Alternative formula for the Hamiltonian

  • Context: Undergrad 
  • Thread starter Thread starter grzz
  • Start date Start date
  • Tags Tags
    Formula Hamiltonian
Click For Summary

Discussion Overview

The discussion revolves around the derivation of a relativistic Hamiltonian function ##W_a## for a point charge as presented by Fermi in his 1932 article 'Quantum theory of radiation'. Participants explore the relationship between Fermi's equation and a proposed equation involving energy, momentum, and mass in the context of electromagnetic fields.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents Fermi's equation and asks if it can be derived from another equation involving energy and momentum.
  • Another participant attempts a derivation but expresses uncertainty about the inclusion of rest mass energy in ##W_a##.
  • A participant notes that a constant change in total energy does not affect the physics, provided gravity is not involved.
  • There is a request for clarification on the origin of the second equation used in the derivation.
  • One participant explains that they are using the equation ##E^2 = p^2 c^2 + {m_o}^2c^4##, substituting ##E## with ##W_a## and incorporating potentials due to electromagnetic fields.
  • Another participant suggests that Fermi defines ##W_a## as energy relative to rest mass energy, which could explain the form of his equation.
  • A later reply questions the assumption that ##W_a## includes rest mass energy, suggesting that it may not be part of the total energy.
  • One participant acknowledges their earlier mistake regarding the assumption about ##W_a## and indicates that their derivation aligns with Fermi's equation after reevaluation.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of ##W_a## and its relationship to rest mass energy. There is no consensus on the derivation or the assumptions made regarding the Hamiltonian function.

Contextual Notes

Participants note potential limitations in their assumptions about the definitions of energy and the role of rest mass energy in the equations discussed.

grzz
Messages
1,035
Reaction score
28
In his article 'Quantum theory of radiation', Reviews of Modern Physics, Jan 1932, volume 4, Fermi gives the relativistic hamiltonian function ##W_a## for a point charge by equation (13),

## 0 = - \frac 1 {2m} \left( \left[ mc +\frac {W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c \right ]^2 \right) + \frac {mc^2} 2. ##

Is it possible to derive the above equation from,

## \left[ {W_a - ~eV} \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 c^2 + m^2c^4? ##

Thanks.
 
Physics news on Phys.org
I tried the following.

## \left[ {W_a - ~eV} \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 c^2 + m^2c^4 ##

## \left[ \frac{W_a - ~eV} c \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 + m^2c^2 ##

## 0 = -\frac 1 {2m}\left(\left[ \frac{W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c \right ] ^2 \right) + \frac {mc^2} 2. ##

This was as far as I could go. I am assuming that ##W_a## includes the rest mass energy ##mc^2##.
Perhaps I am missing something.

Any help?

Thanks.
 
Last edited:
A constant change in the total energy (e.g. adding mc2) doesn't change the physics (as long as we are not dealing with gravity).
 
I am still wondering from where did Fermi got his equation (13).
 
grzz said:
In his article 'Quantum theory of radiation', Reviews of Modern Physics, Jan 1932, volume 4, Fermi gives the relativistic hamiltonian function ##W_a## for a point charge by equation (13),

## 0 = - \frac 1 {2m} \left( \left[ mc +\frac {W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c \right ]^2 \right) + \frac {mc^2} 2. ##

Is it possible to derive the above equation from,

## \left[ {W_a - ~eV} \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 c^2 + m^2c^4? ##

Thanks.
Where did you get the second equation from?
 
kimbyd said:
Where did you get the second equation from?

Thanks 'Kimbyd' for your interest.

I am using the usual

##E^2 = p^2 c^2 + {m_o}^2c^4,##

and replacing E by ##W_a##, assuming ##W_a## stands for the total energy of the particle.

I am also putting in the scalar potential V and vector potential U since the charged particle is in an electromagnetic field.
 
grzz said:
In his article 'Quantum theory of radiation', Reviews of Modern Physics, Jan 1932, volume 4, Fermi gives the relativistic hamiltonian function ##W_a## for a point charge by equation (13),

## 0 = - \frac 1 {2m} \left( \left[ mc +\frac {W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c \right ]^2 \right) + \frac {mc^2} 2. ##

Is it possible to derive the above equation from,

## \left[ {W_a - ~eV} \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 c^2 + m^2c^4? ##

Thanks.
It looks as if Fermi defines ##W_a## as the energy relative to the rest mass energy, so

## E_{rel} \equiv W_a-eV +mc^2 ##

Using this, his equation follows. I think he made this choice so that upon expansion, the constant term ##mc^2/2## disappears.
 
grzz said:
and replacing E by ##W_a##, assuming ##W_a## stands for the total energy of the particle.
That assumption seems unlikely. I believe your math above is correct, which would indicate that the relativistic Hamiltonian does not include the mass as part of the energy. So you'd have to include an extra ##mc^2## to get the total energy. I believe that resolves the discrepancy.

Does the paper define ##W_a## explicitly? That would probably be your best bet in determining whether this is correct or not.
 
That assumption seems unlikely.

Thank you both, 'nrqed' and 'kimbyd'!

Yes, my mistake was assuming that ##W_a## includes the rest mass energy.

I derived it again and it turned out as the one given by Fermi, as 'nrqed' said.
 
Last edited by a moderator:

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K