Understanding particle physics: 4) waves, the classical equation of motion
1. A ball on a spring, the Newtonian version
2. Quantum ball on spring
3. Waves, classic look
4. Waves, the classical equation of motion
5. Quantum waves
6. Fields
7. Particles are quanta
8. How particles interact with fields
In one of the first articles of the cycle, we first derived a formula for the oscillatory motion of a ball.
And then we found the equation of motion for which this formula was a solution
Here
• d 2 z / dt 2 denotes the time variation of the time variation z (t).
• K is the spring force, M is the mass of the ball, z 0 is the equilibrium position.
• ν = √ K / M / 2π
The key step for obtaining the last frequency equation, expressed in terms of K and M, was the calculation of d 2 z / dt 2 for the oscillatory motion of the ball z (t) = z 0 + A cos [2 π ν t]. We found that
Now we want to do the same for the waves. We have found a formula for the shape and motion of a wave, oscillating both in space and in time.
Among the solutions of which equation of motion is there such a formula? You can imagine the answer. Obviously, it includes:
1. d 2 Z / dt 2 , change over time, change over time Z (x, t).
2. d 2 Z / dx 2 , change in space change in the space Z (x, t).
Naturally, we can guess that the equation should look something like this:
Where C t , C x and C 0 are constants. I note that if Where C t = 1, C x = 0, and C 0 = -K / M, you will return to the equation of oscillation of the ball on the spring. What are the constants in our case?
We can always put C t = 1. If you wanted to, say, put C t = 5, I would just ask you to divide the whole equation by 5, which would give you the equivalent of the variant in which C t = 1, just with other values of the other constants.
After this, it turns out that the values of C x and C 0 turn out to be different in different physical systems. We will study two different classes of waves with different constants.
Both classes will have a C x negative, Cx=−c2w (here c w denotes the speed of movement of high-frequency waves).
These classes will differ in that the first class, Class 1, C 0 will be negative, and will be - (2 π μ) 2, and the second, Class 0, C 0 will be equal to zero.
We now study the properties of waves of these two classes of equations. But before that, we need to carry out another calculation, which we have already done earlier.
For our endless wave
We will need to know d 2 Z / dt 2 and d 2 Z / dx 2 . In the previous article we have already shown that for a ball on a spring moving according to z (t) = z 0 + A cos [2 π ν t], it turns out that d2z/dt2=−(2 pi nu)2(z−z0) . A change in time gives us a factor of 2 π ν, and a change in time in a change in time gives us two factors. In addition, there is a common minus sign. Therefore, you will not be surprised that:
• d2Z/dt2=−(2 pi nu)2(Z−Z0)
• d2Z/dx2=−(2 pi/ lambda)2(Z−Z0)
Each time change gives us a factor ν = 1 / T (the longer the period, the slower the time change goes), and each change in space gives us a factor 1 / λ (the longer the wave, the slower the change in space).
For an infinite wave, we have the basic equation
And we want to show that
Some facts:
• Z - Z 0 = A cos (2π [ν t - x / λ]) (just in the basic equation, Z 0 was moved to the left)
• Since Z 0 is a constant independent of time and space, dZ 0 / dt = 0 and dZ 0 / dx = 0.
• d (cos t) / dt = - sin t, and d (sin t) / dt = + cos t
• d (F [at + bx]) / dt = ad (F [at + bx]) / d (a t + bx), where a and b are constants and F is any function of (at + bx).
• d [A f (t)] / dt = A d [f (t)] / dt, where f (t) is any function of t, and A is a constant
All together, this means that:
and
Since the basic formula for the wave does not change when replacing (ν t) with (-x / λ), the calculation of d 2 Z / dx 2 does not differ from the calculation of d 2 Z / dt 2 , just instead of d / dt, giving a multiplier (2π ν ), we will have d / dx giving the multiplier (- 2π / λ). But, since there are two such factors in the answer, we simply replace (2π ν) 2 with (- 2π / λ) 2 = (+ 2π / λ) 2 ; minus does not matter (total minus in addition remains). As we were required to prove
Fine print: all the above derivatives are in fact partial derivatives.
In this class of waves, the equation of motion will be:
By connecting the formula Z (x, t) for an infinite wave and using the calculations we have just made, we find that:
Let's divide the equation by −(2 pi)2(Z−Z0) , we'll get
Since the frequencies, speeds and wavelengths are positive, you can extract the root and get
ν = c w / λ, or, if you wish, λ = c w / ν = c w T
From this formula, we learn that:
• Initially, our wave, as we recorded it, could have any frequency and any wavelength. But the equation of motion makes them depend on each other. For waves of class 0, you can choose any frequency, but after that the wavelength is determined through λ = c w / ν.
