Electromagnetic hamiltonian factor of 1/c question

In summary, the EM Hamiltonian is written as $$H=\frac1{2m}\left(\vec p-\frac ec\vec A\right)^2+e\phi,$$ but this confuses me because it doesn't seem to have the right units. Shouldn't it just be $$H=\frac1{2m}\left(\vec p-e\vec A\right)^2+e\phi,$$ since the vector potential has units of momentum per unit charge? And if so, why do so many authors put in the factor of 1/c?
  • #1
copernicus1
99
0
I often see the EM Hamiltonian written as $$H=\frac1{2m}\left(\vec p-\frac ec\vec A\right)^2+e\phi,$$ but this confuses me because it doesn't seem to have the right units. Shouldn't it just be $$H=\frac1{2m}\left(\vec p-e\vec A\right)^2+e\phi,$$ since the vector potential has units of momentum per unit charge? And if so, why do so many authors put in the factor of 1/c?
 
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  • #2
You're missing the Hamiltonian for the EM field. It's obvious the 1/c factor comes from the convention to describe the 4 potential, either as A0=ϕ/c or just ϕ. If the second option is chosen, you get the 1/c inside the brackets.
 
  • #3
dextercioby said:
You're missing the Hamiltonian for the EM field. It's obvious the 1/c factor comes from the convention to describe the 4 potential, either as A0=ϕ/c or just ϕ. If the second option is chosen, you get the 1/c inside the brackets.

Wow thanks for pointing out how obvious it was to you! But I'm not convinced you're correct. I didn't write the Hamiltonian in terms of the 4-potential, in which case I agree a factor of c would be required; I wrote it (and derived it) in terms of the ordinary magnetic vector potential and the scalar potential, just as its written here. It's very easy to check and see that without the factor of c each term has units of energy. With the factor of c, the vector potential term has units of energy over velocity.
 
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  • #4
This is just a convenient choice of definition of ##\mathbf A##, introduced already in classical electromagnetic theory. In this definition, the electromagnetic force in field ##\mathbf E,\mathbf B = \nabla \times \mathbf A## is given by

$$
\mathbf F = q\mathbf E + q \frac{\mathbf v}{c}\times \mathbf B.
$$
Then E and B have the same units and the velocity v appears always in the companionship of speed of light ##c## as the ratio ##\frac{\mathbf v}{c}##.

This has its practical advantages in relativistic theory. For example, it is very convenient to use in description of the motion of a particle in an external EM wave. In this convention, E and B have usually magnitudes of the same order of magnitude. The magnitude of the magnetic force term is then easily estimated from the value of v/c. Also approximate low-velocity approximations are best formulated in terms of v/c.
 
  • #5
Unfortunately, in electromagnetism there are still (at least) three systems of units in use.

The oldest are the Gaussian units, where the Lagrangian reads
[tex]\mathcal{L}=-\frac{1}{16 \pi} F_{\mu \nu} F^{\mu \nu} - \frac{1}{c} j_{\mu} A^{\mu},[/tex]
where [itex](A^{\mu})=(c \Phi,\vec{A})[/itex] is the four-vector potential of the electromagnetic field [itex]F_{\mu \nu} =\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}[/itex], and [itex]j^{\mu}=(c \rho,\vec{j})[/itex] the four-dimensional current density. I've used the west-coast convention [itex](\eta_{\mu \nu})=\text{diag}(1,-1,-1,-1)[/itex] and the four-vector [itex](x^{\mu})=(c t,\vec{x})[/itex], keeping all factors of [itex]c[/itex].

Because of the "irrational" factor [itex]1/(4 \pi)[/itex] in front of the kinetic term of the gauge field, this system of units is called irrational CGS system (CGS standing for centimeters, grams, seconds, which form the basic units in this system).

Another CGS system just differs by this factor [itex]1/(4 \pi)[/itex]. This is the rationalized Heaviside-Lorentz system of units and usually used in relativistic quantum field theory and thus theoretical high-energy physics. It has the advantage to put the factors [itex]1/(4 \pi)[/itex] where they belong and to reflect the physical dimensions of the quantities best. Of course, electromagnetics is relativistic and thus this system of units is the most natural one. Here, the Lagrangian reads
[tex]\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}-\frac{1}{c} j_{\mu} A^{\mu}.[/tex]

From this Lagrangian it follows that the force on a point charge without magnetic moment is given by
[tex]\vec{F}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ).[/tex]

The official system of units, called SI (for Systeme International), in use in experimental physics and engineering everywhere on the world is taylormade for practictal purposes and provides welldefined accurate realizations of the units. In theoretical electromagnetics it's on the other hand a disease, if you ask me, because the beautiful Lorentz symmetry of the relativistic theory is spoiled. Of course, there is no principle problem to use it also in theory, but then you always get questions like, what are [itex]\epsilon_0[/itex] and [itex]\mu_0[/itex]? The answer is they are conversion factors to transform from the SI units to the more natural Gauß or Heaviside-Lorentz units. Also the SI adds a fourth basis unit to the three mechanical units (the SI concerning the mechanics is an MKS system, using metre, kilogram, second), the Ampere for the electric current. The Lorentz force in this units reads
[tex]\vec{F}=q \left (\vec{E}+\vec{v} \times \vec{B} \right).[/tex]
The only physical universal constant in electromagnetics is the velocity of light, [itex]c[/itex], and it's related to the conversion factors of the SI by
[tex]c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}.[/tex]
The Lagrangian reads
[tex]\mathcal{L}=-\frac{1}{4 \mu_0} F_{\mu \nu} F^{\mu \nu} -j_{\mu} A^{\mu}.[/tex]
 

FAQ: Electromagnetic hamiltonian factor of 1/c question

What is the electromagnetic hamiltonian factor of 1/c?

The electromagnetic hamiltonian factor of 1/c is a term used in the equations of special relativity to account for the effects of electromagnetic fields on the motion of particles. It is equivalent to the speed of light, c, in a vacuum. This factor is important in understanding the behavior of charged particles in the presence of electromagnetic fields.

Why is 1/c used in the electromagnetic hamiltonian factor?

The use of 1/c in the electromagnetic hamiltonian factor is a result of the relationship between electricity and magnetism, as described by Maxwell's equations. These equations show that the speed of light, c, is the fundamental speed at which electromagnetic fields propagate. Therefore, it is necessary to use 1/c in order to accurately describe the behavior of charged particles in these fields.

How does the electromagnetic hamiltonian factor of 1/c affect the motion of charged particles?

The electromagnetic hamiltonian factor of 1/c plays a crucial role in determining the trajectory of charged particles in the presence of electromagnetic fields. This factor accounts for the change in the particle's mass and energy due to its interaction with the fields, which in turn affects its acceleration and velocity.

Can the electromagnetic hamiltonian factor of 1/c be altered?

The electromagnetic hamiltonian factor of 1/c is a fundamental constant and cannot be altered. It is a key component in the equations of special relativity and has been extensively tested and confirmed through experiments and observations.

How does the electromagnetic hamiltonian factor of 1/c relate to the speed of light?

The electromagnetic hamiltonian factor of 1/c is equivalent to the speed of light, c, in a vacuum. This means that in the absence of any other factors, the speed of light is the maximum speed at which electromagnetic fields can propagate. It is a fundamental constant and plays a crucial role in special relativity and our understanding of the behavior of electromagnetic fields and charged particles.

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