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I know that electromagnetic radiation is synchronized by oscillations of electric and magnetic fields, but what causes the disturbance in the fields to create the waves in the first place?

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    $\begingroup$ Moving charges. $\endgroup$ – DanielSank Feb 25 '15 at 16:53
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    $\begingroup$ Nope. Accelerating charges. A changing dipole moment. A changing monopole moment is not enough. $\endgroup$ – Jens Feb 25 '15 at 20:40
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There are several mechanisms to create/cause electromagnetic waves:

  • Macroscopically: accelerating charge (just moving at constant velocity is not enough; this is why we drive antennae with an alternating current pushing electrons back and forth.)
  • Microscopically: spontaneous (i.e. without cause); think of exicted atoms emitting a photon. This is what happens in phosphorous materials that you shone light on.
  • Microscopically: stimulated by an already present wave; this is what happens in a laser when a photon causes an excited atom to emit another photon

There are likely other mechanisms that my physicist colleagues will enumerate.

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Classically speaking, electromagnetic radiation is an oscillation of electric and magnetic fields which propagates.

The acceleration of charged particles causes those oscillations - the motion of charges and the electric and magnetic fields are coupled - see Maxwell's equations.

For example, an antenna broadcasts electromagnetic radiation when an alternating current flows in the conductor of the antenna; i.e. when charges move inside the antenna.

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  • $\begingroup$ The acceleration of charges. $\endgroup$ – Rob Jeffries Jan 29 '17 at 14:59
  • $\begingroup$ Good clarification - editing... $\endgroup$ – Brionius Jan 29 '17 at 21:39
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I am guessing that what you call disturbances in the fields, are the oscillations of electric and magnetic fields that you mention just before. (Well, these disturbances propagate. The solution of the Maxwell's equation are traveling fields. Let me for top simplicity the case without charges (the case with charges are covered in my answer to this question, and also in the answer to a former question.

So, starting with the Maxwell equations

$$\frac {\partial \vec B}{\partial t} = -\nabla \times \vec E , \tag{i}$$

$$\frac {\partial \vec E}{\partial t} = \frac {1}{\epsilon_0 \mu_0} \nabla \times \vec B , \tag{ii}$$

where we know that $1/(\epsilon_0 \mu_0) = c^2$. By the time-derivation of the equation $\text {(ii)}$ and then introducing $\text {(i)}$,

$$\frac {\partial ^2\vec E}{\partial ^2 t} = -c^2 \nabla \times \left(\nabla \times \vec E , \right) \tag{iii}$$

We know how to deal with the twice $\nabla$,

$$\frac {\partial ^2\vec E}{\partial ^2 t} = c^2 \left(\nabla ^2 \vec E - \nabla (\nabla \cdot \vec E) \right)$$

But, since we supposed that there are no charges, and by Maxwell's equation $(\nabla \cdot \vec E) = \rho / \epsilon_0$, the 2nd term in the RHS is zero. So we het a homogeneous equation, which is the wave-equation,

$$\frac {\partial ^2\vec E}{\partial ^2 t} - c^2 \nabla ^2 \vec E = 0, \tag{iv}$$

whose solution is of the form

$$vec E = \vec E_1 e^{i(\vec k \cdot \vec r - \omega t)} + \vec E_2 e^{-i(\vec k \cdot \vec r + \omega t)}. \tag{v}$$,

So, we got traveling waves.

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