Intensity, pressure and impulse of an electromagnetic wave. Wave optics

Learning Objectives: introduce and formulate the concepts of intensity, pressure and impulse electromagnetic wave; theoretically and experimentally substantiate these concepts.

Development Goals: improve critical thinking and the ability to reason by analogy; ability to apply theoretical knowledge to explain physical phenomena.

Educational goals: develop strong-willed, motivational and tolerant personality characteristics.

Didactic tools:

1. Myakishev G.Ya. Physics: Textbook. for 11th grade general education institutions/ G.Ya.Myakishev, B.B.Bukhovtsev.– M.: Education, 2004.
2. Kasyanov V.A. Physics. 11th grade: Educational. for 11th grade general education textbook establishments. – M.: Bustard, 2002.
3. Electronic version of the lesson summary; video clips of demonstration experiments.
4. A set for studying electromagnetic waves (produced by JSC NPK Computerlink), voltmeter, milliammeter, adjustable voltage source.

5.1. Introduction

Teacher. Today we will continue our acquaintance with the most important characteristics electromagnetic wave as a material object. Energy transfer by a wave is characterized by a special quantity called intensity. An electromagnetic wave falling on an obstacle exerts pressure on it. In this case, the obstacle acquires momentum; therefore, the electromagnetic radiation itself has momentum. The pressure and momentum of the electromagnetic wave available to us are negligible, so we will not be able to measure them in educational experiments. However, we will be able to explain their existence and estimate the values ​​of the corresponding quantities.

5.2. Electromagnetic wave intensity

Teacher. Remember how a harmonic wave is written mathematically and how its energy is expressed.

Students. Equation for tension electric field in a harmonic electromagnetic wave has the form Where

and its energy density:

ω = ε 0 ε E 2 . (5.2)

Teacher. The product of energy density and wave speed is called surface energy flux density j= ω υ .

Students. Do we really have to remember this long term?!

Teacher. Of course not. But for some reason the authors of school textbooks love it very much, so if you want to get higher education, get acquainted with this term and its familiar variant “energy flow density”, whether you want it or not, you will have to remember it.

Students. Then you need to at least understand where it came from.

Teacher. A wave passing normally through an area S during t, occupies volume V = sυt(Fig. 5.1). Since the energy density is equal to the energy per unit volume: ω = W/V, – then the surface energy flux density can be written as:

Wave Energy Ratio W In time t during which it passes through the surface is called flow of energy. And the ratio of the energy flow to the surface area through which it passes can naturally be called surface energy flux density.

Students. Now it is clear that this is simply the energy transferred by a wave per unit time through a unit area, or the radiation power passing through a unit area.

Teacher. Find out how the surface energy flux density of an electromagnetic wave depends on its frequency.

Students. From formulas (5.1), (5.2) and (5.3) we obtain:

Since the cosine here is squared, the surface energy flux density of the electromagnetic wave oscillates at a frequency twice the frequency of the wave. How to measure this value?

Teacher. They measure not the instantaneous, but the time-average value of the energy flux density, which is called wave intensity. You know well that the average value of the square of the cosine is 1/2. Substituting it into the previous formula and taking into account the expressions for E m(5.1) and for after small transformations it can be obtained that the intensity of the harmonic wave is equal to

Where K– constant coefficient. Analyze this result.

Students. From formula (5.4) it follows that the intensity of the electromagnetic wave emitted by a harmonic oscillator, other things being equal, is proportional to the fourth power of its frequency and inversely proportional to the square of the distance traveled by the wave.

Teacher. Give another option for defining wave intensity and qualitatively explain why the intensity of an electromagnetic wave is proportional to the fourth power of its frequency.

Students. Wave intensity is the time-average energy W cp passing through a unit area per unit time:

This means that the intensity is proportional to the wave energy J ~ W cf. And the energy is proportional to the square of the electric field strength W Wed ~ E m 2. In turn, the electric field strength is proportional to the acceleration of the charge emitting the wave E m ~ a m, and the acceleration is proportional to the square of the charge oscillation frequency a m~ ω 2 . It follows that intensity is proportional to the fourth power of frequency:

J ~ W Wed ~ E m 2 ~ a m 2 ~ ω 4. (5.6)

Teacher. Clarify what values ​​of tension and acceleration you mean.

