Question

the same for the magnetic field vector B (in GR units, both vectors have the same units where [E, E,, E.) are the components of the electric field vector É and [B,,B,,B.] are of kg . c-'m-': see box 4.2). Using this tensor, we can write a relativistically valid ver- sion of the Lorentz force law (which describes the total electromagnetic force acting on particle with charge q moving through an electromagnetic field), dp t = qFHV Nvalla and Gauss's law and the Ampere-Maxwell relation become the single equation OFW = 41kJ4 (4.15) dx' (4.16) P4.7 Write out equation 4.15 in the low-velocity limit when u = t. What is the physical meaning of the resulting equation? (Hint: Remember that the time component of the four-momentum is relativistic energy.)

Answer 1

In the low velocity limit, the proper time "tau" is just the coordinate time "t"

$d\tau =\sqrt{1-v^2/c^2}\,\,\,dt\simeq dt$

since v << c in the low velocity limit.

and hence the space component of the 4-velocity is just the 3-velocity

$u^i=\frac{dx^i}{d\tau}\simeq \frac{dx^i}{dt}=v^i$

for i = x,y,z (space components)

Also the time component of the 4-momentum is relativistic energy. In the low velocity limit, the relativisitic energy is simply kinetic energy divided by speed of light which is constant.

$p^t =\mathcal E /c$

Since

$F^{tx}=E^x/c,\quad F^{ty}=E^y/c,\quad F^{tz}=E^z/c,\quad$

where E is the electric field.

this means

$F^{\mu\nu}\eta_{\nu\alpha}u^\alpha$

can be written as

$F^{tx}\eta_{xx}u^x+F^{ty}\eta_{yy}u^y+F^{ty}\eta_{zz}u^z$

which is equal to

$(E^x v_x+E^y v_y+E^zv_z )/c=\vec E.\vec v/c$

Therefore the equation becomes

$\frac{d}{dt}(\mathcal E/c)=\vec E.\vec v/c$

or

$\frac{d}{dt}(\mathcal E)=\vec E.\vec v$

This means the rate of change of the total kinetic energy depends on the dot product of electric field and the particle's velocity. The kinetic energy remains constant if there is no electric field. And the kinetic energy does not change if only magnetic field is applied.