Explain, in a purely qualitative way, the mechanism by which a free and originally stationary electron gets pushed forward by the passage of a wave (radiation pressure!). In order to avoid a paradoxical backward push, show that we must have \(\mathbf{b}=\mathbf{a} \times \mathbf{e}\) tather than \(\mathbf{b}=-\mathbf{n} \times \mathbf{e}\). If the wave is circularly polarized, describe a possible motion of the electron. Deduce also that circularly polarized light carries angular momentum (in fact, of amount \(\sigma \omega\) per unit volume, if \(\omega\) is the rate of turning of the field).

Short Answer

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Question: Explain the mechanism of radiation pressure on a free and originally stationary electron and how it relates to the angular momentum of circularly polarized light. Answer: When an electromagnetic wave passes through a free electron, the electron experiences a force due to the electric field of the wave, causing the electron to oscillate. This oscillating electron generates a magnetic field, which must be related to the electric field and wave vector to prevent backward push on the electron. In the case of circularly polarized light, the electron moves in a helical path due to the changing direction of the electric field, gaining both linear and angular momentum. The angular momentum per unit volume of circularly polarized light is given by \(\sigma \omega\), where \(\sigma\) represents the energy density of the wave, and \(\omega\) is the angular frequency of the wave.

Step by step solution

01

Radiation pressure on electron

As the electromagnetic wave passes through the free electron, it experiences a force due to the electric field \(\mathbf{e}\) of the wave. This force makes the electron oscillate, and the oscillating electron in turn generates a magnetic field \(\mathbf{b}\) which interacts with the initial electric field.
02

Avoid paradoxical backward push

In order to avoid a backward push on the electron, the magnetic field of the wave, \(\mathbf{b}\), must be related to the electric field \(\mathbf{e}\) and wave vector \(\mathbf{a}\) (which points in the direction of propagation) in a way that prevents it. If we had \(\mathbf{b} = -\mathbf{n} \times \mathbf{e}\), it would imply that the magnetic field would be in the opposite direction of the correct orientation needed to produce a forward force on the electron. Therefore, we must have \(\mathbf{b} = \mathbf{a} \times \mathbf{e}\).
03

Electron motion in circularly polarized light

In the case of circularly polarized light, the electric field \(\mathbf{e}\) changes its direction continuously as the wave propagates, tracing a circular path. As a result, the force on the electron will also change its direction, causing the electron to move along a helical path parallel to the direction of the wave propagation.
04

Angular momentum of circularly polarized light

As the electron moves in a helical path, it gains both linear and angular momentum. Based on the symmetry of circularly polarized light, we can deduce that the wave carries angular momentum as well. The angular momentum per unit volume is given by \(\sigma \omega\), where \(\sigma\) represents the energy density of the wave, and \(\omega\) is the angular frequency of the wave (or the rate of turning of the field).

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Most popular questions from this chapter

The apparent brightness \(b\) of a distant luminous source (which can be idealized as a point source) is the rate at which radiant energy from it is received on a unit area perpendicular to the line of sight; the absolute brightness \(B\) is defined as the apparent brightness at unit distance from the source in its rest frame. Prove that for two momentarily coincident observers the apparent brightnesses of the same source are in the ratio of the squares of the observed frequencies: \(b^{\prime} / b=v^{\prime 2} / v^{2}\). Hence or otherwise prove that \(b=\left(B / R^{2}\right)\left(v^{4} / v_{0}^{4}\right)\). where \(R\) is the distance of the source from the oberserver in the observer's inertial frame at the time the light he measures was emitted, and \(v_{0}\) is its proper frequency. [Hint: for the two frames of Fig. \(10(b)\), \(\left.R \sin \alpha=R^{\prime} \sin \alpha^{\prime} .\right]\)

Obtain the Liénard-Wiechert potentials \((40.8)\) of an arbitrarily moving charge \(q\) by the following alternative method: Assume, first, that the charge moves uniformly and that in its rest frame the potential is given by \((40.2)\). Then transform this to the general frame, using the four-vector property of \(\Phi^{\mu}\). Finally extend the result to an arbitrarily moving charge by the argument we used after (40.2). [Hint: if the separation \((c t, r)\) between two events satisfies \(r=c t\) in one frame, it does so in all frames.]

In a frame \(S\) there is a uniform electric field \(e=(0, a, 0)\) and a uniform magnetic field \(c \mathbf{b}=(0,0,5 a / 3)\). A particle of rest mass \(m\) and charge \(q\) is released from rest on the \(x\)-axis. What time elapses before it returns to the \(x\)-axis? [Answer: \(74 \pi \mathrm{cm} / 32 \mathrm{aq}\). Hint: look at the situation in a frame in which the electric field vanishes.]

14\. Prove that the electromagnetic energy tensor satisfies the following two identities: $$ M_{\mu}^{\mu}=0, \quad M_{\sigma}^{\mu} M_{v}^{a}=\left(I \varepsilon_{0} / 2\right)^{2} \delta_{v}^{\mu}, $$ where \(I^{2}=\left(c^{2} b^{2}-e^{2}\right)^{2}+4 c^{2}(e \cdot b)^{2}\). [Hint: It may be easiest to establish the second identity in a particular frame, e.g. if \(I \neq 0\), in one in which e is parallel to b. (We regard the vanishing of e or b as a particular case of this.) For the case \(I=0\), appeal to continuity. \(]\)

Obtain the field (41.5), (41.6) of a uniformly moving charge \(q\left[=\left(4 \pi \varepsilon_{0}\right)^{-1} Q\right]\) by the following alternative method: Assume that the field in the rest frame \(S^{\prime}\) of the charge is given by $$ \mathbf{e}^{\prime}=\left(Q / r^{\prime 3}\right)\left(x^{\prime}, y^{\prime}, z^{\prime}\right), \quad \mathbf{b}^{\prime}=0, \quad r^{\prime 2}=x^{\prime 2}+y^{\prime 2}+z^{\prime 2} $$ then transform this field to the usual second frame \(\mathrm{S}\) at \(t=0\). [Hint: obtain \(\mathbf{b}=\mathbf{u} \times \mathbf{e} / c^{2}\) from \((39.2)\); from the inverse of \((39.1)\) obtain \(\mathbf{e}=\left(Q \gamma / r^{3}\right)(x, y, z) ;\) finally use (41.7). ]

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