• All waves of class 0, regardless of frequency, move with speed c w . This follows from the formula λ = c w T and fig. 3 previous articles . Observe how the wave travels one cycle of oscillations during a single T period. What happens? The wave looks exactly the same after T, but each ridge has shifted to where its neighbor was - by a distance of λ. This means that the ridge moves a distance λ in time T - one wavelength in one oscillation period - and this means that the ridges move with a speed λ / T = c w . This is true for all frequencies and their periods, and all wavelengths!
• As in the case of a ball on a spring, the amplitude A of these waves can be any arbitrarily large or small. And this is true for all frequencies.
For this class of waves, our equation of motion will be:
Substituting the formula Z (x, t) for an infinite wave and using the fast calculation mentioned above, we find that
Dividing the equation by −(2 pi)2(Z−Z0) , we'll get
Since the frequencies, speeds and wavelengths are positive, we can take the square root and get
Recall that y 1/2 is the same as √y.
This formula is very different from the formula for class 0 waves, as well as the consequences of its application.
Firstly, the equation of motion indicates the presence of the minimum permissible frequency. Since (cw / λ) 2 is always positive,
To get closer to ν = μ, it is necessary to increase λ. For very long wavelengths, the frequency approaches μ, but it cannot become less. For class 0 waves, this was not the case. They had ν = cw / λ, so for them, the more you make λ, the more ν approaches zero. For class 1 waves, any value of ν greater than μ is possible.
Secondly, we have found evidence that all waves of class 0 have the same speed, but it does not work for waves of class 1. The only option in which it can work if we take ν is much more than μ; for this we need to make λ very small (and, accordingly, 1 / λ very large). In this case
That is, at very high frequencies and short wavelengths, waves of class 1 will have approximately the same ratio between frequency and wavelength as waves of class 0, therefore, for the same reasons as waves of class 0, such waves will move with speed , (approximately) equal to cw.
What is true for the waves of both classes is that the amplitude A can be either arbitrarily small or large, and does not depend on frequency.
Fig. 1. For class 0 and 1 waves, the equation of motion gives the relationship between frequency, or period, and wavelength, or 1 / wavelength. On each of the graphs shows the relationship of these values depending on the equation of motion. Three graphs show the same thing, but they are built on different variables. Blue lines belong to class 0 waves. Red ones represent class 1 waves, the speed of which is the same at very high frequencies and short wavelengths, when they coincide with blue lines. But at the minimum frequency μ (and with a maximum period of 1 / μ), marked green, the two curves diverge with increasing wavelengths.
Small print: you may have noticed that I cheated a little. I did not count the speed of the waves of class 1. The fact is that there was a very cunning trick here. For waves of class 0, I counted their speed, watching the movements of the crests. This works because in class 0 the waves of all frequencies move at the same speed. But in class 1, or in any other, where waves of different frequencies move at different speeds, the speed of a real wave is not given by the speed at which its crests move! It turns out that the crests move faster than cw, but the wave velocity is less than cw. To understand this, it is necessary to use a very unobvious logic and the difference between the “group” and “phase” speeds. I'm going around this dirty trick for now; just wanted to draw your attention to its existence, so that you don't get the wrong idea.
Many familiar examples of class 0 waves can be found, including sound in air, water or metal (where cw is the speed of sound waves in a material), light, and other electromagnetic waves (where cw = c in a vacuum), and waves on ropes or strings, as in fig. 2 in the previous article. Therefore, class 0 waves are taught in elementary physics courses. I cannot give an example of class 1 waves in ordinary life, but soon we will see that these waves are just as important for the Universe.
We have a convenient formula E = 2 π 2 ν 2 A 2 M for the energy of the ball of mass M on the spring. The formulas for other oscillators depend on their nature, but their shape is about the same. But in the case of waves, we did not mention their energy. In particular, due to the fact that in order to simplify mathematics we studied waves with an infinite number of crests. It is intuitively clear that some energy must be stored in motion and in the shape of each ridge and trough, and with an infinite number of crests and valleys, the amount of energy in a wave will be infinite. This can be circumvented in two ways. The exact formulas depend on the type of wave, but let's consider the waves of class 0 on a rope.
• The amount of energy per wavelength (stored between the point x and the point x + λ) is, of course, 2 π 2 ν 2 A 2 M λ , where M λ is the mass of the segment of the rope with length λ.
