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Max von Laue

Max Theodor Felix von Laue

A course in two strands: the life of the theorist who showed that crystals diffract X-rays and who defended Einstein’s physics in Nazi Berlin, and the physics of lattices, relativistic continua and superconductors that he helped found.

Born 9 October 1879, Pfaffendorf near Koblenz Died 24 April 1960, West Berlin Nobel Prize Physics, 1914 Level advanced undergraduate → graduate
A Laue pattern computed for zinc blende (face-centred cubic, a = 5.41 Å) with a white X-ray beam along [001]. Spot size follows a rough structure-factor weight; the fourfold symmetry is the one Friedrich and Knipping recorded in 1912.
Lesson 00

How to use this course

Max von Laue is remembered for one experiment, and he did not perform it himself. In 1912 he proposed that a crystal should act as a three-dimensional diffraction grating for X-rays. Two experimenters in Munich tried it, and within weeks the result settled two questions at once: X-rays are waves, and crystals are lattices of atoms. Crystallography, solid-state physics and structural biology all start there.

Laue was also one of the first physicists to take up Einstein’s relativity. He wrote its first textbook, gave the first relativistic treatment of extended bodies, and later contributed to the theory of superconductivity. Under the Nazi regime he was among the very few German physicists who opposed the attacks on Einstein openly. This course follows both the science and the life.

Prerequisites for Part II
  • Special relativity: Lorentz transformations, four-vectors and index notation.
  • Wave optics: interference, gratings and the Fourier picture of diffraction.
  • Basic crystallography helps: Bravais lattices, Miller indices.
  • For Lesson 10, Maxwell’s equations and magnetostatics with boundary conditions.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Lessons 04 and 05 cover relativity (1907–1911), Lessons 06 to 09 the diffraction of X-rays (1912–1949), and Lesson 10 superconductivity (1932–1947). Formulas use SI units unless stated otherwise.

Reference

Timeline

1879Born 9 October in Pfaffendorf, near Koblenz, son of Julius Laue, an official in the military administration, and Minna Zerrenner.
1898Abitur in Strassburg. Military service; begins studying at the University of Strassburg.
1899–1902Göttingen, under the influence of Woldemar Voigt; a semester in Munich.
1902–03Berlin. Doctorate under Max Planck (July 1903) on interference in plane-parallel plates.
1903–05Back in Göttingen; passes the state examination for secondary-school teaching.
1905Planck’s assistant at the Institute for Theoretical Physics, Berlin.
1906Habilitation in Berlin, on the entropy of interfering light beams.
1907Derives Fresnel’s drag coefficient from Einstein’s velocity addition.
1909Privatdozent at Arnold Sommerfeld’s Institute for Theoretical Physics, Munich.
1910Marries Magdalene Degen in Munich.
1911Das Relativitätsprinzip, the first monograph on special relativity. Paper on the dynamics of extended systems.
1912Early in the year, conversations with Paul Peter Ewald. April: Walter Friedrich and Paul Knipping record the first X-ray diffraction patterns. June: results presented to the Bavarian Academy. Extraordinary professor at Zurich.
1913His father is raised to the hereditary nobility; the family becomes “von Laue”.
1914Full professor of theoretical physics at the new University of Frankfurt. Nobel Prize in Physics for 1914.
1916War work at Würzburg on high-vacuum tubes for telephony and radio.
1917Joins the Kaiser Wilhelm Institute for Physics, handling its administration for Einstein.
1919Exchanges chairs with Max Born and returns to Berlin as professor, where he stays until 1943.
1925Becomes theoretical adviser to the Physikalisch-Technische Reichsanstalt.
1932Max Planck Medal. Explains the shape dependence of the critical field of superconductors.
193318 September: as chairman of the German Physical Society, opens the Würzburg meeting with a defence of relativity. December: removed as adviser to the Reichsanstalt.
193529 January: helps organize the memorial for Fritz Haber in Berlin-Dahlem. Paper with Fritz and Heinz London on superconductivity.
1941Röntgenstrahl-Interferenzen, his monograph on X-ray diffraction.
1943Retires from the Berlin chair and moves with the Kaiser Wilhelm Institute to Hechingen.
1945–46Taken into custody by the Alsos mission; interned at Farm Hall, England, with nine other German scientists.
1946Göttingen. Geschichte der Physik. Only German invited to the international crystallography conference in England.
1947Theorie der Supraleitung.
1948Honorary president of the International Union of Crystallography.
1951–59Director of the Max Planck Institute for Physical Chemistry in Berlin-Dahlem, renamed the Fritz Haber Institute in 1953 at his request.
19608 April: his car is struck by a motorcyclist in Berlin. Dies of his injuries on 24 April, aged 80.
Part I · Formation
Lesson 01

A wandering childhood, 1879–1903

Max Laue was born on 9 October 1879 in Pfaffendorf, a village on the Rhine opposite Koblenz. His father, Julius Laue, was a civil official in the military administration, and his postings moved the family across the German Empire: Brandenburg, Altona, Posen, Berlin and finally Strassburg. Max went to school in Posen, Berlin and Strassburg. At the Protestant gymnasium in Strassburg a teacher named Goering introduced him to the exact sciences, and he never looked back.

He took the Abitur in 1898 and began his studies of mathematics, physics and chemistry at the University of Strassburg during his year of military service. In 1899 he moved to Göttingen. There Woldemar Voigt, the theorist of crystal physics, drew him to theoretical physics, and Max Abraham, the electron theorist, made a lasting impression. After a semester in Munich he went to Berlin in 1902 to study with Max Planck. He also attended Otto Lummer’s lectures on interference spectroscopy and heat radiation, which shaped his first research topic.

His doctoral thesis, Über die Interferenzerscheinungen an planparallelen Platten, dealt with the interference fringes produced by plane-parallel glass plates, the basis of the Lummer–Gehrcke interferometer. Planck accepted it in July 1903. Laue later called Berlin his true intellectual home, and Planck remained the central figure of his scientific life.

Lesson 02

Planck’s assistant, 1903–1909

After the doctorate Laue spent two years in Göttingen and passed the state examination to teach in secondary schools. In 1905 Planck called him back to Berlin as his assistant at the Institute for Theoretical Physics. His habilitation of 1906 addressed a question at the edge of Planck’s radiation theory: what is the entropy of light beams that can interfere? Laue showed that the entropies of two coherent beams cannot simply be added, because the beams are not statistically independent. Thermodynamics, he argued, must take coherence into account. The problem kept him close to optics, which would matter in 1912.

The year 1905 also brought Einstein’s paper on the electrodynamics of moving bodies. Planck took it up at once, and Laue, after initial hesitation, became one of its first convinced supporters. In 1907 he published a short paper showing that Einstein’s rule for adding velocities explains Fizeau’s 1851 measurement of the speed of light in flowing water (Lesson 04). This was one of the first experimental arguments for the new theory, and it was achieved without any ether.

Between 1907 and 1911 he wrote eight papers on relativity. They established him as one of the few physicists who could use the theory as a working tool.

