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JayKo
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is hv as we all know, but how do with derive it from Einstein's special theory of relativity? any pointer to help me kickstart off? thanks ;)
JayKo said:is hv as we all know, but how do with derive it from Einstein's special theory of relativity? any pointer to help me kickstart off? thanks ;)
It doesn't come from anything. When you will have found its derivation (as long with the value of h) you will won the Nobel prize.Avodyne said:[itex]E=h\nu[/itex] comes from quantum mechanics, not special relativity, and is true for all particles, not just photons. (Photons are special because they can be built up into a coherent electromagnetic wave whose frequency is directly measurable.)
nicksauce said:If you REALLY want to use special relativity, you can start with p=h/lambda, and then use E = sqrt(p^2c^2 - m^2c^4), and then use m=0.
JayKo said:Energy of a single photon is hv as we all know, but how do with derive it from Einstein's special theory of relativity?
Beto Pimentel said:We don't, I am afraid.
Avodyne said:[itex]E=h\nu[/itex] comes from quantum mechanics, not special relativity
lightarrow said:It doesn't come from anything. When you will have found its derivation (as long with the value of h) you will won the Nobel prize.
This seems to show that if the proportionality holds in a frame, then it must hold in every frame, but how do you show that it holds in the first frame?cesiumfrog said:For any particle (e.g. photon) that you wish to express relativistically as a wave packet:
The phase of a plane wave at different points in space-time is a scalar tensor and it can obviously be expressed as the contraction of the position tensor with what we shall call the wave four-vector. Therefore that vector is a tensor, its scalar magnitude is frame invariant. The magnitude equation constitutes a dispersion relation, for the class of relativistic plane waves that are "the same as this one" in another frame, and from this you can trivially calculate the velocity of packets of such waves. This will show that the four-momentum tensor of the "particle" wave-packet is co-linear with the mean wave tensor, from which we can read off the fact that in every frame (and for all same particles) the energy is proportional to frequency. QED. https://www.physicsforums.com/showthread.php?t=243135"
lightarrow said:This seems to show that if the proportionality holds in a frame, then it must hold in every frame, but how do you show that it holds in the first frame?
cesiumfrog said:It shows the four-velocity of a wave-packet is always parallel to the four wave-vector.
Of course (in fact, the line you quoted is even true for sound waves and such, it was the following stage of the argument that specifically pertains to EM and de Broglie matter waves).jtbell said:But this is also true for a classical electromagnetic wave, isn't it?
Do you dispute that E=nhw and p=nhk for any classical EM wave-packet?jtbell said:In a classical em wave, the energy and momentum don't depend on the frequency and wavenumber; they depend instead on the square of the amplitude of the wave.
cesiumfrog said:For any particle (e.g. photon) that you wish to express relativistically as a wave packet:
The phase of a plane wave at different points in space-time is a scalar tensor and it can obviously be expressed as the contraction of the position tensor with what we shall call the wave four-vector. Therefore that vector is a tensor, its scalar magnitude is frame invariant. The magnitude equation constitutes a dispersion relation, for the class of relativistic plane waves that are "the same as this one" in another frame, and from this you can trivially calculate the velocity of packets of such waves. This will show that the four-momentum tensor of the "particle" wave-packet is co-linear with the mean wave tensor, from which we can read off the fact that in every frame (and for all same particles) the energy is proportional to frequency. QED. https://www.physicsforums.com/showthread.php?t=243135"
Beto Pimentel said:The fact that the formalism states that E IS PROPORTIONAL to the frequency does not mean the proportionality constant is h, which still needs to be determined experimentally or through assumptions based on Planck's hypothesis, right?
cesiumfrog said:It shows the four-velocity of a wave-packet is always parallel to the four wave-vector. That implies E=hf and p=h/[itex]\lambda[/itex] (where h is just some constant).
Plus, if one particle can be represented relativistically by a wave-packet then the same proportionality constant must apply, not only for the same particle in another frame (as you acknowledged), but also for identical particles (of different velocities) in the first frame. The key concept is that if a law obeyed by one plane wave in some frame is relativistic, that means that there exists a whole class of different plane waves (and hence packets) obeying the same law (because in some other frame they would be equivalent). The argument doesn't prove that the *same* proportionality constant will apply to photons as does for electrons.
As for that Nobel prize you mentioned, this derivation is basically from de Broglie's PhD thesis.
Did you follow the link? Let me know if there's a specific detail you don't follow. I found the argument in the RQM texts by Crewther or Dyson.lightarrow said:Cesiumfrog, could you please elaborate that in detail? I have no books on which to study these computations, it would be very much appreciated!
Thanks
lightarrow.
The energy of a single photon is determined by its frequency, which is directly proportional to its energy. This relationship is described by the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency.
The energy of a single photon can be measured using a variety of methods, including spectroscopy, photoelectric effect, and Compton scattering. These methods involve the interaction of the photon with matter, which allows for the determination of its energy.
The energy of a single photon is typically measured in electron volts (eV), which is a unit of energy commonly used in atomic and nuclear physics. One eV is equivalent to the energy gained by an electron when it is accelerated through a potential difference of one volt.
Yes, the energy of a single photon is constant and does not change as it travels through space. However, the frequency and wavelength of the photon may change depending on the medium it is traveling through.
The energy of a single photon is inversely proportional to its wavelength. This means that as the wavelength of a photon decreases, its energy increases. This relationship is described by the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength.