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x). (b) Show that if ψ (1) n , then ˆAψ (1) (1) is an n n (x) is an eigenstate of ˆH1 with eigenstate of the same eigenvalue. The two Hamiltonians therefore have essentially identical spectra. ˆH2 with the same eigenvalue. Similarly, show that if ψ (2) n , then ˆA†ψ (2) (2) ˆH2 with eigenvalue E is an eigenstate of ˆH1... |
and the second is a function only of φ, so each must be a constant. This time3 I’ll call the separation constant m2: sin θ d dθ sin θ d dθ + ( + 1) sin2 θ = m2; (4.20) 1 The φ equation is easy: 1 d2 dφ2 = −m2. d2 dφ2 = −m2 ⇒ (φ) = eimφ. (4.21) (4.22) 2 Note that there is no loss of generality here—at this stage could ... |
2/2 + x 4/4! − · · · ), j0(x) ≈ 1; n0(x) ≈ 1 x ; j1(x) ≈ x 3 ; j2(x) ≈ x 2 15 ; etc. Notice that Bessel functions are finite at the origin, but Neumann functions blow up at the origin. Accordingly, B = 0, and hence R(r ) = A j(kr ). (4.47) There remains the boundary condition, R (a) = 0. Evidently k must be chosen such... |
adial wave function has N − 1 nodes. we have But ρ0 determines E (Equations 4.54 and 4.55): ρ0 = 2n. E = − 2κ 2 2m = − mee4 8π 22 0 2ρ2 0 , so the allowed energies are 4.2 The Hydrogen Atom 147 (4.68) (4.69) En = − 2 e2 4π 0 me 22 1 n2 = E1 n2 , n = 1, 2, 3, . . . . (4.70) This is the famous Bohr formula—by any measure... |
. Problem 4.14 (a) Using Equation 4.88, work out the first four Laguerre polynomials. (b) Using Equations 4.86, 4.87, and 4.88, find v(ρ), for the case n = 5, = 2. (c) Find v(ρ) again (for the case n = 5, = 2), but this time get it from the recursion formula (Equation 4.76). ∗ Problem 4.15 (a) Find r and (b) Find x and r... |
b Figure 4.11: The “ladder” of angular momentum states. is an eigenfunction of L z with the new eigenvalue μ ± . We call L + the raising so L ± f operator, because it increases the eigenvalue of L z by , and L − the lowering operator, because it lowers the eigenvalue by . For a given value of λ, then, we obtain a “ladd... |
cos θ eiφ. Apply the raising operator to find Y 2 2 (θ, φ). Use Equation 4.121 to get the normalization. 34 For an interesting discussion, see I. R. Gatland, Am. J. Phys. 74, 191 (2006). 4.4 Spin 165 ∗∗ Problem 4.27 Two particles (masses m1 and m2) are attached to the ends of a massless rigid rod of length a. The syste... |
.e. the spinor must be normalized: χ †χ = 1).42 But what if, instead, you chose to measure Sx ? What are the possible results, and what are their respective probabilities? According to the generalized statistical interpretation, we need to know the eigenvalues and eigenspinors of Sx . The characteristic equation is −λ ... |
Thus S is tilted at a constant angle α to the z axis, and precesses about the field at the Larmor frequency ω = γ B0, (4.167) 46 This does assume that a and b are real; you can work out the general case if you like, but all it does is add a constant to t. 174 CHAPTER 4 Quantum Mechanics in Three Dimensions ω z α 〈S〈 y x... |
m will be some linear combination of the composite states |s1 s2 m1 m2: C s1s2s |s m = m1+m2=m m1m2m |s1 s2 m1 m2 (4.183) (because the z-components add, the only composite states that contribute are those for which m1 + m2 = m). Equations 4.175 and 4.176 are special cases of this general form, 52 I say spins, for simp... |
discussion see Leslie E. Ballentine, Quantum Mechanics: A Modern Development, World Scientific, Singapore (1998), Section 11.3. 61 See, for example, Griffiths (footnote 43), Section 10.1.2. 4.5 Electromagnetic Interactions 183 ≡ eiq/ (4.197) satisfies Equation 4.191 with the gauge-transformed potentials ϕ and A (Equation ... |
that 8 J · da = − d dt where V is a (fixed) volume and S is its boundary surface. In words: The flow of probability out through the surface is equal to the decrease in probability of finding the particle in the volume. ||2 d3r, (4.222) V S (b) Find J for hydrogen in the state n = 2, l = 1, m = 1. Answer: 64π ma5 r e−r/a s... |
