Symmetry Arguments and the Role They Play in Using Gauss Law

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1 Smmetr Arguments and the Role The la in Using Gauss Law K. M. Westerberg (9/2005) Smmetr plas a ver important role in science in general, and phsics in particular. Arguments based on smmetr can often simplif calculations b allowing us to take shortcuts (e.g., not computing some component of the electric field because we can prove using smmetr that it must be zero). Smmetr also plas a vital role in much of the current theoretical phsics research. Aside from a few general principles (and perhaps a few parameters which need to be fit to eperimental data), the onl assumptions that are made in these modern Quantum Field Theories or String Theories are the smmetries that the universe is assumed to obe. It is important to understand smmetr arguments not onl how the work, but also wh the work (i.e., wh such arguments can be considered to be rigorousl valid, and not cheating ). In this short note, we discuss smmetr arguments used in calculating the electric field due to various charge distributions, and in particular, the role plaed b smmetr in the use of Gauss Law to calculate the electric field due to highl smmetric charge distributions. Underling principles The validit of smmetr arguments involved in the calculation of the electric field due to some charge distribution is based on two principles: 1. For an given charge distribution and given point, there can onl be one possible value for the electric field at due to the given charge distribution. 2. If the charge distribution is moved either b translation, rotation around some fied ais, mirror reflection across some plane, or an combination of these operations 1 then the electric field must move with the charge distribution. The first principle should be self-evident. We can calculate the electric field at using Coulomb s Law and superposition once we are given the charge distribution and the point where we are evaluating the electric field. To help illustrate the second principle, consider the following charge distribution and point illustrated in Figure 1(a) below. 1 The common propert shared b all three of these operations (translation, rotation, and mirror reflection) is that the are all isometric the preserve all distances and angles among the various charges. 1

2 Figure 1(a) (b) q 2 q 1 Suppose we calculate the electric field at due to this charge distribution (presumabl using Coulomb s Law and superposition), and the result is as indicated in Figure 1(a). Now consider what happens when we move all three charges b the same displacement vector (Fig. 1(b)). If we also move the point b the same displacement, it seems reasonable that the electric field E at the translated point due to the translated charge distribution should be equal to the original electric field E at due to the original charge distribution. That is precisel what happens: E = E. 2 The same idea can be applied to rotation and mirror reflection. In Figure 2 below, we appl a 90 counterclockwise rotation around the z-ais (perpendicular to the plane of the page) to our original charge distribution. After rotating the point as well, we conclude that, the electric field at the rotated point due to the rotated charge distribution, should be related to the electric field E at due to the original charge distribution. In this case, is not equal to E, but is the result of rotating E b the same 90 angle (in particular, the magnitudes of E and E are equal). Figure 2(a) (b) q 1 q 2 q 1 q 2 Likewise, in Figure 3 below, we mirror-reflect the charge distribution and the point through the -z plane (changing the sign of all coordinates but not changing an or z coordinates). The resulting electric field E should be the mirror reflection of the original electric field E (once again, the magnitudes will be equal). 2 One wa to see that this is reasonable is to ask what would happen if, instead of moving the charges and the point, ou moved the coordinate sstem in the opposite direction. 2

3 Figure 3(a) (b) q 1 q 1 q 2 q 2 In all three cases, the fact that the electric fields before and after the transformation are related as described above can be proved rigorousl b Coulomb s Law and superposition, with the help of some linear algebra. Smmetric charge distributions In each case above, we were able to relate the electric field at some point due to some charge distribution to the electric field at some other point due to a different charge distribution. However, we were unable to learn anthing about the electric field at due to the original charge distribution, nor were we able to compare the electric fields at two different points due to the same charge distribution. This is because the charge distribution considered in the previous section was not smmetric under translation, rotation, or mirror reflection. It turns out that when we consider smmetric charge distributions, we will be able to gain more useful information b moving the charges around. Consider the smmetric charge distribution shown in Figure 4 below. Figure 4(a) (b) If we mirror-reflect the charge distribution and the point across the -z plane, as we did in Figure 3 in the previous section, we will find that the electric field E at the mirrorreflected point due to the mirror-reflected charge distribution is, itself, a mirror reflection of the electric field E at the original point due to the original charge distribution. 3

