npsm 새물리 New Physics : Sae Mulli

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Research Paper

New Phys.: Sae Mulli 2021; 71: 771-774

Published online September 30, 2021 https://doi.org/10.3938/NPSM.71.771

Copyright © New Physics: Sae Mulli.

Error in Handling Significant Figures in Polar Coordinates in University Physics Textbooks

Chang Uk Jung*

Department of Physics and Memory and Catalyst Research Center, Hankuk University of Foreign Studies, Yongin 17035, Korea

Correspondence to:cu-jung@hufs.ac.kr

Received: July 26, 2021; Revised: August 19, 2021; Accepted: August 19, 2021

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

“Significant figures” are basic elements of science. The convenient rule-of-thumb in handling significant figures is to meet the top constraint; that is, “the answer you obtain should not be more precise than the numbers you started with.” I previously demonstrated that applying this rule-of-thumb in calculating the areas of squares and the volumes of cubes could easily and widely violate the top constraint of significant figures. In most textbooks dealing with significant figures in a polar coordinate system with rotational symmetry, I recently found a “more significant” error absent in an ordinary rectangular coordinate system. Herein, I suggest a simple prescription in the notation of angles in polar coordinates.

Keywords: Significant figures, Rule-of-thumb, Error area, Polar coordinate system, Rotational symmetry

The “Wikipedia” page on “significant figures (SFs)” show the “top constraint” rule-of-thumb (ROT) in the handling of SF: “For quantities created from measured quantities by multiplication and division, the calculated result should have as many significant figures as the measured number with the least number of significant figures. For example, 1.234 × 2.0 = 2.468… ≈ 2.5.” The first chapter of most undergraduate study textbooks at universities of natural science contains the ROT in the handling of SF. This ROT is also observed in nearly all chemistry and elementary physics textbooks. [14] Recently, I demonstrated that the application of convenient ROT resulted in a serious violation of the top constraint in calculating the area of squares and the volume of cubes.

I have found another significant error in SF calculations given in most textbooks. When we report that a force F1 along the north is 15.2 N, the reader believes that the real force vector is located at a linear line along the exact y-axis given by 15.15 ≤ F1 < 15.25 N. The “north ” is a mathematical/ideal expression and means zero uncertainty or equivalently infinite precision. More professional researchers may write F1 = (15.2±0.05) N or ΔF1 = 0.1 N. The “ΔF1” is usually called the size of the error bar and gives an estimation on how much we can rely on the measurement value F1. Similarly, when we report that another force F2 along the east is 26.3 N, the reader believes that the real force vector is located at a linear line along the exact x-axis given by 26.25 N ≤ F2 < 26.35 N and ΔF2 = 0.1 N. Imagine that both forces are simultaneously applied to a particle. Then, what is the magnitude and direction of the total force, and what is the error range for the estimated magnitude and direction? Nearly all textbooks use polar coordinates to solve this question. Let us define “error area” such that ΔF1ΔF2 = (0.1 N)2 gives the accuracy of the total force vector in a rectangular (x-y) coordinate system.

The special theory of relativity is based on a hypothesis that suggests that physics laws are the same among all observers who move at a constant velocity vector with each other. Similarly, the accuracy of the total force vector, equivalently, the error area, should nearly be the same and “single-valued” when using either the rectangular coordinate system or polar coordinate system.

Typical exercise problems in early chapters of nearly all elementary physics (or chemistry) university textbooks are presented in the followings. [13] For example, a hockey puck received two forces simultaneously. One vector F1 has a magnitude of 15.2 N along the north. The other vector F2 has a magnitude of 26.3 N along the east. Let us calculate the total force and its direction. The solution provided in the textbooks using conventional ROT estimates the total force and its direction as follows:

Exercise 1)

F1+2=F1+F2=0.0000..., 15.2+26.3, 0. 0000.=26.3, 15.2F1+2=15.22+26.32=30.376 F1+2=30.4θ1+2=arctan15.226.3=30.0256...°=30.0°

Most textbooks state that the angle of the total force is 30.0°, which has three SFs with an error angle of Δθ = 0.1°. Now, let us try to calculate a very similar exercise problem of F1+3 = F1 + F3 with F3 = F2. Here, the final angle is θ = 1.50×102° with an angular uncertainty of Δθ = 1°, as shown in Fig. 1.

Figure 1. (Color online) The paradox of significant figures of an angle in polar coordinates by comparing F1+2 in Exercise 1-1) and F1+3 in Exercise 1-2). F1+2 = F1 + F2 = (0.0000…, 15.2)+(26.3, 0.0000.) = (26.3, 15.2). F1+3 = F1 + F3 = (0.0000…, 15.2) + (-26.3, 0.0000…) = (-26.3, 15.2). Both vectors have the same error area in a rectangular system since F3 = F2. In a polar system, F1+3 produces ten times less angle precision than that in F1+2.

The physical systems in the two exercise problems, i.e., F1 + F2 vs. F1 + F3, have exact mirror symmetry with respect to the y-axis and the same precision in the rectangular coordinate system, as shown in Fig. 1. However, the uncertainty of estimated angle in the polar coordinate system has 10-times different with each other. Thus, the application of the ROT fails. The origin of this paradox is the peculiar rotational symmetry in the polar coordinate system. Rotation of the polar coordinate system around the origin by any angle should not change the number of SFs (or equivalently the error bar and error area).

