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How can magnetic field be a relativistic phenomenon?

  1. Apr 5, 2006 #1
    How can magnetic field be a relativistic phenomenon??

    According to Purcell (in his book: Electricity and Magnetism) the
    magnetic field force affecting a moving charge parallel to a current
    carrrying conductor is merely resulting from the electric field in the
    lab frame obtained when Lorentz transformation of the elecrtic field in
    the charge frame is done.

    If Magnetic phenomenon is a relativistic phenomenon, then we have to
    exclude the magnetic fields from the field transformation relations and
    keep only the electric field which say that the normal components of the
    electric field are related by gamma factor , while parallel components
    (to the direction of relative velocity between the 2 frames) are equal.

    This should be everything, But does this account for everything? Let's

    Consider two single charges in relative motion in two orhtogonal
    directions at the instant they were along the directions of one
    of the charges. In the lab frame, there are coulomb force in the radial
    direction, and Biot-Savart force in the normal directions direction.
    How can this Biot Savart force be derived from a relativistic effect
    like what Purcell did for the currents??
  2. jcsd
  3. Apr 6, 2006 #2


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    Watch out for words like "merely" in textbooks.
    It's not sensible to say that magnetic fields are merely the result of electric fields plus relativity. One could as well say that electric fields are magnetic fields modified by relativistic effects. It is more accurate and less confusing to say that electric and magnetic fields are two aspects of a larger phenomenon. One transforms into another depending upon the velocity of one's reference frame.
  4. Apr 6, 2006 #3

    Hans de Vries

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    The magnetic field is a relativistic side effect of the electric field but not
    exactly in the sense you are describing above.

    Remember that a particle in it's rest frame only feels the electric force.
    When you give it a boost to a speed v it again feels only the electric
    force but now from the E-field in the new reference frame.

    To do the Lorentz transforms with the E-field alone however you need
    to know the positions, speeds, and directions of speed of the individual
    charges that together make up the E-field, calculate all the individual
    contributions, and sum up the result to get the new E field.

    You need to do this even in the case of a simple wire because at one
    side the electrons move away and at the other they move towards you.
    If you start moving all the relative motions change.

    Now this is why B comes in so handy. All the effects of all the moving
    charges may simply be added into one vector and you can forget about
    all the individual speeds. The second big simplification is that while the
    total Lorentz Transformation can be written as:

    \textbf{E}'\ \Rightarrow \ \gamma(\textbf{E}\ +\ \textbf{v} \times
    \textbf{B}) - \frac{\gamma^2}{\gamma+1}\ \frac{\textbf{v}}{c} (\textbf{v}
    \cdot \textbf{B})

    where [itex]\gamma[/itex] is given by:

    \gamma \ =\ \frac{1}{\sqrt{1-v^2/c^2} }

    but because at non-relativistic speeds [itex]\gamma \approx 1[/itex], the transformation becomes:

    \textbf{E}'\ \Rightarrow \ \textbf{E}\ +\ \textbf{v} \times

    Which is the well known form.

    Regards, Hans
    Last edited: Apr 6, 2006
  5. Apr 6, 2006 #4

    Hans de Vries

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    This looks awkward but that's more because it's written in a form to
    arrive at the Lorentz Force:

    \textbf{F}\ =\ q(\textbf{E}\ +\ \textbf{v} \times

    A boost only in the x direction gives the following transformation:

    [tex] \textbf{E}_x'\ \Rightarrow \ \textbf{E}_x [/tex]

    [tex] \textbf{E}_y'\ \Rightarrow \ \gamma(\textbf{E}_y\ +\ v\textbf{B}_z) [/tex]

    [tex] \textbf{E}_z'\ \Rightarrow \ \gamma(\textbf{E}_z\ -\ v\textbf{B}_y) [/tex]

    That doesn't look to bad does it.

    Regards, Hans
    Last edited: Apr 6, 2006
  6. Apr 6, 2006 #5
    What about the special case mentioned in my message? How can we get the Lorentz force in the lab frame from such a lorentz transformation of the electric field in the charge rest frame ?
  7. Apr 7, 2006 #6


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    Let me add a short introduction / overview to my post.

    Before relativity, one would treat the electric and magnetic fields separately,as one treated space and time separately.

    After relativity, one unites all the electric and magnetic fields into a single entity, called the Faraday tensor, that transforms as a unit, according to the Lorentz transformation.

    This single entity is not a vector, though - it is a matrix, aka a rank 2 tensor. Thus it may not be immediately obvious how the Lorentz transforms apply to such an entity.

    The manner in which one combines the E&B fields into a tensor, and the manner in which this tensor transforms under a boost (specifically a boost in the z direction) can be found online at:


    Unfortunately the link forgets to mention that the transformation laws are for a "boost" in the z direction.

    To address your specific question from the above formulas:

    So if we have a charge sitting in the lab frame, we know that there are only Ex, Ey, and Ez components in the field.