Students. We are talking about the amplitude of the electric field strength E m electromagnetic wave and acceleration amplitude a m harmonically oscillating charge.

Teacher. Why is intensity inversely proportional to the square of the distance?

Students. Because the electric field strength of an electromagnetic wave created by an oscillating charge is inversely proportional to the distance to the charge, and the intensity of the wave is proportional to the square of the intensity.

5.3. Experimental study of dipole radiation

Teacher. We will experimentally study the dependence of the intensity of an electromagnetic wave on the distance to the radiating vibrator. To do this, next to the receiving dipole lamp (2.5 V; 0.15 A) we will place exactly the same incandescent lamp, connect it through an ammeter to an adjustable constant voltage source and turn on a voltmeter in parallel with this reference lamp. Let's set the distance between the emitting and receiving dipoles to 10 cm and, by adjusting the source voltage, ensure that the brightness of the reference lamp becomes equal to the brightness of the receiving lamp (Fig. 5.2, A). Then we can say that the same power is released in the reference lamp as in the receiving lamp. Calculate it.

Students. The instruments show that the current and voltage across the reference lamp are respectively equal I 1 = 0.111 A and U 1 = 1.8 V, which means the required power P 1 = U 1 I 1 = 0.20 W.

Teacher. Now let’s remove the receiving dipole to a distance of 20 cm from the emitting dipole, repeat the measurements and draw conclusions.

Students. Happened I 2 = 0.087 A and U 2 = 1.2 V (Fig. 5.2, b), That's why P 2 = U 2 I 2 = 0.10 W. Attitude P 1 / P 2 equals two, not four, as you might expect! Is there really a mistake in the theory?

Teacher. Before changing the theory, let's see whether the experimental conditions correspond to its initial data. Let us remember that when considering the propagation of energy from a radiating dipole, we tacitly assumed that it was radiated equally in all directions. In other words, we assumed that the dipole is isotropic source. In this case, electromagnetic energy is evenly distributed over the spherical surface. Since the area of ​​the sphere S= 4π r 2 is proportional to the square of its radius, then the power per unit area, i.e. The intensity of the wave is inversely proportional to the square of the distance.

Students. It is necessary to investigate how the dipole radiates along different directions, and then draw a conclusion about the radiation intensity.

Teacher. I place the receiving dipole parallel to the emitting dipole so that the brightness of its lamp becomes maximum, and move it around a circle with the center at the center of the emitting dipole (Fig. 5.3). Draw a conclusion from the result of the experiment.

Students. At all points of the circle, the receiving dipole lamp burns with the same intensity. This means that in all directions perpendicular to the radiating dipole, the intensity of the electromagnetic wave is the same.

Teacher. Now I move and rotate the receiving dipole in a plane passing through the emitting dipole (Fig. 5.4). I do this so that the receiving dipole, moving in a circle with the center in the emitting dipole, is directed tangentially to this circle. What do you observe and what conclusion do you come to?

Students. The lamp burns less and less as the receiving dipole rotates relative to the emitting dipole. This means that a dipole connected to the generator produces maximum radiation in the direction perpendicular to the dipole, and does not radiate at all in the direction of the dipole itself.

Teacher. If in a polar coordinate system we plot the dependence of the intensity of an electromagnetic wave on the angle between the dipole and the direction of radiation, we will obtain a radiation pattern of a half-wave dipole similar to that shown in Fig. 5.4 (the length of the arrows is proportional to the intensity). Now return to the experiment in which we measured the dependence of the intensity of an electromagnetic wave on distance, and try to explain its result.

Students. The experiment just performed shows that the dipole is not an isotropic source of an electromagnetic wave: the radiation propagates mainly in a plane perpendicular to the radiating dipole and passing through its center. This means that the radiated energy near the dipole falls not on a spherical, but on a cylindrical surface. The lateral surface area of ​​a cylinder is proportional to its radius. Therefore, the intensity of the dipole radiation is inversely proportional not to the square of the distance, but simply to the distance to the source.

