-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathindex.html
192 lines (149 loc) · 15.4 KB
/
index.html
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8" />
<meta name="viewport" content="width=device-width, initial-scale=1" />
<link rel="preconnect" href="https://fonts.googleapis.com">
<link rel="preconnect" href="https://fonts.gstatic.com" crossorigin>
<link href="https://fonts.googleapis.com/css2?family=Inter:ital,opsz,wght@0,14..32,100..900;1,14..32,100..900&display=swap" rel="stylesheet">
<link href="style.css" rel="stylesheet" />
<link rel="icon" type="image/x-icon" href="favicon.ico" />
<title>Ramunas Girdziusas</title>
<script>
MathJax = {
tex: {
inlineMath: [['$', '$'], ['\\(', '\\)']]
},
svg: {
fontCache: 'global'
},
loader: {load: ['[tex]/mathtools']},
tex: {packages: {'[+]': ['mathtools']}}
};
</script>
<script type="text/javascript" id="MathJax-script" async
src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-svg.js">
</script>
</head>
<body>
<div class="content">
<div style="text-align:center; font-size: 2.5rem; margin-top: 3rem; margin-bottom: 1.6rem;">
<div style="font-size: 2rem; margin-top: 2rem; margin-bottom: 2rem;">Notes on <a href="https://aapt.scitation.org/doi/10.1119/1.1976018">Shankland (1970)</a></div>
<a style="font-size: 1.5rem; margin-top: 1.5rem; margin-bottom: 1.5rem;" href="https://aabbtree77.github.io/">Ramūnas Girdziušas</a>
<div style="font-size: 1rem; margin-top: 1rem; margin-bottom: 1rem;">Last Update: September 26, 2024</div>
</div>
<h2>Tensor Freedom</h2>
<p>A tensor is anything that transforms like a tensor.</p>
<p><strong>Exercise 1.</strong> Show that tensors allow the following index positioning freedom:</p>
<p><span class="math display">\[
X{^i}{^j}{_k} = X{^i}{_k}{^j} = X{_k}{^i}{^j}.
\]</span></p><p>Hint: Define a tensor as a weighted sum of the Kronecker products of the basis vectors:</p>
<p><span class="math display">\[
\begin{align}
X{^i}{^j}{_k}\equiv \sum_{ijk}\,c_{ijk}\,e^{i}\otimes e^{j}\otimes e_{k}\,,
\end{align}
\]</span></p><p>where <span class="math inline">\(c_{ijk}\)</span> are some constants. Assume that <span class="math inline">\(e^{i}\)</span> is a unit row-vector, and <span class="math inline">\(e_{k}\)</span> is a unit column-vector. Note the property:</p>
<p><span class="math display">\[
\begin{align}
a^{i}\otimes b^{j} &\neq b^{j}\otimes a^{i}\,,\\
a^{i}\otimes b_{j} &= b_{j}\otimes a^{i}\,.
\end{align}
\]</span></p><p>It holds for any row-vector <span class="math inline">\(a^{i}\)</span> and column-vector <span class="math inline">\(b_{j}\)</span>.</p>
<p>A good book on tensors is <a href="https://kishorekoduvayur.wordpress.com/wp-content/uploads/2017/12/schaums-tensor-calculus-238.pdf">Schaum’s Tensor Calculus</a> by David C. Kay. Sadly, we do not have anything like it written for spinors and Lie groups.</p>
<h2>Shankland’s Tensor Algebras</h2>
<p>A problem that ChatGPT cannot solve: Given the <a href="https://en.wikipedia.org/wiki/Four-vector">four-vector</a> <span class="math inline">\(k_{\mu}\)</span> and <a href="https://en.wikipedia.org/wiki/Metric_tensor">the metric tensor</a> <span class="math inline">\(g_{\mu\nu}\)</span>, write down the most general dimensionless tensor <span class="math inline">\({T_{\mu\nu}}^{\rho \sigma}\)</span> symmetric under the permutations of the covariant indices <span class="math inline">\((\mu, \nu)\)</span>, and also symmetric w.r.t. the permutations of contravariant indices <span class="math inline">\((\rho, \sigma)\)</span>. It should be a sum of linearly independent terms, each with a manifest symmetry, and at most fourth order in <span class="math inline">\(k_{\mu}\)</span>.</p>
<p><a href="https://aapt.scitation.org/doi/10.1119/1.1976018">Shankland (1970)</a> jumps into the answer, which is a linear combination of</p>
<p><span class="math display">\[
\begin{align}
X_{1} & = \frac{1}{4} {g_{(\mu}}^{(\rho} {g_{\nu)}}^{\sigma)}\,,\\
X_{2} & = g_{\mu \nu} g^{\rho \sigma}\,, \\
X_{3} & = \frac{1}{k^2} g_{\mu \nu} k^{\rho} k^{\sigma}\,, \\
X_{4} & = \frac{1}{k^2} k_{\mu} k_{\nu} g^{\rho \sigma}\,, \\
X_{5} & = \frac{1}{k^4} k_{\mu} k_{\nu} k^{\rho} k^{\sigma}\,, \\
X_{6} & = \frac{1}{k^2} k_{(\mu} {g_{\nu)}}^{(\rho} k^{\sigma)}\,.