3 (|↑↓ − |↓↑) |↑ . 1 2 1 2 = 1√ 2 1 1 2 . Note: the Take it from there two doublets are not uniquely determined—any linear combination of them would still carry spin 1/2. The point is to construct two independent doublets. 2 is orthogonal to 1 2 1 2 1 and to 1 2 3 2 make sure 1 2 Problem 4.66 Deduce the condition for m... |
n the charge confined to the ring can be written ω 2 π where ω is the angular velocity of the particle. dW d = −q , (b) Calculate the z component of the mechanical angular momentum,77 Lmechanical = r × m v = r ×(p − q A) , for a particle in the state ψn in Example 4.6. Note that the mechanical angular momentum is not qu... |
It seems strange that relativity should have anything to do with it, and there has been a lot of discussion as to whether it might be possible to prove the spin-statistics connection in other ways. See, for example, Robert C. Hilborn, Am. J. Phys. 63, 298 (1995); Ian Duck and E. C. G. Sudarshan, Pauli and the Spin-Sta... |
r, if you prefer, we assumed the particles are in the same spin state). (a) Do it now for particles of spin 1/2. Construct the four lowest-energy configurations, and specify their energies and degeneracies. Suggestion: Use the notation |s m, where ψn1n2 is defined in Example 5.1 and |s m in Section 4.4.3.9 ψn1n2 6 To con... |
ψ(r1, r2, . . . , rZ ).15 Unfortunately, the Schrödinger equation with Hamiltonian in Equation 5.36 cannot be solved exactly (at any rate, it hasn’t been), except for the very simplest case, Z = 1 (hydrogen). In practice, one must resort to elaborate approximation methods. Some of these we shall explore in Part II; for... |
the next row (n = 5), and wait until later to slip in the = 2 and = 3 orbitals from the n = 4 shell. For details of this intricate counterpoint I refer you to any book on atomic physics.18 I would be delinquent if I failed to mention the archaic nomenclature for atomic states, because all chemists and most physicists ... |
(5.52) (5.53) is the free electron density (the number of free electrons per unit volume). The boundary separating occupied and unoccupied states, in k-space, is called the Fermi surface (hence the subscript F). The corresponding energy is the Fermi energy, E F ; for a free electron gas, E F = 2 2m 3ρπ 2 2/3 . (5.54) ... |
thing to notice is that f (z) strays outside the range (−1, +1), and in such regions there is no hope of solving Equation 5.71, since |cos (qa)|, of course, cannot be greater than 1. These gaps represent forbidden energies; they are separated by bands of allowed energies. Within a given band, virtually any energy is al... |
5.56) in terms of the radius, the number of nucleons (protons and neutrons) N , the number of electrons per nucleon d, and the mass of the electron m. Beware: In this problem we are recycling the letters N and d for a slightly different purpose than in the text. (b) Look up, or calculate, the gravitational energy of a ... |
ndaries of a two-dimensional infinite square well. In that case the potential energy would have the same rotational symmetries as the piece of paper and (because the kinetic energy is unchanged by a rotation) the Hamiltonian would also be invariant. In quantum mechanics, when we say that a system has a symmetry, this is... |
mplies that V (x + a) = V (x) . (6.10) Now, there are two very different physical settings where Equation 6.10 might arise. The first is a constant potential, where Equation 6.10 holds for every value of a; such a system is said to have continuous translational symmetry. The second is a periodic potential, such as an el... |
tors with the same eigenvalue qn : ˆQ | f (i) n = , then we need to sum over those states: P(qn ) = 0 (i) n f (t) 1 2 . i Except for the sum over i the proof proceeds unchanged. |(t) = m e−i Em t/ cm |ψm , 6.4 Parity 243 2 . 2 where the |ψn are the eigenstates of ˆH , and therefore P(qn) = e−i Em t/ cm fn|ψm m Now the ... |
er this question we use the infinitesimal form of the operator: ˆRz(δ) ≈ 1 − i δ ˆL z. Then the operator ˆx transforms as ˆx = ˆR† ˆx ˆR ≈ # ≈ ˆx + i δ ˆL z 1 − i δ ˆx ˆL z 1 + i δ $ ˆL z, ˆx ≈ ˆx − δ ˆy (I used Equation 4.122 for the commutator). Similar calculations show that ˆy = ˆy + δ ˆx and ˆz = ˆz. We can combine... |