Lesson 03

Munich and the idea of 1912

In 1909 Laue moved to Munich as Privatdozent at Arnold Sommerfeld’s Institute for Theoretical Physics, where he taught optics, thermodynamics and relativity. In 1910 he married Magdalene Degen. In 1911 he published Das Relativitätsprinzip, the first textbook of special relativity. It used Minkowski’s four-dimensional formalism throughout and went through many editions; a volume on general relativity followed in 1921.

Munich in 1912 held three ingredients in one building complex. Wilhelm Röntgen, discoverer of X-rays, directed the experimental institute. Sommerfeld had worked on the theory of X-ray emission, and his group believed X-rays were very short electromagnetic pulses or waves, with wavelengths estimated at roughly \(10^{-9}\) to \(10^{-8}\) cm. Crystallographers in Munich, following Paul Groth, took for granted that crystals were regular arrays of atoms, with spacings of about \(10^{-8}\) cm, though nobody had seen the lattice.

Early in 1912, Sommerfeld’s doctoral student Paul Peter Ewald consulted Laue about his thesis on the optics of a lattice of dipoles. While discussing it, Laue asked what would happen if the waves were much shorter than the lattice spacing. The thought followed: if X-rays have wavelengths comparable to atomic spacings, a crystal should diffract them as a grating diffracts light. Sommerfeld was sceptical, partly because thermal motion of the atoms was expected to wash out the pattern. Walter Friedrich, Sommerfeld’s assistant, and Paul Knipping, a young physicist from Röntgen’s institute, agreed to try.

The first exposure, through a crystal of copper sulphate, showed spots around the direct beam. A thin plate of zinc blende cut perpendicular to a cube axis then gave a sharp, fourfold-symmetric pattern. In June 1912 the results went to the Bavarian Academy of Sciences as Interferenz-Erscheinungen bei Röntgenstrahlen, by Friedrich, Knipping and Laue, with Laue’s theory alongside. Laue was by then leaving for an extraordinary professorship in Zurich.

Exercise 3.1Ewald’s thesis concerned visible light, of wavelength about \(5\times10^{-5}\) cm, in a crystal with spacing about \(3\times10^{-8}\) cm. Explain why such light shows no diffraction spots, while X-rays of wavelength \(10^{-8}\) cm can.

Show solution
A diffracted beam from planes of spacing \(d\) requires \(\lambda\le 2d\) (Lesson 06). For visible light \(\lambda/2d\approx 800\), so no direction satisfies the condition and the crystal acts as a continuous medium with a refractive index. That was exactly Ewald’s topic. For X-rays \(\lambda/2d\) is of order 0.2, so many lattice planes can diffract.
Part II · The physics
Lesson 04

Fizeau’s experiment and velocity addition

For the general reader. In 1851 Hippolyte Fizeau sent light through moving water and found that the water drags the light along, but only partly. For fifty years the explanation required an ether that was partly carried by matter. Laue showed in 1907 that Einstein’s rule for adding velocities gives the measured result directly, with no ether at all.

Goals
  • Derive the Fresnel drag coefficient from relativistic velocity addition.
  • Estimate the size of the effect in Fizeau’s apparatus.
  • See why the effect is invisible in air.

The experiment

Fizeau split a beam, sent the two halves through parallel tubes of water flowing in opposite directions, and recombined them. The fringe shift measured the difference in light speed with and against the flow. Augustin Fresnel had predicted in 1818 that a medium of refractive index \(n\) moving at speed \(v\) changes the light speed by \(f v\), with the drag coefficient \(f=1-1/n^2\). Fizeau confirmed this, and Albert Michelson and Edward Morley repeated it with greater precision in 1886.

Result (Laue, 1907)

Let light travel with speed \(c/n\) relative to a medium, and let the medium move with speed \(v\) parallel to the light in the laboratory. By Einstein’s addition theorem the laboratory speed is

\[ u=\frac{c/n+v}{1+\dfrac{v}{nc}} = \frac{c}{n}+v\Big(1-\frac{1}{n^2}\Big)+O\!\Big(\frac{v^2}{c}\Big). \]

The first-order term is Fresnel’s drag, derived as kinematics.

Derivation

Expand the denominator: \(\big(1+\tfrac{v}{nc}\big)^{-1}=1-\tfrac{v}{nc}+O(v^2/c^2)\). Then

\[ u=\Big(\frac cn+v\Big)\Big(1-\frac{v}{nc}\Big)+O\!\Big(\frac{v^2}{c}\Big)=\frac cn+v-\frac{v}{n^2}+O\!\Big(\frac{v^2}{c}\Big). \]

The expansion parameter \(v/c\) is about \(10^{-8}\) for Fizeau’s apparatus, so the second-order terms are far below any measurement. The ether theories needed a physical hypothesis about how matter drags the ether. Relativity needs only the transformation law of velocities, and it treats source, medium and observer symmetrically. Lorentz had already derived the drag from his electron theory, and later added a correction for dispersion: since the light frequency in the moving water is Doppler-shifted, \(n\) should be evaluated at the frequency seen by the water, giving \(f=1-\frac1{n^2}-\frac{\lambda}{n}\frac{dn}{d\lambda}\). Zeeman’s measurements of 1914–15 confirmed this term.

Exercise 4.1In Fizeau’s experiment the water moved at about 7 m/s and \(n=1.333\). Find the change in light speed in each tube, and the difference between the two tubes.

Show solution
\(1-1/n^2=1-0.563=0.437\), so each tube changes the light speed by \(0.437\times7\approx3.1\) m/s. With and against the flow, the difference is about 6.1 m/s, out of \(c/n\approx2.25\times10^8\) m/s. The fringe shift accumulates this tiny difference over the length of the tubes, which is why an interferometer was needed.

Exercise 4.2Show that for a gas with \(n=1+\epsilon\), \(\epsilon\ll1\), the drag coefficient is approximately \(2\epsilon\). For air, \(\epsilon\approx3\times10^{-4}\). What does this imply for experiments that compare light paths through moving and resting air?

Show solution
\(1/n^2=(1+\epsilon)^{-2}\approx1-2\epsilon\), so \(f\approx2\epsilon\approx6\times10^{-4}\). Moving air drags light by less than a thousandth of its own speed, so effects of moving air on light are almost undetectable, and ether-drift experiments in air cannot be explained by air dragging the ether along.

Exercise 4.3Show that the relativistic formula never gives \(u>c\) for \(n\ge1\) and \(0\le v<c\), while the Fresnel first-order formula can.

Show solution
\(c-u=\dfrac{c(1-1/n)(1-v/c)}{1+v/(nc)}\ge0\), with equality only for \(n=1\). The first-order formula \(c/n+v(1-1/n^2)\) exceeds \(c\) when \(v>c\,n/(n+1)\), so at large speeds it is unphysical: it is only the first term of an expansion.
Lesson 05

Relativistic continua and Laue’s theorem

For the general reader. Relativity says energy has inertia. But a body is made of parts under stress: a gas pushing on its container, a charge held together against its own repulsion. Laue showed in 1911 that stresses carry momentum when the body moves, and that the familiar rule \(E=mc^2\) holds for a body as a whole only when its internal stresses balance out. This solved a puzzle that had dogged the theory of the electron.