4 However, in this case, the original charge distribution is smmetric under the mirror reflection we just performed. How does this help us? When we mirror-reflect the original smmetric charge distribution, the result is eactl the same charge distribution again (see Fig. 4(b)). Therefore, the electric fields E and E at the points and, respectivel, are due to the same charge distribution. Evidentl, when the charge distribution is smmetric, we can use a smmetr argument, such as the one given above, to compare the electric field at two distinct (but smmetricall related) points due to that same charge distribution. If we were to appl the same argument to a number of points in the vicinit of this charge distribution, we would find the entire electric field pattern to be smmetric. If we were to draw an electric field line diagram for these charges, that diagram would have to be smmetric as well. The electric field must obe the same smmetries that are obeed b the charge distribution. This is not the result of a hand-waving argument this principle is based on rigorous mathematical reasoning. For another eample, this one involving rotation, consider the circular (not spherical) charge distribution shown below in Figure 5 (the circle and the point both lie in the - plane). Figure 5(a) (b) If we rotate the charge distribution counterclockwise b 90, we conclude that the electric field E at due to the rotated charge distribution must be the result of a 90 counterclockwise rotation of the electric E at due to the original charge distribution. Now this is the ke if the circular charge distribution is uniform, or otherwise smmetric under 90 rotation, then the rotated charge distribution will coincide with the original charge distribution, and the electric fields E and E will both be due to the same (original) charge distribution. In fact, an arbitrar-angle rotation smmetr (which is obeed b the circular rod if its linear charge densit is uniform) allows us to compare the electric field at to the electric field at all points in the - plane which are the same distance from the center of the circle as the point itself. However, if the linear charge densit of the circle is not uniform and is not otherwise 90 -rotation smmetric, then the comparison of electric fields at the points and will not work, as those fields would be due to different charge distributions. For an eample involving translational smmetr, consider a uniforml charged, straight, infinite rod. The translational smmetr of the rod will allow a comparison of the electric 4

5 field at points and which are related b a displacement pointing in the same direction as the rod. The details are left to the reader (ou will see this argument presented in class when Gauss Law is used to calculate the electric field due to an infinite rod or clinder). Smmetricall located points As we have seen from last section, a smmetr argument can be used to compare the electric field at two distinct points due to a smmetric charge distribution. The same tpe of argument can also be used to gain information about the electric field at a single point, provided that the point is smmetricall located. For our first eample, consider the three-charge smmetrical charge distribution. Suppose we wish to evaluate the electric at a point located on the -ais. Our gut instinct tells us that the electric field must point in a smmetrical direction (either the + or direction), and must therefore resemble Figure 6, shown below. Figure 6 However, gut instincts can sometimes be wrong, so can we actuall prove that the electric field must point in one of those two directions? It turns out that we can, using the same tpe of smmetr argument that we used in the last section. Suppose for a moment that the electric field is not smmetrical, in that it has a non-zero -component, as shown in Figure 7(a) below. Figure 7(a) (b) If we mirror-reflect the charge distribution and the point across the -z plane, as we have alread done several times, we find that the electric field E at the reflected point 5

