A Really Big Bang?

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21 years 5 months ago #5802 by tvanflandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[SpaceMan]: If a photon is a collection of (freely moving?) particles that collide with one another, then what about the 2-slit defraction patterns? What's binding those particles together to act as a cohesive quantum unit?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The 2-slit experiment is standard wave behavior. When one slit is covered, a single wave gets through the slit, so no diffraction bands are produced. When both slits are open, waves pass through both which then reinforce or interfere at different locations on the screen. Where the waves cancel, no photoelectrons are ejected. Where they combine, the probability of a photoelectron ejection becomes much higher.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Are saying you can reduce the intensity (brightness)of a single photon?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Intensity is wave amplitude squared. If there is but a single wave, its amplitude (wave height) can still take on any value. Note that the photoelectric effect depends on frequency, not intensity. -|Tom|-


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21 years 5 months ago #6154 by SpaceMan
Replied by SpaceMan on topic Reply from Tyler Keys
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>If there is but a single wave, its amplitude (wave height) can still take on any value.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>No, that's not true. A photon is a single wave and it can not change it's amplitude. It's energy is simply h<img src=icon_nu.gif border=0 align=middle>. Photons only differ in frequency. There is a thing called a quantum amplitude , but that refers to the PROBALITY of the photon hitting the screen somewhere. If a photon could change its amplitude then you could have bright photons and dim photons but that's not what we see out there.

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21 years 5 months ago #6023 by tvanflandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[SpaceMan]: A photon is a single wave and it can not change it's amplitude.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

It is easier in any discussion to disagree about opinions. But when parties disagree about the facts, one must be correct and the other incorrect. There is no middle ground.

Pick up any introductory physics text that covers wave theory. (Resnick & Halliday is my favorite; but then, my sources are dated in comparison with many that are now available. I also have Lerber & Trigg's <i>Encyclopedia of Physics</i> on hand, and it says essentially the same thing.)

Turn first to the definition of "amplitude" in wave motion. (I think you may have confused it with wavelength, when in reality there is no connection with wavelength.) You will find words to the effect that, for all transverse waves in space such as light, amplitude measures from the wave center to the wave's peak height in the direction perpendicular to the direction of propagation. It is a free parameter, and can be indefinitely large or indefinitely small for a wave of any wavelength (l) or frequency (nu), where l x nu = c.

Next, turn to the discussion of intensity. My book says (p. 416): "In a three-dimensional wave such as a light wave or a sound wave from a point source, it is more significant to speak of the <i>intensity</i> of the wave. Intensity is defined as the power transmitted across a unit area normal to the direction in which the wave is traveling. ... The intensity of a space wave is always proportional to the square of the amplitude."

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>It's energy is simply h<img src=icon_nu.gif border=0 align=middle>.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

True. But energy and intensity are unrelated. Any frequency of wave (specifying a particular energy) can have any intensity and corresponding amplitude.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Photons only differ in frequency.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Incorrect. Light is a <b>transverse</b> wave, which requires two parameters to describe; e.g., wavelength and amplitude. (Actually, three are required if polarization is considered too.)

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>If a photon could change its amplitude then you could have bright photons and dim photons but that's not what we see out there.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Again, you are confusing energy and power per unit area when these two are independent properties.

Look on this as an opportunity to improve your own understanding of nature, or to teach me something new (if you can show that I'm the one that learned it wrongly). IMO, misconceptions like this are one of the greatest obstacles to correct theorizing about nature. We all need to work hard on rooting them out. -|Tom|-


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21 years 5 months ago #6218 by SpaceMan
Replied by SpaceMan on topic Reply from Tyler Keys
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Pick up any introductory physics text that covers wave theory.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>Try "An Introduction to Quantum Physics" by French and Taylor. I'm looking at the index page:
"Amplitudes, see Probability amplitude; Projection amplitudes; Quantum amplitudes."
You're quoting from a first year text book that's discussing the macroscopic properites of waves (yes, including light). The wave properties you're refering to, such as intensity and amplitude, simply don't EXIST in the quantum world. I just checked in the index for "intensity", but not surprisingly, there's nothing there either. This is not an oversight by French and Taylor. These properites simply have no meaning when looking at individual particles.
A photon is a quantum particle. You can't have big photons and little photons. They come in different colors, but only one size.

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21 years 5 months ago #5939 by Jim
Replied by Jim on topic Reply from
The way the photon is assumed to exist is E=hf. But, isn't that the Planck bundle? The bundle and the photon are not the same thing. The bundle needs to be researched to see how it came into existance from the prior understanding(before Planck)

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21 years 5 months ago #5940 by tvanflandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[SpaceMan]: The wave properties you're refering to, such as intensity and amplitude, simply don't EXIST in the quantum world.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

As I said, when peeple differ about facts...

Those two properties exist for all waves. They measure the wave height. Modern QM has abandoned the essential wave nature of light in favor of a physically undefined "dual" nature of photons. In MM, light is a pure wave, and wave explanations exist for the two experiments (photoelectric effect and Compton effect) that supposedly show particle-only properties.

Your writing suggests that you also prefer a wave explanation of light. But you can't have it both ways. For exanple, how would you propose to explain the photoelectric effect, and why it is independent of intensity? -|Tom|-

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