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louvig
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I Help Understanding Equation 3.6 in Covariant Physics by Moataz H. Emam
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- Thread starter louvig
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- Covariant Transformation
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The discussion centers on understanding Equation 3.6 in "Covariant Physics" by Moataz H. Emam, specifically regarding the transformation of a displacement vector. Participants highlight the use of Einstein index notation to demonstrate the covariance of classical mechanics, noting that the equation involves both the transformation of the position vector and its derivative. A key point is that Equation 3.6 contains a typo, where an index should read "j." The derivation is described as straightforward, involving the transformation of coordinates and the application of the chain or product rule. Overall, the equation illustrates how the transformation behaves like a tensor, maintaining invariance.
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It's a bit difficult to read. Also, perhaps needs some context re the author's notation.
- #3
louvig
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Sorry. I tried a screenshot from Kindle instead. I am able to click on it in my smartphone and make it full screen and is legible. The author is showing the covariance of classical mechanics using Einstein index notation. In this instance he is showing the transformation of the position vector which is straightforward and then the transformation of the derivative of the position vector. His point is to show ot transforms like a tensor and is therefore invariant.PeroK said:
- #4
malawi_glenn
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eq 3.6 has a typo, this index should read ##j## https://web.cortland.edu/moataz.emam/
The derivation is straight-forward:
Use that ##\hat{ \textbf{g}}_{i'} = \lambda^k_{i'} \hat{ \textbf{e}}_k ## and ##x^{i'} = \lambda^{i'}_j x^j##.
We get ## d\hat{ \textbf{g}}_{i'} = \hat{ \textbf{e}}_k d \lambda^k_{i'} ## and ##x^{i'} = x^j d\lambda^{i'}_j + \lambda^{i'}_j dx^j##.
And you will obtain the final step in that equation.
The derivation is straight-forward:
Use that ##\hat{ \textbf{g}}_{i'} = \lambda^k_{i'} \hat{ \textbf{e}}_k ## and ##x^{i'} = \lambda^{i'}_j x^j##.
We get ## d\hat{ \textbf{g}}_{i'} = \hat{ \textbf{e}}_k d \lambda^k_{i'} ## and ##x^{i'} = x^j d\lambda^{i'}_j + \lambda^{i'}_j dx^j##.
And you will obtain the final step in that equation.
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louvig
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Thank you so much. Makes sense.
- #6
Moataz Emam
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louvig said:
Everything with primed coordinates was replaced with its transformation. So x’=lambda x and so on.