Goals
  • Write the energy-momentum tensor and its conservation law.
  • Prove Laue’s theorem for static closed systems.
  • Explain the 4/3 problem of the electromagnetic electron and its resolution.

The energy-momentum tensor

Minkowski had combined the electromagnetic energy density, momentum density and Maxwell stresses into one four-dimensional object. Laue’s 1911 paper Zur Dynamik der Relativitätstheorie extended this to any continuous medium. Using units with \(c=1\), the symmetric tensor \(T^{\mu\nu}\) has \(T^{00}\) the energy density, \(T^{0i}\) the momentum density (equal to the energy flux), and \(T^{ij}\) the stress, the flux of \(i\)-momentum across a surface normal to \(j\). For a complete system with no external forces,

\[ \partial_\mu T^{\mu\nu}=0. \]

The four integrals \(P^\nu=\int T^{0\nu}\,d^3x\) are the total energy and momentum. They are conserved, but whether they form a four-vector is a separate question.

Theorem (Laue, 1911)

Let \(T^{\mu\nu}\) be symmetric, conserved, and static in some frame (\(\partial_0T^{\mu\nu}=0\), \(T^{0i}=0\)), and let it vanish outside a bounded region, or fall off fast enough at infinity. Then in that rest frame

\[ \int T^{ij}\,d^3x=0\qquad(i,j=1,2,3), \]

and the total energy and momentum, computed at fixed time in any inertial frame, form a four-vector \((E,\mathbf P)=(\gamma M,\gamma M\mathbf v)\) with \(M=\int T^{00}\,d^3x\).

Proof

For static fields the conservation law reduces to \(\partial_jT^{jk}=0\). Then

\[ \partial_j\big(x^iT^{jk}\big)=\delta^i_j\,T^{jk}+x^i\,\partial_jT^{jk}=T^{ik}, \]

so \(\int T^{ik}\,d^3x\) is the integral of a divergence. By Gauss’s theorem it equals a surface integral that vanishes for a bounded system. For the second part, boost the body to velocity \(v\) along \(x\). The tensor transforms as \(T'^{00}=\gamma^2(T^{00}+\beta^2T^{11})\) and \(T'^{01}=\gamma^2\beta(T^{00}+T^{11})\) (with \(T^{01}=0\) in the rest frame), while the volume element at fixed laboratory time contracts by \(1/\gamma\). Writing \(S=\int T^{11}\,d^3x\),

\[ E'=\gamma\,(M+\beta^2S),\qquad P'=\gamma\beta\,(M+S). \]

These are the components of a four-vector of mass \(M\) exactly when \(S=0\), which the first part guarantees.

The 4/3 problem

Before relativity, Max Abraham and Hendrik Lorentz had computed the momentum of the electric field of a moving charged sphere and found an electromagnetic mass \(\tfrac43\,E_{\text{em}}/c^2\) rather than \(E_{\text{em}}/c^2\). The electromagnetic field alone is not a closed system: its stresses push the charge outward, and the field tensor is traceless, \(-T^{00}+\sum_iT^{ii}=0\). For a spherically symmetric field each diagonal stress then integrates to \(S=E_{\text{em}}/3\), and the boost formula gives \(P'=\gamma\beta\cdot\tfrac43E_{\text{em}}\). Laue showed that once the cohesive stresses that hold the electron together, proposed by Henri Poincaré in 1905–06, are included, the total system satisfies his theorem and the factor disappears. The same argument handles a gas in a box, black-body radiation in a cavity and stressed solids. It was a general demonstration that the inertia of every form of energy obeys \(E=Mc^2\) for a closed system.

A note on the twin paradox

Laue also wrote on the clock problem. In 1913, in the second edition of his book, he explained the asymmetric ageing of the travelling twin by the fact that the traveller occupies two different inertial frames, outward and back, while the stay-at-home twin occupies one. The point is that proper time \(\int\sqrt{1-v^2/c^2}\,dt\) is a property of the world-line, and the inertial world-line between two events is the longest.

Exercise 5.1A cubic box of volume \(V\) contains black-body radiation with energy density \(u\) and pressure \(p=u/3\). Compute \(\int T^{11}\,d^3x\) over the radiation alone. What must the walls contribute, and why does a moving box of radiation have inertia \(E/c^2\) with \(E\) the total rest energy?

Show solution
In the rest frame \(T^{11}=p=u/3\) throughout the gas, so \(\int T^{11}=uV/3\). The radiation alone would have momentum \(\gamma\beta(uV+uV/3)\), the 4/3 factor again. The walls are in tension to resist the pressure; by Laue’s theorem their integrated stress is \(-uV/3\), so the total \(S\) vanishes. The complete box then has \(P=\gamma\beta(E_{\text{rad}}+E_{\text{walls}})\), with all rest energy contributing \(1/c^2\) to the inertia.

Exercise 5.2Show that the trace of the electromagnetic energy-momentum tensor vanishes, using \(T^{00}=\tfrac12(E^2+B^2)\) and \(T^{ij}=-\big(E_iE_j+B_iB_j\big)+\tfrac12\delta_{ij}(E^2+B^2)\) in Heaviside–Lorentz units with \(c=1\).

Show solution
\(\sum_iT^{ii}=-(E^2+B^2)+\tfrac32(E^2+B^2)=\tfrac12(E^2+B^2)=T^{00}\). With signature \((-,+,+,+)\), \(T^\mu{}_\mu=-T^{00}+\sum_iT^{ii}=0\). For a spherically symmetric configuration the three diagonal integrals are equal, so each is \(E_{\text{em}}/3\).

Exercise 5.3Twin A stays home; twin B travels at \(0.8c\) to a star 4 light-years away and returns at once. Compute each twin’s elapsed time and identify the two inertial frames that Laue’s argument uses.

Show solution
A’s time is \(2\times4/0.8=10\) years. B’s time is \(10\sqrt{1-0.64}=6\) years. B’s outbound and inbound legs are at rest in two different inertial frames; the change of frame at the star shifts B’s notion of “simultaneous at home” by 6.4 years, which accounts for the asymmetry. A never changes frame.
Lesson 06

The Laue equations and the reciprocal lattice

For the general reader. A ruled grating spreads light into a few sharp directions because the waves from each groove add up in step only in those directions. A crystal is a grating in three dimensions. Laue wrote down the three conditions that the scattered X-rays must satisfy to add up in step, one for each lattice direction. All of crystallography rests on these conditions.

Goals
  • Derive the Laue conditions from a lattice sum.
  • Rewrite them with the reciprocal lattice as \(\Delta\mathbf k=\mathbf G\).
  • Show that they are equivalent to Bragg’s law.

