I recently asked a related question here, but I think I failed to convey my main concerns and questions well, so I have decided to rewrite it in another way. For background, please read my aforementioned question, because I think it works well with this one to clarify my concerns and questions.

My textbook, Fundamentals of Photonics, Third Edition, by Saleh and Teich, gives the following:

enter image description here

This seems to be mathematically incorrect to me?

Firstly, the author stated that $\phi = \psi - \theta_1 \approx \dfrac{y}{-R} - \theta_1$, and then substitutes this into $\theta_2 = 2\phi + \theta_1$ to get $\theta_2 = 2\phi + \theta_1 = 2\left[ \dfrac{y}{-R} - \theta_1 \right] + \theta_1$. But shouldn't this be $\theta_2 = 2\phi + \theta_1 \approx 2\left[ \dfrac{y}{-R} - \theta_1 \right] + \theta_1$?

And lastly, the author stated that $y \approx y_1 + \theta_1 z_1$, and then substitutes this into $\dfrac{2y}{-R} - \theta_1$ to get $\dfrac{2y}{-R} - \theta_1 = \dfrac{2(y_1 + \theta_1 z_1)}{-R} - \theta_1$. But shouldn't this be $\dfrac{2y}{-R} - \theta_1 \approx \dfrac{2(y_1 + \theta_1 z_1)}{-R} - \theta_1$?

Taking all of this into account, the result would be

$$\theta_2 = 2\phi + \theta_1 \approx 2\left[ \dfrac{y}{-R} - \theta_1 \right] + \theta_1 = \dfrac{2y}{-R} - \theta_1 \approx \dfrac{2(y_1 + \theta_1 z_1)}{-R} - \theta_1,$$

which, if my understanding is correct, is a very different mathematical result, in terms of the mathematical conclusions we can draw from this, than what the author has, due to the differences between $=$ and $\approx$ and how we treat them in mathematical calculations.

I'm not familiar with how physicists go about their calculations, but, if my understanding of the mathematics is correct, $\approx$ is not necessarily transitive, so if we have that $A \approx B$ and $B \approx C$, it is not necessarily true that we therefore have $A \approx C$?

The author illustrates what I mean here in their next conclusion, which is a consequence of the, what I believe to be, erroneous mathematics of the last result:

enter image description here

If my understanding of the mathematics is correct, then due to the differences between $=$ and $\approx$, we cannot simply treat $\approx$ as $=$ and draw conclusions in a "chain" of equations and approximations such as $\theta_2 = 2\phi + \theta_1 \approx 2\left[ \dfrac{y}{-R} - \theta_1 \right] + \theta_1 = \dfrac{2y}{-R} - \theta_1 \approx \dfrac{2(y_1 + \theta_1 z_1)}{-R} - \theta_1$. Here, we have a case of $A = \theta_2 \approx B = 2\left[ \dfrac{y}{-R} - \theta_1 \right] + \theta_1$ and $B \approx C = \dfrac{2(y_1 + \theta_1 z_1)}{-R} - \theta_1$, and the author assumes that $\approx$ is transitive so that, logically, we have $(A \approx B) \land (B \approx C) \Rightarrow (A \approx C)$. I do not think this is correct?

Is my understanding of the mathematics here correct?

I would greatly appreciate it if people could please take the time to clarify this.


$\approx$ is not commonly used in mathematics, and it doesn't have a broadly accepted definition among mathematicians.

However, strictly speaking, if you are going to write $x\approx y$ whenever $x$ and $y$ are so close that you don't care about what their difference exactly is, and it might as well be $0$, then clearly $\approx$ is not an equivalence relation, otherwise any two numbers would be in the same equivalence class, including pairs that are too far apart to be reasonably considered as equivalent.

This being said, it doesn't look like complete nonsense to apply transitivity if you're going to apply it once. If $x$ and $y$ are ridiculously close, and so are $y$ and $z$, then $x$ and $z$ are at most twice as far apart, which can't really be that far. The difference would be of the same order of magnitude.

  • $\begingroup$ Ok, this makes sense. So $\approx$, as it is "colloquially" used (such as in physics), and not within the context of rigorous definitions formulated by mathematicians (such as those involving tying the definition of $\approx$ to some metric), is actually a sort of hand-wavy convention used for ease/simplicity of calculation, rather than some well-defined mathematical construct (as the relation $=$ would be)? $\endgroup$ – The Pointer Nov 17 '19 at 14:56
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    $\begingroup$ @ThePointer That's a very accurate way to say it! $\endgroup$ – Arnaud Mortier Nov 17 '19 at 14:58
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    $\begingroup$ @ThePointer For completeness (you probably know about that), mathematics have the rigorously defined $\sim$ where $a\sim b$ means that the quotient $\frac ab$ has limit $1$ when some variable tends to some limit. In your case, the variable would be $\theta_1$ and it would tend to $0$. You can rewrite everything in terms of $\sim$ to make it rigorous, but the point is it would be cumbersome and physicists "know" that they could if they wanted to (which they don't :) ). $\endgroup$ – Arnaud Mortier Nov 17 '19 at 15:07

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