Teacher. Note that the receiver is not isotropic: its sensitivity also depends on the direction in which the wave falls on it. IN theoretical model we assumed the source and receiver to be pointlike and isotropic. It is not difficult to realize that the conditions of this model will be met if the distance between the source and the receiver significantly exceeds their sizes.

5.4. Electromagnetic Wave Pressure and Momentum

Teacher. Experiments show that an electromagnetic wave transfers energy, which means that when falling on obstacles, it must exert pressure on them. It is quite difficult to correctly derive the corresponding formula, so we will use hydrodynamic analogy. Imagine that water flows through a pipe with cross-sectional area S at speed u (Fig. 5.5). The energy density in moving water is obviously equal to ω = W/V = mu 2 /(2V) = ρ u 2 /2, where ρ is the density of water. Suddenly the pipe opening is closed with a valve. What happens?

Students. The water near the valve stops and contracts. The compression front propagates with the speed of elastic deformation υ towards the moving water. Speed υ is the speed of an elastic wave or the speed of sound in water.

Teacher. Right. Let us apply the law of conservation of momentum to the phenomenon under consideration. In a short time τ the valve stops the volume of water Sυτ with mass ρ Sυτ, which transmits impulse ρ to the damper Sυτ u. In this case, a force acts on the damper F, whose momentum is equal Fτ. Equating the last two expressions, after reducing by time τ we obtain the equality ρ Sυu = F. Hence the pressure of the suddenly stopped flow of water is equal to P = F/S = ρ uυ.

Students. But the speed of sound in water is 1500 m/s, does the pressure really increase that much?

Teacher. That's right, and this phenomenon is called hydrodynamic shock. By the way, his theory was created by our compatriot N.E. Zhukovsky. But let's not get distracted. Let us assume that water in the pipe flows with the speed of an elastic wave u = υ . What follows from this?

Students. Then the resulting pressure is P = ρ uυ = ρ u 2. Since the energy density in flowing water is ω = ρ u 2/2, then we must conclude that the pressure when the water suddenly stops is P= 2ω.

Teacher. You have just found a formula for the pressure exerted on a completely reflective obstacle by a normally elastic wave incident on it. But if this formula is valid for elastic waves, then why not assume that it will also be valid for electromagnetic waves?

Students. Then we can assume that the electromagnetic wave exerts on the obstacle or mirror reflecting it a pressure equal to twice the energy density of the incident wave. If a wave propagates in a vacuum, then its speed υ = c and taking into account the expression for intensity J= ω cр υ = ω cр With. (5.5) we have:

P= 2ω cр = 2 J/c. (5.7)

Teacher. Since an electromagnetic wave exerts pressure, it must have momentum. Try to find the formula for the pulse of electromagnetic radiation. To do this, consider the reflection of a short burst of electromagnetic radiation from a mirror.

Students. If the pulse of the electromagnetic wave is p, then at full reflection from the mirror in time t the change in momentum is 2 p. Mirror at the same time t gets momentum Ft = PSt = 2p. Since the pressure P = 2J/c(5.7), then, substituting this expression into the previous formula, we obtain that the impulse of the electromagnetic wave p = J/c St.

Teacher. Once again recalling the expression for intensity J = W cf / St(5.5), we get

p = W cf / With. (5.8)

Thus, the momentum of an electromagnetic wave propagating in a vacuum is equal to the time-averaged energy of the wave divided by the speed of light in a vacuum.

5.5. Why does an electromagnetic wave exert pressure?

Teacher. We now need to establish the physical reason why an electromagnetic wave exerts pressure. Opposite the emitting dipole, I place a receiving dipole with an incandescent lamp. Prove that in an electromagnetic field a force acts on a dipole in the direction of wave propagation.

Students. Under the influence of the electric field of the wave, electrons in the receiving dipole come into oscillatory motion. In this case, there is a variable across the dipole electricity, as evidenced by the glow of the lamp. But where does the strength come from?

Teacher. Do not forget that in an electromagnetic wave, in addition to the electric one, there is a magnetic field.

