/2 as fast as their own clocks.
SO WHOSE CLOCKS are actually going slower???
Well the answer is that this is completely the WRONG question. It is really the case that NEITHER is going slower or both is going slower if you like and what makes it all make sense in the end is the relativity of simultaneity. To explain, let the guy on the ship be called A and the guy on the earth be called B.
Let's imagine that the all the points of space has clocks at rest with respect to B and synchronized according to B. Then according to A they are not only all like B's watch, one second passing for every two seconds on his own watch, but they don't even show the same time. Those ahead of him in the direction he is moving with respect to B are ahead of the ones behind, and the farther ahead he looks the farther ahead is the time which those clocks read.
Now suppose instead of looking at a single clock (or his own watch), A just takes the time from the nearest clock he happens to be passing (lets call this "passing clock" time). What he will get is a reading of time which is not slower but faster -- in fact, for second of this "passing clock" time, his own watch will only show only half a second has passed.
Now consider what happens if A slows down to be at rest with respect to B. What happens to all the clocks at all the points of space at rest with respect to B? The time they read all change until they all agree with the clocks that are nearest A's position. The result is that B's watch also changes until it agrees with the "passing clock" time, and that means that A will now come to the conclusion that B was correct in saying that A's watch was slower than B's. But the only reason B turned out to be correct is that A changed his velocity to match that of B. If it was B who accelerates to match the velocity of A, the opposite happens -- namely it seems that the time shown on A's watch jumps ahead so that B finds out that it was actually his own watch that was running slower than A's watch.
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