7. Symmetry, Infinite Number of Coupled Oscillators

[@hyprk5590]

I’m South korean and I majored in EE. I took several physics courses at that time, especially related to applied physics. But Whenever symmetry was poped up in the courses, I didn’t understand that physical meaning exactly and simply calculated matrix problem. Now I got that physical meaning after this lecture. Thank you for your explanation.

[@joshuaronisjr]

For people saying they don't understand the first part, here is what I got from it. Say that x(t) is a normal-mode solution. What does that mean? That means that all the components of the system are oscillating in phase with one another, at the same frequency, and at constant amplitudes. The normal mode solution x(t) can be written as αA cos (wt+φ), where A is a vector containing the relative amplitudes of the components, α is the smallest amplitude (the amplitude associated with the component of the system that has a "1" in A), ω is the frequency, and φ is the phase-shift. x(t)=αA cos (wt+φ) Now, Sx(t) will be that same normal mode, but looking at it from some other point of view (in this case, from the other side, since we have mirror symmetry). Since its the same normal mode, the frequencies of the components won't change, the phase shifts of the components won't change, and, most importantly, the relative amplitudes of the components won't change. All that will change is the coordinates of the components (in this example, x_1 ->- x_2 and x_2 -> - x_1). Sx(t) is also a normal mode...how can we write it? Well, it'll have the same α, the same φ, and the same ω. All that will change is A, since the coordinates of each component changed. That means that we can write: Sx(t)=αB cos(ωt+φ). In the above, B is the new vector, holding the relative amplitudes of the components of the system from this new point of view. Here's the key part - even though since we changed our point of view, and the components got assigned different variables, and thus from this new normal mode B isn't the same as A, its STILL the same normal mode. If we had a system where the bigger mass was oscillating with half the amplitude as it was from the original point of view, the bigger mass will also be oscillating with half the amplitude as it is from the new (symmetric) point of view. In this specific case, all the symmetric matrix did was apply negative signs...but, the idea holds regardless of what our system is. After we apply the symmetry transformation and switch to the new point of view, THE RELATIVE AMPLITUDES OF THE NEW COORDINATES OF THE COMPONENTS MUST STAY THE SAME! THE PROPORTIONS BETWEEN THE COMPONENTS OF A AND B MUST STAY THE SAME. That must mean that A and B must be PROPORTIONAL to one another. A~B So, we have that αB cos(ωt+φ) and αA cos (wt+φ) are proportional to one another. αB cos(ωt+φ)~αA cos(ωt+φ) Or, written out with the transformation matrix, that Sx(t) and x(t) are proportional to one another, aka SαA cos (wt+φ) and αA cos (wt+φ) are proportional to one another. S αA cos(ωt+φ)~αA cos(ωt+φ) Cancelling out all the terms that are the same on both sides, we get that SA and A are proportional. SA ~ A The only way this is true is if A is an eigenvector of S!!! So, by finding the eigenvectors of the symmetric change-of-point-of-view matrix for our specific system, we have found the vectors which hold information about the relative amplitudes of the components in the full solution! Then, once we've found those eigenvectors, we can find the eigenvalues (the frequencies of the normal modes) by using (M^-1 K) - it'll be much easier to find those frequencies knowing the eigenvectors (the A^(1) and A^(2)) already! Especially if (M^-1 K) is big (for example, if we have 4 components in the system, then finding the determinant of (M^-1 K) - Λ) and setting it equal to zero to find the eigenvalues (the frequencies) would've been quite messy). I hope that made sense!

[@hershyfishman2929]

Thank you! I still don't get it though. If A = , and B = , this system still has reflectional symmetry, because it shouldn't matter which side you look at it from, but how are A and B proportional?

[@tadaiyoradima]

THANK YOU SO MUCH

[@abdurrahmanlabib916]

@@hershyfishman2929 if there is reflection symmetry, B can not be a vector as such

[@hershyfishman2929]

@@abdurrahmanlabib916 can you elaborate?