Students. Got it! On the current in the conductor from the side magnetic field Ampere force acts (Fig. 5.6). To determine its direction, we apply the left-hand rule. It turns out that strength F the dipole acts in the direction of propagation of the electromagnetic wave. In the next half period alternating current in a dipole, the direction of induction will change to the opposite, but the direction of the Ampere force will not change.

Teacher. Calculations, which we will not carry out, show that the time-average value of the Lorentz force acting on electrons, which is per unit area of ​​the reflecting conductor, exactly coincides with expression (5.7). Therefore, the hydrodynamic analogy (Fig. 5.5), which we used in the theoretical model, is quite appropriate.

5.6. Conclusion

Teacher. What new did you learn in this lesson? What have you learned? What impressed you the most?

Students. We learned what intensity, pressure and momentum of an electromagnetic wave are and how they relate to each other. We found out how the intensity depends on the frequency and distance traveled by the wave. We learned to experimentally determine the intensity of electromagnetic radiation. The analogy between the flow of water and the propagation of a wave is very interesting. Experiments in which the spatial distribution of the intensity of electromagnetic radiation from a dipole is determined are convincing.

Teacher. As usual, homework given to those who are interested in doing it, or to those who want to repeat what they have learned, learn new things, deepen their knowledge and skills. You will find material for completing the task in physics textbooks and in the electronic version of the lesson summary.

The article was prepared with the support of the lecture bank www.Siblec.Ru. If you decide to acquire or expand your knowledge in various fields of science and technology, then optimal solution will go to the website www.Siblec.Ru. By clicking on the link: “lectures on physics”, you can, without spending a lot of time, gain access to lectures on physics and other scientific disciplines. The bank of lectures www.Siblec.Ru is constantly updated, so you can always find fresh and relevant material.

1. Define the surface radiation flux density. What is meant by a point source of electromagnetic radiation? How does radiation flux density depend on frequency and distance to the source? [ G.Ya.Myakishev, § 50; V.A.Kasyanov, § 49.]
2. What is the intensity of an electromagnetic wave? How does intensity depend on wave frequency? According to what law does the intensity of an electromagnetic wave emitted by a point source decrease? [ G.Ya.Myakishev, § 50; V.A.Kasyanov, § 49.]
3. How are the pressure and momentum of an electromagnetic wave determined? What is the essence of P.N. Lebedev’s experiments to determine the pressure of light? [ G.Ya.Myakishev, § 92; V.A.Kasyanov, § 50.]
4. Derive formula (5.4) for the intensity of a harmonic electromagnetic wave. [OK.]
5. How to experimentally prove that a radiating dipole is not an isotropic source of an electromagnetic wave? [OK.]
6. The radiation power of a point isotropic source of electromagnetic wave is 2 W. What is the intensity at a distance of 1 m from the source?
7. In a certain area, the intensity of electromagnetic radiation is 1 W/m2. What are the electric field strength and magnetic field induction in this area?

It is very difficult to replace the sun for a plant. Try turning on a lamp in a room on a sunny day, and you will understand how little light it can give to plants.

To the human eye, light is energy waves ranging from 380 nanometers (nm) (violet) to 780 nm (red). The wavelengths important for photosynthesis lie between 700 nm (red) and 450 nm (blue). This is especially important to know when using artificial lighting, because in this case there is no uniform distribution of waves different lengths like in sunlight. Moreover, due to the design of the lamp, some parts of the spectrum may be more intense than others. In addition, the human eye is better at perceiving waves of wavelengths that are not very suitable for plants. As a result, it may turn out that some lighting seems pleasant and bright to us, but for plants it will be inappropriate and weak.

Light intensity indoors and outdoors

The intensity of light falling on a certain plane is measured in the unit “lux”. In summer, at sunny noon, the light intensity in our latitudes reaches 100,000 lux. In the afternoon, the light brightness decreases to 25,000 lux. At the same time, in the shade, depending on its density, it will be only a tenth of this value or even less.

In houses, the lighting intensity is even less, since the light does not fall there directly, but is weakened by other houses or trees. In summer, on the south window, directly behind the glass (that is, on the windowsill), the light intensity reaches best case scenario from 3000 to 5000 lux, and quickly decreases towards the middle of the room. At a distance of 2-3 meters from the window it will be about 500 lux.

