Lecture Note
3.60 Symmetry, Structure, and Tensor Properties of Materials: Notes
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University:
Massachusetts Institute of Technology -
Course:
3.60 | Symmetry, Structure, and Tensor Properties of Materials Academic year:
2019
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109
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65
Author:
scannalxa
3.60 Symmetry, Structure and Tensor Properties of Materials
PROPAGATION OF WAVES ALONG ONE-DIMENSIONAL CRYSTAL WITH TWO KINDS OF ATOMS
CONSIDER A ONE-DIMENSIONAL CRYSTAL with 2 KINDS of ATOMS (M_A and M_B). NUMBER THE ATOMS along the chain such that A ATOMS ARE LOCATED EVEN (2n) and B ATOMS ODD (2n+1).
ONLY A SINGLE FORCE CONSTANT, β, REMAINS NECESSARY TO DESCRIBE NEAREST NEIGHBOR INTERACTIONS, AS AN A ATOM INTERACTS with B THE SAME WAY AS B with AN A.
IN SETTING UP A FORCE-BALANCE EQUATION WE MUST NOW WRITE A SEPARATE EQUATION FOR THE A AND B ATOMS - THEIR MASSES ARE DIFFERENT.
m_A d²u_2n/dt² = β(u_2n+1 - 2u_2n + u_2n-1)
m_B d²u_2n+1/dt² = β(u_2n+2 - 2u_2n+1 + u_2n)
LOOK FOR TRIAL SOLUTION
u_2n = u_A e^[i(ωt-2nka)]
u_2n+1 = u_B e^[i(ωt-(2n+1)ka)]
NO LONGER POSSIBLE TO ASSUME THE SAME AMPLITUDE FOR A & B ATOMS AS THEIR MASSES ARE DIFFERENT.
SUBSTITUTING TRIAL SOLUTIONS INTO FORCE EQUATIONS
-M_Aω²u_A e^[i(ωt-2nka)] = β[u_B e^[i(ωt-(2n+1)ka)] - 2u_A e^[i(ωt-2nka)] + u_B e^[i(ωt-(2n-1)ka)]]
-M_Bω²u_B e^[i(ωt-(2n+1)ka)] = β[u_A e^[i(ωt-2nka)] - 2u_B e^[i(ωt-(2n+1)ka)] + u_A e^[i(ωt-2(n+1)ka)]]
NOTE THAT AMPLITUDES REMAIN DIFFERENT BUT THINGS ARE NOT QUITE AS TIDY AS BEFORE
e^[i(x/2)] + e^[i(-x/2)] = 2cos(x)
-M_Aω²u_A = β[u_B e^[ika] - 2u_A + u_B e^[ika]]
-M_Bω²u_B = β[u_A e^[ika] - 2u_B + u_A e^[ika]]
REARRANGING
(2β - M_Aω²)u_A - (2β cos(ka))u_B = 0
-(2β cos(ka))u_A + (2β - M_Bω²)u_B = 0
REGARD THE AMPLITUDES U_A & U_B AS UNKNOWNS. AND WE HAVE A SET OF HOMOGENEOUS EQUATIONS FOR WHICH A NON-TRIVIAL SOLUTION EXISTS ONLY IF THE DETERMINANT OF COEFFICIENTS OF THE VARIABLES VANISHES.
Expanding
(2β - M_Aω²)² - (-2βcoska)² = 0
4β² - 2βω²(M_A + M_B) + M_AM_Bω⁴ + 4β²cos²ka = 0
M_AM_Bω⁴ - 2β(M_A + M_B)ω² + 4β²(1 - cos²ka) = 0
Solving, using Binomial Theorem
(ax² + bx + c = 0 ⇒ x = (-b ± √(b² - 4ac))/2a)
ω² = [2β(1/M_A + 1/M_B) ± √{4β²(1/M_A + 1/M_B)² - 4(M_AM_B/M_AM_B)4β²(1 - cos²ka)}]/2
ω² = β(1/M_A + 1/M_B) ± β√{(1/M_A + 1/M_B)² - 4(1 - cos²ka)/M_AM_B}
AGAIN WE HAVE THE CONDITION THAT ω BE A FUNCTION OF k. SO A SOLUTION IS ONLY ACCEPTABLE IF WE CAN FIND k WHICH MAKES IT REAL.