eigenvalues En and qm respectively. Since ˆH and ˆ commute we know that the state |g = ˆ |ψ is also an ˆH with eigenvalue En. Since ˆQ and ˆ do not commute we know (Section A.5) eigenstate of ˆQ, ˆ that there cannot exist a complete set of simultaneous eigenstates of all three operators and ˆH . Therefore, there must ... |
8 CHAPTER 6 Symmetries & Conservation Laws Problem 6.20 Show that the commutator $ ˆL +, ˆf tion 6.46, as does the commutator # # $ ˆL −, ˆf = 0. = 0 leads to the same rule, Equa- ∗ Problem 6.21 For an electron in the hydrogen state ψ = 1√ 2 (ψ211 + ψ21−1) , find r after first expressing it in terms of a single reduced m... |
is this analysis limited to the case where ˆH is independent of time? Whether or not ˆH depends on time, ˆH is time dependent then the second derivative of is Schrödinger’s equation says i ˙ = ˆH . However, if given by ∂2 ∂t 2 = ∂ ∂t 1 i ˆH = 1 i ∂ ˆH ∂t − 1 2 ˆH 2 and higher derivatives will be even more complicated. ... |
n page 123) this means ˆT (a) ψ(x) = e−i a p/ f p(x) ( p) dp, (6.81) ∞ −∞ where ( p) is the momentum space wave function corresponding to ψ(x) and f p(x) is defined in Equation 3.32. Show that the operator ˆT (a), as given by Equation 6.81, applied to the function √ ψ(x) = λ e−λ |x| (whose first derivative is undefined at... |
ime-reversal invariance requires # $ ˆ, ˆH = 0. (6.83) (c) Show that, for a time-reversal invariant Hamiltonian, if ψn(x) is a stationary state with energy En, then ψ ∗ (x) is also a stationary state with the same energy En. If n the energy is nondegenerate, this means that the stationary state can be chosen as real. (... |
7.13. ∗ Problem 7.1 Suppose we put a delta-function bump in the center of the infinite square well: H = αδ(x − a/2), where α is a constant. (a) Find the first-order correction to the allowed energies. Explain why the energies are not perturbed for even n. (b) Find the first three nonzero terms in the expansion (Equation 7... |
the “good” states when we can’t solve the system exactly? For the moment let’s just write the “good” unperturbed states in generic form (Equation 7.17), keeping α and β adjustable. We want to solve the Schrödinger equation, H ψ = Eψ, (7.24) with H = H 0 + λH and E = E 0 + λE 1 + λ2 + λψ 1 + λ2ψ 2 + · · · . (7.25) Plugg... |
rder corrections ⎞ ⎠ , (7.41) and the “good” states are the corresponding eigenvectors:7.42) Once again, if you can think of an operator A that commutes with H 0 and H , and use the simultaneous eigenfunctions of A and H 0, then the W matrix will automatically be diagonal, and you won’t have to fuss with calculating th... |
se of p4 is more subtle. The Laplacian of 1/r picks up a delta function (see, for example, D. J. Griffiths, Introduction to Electrodynamics, 4th edn, Eq. 1.102). Show that $ # e−kr = ∇4 − 4k3 r + k4 e−kr + 8π k δ3 (r) , and confirm that p4 is hermitian.13 7.3.2 Spin-Orbit Coupling Imagine the electron in orbit around the... |
is the red Balmer line, coming from the transition n = 3 to n = 2. First of all, determine the wavelength and frequency of this line according to the Bohr theory. Fine structure splits this line into several closely-spaced lines; the question is: How many, and what is their spacing? Hint: First determine how many suble... |
eeman splitting. Express each answer as the sum of three terms: the Bohr energy, the fine-structure (proportional to α2), and the Zeeman contribution (proportional to μB Bext). If you ignore fine structure altogether, how many distinct levels are there, and what are their degeneracies? Problem 7.28 If = 0, then j = s, m ... |
ace: ψ 0 b ψ 0 a PD = + (7.100 The Hamiltonian can be written H = H 0 + H = ˜H 0 + ˜H (7.101) where ˜H 0 = H 0 + PD H PD, (7.102) ˜H 0 as the “unperturbed” Hamiltonian and ˜H as the perturbation; ˜H 0 is nondegenerate, so we can use ordinary nondegenerate ˜H = H − PD H PD. The idea is to treat as you’ll soon discover, ... |