6 due to the reflected charge distribution must be the result of mirror-reflecting the original electric field E at the original point due to the original charge distribution. However, both the charge distribution and the point where we are evaluating the field are smmetrical, and therefore coincides with and the reflected charge distribution coincides with the original charge distribution. It follows that the two electric fields, E and E, must both be the electric field at the same point due to the same charge distribution. According to the first principle, there can onl be one possible value for the electric field at this point, so E and E must be equal. However, this is not the case if E is not smmetrical (see Fig. 7). It follows that the electric field cannot be asmmetrical, as shown in Fig. 7(a), and must therefore be smmetrical, as shown in Fig. 6. There is another argument that the electric field at must be smmetrical, with which the reader ma be more familiar. The electric field due to each individual charge is calculated using Coulomb s Law and then added together according to the Superposition rinciple to get the total electric field. For ever charge on the left side of the smmetr plane, there is a corresponding charge on the right side (e.g., the two +4q charges in the present case). Because of the smmetric placement of those charges, and of the point, the components of the electric field contributions from the two charges must cancel. When all such electric field contributions are added together, all of the electric field components cancel out, and so E must be zero. There is nothing wrong with that argument. However, it is quite different from the smmetr argument we presented using Figure 7. In that figure, the vector E is intended to represent the total electric field, presumabl having been calculated b summing up contributions from all of the charges. We hpothesize that, despite the smmetr of the charge distribution and the point, this asmmetric vector E indeed represents the total electric field at. We then use mirror reflection to conclude that the mirror-reflected vector must also represent the total electric field at this same point due to the same charge distribution. We are in no wa asserting that E and E are to be combined in an wa (e.g., b adding them) in order to produce the actual electric field at. Instead, we reason that since E is not equal to E, our original assumption that E is the (total) electric field at must have been false. The actual electric field at must coincide with its own mirror reflection, an eample of which is the electric field shown in Figure 6. Another eample of this same tpe of smmetr argument is the argument that the electric field due to a uniform circle of charge ling in the - plane must be radial at all points in the - plane, as shown in Figure 8 below. 6

7 Figure 8 To make the case, we take advantage of the 180 flip rotation smmetr of the circular charge around the dashed-line ais passing through the point (the dashed line in the above figure lies in the - plane it is not a perspective drawing of the z ais). The 180 rotation essentiall flips the circle over like a pancake, moving the upper-left portion of the circle to the lower-right, and vice versa. The point lies on the ais of rotation (the ais was chosen so that this was true), and thus is not affected b the rotation. Suppose that the electric field were not radial, but instead were directed as shown in Figure 9, below. Figure 9(a) (b) Appling the flip rotation to the circular charge distribution and the point leaves both unchanged, but causes the electric field E to flip over to E. Since there can onl be one value of the electric field at due to the uniform circular charge distribution, E would have to equal E. However, that cannot happen if E is not radial, as shown in Figure 9. Therefore, the total electric field must in fact be radial, as shown in Figure 8 (note that the electric field in Figure 8 would be unaffected b the flip rotation). Note that if the charge distribution did not obe this flip rotation smmetr (e.g., because the linear charge densit were not uniform), the smmetr argument would fall apart the electric fields E and E, both at the point, would be due to different charge distributions, and no contradiction results even if E were not radial. There is no translation-smmetric eample appropriate for this section, since an translation will move a point to a different point. 7

8 Gauss Law applications Gauss Law asserts that the net electric flu (in the outward direction) through an closed surface must be proportional to the charge enclosed b that surface: d A = Q encl ɛ 0. For eample, in Figure 10 below, the electric flu through the indicated surface must be equal to ( + + )/ɛ 0. Figure 10 E It should be pointed out that the validit of Gauss Law has nothing to do with smmetr. Gauss Law is still true, even when the charge distribution and the closed surface are not smmetric in an wa. Where smmetr arguments come into pla is when we attempt to use Gauss Law to evaluate the electric field at an individual point. Suppose we wanted to evaluate the electric field at in Figure 10 using Gauss Law. The problem is that the electric flu not onl depends on the electric field at, it also depends on the electric field at, and at ever other point on the Gaussian surface. Because of the lack of smmetr, there is no obvious relationship between the electric field at and at, and so no useful information about the electric field at individual points can be gained b using Gauss Law to calculate E d A. Now consider a sphericall smmetric charge distribution, illustrated in Figure 11 below. Figure 11 Can Gauss Law be used to calculate the electric field at due to this charge distribution? The answer turns out to be es, provided that we choose our Gaussian surface wisel. Is the 8