\end{align}
\]</span></p><p>Here the division by 4 of the first basis element is not important, but it turns <span class="math inline">\(X_{1}\)</span> into an index symmetrization operator if one acts with it on any two-index tensor.</p>
<p><strong>Magically, these tensor basis expressions form an algebra under the product defined as</strong></p>
<p><span class="math display">\[
A\, B \equiv {A_{\mu \nu}}^{\alpha \beta} {B_{\alpha \beta}}^{\rho \sigma}\,,
\]</span></p><p>with the implied summation over repeated indices.</p>
<p><strong>Exercise 2.</strong> Show that</p>
<p><span class="math display">\[
X_{6} X_{6} = 8 X_{5} + 2 X_{6},
\]</span></p><p>which is not <span class="math inline">\(8 X_{4} + 2 X_{6}\)</span> stated by <a href="https://aapt.scitation.org/doi/10.1119/1.1976018">Shankland (1970)</a>.</p>
<p>Hint:</p>
<p>The definitions of the product and index symmetrization <span class="math inline">\((\cdot, \cdot)\)</span> lead to <span class="math inline">\(X_{6} X_{6}\)</span> expressed as</p>
<p><span class="math display">\[
\begin{align}
& \equiv \frac{1}{k^2} k_{(\mu} {g_{\nu)}}^{(\alpha} k^{\beta)} \frac{1}{k^2} k_{(\alpha} {g_{\beta)}}^{(\rho} k^{\sigma)}\\
&= \frac{1}{k^4} \Big( k_{\mu} {g_{\nu}}^{\alpha} k^{\beta} + k_{\mu} {g_{\nu}}^{\beta} k^{\alpha} \\
& \;\;\;\;+ k_{\nu} {g_{\mu}}^{\alpha} k^{\beta} + k_{\nu} {g_{\mu}}^{\beta} k^{\alpha} \Big)\\
& \;\;\;\;\cdot \Big( k_{\alpha} {g_{\beta}}^{\rho} k^{\sigma} + k_{\alpha} {g_{\beta}}^{\sigma} k^{\rho} \\
& \;\;\;\;+ k_{\beta} {g_{\alpha}}^{\rho} k^{\sigma} + k_{\beta} {g_{\alpha}}^{\sigma} k^{\rho} \Big).
\end{align}
\]</span></p><p>This will produce 16 terms, but they will further simplify with the use of</p>
<p><span class="math display">\[
\begin{align}
k_{\alpha} k^{\alpha} & = k^2 \\
{g_{\nu}}^{\alpha}k_{\alpha} &= k_{\nu} \\
{g_{\nu}}^{\beta} {g_{\beta}}^{\sigma} &= {g_{\nu}}^{\sigma}\;,
\end{align}
\]</span></p><p>e.g.</p>
<p><span class="math display">\[
\begin{align}
t_{1} & = k_{\mu} {g_{\nu}}^{\alpha} k^{\beta} \cdot k_{\alpha} {g_{\beta}}^{\rho} k^{\sigma} \\
& = k_{\mu} \big({g_{\nu}}^{\alpha} k_{\alpha}\big) \big({g_{\beta}}^{\rho} k^{\beta}\big) k^{\sigma} \\
&= k_{\mu} k_{\nu} k^{\rho} k^{\sigma}.
\end{align}
\]</span></p><p>A page later ;) there will be 8 such terms and a doubled <span class="math inline">\(X_{6}\)</span> left.</p>
<div class="imgcontainer">
<img src="imgs/lifetime.jpg" title="Volumetric Lighting in Max Payne (2008)" width=100%>
</div>
<h2>The Spectrum of a Tensor Field</h2>
<p>The products <span class="math inline">\(X_{i}X_{j}\)</span> and the traces <span class="math inline">\(\textit{tr} X_{i}\)</span> determine the spectrum of the algebra. Shankland does not define a tensor field, but it is assumed that <span class="math inline">\(X_{i}\)</span> will form an operator applied to build a quadratic form for a field, where <span class="math inline">\(k_{i}\)</span> becomes a four-nabla. The PhD thesis of <a href="https://spiral.imperial.ac.uk/bitstream/10044/1/13413/2/Barnes-KJ-1963-PhD-Thesis.pdf">K.J. Barnes (1963)</a> sheds more light here, but his notation requires getting used to.</p>
<p>The traces are defined as</p>
<p><span class="math display">\[
\begin{align}
\textit{tr}\,{X_{\mu \nu}}^{\rho \sigma} \equiv {X_{\mu \nu}}^{\mu \nu}\,.
