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The Lorentz force in the MKS and CGS measurement systems

  1. Nov 11, 2004 #1
    Hi there!

    I have a problem deriving the formula for the Lorenz force that I found in my lecture notes in Theoretical Mechanics.

    The formula is:

    [tex]\vec{F}=q \cdot \vec{E} + \frac{q \cdot (\vec{v} \times \vec{B})}{c}[/tex]

    [tex]\vec{F}[/tex] is the Lorentz force
    [tex]\vec{E}[/tex] is the electric field
    [tex]\vec{v}[/tex] is the velocity of the charged particle
    [tex]\vec{B}[/tex] is the magnetic field
    [tex]q[/tex] is the charge of the particle
    [tex]c[/tex] is the speed of light

    The problem is that this relation is supposed to be formulated with respect to the "Meter, Kilogram, and Second" (MKS) measurement system (as stated in the lecture). According to the information I found, the expression for the magnetic field :

    [tex]\frac{q \cdot (\vec{v} \times \vec{B})}{c}[/tex]

    is written in terms of the "Centimeter, Gram, and Second" (CGS) measurement system ! It differs from the same expression in terms of the MKS system by the factor [tex]1/c[/tex] .
    At the same time the expression for the electric field is in terms of the MKS system.

    I have three problems:
    1. I cannot understand whether the formulation of the Lorentz force, given in the lecture, is a correct one.
    2. Is it possible to mix epressions written in terms of different measurement systems in one physical relation?
    3. Where does the factor [tex]1/c[/tex], in the expression for the magnetic field in CGS, come from?
    According to my calculation, this factor should be [tex]1/c^2[/tex] as it depends on the product of the permeability of free space and the permittivity of free space.

    Permittivity of Free Space

    If someone is able to help - thank you in advance!

    Last edited: Nov 11, 2004
  2. jcsd
  3. Nov 11, 2004 #2


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    1. The formula is for cgs units
    2. I don't know what you mean. Are asking whether for example 1 foot plus 1 inch equals 2 feet?
    3. The factor is there as a convention. CGS has the feature that then electric and magnetic fields have the same units. One is free to define any units for magnetic fields. I could for example have [itex]\vec{F}=q\vec{E}+q\vec{v}\times\vec{B}/\pi[/itex], where the units of B is the krab and [itex]\pi[/itex] krabs = 1 Tesla, and the other units are as in MKS.
    Last edited: Nov 11, 2004
  4. Nov 11, 2004 #3


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    I think Krab has given a concise correct response.
    here is some more detail:
    If you are near a College Physics department library they will have
    Jackson Classical Electrodynamics
    and at the end of the book in an appendix he explains the many systems of units
    and he gives the correct Lorentz force formula in each system

    In MKS---the dominant SI version---the Lorentz force is written

    [tex]\vec{F}=q \cdot \vec{E} + q \cdot (\vec{v} \times \vec{B})[/tex]

    and SI uses the Tesla unit for Bfield. But some other systems use the Lorentz force formula which you wrote

    [tex]\vec{F}=q \cdot \vec{E} + \frac{q \cdot (\vec{v} \times \vec{B})}{c}[/tex]

    As Krab says, if you are using a system in which it is written thus, then the electric and magnetic fields are "commensurable" in other words they are measured in the same unit.

    this is nicely compatible with Special Rel, because magnetic force is a relativistic partner of electrostatic----which is which depends on the frame---and "should" be measured in the same unit.

    in the "good" form of Lorentz force you will see the beta, v/c, which is familiar in all Special Rel formulas.

    SI, with its Tesla unit, sucks.
  5. Nov 11, 2004 #4


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    The only thing I have to add to the correct responses are a few web resources.

    You can find the formula for the Lorentz force in both cgs units and mks units here

    It's also useful to know what Maxwell's equations look like in both systems, that's in the same family of web pages, but it's located here2
  6. Nov 12, 2004 #5
    Thank you for the quick reply!

    Thank you for the quick reply, but I still have a problem deriving the formula in CGS.

    Does this formula completely describe the motion of a charged particle in constant homogeneous electric and magnetic fields? Did I understand it correctly - the magnetic field orientation is perpendicular to that of direction of motion of the particle, and the electric field orientation is perpendicular to both the orientation of the magnetic field and that of the direction of motion of the particle.I need that in order to write the equations of motion of the particle.

  7. Nov 12, 2004 #6


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    The formula gives you the magnitude and direction of the force. It says nothing about the directions of the fields or the particle. These 3 directions are the given conditions, and so can be anything. So for example, the magnetic field can be parallel to the velocity, in which case the particle experiences zero magnetic force. The component of the force coming from the electric field is always parallel to the electric field.
    Last edited: Nov 12, 2004
  8. Nov 12, 2004 #7


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    Yes,it completey describes at classical level the motion of the particle.That is,the particle (charged one)does not cause those fields,but they are given as exterior factors acting on that particle.Usually,it takes very particular field configurations,to be able to integrate the 3 COUPLED differential equations which are to be found by projecting the vectorial relationship on a triedrum of ortonormal axes.For general fields (a general em field) one cannot solve the system.
  9. Nov 12, 2004 #8

    Here is the exercise that caused all the trouble:

    " Consider a charged particle in a constant electric and a constant magnetic field.
    a) Derive the equations of motion.
    b) Determine the solution for the velocity.
    c) Determine the solution for the position.
    d) Make a qualitative drawing of the trajectory. "

    I had a problem solving the exercise as I didn't know how to treat the electric field . In my opinion, it is not made clear in the exercise what is the direction of the electric field - that is why I assumed that it should be perpendicular to the orientation of both the magnetic field and the velocity. The first thing I did while trying to solve the exercise was to choose a coordinate system, such that the orientation of the magnetic field is paralel to the z-axis.
    As I was confused about the direction of the electric field I had prolems writing the equations of motion in terms of (x,y,z).
    What suggestions do you have for the interpretation of this exercise?
  10. Nov 13, 2004 #9


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    Usually the choise for the magnetic field is along the z axis,and that's because it enters in that awful cross product involving the speed.As for the electric field,there's no a priori condition.U can simplify things to an extreme,by chosing it along one of the other (2) axes.In this case,the particle describes a cycloide,becuse the electric field would try to accelerate it on his own direction,but the perpendicular magnetic field would try to make it move round a circle.Composing the two tendencies,u get a sliding circle which generates a cycloide...
  11. Nov 13, 2004 #10

    Yes, I agree. But what about the equations of motion? If , for example, we assume that the electric field is along the y-axis and is perpendicular to the velocity, does it mean that the electric field should be only included in the equation of motion along the y-axis. If so how should the expression for the electric field look like in accordance to what is stated in the exercise. We have neither the potential, nor the force, describing the electric field.
  12. Nov 15, 2004 #11


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    That's right.

    [tex] \vec{E}=(0,E,0) [/tex].

    What do you mean?Lorentz's force is given:
    [tex]\vec{F}=q \cdot \vec{E} + q \cdot (\vec{v} \times \vec{B})[/tex]

    I already gave you the answer to your problem.You just need to decouple those 2 differential equations,thats's all.
  13. Nov 16, 2004 #12
    Ok Thank you!

    I will give it a try
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