• In reality, waves are never infinite. As a pulse of several ridges and troughs, shown in Fig. 2 in the last article, any wave will be finite, it will have a finite number of ridges and hollows. If it stretches over the length L, that is, it has L / λ crests, then the energy transferred to it will be 2 π 2 ν 2 A 2 M L , where M L is the mass of the length of the rope length L. It’s just L / λ, multiplied by energy by one wavelength.
For waves propagating not along the ropes, the details of the equations will differ, but the energy per wavelength of a simple oscillatory system will always be proportional to ν 2 A 2 .
In class 1 there is a very interesting wave, which does not exist in class 0. This is the case when ν = μ, the minimum value, and λ = infinity. In this case, the wave takes the form
This wave does not depend on x at any time, that is, Z (x, t) will be constant over the entire space, and Z oscillates in time exactly like a ball on a spring with a frequency μ. Such a stationary wave shown in Fig. 2, will be very important in further discussion.
Fig. 2
For a ball on a spring, the difference between the classical and quantum systems was that in the first case, the amplitude could take arbitrary values, like energy, and in the quantum case, the amplitude and energy were quantized. For any similar oscillating system, it works the same way. Perhaps we can guess that this is also true for the waves ...
2. Quantum ball on spring
3. Waves, classic look
4. Waves, the classical equation of motion
5. Quantum waves
6. Fields
7. Particles are quanta
8. How particles interact with fields
Let us return to the equation of oscillations of the ball on the spring
In one of the first articles of the cycle, we first derived a formula for the oscillatory motion of a ball.
z(t)=z0+Acos[2 pi nut]
And then we found the equation of motion for which this formula was a solution
d2z/dt2=−K/M(z−z0)
Here
• d 2 z / dt 2 denotes the time variation of the time variation z (t).
• K is the spring force, M is the mass of the ball, z 0 is the equilibrium position.
• ν = √ K / M / 2π
The key step for obtaining the last frequency equation, expressed in terms of K and M, was the calculation of d 2 z / dt 2 for the oscillatory motion of the ball z (t) = z 0 + A cos [2 π ν t]. We found that
d2z/dt2=−(2 pi nu)2(z−z0)
Wave motion equation
Now we want to do the same for the waves. We have found a formula for the shape and motion of a wave, oscillating both in space and in time.
Z(x,t)=Z0+Acos(2 pi[ nut−x/ lambda])=Z0+Acos(2 pi[t/T−x/ lambda])
Among the solutions of which equation of motion is there such a formula? You can imagine the answer. Obviously, it includes:
1. d 2 Z / dt 2 , change over time, change over time Z (x, t).
2. d 2 Z / dx 2 , change in space change in the space Z (x, t).
Naturally, we can guess that the equation should look something like this:
Ctd2Z/dt2+Cxd2Z/dx2=C0(Z−Z0)
Where C t , C x and C 0 are constants. I note that if Where C t = 1, C x = 0, and C 0 = -K / M, you will return to the equation of oscillation of the ball on the spring. What are the constants in our case?
We can always put C t = 1. If you wanted to, say, put C t = 5, I would just ask you to divide the whole equation by 5, which would give you the equivalent of the variant in which C t = 1, just with other values of the other constants.
After this, it turns out that the values of C x and C 0 turn out to be different in different physical systems. We will study two different classes of waves with different constants.
Both classes will have a C x negative, Cx=−c2w (here c w denotes the speed of movement of high-frequency waves).
These classes will differ in that the first class, Class 1, C 0 will be negative, and will be - (2 π μ) 2, and the second, Class 0, C 0 will be equal to zero.
We now study the properties of waves of these two classes of equations. But before that, we need to carry out another calculation, which we have already done earlier.
Quick count
For our endless wave
Z(x,t)=Z0+Acos(2 pi[ nut−x/ lambda])=Z0+Acos(2 pi[t/T−x/ lambda])
We will need to know d 2 Z / dt 2 and d 2 Z / dx 2 . In the previous article we have already shown that for a ball on a spring moving according to z (t) = z 0 + A cos [2 π ν t], it turns out that d2z/dt2=−(2 pi nu)2(z−z0) . A change in time gives us a factor of 2 π ν, and a change in time in a change in time gives us two factors. In addition, there is a common minus sign. Therefore, you will not be surprised that:
• d2Z/dt2=−(2 pi nu)2(Z−Z0)
• d2Z/dx2=−(2 pi/ lambda)2(Z−Z0)
Each time change gives us a factor ν = 1 / T (the longer the period, the slower the time change goes), and each change in space gives us a factor 1 / λ (the longer the wave, the slower the change in space).