Scattering from a lattice

Let a plane wave \(e^{i\mathbf k_0\cdot\mathbf r}\) fall on point scatterers at the lattice sites \(\mathbf R=n_1\mathbf a_1+n_2\mathbf a_2+n_3\mathbf a_3\). Far away in the direction of wavevector \(\mathbf k\), with \(|\mathbf k|=|\mathbf k_0|=2\pi/\lambda\) (elastic scattering), the amplitude is proportional to the lattice sum

\[ A(\mathbf q)=\sum_{\mathbf R}e^{-i\mathbf q\cdot\mathbf R},\qquad \mathbf q=\mathbf k-\mathbf k_0. \]

For a crystal of \(N_1\times N_2\times N_3\) cells the sum factorizes into three geometric series:

\[ |A(\mathbf q)|^2=\prod_{j=1}^{3}\frac{\sin^2\!\big(\tfrac12N_j\,\mathbf q\cdot\mathbf a_j\big)}{\sin^2\!\big(\tfrac12\,\mathbf q\cdot\mathbf a_j\big)}. \]

Each factor has sharp maxima of height \(N_j^2\) where \(\mathbf q\cdot\mathbf a_j\) is a multiple of \(2\pi\), and is small elsewhere.

Laue conditions (1912)

A diffracted beam of appreciable intensity occurs only when there are integers \(h,k,l\) with

\[ \mathbf a_1\cdot(\mathbf k-\mathbf k_0)=2\pi h,\qquad \mathbf a_2\cdot(\mathbf k-\mathbf k_0)=2\pi k,\qquad \mathbf a_3\cdot(\mathbf k-\mathbf k_0)=2\pi l. \]

Equivalently, \(\mathbf k-\mathbf k_0=\mathbf G_{hkl}=h\mathbf b_1+k\mathbf b_2+l\mathbf b_3\), where the reciprocal basis satisfies \(\mathbf a_i\cdot\mathbf b_j=2\pi\delta_{ij}\).

Laue himself wrote the conditions with unit vectors \(\mathbf s_0,\mathbf s\) as \(\mathbf a_1\cdot(\mathbf s-\mathbf s_0)=h\lambda\), and so on. Each condition alone fixes a cone of directions around \(\mathbf a_j\); a diffracted beam appears only where three cones meet in a common line, which for a fixed wavelength is exceptional. This is why a single-wavelength beam on a fixed crystal usually gives no spots, and why the white (continuous) spectrum of an X-ray tube was essential in 1912. The reciprocal basis is \(\mathbf b_1=2\pi\,\mathbf a_2\times\mathbf a_3/V_c\) and cyclically, with \(V_c=\mathbf a_1\cdot(\mathbf a_2\times\mathbf a_3)\).

Equivalence with Bragg’s law

The vector \(\mathbf G_{hkl}\) is perpendicular to the lattice planes with Miller indices \((hkl)\), and if \(h,k,l\) have no common factor, \(|\mathbf G_{hkl}|=2\pi/d_{hkl}\), with \(d_{hkl}\) the spacing of these planes. For elastic scattering through angle \(2\theta\), \(|\mathbf k-\mathbf k_0|=2k\sin\theta=(4\pi/\lambda)\sin\theta\). Setting this equal to \(|\mathbf G_{nh,nk,nl}|=2\pi n/d_{hkl}\) gives

\[ 2d_{hkl}\sin\theta=n\lambda, \]

with \(\mathbf k-\mathbf k_0\) normal to the planes, so the beam is mirror-reflected by them. This is the form that Lawrence Bragg found in the autumn of 1912. The two descriptions are the same mathematics; Bragg’s picture made the geometry intuitive and opened the way to structure determination.

Exercise 6.1Show that the reciprocal lattice of a face-centred cubic lattice with cubic edge \(a\) is body-centred cubic with cubic edge \(4\pi/a\).

Show solution
Take \(\mathbf a_1=\tfrac a2(0,1,1)\), \(\mathbf a_2=\tfrac a2(1,0,1)\), \(\mathbf a_3=\tfrac a2(1,1,0)\), so \(V_c=a^3/4\). Then \(\mathbf b_1=2\pi\,\mathbf a_2\times\mathbf a_3/V_c=\tfrac{2\pi}{a}(-1,1,1)\), and cyclically \(\mathbf b_2=\tfrac{2\pi}a(1,-1,1)\), \(\mathbf b_3=\tfrac{2\pi}a(1,1,-1)\). These are the primitive vectors of a bcc lattice with cubic edge \(4\pi/a\). In cubic indices, the allowed \(\mathbf G\) are \(\tfrac{2\pi}{a}(h,k,l)\) with \(h,k,l\) all even or all odd.

Exercise 6.2Show from the Laue conditions that no diffraction occurs if \(\lambda>2d_{\max}\), where \(d_{\max}\) is the largest interplanar spacing. Relate this to Exercise 3.1.

Show solution
\(|\mathbf k-\mathbf k_0|\le2k=4\pi/\lambda\), while every non-zero \(\mathbf G\) has \(|\mathbf G|\ge2\pi/d_{\max}\). A solution needs \(2\pi/d_{\max}\le4\pi/\lambda\), that is \(\lambda\le2d_{\max}\). Visible light violates this by a factor of several hundred, so a crystal only refracts it.

Exercise 6.3For a one-dimensional row of \(N\) scatterers, find the angular half-width of a diffraction maximum in \(q=\mathbf q\cdot\mathbf a/|\mathbf a|\), and explain why larger crystals give sharper spots.

Show solution
The factor \(\sin^2(Nqa/2)/\sin^2(qa/2)\) first vanishes at \(qa=2\pi m\pm2\pi/N\). The peak width in \(q\) is therefore about \(2\pi/(Na)\), inversely proportional to the crystal length. This is the basis of Scherrer’s later estimate of crystallite size from line widths.
Lesson 07

Reading a Laue photograph

For the general reader. The 1912 photographs show a central black spot surrounded by smaller spots arranged with the symmetry of the crystal. Each spot is a beam reflected from one family of atomic planes, picking out from the white X-ray beam just the wavelength those planes can reflect. Reading the pattern means working back from spot positions to plane orientations.

Goals
  • Use the Ewald sphere to see which reflections a white beam produces.
  • Locate a spot on a flat plate from its Miller indices.
  • Understand the zinc blende photograph and Laue’s first, imperfect analysis.

The Ewald construction

Draw \(\mathbf k_0\) ending at the origin of reciprocal space. The condition \(\mathbf k=\mathbf k_0+\mathbf G\) with \(|\mathbf k|=|\mathbf k_0|\) says that \(\mathbf G\) lies on a sphere of radius \(k\) centred at the tail of \(\mathbf k_0\). Ewald introduced this construction in 1913. A monochromatic beam gives a single sphere, which rarely passes through reciprocal lattice points. A white beam with \(\lambda_{\min}\le\lambda\le\lambda_{\max}\) gives a family of spheres filling a crescent-shaped region, and every lattice point inside the region produces a spot.

