Energy of Single Photon

In summary, the concept of energy being proportional to frequency (E=h\nu) comes from quantum mechanics, not special relativity. However, through the analysis of relativistic wave-packets, it can be shown that this proportionality holds true in all frames for a particle represented by a wave-packet. This was first proposed by de Broglie and is now known as the de Broglie wavelength.
  • #1
JayKo
128
0
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 ;)
 
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  • #2
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 ;)

We don't, I am afraid. I suppose E=hv is an experimental result from Einstein's ananlysis of the photoelectric effect through Planck's idea of quantized energy in a blackbody radiator.
 
  • #3
[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.)
 
  • #4
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.
 
  • #5
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.)
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.
 
  • #6
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.

Err should be E = sqrt((pc)^2 + (mc^2)^2)
 
  • #7
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?

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:
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.
:smile:
 
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  • #8
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"
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?
 
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  • #9
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?

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.
 
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  • #10
cesiumfrog said:
It shows the four-velocity of a wave-packet is always parallel to the four wave-vector.

But this is also true for a classical electromagnetic wave, isn't it?

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.

In the QM wave function, it's the other way around: the energy and momentum are proportional to the frequency and wavenumber, and they don't depend on the amplitude, which is fixed by the normalization condition.
 
  • #11
jtbell said:
But this is also true for a classical electromagnetic wave, isn't it?
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:
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.
Do you dispute that E=nhw and p=nhk for any classical EM wave-packet?
 
  • #12
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"




:smile:

Point acknowledged. I wouldn't have been able to even think of this line of reasoning, and would vote for your Noble Prize, were it not for you sharing with us that it was De Broglie's reasoning in the first place. I would vote for him anyway, but he's got one Nobel already, so...
Anyway, this does not invalidate the previous affirmatives (that E=hv cannot be deduced from Relativity alone). 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?
 
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  • #13
First, everyone is of course right that E=hv is experimentally derived and cannot simply be derived from special relativity alone.

What is derivable from special relativity alone is the idea that massless particles still have momentum and energy: E=pc. That comes from the larger E^2 = (mc^2)^2 + (pc)^2 where m is equal to zero.

It can also be shown in SR, though, that potential energy is inversely proportional to distance, so E proportional to 1/lambda and therefore proportional to frequency is inferable from SR. But the exact relation - involving h - must be derived experimentally.
 
  • #14
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?

Historically, I understand that at the time of de Broglie's work, the electron diffraction experimental results already existed to determine the value of h for electron wave-packets, but nobody had thought to interpret it in that way yet (this being only... 7? years after Planck and Einstein attributed a particle nature to light waves). Instead, de Broglie noted that fitting the same h for electrons as for photons would reproduce (and justify) the Hydrogen model of the day.

People forget how much QM was influenced by SR. This relativity argument explained E=ihd/dt and p=ihd/dx, which became the starting point for Schroedinger (about a decade later), and if BECs had been familiar at the time then we would have had relativistic QM from the very start.
 
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  • #15
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.

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.
 
  • #16
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.
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.
 

What is the energy of a single photon?

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.

How is the energy of a single photon measured?

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.

What is the unit of measurement for the energy of a single photon?

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.

Is the energy of a single photon constant?

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.

What is the relationship between the energy of a single photon and its wavelength?

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.

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