ক্রিস্টোফেল সংকেত [ সম্পাদনা ] সমস্যা: প্রমাণ কর যে, ক্রিস্টোফেল সংকেত টেন্সর নয় কিন্তু দুটি ক্রিস্টোফেল সংকেতের বিয়োগফল একটি টেন্সর।
সমাধান: তিন সূচকের ক্রিস্টোফেল সংকেতকে এভাবে লেখা যায়,
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{\displaystyle {\begin{aligned}\Gamma _{jk}^{i}&={\frac {\partial x^{i}}{\partial \xi ^{l}}}{\frac {\partial ^{2}\xi ^{l}}{\partial x^{j}\partial x^{k}}}={\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial }{\partial x^{j}}}\left[{\frac {\partial \xi ^{l}}{\partial x^{k}}}\right]\\&={\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial }{\partial x^{j}}}\left[{\frac {\partial \xi ^{l}}{\partial {\bar {x}}^{n}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{k}}}\right]\\&={\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial \xi ^{l}}{\partial {\bar {x}}^{n}}}{\frac {\partial ^{2}{\bar {x}}^{n}}{\partial x^{j}\partial x^{k}}}+{\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{k}}}{\frac {\partial ^{2}\xi ^{l}}{\partial x^{j}\partial {\bar {x}}^{n}}}={\frac {\partial x^{i}}{\partial {\bar {x}}^{n}}}{\frac {\partial ^{2}{\bar {x}}^{n}}{\partial x^{j}\partial x^{k}}}+{\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{k}}}{\frac {\partial {\bar {x}}^{o}}{\partial x^{j}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial ^{2}\xi ^{l}}{\partial {\bar {x}}^{o}\partial {\bar {x}}^{n}}}\\&={\frac {\partial x^{i}}{\partial {\bar {x}}^{n}}}{\frac {\partial ^{2}{\bar {x}}^{n}}{\partial x^{j}\partial x^{k}}}+{\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{k}}}{\frac {\partial {\bar {x}}^{o}}{\partial x^{j}}}\Gamma _{no}^{m}\end{aligned}}}
দেখা যাচ্ছে, ক্রিস্টোফেল সংকেতকে ভিন্ন একটি স্থানাংক ব্যবস্থায় রূপান্তরিত করার পর দুটি টার্ম আসছে। দ্বিতীয় টার্ম নিয়ে কোন সমস্যা নেই, এই টার্মে রূপান্তরিত ক্রিস্টোফেল সংকেত আছে, তার আগে আছে তিনটি রূপান্তর মেট্রিক্স যার সবগুলোই টেন্সরের মত আচরণ করছে। কেবল এই টার্মটি থাকলে ক্রিস্টোফেল সংকেতের টেন্সর হতে কোন বাঁধা থাকত না। কিন্তু প্রথম টার্মটি থাকার কারণে এটি টেন্সরের মর্যাদা হারিয়েছে।
কিন্তু দুটি ক্রিস্টোফেল সংকেতের জন্য এমন রূপান্তর মেট্রিক্স লেখার পর একটি থেকে অন্যটি বিয়োগ করলে প্রথম টার্মটি বিয়োগ চলে যাবে। থাকবে কেবল বিশুদ্ধ টেন্সরের বৈশিষ্ট্য সম্পন্ন দ্বিতীয় টার্মটি। এজন্যই দুটি ক্রিস্টোফেল সংকেতের বিয়োগফল একটি টেন্সর।
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{\displaystyle \Gamma _{fg}^{e}={\frac {\partial x^{e}}{\partial {\bar {x}}^{n}}}{\frac {\partial ^{2}{\bar {x}}^{n}}{\partial x^{f}\partial x^{g}}}+{\frac {\partial x^{e}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{g}}}{\frac {\partial {\bar {x}}^{c}}{\partial x^{f}}}\Gamma _{bc}^{a}}
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{\displaystyle \Gamma _{jk}^{i}-\Gamma _{fg}^{e}={\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{k}}}{\frac {\partial {\bar {x}}^{o}}{\partial x^{j}}}\Gamma _{no}^{m}-{\frac {\partial x^{e}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{g}}}{\frac {\partial {\bar {x}}^{c}}{\partial x^{f}}}\Gamma _{bc}^{a}}
![{\displaystyle {\begin{aligned}\Gamma _{jk}^{i}&={\frac {\partial x^{i}}{\partial \xi ^{l}}}{\frac {\partial ^{2}\xi ^{l}}{\partial x^{j}\partial x^{k}}}={\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial }{\partial x^{j}}}\left[{\frac {\partial \xi ^{l}}{\partial x^{k}}}\right]\\&={\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial }{\partial x^{j}}}\left[{\frac {\partial \xi ^{l}}{\partial {\bar {x}}^{n}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{k}}}\right]\\&={\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial \xi ^{l}}{\partial {\bar {x}}^{n}}}{\frac {\partial ^{2}{\bar {x}}^{n}}{\partial x^{j}\partial x^{k}}}+{\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{k}}}{\frac {\partial ^{2}\xi ^{l}}{\partial x^{j}\partial {\bar {x}}^{n}}}={\frac {\partial x^{i}}{\partial {\bar {x}}^{n}}}{\frac {\partial ^{2}{\bar {x}}^{n}}{\partial x^{j}\partial x^{k}}}+{\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{k}}}{\frac {\partial {\bar {x}}^{o}}{\partial x^{j}}}{\frac {\partial {\bar {x}}^{m}}{\partial \xi ^{l}}}{\frac {\partial ^{2}\xi ^{l}}{\partial {\bar {x}}^{o}\partial {\bar {x}}^{n}}}\\&={\frac {\partial x^{i}}{\partial {\bar {x}}^{n}}}{\frac {\partial ^{2}{\bar {x}}^{n}}{\partial x^{j}\partial x^{k}}}+{\frac {\partial x^{i}}{\partial {\bar {x}}^{m}}}{\frac {\partial {\bar {x}}^{n}}{\partial x^{k}}}{\frac {\partial {\bar {x}}^{o}}{\partial x^{j}}}\Gamma _{no}^{m}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85a9cad2fcbb2194dc58457bb3a769117b2f7055)