reduces to the Mathieu equation. 37 C. Sánchez del Rio, Am. J. Phys., 50, 556 (1982); H. S. Valk, Am. J. Phys., 54, 921 (1986). 38 In part (b) we treat as a continuous variable; n becomes a function of , according to Equation 4.67, because N , which must be an integer, is fixed. To avoid confusion, I have eliminated n,... |
ion exactly, treating separately the regions 0 ≤ x < x0 and x0 < x < a, and imposing the boundary conditions at x0. Derive the transcendental equation for the energies: un sin(un) + sin( pun) sin Here ≡ 2maλ/2, un ≡ kna, and kn ≡ reproduces your result from part (a), in the appropriate limit. (1 − p) un √ = 0 (E > 0) .... |
the form of the actual ground state (Equation 2.60). But the gaussian is very easy to work with, so it’s a popular trial function, even when it bears little resemblance to the true ground state. Example 8.2 Suppose we’re looking for the ground state energy of the delta function potential: 8.1 Theory 329 2 d2 d x 2 Agai... |
e is equal to e2 4π 0 8 πa3 1 + 2r1 a 1 − e−4r1/a . e−4r1/a e−4r1/ar1 sin θ1 dr1dθ1dφ1. The angular integrals are easy (4π ), and the r1 integral becomes ∞ r e−4r/a − 0 r + 2r 2 a e−8r/a dr = 5a2 128 . (8.24) (8.25) 8.2 The Ground State of Helium 335 Finally, then, and therefore Vee = 5 4a e2 4π 0 = − 5 2 E1 = 34 eV, H... |
brium separation of the protons is about 2.4 Bohr radii, or 1.3 Å (the experimental value is 1.06 Å). The calculated binding energy is 1.8 eV, whereas the experimental value is 2.8 eV (the variational principle, as always, over estimates the ground state energy—and hence under estimates the strength of the bond—but nev... |
bital (Equation 8.38) and one in the anti-bonding orbital (Equation 8.53). Problem 8.13 Verify Equation 8.63 for the electron–proton potential energy. ∗∗∗ Problem 8.14 The two-body integrals D2 and X2 are defined in Equations 8.65 and 8.66. To evaluate D2 we write ψ0 r 2 2 (r2) d3r2 D2 = = −2 e R2+r 2 2 −2 R r2 cos θ2/a... |
any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say, x a), and solve the Schrödinger equation by separation of variables; if the wave function propagates out to infinity, the dependence on x must take the form exp (ikx x) with kx > 0. 21 The classic paper on muon-catalyzed fusi... |
we can express it in terms of its amplitude, A(x), and its phase, φ(x)—both of which are real: Using a prime to denote the derivative with respect to x, ψ(x) = A(x)eiφ(x). = A + i Aφ eiφ, dψ d x and # = d2ψ d x 2 A + 2i Aφ + i Aφ − A $ 2 φ eiφ. Putting this into Equation 9.1: A + 2i Aφ + i Aφ − A φ 2 = − p2 2 A. (9.3) ... |
easured lifetime against 1/ E, the result is a beautiful straight line (Figure 9.6),8 just as you would expect from Equations 9.26 and 9.29. (9.29 12 8 4 0 −4 −8 4 U238 Th232 236 230 234 5 E [MeV] 232 228 230 228 226 6 7 Figure 9.6: Graph of the logarithm of the half-life energy of the emitted alpha particle), for isot... |
int. We’re done with the patching wave function now—its only purpose was to bridge the gap. Expressing everything in terms of the one normalization constant D, and shifting the turning point back from the origin to an arbitrary point x2, the WKB wave function (Equation 9.32) becomes ψ(x) ≈ ⎧ ⎪⎨ ⎪⎩ # 2D√ p(x) sin D√ | p... |
(0) = 0. Show that this leads to the following quantization condition: where tan θ = ±2eφ, φ ≡ 1 x1 −x1 p x d x . (9.60) (9.61) Equation 9.60 determines the (approximate) allowed energies (note that E comes into x1 and x2, so θ and φ are both functions of E). (c) We are particularly interested in a high and/or broad ce... |
time (the event rate, d N ), divide by d, and normalize to the luminosity of the incident beam. 10.1 Introduction 379 ∗∗∗ Problem 10.1 Rutherford scattering. An incident particle of charge q1 and kinetic energy E scatters off a heavy stationary particle of charge q2. (a) Derive the formula relating the impact paramete... |
j(z) + in(z) ≈ −i j(z) n(z) ≈ −i 2!z/ (2 + 1)! − (2)!z−−1/2! = i 2 + 1 2 2! (2)! z2+1, (10.35) and hence σ ≈ 4π k2 ∞ =0 1 2 + 1 4 2! (2)! (ka)4+2 . But we’re assuming ka 1, so the higher powers are negligible—in the low-energy approximation the scattering is dominated by the = 0 term. (This means that the differential... |
