1
$\begingroup$

Maxwell's equations allow for wave solutions via oscillations between electric and magnetic field content.

Couldn't we generate electric waves also if that solution didn't exists? Imagine there was no magnetic field, only the electric field. If at one point in space we generated an electric field nicely varying in a sinus pattern of constant frequency, wouldn't this propagate with light speed, just like a plane EM-wave of that frequency, even though there was no magnetic field?

Improve this question
$\endgroup$
1
  • $\begingroup$ Thanks for the answers. I see now that it wasn't a good idea to phrase the question in terms of an imaginary electric field w/o magnetic field. Either we talk about EM or not. I was more wondering about varying masses and their gravitational fields, maybe I'll ask about that separately. Sorry,this was my first try at a SE question. $\endgroup$
    – Wave
    Commented Jan 12, 2013 at 19:06

2 Answers 2

Reset to default
0
$\begingroup$

No, it does not propagate. A low frequency electric field is a near field - it may depend on time but its distance-dependence is such that this field decays as $1/R^2$ at best. The propagating part of a low-frequency field ($\propto 1/R$) has a magnetic component too.

EDIT: The time-dependent field $E(t)$ has two terms: one is a near field and another is a propagating field $$E(t)=A_1/R^2+A_2\omega/R.$$ The propagating field strength is proportional to the frequency $\omega$ and is accompanied with a propagating magnetic field of the same strength. They can be small. A near magnetic field is smaller than a near electric field due to low frequency. So it is possible to have a variable electric field without variable magnetic field, but it does not propagate in reality. Light is a propagating EMF.

Improve this answer
$\endgroup$
1
  • $\begingroup$ So, the light that reaches us from a star fell off with 1/r^2, too, just because the area of a sphere increases like that, but the amplitude of the light waves pointing in our direction didn't change unlike the amplitude of the imagined oscillating e-field? $\endgroup$
    – Wave
    Commented Jan 12, 2013 at 18:43
0
$\begingroup$

No. Why? Because the electromagnetic field is controlled by Maxwell's equations!

When you produce an oscillating electric field, it happens that the last of Maxwell's equation produces itself a magnetic field:

$$ \nabla \times \mathbf{B} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t} +\;...$$

this means that whenever an electric field oscillates, a B field is generated in a certain way (and vice versa, as stated in another Maxwell's equation).

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.