Spot positions

Let the beam run along \(\mathbf a_3\) of a cubic crystal of edge \(a\), so \(\mathbf k_0=(0,0,2\pi/\lambda)\) and \(\mathbf G=\tfrac{2\pi}{a}(h,k,l)\). The condition \(|\mathbf k_0+\mathbf G|=|\mathbf k_0|\) selects the wavelength

\[ \lambda_{hkl}=\frac{-2la}{h^2+k^2+l^2}\qquad(l<0), \]

and a flat plate at distance \(D\) behind the crystal records the spot at

\[ (x,y)=\frac{2l\,D}{l^2-h^2-k^2}\,(h,k). \]

All harmonics \((nh,nk,nl)\) fall on the same spot, since the plate sees only the direction. Each spot is thus a superposition of orders, which made intensities hard to interpret.

Zinc blende and Laue’s analysis

The zinc blende photograph, with the beam along a cube axis, showed fourfold symmetry. Laue indexed the spots on a simple cubic lattice with one ZnS unit per cube, whose edge he computed from the density. He found he could account for the observed spots only by supposing that the beam contained a few discrete wavelengths, as if the crystal emitted characteristic radiation. In November 1912 Lawrence Bragg showed that a continuous spectrum and a face-centred cubic lattice explain the pattern: the missing spots are exactly those forbidden by the fcc lattice (Exercise 6.1). Laue’s geometric insight was right; his model of the crystal and of the beam needed correction.

Exercise 7.1Laue used density \(4.06\) g/cm³, molar mass \(97.4\) g/mol and Avogadro’s constant \(6.17\times10^{23}\) /mol, assuming one ZnS unit per cube. Reproduce his lattice constant. Then redo the calculation for four ZnS units per cubic cell, with the modern value \(N_A=6.022\times10^{23}\).

Show solution
Volume per ZnS unit \(=97.4/(4.06\times6.17\times10^{23})=3.89\times10^{-23}\) cm³, so \(a=3.39\times10^{-8}\) cm, Laue’s 3.38 Å. With modern \(N_A\) the volume per unit is \(3.98\times10^{-23}\) cm³; four units per cell give \(a=(4\times3.98\times10^{-23})^{1/3}=5.42\times10^{-8}\) cm, the accepted 5.41 Å.

Exercise 7.2Planes belonging to a zone share a common direction \(\mathbf u\) (the zone axis). Show that the beams they reflect from a white incident beam all lie on a cone with axis \(\mathbf u\) that contains the incident direction, so their spots lie on a conic through the central spot.

Show solution
Every plane of the zone contains \(\mathbf u\), so its normal \(\hat{\mathbf n}\) satisfies \(\hat{\mathbf n}\cdot\mathbf u=0\). Mirror reflection gives \(\hat{\mathbf s}=\hat{\mathbf s}_0-2(\hat{\mathbf s}_0\cdot\hat{\mathbf n})\hat{\mathbf n}\), so \(\hat{\mathbf s}\cdot\mathbf u=\hat{\mathbf s}_0\cdot\mathbf u\) whatever the wavelength. All reflected directions make the same angle with \(\mathbf u\) as the incident beam: they lie on a cone that contains \(\hat{\mathbf s}_0\). A plane cuts the cone in a conic passing through the point where \(\hat{\mathbf s}_0\) meets it. These are the “intersecting circles” of spots noted in 1912.

Exercise 7.3For zinc blende with \(a=5.41\) Å and the beam along [001], at what wavelength does the \((2\,0\,\bar2)\) reflection occur, and at what angle from the beam?

Show solution
\(\lambda=2\cdot2\cdot5.41/(4+0+4)=2.71\) Å, rather long for a tungsten or platinum tube. The angle: \(\tan2\theta=x/D=2l\,h/(l^2-h^2-k^2)\) has a zero denominator, so \(2\theta=90^\circ\). Indeed the (101) planes sit at 45° to the beam. So this reflection would miss a plate behind the crystal. Spots near the centre come from planes nearly parallel to the beam, with \(|l|\) small compared with \(\sqrt{h^2+k^2}\), reflecting short wavelengths.
Lesson 08

Structure factors and Friedel’s law

For the general reader. The lattice decides where the spots are. The arrangement of atoms inside each cell decides how bright they are. Laue’s theory of 1912 already contained this split, and it is the basis of every crystal structure determination since, from salt to proteins.

Goals
  • Define the structure factor and derive systematic absences.
  • Prove Friedel’s law and see why a Laue photograph looks centrosymmetric.
  • Understand the phase problem.

The kinematic theory

Suppose each wave is scattered once and the incident beam is not depleted. If the cell contains atoms at positions \(\mathbf r_j\) with scattering factors \(f_j(\mathbf q)\), the amplitude factorizes as \(A(\mathbf q)=F(\mathbf q)\sum_{\mathbf R}e^{-i\mathbf q\cdot\mathbf R}\), with

Structure factor \[ F_{hkl}=\sum_j f_j\,e^{-i\mathbf G_{hkl}\cdot\mathbf r_j}=\sum_j f_j\,e^{-2\pi i(hx_j+ky_j+lz_j)}, \]

where \((x_j,y_j,z_j)\) are fractional coordinates. The integrated intensity of reflection \(hkl\) is proportional to \(|F_{hkl}|^2\), times geometric and polarization factors.

The scattering factor \(f_j\) is the Fourier transform of the atom’s electron density: it equals the number of electrons at \(\mathbf q=0\) and decreases with angle. Thermal vibration further reduces intensities by the Debye–Waller factor \(e^{-2W}\), where \(2W\) is proportional to the mean-square displacement times \(q^2\). This is why Sommerfeld’s worry about thermal motion was misplaced: vibrations weaken the spots, especially at large angles, but do not blur them away.

Friedel’s law

Theorem (Friedel, 1913)

If all \(f_j\) are real, then \(F_{\bar h\bar k\bar l}=F_{hkl}^{*}\), so \(|F_{\bar h\bar k\bar l}|=|F_{hkl}|\). Diffraction intensities are centrosymmetric even when the crystal is not.

The proof is one line: complex conjugation reverses the sign of every exponent. As a consequence, X-ray diffraction alone, in this approximation, cannot distinguish a crystal from its inverted image, and a Laue photograph shows one of the 11 centrosymmetric Laue classes rather than one of the 32 point groups. Near an absorption edge, \(f_j\) acquires an imaginary part and Friedel’s law fails; this “anomalous scattering” is now used to determine absolute configurations and to solve protein phases.

The phase problem

Since the electron density is \(\rho(\mathbf r)=\frac1{V_c}\sum_{hkl}F_{hkl}\,e^{i\mathbf G\cdot\mathbf r}\), knowing every complex \(F_{hkl}\) would give the structure at once. Experiments measure only \(|F_{hkl}|^2\). Recovering the lost phases is the central problem of crystallography, solved by trial structures in the Braggs’ era, then by Patterson maps, direct methods and heavy-atom substitution.

