1. 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?  2.

3. Originally Posted by PritishKamat
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.  4. E = p*c
isnt it 1/2p*c?  5. Originally Posted by PritishKamat
E = p*c
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?   6. how's it in relativity?  7. Originally Posted by PritishKamat
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((mc²)²+c²p²)
For a massless object this becomes E = pc
For a massive object this becomes E = γmc²
(this is because for a massive particle p = γmv and when you plug this
into the first equation you get E = mc² sqrt(1+γ²v²/c²) = γmc²)

Thus in general, kinetic energy = sqrt((mc²)²+c²p²) - mc²
For massless particles this becomes kinetic energy = total energy = c p
For a massive particle this becomes kinetic energy = γmc² - mc² = (γ-1)mc²  8. 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]  9. Originally Posted by PritishKamat
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.  10. Originally Posted by PritishKamat
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.  11. 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?! :?  12. Originally Posted by PritishKamat
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.  13. 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.  14. Originally Posted by PritishKamat
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.  15. How come? I still can't digest this fact. By the way, how is time-keeping done in physics experiments on relativity?  16. Originally Posted by PritishKamat
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.  17. I've actually done some reading on relativity since my last post, so I now kinda understand time dilation and simultaneity better. But my actual question is, the thing about clocks going slower, do the hands of a mech watch actually obey relativity? cuz I know that we see time dilation on the basis of observing common events from non inertial frames having different velocities.  18. Originally Posted by PritishKamat
I've actually done some reading on relativity since my last post, so I now kinda understand time dilation and simultaneity better. But my actual question is, the thing about clocks going slower, do the hands of a mech watch actually obey relativity? cuz I know that we see time dilation on the basis of observing common events from non inertial frames having different velocities.
Your question is not clear to me.  19. Okay, one last time, my question is, when we say clocks run slower in Relativity, do they PHYSICALLY run slower, or not?  20. Originally Posted by PritishKamat
Okay, one last time, my question is, when we say clocks run slower in Relativity, do they PHYSICALLY run slower, or not?
No they do not. The physics in all inertial frames is exactly the same and so a watch behaves the same no matter which inertial frame it is in. The only problem is when you try to compare watches in different intertial frames and the result of your comparison depends on what inertial frame YOU are in.

But even though time dilation is a purely relative thing, this doesn't mean that it isn't real. Muons traveling at a high velocity have a measurably slower decay rate because of time dilation.

If this seems contradictory, you need to go back to my post which explains how this contradiction is explained using the relativity of simultaneity. This really is the key as is the lorentz transformations.  21. I'm glad I got my answer . Now, if somebody saw a clock in some other inertial reference frame moving with respect to the observer, would it seem to go slower?
I know my questions maybe silly, but they contribute to my complete understanding relativity properly. By the way, thanks for answering them.   22. Originally Posted by PritishKamat
Now, if somebody saw a clock in some other inertial reference frame moving with respect to the observer, would it seem to go slower?
What do you mean by see? Seeing usually requires the movement of photons from the thing you are seeing to your eyes. This means that what you see will depend on whether the clock is coming towards you or going away from you.

But if you do a calculation to take this into account in order to determine how fast the clock is "really" going, the result is time dilation.

To actually show this in an example, then since we have no relativistic space ships, we must do just the opposite: assume time dilation and then compute the time it takes light to travel to the observer in order to determine what is actually seen.

So for example if B is traveling towards A at 86.6% of the speed of light starting clocks at 0 when B is 10 light seconds from A then lets do the calculation to see what A sees. Using the formula for gamma 86.6% of the speed of light gives a time dilation/length contraction factor of gamma=2.

1. When A's clock reads 0s, B's clock reads 0s at a distance of 10 ls so it takes 10s for the light to get to A and A sees B's clock read 0s when A's clock reads 10s.

2. When A's clock reads 2s, B's clock reads 1s at a distance of 10-2*.866 ls so it takes 8.268s for the light to get to A and A sees B's clock read 1s when A's clock reads 10.268s.

3. When A's clock reads 4s, B's clock reads 2s at a distance of 10-4*.866 ls so it takes 6.536s for the light to get to A and A sees B's clock read 2s when A's clock reads 10.536s.

4. When A's clock reads 6s, B's clock reads 3s at a distance of 10-6*.866 ls so it takes 4.804s for the light to get to A and A sees B's clock read 3s when A's clock reads 10.804s.

5. When A's clock reads 8s, B's clock reads 4s at a distance of 10-8*.866 ls so it takes 3.072s for the light to get to A and A sees B's clock read 4s when A's clock reads 11.072s.

6. When A's clock reads 10s, B's clock reads 5s at a distance of 10-10*.866 ls so it takes 1.34s for the light to get to A and A sees B's clock read 5s when A's clock reads 11.34s.