The minimum amount of light each plant requires to survive is approximately 500 lux. In weaker light it will inevitably die. For normal life and growth even unpretentious plants with little light requirement, a minimum of 800 lux is required.

How to measure illumination?

The human eye is not able to determine the absolute intensity of light, since it is endowed with the ability to adapt to lighting. In addition, the human eye better perceives waves of such lengths that are not very suitable for plants.

What to do? A special device - a lux meter - can help. When purchasing it, it is very important to pay attention to what range of the light spectrum (wavelength) it is able to measure. Otherwise, it may happen that when measuring you end up at a wavelength unsuitable for plants. Remember - a lux meter, although more accurate than the human eye, also perceives a limited range of light waves.

A camera or photo exposure meter is suitable for assessing light intensity. But since when photographing, illumination is not measured in “lux,” you will have to carry out an appropriate recalculation.

The measurement is carried out as follows:

1.Set ISO to 100 and Aperture to 4.

2. Put White list paper in the place where you want to measure the light intensity, and point the camera at it.

3. Determine shutter speed.

4. The shutter speed denominator multiplied by 10 will give an approximate lux value.

Example: if the exposure time was 1/60 second, this corresponds to 600 lux.

Based on materials:

Paleeva T.V. “Your flowers. Care and treatment", M.: Eksmo, 2003;

Anita Paulisen “Flowers in the House”, M.: Eksmo, 2004;

Vorontsov V.V. “Care for indoor plants. Practical advice for flower lovers”, M.: ZAO “Fiton+”, 2004;

Bespalchenko E. A. “Tropical ornamental plants for home, apartment and office", LLC PKF "BAO", Donetsk, 2005;

D. Gosse, “Even the sun needs help”, magazine “Vestnik Florist”, No. 3, 2005.

Thus, in geometric optics, a light wave can be considered as a beam of rays. The rays, however, themselves determine only the direction of propagation of light at each point; The question remains about the distribution of light intensity in space.

Let us select an infinitesimal element on any of the wave surfaces of the beam under consideration. From differential geometry it is known that every surface has at each point two, generally speaking, different principal radii of curvature.

Let (Fig. 7) be the elements of the main circles of curvature drawn on a given element of the wave surface. Then the rays passing through points a and c will intersect each other at the corresponding center of curvature, and the rays passing through b and d will intersect at another center of curvature.

For given opening angles, the rays emanating from the length of the segments are proportional to the corresponding radii of curvature (i.e., the lengths and); the area of ​​a surface element is proportional to the product of lengths, i.e., proportional. In other words, if we consider an element of a wave surface limited by a certain number of rays, then when moving along them, the area of ​​this element will change proportionally.

On the other hand, intensity, i.e., energy flux density, is inversely proportional to the surface area through which a given amount of light energy passes. Thus, we come to the conclusion that the intensity

This formula should be understood as follows. On each given ray (AB in Fig. 7) there are certain points and , which are the centers of curvature of all wave surfaces intersecting this ray. The distances from point O of the intersection of the wave surface with the ray to the points are the radii of curvature of the wave surface at point O. Thus, formula (54.1) determines the intensity of light at point O on a given ray as a function of the distances to certain points on this ray. We emphasize that this formula is not suitable for comparing intensities at different points of the same wave surface.

Since the intensity is determined by the square of the field modulus, to change the field itself along the ray we can write:

where in the phase factor R can be understood as both and the quantities differ from each other only by a constant (for a given beam) factor, since the difference , the distance between both centers of curvature, is constant.

If both radii of curvature of the wave surface coincide, then (54.1) and (54.2) have the form

This occurs, in particular, always in those cases when light is emitted by a point source (wave surfaces are then concentric spheres, a R - distance to the light source).

From (54.1) we see that the intensity goes to infinity at points, i.e., at the centers of curvature of the wave surfaces. Applying this to all rays in a beam, we find that the intensity of light in a given beam goes to infinity, generally speaking, on two surfaces - the geometric locus of all centers of curvature of the wave surfaces. These surfaces are called caustics. In the particular case of a beam of rays with spherical wave surfaces, both caustics merge into one point (focus).