Exercise 8.1Rock salt has Na at \((0,0,0)\) and Cl at \((\tfrac12,0,0)\), each repeated by the fcc translations. Show that \(F_{hkl}=4(f_{\text{Na}}+f_{\text{Cl}})\) for \(h,k,l\) all even and \(4(f_{\text{Na}}-f_{\text{Cl}})\) for all odd. Why does KCl look like a simple cubic crystal of edge \(a/2\)?

Show solution
The fcc translations contribute \(1+e^{-\pi i(h+k)}+e^{-\pi i(k+l)}+e^{-\pi i(h+l)}\), which is 4 for unmixed indices and 0 otherwise. The Cl atom adds the phase \(e^{-\pi ih}=(-1)^h\). K⁺ and Cl⁻ both have 18 electrons, so \(f_{\text{K}}\approx f_{\text{Cl}}\) and the all-odd reflections almost vanish. The surviving all-even reflections are those of a simple cubic lattice of edge \(a/2\) with one atom per site. This puzzle appeared in the Braggs’ first analyses of the alkali halides.

Exercise 8.2Zinc blende has Zn on an fcc lattice and S displaced by \((\tfrac14,\tfrac14,\tfrac14)\). Find \(|F_{hkl}|\) for unmixed indices according to \(h+k+l\bmod 4\).

Show solution
\(F=4\big(f_{\text{Zn}}+f_{\text{S}}\,e^{-\pi i(h+k+l)/2}\big)\). For \(h+k+l\equiv0\): \(4(f_{\text{Zn}}+f_{\text S})\). For \(\equiv2\): \(4|f_{\text{Zn}}-f_{\text S}|\). For odd: \(4\sqrt{f_{\text{Zn}}^2+f_{\text S}^2}\). The hero figure uses these three weights. Because \(F_{\bar h\bar k\bar l}=F^*_{hkl}\), the photograph shows the centrosymmetric symmetry \(m\bar3m\), although zinc blende itself, of point group \(\bar43m\), lacks a centre of inversion.

Exercise 8.3For a body-centred cubic crystal with identical atoms at \((0,0,0)\) and \((\tfrac12,\tfrac12,\tfrac12)\), which reflections are absent?

Show solution
\(F=f\big(1+e^{-\pi i(h+k+l)}\big)\), which vanishes when \(h+k+l\) is odd. So (100), (111) and (210) are absent, while (110), (200) and (211) are present.
Lesson 09

Dynamical theory and anomalous transmission

For the general reader. In a large perfect crystal a reflected X-ray beam is strong enough to be reflected again, back into the forward direction. The two beams trade energy back and forth as they travel. Laue reformulated this “dynamical” theory in 1931, and after the war he explained a striking effect: under exact reflection conditions, a thick crystal can become much more transparent to X-rays.

Goals
  • Set up the two-beam equations of dynamical diffraction.
  • Derive the Pendellösung period.
  • Explain the Borrmann effect by standing waves.

From Darwin and Ewald to Laue

Charles Galton Darwin gave the first dynamical theory in 1914, treating the crystal as a stack of reflecting planes. Ewald, in his work of 1916–17, solved the problem with a lattice of dipoles. In 1931 Laue recast the theory with a continuous, periodic dielectric susceptibility, \(\chi(\mathbf r)=\sum_{\mathbf G}\chi_{\mathbf G}e^{i\mathbf G\cdot\mathbf r}\), which is the formulation used today. For X-rays \(\chi_{\mathbf G}=-\dfrac{r_e\lambda^2}{\pi V_c}F_{\mathbf G}\), of order \(10^{-5}\) to \(10^{-6}\), where \(r_e\) is the classical electron radius.

Two-beam theory

Inside the crystal, write the field as a Bloch wave \(\mathbf D=\sum_{\mathbf G}\mathbf D_{\mathbf G}e^{i(\mathbf K_0+\mathbf G)\cdot\mathbf r}\). When only \(\mathbf G=0\) and one reflection \(\mathbf G=\mathbf h\) are near the Ewald sphere, Maxwell’s equations reduce, for one polarization, to

\[ \begin{aligned} \big(K_0^2-k^2(1+\chi_0)\big)D_0&=k^2\chi_{\bar h}D_h,\\ \big(K_h^2-k^2(1+\chi_0)\big)D_h&=k^2\chi_{h}D_0, \end{aligned} \]

with \(\mathbf K_h=\mathbf K_0+\mathbf h\). A solution exists only if the determinant vanishes, which defines the dispersion surface, a hyperboloid with two branches separated near the Bragg condition. Each incident wave excites one point on each branch.

Pendellösung (symmetric Laue case)

At the exact Bragg angle, the two wavefields excited in a crystal slab beat with each other, so energy oscillates between the forward and diffracted beams with depth \(z\) as \(I_h\propto\sin^2(\pi z/\Lambda)\), where for \(\sigma\) polarization

\[ \Lambda=\frac{\pi V_c\cos\theta_B}{r_e\,\lambda\,|F_h|}. \]

The name, “pendulum solution”, is Ewald’s, by analogy with two coupled pendulums exchanging energy.

The Borrmann effect

In 1941 Gerhard Borrmann observed that a thick, nearly perfect crystal, set at a Bragg angle in transmission, lets through far more X-rays than its absorption coefficient allows. Laue gave the explanation in a paper he wrote while interned at Farm Hall and published in Acta Crystallographica. At the Bragg condition the two wavefields are standing waves across the reflecting planes. In one, the nodes sit on the atomic planes, where absorbing electrons are concentrated, so it is barely absorbed. In the other, the antinodes sit on the planes, and it is absorbed strongly. Through a thick crystal only the first survives. At the Fritz Haber Institute in the 1950s Laue continued this work with Borrmann. Anomalous transmission is now a standard tool in X-ray topography and standing-wave methods.

Exercise 9.1Superpose two plane waves of equal amplitude, \(e^{i\mathbf K_0\cdot\mathbf r}\) and \(\pm e^{i(\mathbf K_0+\mathbf h)\cdot\mathbf r}\). Show that the intensity is \(4\cos^2(\mathbf h\cdot\mathbf r/2)\) or \(4\sin^2(\mathbf h\cdot\mathbf r/2)\). Where are the nodes relative to planes \(\mathbf h\cdot\mathbf r=2\pi m\)?

Show solution
\(|1\pm e^{i\mathbf h\cdot\mathbf r}|^2=2\pm2\cos(\mathbf h\cdot\mathbf r)\), which equals \(4\cos^2(\mathbf h\cdot\mathbf r/2)\) for + and \(4\sin^2(\mathbf h\cdot\mathbf r/2)\) for −. The − field vanishes on the lattice planes \(\mathbf h\cdot\mathbf r=2\pi m\) and is weakly absorbed; the + field peaks there and is strongly absorbed. This is the Borrmann mechanism (for atoms sitting on those planes).

Exercise 9.2Estimate \(\Lambda\) for the silicon 220 reflection with Mo Kα radiation (\(\lambda=0.711\) Å). Use \(a=5.431\) Å, eight atoms per cubic cell, \(f_{\text{Si}}\approx8.5\) at this angle, and \(r_e=2.82\times10^{-5}\) Å.