mbining the two terms into a single integral, and using Plancherel’s theorem, Equation 2.103. 390 CHAPTER 10 Scattering Im(s) Figure 10.9: Skirting the poles in the contour integral (Equation 10.61). s = −k s = +k Re(s) Figure 10.10: Closing the contour in Equations 10.63 and 10.64. (a) (b) if z0 lies within the contou... |
ical scattering theory. In the impulse approximation we begin by pretending that the particle keeps going in a straight line (Figure 10.12), and compute the transverse impulse that would be delivered to it in that case: I = F⊥ dt. (10.95) If the deflection is relatively small, this should be a good approximation to the ... |
es are identical (spinless) bosons, then ψr (r) must be an even function of r (Figure 10.15). (b) By symmetrizing Equation 10.12 (why is this allowed?), show that the scattering amplitude in this case is f B(θ ) = f (θ ) + f (π − θ ) where f (θ ) is the scattering amplitude of a single particle of mass μ from a fixed ta... |
rder values on the right side of Equation 11.17: First Order: (1) dc a dt = 0 ⇒ c(1) a = − i baeiω0t ⇒ c H (1) b (1) dc b dt (t) = 1; = − i 0 t H ba t eiω0t dt . (11.21) Now we insert these expressions on the right side of Equation 11.17 to obtain the second-order approximation: Second Order: (2) dc a dt (t(2) a abe−iω... |
ION OF RADIATION 11.2.1 Electromagnetic Waves An electromagnetic wave (I’ll refer to it as “light”, though it could be infrared, ultraviolet, microwave, x-ray, etc.; these differ only in their frequencies) consists of transverse (and mutually perpendicular) oscillating electric and magnetic fields (Figure 11.6). An atom... |
, Pb→a(t) ≈ π |℘|2 02 ρ(ω0)t. (11.48) (11.49) This time the transition probability is proportional to t. The bizarre “flopping” phenomenon characteristic of a monochromatic perturbation gets “washed out” when we hit the system with an incoherent spread of frequencies. In particular, the transition rate (R ≡ d P/dt) is n... |
he classical oscillator frequency. The transition rate (Equation 11.63) is and the lifetime of the nth stationary state is A = nq2ω2 6π 0mc3 , τn = 6π 0mc3 nq2ω2 . (11.70) (11.71) Meanwhile, each radiated photon carries an energy ω, so the power radiated is Aω: P = q2ω2 6π 0mc3 (nω) , or, since the energy of an oscilla... |
a solid angle d. We want to count the number of states with energies between E and E + d E, with wave vectors k lying inside d. In k space these states occupy a section of a spherical shell of radius k and thickness dk that subtends a solid angle d; it has a volume k2 dk d and contains a number of states27 ρ(E) d E = k... |
cylinder of gas. By contrast, if the well expands suddenly, the resulting state is still ψ i (x) (Figure 11.15c), which is a complicated linear combination of eigenstates of the new Hamiltonian (Problem 11.18). In this case energy is conserved (at least, its expectation value is); as in the free expansion of a gas (in... |
the wave functions satisfy (t) = ei ˆH0t/ ˆH (t)e −i ˆH0t/ ˆH I |I (t) = ei ˆH0t/ |(t) . If you apply the Dyson series to the Schrödinger equation in the interaction picture, you end up with precisely the perturbation series derived in Section 11.1.2. For more details see Ramamurti Shankar, Principles of Quantum Mechan... |
fetime of the triplet state. Answer: 1.1 × 107 years. |1| Se |0|2 . A = 41 George B. Arfken and Hans J. Weber, Mathematical Methods for Physicists, 7th edn, Academic Press, San Diego (2013), p. 744. 42 Electric and magnetic dipole moments have different units—hence the factor of 1/c (which you can check by dimensional ... |
of order (t)2. Verify that the matrix in Equation 11.153 is unitary. 2 i t As an example, consider a particle of mass m moving in one dimension in a simple harmonic oscillator potential. For the numerical part set m = 1, ω = 1, and = 1 (this just defines the units of mass, time, and length). (b) Construct the Hamiltoni... |