\end{align}
\]</span></p><p>They demand the values of the metric tensor <span class="math inline">\(g_{ij}\)</span>, along with the mixed tensor</p>
<p><span class="math display">\[
\begin{align}
{g_{i}}^{j}={\delta_{i}}^{j}=\begin{cases}
0 & i \ne j \\
1 & i = j
\end{cases}\;\;.
\end{align}
\]</span></p><p>There is no need to know these values when getting the product tables <span class="math inline">\(X_{i}X_{j}\)</span>.</p>
<p><a href="https://aapt.scitation.org/doi/10.1119/1.1976018">Shankland (1970)</a> applies <a href="https://en.wikipedia.org/wiki/Faddeev%E2%80%93LeVerrier_algorithm">the Faddeev - LeVerrier algorithm</a>, or rather its advanced variant extended to tackle multiple eigenvalues, see e.g. <a href="https://core.ac.uk/download/pdf/81192811.pdf">Helmberg and Wagner (1993)</a>. Here there are no eigenvectors in a traditional sense, they are weighted sums of the basis <span class="math inline">\(X_{i}\)</span>, not some matrix columns.</p>
<p><a href="https://spiral.imperial.ac.uk/bitstream/10044/1/13413/2/Barnes-KJ-1963-PhD-Thesis.pdf">K.J. Barnes (1963)</a> seeks the spectrum differently, with the matrix projection operators.</p>
<p>Mysteriously, the eigenvalues will have multiplicities which can be deduced independently from the Lorentz group theory (Lorentz with “t”), without any iterations and polynomial equations. The group theory alone, however, will not get us to the eigenvector equations leading to the Lorenz gauge condition for spin 1 (Lorenz without “t”).</p>
<p><strong>Exercise 3.</strong> Verify Shankland’s spectral results, esp. the case with one vector and one spinor index: “… we find, together with their antiparticles, the following groups of particles: a quadruplet, and two doublets.”</p>
<p>Hint:</p>
<p>According to group theory, combining indices means taking “tensor products <span class="math inline">\((m,n)\otimes (k,l)\)</span> of the Lorentz irreps”, clf. Weinberg’s QFT, Vol. 1, pages 229-233. What is relevant here is that each such an irrep constitutes a set of subspaces with multiplicities <span class="math inline">\(2s+1\)</span> for <span class="math inline">\(s=|m-n|, |m-n+1|,\,\ldots, m+n\)</span>.</p>
<ul>
<li><p><span class="math inline">\((0,0)\)</span>: A scalar. Shankland’s singlet: A single subspace with eigenvalue multiplicity <span class="math inline">\(2\cdot 0+1=1\)</span>.</p></li>
<li><p><span class="math inline">\((\frac{1}{2},\frac{1}{2})\)</span>: A single four-vector index. Shankland’s singlet and triplet: two subspaces <span class="math inline">\(0, 1\)</span> with multiplicities 1 and 3.</p></li>
<li><p><span class="math inline">\((\frac{1}{2},0)\oplus (0,\frac{1}{2})\)</span>: A full single spinor index. Shankland’s doublet and its antidoublet: <span class="math inline">\(\frac{1}{2}, \frac{1}{2}\)</span> subspaces with multiplicites 2 and 2.</p></li>
<li><p><span class="math inline">\((1,1)\)</span>: Two symmetric tensor indices. A mismatch with Shankland’s pentuplet, triplet, and two singlets: Subspaces <span class="math inline">\(0, 1, 2\)</span> with multiplicities 1, 3, and 5. Where is the missing singlet? In a symmetric two-index tensor case, to remove a singlet also means to make the tensor traceless, so the group theory still matches Shankland with this assumption.</p></li>
<li><p><span class="math inline">\((1,0)\oplus (0,1)\)</span>: Two asymmetric tensor indices. Shankland’s two particle triplets: Subspaces <span class="math inline">\(1\)</span> and <span class="math inline">\(1\)</span> with multiplicities 3 and 3.</p></li>
<li><p><span class="math inline">\((\frac{1}{2},\frac{1}{2}) \otimes \Big((\frac{1}{2},0)\oplus (0,\frac{1}{2})\Big)\)</span>, i.e. combining a vector and a spinor index?</p></li>
</ul>
