Twin Paradox 101

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote> An atom can absorb a bit of the energy from a spreading wave<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
The spreading of the wave imply a 1/r^2 factor in the energy. I don't see that factor in E = h nu. If you take a laser light, you don't see the beam spreading. I think the spreading shell is not a good analogy to describe the propagation of photons. Somebody told me that you can imagine a photon like a packet of waves...

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[jacques]: The spreading of the wave imply a 1/r^2 factor in the energy.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

No, it implies 1/r^2 in the intensity, which is what we see. The energy is a function of frequency. A high-frequency wave (say, x-ray frequency) will be just as penetrating regardless of intensity, with the latter determining how much damage it can do.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>If you take a laser light, you don't see the beam spreading.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

On the contrary, laser beams do spread with distance traveled. Laser beams bounced off the Moon have a spread of several kilometers by the time they return.

Picture a spherical water wave spreading in a pond. If the wave comes to a gateway, only a small portion of the whole spherical wave, a portion as big as the gateway, passes through. On the other side, you see what looks like a linear wave with the width of the gateway and very litle additional spread. Lasers work in a similar way, by allowing only a portion of spherical wavefronts to pass through a gateway.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>I think the spreading shell is not a good analogy to describe the propagation of photons. Somebody told me that you can imagine a photon like a packet of waves...<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

If you did that, then the packets should get farther and farther apart within any single wavefront as the packet of waves travels. But light does not "thin out" in that way. The wavefronts remain continuous, but their amplitude (intensity) keeps decreasing. -|Tom|-

OK for the laser.
What is your definition of intensity?
It must be something by unit area. What is the something?

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>What is your definition of intensity? It must be something by unit area. What is the something?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Intensity is power per unit area. It is the square of wave amplitude. For light, this is usually equated with brightness. Intensity can be high or low for waves of either high or low frequency (energy).

Power, of course, is like work (energy) per unit time. -|Tom|-

Let a source emit 1 photon of energy E=h nu. The source is an exited atom let say of H. The H atom have an electron on the second orbital and the electron emit a photon when he jump to a lower level.
The photon propagate like wave and the energy is constant but spreaded all over the sphere. At some distance their an other H atom which get excited. His electron is in the lower state and jump to the second orbital. This take the same amount of energy as the energy emitted by the first atom. But the energy is spreaded all over the sphere. How can this work?

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Let a source emit 1 photon of energy E=h nu. The source is an exited atom let say of H. The H atom have an electron on the second orbital and the electron emit a photon when he jump to a lower level.
The photon propagate like wave and the energy is constant but spreaded all over the sphere. At some distance their an other H atom which get excited. His electron is in the lower state and jump to the second orbital. This take the same amount of energy as the energy emitted by the first atom. But the energy is spreaded all over the sphere. How can this work?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Sorry. I'm dealing with the "speed of gravity" press release at the moment. Read up a bit about the "photoelectric effect". It answers the exact questions you pose. Light does not behave as classical theory expects. But this strange behavior has explanations in both the photon (particle) model and the wave model (although the latter is not in many textbooks).

If you don't answer your own questions, get back to me in a few days, when things settle down again. -|Tom|-