onal infinite square well, if you like). It’s in the ground state, when an impenetrable partition is introduced, dividing the box into separate halves, B1 and B2, in such a way that the particle is equally likely to be found in either one.6 Now the two boxes are moved very far apart, and a measurement is made on B1 to s... |
ty, in the form of the instantaneous collapse of the wave function (and for that matter also in the symmetrization requirement for identical particles) had always been a feature of the orthodox interpretation, but before Aspect’s experiment it was possible to hope that quantum nonlocality was somehow a nonphysical arti... |
1√ 2 1 , , , , and hence ρ = 1/2 1/2 . 1/2 1/2 Or, more efficiently, ρ = || = √ √ 1/ 1/ 2 2 √ 1/ √ 2 1/ 2 = 1/2 1/2 . 1/2 1/2 Note that ρ is hermitian, its trace is 1, and 1/2 1/2 1/2 1/2 1/2 1/2 ρ2 = 1/2 1/2 = 1/2 1/2 1/2 1/2 = ρ. (12.23) (12.24) (12.25) Problem 12.4 (a) Prove properties 12.17, 12.18, 12.19, and 12.20... |
that their nemesis, Eve, might try to intercept this communication, and thereby crack the code, without their knowledge. Alice prepares a string of photons in four different states: linearly polarized (horizontal |H and vertical |V ), and circularly polarized (left |L and right |R), which she sends to Bob. Eve hopes t... |
y vector |α. And for every vector |α there is an associated inverse vector (| − α),4 such that |α + |−α = |0 . (A.5) 1 For our purposes, the scalars will be ordinary complex numbers. Mathematicians can tell you about vector spaces over more exotic fields, but such objects play no role in quantum mechanics. Note that α, ... |
with components a1, a2, . . . , an into a vector with components7 n a i = Ti j a j . (A.33) j=1 Thus the n2 elements Ti j uniquely characterize the linear transformation ˆT (with respect to a given basis), just as the n components ai uniquely characterize the vector |α (with respect to that basis): ˆT ↔ (T11, T12, . .... |
as the property (see Problem A.17) that Tr(T1T2) = Tr(T2T1), = detTe. (A.65) (A.66) (A.67) 14 Notice, however, the radically different perspective: In this case we’re talking about one and the same vector, referred to two completely different bases, whereas before we were thinking of a completely different vector, refe... |
constant) a(i) itself. If we call the constant νi , b(i) = V a(i) = νi a(i) , (A.87) so a(i) is an eigenvector of V. All that remains is to relax the assumption of nondegeneracy. Assume now that T has at least one degenerate eigenvalue such that both a(1) and a(2) are eigenvectors of T with the same eigenvalue λ0: T a(... |
A.99) (a) Find exp (M), if (iii) M = 0 −θ . θ 0 (b) Show that if M is diagonalizable, then eM det = eTr(M). (A.100) Comment: This is actually true even if M is not diagonalizable, but it’s harder to prove in the general case. (c) Show that if the matrices M and N commute, then eM+N = eMeN. (A.101) Prove (with the simpl... |
etation 102–105 generalized symmetrization principle 207 generalized uncertainty principle 105–108 generating function 54 generator rotations 248–249 translations in space 235 translations in time 263 geometric phase 429–430 g-factor 301 deuteron 320 electron 301, 311 Landé 306 muon 313 positron 313 proton 311 Golden R... |
tion space wave function 104, 121–123 positronium 200, 313 potential 25 Coulomb 143 delta-function 61–70 Dirac comb 221 effective 138 634 504 finite square well 70–76 hydrogen 143 infinite square well 31–39 Kronig–Penney 221–222 power law 371 reflectionless 81 scalar 181 sech-squared 81, 371, 375 spherically symmetrica... |
ctor 464–465 Zurek, W. 459 644 514 |
ions of structure for a spherical star clus ter. Isothermal star clusters. 685 683 Eigenvalue problem and variational principle for normal-mode pulsations. 695 26.2. Critical adiabatic index for nearly Newtonian stars. 697 27.1. 1 Cosmology in brief. 27.2. 704 The 3-geo metry of hypersu rfaces of ho mogene ity. Friedma... |