<p>The last case, spin <span class="math inline">\(\frac{3}{2})\)</span>, splits into a spinor and <span class="math inline">\((1,\frac{1}{2}) \oplus (\frac{1}{2},1)\)</span>, clf. Weinberg’s QFT, Vol. 1, page 232. The latter brings subspaces <span class="math inline">\(\frac{1}{2}\)</span> and <span class="math inline">\(\frac{3}{2}\)</span> with multiplicities 2 and 4, along with their “antisubspaces”. All of this combined perfectly matches the result of Shankland.</p>
<p>Note that the construction of algebras is skipped, but it is not trivial. For spin <span class="math inline">\(\frac{3}{2}\)</span>, Shankland had to spot that the combination <span class="math inline">\(\gamma_{\mu}p^{\mu}\)</span> acted independently of <span class="math inline">\(p\)</span>, <span class="math inline">\(\gamma\)</span>, and <span class="math inline">\(g\)</span>. This has effectively doubled the basis dimension of the vector-spinor algebra from 5 to 10.</p>
<h2>Other Relevant Algebras</h2>
<p>One can find some other mildly successful uses/hints of tensor algebras in <a href="https://journals.aps.org/pr/abstract/10.1103/PhysRev.106.1345">Phys. Rev. 106, 1345 (1957)</a>; <a href="https://link.springer.com/article/10.1007/BF02752873">Nuovo Cimento, 43, 475 (1966)</a>; <a href="https://link.springer.com/article/10.1007/BF02818340">Nuovo Cimento 47, 145 (1967)</a>; <a href="https://journals.aps.org/pr/abstract/10.1103/PhysRev.153.1652">Phys. Rev. 153, 1652 (1967)</a>; <a href="https://journals.aps.org/pr/abstract/10.1103/PhysRev.161.1631">Phys. Rev. 161, 1631 (1967)</a>; <a href="https://journals.aps.org/prd/abstract/10.1103/PhysRevD.8.2650">Phys. Rev. D 8, 2650 (1973)</a>; <a href="https://inspirehep.net/literature/98459">Nuovo Cimento 28, 409 (1975)</a>; <a href="https://arxiv.org/abs/hep-th/9212008">Phys. Lett. B 301 4 339 (1993)</a>; <a href="https://arxiv.org/abs/hep-ph/0103172">Phys. Rev. C 64, 015203 (2001)</a>; <a href="https://journals.aps.org/prd/abstract/10.1103/PhysRevD.64.125013">Phys. Rev. D 64, 125013 (2001)</a>; <a href="https://www.imath.kiev.ua/~nikitin/PAPER26.pdf">Hadronic J. 26, 351 (2003)</a>; <a href="https://journals.aps.org/prd/abstract/10.1103/PhysRevD.67.085021">Phys. Rev. D 67, 085021 (2003)</a>; <a href="https://journals.aps.org/prd/abstract/10.1103/PhysRevD.67.125011">Phys. Rev. D 67, 125011 (2003)</a>; <a href="https://arxiv.org/abs/hep-th/0505255">Nucl. Phys. B724, 453 (2005)</a>; <a href="https://journals.aps.org/prd/abstract/10.1103/PhysRevD.74.084036">Phys. Rev. D 74, 084036 (2006)</a>; <a href="https://birdtracks.eu/">P. Cvitanović (2008)</a>; <a href="https://royalsocietypublishing.org/doi/10.1098/rspa.2010.0149">V. Monchiet and G. Bonnet (2010)</a>; <a href="https://journals.aps.org/prd/abstract/10.1103/PhysRevD.97.115043">Phys. Rev. D 97, 115043 (2018)</a>; <a href="https://news.stonybrook.edu/facultystaff/qa-with-breakthrough-prize-winner-peter-van-nieuwenhuizen/">SUGRA and CDC</a>…</p>
<p>It is tough to read this literature, and the results may not always justify the complexity.</p>
<h2>Why Shankland?</h2>
<p>To sum up, we are given a field with tensor/spinor indices and their permutation symmetries. A Lorentz-invariant operator is then constructed. It may serve as a quadratic form for the field, which in turn may serve later in building invariant physics. A spin content of the field is discovered as the eigenvalue multiplicities of that operator. One test of this formalism confirms that removing spin 0 from a vector field leads to “apesanteur” <span class="math inline">\(A\)</span> aka vector potential.</p>
<p>Considering the vast literature on group theory, irreducible representations, angular momentum, higher-spin field theories, spin projection operators, tensors, spinors, Weyl, Wigner, and Weinberg… Shankland’s system is the closest thing to the assembly language of nature.</p>
<div class="imgcontainer">
<a style="font-size: 1.5rem;" href="https://youtu.be/Y183gJQ9yCY?t=20">Sign the contract big boy...</a>
<br><br>
<a href="https://voi.id/en/memori/185165">
<img src="imgs/IronMike-min-rs.png" title="Iron Mike" width="60%">
</a>
</div>
</div>
</body>
</html>