Show solution
\(V_c=160.2\) ų and \(d_{220}=a/\sqrt8=1.920\) Å, so \(\sin\theta_B=0.185\), \(\cos\theta_B=0.983\). For the diamond structure with \(h+k+l\equiv0\pmod4\), \(|F|=8f\approx68\). Then \(\Lambda\approx\pi\cdot160.2\cdot0.983/(2.82\times10^{-5}\cdot0.711\cdot68)\approx3.6\times10^5\) Å \(\approx36\) μm, close to measured values. Crystals much thinner than \(\Lambda\) obey the kinematic theory; thicker perfect crystals do not.
Lesson 10

Superconductors in a magnetic field

For the general reader. A superconductor carries current with no resistance, but a strong enough magnetic field destroys the effect. In the early 1930s experimenters found that the destroying field seemed to depend on the shape of the sample, which made no sense for a property of the material. Laue explained it: the field at the surface of a sample is not the applied field, and it depends on the shape. He spent much of the next fifteen years on the electrodynamics of superconductors.

Goals
  • Compute the surface field on a perfectly diamagnetic sphere and cylinder.
  • State the London equations and derive the penetration depth.
  • See where Laue’s classical approach fits in the history.

Shape and the critical field

In 1932 Laue, in contact with Walther Meissner at the Reichsanstalt, interpreted a puzzling measurement by Wander de Haas. He pointed out that what matters is the field at the surface of the sample, which differs from the applied field by an amount set by the shape. The argument helped prepare the discovery by Meissner and Robert Ochsenfeld in 1933 that a superconductor expels magnetic flux: inside, \(\mathbf B=0\). A sample in an applied field \(\mathbf H_0\) therefore distorts the field around it, concentrating it at some points of the surface. Superconductivity first fails where the local surface field reaches the critical value \(H_c\).

Perfect diamagnet in a uniform field

For a sphere with \(\mathbf B=0\) inside, placed in a uniform field \(H_0\hat{\mathbf z}\), the field outside is the applied field plus a dipole, and on the surface it is tangential with magnitude

\[ H_{\text{surface}}=\tfrac32H_0\sin\vartheta, \]

where \(\vartheta\) is the polar angle. The sphere begins to lose superconductivity at its equator when \(H_0=\tfrac23H_c\). For a long cylinder in a transverse field the surface field is \(2H_0\sin\varphi\), and the threshold is \(H_0=\tfrac12H_c\).

Derivation. Outside, \(\mathbf H=-\nabla\Phi\) with \(\Phi=-H_0\big(r+\frac{R^3}{2r^2}\big)\cos\vartheta\). This satisfies Laplace’s equation and gives \(H_r=-\partial_r\Phi=H_0\big(1-\frac{R^3}{r^3}\big)\cos\vartheta\), which vanishes at \(r=R\), as required by \(B_r\) continuous and zero inside. The tangential component at \(r=R\) is \(H_\vartheta=-\frac1R\partial_\vartheta\Phi=-\frac32H_0\sin\vartheta\). Between \(\tfrac23H_c\) and \(H_c\) the sphere is in an “intermediate state” of normal and superconducting domains.

The London equations

In 1935 Fritz London and Heinz London proposed that the supercurrent density \(\mathbf J_s\) obeys

\[ \frac{\partial}{\partial t}(\Lambda\mathbf J_s)=\mathbf E,\qquad \nabla\times(\Lambda\mathbf J_s)=-\mathbf B,\qquad \Lambda=\frac{m}{n_se^2}. \]

Combined with Ampère’s law \(\nabla\times\mathbf B=\mu_0\mathbf J_s\), the second equation gives \(\nabla^2\mathbf B=\mathbf B/\lambda_L^2\), with

\[ \lambda_L=\sqrt{\frac{m}{\mu_0n_se^2}}, \]

so fields decay exponentially within a penetration depth of the surface. The same year Laue published a joint paper with the London brothers in the Zeitschrift für Physik. From 1937 to 1947 he wrote twelve papers and the book Theorie der Supraleitung (1947), treating superconductors within Maxwell’s phenomenological electrodynamics and thermodynamics. The microscopic explanation came with the theory of John Bardeen, Leon Cooper and Robert Schrieffer in 1957.

Exercise 10.1Estimate \(\lambda_L\) for \(n_s=10^{28}\) m⁻³, using \(m=9.11\times10^{-31}\) kg and \(e=1.60\times10^{-19}\) C.

Show solution
\(\mu_0n_se^2=4\pi\times10^{-7}\cdot10^{28}\cdot2.56\times10^{-38}=3.22\times10^{-16}\). Then \(\lambda_L^2=9.11\times10^{-31}/3.22\times10^{-16}=2.83\times10^{-15}\) m², so \(\lambda_L\approx53\) nm. Real superconductors have penetration depths of tens to hundreds of nanometres.

Exercise 10.2Verify the cylinder result. A long superconducting cylinder of radius \(R\) lies along \(z\) in a field \(H_0\hat{\mathbf x}\). Use \(\Phi=-H_0\big(r+\frac{R^2}{r}\big)\cos\varphi\) and find the surface field.

Show solution
\(H_r=-\partial_r\Phi=H_0\big(1-\frac{R^2}{r^2}\big)\cos\varphi\), which vanishes at \(r=R\). The tangential field is \(H_\varphi=-\frac1r\partial_\varphi\Phi=-H_0\big(1+\frac{R^2}{r^2}\big)\sin\varphi\), so at \(r=R\) its magnitude is \(2H_0|\sin\varphi|\). The maximum \(2H_0\) reaches \(H_c\) when \(H_0=H_c/2\). A long cylinder parallel to the field has no field enhancement and stays fully superconducting up to \(H_c\): the critical field “depends on shape”.

Exercise 10.3Solve \(B''=B/\lambda_L^2\) for a superconductor filling \(x>0\) with \(B=B_0\hat{\mathbf z}\) at \(x=0\), and find the supercurrent.

Show solution
The bounded solution is \(B_z=B_0e^{-x/\lambda_L}\). Ampère’s law gives \(\mu_0J_y=-\partial_xB_z=\frac{B_0}{\lambda_L}e^{-x/\lambda_L}\). The screening current flows in a surface sheet of thickness \(\lambda_L\), with total current per unit length \(B_0/\mu_0\).
Part III · Life and legacy
Lesson 11

Berlin under two regimes, 1919–1945

Laue’s Zurich post lasted two years. In 1914 he took the chair of theoretical physics at the newly founded University of Frankfurt, and he received the Nobel Prize in Physics for 1914, for the discovery of the diffraction of X-rays by crystals. In 1913 his father had been ennobled, and from then on he was Max von Laue. During the war he worked at Würzburg on vacuum tubes for military telephony and radio. In 1917 he joined the new Kaiser Wilhelm Institute for Physics in Berlin, where he took over most of the administration from its director, Albert Einstein, a close friend. In 1919 he and Max Born exchanged chairs, so that Laue could return to Berlin and to Planck. He held the Berlin professorship until 1943, and in 1925 he became theoretical adviser to the Physikalisch-Technische Reichsanstalt. Leo Szilard took his doctorate under him in 1922.

