# Clarification on factors of $c$ in the Lorentz' force

I was told that Lorentz' force is given by

$${\bf F}= q{\bf v} \times {\bf B}.$$

But I have read that it is given by $${\bf F}= \frac{q}{c}{\bf v} \times {\bf B}.$$

Why have these two relations different forms if they represent the same force? Thanks for any help!

• These two expressions represent the same force in different systems of units (SI and Gaussian, respectively). Have a look at my answer to this question: Hamiltonian for a charged particle in electromagnetic field and at the wikipage which mentions the distinction between SI and Gaussian (cgs) units: Lorentz force. – Wouter Feb 8 '13 at 17:52
• Also related: coulomb force in SI and cgs, and several other questions on the seemingly random appearance and disappearance of various factor in E&M relationships. – dmckee --- ex-moderator kitten Feb 8 '13 at 18:18
• I am tempted to close this as a duplicate of the question @Wouter links as the underlying issue is exactly the same, but I suspect this formulation may be more accessible to beginners. Thoughts from the peanut gallery? – dmckee --- ex-moderator kitten Feb 8 '13 at 18:20
• @dmckee I agree that this question, per se, is exactly the same as the linked questions. At the same time, the titles and question-phrasings of the other instances are far from unambiguously related to this one for a naive search. I think preserving this question will make similar questions more easily resolved. – DilithiumMatrix Feb 8 '13 at 19:00
• @dmckee I don't think this necessarily needs to be closed. While the underlying issue is the same one asked about elsewhere, this question itself is not exactly the same as any of the linked questions, so arguably not really a duplicate. – David Z Feb 8 '13 at 19:22

Both relations you mention are completely equivalent, the only difference being the system of units in which they are expressed. 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 the factor of $c$ does not constitute a conflict, which can be seen when the equations are transformed. This is shown in this useful section on the wikipage of the Lorentz force.
Looking at the full Lorentz force expression (of which your expressions are a special case with no electric field), the first one you mention, $$\vec{F} = q\left(\vec{E} + \vec{v}\times\vec{B}\right),$$ is expressed in SI units.$^1$ The second relation, $$\vec{F} = q\left(\vec{E} + \frac{1}{c}\vec{v}\times\vec{B}\right),$$ is expressed in Gaussian units. So both relations are equally valid, as long as you use the correct expression consistent with any other expressions - meaning you should at all times stay within the same system of units. Consistency is key.
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 or contradicting results, but there is no problem if you remember the conventional differences and, again, stay consistent.
$^1$ Note that the expression for the Lorentz force also looks like this in natural units, which is another widely used system of units. Here the units are chosen such that certain natural constants such as the speed of light $c$ have a numerical value of 1. It is then common practice to omit those constants from all equations, for sake of simplicity.