    When the charge moves in the z direction, we see that the y component of the electric field in the lab frame generates an x component of the magnetic field in the moving frame, and the x component of the electric field in the lab frame generates a y component of the magnetic field in the moving frame via the relations for Bx and By on the webpage.

    How are these derived from the Lorentz transform? Because this is a tensor quantity, it's not at all obvious at first how it transforms.

    The Lorentz transform for familiar 4-vectors can be written as

    a \prime^i = \Lambda^i{}_j a^j

    where t=a^0, x=a^1, y=a^2, z=a^3

    Since z' = [itex]\gamma*(z-\beta t)[/itex], we see that for instance

    [itex]\Lambda^3{}_3 = \gamma[/itex] [itex]\Lambda^3{}_0 = \gamma \beta[/itex]

    The Faraday tensor transforms according to the law

    F \prime ^{ij} = \Lambda^i{}_k \Lambda^j{}_l F^{kl}

    because it is a rank 2 tensor - that's how matrix / tensor quantites transform. (You are probably more familiar with how vectors transform, the above expression is the equivalent one for matrices/rank 2 tensors)
    Last edited: Apr 7, 2006
  8. Apr 7, 2006 #7


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    here is a repost that might be relevant:

    the electromagnetic action is nothing other than the sole electrostatic action, but with special relativity taken into consideration. take a look at http://en.wikipedia.org/wiki/Magnetic_field :

    The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration. The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of [itex] \lambda \ [/itex] ) and some non-zero mass per unit length of [itex] \rho \ [/itex] separated by some distance [itex] R \ [/itex]. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static non-relativistic repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance [itex] R \ [/itex]) for each infinite parallel line of charge would be:

    [tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex]

    If the lines of charge are moving together past the observer at some velocity, [itex] v \ [/itex], the non-relativistic electrostatic force would appear to be unchanged and that would be the acceleration an observer traveling along with the lines of charge would observe.

    Now, if special relativity is considered, the in-motion observer's clock would be ticking at a relative rate (ticks per unit time or 1/time) of [itex] \sqrt{1 - v^2/c^2} [/itex] from the point-of-view of the stationary observer because of time dilation. Since acceleration is proportional to (1/time)2, the at-rest observer would observe an acceleration scaled by the square of that rate, or by [itex] {1 - v^2/c^2} \ [/itex], compared to what the moving observer sees. Then the observed outward acceleration of the two infinite lines as viewed by the stationary observer would be:

    [tex] a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex]


    [tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho} [/tex]

    The first term in the numerator, [itex] F_e \ [/itex], is the electrostatic force (per unit length) outward and is reduced by the second term, [itex] F_m \ [/itex], which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors).

    The electric current, [itex] i_0 \ [/itex], in each conductor is

    [tex] i_0 = v \lambda \ [/tex]

    and [tex] \frac{1}{\epsilon_0 c^2} [/tex] is the magnetic permeability

    [tex] \mu_0 = \frac{1}{\epsilon_0 c^2} [/tex]

    because [tex] c^2 = \frac{1}{ \mu_0 \epsilon_0 } [/tex]

    so you get for the 2nd force term:

    [tex] F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R} [/tex]

    which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by [itex] R \ [/itex], with identical current [itex] i_0 \ [/itex].
    Last edited: Apr 7, 2006
  9. Apr 9, 2006 #8
    How can we apply the same logic on my special case of two charges travelling orthogonal to each other at this instant?
    Last edited: Apr 9, 2006
  10. Apr 9, 2006 #9


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    Pick a charge (let's call this charge #1) and do a Lorentz boost so that you work in an inertial frame of reference where that charge (charge #1) is stationary.

    Then the force on charge #1 will be given by the electric field of charge #2 at the position of charge #1. (Because the velocity of the charge #1 is zero, the only force components will be due to the electric field, the magnetic field due to charge #2 will be irrelevant).

    You'll find the second charge is moving on some diagonal when you peform the boost.
    The electric field of this diagonally moving charge (charge #2) can be computed from the formulas at


    It will point towards charge #2 in this frame, with a magnitude given by the formula on the webpage.

    If you want the force on charge #1 in the original frame, you'll have to transform the force back. The easiest way to do this is to convert the force into a 4-force, transform it like any 4-vector, and then covert the 4-force back into a 3-force. Or you can use the relationship.

    force = dp/dt
    4-force = dp/dtau

    where t is coordinate time and tau is "proper time".

    You can repeat the process to find the force on the other charge. I'm not sure offhand whether or not any momentum is being carried away by the fields with this configuration, so I'm not sure whether or not the two forces will sum to zero.

    Remember that you will be changing your defintion of simultaneity when you change frames. The 4-force on the charge will be independent of the particular coordinate system you use, however, as long as you are careful to compute it at the same space-time location.
    Last edited by a moderator: Apr 22, 2017
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