When the Nazis came to power in 1933, Laue resisted openly and often alone. When Einstein resigned from the Prussian Academy, Laue protested in the Academy and found almost no support. In September 1933, as chairman of the German Physical Society, he opened its Würzburg meeting by comparing the persecution of Galileo with the campaign against relativity. He blocked the election of Johannes Stark, champion of “German physics”, to the Prussian Academy. Stark, now president of the Reichsanstalt, removed him as adviser in December 1933.

In 1934 Laue published an obituary of the chemist Fritz Haber, who had died in exile; it was read as a provocation. On 29 January 1935 he, Planck and Otto Hahn held a memorial for Haber in Berlin-Dahlem although the ministry had forbidden university staff to attend. With Hahn he quietly helped persecuted colleagues to emigrate. In 1937 his son Theodor left for the United States. When Germany occupied Denmark in 1940, George de Hevesy dissolved the Nobel medals of Laue and James Franck in aqua regia at Niels Bohr’s institute to keep them from the Nazis; the Nobel Foundation later recast them from the recovered gold.

Laue remained deputy director of the Kaiser Wilhelm Institute for Physics, where Peter Debye was director from 1935 to 1939. Laue did not take part in the uranium project. In 1943 he retired early from his chair and moved with the institute to Hechingen in Württemberg, where he wrote a short history of physics.

DiscussionLaue kept his post and stayed in Germany, yet opposed the regime publicly on scientific matters. Compare his choices with those of Planck, Heisenberg and Einstein. What could a scientist in his position achieve, and at what cost?

Lesson 12

Farm Hall, rebuilding, legacy

French troops reached Hechingen on 23 April 1945. The Allied Alsos mission took Laue into custody, and he was interned with nine other German scientists, among them Werner Heisenberg and Otto Hahn, at Farm Hall, near Huntingdon in England. Their conversations were secretly recorded, including their reactions to the news of Hiroshima. Laue, who had no part in nuclear research, used the time to write his paper on the absorption of X-rays under diffraction conditions (Lesson 09).

He returned to Germany early in 1946 and settled in Göttingen, where he helped re-establish the Kaiser Wilhelm Institute for Physics as a Max Planck Institute. In July 1946 he was the only German invited to an international crystallography conference in England. He helped found the German Physical Society in the British zone, helped reunite the Reichsanstalt as the Physikalisch-Technische Bundesanstalt in Braunschweig, and in 1950 helped re-establish the federation of German physical societies. In 1948 the new International Union of Crystallography made him its honorary president. In 1951, aged 71, he became director of the Max Planck Institute for Physical Chemistry and Electrochemistry in Berlin-Dahlem; at his request it was renamed the Fritz Haber Institute in 1953. He retired in the late 1950s.

Laue loved fast driving. On 8 April 1960, on his way to the institute, his car was struck by a motorcyclist. The motorcyclist was killed and the car overturned. Laue died of his injuries on 24 April 1960, aged 80, and was buried in Göttingen.

Legacy

  • Diffraction: the 1912 experiment founded X-ray crystallography and, through it, solid-state physics, mineralogy, metallurgy and structural biology. The Laue method is still used with white synchrotron and neutron beams, notably for fast time-resolved studies.
  • Relativity: the first textbook, the derivation of Fresnel drag, and Laue’s theorem on the inertia of stressed systems.
  • Dynamical diffraction: his 1931 formulation and his explanation of the Borrmann effect underlie X-ray topography and the study of perfect crystals.
  • Honours: the Nobel Prize in Physics for 1914, the Max Planck Medal (1932), foreign membership of the Royal Society, the Pour le Mérite for Sciences and Arts, and many academies. The Institut Laue–Langevin in Grenoble, the European neutron source, bears his name together with that of Paul Langevin.

Final projectChoose one result from Part II. Write a two-page account for a strong undergraduate that states it, works one example numerically, and explains what problem it solved for the physicists of Laue’s time.

Reference

Connections

People linked to Max von Laue across this collection. A name without a link has no page here yet. See Max von Laue in the network of all 337 lives →

Teachers and advisors

Students

  • Leo Szilard 1922

Influenced by

Influenced

Colleagues and collaborators

  • Arnold Sommerfeld 1909-1912
  • Max Born
  • Peter Debye 1935-1939
  • Fritz London 1935
  • Gerhard Borrmann 1950s
  • Heinz London 1935
  • Paul Knipping 1912
  • Paul Peter Ewald 1912
  • Walter Friedrich 1912
  • Walther Meissner
  • Wilhelm Röntgen 1909-1912

Rivals

  • Johannes Stark

Friends

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. Max von Laue: Biographical, NobelPrize.org.
  2. Max von Laue, Wikipedia (with extensive references).
  3. A. Hermann, Laue, Max von, Neue Deutsche Biographie 13 (1982).
  4. International Union of Crystallography, Max von Laue, including his own account of 1912.
  5. P. P. Ewald, Fifty Years of X-ray Diffraction, IUCr, 1962.
  6. P. P. Ewald, Max von Laue 1879–1960, Biographical Memoirs of Fellows of the Royal Society 6 (1960).
  7. K. Hentschel (ed.), Physics and National Socialism: An Anthology of Primary Sources, Birkhäuser, 1996.

Physics

  1. M. Laue, Die Mitführung des Lichtes durch bewegte Körper nach dem Relativitätsprinzip, Annalen der Physik 23 (1907).
  2. M. Laue, Das Relativitätsprinzip, Vieweg, Braunschweig, 1911.
  3. M. Laue, Zur Dynamik der Relativitätstheorie, Annalen der Physik 35 (1911).
  4. D. Giulini, Laue’s theorem revisited: energy-momentum tensors, symmetries, and the habitat of globally conserved quantities, arXiv:1808.09320.
  5. W. Friedrich, P. Knipping, M. Laue, Interferenz-Erscheinungen bei Röntgenstrahlen, Sitzungsberichte der Bayerischen Akademie der Wissenschaften (1912) 303.
  6. M. von Laue, Concerning the detection of X-ray interferences, Nobel lecture, 1920.
  7. M. von Laue, Die dynamische Theorie der Röntgenstrahlinterferenzen in neuer Form, Ergebnisse der exakten Naturwissenschaften 10 (1931).
  8. M. von Laue, Röntgenstrahl-Interferenzen, Akademische Verlagsgesellschaft, Leipzig, 1941.
  9. F. London, H. London, The electromagnetic equations of the supraconductor, Proc. R. Soc. A 149 (1935).
  10. M. von Laue, Theorie der Supraleitung, Springer, 1947.
  11. A. Authier, Dynamical Theory of X-ray Diffraction, Oxford University Press, 2001.