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Post by mondo on Sept 24, 2022 19:48:03 GMT
On page 125, right above equation (4) author says that the length from $q$ to $z$ is $p$ -> $(q-z) = p$ which is ok, similarly $(\tilde{z} - q) = \frac{R^2}{p}$. So their product is $(\tilde{z} - q)(z-q) = R^2$, however author wrote is as $(\tilde{z} - q)\bar{(z-q)}$ which is $\frac{R^2}{p^2}$. Is it a typo?
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Post by Admin on Sept 24, 2022 21:34:50 GMT
On page 125, right above equation (4) author says that the length from $q$ to $z$ is $p$ -> $(q-z) = p$ which is ok, similarly $(\tilde{z} - q) = \frac{R^2}{p}$. So their product is $(\tilde{z} - q)(z-q) = R^2$, however author wrote is as $(\tilde{z} - q)\bar{(z-q)}$ which is $\frac{R^2}{p^2}$. Is it a typo? Mondo No it's not a typo. Vasco |
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Post by mondo on Sept 25, 2022 7:25:31 GMT
On page 125, right above equation (4) author says that the length from $q$ to $z$ is $p$ -> $(q-z) = p$ which is ok, similarly $(\tilde{z} - q) = \frac{R^2}{p}$. So their product is $(\tilde{z} - q)(z-q) = R^2$, however author wrote is as $(\tilde{z} - q)\bar{(z-q)}$ which is $\frac{R^2}{p^2}$. Is it a typo? Mondo No it's not a typo. Vasco $\bar{(z-q)}$ is the inverse of $(z-q)$ right? If so it is $\frac{1}{\rho}$ and when multiplied by the other length it will give $\frac{R^2}{\rho^2}$ instead of $R^2$ as in book. |
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Post by Admin on Sept 25, 2022 7:33:23 GMT
Mondo
$\overline{(z-q)}$ is not the inverse of $(z-q)$.
You can see that $\overline{(z-q)}(z-q)=\frac{R^2}{\rho^2}$ is wrong: how can two lengths multiplied together give a quantity which is dimensionless?
The dimensions of the RHS must be length squared.
Vasco
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Post by mondo on Sept 25, 2022 15:30:12 GMT
Mondo $\overline{(z-q)}$ is not the inverse of $(z-q)$. You can see that $\overline{(z-q)}(z-q)=\frac{R^2}{\rho^2}$ is wrong: how can two lengths multiplied together give a quantity which is dimensionless? The dimensions of the RHS must be length squared. Yes I see that. I never said I think $\frac{R^2}{\rho^2}$ is the correct answer, this is just what I get when I treat $\overline{(z-q)}$ as an inverse of $(z-q)$. So if this is not an inverse, what is this top bar then? That would answer my question fully. |
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Post by Admin on Sept 25, 2022 16:20:13 GMT
Mondo
well what do you think the top bar means?
Vasco
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Post by Admin on Sept 25, 2022 16:22:11 GMT
Mondo
It's standard notation.
Vasco
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Post by mondo on Sept 25, 2022 16:26:07 GMT
If that means a line segment why wasn't it used for $(\tilde{z} - q)$?
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Post by Admin on Sept 25, 2022 16:39:37 GMT
Mondo
The bar notation is explained on page 6 of the book in the table.
Vasco
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Post by Admin on Sept 25, 2022 16:46:37 GMT
Mondo
The bar does not mean line segment.
Vasco
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Post by mondo on Sept 25, 2022 16:53:37 GMT
Mondo The bar notation is explained on page 6 of the book in the table. Vasco Hmm if that is complex conjugation then I don't see its use in this context. We just multiply two lengths that lie on the same line. Both of them can be obtained by subtraction of endpoints $\tilde{z}, z, q$ so why do we need complex conjugate here? |
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Post by Admin on Sept 25, 2022 17:12:08 GMT
Mondo The bar notation is explained on page 6 of the book in the table. Vasco Hmm if that is complex conjugation then I don't see its use in this context. We just multiply two lengths that lie on the same line. Both of them can be obtained by subtraction of endpoints $\tilde{z}, z, q$ so why do we need complex conjugate here? Mondo Work it out. Hint: $R^2$ is a real number. Vasco |
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Post by Admin on Sept 25, 2022 17:47:00 GMT
Mondo
If $(\widetilde{z}-q)=\rho e^{i\phi}$ then $\overline{(z-q)}=(R^2/\rho)e^{-i\phi}$ and so
$(\widetilde{z}-q)\overline{(z-q)}=\rho e^{i\phi}(R^2/\rho)e^{-i\phi}=R^2$
Vasco
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Post by Admin on Sept 25, 2022 17:55:56 GMT
Mondo
Here's a piece of advice: Why don't you do the exercises at the end of each chapter?
It's the only way to make sure you really understand what you have read.
Most people start at chapter 1 and then do chapter 2 and then chapter 3 and so on.
You started at chapter 10 or 11 and then chapter 2 and now chapter 3 without doing any exercises.
This is not a good idea as your questions demonstrate.
Why do you do things like this? Are you in a hurry?
Vasco
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Post by Admin on Sept 25, 2022 18:04:39 GMT
Mondo
Don't try to run before you can walk.
Vasco
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