l redshift on the Earth. Brault's determination of the redshift of the 0 1 line of sodium from the sun. The Turner-Hill search for a dependence of proper clock rate on velocity relative to distant mat ter. 1059 1057 1065 40.1.1 Bending of trajectory near the sun. 40.2. Coordinates used in calculating the deflection of ... |
ana tion for 'gravi ta tion.''' Then the dinner bell was quiet. and he was gone, with book, magnifying glass-and apple. §1.2. SPACETIME WITH AND WITHOUT COORDINATES Now it came to me: ... the independence of the gravitational acceleration from the nature of the falling substance, may be expressed as follows: In a gravi... |
Einstein summation com'ention is used here: any index that is repeated in a product is automatically slimmed on .\a(~.p) or just x a, X';'(:1') or just xa '. x"(:'p) or just x,i. , Xl, x 2• x 3 ). XO(X O X1(Xo. xl, x 2. x 3 ), x2(XO• x', x 2• x 3 X3(X O• x', x 2• x 3 xa(x B). ). ), u or v or (. or :'1' - i!. ~" == .\"... |
stone away from its natural "world line," (Le., its natural track through space time). That could lead to an interesting issue of solid-state physics, but that is not the topic of concern here. Fall is synonymous with weightlessness: absence of any force to drive the object away from its normal track through space time... |
cally exactly equivalent, that is, if we assume that we may just as well regard the system K as being in a space free from gravitational fields, if we then regard K as uniformly accelerated. This assumption of exact physical equivalence makes it impossible for us to speak of the absolute acceleration of the system of r... |
cal Lorentz frame where particle remains at rest: 1(nl = t2 - X2 = (t - X)(t + x) = 1 (/'}'1(/Z. B. Language ofcoordinates (Lorentz, Poincare, Minkowski, Einstein): From any event d to any other nearby event ~i3, there is a proper distance Sd<tJ or proper time 1(/,,! given in suitable (local Lorentz) coordinates by S(/... |
emed to fall by day. For a profound darkness arose so that stars even appeared in the sky. This happened in the eastern sky when the Sun dwelt in Capricorn" [from Westermann and Boissonade (1878»). Does this 30° for this eclipse, IOgether with corresponding amounts for other eclipses, represent the "right" correction? ... |
cles move in a straight line with constant velocity. Where is any gravitational deflection to be seen in that? For answer, turn back to the apple (Figure 1.1). Inspect again the geodesic tracks of the ants on the surface of the apple. Note the reconvergence of two nearby geodesics that originally diverged from a common... |
more reflection produces a renewed sense of concern. Curvature with respect to what? Not with respect to the labo- Photograph of stars when sun (eclipsed bv moon) lies " a~ indicated .. ." Photograph of stars when sun swims elsewhere t I I I I l' I I I I A. Bending of light by the sun depicted as a conse quence of the ... |
he same density, p == 5.52 g/cm3 , as the average density of the Earth. A hole is bored through this sphere. Two test particles, A and E, execute simple harmonic motion in this hole, with an 84-minute period. Therefore their geodesic separation (, however it may be oriented, undergoes a simple periodic motion with the ... |
shows how the stress-energy of matter generates an average curvature (Einstein = G) in its neighborhood. Simultaneously, the field equation is a propagation equation for the remaining, anisotropic part of the curvature: it governs the external spacetime curvature of a static source (Earth); it governs the generation of... |
he next few sections will introduce several geometric objects, and show the roles they playas representatives of physical quantities in flat spacetime. Every physical quantity can be described by a geometric object laws of physics can be All expressed geometrically 49 §2.3. VECTORS A .~ Two events 9('\ = 0.7) C/,\=l cu... |
cance. Diffract this wave from a crystal lattice. From the pattern of diffraction, one can determine not merely the length of the de Broglie waves, but also the pattern in space made by surfaces of equal, integral phase ep = 7, ep = 8, ep = 9, .... This The 1-form illustrated by de Broglie waves 54 2. FOUNDATIONS OF SP... |