Evidence
For an infinite wave, we have the basic equation
Z(x,t)=Z0+Acos(2 pi[ nut−x/ lambda])=Z0+Acos(2 pi[t/T−x/ lambda])
And we want to show that
d2Z/dt2=−(2 pi nu)2(Z−Z0)
d2Z/dx2=−(2 pi/ lambda)2(Z−Z0)
Some facts:
• Z - Z 0 = A cos (2π [ν t - x / λ]) (just in the basic equation, Z 0 was moved to the left)
• Since Z 0 is a constant independent of time and space, dZ 0 / dt = 0 and dZ 0 / dx = 0.
• d (cos t) / dt = - sin t, and d (sin t) / dt = + cos t
• d (F [at + bx]) / dt = ad (F [at + bx]) / d (a t + bx), where a and b are constants and F is any function of (at + bx).
• d [A f (t)] / dt = A d [f (t)] / dt, where f (t) is any function of t, and A is a constant
All together, this means that:
dZ/dt=d[Acos(2 pi[ nut−x/ lambda])]/dt=Ad[cos(2 pi[ nut−x/ lambda])]/dt=A(2 pi nu)d[cos(2 pi[ nut−x/ lambda])]/d(2 pi[ nut−x/ lambda])=−(2 pi nu)Asin(2 pi[ nut−x/ lambda])
and
d2Z/dt2=d[−(2 pi nu)Asin(2 pi[ nut−x/ lambda])]/dt=−(2 pi nu)Ad[sin(2 pi[ nut−x/ lambda])]/dt=−(2 pi nu)2Acos(2 pi[ nut−x/ lambda])=−(2 pi nu)2(Z−Z0)
Since the basic formula for the wave does not change when replacing (ν t) with (-x / λ), the calculation of d 2 Z / dx 2 does not differ from the calculation of d 2 Z / dt 2 , just instead of d / dt, giving a multiplier (2π ν ), we will have d / dx giving the multiplier (- 2π / λ). But, since there are two such factors in the answer, we simply replace (2π ν) 2 with (- 2π / λ) 2 = (+ 2π / λ) 2 ; minus does not matter (total minus in addition remains). As we were required to prove
d2Z/dt2=−(2 pi/ lambda)2(Z−Z0)
Fine print: all the above derivatives are in fact partial derivatives.
Class 0: waves of any frequency and equal speeds
In this class of waves, the equation of motion will be:
d2Z/dt2−c2wd2Z/dx2=0
By connecting the formula Z (x, t) for an infinite wave and using the calculations we have just made, we find that:
−(2 pi nu)2(Z−Z0)−(−c2w)(2 pi/ lambda)2(Z−Z0)=0
Let's divide the equation by −(2 pi)2(Z−Z0) , we'll get
nu2−c2w/ lambda2=0
Since the frequencies, speeds and wavelengths are positive, you can extract the root and get
ν = c w / λ, or, if you wish, λ = c w / ν = c w T
From this formula, we learn that:
• Initially, our wave, as we recorded it, could have any frequency and any wavelength. But the equation of motion makes them depend on each other. For waves of class 0, you can choose any frequency, but after that the wavelength is determined through λ = c w / ν.
• All waves of class 0, regardless of frequency, move with speed c w . This follows from the formula λ = c w T and fig. 3 previous articles . Observe how the wave travels one cycle of oscillations during a single T period. What happens? The wave looks exactly the same after T, but each ridge has shifted to where its neighbor was - by a distance of λ. This means that the ridge moves a distance λ in time T - one wavelength in one oscillation period - and this means that the ridges move with a speed λ / T = c w . This is true for all frequencies and their periods, and all wavelengths!
• As in the case of a ball on a spring, the amplitude A of these waves can be any arbitrarily large or small. And this is true for all frequencies.
Class 1: waves with a frequency greater than the minimum, with different speeds
For this class of waves, our equation of motion will be:
d2Z/dt2−cw2d2Z/dx2=−(2 pi mu)2(Z−Z0)
Substituting the formula Z (x, t) for an infinite wave and using the fast calculation mentioned above, we find that
−(2 pi nu)2(Z−Z0)−(−cw2)(2 pi/ lambda)2(Z−Z0)=−(2 pi mu)2(Z−Z0)
Dividing the equation by −(2 pi)2(Z−Z0) , we'll get
nu2−cw2/ lambda2= mu2
Since the frequencies, speeds and wavelengths are positive, we can take the square root and get
nu=[(cw/ lambda)2+ mu2]1/2
Recall that y 1/2 is the same as √y.
This formula is very different from the formula for class 0 waves, as well as the consequences of its application.