9 bo depicted in Figure 11 a good choice? No it isn t. Once again, the electric flu through the bo depends on the electric field at, at, and at ever other point on the surface of the bo. In order to have an chance of being able to evaluate the electric field at from the integral E d A, we need to be able to relate the electric fields at and, presumabl b taking advantage of the rotation smmetr of the sphericall smmetric charge distribution. However, the points and are at different radii (i.e., different distances from the center of the charge distribution), and so it is impossible to rotate into. There is no obvious relationship between the electric fields at and (assuming we haven t alread solved the problem b some other means), and so once again, we conclude that no useful information about the electric field at individual points can be gained b calculating E d A for the bo. Again, the validit of Gauss Law is not being questioned we ma still conclude bo d A = Q encl ɛ 0. Knowing this, however, does not help us calculate the electric field at individual points, such as. (B the wa, can a smmetr argument be used to evaluate E d A over just a single face of the bo? Think about it.) To calculate the electric field at due to the sphericall smmetric charge distribution, we need to be able to take advantage of the rotation smmetr of the charge distribution. In general, this means that the Gaussian surface must have the same tpe of smmetr as the charge distribution. The bo proved to be a poor choice for the Gaussian surface because it had the wrong smmetr. Let s now consider the spherical Gaussian surface depicted in Figure 12 below. Figure 12 Can we calculate the electric field at individual points (such as ) b appling Gauss Law to the sphere? Yes, this will work. The reason it works is because the charge distribution and the Gaussian surface are both highl smmetrical the both possess full (arbitrar-angle) rotation smmetr about an ais passing through the center of the charge distribution. We ma first appl the 180 flip rotation smmetr to conclude that the electric field at all points must be radiall directed. The argument is eactl the same as for the uniform circle (see Figs. 8 and 9). The reason wh this first step is necessar is that the electric flu onl depends on one component of the electric field at an given point namel the 9

10 component perpendicular to the Gaussian surface (the radial component, in the case where the Gaussian surface is a sphere). We need to find some other means to evaluate the electric field components which are parallel to the surface. In the present case, those non-radial components must be zero, b smmetr. At this point, we can simplif the electric flu integral as follows: Flu = d A = E r da. However, the flu integral still depends on the electric field not onl at, but also at, and at all other points on the Gaussian sphere. How do we get past this problem? We need to be able to relate the electric field at and at, and at all other points on the sphere. Since all points on the Gaussian sphere are at the same radius, we can rotate to an one of the other points on the surface. For eample, we can map to using a 90 counterclockwise rotation around the z-ais. We ma use eactl the same smmetr argument given in Figure 5 for the circle to conclude that E is the result of rotating E through the same 90 angle that we used to rotate into. In particular, E and E must have the same radial component. It follows that the radial component in the flu integral above is constant and can be brought out of the integral. The flu integral can be further simplified: Flu = E r da = E r da = E r (4πr 2 ). The value of E r that we pulled out of the integral is the radial component of the electric field at an individual point on the Gaussian surface, including the point. Setting the flu equal to Q encl /ɛ 0 (Gauss Law) and solving for E r ields E r = 1 4πɛ 0 Q encl r 2. This formula can be applied to an sphericall smmetric charge distribution. It is important that ou understand the argument above, and in particular, the large role that smmetr plaed in the calculation of the electric field. Gauss Law itself was involved onl briefl (towards the end when we set the flu equal to Q encl /ɛ 0 ). Most of the calculation was based on simplifing the electric flu integral with the help of smmetr arguments. In general, we require two smmetr arguments: (1) to determine the direction of the electric field (in order to establish that all but one component of the electric field is zero, and allow us to simplif the dot product E d A), and (2) to prove that the non-zero component of the electric field is constant over the surface, thereb allowing us to pull that component out of the flu integral. Onl b pulling the electric field component out of the integral can we then solve for the electric field at individual points when we finall set the flu equal to Q encl /ɛ 0. Other eamples of Gauss Law calculations will be presented in class. 10