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Let's say you place an electric field meter some distance from a light bulb.

As a function of time the output of the meter would be $\mathbf{E}(t)$. I would guess that the electric field will be some very rapidly fluctuating random-looking function.

My question is, what determines the timescale of these fluctuations in $\mathbf{E}(t)$?

In my understanding, the light from a light bulb is emitted by electrons in the atoms in the filament jumping up and down between energy levels. Each de-excitation emits a photon. I am imaging each of these photons as a traveling localized wavepacket. The filament is made up of $\sim10^{23}$ atoms and they're all emitting these wavepackets independently of one another. So I guess the total E-field is the sum of a prodigious number of these wavepackets added together with random phases.

What determines the timescale of the individual wavepackets? The average lifetime of an excited state? What's the back-of-the-envelope way of estimating that? How many of these wavepackets does one atom emit per second? Finally, what determines the time scale of fluctuations of the sum of a ton of these wavepackets?

I'm looking for a solid physical picture of what's going on that can let me calculate some numbers (order of magnitude estimates). Also, what is right terminology that people use when talking about this?

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Your scenario is correct. If you could measure the electric field rapidly enough then it would show very rapid changes in its strength and polarisation caused by the superposition of wave packets from many radiative "events" within the bulb filament.

A key point then is the length of a wave packet. This will be determined by the radiative lifetime of the excited state that produced it. There is only so much generalisation that can be made here. The light of the bulb arises in a variety of ways from both continuum and discrete transitions.

Of the former, I do not know of any fundamental timescale and suspect this is also a continuum. Of the latter it is the inverse of the Einstein A coefficient for the transition.

Many transitions at visible wavelengths have radiative lifetimes that are close to the classical radiative lifetime (see for example https://physics.stackexchange.com/a/142387/43351 ) , unless quantum mechanics conspires to demand they proceed by other than an electric dipole transition. The timescale is $$ \tau = \frac{6\pi\, \epsilon_{0}\,m_{e}\, c^{3}}{e^{2}\, \omega^{2}}\, , $$ where $m_e$ is the electron mass and $\omega$ the angular frequency of the radiation $(2\pi f)$. A more convenient statement of this is $$\tau = 3.5\times 10^{-8} \left(\frac{\lambda}{\rm 500 nm}\right)^2 {\rm s},$$ where the wavelength $\lambda$ is in nm.

Each wave packet carries a photon of energy,so that gives you a number per second emitted by a light bulb of specified radiative power.

The packets are emitted in all directions, so their density is diluted by the inverse square law.

The connection to an electric field is via the classical electromagnetic energy density, which is proportional to the square of the electric field.

I think those are the ingredients, but am unsure how to proceed other than by simulation. However the answer will be unrealistic because of the range of radiative lifetimes from the real emission processes.

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  • $\begingroup$ "The connection to an electric field is via the classical electromagnetic energy density, which is proportional to the square of the electric field." Do you mean something like: the energy one of these wavepackets is $\hbar \omega$ so the amplitude of the electric field of the wavepacket is $\sqrt{\hbar \omega}$? $\endgroup$ – Alex Oct 4 '18 at 14:24
  • $\begingroup$ @Alex, no I don't. The energy density of the electromagnetic field is $\epsilon_0 E^2/2 + B^2/2\mu_0$ and for an EM wave the amplitude of the electric and magnetic field are related by $E_0 = cB_0$ (in vacuum). The energy density in a photon model would be the number of photons per unit volume multiplied by their energies. $\endgroup$ – Rob Jeffries Oct 4 '18 at 15:54
  • $\begingroup$ Ah ok. But this is starting to look like counting particles in a region and then working backwards to the E-field rather than starting with the E-field's of all the wavepackets and summing up to get the total. $\endgroup$ – Alex Oct 4 '18 at 16:10
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Let's say you place an electric field meter some distance from a light bulb.

italics mine

This is a science fiction scenario because no physical simple detector measuring a field can detect the varying electric field of light which in its instantaneous position, because of the size of the velocity c , will have moved away from the detector, even if it is from a coherent source, as a laser beam. This is what the electric field and magnetic field of a coherent beam looks like.

elm

Look at this link of how electromagnetic field detectors measure electric field. This set up is not adaptable to the frequency of light from an incoherent source.