Firstly, the equation of motion indicates the presence of the minimum permissible frequency. Since (cw / λ) 2 is always positive,
nu=[(cw/ lambda)2+ mu2]1/2≥ mu
To get closer to ν = μ, it is necessary to increase λ. For very long wavelengths, the frequency approaches μ, but it cannot become less. For class 0 waves, this was not the case. They had ν = cw / λ, so for them, the more you make λ, the more ν approaches zero. For class 1 waves, any value of ν greater than μ is possible.
Secondly, we have found evidence that all waves of class 0 have the same speed, but it does not work for waves of class 1. The only option in which it can work if we take ν is much more than μ; for this we need to make λ very small (and, accordingly, 1 / λ very large). In this case
nu=[(cw/ lambda)2+ mu2]1/2≈cw/ lambda
That is, at very high frequencies and short wavelengths, waves of class 1 will have approximately the same ratio between frequency and wavelength as waves of class 0, therefore, for the same reasons as waves of class 0, such waves will move with speed , (approximately) equal to cw.
What is true for the waves of both classes is that the amplitude A can be either arbitrarily small or large, and does not depend on frequency.
Fig. 1. For class 0 and 1 waves, the equation of motion gives the relationship between frequency, or period, and wavelength, or 1 / wavelength. On each of the graphs shows the relationship of these values depending on the equation of motion. Three graphs show the same thing, but they are built on different variables. Blue lines belong to class 0 waves. Red ones represent class 1 waves, the speed of which is the same at very high frequencies and short wavelengths, when they coincide with blue lines. But at the minimum frequency μ (and with a maximum period of 1 / μ), marked green, the two curves diverge with increasing wavelengths.
Small print: you may have noticed that I cheated a little. I did not count the speed of the waves of class 1. The fact is that there was a very cunning trick here. For waves of class 0, I counted their speed, watching the movements of the crests. This works because in class 0 the waves of all frequencies move at the same speed. But in class 1, or in any other, where waves of different frequencies move at different speeds, the speed of a real wave is not given by the speed at which its crests move! It turns out that the crests move faster than cw, but the wave velocity is less than cw. To understand this, it is necessary to use a very unobvious logic and the difference between the “group” and “phase” speeds. I'm going around this dirty trick for now; just wanted to draw your attention to its existence, so that you don't get the wrong idea.
Final Comments on Classic Waves
Many familiar examples of class 0 waves can be found, including sound in air, water or metal (where cw is the speed of sound waves in a material), light, and other electromagnetic waves (where cw = c in a vacuum), and waves on ropes or strings, as in fig. 2 in the previous article. Therefore, class 0 waves are taught in elementary physics courses. I cannot give an example of class 1 waves in ordinary life, but soon we will see that these waves are just as important for the Universe.
We have a convenient formula E = 2 π 2 ν 2 A 2 M for the energy of the ball of mass M on the spring. The formulas for other oscillators depend on their nature, but their shape is about the same. But in the case of waves, we did not mention their energy. In particular, due to the fact that in order to simplify mathematics we studied waves with an infinite number of crests. It is intuitively clear that some energy must be stored in motion and in the shape of each ridge and trough, and with an infinite number of crests and valleys, the amount of energy in a wave will be infinite. This can be circumvented in two ways. The exact formulas depend on the type of wave, but let's consider the waves of class 0 on a rope.
• The amount of energy per wavelength (stored between the point x and the point x + λ) is, of course, 2 π 2 ν 2 A 2 M λ , where M λ is the mass of the segment of the rope with length λ.
• In reality, waves are never infinite. As a pulse of several ridges and troughs, shown in Fig. 2 in the last article, any wave will be finite, it will have a finite number of ridges and hollows. If it stretches over the length L, that is, it has L / λ crests, then the energy transferred to it will be 2 π 2 ν 2 A 2 M L , where M L is the mass of the length of the rope length L. It’s just L / λ, multiplied by energy by one wavelength.
For waves propagating not along the ropes, the details of the equations will differ, but the energy per wavelength of a simple oscillatory system will always be proportional to ν 2 A 2 .
In class 1 there is a very interesting wave, which does not exist in class 0. This is the case when ν = μ, the minimum value, and λ = infinity. In this case, the wave takes the form
Z(x,t)=Z0+Acos(2 pi mut)
This wave does not depend on x at any time, that is, Z (x, t) will be constant over the entire space, and Z oscillates in time exactly like a ball on a spring with a frequency μ. Such a stationary wave shown in Fig. 2, will be very important in further discussion.
Fig. 2
Quantum waves
For a ball on a spring, the difference between the classical and quantum systems was that in the first case, the amplitude could take arbitrary values, like energy, and in the quantum case, the amplitude and energy were quantized. For any similar oscillating system, it works the same way. Perhaps we can guess that this is also true for the waves ...
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