Note that, according to the properties of the locus of the centers of curvature of a family of surfaces known from differential geometry, the rays touch the caustics.

It must be borne in mind that (with convex wave surfaces) the centers of curvature of the wave surfaces may turn out to lie not on the rays themselves, but on their extensions beyond the optical system from which they emanate. In such cases we speak of imaginary caustics (or imaginary focuses). In this case, the light intensity does not reach infinity anywhere.

As for turning the intensity to infinity, in reality, of course, the intensity at the points of the caustic becomes large, but remains finite (see the problem in § 59). The formal conversion to infinity means that the geometric optics approximation becomes in any case inapplicable near caustics. The same circumstance is also related to the fact that the change in phase along the ray can be determined by formula (54.2) only in sections of the ray that do not include points of contact with caustics. Below (in § 59) it will be shown that in reality, when passing past a caustic, the field phase decreases by . This means that if in the section of the ray before it touches the first caustic the field is proportional to the multiplier - the coordinate along the ray), then after passing the caustic the field will be proportional. The same will happen near the point of contact of the second caustic, and beyond this point the field will be proportional

A person needs lighting not only for orientation and performing any actions in the dark, but also for maintaining psychological health, comfort. Besides, artificial lighting allows workers to continue to perform their duties in the evening and at night. However, you should choose luminaires and lamps based on their characteristics, the most important of which is luminous efficiency, which is measured in lumens per watt (lm/W). In the room itself, it is also necessary to control the level of illumination and, taking this into account, select its sources.

Types of light

The most useful and safe lighting is, of course, natural. It has a warm tint and does not harm the eyes.

Note! According to its parameters, it is closest to this type There were incandescent lamps, which were characterized by a reddish glow. They did not cause eye irritation and were almost identical in the emitted spectrum natural light from the sun entering the premises through windows.

The development of technology has led to the emergence of many options for lighting devices, so when purchasing, you should pay attention to the characteristics that are indicated on the lamp packaging.

Additional Information. Thus, warm light is recommended to be placed in apartments or residential buildings, neutral – for lighting offices and production workshops. Cold - effectively used in rooms where work with small parts is carried out. It is also often used in subtropical climates, where this shade creates a feeling of coolness.

Thus, the choice of light bulb affects not only the illumination of the space, but also the moral and psychological state of an employee at work or a person in an apartment.

Luminous flux characteristics

When purchasing light bulbs, buyers often do not know or do not think about the answer to the question of how light is measured, and yet there are quite a lot of such indicators:

• Light output;
• The power of light;
• Intensity;
• Brightness.

All this physical properties luminous flux, which can be measured with special instruments, they should be taken into account in mandatory when planning room lighting (calculating the required number of lighting devices in each room or office), because this affects the health of the eyes and nervous system.

Luminous output

Luminous output is the most important parameter. It reflects the ratio of the luminous flux emitted by a light bulb or other device to the power it consumes. Accordingly, its units of measurement are lumens per watt (lm/W). This parameter allows you to evaluate economic efficiency lighting method.

The higher the luminous efficiency, the more efficiently energy is used, which means the costs for public utilities, which becomes especially relevant in the context of constant growth of tariffs. For this reason, energy-saving lamps, which provide one of the highest lm/W ratios, are very popular.

The power of light

The characteristic of radiation is not only the light output, but also the force with which its energy moves from one point in space to another over a certain time period. It must be taken into account that the light intensity can change the direction of movement depending on the conditions set by the device that generates the flow.

This parameter can be measured in candelas.

Important! When choosing a lamp, you should also pay attention to the described parameter, only the relationship is not as direct as in the case of luminous efficiency. The strength level should be selected based on the standard value that a unit of brightness of a luminous surface should have. This indicator can be found in various standards, as well as building codes and rules. It varies depending on the purpose of the room, its configuration, and so on.

Light intensity

This characteristic is often called illumination or saturation. It represents the ratio of the luminous flux to the area of ​​the object on which it falls. This unit of brightness of a luminous surface is measured in lux.

