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The nature of force
20 years 3 months ago #10914
by Jim
Replied by Jim on topic Reply from
So then the common definition of a photon as a packet or particle of energy equal to E=hf is wrong in MM? Can the MM photon be understood in terms like the ones usually used when the term is used. Does MM reject Planck's Constant since it is the corner stone of QM?
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- tvanflandern
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20 years 3 months ago #10278
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />So then the common definition of a photon as a packet or particle of energy equal to E=hf is wrong in MM?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Just the opposite. The energy packets (hf) are singlet waves. This explains why the energy depends on the frequency, which could not be true if the energy packets were pure particles. (Frequency is a wave property.) In QM, light is some sort of dual particle-wave entity, not carefully defined. In MM, this vagueness goes away. Light is a pure wave, and non-standing waves of all kinds carry both energy and momentum. -|Tom|-
<br />So then the common definition of a photon as a packet or particle of energy equal to E=hf is wrong in MM?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Just the opposite. The energy packets (hf) are singlet waves. This explains why the energy depends on the frequency, which could not be true if the energy packets were pure particles. (Frequency is a wave property.) In QM, light is some sort of dual particle-wave entity, not carefully defined. In MM, this vagueness goes away. Light is a pure wave, and non-standing waves of all kinds carry both energy and momentum. -|Tom|-
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20 years 3 months ago #10217
by Jim
Replied by Jim on topic Reply from
So, except for details both MM & QM use the same photon value given by the statement: E=hf. Is that right? You also say there is momentum in photons and add mass by mixing in E=mc^2 to make the statements balance. So, why is it the photon has zero rest mass? I suppose both MM & QM hold that idea.
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20 years 3 months ago #10219
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />So, except for details both MM & QM use the same photon value given by the statement: E=hf. Is that right?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes, but for clarity MM would prefer to say "lightwave" instead of "photon" to avoid the implication that the energy packet is localized in space instead of along an entire wavefront. Just as for a water wave, every wavefront has energy that depends on its frequency.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">You also say there is momentum in photons and add mass by mixing in E=mc^2 to make the statements balance. So, why is it the photon has zero rest mass? I suppose both MM & QM hold that idea.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Whether photons have mass in QM, or not, is still debated, although the consensus view is that they do not.
In MM, a wavefront is not normally characterized by a "mass" because its extent parallel to the front is arbitrary. In other words, would the "mass" represent one millimeter of wave along the wavefront, or a million kilometers along the same wavefront? I merely pointed out that waves are not mystical, but are ultimately composed of particles that do the actual energy and momentum transfers.
For light, those constituent particles are named "elysons" in MM. But the mass of an elyson is as arbitrary to a lightwave as the mass of an air molecule is arbitrary to the force of a wind gust, even though the wind's thrust is ultimately dependent of the mass of air molecules. But since there are something like 10^20 air molecules in a cubic centimeter of air, we do not attempt to count molecules to determine the thrust of the wind. We just treat the total air mass by average statistical properties such as speed, density, temperature, and barometric pressure. We use the same kinds of average statistical approximations for the gross properties of lightwaves, so we are not normally concerned with the mass of elysons. -|Tom|-
<br />So, except for details both MM & QM use the same photon value given by the statement: E=hf. Is that right?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes, but for clarity MM would prefer to say "lightwave" instead of "photon" to avoid the implication that the energy packet is localized in space instead of along an entire wavefront. Just as for a water wave, every wavefront has energy that depends on its frequency.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">You also say there is momentum in photons and add mass by mixing in E=mc^2 to make the statements balance. So, why is it the photon has zero rest mass? I suppose both MM & QM hold that idea.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Whether photons have mass in QM, or not, is still debated, although the consensus view is that they do not.
In MM, a wavefront is not normally characterized by a "mass" because its extent parallel to the front is arbitrary. In other words, would the "mass" represent one millimeter of wave along the wavefront, or a million kilometers along the same wavefront? I merely pointed out that waves are not mystical, but are ultimately composed of particles that do the actual energy and momentum transfers.
For light, those constituent particles are named "elysons" in MM. But the mass of an elyson is as arbitrary to a lightwave as the mass of an air molecule is arbitrary to the force of a wind gust, even though the wind's thrust is ultimately dependent of the mass of air molecules. But since there are something like 10^20 air molecules in a cubic centimeter of air, we do not attempt to count molecules to determine the thrust of the wind. We just treat the total air mass by average statistical properties such as speed, density, temperature, and barometric pressure. We use the same kinds of average statistical approximations for the gross properties of lightwaves, so we are not normally concerned with the mass of elysons. -|Tom|-
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20 years 3 months ago #10221
by Jim
Replied by Jim on topic Reply from
Well then the elyson has mass but that mass is different for different packets of energy? Either there is mass or there ain't mass at some scale or at all scales. In QM the issue is treated just as you are doing.
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- Larry Burford
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20 years 3 months ago #10916
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
Jim,
What is the mass of a sound wave? Middle C, for instance.
What is the mass of a sound wave? Middle C, for instance.
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