Help Understanding Equation 3.6 in Covariant Physics by Moataz H. Emam

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SUMMARY

The discussion focuses on Equation 3.6 from "Covariant Physics" by Moataz H. Emam, specifically regarding the transformation of displacement vectors in the context of classical mechanics. The equation illustrates how the transformation of the position vector and its derivative behaves like a tensor, demonstrating invariance. The derivation involves using the transformation equations for coordinates and basis vectors, leading to the conclusion that the equation is straightforward once the correct notation and transformations are applied. A noted typo in the equation indicates that the index should read "j".

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Physics students, educators, and enthusiasts interested in understanding tensor transformations and the mathematical foundations of classical mechanics as presented in "Covariant Physics" by Moataz H. Emam.

louvig
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I am a physics enthusiast reading Covariant Physics by Moataz H. Emam. In his chapter about Point Particle mechanics there is a transformation equation for a displacement vector. I don't see how he arrived at the final equation 3.6. Is it a chain rule or product rule? Can't seem to figure it out. See attachment. Thanks in advance for any insight.
 
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It's a bit difficult to read. Also, perhaps needs some context re the author's notation.
 
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PeroK said:
It's a bit difficult to read. Also, perhaps needs some context re the author's notation.
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.
 
eq 3.6 has a typo, this index should read ##j## https://web.cortland.edu/moataz.emam/
1687583372935.png


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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Thank you so much. Makes sense.
 
louvig said:
View attachment 328307I am a physics enthusiast reading Covariant Physics by Moataz H. Emam. In his chapter about Point Particle mechanics there is a transformation equation for a displacement vector. I don't see how he arrived at the final equation 3.6. Is it a chain rule or product rule? Can't seem to figure it out. See attachment. Thanks in advance for any insight.

Everything with primed coordinates was replaced with its transformation. So x’=lambda x and so on.
 
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