Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I've been looking for the Hamiltonian of a charged particle in an electromagnetic field, and I've found two slightly different expressions, which are as follows:

$$H=\frac{1}{2m}(\vec{p}-q \vec{A})^2 + q\phi $$

and also

$$H=\frac{1}{2m}(\vec{p}-\frac{q}{c}\vec{A})^2 + q\phi $$

with $\vec{p}$ the momentum, $q$ the electric charge, $\vec{A}$ the vector potential, $\phi$ the scalar potential and $c$ the speed of light.

So basically the difference is in the term $1/c$ multiplying $\vec{A}$, present in the second form (which I use in my lectures) but not in the first one (used by Griffiths in Introduction to quantum dynamics to treat the Aharonov-Bohm effect). Why does this difference exist and what does it mean? And how does the term $1/c$ affect the dimensional analysis (the units) of the problem?

share|improve this question
3  
The first is the expression in the SI units (which most modern books are written in). The second is the expression in Gauss units (which still most scientific literature and old books use). –  Fabian Jan 27 '13 at 18:39
    
Thank you! Now everything makes sense! I'll look for the details of how to pass from one system to another. Thanks!!! –  Ajayu Jan 27 '13 at 20:27
    
Ajayu, you should probably accept the answer you like best by clicking on the gray check symbol next to the answer. –  Rafael Reiter Jan 28 '13 at 12:56

3 Answers 3

up vote 3 down vote accepted

The missing $1/c$ in your first expression is simply a consequence of the units used. The second expression is in Gaussian units while the first one is in either SI units or in natural units. In the latter system of units (natural units) certain constants like $\hbar$ and $c$ have a numerical value of 1, so they can be left out of the equations.$^1$ This is common practice in physics and it doesn't change anything about the dimensional analysis of the problem, as long as you keep in mind that you're working with those natural units.

The same goes for any other system as well. Every system of units $A$ is consistent with any other system of units $B$ as long as you yourself are consistent in their usage and correctly transform everything between $A$ and $B$ when desired.
So there is no fundamental difference between a dimensional analysis in SI, Gaussian or natural units, as long as you keep in mind what units you're working with. The units themselves will (obviously) vary between systems, but dimensional analysis in one system will be entirely consistent with dimensional analysis in another.$^2$


$^1$ Note that this is not the case for SI units. As is rather well-known, the numerical value of $c$ in SI units is about $3\times10^8$ $(\mathrm{m/s})$. The reason for the absence of $1/c$ in SI units is a conventional difference. Wikipedia has a comparison between Gaussian and SI units explaining the major differences here.

$^2$ Perhaps one important note concerning Gaussian and SI units here is that due to the different conventions, it can be more difficult to transform between them. E.g. making an equation dimensionless in SI units, might yield a non-dimensionless equation when transformed into Gaussian units.
One example is when we consider Gauss's law in Gaussian units divided by the free charge density: $(1/\rho)\vec{\nabla}\cdot\vec{E} = 4\pi$. The quantity on the left-hand side is dimensionless in Gaussian units, but not in SI units, where it is $(1/\rho)\vec{\nabla}\cdot\vec{E} = 1/\epsilon_0$. So you have to watch out for that when transforming your equations. Dimensional analysis may therefore also yield seemingly different results in SI or Gaussian units, but there is no problem if you remember the conventional differences and, again, stay consistent.

share|improve this answer
    
Thank you for the answer! I thought about it at first, but it still gave me some problems (this is just a part of an excercise and I had more problems with the units later), so I don't think it's the natural units, and the book I took it from explicitly says it's in SI units. It seems to me that it's like Fabian says in another comment, the second expresion is in Gaussian units. However, thanks for your help! –  Ajayu Jan 27 '13 at 20:30
    
@Ajayu Indeed, if you know the first expression is in SI units, that's your answer. I edited my answer to incorporate that possibility. I left the other possibility of natural units in there because that's something you come across often as well. –  Wouter Jan 28 '13 at 17:06
    
Maybe clarify that only natural units have $c = 1$, because now it sounds like both SI and natural units set $c$ and $\hbar$ as $1$. –  Kitchi Jan 28 '13 at 19:30
1  
@Kitchi I've added some clarification. In the mean time I've also elaborated on the differences between SI and Gaussian units, for the sake of completeness. And I think I've hammered the point of consistency to death. I hope it's all correct and clear. –  Wouter Jan 28 '13 at 21:12

The $\frac{1}{c}$ makes the electric field and magnetic field have the same units. Along with the other answers provided, this makes second equation very useful.

share|improve this answer
1  
Thank you very much! –  Ajayu Jan 27 '13 at 20:47

As in any problem in physics you can use whatever units you like the most. In electromagnetism factor like $\frac{1}{4\pi \epsilon_{0}}$ or $\frac{1}{c}$ appear a lot. So you can define a new system of units in which $c=1$ such as the Natural units or the Planck units. See http://en.wikipedia.org/wiki/Natural_units

share|improve this answer
1  
Thank you for the answer! I thought about it at first, but it still gave me some problems (this is just a part of an excercise and I had more problems with the units later), so I don't think it's the natural units, and the book I took it from explicitly says it's in SI units. It seems to me that it's like Fabian says in another comment, the second expresion is in Gaussian units. However, thanks for your help! –  Ajayu Jan 27 '13 at 20:31

protected by Qmechanic Jan 28 '13 at 12:56

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.