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Title: Levi-Civita Symbol
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Series: Advent of Mathematical Symbols
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YouTube-Title: Levi-Civita Symbol
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Bright video: https://youtu.be/H8_cia855zg
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Dark video: https://youtu.be/5rx_NK_iz3Q
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Ad-free video: Watch Vimeo video
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Print-PDF: Download printable PDF version
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: aoms02_sub_eng.srt
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Timestamps (n/a)
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Subtitle in English
1 00:00:00,543 –> 00:00:04,044 The mathematical symbol of today is the Levi-Civita symbol.
2 00:00:04,686 –> 00:00:09,755 Which is written as a lower case epsilon, with 3 indices “i”, “j” and “k”.
3 00:00:10,643 –> 00:00:13,700 This symbol represents exactly 3 numbers.
4 00:00:13,772 –> 00:00:17,252 It’s either 1, -1 or 0.
5 00:00:17,986 –> 00:00:25,616 and we distinguish these 3 cases depending in which order the numbers 1, 2 and 3 occur in the indices here.
6 00:00:26,529 –> 00:00:28,207 If its in the correct order,
7 00:00:28,208 –> 00:00:31,370 which means like you would count. 1, 2, 3.
8 00:00:31,586 –> 00:00:33,554 Then we get out 1.
9 00:00:34,486 –> 00:00:36,649 On the other hand, if its backwards.
10 00:00:36,849 –> 00:00:41,003 So like you would count for a countdown. 3, 2, 1.
11 00:00:41,286 –> 00:00:43,201 Then we get out -1.
12 00:00:44,071 –> 00:00:50,042 In fact in this counting order i described, also cyclic permutations are allowed.
13 00:00:50,914 –> 00:00:58,445 For example in the first case the last position could be the 1, if then the first position is 2 and then comes 3.
14 00:00:59,057 –> 00:01:02,457 So this counting here goes in a cyclic way.
15 00:01:03,271 –> 00:01:07,117 Hence the third possibility here, would be 3, 1, 2.
16 00:01:07,857 –> 00:01:10,171 This is also counting forwards.
17 00:01:11,229 –> 00:01:16,678 On the other hand another possibility for counting backwards would be 2, 1, 3.
18 00:01:17,257 –> 00:01:20,124 Also here you should think in a cyclic way.
19 00:01:20,914 –> 00:01:25,320 and finally then the last possibility would be 1, 3, 2.
20 00:01:26,143 –> 00:01:31,369 So what you can remember is, the cyclic order here gives us a sign for the symbol.
21 00:01:32,229 –> 00:01:38,788 and then in all other cases, maybe all the indices here are equal to 1. We get out 0.
22 00:01:39,429 –> 00:01:43,995 Ok, that’s the whole definition of the Levi-Civita symbol, with 3 indices.
23 00:01:44,871 –> 00:01:48,143 There are also generalisations with more indices,
24 00:01:48,144 –> 00:01:53,114 but this one is the most important one, because we can use it in 3 dimensions.
25 00:01:53,743 –> 00:02:01,247 For example it’s helpful when we deal with the cross product of two 3-dimensional vectors, “a” and “b”.
26 00:02:02,200 –> 00:02:07,226 So the result of this crossproduct is again a vector with 3 components.
27 00:02:07,900 –> 00:02:10,929 So we can ask, what is the i-th component of it.
28 00:02:11,643 –> 00:02:16,407 and it turns out, this can be calculated using the Levi-Civita symbol.
29 00:02:17,514 –> 00:02:21,698 It’s the double sum over all possibilities for “j” and “k”
30 00:02:21,914 –> 00:02:25,870 and then we take the Levi-Civita symbol, epsilon_i_j_k.
31 00:02:26,843 –> 00:02:32,788 times the j-th component of “a” times the k-th component of “b”.
32 00:02:33,786 –> 00:02:40,278 In fact this is a formula some people use to define the cross product of two 3-dimensional vectors.
33 00:02:41,443 –> 00:02:44,372 Ok, so this is the Levi-Civita symbol.
34 00:02:44,572 –> 00:02:46,178 Thanks for listening!
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Quiz Content
Q1: What is the value of $\varepsilon_{123}$?
A1: 1
A2: 0
A3: -1
A4: 2
Q2: What is the value of the Levi-Civita-Symbol $\varepsilon_{323}$?
A1: 0
A2: 1
A3: -1
A4: 2
Q3: Consider two vectors $a, b \in \mathbb{R}^3$ and the canonical unit vectors $\mathbf{e}_i \in \mathbb{R}$ where $\mathbf{e}_1 = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}$, $\mathbf{e}_2 = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}$, and $\mathbf{e}_3 = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}$. What is correct for the cross product?
A1: $ a \times b = \sum_{ijk} \varepsilon_{ijk} a_i b_j \mathbf{e}_k $
A2: $ a \times b = \sum_{ijk} \varepsilon_{ijk} a_i b_j \mathbf{e}_j $
A3: $ a \times b = \sum_{ijk} \varepsilon_{ijk} a_k b_i \mathbf{e}_i $
A4: $ a \times b = \sum_{i} \varepsilon_{iii} a_i b_i \mathbf{e}_i $
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Last update: 2024-11
