E=hf as energy of a particle (other than photon)

  • #1
131
2

Main Question or Discussion Point

In de Broglie's matter wave equation, a book(namely Concepts of Modern Physics) derived the matter wavelength by putting ##E=hf## and ##E=\gamma mc^2## as equals. However I thought that ##hf## was energy of a photon, not a particle. Aren't these two very different concepts?
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,693
6,467
No, the concepts are not very different. A photon is a particle in the modern quantum nomenclature (however, note that this is quite different from the classical particle concept, i.e., essentially small billiard balls). The description by de Broglie was a very early contribution that was an important step towards our current understanding. It associates a frequency and wavelength to all particles related to its energy and momentum, respectively, by Planck's constant.
 
  • #3
131
2
No, the concepts are not very different. A photon is a particle in the modern quantum nomenclature (however, note that this is quite different from the classical particle concept, i.e., essentially small billiard balls). The description by de Broglie was a very early contribution that was an important step towards our current understanding. It associates a frequency and wavelength to all particles related to its energy and momentum, respectively, by Planck's constant.
What other equation is there that applies to the photon as well as other particles? Can I treat these two as a same thing?
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,693
6,467
Photons are not the same thing as other particles, if it was then why make a distinction. However, they do have similar properties and the de Broglie relations are of importance for both.

One important difference is how the energy (and thus frequency) relates to the momentum (and thus wave length) through ##E^2 = m^2 + p^2##, where ##m## is the mass. Photons are massless and so you get ##E = p##.
 
  • #5
131
2
So ##E=hf## includes the particles mass energy as well? Also can you put ##E## in ##E^2=m^2c^4+p^2c^2## and ##E## in ##E=\gamma mc^2## as equals?
 
  • #6
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,693
6,467
So ##E=hf## includes the particles mass energy as well? Also can you put ##E## in ##E^2=m^2c^4+p^2c^2## and ##E## in ##E=\gamma mc^2## as equals?
Yes, ##E = hf## is the total energy of the particle, just as the ##E## in the relativistic dispersion relation. From ##E = m\gamma## follows ##v = p/E##. (Note, I am working in units where ##c = 1##.)
 
  • #7
131
2
So all those ##E##s are the same in a particle, right? And why is ##E^2=m^2c^4+p^2c^2## expressed in that way, I mean why can't you write it as ##E=mc^2+pc##? How was the first form derived?
 
  • #8
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,693
6,467
I mean why can't you write it as ##E=mc^2+pc##?
You cannot write it like that because it is not the same thing. If you want to solve for ##E## you get ##E = \sqrt{m^2 + p^2}##, not ##E = m + p##.
 
  • #9
131
2
I just want to make sure that those three ##E##s are the same. Plus, I put
##E=\gamma mc^2## and ##E^2=m^2c^4+p^2c^2## as equals and got
##\gamma^2m^2c^4=m^2c^4+p^2c^2##
and ##(\gamma^2-1)m^2c^4=p^2c^2## and because ##p^2=m^2v^2##
##(\gamma^2-1)m^2c^4=m^2v^2c^2## which is
##(\gamma^2-1)=v^2/c^2##
therefore ##\gamma^2=1+v^2/c^2##
and got ##\gamma=\sqrt{1+v^2/c^2}## which is clearly wrong
What did I do wrong here?
 
  • #10
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,693
6,467
I just want to make sure that those three ##E##s are the same. Plus, I put
##E=\gamma mc^2## and ##E^2=m^2c^4+p^2c^2## as equals and got
##\gamma^2m^2c^4=m^2c^4+p^2c^2##
and ##(\gamma^2-1)m^2c^4=p^2c^2## and because ##p^2=m^2v^2##
##(\gamma^2-1)m^2c^4=m^2v^2c^2## which is
##(\gamma^2-1)=v^2/c^2##
therefore ##\gamma^2=1+v^2/c^2##
and got ##\gamma=\sqrt{1+v^2/c^2}## which is clearly wrong
What did I do wrong here?
In relativity, ##p## is not equal to ##mv## but to ##mv\gamma##.
 
  • #11
131
2
Oh yeah I forgot about that :)
 
  • #12
131
2
I put ##hf## and ##\gamma mc^2## as equals so
##hv/\lambda=\gamma mc^2## and
##pv=\gamma mc^2## and putting ##p=\gamma mv##
I get ##\gamma mv^2=\gamma mc^2## so ##v=c##
But this doesn't reply to matterwaves do they? They can't move at light speed, also I remember (not fully) that the result of putting the above two equations
was ##v## being faster than ##c##. But I didn't get any of that as well.
1. What did I do wrong here?
2. How do I get the result ##v>c##?
3. I didn't say it here but I just want to make sure for the last time (I promise), ##E##s in the above equations are the same right?
 
  • #13
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,693
6,467
1. What did I do wrong here?
You are using v to denote two different things. In the first step, you use it as the phase velocity of the matter wave and in the second you are using it as the velocity of the particle. These are not the same velocities and in fact they are related as ##v_{\rm phase} = c^2/v_{\rm particle}##. If you are careful, instead of arriving at ##v^2 = c^2##, you will arrive at ##v_{\rm phase} v_{\rm particle} = c^2##, which for ##v_{\rm particle} < c## implies phase velocities larger than c. Yes, the Es are the same.
 
