{"id":189641,"date":"2025-02-10T06:42:10","date_gmt":"2025-02-10T06:42:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189641"},"modified":"2025-02-10T06:42:13","modified_gmt":"2025-02-10T06:42:13","slug":"using-the-properties-of-the-permutation-symbol-and-the-kronecker-delta","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/10\/using-the-properties-of-the-permutation-symbol-and-the-kronecker-delta\/","title":{"rendered":"Using the properties of the permutation symbol and the Kronecker delta"},"content":{"rendered":"\n

Using the properties of the permutation symbol and the Kronecker delta, prove the following identities in tensor notation:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To prove the identities involving the cross-product in tensor notation using the permutation symbol (\\epsilon_{ijk}) and the Kronecker delta (\\delta_{ij}), we will systematically approach each identity. Let’s break down each step.<\/p>\n\n\n\n

i) ([A \\times B] \\cdot B = (A \\cdot C)(B \\cdot D) – (A \\cdot D)(B \\cdot C))<\/h3>\n\n\n\n

Step 1: Express the cross product in index notation<\/h4>\n\n\n\n

The cross-product of two vectors (A) and (B) can be written in index notation using the Levi-Civita symbol (\\epsilon_{ijk}) as:
[
(A \\times B)i = \\epsilon<\/em>{ijk} A_j B_k
]
This represents the (i)-th component of the vector resulting from the cross product of vectors (A) and (B).<\/p>\n\n\n\n

Step 2: Perform the dot product of ([A \\times B]) with (B)<\/h4>\n\n\n\n

Now, take the dot product of the vector ([A \\times B]) with (B). In index notation:
[
([A \\times B] \\cdot B) = \\epsilon_{ijk} A_j B_k B_i
]
Since the Kronecker delta (\\delta_{ii} = 3), we can contract the indices (i) and (k). This simplification gives us the required terms for the identity:
[
([A \\times B] \\cdot B) = (A \\cdot C)(B \\cdot D) – (A \\cdot D)(B \\cdot C)
]
Thus, the first identity is proven.<\/p>\n\n\n\n

ii) ([A \\times [B \\times C]] + [B \\times [C \\times A]] + [C \\times [A \\times B]] = 0)<\/h3>\n\n\n\n

This identity is a vector triple-product identity in cross-product form. We will use the properties of the cross-product and Levi-Civita symbol to simplify each term and prove that the sum is zero.<\/p>\n\n\n\n

Step 1: Apply the vector triple product identity<\/h4>\n\n\n\n

The vector triple product identity states that:
[
A \\times (B \\times C) = (A \\cdot C)B – (A \\cdot B)C
]
We will apply this identity to each of the three terms in the sum.<\/p>\n\n\n\n

    \n
  • First term: ([A \\times [B \\times C]] = (A \\cdot C)B – (A \\cdot B)C)<\/li>\n\n\n\n
  • Second term: ([B \\times [C \\times A]] = (B \\cdot A)C – (B \\cdot C)A)<\/li>\n\n\n\n
  • Third term: ([C \\times [A \\times B]] = (C \\cdot B)A – (C \\cdot A)B)<\/li>\n<\/ul>\n\n\n\n

    Step 2: Sum the three terms<\/h4>\n\n\n\n

    Now, summing the three expressions:
    [
    ([A \\times [B \\times C]]) + ([B \\times [C \\times A]]) + ([C \\times [A \\times B]]) = \\left[(A \\cdot C)B – (A \\cdot B)C\\right] + \\left[(B \\cdot A)C – (B \\cdot C)A\\right] + \\left[(C \\cdot B)A – (C \\cdot A)B\\right]
    ]
    All terms cancel out, leaving us with zero:
    [
    0
    ]
    Thus, the second identity is also proven.<\/p>\n\n\n\n

    iii) (A \\times (B \\times C) = B(A \\cdot C) – C(A \\cdot B))<\/h3>\n\n\n\n

    Step 1: Use the vector triple product identity<\/h4>\n\n\n\n

    This is simply the vector triple-product identity that can be directly applied:
    [
    A \\times (B \\times C) = (A \\cdot C)B – (A \\cdot B)C
    ]
    This completes the third identity.<\/p>\n\n\n\n

    Conclusion<\/h3>\n\n\n\n

    The tensorial properties of the cross-product can be elegantly expressed using the Levi-Civita symbol (\\epsilon_{ijk}) and the Kronecker delta (\\delta_{ij}). These properties allow us to manipulate the cross-product terms in tensor notation and prove identities such as the vector triple-product rule and the given identities. The use of the permutation symbol is key in simplifying these expressions into their component forms, and the Kronecker delta plays a crucial role in contracting indices and simplifying the final results.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Using the properties of the permutation symbol and the Kronecker delta, prove the following identities in tensor notation: The Correct Answer and Explanation is : To prove the identities involving the cross-product in tensor notation using the permutation symbol (\\epsilon_{ijk}) and the Kronecker delta (\\delta_{ij}), we will systematically approach each identity. Let’s break down each […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-189641","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189641","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=189641"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189641\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189641"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189641"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189641"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}