[@NN-lt3kh]

@@hershyfishman2929 If the system is symmetrical A and B cannot be equal to the values you gave above. Using the two oscillator case as an example, the amplitudes of each oscillator must be equal, so the only possible amplitude matrices are and (and their multiples). The same logic would hold for more oscillators, each oscillator (except for the middle one if there is an odd number) will have a counterpart with the same amplitude. When you apply the symmetry matrix and reflect the system, The amplitude matrix is either the same as before or multiplied by -1.

[@sultanmahmudsabuj4246]

thnx for providing such quality lectures.. keep doing MIT

[@BlueRaja]

Where does S come from? He mentions the equations x1' = -x2 x2' = -x1 We can rewrite these slightly as x1' = 0*x1 + -1*x2 x2' = -1*x1 + 0*x2 which can then be written using matrix notation: (x1') = (0 -1) (x1) (x2') = (-1 0) (x2) (pretend those little parentheses are one large vertical parenthesis) then if we call the matrix "S", we can rewrite that as a vector equation: x' = Sx

[@eamon_concannon]

24:30 If an nXn matrix has n distinct eigenvalues then the corresponding eigenvectors are linearly independent and form a basis for R^n. Hence, eigenvectors corresponding to the same eigenvalue are scalar multiples of each other, otherwise there would be more than n linearly independent eigenvectors in R^n which is impossible.

[@BlueRaja]

How did he get sin(x)? The equation he gives is actually one definition of sin. You can use Euler's formula and check for yourself that sin(x) = 1/2i (e^ix - e^-ix) However, his explanation was not great. He claims a linear combination of valid eigenvalues is also a valid eigenvalue, but actually in this case that's not true since we're limiting ourselves to eigenvalues where |β| = 1. Thus to keep that true, he divides by two (and justifies the 'i' with "to make it real"). A better explanation would've been to take Re[e^ix] (a linear combination with e^-ix adds nothing, since e^ix can already generate all possible imaginary values with length 1), which is cos(x). Then since sin(x) and cos(x) only differ by phase, which we don't care about (since we didn't specify initial conditions), we can just as easily label this sin(x). The reason we want sin is because it will satisfy our initial conditions later, when we add them. Also note the possibly confusing fact that, for a given mass, sin(jka) is CONSTANT. It represents the maximum amplitude of that mass. The fact that it's a sin (or cos) wave is what makes the combination of all the masses behave like a standing wave over time. The value changes for different masses, but for a given mass it does not change over time. Why sin(jka) instead of sin(jx)? Where did "ka" come from? He simply pulled a factor of "a" out of x and called the result "k". Why do that? He never explained that in this lecture, but he explains it in the next one.

[@ashutoshpanigrahy7326]

This is what I understood from the lecture. a is the natural length of the spring between the masses k is the wavenumber ie 2*pi/wavelength the body is j masses away from the 0th mass x =j*a it is the position of the jth mass from x=0 (where the 0th mass of the infinite oscillator is present) A=e^ix and B=e^(-ix) both lead to the same omega, One can assume them to basis vectors and any linear combination of these basis vectors should also produce an A_j such that the omega remains unchanged. sin(kx) and cos(kx) are few examples of linear combinations of A and B.

[@luke9771]

could not understand the first 28 minutes covering symmetry.

[@abdurrahmanlabib916]

29:10 infinite coupled oscillators 56:40 sine form

[@xiangli221]

If cos(ka)=-1 then ka=(2n+1)*pi, n is integer, so sin(jka) is equal to zero for all j since it’s integer, then all the amplitude of the system is zero, how does it oscillate?

[@ayushwithasingle_a]

Difficult Topic.Could not understand. I guess I need to focus more.

[@andrewzhao1976]

This topic is so beautiful.

[@heiheihei60]

Thank you so much

[@kingarth0r]

I was hoping we'd talk about group theory. While not explicitly mentioned, there were elements from it like commutators and such.

[@ricardocesardasilvagomes9549]

Espetacular.....essas aulas são demais!

[@idontknowman-0-]

i think latter part is harder

[@LydellAaron]

29:30 I am curious about that system as well.

[@pmemoli9299]

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