PHY146_ExtraCredit

8/13/2019 PHY146_ExtraCredit

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J.T. Rubin [email protected]

PHY 146 Extra credit

Part I Principle of Least Time

We discussed in class that the laws of propagation of light were understood long before theelectromagnetic or even wave nature of visible light were known. In the first century A.D., Heroof Alexandria described the law of reflection. Around 1000 A.D., the Persian scientist Al-hazenpublished a massive study of optics in which he described the law of refraction (Al-hazen hadplenty of time to carry out his scientific research- he was under house arrest for 10 years due toa dispute with the ruler of Eqypt.) Then , in the 1650s, Pierre de Fermat (a French lawyer whostudied mathematics in his spare time) described a more general law called the principle ofleast time .

Fermats principle states that a light ray trave ling between two points always takes the path of

minimum time. To use the principle mathematically, the time of travel between two points canbe expressed in terms of unknown variables. From basic calculus, the extrema of a functionoccur where its first derivative is zero. The extremum is a minimum if the second derivative ispositive.

Consider the situation below, where a ray propagates across the boundary of two media, eachwith different refractive index:

In this picture, the y lengths are constant for the two given points, but the x lengths can bevaried to produce different path lengths (and thus times). The total distance of travel is:


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The velocity of propagation in a medium with refractive index n is v = c/n , from which the traveltime can be determined:

Question I.A

Use Fe rmats principle to derive Snells law. Since the only variable is , you need to showthat when , the resulting equation can be re-arranged to give the law ofrefraction. Make sure you express the law in terms of angles defined with respect to thenormal. You do not need to evaluate the second derivative since it is obvious that thisfunction is a minimum and not a maximum (any small change of path will extend the lengthin the slower medium and shorten it in the faster medium, increasing the time) .

Part II A different route to the laws of light propagation

A different approach to determining the laws of light propagation is to start directly from thewave equation:

A plane wave solution to this equation is:

Where the wave velocity is .

For a wave propagating in two dimensions the wave equation is:

The general solution is:

| |


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This is the simplest form of a wave, but it is only valid for waves in a single medium. When lightpasses through or reflects off boundaries of different substances it moves in more complicatedpaths. To derive these paths, the wave equation must be solved. However, if we are interestedonly in ray propagation and not interference effects, simplifications can be made. One method

is to assume that the equation of a ray for a non-straight path is similar to the simple rayequation, but with a different position dependence than in Eq. (5). For example:

Where is some function of position. This is known as the eikonal approximation, fromthe Greek word eikon meaning image or icon.

To find the unknown function, we use the wave equation, Eq. (4).

Question II.A

Show that when Eq. (6) is plugged into the wave equation Eq. (4), the result is:

() ( ) Eq. (7)

For large frequencies, the second term in Eq. (7), proportional to , is large compared to the

first. Additionally, in the limit that the ray travels in straight lines , the ray equation takes theform of Eq. (5), in which case , and the second derivatives of are zero.

For small deviations from straight paths, we expect the second derivatives in Eq. (7) to be small,so that the first term can be neglected:

() ( ) Dividing by the overall factors gives:

( ) ( ) This can be re-written. Recall that:


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Therefore:

( ) () So that Eq. (8) becomes:

This equation is simpler if we get rid of the square. Since is a vector, the right hand side ofthe equation must be a vector too. To find the vector direction, note that the operatorapplied to a function points in the direction in which the function is changing. Since changes

along the propagation direction of the ray, | | , where the vector is a unit vectorparallel to the ray. Therefore:

This equation can be converted to integral form. The fundamental theorem of calculus statesthat the line integral of the derivative of a function along a closed path is zero.

As a result:

This equation will be used to derive Snell s law.

Consider the situation below, where a light ray passes through from one medium to anotherwith the boundary along the z axis.

4 3

21


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To integrate Eq. (11), well use the blue dotted path:

The integrals along the horizontal paths (1 2 and 3 4) are equal and opposite, and cancel:

Since the ray vector and refractive index are different on each side of the boundary, the integralcan be written:

The limits of these integrals are equal and opposite, so they can be combined. The integration

is along the z axis, which we indicate with the unit vector , so that :

This equation must hold for any arbitrary integration path, so

Note that , since and are unit vectors. The angle is between and . Thus:

This almost looks like Snell s law, but is parallel to the boundary, so and are not theconventional angles formed with respect to the normal line.

Question II.B

Use Eq. (13) to derive Snells law. To do this, draw a normal line and label the anglesmeasured with respect to the normal, and express these angles in terms of the and .Then use trigonometric relationships to simplify and arrive at Snells Law.