We may express θ1+3 as follows: R = 90.00000° has an infinite number of SFs and rotation of the vector/angle simply changes the angle into the next quadrant (from the first to second quadrant in the polar coordinate system). Then, the same three SFs have an error angle of Δθ = 0.1° by introducing an integer multiple of R ≡ π/2

θ1+3=arctan(15.226.3)=149.97437...°=R+59.97437...=R+60.0°

Furthermore, Fig. 2 shows the obscurity in handling SFs in the polar coordinate system by calculating the error area. First, we start with Exercise 1. The vector F1 has an error bar of ΔF1 = 0.1 N and F2 has the same error bar. The error area in the N2 unit should be 0.01 in the rectangular coordinate system. Notably, the shape of the error-area analysis in Fig. 2 is quite different from that in my previous report. [5] Here, I am interested in estimating the “endpoint” of the added vector of the two forces. Therefore, the endpoint is likely to be located at the rectangular area with ΔF1ΔF2 = (0.1 N)2.

Figure 2. (Color online) (a) The error area in N2 units is ΔA (for F1+2) = 0.1 × 0.1 = 0.01 in the rectangular coordinate system in F1+2 for Exercise 1-1. (b) The error area in N2 units in the polar coordinate system is ΔA (for F1+2) = dr × rdθ = 0.1 × 30.4 × 0.1° × π180° = 0.005, which is half of that in the rectangular coordinate system.

How about the polar coordinate system? As shown in Fig. 2(b), the ROT is F1+2 = 15.22+26.32 = 30.4 and θ1+2 = 30.0°. The error area of the truncated cone is given as follows: ΔA = dr×rdθ = 0.1×30.4×0.1° = (π/180°) = 0.0053 N2. Therefore, the error area for the-two-dimensional (2D) vector of F1+2 is reduced to half by simply changing the rectangular coordinate system to the polar coordinate system. For comparison, the error area for the 2D vector of F1+3 increases by five times by changing the coordinate transformation. The area of a rectangle spanned by dx and dy is irrespective of x and y and is equal to ΔA = dx × dy. However, the area of a truncated cone given by dr and is not independent of r but is proportional to r, i.e., ΔA = dr×rdθ.

Exercise 1-1)

F1+2=F1+F2=0.0000...,15.2+26.3,0.0000...=26.3,15.2ΔAfor F1+2=0.1×0.1=0.01in the rectangular coordinate systemΔAfor F1+2=dr×rdθ=0.1×30.4×0.1°×π180°=0.0053in the polar coordinate system

Exercise 1-2)

F1+3=F1+F3=0.0000...,15.2+26.3,0.0000...=26.3,15.2ΔAfor F1+3=0.1×0.1=0.01in the rectangular coordinate systemΔAfor F1+3=dr×rdθ=0.1×30.4×1°×π180°=0.053in the polar coordinate system

Radians are more suitable for dealing with error areas in polar coordinates. Therefore, I can start from the radian angle, as shown in Exercise 1-1. By simply changing angle from degrees (30.0°) to radian (0.524), the error-area changes significantly.

Exercise 1-1)

F1+2=F1+F2=0.0000...,15.2+26.3,0.0000...=26.3,15.2θ=arctan15.226.3=30.0256...°=30.0°θ=arctan15.226.3=30.0256...°=30.0°=0.524radΔAfor F1+2=0.1×0.1=0.01in the rectangular coordinate systemΔAfor F1+2=dr×rdθ=0.1×30.4×0.001=0.003

If I exchange the magnitude of forces between F1 and F2, the radian angle (using the ROT) changes from 0.524 to 1.05, and ΔA increases by tenfold with the same error area as in the rectangular coordinate system, as shown in Exercise 2 and Fig. 3.

Figure 3. (Color online) Ten times different error areas in the polar coordinate system between F1+2 = (26.3,15.2) and F4+5 = (15.2,26.3) due to 10 times different precision in the angle between θ1+2 and θ4+5. The error area for both exercises in the rectangular coordinate system is the same, which is clear in Fig. 2(a).

Exercise 2)

F4+5=F4+F5=15.2,0.0000...+0.0000...,26.3θ=arctan15.226.3=30.0256...°=0.524045...=1.05radΔAfor F4+5=0.1×0.1=0.01in the rectangular coordinate systemΔAfor F4+5=dr×rdθ=0.1×30.4×0.1=0.03

Therefore, handling SF during the conversion from the rectangular coordinate system to the polar coordinate system is very misleading and can easily violate the top constraint on SF if the ROT is used. When we change the position of some vector from a rectangular coordinate system to a polar coordinate system, we must carefully use the ROT on SF. One simple prescription for the examples given in this report is that the angle should be expressed in the following form: θ = /2 ± ϕ, where n is an integer and 0 ± ϕ < 1. Here “/2” plays the same role as “10−3” in the 0.00450 = 4.50 × 10−3.

The convenient ROT in handling SF uses the top constraint that “the answer you obtain should not be more precise than the numbers you measured with.” Each undergraduate student in the department of Physical Education after graduation will educate much more students in high school but the undergraduate students were found to have difficulty in handling SF even in conventional way. [6] I want to alert students that transforming a vector from the rectangular to the polar coordinate system results in unexpected errors. The simple ROT results in very unreasonable precision when a position vector is transformed from the rectangular to the polar coordinate system. I suggest a simple method for the notation of angles in polar coordinates.

The author appreciates H. J. Choi, G. C. Yi, J. Y. Park, and M. Ryu for reading and commenting on the manuscript. The author appreciates Dr. Venkata Raveendra Nallagatla for his assistance in preparing figures and discussions. This work was supported by Hankuk University of Foreign Studies Research Fund of 2021. This work was partly supported through the National Research Foundation of Korea (NRF) grant funded by the Korean government (MSIT; No. 2020R1A2C2006187).

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