  • #14
433
7
Yes, ##E = hf## is the total energy of the particle, just as the ##E## in the relativistic dispersion relation. From ##E = m\gamma## follows ##v = p/E##. (Note, I am working in units where ##c = 1##.)
So then ## E^2 = m^2c^4 + p^2c^2 = (hf)^2 ## ?
 
  • #15
640
15
So then ## E^2 = m^2c^4 + p^2c^2 = (hf)^2 ## ?
Yes. To revisit de Broglie's derivation, starting with the well known relations [itex]E = \gamma mc^2[/itex] and [itex]p = \gamma mv[/itex]
de Broglie arranged them as [itex]\frac{E}{\gamma c^2} = m[/itex] and [itex]\frac{p}{\gamma v} = m[/itex] or [itex]\frac{E}{\gamma c^2} = \frac{p}{\gamma v}[/itex]

Eliminating the mass from the equation in that way gives us a universal relationship between energy and momentum ([itex]p = \frac{Ev}{c^2}[/itex])

That can also be expressed [itex]pc^2 = Ev[/itex] which becomes [itex]pc = E[/itex] for a photon

One thing that the de Broglie relations give you, I venture to say, that standard relativity mathematics doesn't is the full dynamics of mass-to-energy translation (and vice versa). That facilitates the development of momentum descriptions.
 
  • #16
433
7
Yes. To revisit de Broglie's derivation, starting with the well known relations [itex]E = \gamma mc^2[/itex] and [itex]p = \gamma mv[/itex]
de Broglie arranged them as [itex]\frac{E}{\gamma c^2} = m[/itex] and [itex]\frac{p}{\gamma v} = m[/itex] or [itex]\frac{E}{\gamma c^2} = \frac{p}{\gamma v}[/itex]

Eliminating the mass from the equation in that way gives us a universal relationship between energy and momentum ([itex]p = \frac{Ev}{c^2}[/itex])

That can also be expressed [itex]pc^2 = Ev[/itex] which becomes [itex]pc = E[/itex] for a photon

One thing that the de Broglie relations give you, I venture to say, that standard relativity mathematics doesn't is the full dynamics of mass-to-energy translation (and vice versa). That facilitates the development of momentum descriptions.
Thank you for that! So for a massive particle (e.g. electron), this relationship does not hold true then since the cancellation of ## c ## would not occur? Would ##hf = pc = E## if you found ## v/λ = f ##? Is v equivalent to c in this case when considering the wavelengths of a massive particle or is it the actual velocity of the particle (i.e. ##v < c##)?
 
  • #17
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,693
6,467
Thank you for that! So for a massive particle (e.g. electron), this relationship does not hold true then since the cancellation of ## c ## would not occur? Would ##hf = pc = E## if you found ## v/λ = f ##? Is v equivalent to c in this case when considering the wavelengths of a massive particle or is it the actual velocity of the particle (i.e. ##v < c##)?
E is always equal to hf. Also beware in the last relation and do not mix up particle velocity with the de Broglie phase velocity, the de Broglie phase velocity is going to be greater than c and the particle velocity and phase velocity obey ##v_{\rm particle}v_{\rm phase} =c^2##!
 
  • #18
433
7
E is always equal to hf. Also beware in the last relation and do not mix up particle velocity with the de Broglie phase velocity, the de Broglie phase velocity is going to be greater than c and the particle velocity and phase velocity obey ##v_{\rm particle}v_{\rm phase} =c^2##!
Woah, I did not know ##v_{\rm particle}v_{\rm phase} =c^2##. Thanks! Is there a particular link you know where I can read up more about that?

Also, I was under the assumption ## E = hf ## only for massless particles...(e.g. PhilDSP's post only shows ## E = pc## and does not include the ## (mc^2)^2 ## term). Could you possibly lead me to a link for this derivation as well?

Sorry for all the questions! These seem pretty fundamental and I just want to ensure I'm not misinterpreting the equations.
 
  • Like
Reactions: KYY
  • #19
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,693
6,467
E = pc holds only for massless particles. This does not mean that E = hf holds only for massless particles.
 
  • Like
Reactions: KYY
  • #20
640
15
Woah, I did not know ##v_{\rm particle}v_{\rm phase} =c^2##. Thanks! Is there a particular link you know where I can read up more about that?
##v_{\rm particle}## is also given the name ##v_{\rm group}## or group velocity in wave mechanics or wave analysis.
The original text is Lord Rayleigh's "Theory of Sound" from the 1870's where he derives ##v_{\rm group}v_{\rm phase} =c^2##

http://en.wikipedia.org/wiki/John_William_Strutt,_3rd_Baron_Rayleigh

This book is inexpensive and indispensable if you're looking for deeper technical details:

"An Introduction to the Study of Wave Mechanics" Louis de Broglie

Still deeper studies that apply to optics and materials are:

"Wave Propagation and Group Velocity" Leon Brillouin
"Optics (Volume IV)" Arnold Sommerfeld
 

Related Threads on E=hf as energy of a particle (other than photon)

Replies
2
Views
1K
Replies
5
Views
663
  • Last Post
Replies
2
Views
2K
Replies
2
Views
3K
Replies
7
Views
1K
Replies
2
Views
2K
Replies
17
Views
14K
Replies
0
Views
1K
Replies
2
Views
3K
Replies
44
Views
15K
Top