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Part III Emission of electromagnetic waves

In class, I described a hueristic derivation of the emission of waves by a vibrating charge. Itinvolved taking the standard expressions for the potential of a point charge, and then shifting

the time variable by an amount R/c, which is the time it takes for the change in the field topropogate a distance R. The basic idea is that electric and magnetic interactions betweendistant objects do not occur instantaneously, but take some time to propogate through space.Maxwell s equations tell us that these changes propogate at speed c.

To actually determine the waves produced by some source, one needs to solve theinhomogeneous wave equation. For two dimensions this is:

Where is some function of space and time that represents the source of waves(accelerating charges or currents). This equation is more difficult to solve than the free spacewave equation because of the dependence of on space and time. Without thefunction (i.e. in free space), the wave equation contains time and space derivatives ofthe field E, but not the time and space coordinates themselves. If you take a dedicated courseon electromagnetism, you will study methods for solving this equation. For now, we willanalyze another hueristic approach in order to understand the main physical conceptsinvolved in wave emission without having to solve the inhomogeneous wave equation.

A very simple hueristic derivation of the process of wave emission was developed by J.J.Thomson, the discoverer of the electron. His idea is based on the concept, described above,that disturbances in the electromagnetic field propogate at finite velocity. Therefore, if anelectron which was initially at rest is suddenly accelerated and begins to move, the electric fieldat a far away point R does not immediatley change from its stationary value. Instead, it takesan amount of time t = R/c for this field point to receive the message that the electron hasmoved.

Our electron, initially at rest, produces an electric field with magnitude:

The vector direction of the field points radially outward from the electron, as shown below:


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Now suppose that at the electron is suddenly given a push, and has accelerated to somesmall velocity after time , so that its acceleration is .

At some later time, , the charge has moved from its original position by a distance .When the electron is located at its new position, the field is again directed radially outward.However, at this time, the change in field distribution has only propogated a distance .Therefore, distances for still have the original field distribution, while distances have the updated field distribution.

The picture below shows this scenario. The new field ( ) and old field ( are shiftedwith respect to each another, so field lines do not match up. In between them is a thin shell oflength corresponding to the field produced during the short time when the electron was

accelerating. Since electric field lines must be continuous (a discontinuity means the derivativeis undefined, which violates Maxwells equations), the old and new fields must be matched upwithin this region.

(Image taken from High Energy Astrophysics by M.S. Longair)


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A more detailed view is shown below:

The field lines in the thin region are not purely radial, but have a perpendicular component.From the geometry in the picture, the ratio of radial to perpendicular components is:

Problem III.A

Using Eq. (15) for , as well as and , show that is given by:

Show that the intensity, | | is given by:

This intensity has the proper inverse square dependence, so that when integrated over aspherical surface of area , the factors cancel. This means that the total powerradiated outward is the same through any surface, even in the limit that .


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Problem III.B

Finally, for a vibrating charge with position, , show that the averageintensity | | is:

The dependence means that higher frequencies radiate a much greater intensity. This isresponsible for the blue color of sky, the high frequency blue light being much more stronglyscattered than red.

Epilogue

Finally, lets assume that we have a very hot object in which the electrons undergo continualoscillations. The energy of an oscillator, from basic mechanics, is

In classical thermodynamics, the temperature is proportional to the average internal energy permolecule.

Where is Boltzmann s constant and T is the temperature in units of K. When this expressionis inserted into the expression for intensity, we have:

This equation is very strange. It seems to be saying that higher frequencies radiate with largerintensity, for a given temperature. If we were to integrate over to get the total intensity forall frequencies, the result is infinite. This obviously does not occur. In classical physics thisphenomenon was known as the ultraviolet catastrophe, meaning that high freq uenciesseemingly emit an infinite amount of energy.

The actual spectrum of a hot object is finite, and the intensity has a maximum value at afrequency . This spectrum is shown below. This frequency increases with temperature,which responsible for the well-known observation that blue radiation is hotter than redradiation (such as in a flame).


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It is incredible that the classical theory of electromagnetism, which explains circuits, motors,generators, radio broadcasting and many other phenomena, fails when applied to something asapparently simple as the radiation spectrum of a hot object. This is the starting point ofquantum theory.

http://en.wikipedia.org/wiki/File:Black_body.svg
http://en.wikipedia.org/wiki/File:Black_body.svghttp://en.wikipedia.org/wiki/File:Black_body.svghttp://en.wikipedia.org/wiki/File:Black_body.svg

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