# Geometrical optics problem and broader questions about the correct use of $\approx$ in physical calculations

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

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:

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?

I think my understanding of the mathematics is correct here, but I wonder if there is something about physics conventions that I am unaware of that makes this actually acceptable in physical calculations?

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

• This is justifiable because real calculations in physics might make 20 distinct approximations. If we didn’t do this, we would never use any equals signs at all, which would be a waste of a perfectly good symbol. – knzhou Nov 17 '19 at 16:22
• For example, even the use of geometrical optics itself is an approximation, as is the assumption that the mirror is perfectly smooth, and perfectly spherical, and that the object is perfectly point like, and so on. – knzhou Nov 17 '19 at 16:23
• Maybe you're worried because this approximation doesn't have a name, unlike the rest. But it does -- it's called the paraxial approximation, and if you work in it continually, you're using the theory of paraxial optics. – knzhou Nov 17 '19 at 17:36
• Yes, that can happen. In practice, we just have to remember how and when approximations break down, and work around them. For example, if you found that an angle was exactly zero in the paraxial approximation, that would definitely be a case where the approximation breaks down, since you would have to go to second order in the angles to see what actually happens. – knzhou Nov 17 '19 at 17:39
• It is all very ad hoc but that's just what you need to describe reality. Maybe some very mathematical physicists refrain from any approximations -- but these people generally can't tell you how a lens works. – knzhou Nov 17 '19 at 17:40

## 1 Answer

The meaning of $$\approx$$ or $$\sim$$ in physics calculations is often vague and left undefined. However, I think it usually does have a concrete meaning in context (just one that is unstated). In your context for instance, $$\approx$$ means 'to first order in $$\theta$$'. This then is a transitive relation.

The main examples of specific meanings I can think of are:

a) $$\sim$$ being used for scalings. For instance, on dimensional grounds the period of a pendulum must be $$T \sim \sqrt{\frac{L}{g}}$$. This is a transitive relation because if $$a$$ scales linearly with $$b$$ and $$b$$ with $$c$$ then $$a$$ does scale linearly with $$c$$. This can sometimes (as here) be replaced by the proportionality symbol $$\propto$$ but can also refer to asymptotic scalings in a limit.

b) Related to asymptotic scalings are low-order relations. Such as $$\sin(\theta) \approx \theta$$ to mean 'to first order in the Taylor expansion'. These are also transitive. In a given context its possible that second order effects are important and $$\approx$$ then might mean ignoring third order effects.

c) 'Is of the order of magnitude of'. This is not transitive, although it might be termed approximately transitive in that if you know $$a$$ is the same order of magnitude as $$b$$ and $$b$$ the same order of magnitude as $$c$$, then at least $$a$$ and $$c$$ can only be one order apart.

Asked by OP in comments: "What about error propagation and accumulation in this context? What I mean is that, when making these types of approximations repeatedly, what originally may have been an acceptable, relatively minor error, could balloon into an unacceptably large error that makes the entire calculation practically useless. This is the first "practical" problem I see with this sort of interchanging of ≈ and =. How do physicists justify this?"

My point above is that this is avoided by fixing a meaning of $$\approx$$. Yes, if it only means 'is similar to' then errors can compound (see meaning (c)) but if it has a well defined meaning to do with scaling (a and b) then it is not just 'approximately transitive' but actually transitive. One must just remember the meaning of $$\approx$$ in the end result and interpret accordingly.

• Thanks for the answer. What does "to first order in $\theta$" mean? – The Pointer Nov 17 '19 at 16:19
• It means that you truncate the Taylor expansion of the exact value and then truncate after the linear term. Then $\approx$ means equality in the zeroth and first order Taylor co-efficients, so it is transitive. – jacob1729 Nov 17 '19 at 16:25
• For a), I think you mean "$a$ does scale linearly with $c$"? – The Pointer Nov 17 '19 at 16:30
• Yes, I have corrected the mistake. – jacob1729 Nov 17 '19 at 16:34
• I think c) should also say "and $b$ the same order of magnitude as $c$" and then "at least $a$ and $c$ can only be one order apart"? – The Pointer Nov 17 '19 at 17:27