; its description of the forward march of time is not so nice! One often omits the tilde from the I-form p corresponding to a vector p, and uses the same symbol p for both. Such practice is justified by the unique correspond ence (both mathematical and physical) between p and p. Exercise 2.1. Show that equation (2.14) ... |
v = wr, is unchanging in time, 0 are equal, and the magnitudes-but not the directions-of ue and ua 0 and Ua ue are equal: From the geometry of Figure 2.9, one sees that ue makes the same angle with p as does Ua • Consequently, p. ue = p. ua ' and Aabsorbed/Aemitted = 1. There is no redshift ! Notice that this solution ... |
pair by a photon in the field of force of a massive nucleus is 2me. (5) The threshold for the production of an (e+, e-) pair by a photon in an encounter with an electron at rest • is 4me (or 4me f when account is taken of the binding of the e+e-e- system in a very light "molecule"). All these results (topics for indepe... |
hanged. Choose a specific tensor S, of rank cD for explicitness. Into the slots of S, insert the basis vectors and I-forms of a specific Lorentz coordinate frame. The output is a "component of S in that frame": This defines components. Knowing the components in a specific frame, one can easily calculate the output prod... |
c. (validity for all test particles). (5) In this way arrive at the expressions (3.23) for the E and jj in terms of the E and B. The contrast in difficulty is obvious! §3.4. MAXWELL'S EQUATIONS Turn now from the action of the field on a charge, and ask about the action of a charge on the field, or, more generally, ask ... |
CS Equation Name and Discussion Computing components. Computing other components. Reconstructing the rank-@ version of S. Reconstructing the rank-a) version of S. [Recall: one does not usually distinguish between the various versions; see equa tion (3.15) and associated discussion.] Raising an index. Lowering an index.... |
by: E. Cartan (1945); de Rham (1955); Nickerson, Spencer, and Steenrod (1959); Hauser (1970); Israel (1979); especially Flanders (1963, relatively easy, with many applications); Spivak (1965, sophomore or junior level, but fully in tune with modem mathematics); H. Cartan (1970); and Choquet-Bruhat (l968a). §4.1. EXTER... |
tromagnetic 2-form or Faraday: IIFII2 = B2 _ £2. 2. Dual of a pjorm. a. In an n-dimensional space, the dual of a p-form a is the (n - p)-form *a, with components ( *0:) kt ...k n _ p - o:litoo.ipl E il ...i p kt ...kn-p· b. Properties of duals: **a = (-I)P-1a in spacetime; a /\ *a = lIall 2e in general. c. Note: the de... |
roduct of I-forms. Moreover, though it the tube picture, abstract must be if it is to be truthful, also has this advan tage, that the orientation of the elementary tubes (sense of circulation as indicated by arrows in Figures 4.1 and 4.5, for example) lends itself to ready visualization. Let the "walls" of the tubes th... |
s dual, according to the prescription in exercise 3.14, is Maxwell: Maxwell = *F = e sin B dB /\ dfP, (4.16) Pattern of tubes in dual structure Mexwel/ for point-charge at rest (4.17) as illustrated in Figure 4.5. Take a tour in the positive sense around a region of the surface of the sphere illustrated in Figure 4.5. ... |
resent, the situation changes. Tubes of Maxwell take their origin in such a region. The density of endings is described by the 3-form *J = charge, a "collection of eggcrate cells" collected along bundles of world lines. The two equations and dF= 0 d*F=4'iT*J summarize the entire content of Maxwell's equations in geomet... |
back to the original form for forms of odd order, and to the negative thereof for forms of even order. In Euclidean 3-space the operation reproduces the original form, regardless of its order. Duality Plus Exterior Differentiation Start with scalar cf;. Its gradient dcf; is a I-form. Take its dual, to get the 3-forrn ... |
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