In any case, in an incoherent beam,as is the one from a light bulb, there will be superpositions in all directions and the electric field is a vector,and the electric fields from all those frequencies emitted by the bulb, will instantaneously , within errors, average out to zero because of the randomness of the phase differences and the sinusoidal shape of the waves.

Direct measurement of the electric field of coherent light is described here, and it is evident that coherence is the key to such a measurement:

elfield

The key to this measurement was the generation of single 250-attosecond extreme ultraviolet pulses, a feat achieved by the same collaboration a few months ago (Nature, February 26, 2004). The attosecond extreme ultraviolet pulse knocks electrons free from atoms to probe the electric field of a wave consisting of only a few cycles of red laser light. The electric field of red light accelerated or decelerated the electrons set free with respect to the light wave with a 100-attosecond timing precision. The change in the electrons’ energy (shown in units of electron volts, eV, in Fig. 1), measured as a function of delay (shown in units of femtoseconds, fs, in Fig. 1) between the attosecond pulse and the laser light wave clearly exhibits the build-up and disappearance of the laser pulse within a few femtoseconds as well as oscillations with a period of the 2.5-fs wave cycle of 750-nm (red) light. The measured energy change directly yields the variation of the instantaneous strength and direction of the electric field of the few-cycle light wave (Fig. 2).

You ask in the title:

What determines the timescale for fluctuations in the electromagnetic field from a light source?

From the figure above ,of the successful measurement of the electric field of red light, part of the frequencies emitted by a light by a light bulb, one sees that the time is determined by the frequency of the light, and is of the order of a few femptoseconds ($10^−{15}$), and it is not a simple electric field meter that could work at such timescales.

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  • $\begingroup$ Thanks. I am specifically interested in the incoherent situation, not the coherent one. This is a conceptual question (at the moment) -- the "electric field meter" is a fiction as you say. But there is some actual well-defined E-field which is continuous (I imagine) as a function of time. I want to understand the behavior of this actual E-field and not the time averaged E-field. $\endgroup$ – Alex Oct 4 '18 at 12:38
  • $\begingroup$ within delta(time) my view is that due to superposition and the vector nature of electric fields and and the large number of frequencies in the bulb light, the field will be zero if at all possible to measure it. Look at the figure of the pulse, how the field varies and how there will be different vectorial directions coming from the source, and think of the superposition of many frequencies at a given (x,y,z) at time t. $\endgroup$ – anna v Oct 4 '18 at 13:10
  • $\begingroup$ I disagree that the field will be zero. If $E$ is identically 0 there's simply no radiation at all. $E$ will average to zero over a finite spatial or temporal scale, but I want to know about the behavior of $E(t)$ at a single location. Also, we can forget about polarizations. Let's say we only care about the $x$-polarization (replace all my $E$'s with $E_x$'s). $\endgroup$ – Alex Oct 4 '18 at 13:21
  • $\begingroup$ You are confusing energy with the electric field. The energy is proportional to the square of the electric field, the peak energy ~to E^2 farside.ph.utexas.edu/teaching/302l/lectures/node119.html . The electric field itself is a vector quantity, with + and - that sinusoidally fluctuates. If you go to femtoseconds and a single frequency you can measure the fluctuations. Otherwise with a mix of frequencies and larger times even with the laser experiment, the electrons would go up and down and seem to remain in place if the timing is larger that femptoseconds. $\endgroup$ – anna v Oct 4 '18 at 15:49
  • $\begingroup$ I'm not concerned with the practical difficulties of measuring electric fields. I want to understand conceptually what's going on. It seems to me that the electric field $E(t)$ as a physical entity exists even if it's hard to measure. I think it will be some continuous function of time and not identically zero. $\endgroup$ – Alex Oct 4 '18 at 16:16

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