Electromagnetism electric fields

Click For Summary
The discussion revolves around the expression for an electric field in free space, specifically E=Re{Eo exp[j(ωt-κy)]} with confusion about the unit vector and the meaning of "Re." The "Re" signifies the real part of a complex expression, while "j" represents the imaginary unit. The electric field oscillates in the x-direction while propagating in the y-direction, confirming that electromagnetic waves are transverse as dictated by Maxwell's equations. The conversation also touches on the use of Euler's formula to convert the exponential form into sine and cosine functions, questioning whether the expression can remain in its exponential form. The importance of understanding these concepts is crucial for grasping the behavior of electromagnetic waves.
james walshe
Messages
7
Reaction score
0
An Electric field has the following form in free space;

E=Re{Eo exp[j(ωt-κy)\hat{i}

I am confused as to why a unit i vector is in the expression for an electric field oscillating in the Y direction? and also what does the Re mean ? i was reading somewhere about it being to do with rectangular coordinates.
 
Physics news on Phys.org
Re is the real part of a complex expression, and \mathrm{j} is used by electrical engineers as the imaginary unit, i.e., \mathrm{j}^2=-1.

A electromagnetic wave in free space must be transverse (both, the electric and magnetic components of the electromagnetic field are vectors perpendicular to the direction of the wave propagation). So what you have here is a plane wave
\vec{E}(t,\vec{x})=\left [\mathrm{Re} \; E_0 \cos(\omega t-k y)-\mathrm{Im} \; E_0 \sin(\omega t-k y) \right ] \vec{i}.
This is a wave propagating in y direction (if k>0) and oscillating in x direction).

That the wave must be transverse follows from Maxwell's equations for free fields in vacuo. Among them you have
\vec{\nabla} \cdot \vec{E}=0,
and this is fulfilled for the above wave.
 
Hi Vanhees71 thank you for the swift reply.
Did you use the Euler relationships to turn the exponential into the sine and cosine functions? i.e [e^jx=cosx+jsinx] and hence the imaginary part can be ignored? or can the expression be left in the exponential form.
 
Susskind (in The Theoretical Minimum, volume 1, pages 203-205) writes the Lagrangian for the magnetic field as ##L=\frac m 2(\dot x^2+\dot y^2 + \dot z^2)+ \frac e c (\dot x A_x +\dot y A_y +\dot z A_z)## and then calculates ##\dot p_x =ma_x + \frac e c \frac d {dt} A_x=ma_x + \frac e c(\frac {\partial A_x} {\partial x}\dot x + \frac {\partial A_x} {\partial y}\dot y + \frac {\partial A_x} {\partial z}\dot z)##. I have problems with the last step. I might have written ##\frac {dA_x} {dt}...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (Second part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8. I want to understand some issues more correctly. It's a little bit difficult to understand now. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. In the page 196, in the first paragraph, the author argues as follows ...

Similar threads

Replies
1
Views
1K
Replies
2
Views
1K
Replies
11
Views
2K
Replies
21
Views
2K
Replies
1
Views
1K
Replies
30
Views
3K