So while B is traveling towards A, A sees B's clock going 1 second for every .268 seconds of his own which is 3.73 times as fast as his own clock. But lets restart the clocks when B passes A, to see what happens after B passes A

1. When A's clock reads 0s, B's clock reads 0s at a distance of 0 ls so it takes 0s for the light to get to A and A sees B's clock read 0s when A's clock reads 0s.

2. When A's clock reads 2s, B's clock reads 1s at a distance of 2*.866 ls so it takes 1.732s for the light to get to A and A sees B's clock read 1s when A's clock reads 3.732s.

3. When A's clock reads 4s, B's clock reads 2s at a distance of 4*.866 ls so it takes 3.464s for the light to get to A and A sees B's clock read 2s when A's clock reads 7.464s.

4. When A's clock reads 6s, B's clock reads 3s at a distance of 6*.866 ls so it takes 5.196s for the light to get to A and A sees B's clock read 3s when A's clock reads 11.196s.

5. When A's clock reads 8s, B's clock reads 4s at a distance of 8*.866 ls so it takes 6.928s for the light to get to A and A sees B's clock read 4s when A's clock reads 14.928s.

6. When A's clock reads 10s, B's clock reads 5s at a distance of 10*.866 ls so it takes 8.66s for the light to get to A and A sees B's clock read 5s when A's clock reads 18.66s.

So while B is traveling away from A, A sees B's clock going 1 second for every 3.732 seconds on his clock which is only 26.8% as fast as his own clock. But now lets start over at the same distance and take a look at how things look to B. Well first of all when we said they started at a distance of 10 light seconds that is only according to A, because we have to use length contraction, we see that according to B, when B's clock read 0s the distance is only 5 light seconds.

1. When B's clock reads 0s, A's clock reads 0s at a distance of 5 ls so it takes 5s for the light to get to B and B sees A's clock read 0s when B's clock reads 5s.

2. When B's clock reads 1s, A's clock reads .5s at a distance of 5-.866 ls so it takes 4.134s for the light to get to B and B sees A's clock read .5s when B's clock reads 5.134s.

3. When B's clock reads 2s, A's clock reads 1s at a distance of 5-2*.866 ls so it takes 3.268s for the light to get to B and B sees A's clock read 1s when B's clock reads 5.268s.

4. When B's clock reads 3s, A's clock reads 1.5s at a distance of 5-3*.866 ls so it takes 2.402s for the light to get to B and B sees A's clock read 1.5s when B's clock reads 5.402s.

5. When B's clock reads 4s, A's clock reads 2s at a distance of 5-4*.866 ls so it takes 1.536s for the light to get to B and B sees A's clock read 2s when B's clock reads 5.536s.

6. When B's clock reads 5s, A's clock reads 2.5s at a distance of 5-5*.866 ls so it takes .67s for the light to get to B and B sees A's clock read 2.5s when B's clock reads 5.67s.

So while B is traveling towards A, B sees A's clock going .5s for every .134 seconds on his clock which is 3.73 times as fast as his own clock. But lets restart the clocks when B passes A, to see what happens after B passes A

1. When B's clock reads 0s, A's clock reads 0s at a distance of 0 ls so it takes 0s for the light to get to B and B sees A's clock read 0s when B's clock reads 0s.

2. When B's clock reads 1s, A's clock reads .5s at a distance of .866 ls so it takes .866s for the light to get to B and B sees A's clock read .5s when B's clock reads 1.866s.

3. When B's clock reads 2s, A's clock reads 1s at a distance of 2*.866 ls so it takes 1.732s for the light to get to B and B sees A's clock read 1s when B's clock reads 3.732s.

4. When B's clock reads 3s, A's clock reads 1.5s at a distance of 3*.866 ls so it takes 2.598s for the light to get to B and B sees A's clock read 1.5s when B's clock reads 5.598s.

5. When B's clock reads 4s, A's clock reads 2s at a distance of 4*.866 ls so it takes 3.464s for the light to get to B and B sees A's clock read 2s when B's clock reads 7.464s.

6. When B's clock reads 5s, A's clock reads 2.5s at a distance of 5*.866 ls so it takes 4.33s for the light to get to B and B sees A's clock read 2.5s when B's clock reads 9.33s.

So while B is traveling away from A, B sees A's clock going .5s for every 1.866 seconds on his own clock which is only 26.8% as fast as his own clock.

So it is only when A and B correct for the time it takes light to get to them that they actually conclude that the other person's clock is going half as fast as their own clock all the time.  23. One of the things that I think is rather startling about the above example is the mathematical symmetry of the results. Each sees the other's clock as going 3.73 times as fast as his own clock when they are traveling towards each other and each sees the other's clock as going slower than his own clock, by the same factor of 3.73, when they are traveling away from each other. AND YET, and this is what makes it so surprising, this is derived from knowing that the other persons clock is actually always going slower than your own clock by a factor of two whether you are going towards each other or away from each other!

Doesn't anyone else find this surprising?  24. Come on! All the effort to work out a concrete example and there is no comments at all? sheesh! Anybody interested in real physics?  Bookmarks
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