2
$\begingroup$

Is it possible to create a beam of light with frequency of 0?

So this would involve photon(s) that move forward without fluctuating with any frequency.

If yes, how could this be done?

Also, considering that such a beam would not have any known color (since our eyes recognize color based on the frequency of the light beam), would it be possible to create a sensor that senses that beam?

$\endgroup$
4
$\begingroup$

We can represent a monochromatic electromagnetic wave by one of its fields, the $\vec E$ or $\vec B$ field (or $\vec H$ in the case of the diagram further down). For example, we can write $$\vec E (\vec r, t) = \vec E_0 e^{i(\vec k \cdot \vec r - \omega t)}$$ where the relevant part for this question is $\omega$, the angular velocity. When the frequency of a wave approaches 0, the angular velocity also approaches zero ($\omega = 2\pi f$, so $\omega$ and $f$ only differ by a factor of $2\pi$).

enter image description here

For an electromagnetic wave, the wavenumber follows $k = \frac{2\pi}{\lambda}$. From this you can see that with increasing wavelength $\lambda$, $k$ approaches zero as well.

The exponent in the exponential function therefore tends to zero with decreasing frequency and increasing wavelength: $e^0 = 1$.

We are left with $$\vec E (\vec r, t) = \vec E_0$$ which is the equation for a static electric field $\vec E_0$ which does not vary in time or position.

We are able to detect such a field for example by probing it with a test charge or measuring a voltage between two points.

The same derivation can be done with the magnetic field $$\vec B = \vec B_0 e^{i(\vec k \cdot \vec r - \omega t)}$$ and yields the same result (static magnetic field).

As for the question about color, color is merely a definition for specific wavelengths in the visible region of the EM spectrum. Since we also don't assign a color to x-rays or microwaves, the concept of color also does not apply to such a wave.

$\endgroup$
1
$\begingroup$

It will help if your read the answer by Motl in a similar question Virtual photon description of B and E fields.

Is it possible to create a beam of light with frequency of 0?

It will not be light, as you approach 0, it will be a radio wave, i.e. it will be generated by an antenna. Think of starting with a very low frequency and keep on diminishing it to approach 0 . The electrons on the antenna will be moving so slowly on the antenna that what one will see is a static electric field and will have to wait a very long time to see any change.

So this would involve photon(s) that move forward without fluctuating with any frequency.

Now we are talking quantum mechanics and the link above is illuminating. In quantum mechanics photons can be virtual,

the wave function of a single photon has several components - much like the components of the Dirac field (or Dirac wave function) - and this wave function is pretty much isomorphic to the electromagnetic field, remembering the complexified values of E and B vectors at each point. The probability density that a photon is found at a particular point is proportional to the energy density (E2+B2)/2 at this point. But again, the interpretation of B,E for a single photon has to be changed.

So whether the field around an object is electric or magnetic or both is encoded in the "polarization" of the virtual photons.

And virtual photons need studying Feynman diagrams and quantum electrodynamics. At the level you are asking you should take it on trust that quantum electrodynamics blends with classical electrodynamics smoothly .

If yes, how could this be done?

It is already done wherever there are electrostatic fields and magneto-static fields, they are not called beams.

Also, considering that such a beam would not have any known color (since our eyes recognize color based on the frequency of the light beam), would it be possible to create a sensor that senses that beam?

Voltmeters measure electric fields and compasses display magnetic ones.

$\endgroup$
  • $\begingroup$ I apologize to disturb you about a question asked many hours ago, but a photon with 0 energy doesn't carry electrons with it. A photon with non-zero energy is the photon itself, it doesn't carry with it the charges of the antenna. So, if the antenna is reduced to static charges, that is a thing separated of the photon. $\endgroup$ – Sofia Mar 11 '15 at 13:23
  • $\begingroup$ @Sofia approaching the limit of zero, it is a photon. At that limit the sinusoid becomes practically flat as far as an experiment would go. I mean that the acceleration to the electrons is so slow that just the electric field of the electrons is displayed $\endgroup$ – anna v Mar 11 '15 at 13:52
  • $\begingroup$ Anna, it seems to me that we have in mind two different things. The sinusoid on the oscilloscope becomes direct current, that's true, but imagine the photons flying in vacuum, no charges, no current, as the user says "a beam of light with frequency of 0". Their energy is $\hbar \omega$. And when $\omega = 0$ what is their energy? Zero, nothing. They won't produce electric charges out of nothing. $\endgroup$ – Sofia Mar 11 '15 at 20:37
  • $\begingroup$ @Sofia The sinusoid I am talking about is the one in the answer by andy.. above, built up by zillions of photons of frequency nu. In an antenna the electrons are accelerated and give off those photons. The accelerated charge belongs to the electrons. In the limit where the acceleration goes towards zero, the sinusoid becomes very flat.( at zero the photons are virtual , which is what Lubos' answer is saying.) so the E part is really the field of the electrons. $\endgroup$ – anna v Mar 12 '15 at 4:07
0
$\begingroup$

A photon with 0 frequency wouldn't interact with anything, so it is impossible to know if it exists.

$\endgroup$
  • 2
    $\begingroup$ As others have pointed out, it would represent e a static electric field. $\endgroup$ – Kyle Kanos Mar 11 '15 at 0:25
  • 1
    $\begingroup$ @KyleKanos How can be a static electric field? The latter has energy, while a photon of 0 frequency has no energy. $\endgroup$ – Sofia Mar 11 '15 at 13:07
  • $\begingroup$ @Sofia: Did you bother reading the other answers? $\endgroup$ – Kyle Kanos Mar 11 '15 at 13:08
  • $\begingroup$ @KyleKanos yes, of course I read, but I don't agree, simply because a photon with no energy cannot contain energy. Mathematics may elude if we don't control it. E.g. $\vec E = E_0 e^{i(\vec k \cdot \vec r - \omega t)}$ becomes $E_0$ for $\omega = 0, k = 0$. But $E_0$ needs charges. Where they are? Where from should they appear? If we loose the ground (phenomenology) we can get many mathematical miracles. $\endgroup$ – Sofia Mar 11 '15 at 13:18
  • $\begingroup$ @KyleKanos As I am not sure whether my previous comment was clear, I summarize here what I wrote to Anna. When an antenna emits photons, they don't carry with them the charges of the antenna. The charges and the photons separate as two different entities. If the antenna becomes a static charge, it emits no more. The no more are the zero energy photons, not the charges of the antenna. $\endgroup$ – Sofia Mar 11 '15 at 13:42
0
$\begingroup$

When working with nonlinear crystals, you can generate 0 frequency. TWM or Three wave mixing allows you to generate a frequency that can be sum or subtraction of two other frequency: $$\omega_3=\omega_2-\omega_1$$ If $\omega_2=\omega_1$ we have $\omega_3=0$ and then we can generate 0 frequency. when this occurs we have a constant current.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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