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Somebody mentioned on Freenode chatroom for physics that All Electromagnetic Radiation are delivered in form of Photons not just light.

Is it true? Does that mean if we get a THF electrical oscillator to emit the frequencies that are visible to the eye, do we see light coming out of the circuit?

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    $\begingroup$ See this question. The term light is normally used for EM radiation of a wavelength between 400 and 700nm. The only difference between light and other EM radiation is what our eyes are sensitive to. $\endgroup$ Mar 24, 2014 at 18:35
  • $\begingroup$ Photons are light in form of particles. Human eye can see light with 430-790Thz frequency (its called visible spectrum) so if you have device which can emit light with 430-790thz frequency then yes you will see it $\endgroup$
    – user21420
    Mar 24, 2014 at 18:41
  • $\begingroup$ I prefer to say that all EM radiation is light. At least in Physics, where biological characteristics of our eyes should not matter. Otherwise, you paradoxes like: interference of light doesn't give you light but other kind of radiation or saying UV light (when you can't see UV). $\endgroup$
    – jinawee
    Mar 24, 2014 at 18:44
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    $\begingroup$ BTW an electrical circuit is unlikely to be operable at visible light frequencies because those frequencies are higher than the plasma frequencies of (most?) metals. However, in principle an electrical circuit running at visible light frequencies would indeed emit EM radiation that you could see. $\endgroup$ Mar 24, 2014 at 19:07

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In undergraduate physics courses (c.f. MIT 8.03 course on waves), electromagnetic radiation is introduced as a pair of time-dependent oscillating magnetic and electric fields which satisfy the electromagnetic wave equation, as derived by Maxwell:

$$\frac{1}{c^2}\frac{\partial^2 E(x,t)}{\partial t^2} - \frac{\partial^2 E(x,t)}{\partial x^2} = 0$$

Electromagnetic radiation may also be viewed as a particle, i.e. a photon. In quantum electrodynamics, it arises as one quantizes classical electromagnetic field theory, given by the Maxwell Lagrangian,

$$\mathcal{L} = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu}$$

The photon is a gauge boson which in cursory terms acts as a mediator of the electromagnetic force, just as the hypothesized graviton is a mediator of gravitation. The term 'light' refers to electromagnetic radiation, and hence photons. However, in the case of 'visible light,' this restricts us to only photons of a certain wavelength between 400 and 700 nanometers, which through the relation $E = hc/\lambda$ implies an energy range of approximately 1.7 to 3.1 electronvolts.

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