Brightness

The luminous intensity divided by unit area is called luminance. It is measured in candelas per square meter. The source distributes radiation that illuminates a certain area. The higher the area, the correspondingly greater the brightness of the light. This parameter also characterizes the efficiency of the lighting source, and its measurement is required to calculate required amount lighting devices in the room and, accordingly, design their location and wiring.

Thus, the luminous flux has several parameters, and it is not always clear which of them to pay attention to when purchasing lighting devices. It is difficult for the average consumer to understand what luminous efficiency is, how saturation differs from brightness, and so on. Moreover, the units of measurement that are indicated on the boxes are also uninformative for the uninitiated person: lm/W, cd, cd/sq.m, all this looks like hieroglyphs, from which it is not clear how many light bulbs and with what characteristics you need to purchase. Therefore, to calculate the number of lighting devices, it is recommended to either use the services of professionals or a special calculator that can be found on the Internet.

Video

Let us now calculate the total energy emitted by the charge during acceleration. For generality, let us take the case of arbitrary acceleration, considering, however, the motion to be non-relativistic. When the acceleration is directed, say, vertically, electric field radiation is equal to the product of the charge and the projection of the retarded acceleration, divided by the distance. Thus, we know the electric field at any point, and from here we know the energy passing through a unit area in .

The quantity is often found in radio wave propagation formulas. Its inverse value can be called vacuum impedance (or vacuum resistance); it is equal . Hence the power (in watts per square meter) is the average square of the field divided by 377.

Using formula (29.1) for the electric field we obtain

, (32.2)

where is the power at , emitted at an angle . As already noted, it is inversely proportional to distance. Integrating, we get from here full power, radiated in all directions. To do this, we first multiply by the area of ​​the strip of the sphere, then we get the energy flow in the angle interval (Fig. 32.1). The area of ​​the strip is calculated as follows: if the radius is equal to , then the thickness of the strip is equal to , and the length is , since the radius of the annular strip is . Thus, the area of ​​the strip is equal to

(32.3)

Figure 32.1. The area of ​​the ring on the sphere is equal to .

Multiplying the flux [power by , according to formula (32.2)] by the area of ​​the strip, we find the energy emitted in the range of angles and ; Next you need to integrate over all angles from to:

(32.4)

When calculating, we use the equality and as a result we get . From here finally

A few points need to be made about this expression. First of all, since there is a vector, then in formula (32.5) it means, i.e., the square of the length of the vector. Secondly, formula (32.2) for the flow includes acceleration taken into account the delay, i.e., acceleration at the moment in time when the energy now passing through the surface of the sphere was emitted. The idea may arise that the energy was actually emitted at exactly the indicated moment in time. But this is not entirely correct. The moment of emission cannot be determined precisely. It is possible to calculate the result only of such motion, such as oscillations, etc., where the acceleration eventually disappears. Consequently, we can only find the total energy flow for the entire period of oscillation, proportional to the average square of the acceleration over the period. Therefore, in (32.5) should mean the time average of the squared acceleration. For such a motion, when the acceleration at the beginning and at the end becomes zero, the total radiated energy is equal to the time integral of expression (32.5).

Let's see what formula (32.5) gives for an oscillating system, for which the acceleration has the form . The average of the acceleration squared over a period is equal to (when squaring, one must remember that in fact, instead of the exponent, its real part, the cosine, should be included, and the average gives):

Hence,

These formulas were obtained relatively recently - at the beginning of the 20th century. These are wonderful formulas, they had a huge historical meaning, and it would be worth reading about them in old physics books. True, a different system of units was used there, and not the SI system. However, in the final results relating to electrons, these complications can be eliminated using next rule correspondence: the quantity where is the electron charge (in coulombs), previously written as . It is easy to verify that in the SI system the value is numerically equal to , since we know that And . In what follows we will often use the convenient notation (32.7)

If this numerical value is substituted into the old formulas, then all other quantities in them can be considered defined in the SI system. For example, formula (32.5) previously had the form . And the potential energy of a proton and electron at a distance is or , where SI.

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