THERE ARE TWO EIGENVALUES, ω, FOR WHICH SOLUTIONS EXIST. EACH, WHEN SUBSTITUTED BACK INTO THE ORIGINAL EQUATION, WOULD PROVIDE EQS FROM WHICH WE COULD SOLVE FOR U_A, U_B, THUS FOR A GIVEN k, THERE ARE NOW TWO DIFFERENT WAVES OF ANGULAR FREQUENCY ω WHICH MAY BE PROPAGATED IN THE CRYSTAL.
INTERPRETATION OF THE SOLUTIONS
NOTE THAT THE GENERAL FORM OF OUR RESULT IS THAT ω² IS A CONSTANT (β(1/M_A + 1/M_B)) TO WHICH WE EITHER ADD OR SUBTRACT A TERM, WHICH IS GIVEN BY THE SQUARE ROOT OF A CONSTANT LESS A TERM WHICH IS A FUNCTION OF k (SIN²ka).
THE MAXIMUM VALUE OF THE TERM WHICH WE ADD-OR-SUBTRACT OCCURS WHEN THE TERM WHICH IS A FN OF k IS MINIMUM - NAMELY
SIN²ka = 0 ⇒ ka = 0 ⇒ k = 0
THE MINIMUM VALUE OF THE TERM WHICH WE ADD-OR-SUBTRACT OCCURS WHEN THE TERM WHICH IS A FN OF k IS MAXIMUM - NAMELY
SIN²ka = 1 ⇒ ka = π/2 ⇒ k = π/2a
WHAT IS THE REASON FOR THIS? WELL, ALL WE CAN REALLY TALK ABOUT WITH MEANING IS THE RELATIVE DISPLACEMENT OF NEIGHBORING ATOMS
u_n = u_0 e^(i(ωt-kna))
e^(ika)
WE CAN THEREFORE ADD OR SUBTRACT ANY INCREMENT TO k WHICH CHANGES PHASE OF ka BY 2π AND NOTHING WILL HAVE CHANGED. OR WE CAN SAY THAT
k OR k ± 2π/a
ARE EQUIVALENT
THUS IF k IS SUCH THAT |k| > k_max = π/a, WE CAN ALWAYS DEFINE k' IN TERMS OF A NEW k'
k' = k ± p(2π/a)
(p = AN INTEGER)
SUCH THAT |k'| < π/a, EVEN THOUGH, FORMALLY, ONE HAS ω = (β/m)^(1/2) sin(ka/2) DEFINED FOR ARBITRARILY LARGE k
YOU PROBABLY ARE MORE FAMILIAR WITH THIS SITUATION IN CONNECTION WITH X-RAY DIFFRACTION OR ENERGY LEVELS OF ELECTRONS IN CRYSTALS. BUT THEY ARE ALL THE SAME THING.
THE BOUNDARY TO k AT k = π/a IS CALLED THE FIRST BRILLOUIN ZONE.
TOTAL NUMBER OF VIBRATIONAL MODES WHICH MAY BE SUPPORTED BY THE CRYSTAL:
WE HAVE SHOWN THAT THERE IS A LOWER LIMIT, λ = 2a, TO THE WAVES WHICH MAY BE SUPPORTED BY A CRYSTAL COMPOSED OF ATOMS.
FOR A FINITE-SIZED CHUNK OF CRYSTAL THERE WILL ALSO BE AN UPPER LIMIT TO λ DETERMINED BY THE DIMENSIONS OF THE SAMPLE. FOR A ONE-DIMENSIONAL CRYSTAL OF LENGTH L:
λ = 2L, L, 2L/3, 2L/4, 2L/5, ... 2L/m
m = AN INTEGER = 1, 2, 3, ...
WHAT ARE THE VALUES OF k FOR THESE WAVES?
k = 2π/λ = 2π/2L, 2π/L, 2π/2L/3, 2π/2L/4, 2π/2L/5, ... 2π/2L/m
k = mπ/L
NOW, HOW MANY DISTINCT VIBRATIONAL MODES CAN THE CRYSTAL SUPPORT?
k_max = π/a ≈ π/L
m_max = L/a ≈ N
N = THE NUMBER OF ATOMS IN THE CRYSTAL!
THEREFORE, OUR ONE-DIMENSIONAL CRYSTAL WITH ONE KIND OF ATOM MAY SUPPORT A NUMBER OF VIBRATIONAL MODES WHICH IS EQUAL TO THE NUMBER OF ATOMS IN THE CRYSTAL!
OUR PLOT OF ω AS A FUNCTION OF k SHOULD THEREFORE NOT BE REGARDED AS A CONTINUOUS FUNCTION, BUT RATHER A SET OF DISCRETE CLOSELY SPACED STATES, EACH SEPARATED EQUALLY BY AN AMOUNT Δk = π/L.
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