text:
## 1.

The Correct Answer is (B) — This can be written as 3x – 20 = 40, or 3x = 60, so x = 20.

## 2.

The Correct Answer is (C) — We can set the equations equal to each other as 5n + 6 =$n^{2}$ – 2n – 24, then rearrange to get –$n^{2}$ + 7n + 30 = 0. Factoring the left side gives –(n – 10)(n + 3) = 0, meaning that the solutions are n = 10 and n = –3. Of these, only 10 is given as an option.

## 3.

The Correct Answer is (C) — We can write (x - 5)/4 = 6 and multiply both sides by 4 to get x - 5 = 24. Adding 5 to both sides gives x = 29.

## 4.

The Correct Answer is (B) — If x is positive, $\sqrt{9x^{2}}$ – 7x is the same as 3x – 7x, or –4x.

## 5.

The Correct Answer is (A) — Slope is defined as rise (difference in y values) divided by run (difference in x values). The rise is –1, and the run is 2, so the slope is –½.

## 6.

The Correct Answer is (D) — We can approach this problem by noticing that every time y increases by 1, x increases by 3, since the slope (1/3) is the rise (increase in y) divided by the run (increase in x). Since the line passes through (1, 1), we can see that only the point (7, 3) fits these criteria. Alternatively, we can write an equation in standard form: y = 1/3 x + b. Plugging in the given point, (1, 1), for x and y, we find that 1 = 1/3 + b, meaning that b = 2/3. We now have our equation: y = 1/3 x + 2/3. By checking the given points, we can again confirm that D) is the only point that satisfies this equation.

## 8.

The Correct Answer is (B) — If the programmer wants to write 6 websites, she will need to code 6 pages of HTML. Since she can write 3 pages of HTML in 5 hours, it will take her 10 hours to code 6 pages. Similarly, to code the required 6 pages of JavaScript, she will need 9 hours since it takes 3 hours to code 2 pages of JavaScript. 10 + 9 = 19.

## 9.

The Correct Answer is (C) — The formula for the volume of a cone is V = 1/3 $\pi&space;r^{2}h$. Since the radius is half the diameter, r = 8. We can now see that this is a 3-4-5 triangle (keep in mind that the figure is not drawn to scale), so the height is 6. We now have V = 1/3 $\pi&space;8^{2}6$, or 128$\pi$.

## 10.

The Correct Answer is (A) — To avoid mistakes in a problem like this, multiply one term at a time, and then add or subtract as needed: $n^{2}(n^{3}&space;-&space;3n^{^{2}}&space;+&space;5)&space;-&space;n(n^{3}&space;-&space;3n^{2}&space;+&space;5)$.

## 11.

The Correct Answer is (D) — We can rewrite the first equation as x = 6y. Setting the equations equal to each other gives 6y = 8y + 16. Subtracting 8y from both sides gives –2y = 16, or y = –8.

## 12.

The Correct Answer is (C) — The length of an arc is the radius times the central angle, or 6 x 30 degrees. Since the answers are in radians, we must convert 30 degrees to radians: 30 degrees x (2$\pi$radians/360 degrees) = $\pi$/6 radians, so 6 x $\pi$/6 = $\pi$.

## 13.

The Correct Answer is (A) — We can represent the original price as p. Since there was a 15% discount, the discounted price can be written as 0.85p. With tax, the price is (1.06)(0.85)p = d. We can now find p in terms of d: p = d/(1.06)(0.85).

## 14.

The Correct Answer is (C) — We want to find an equation in which e increases as d increases. Answer C) is the only equation that satisfies this; plotting a few points from the other equations confirms this. In Answer A), the term $-12^{d}$ grows faster than d, so e decreases as d increases. In Answer B), $-d^{2}$ grows faster than 100d, so again e decreases as d increases. In answer D), $-10^{d}$grows faster than $10d^{2}$, so e decreases as d increases.

## 15.

The Correct Answer is (D) — The effect of doubling the distance is the same as saying that d2 becomes $(2d)^{2}$, or $4d^{2}$. This means that the denominator is four times larger, meaning that the force of gravity will decrease by 3/4.

## 16.

The Correct Answer is (3/2) — For two lines to be perpendicular, their slopes must be reciprocal and have opposite signs. Since the slope of line k is -2/3, the slope of line j must be its reciprocal (-3/2), with the opposite sign (3/2)

## 17.

The Correct Answer is (1) — We can model Amelie’s position as y = 5x and her brother’s position as y = 15(x – 2), where x is hours. Setting these equations equal to each other gives 5x = 15x – 30, or 10x = 30. So x = 3, meaning that Amelie has been running for 3 hours by the time her brother catches up with her. Since her brother left two hours later, it took him 3 – 2 = 1 hour to reach her.

## 18.

The Correct Answer is (157) — g(3) = f(2 x 3) = f(6) = 63 – 10(6) + 1 = 216 – 60 + 1 = 157.

## 19.

The Correct Answer is (100) — Charlie must travel 15 miles at 12 miles per hour, meaning it will take him 15 miles/12 miles/hour = 1.25 hours = 75 minutes to reach the top. Coming back, he must travel 15 miles at 36 miles per hour, taking 15 miles/36 miles/hour = 0.417 hours = 25 minutes. 75 + 25 = 100.

## 20.

The Correct Answer is (108) — After 5 years, Sarah will have 30 pairs of shoes. After 10 years, she will have 60, and after 15, she will have 120. She donates 10%, or 12 pairs of shoes, leaving her with 120 – 12 = 108 pairs of shoes.


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MathSciNet bibliographic data MR742889 (86i:57036) 57R40 (57M99) Lawson, Terry Detecting the standard embedding of ${\bf R}{\rm P}\sp{2}$${\bf R}{\rm P}\sp{2}$ in $S\sp{4}$$S\sp{4}$. Math. Ann. 267 (1984), no. 4, 439–448. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.

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Paper Info Reviews Meta-review Author Feedback Post-Rebuttal Meta-reviews

# Authors

Ping Wang, Jizong Peng, Marco Pedersoli, Yuanfeng Zhou, Caiming Zhang, Christian Desrosiers

# Abstract

In this paper, we present a Context-aware Virtual Adversarial Training (CaVAT) method for producing anatomically plausible segmentation. The proposed method uses reinforcement learning to include any non-differentiable constraint in a semi-supervised setting. Specifically, we incorporate complex anatomical constraints into the VAT framework, boosting the network’s robustness to adversarial examples that maximize both prediction divergence and constraint violation. Experiments on a well-known cardiac segmentation dataset show the effectiveness of our method in terms of segmentation accuracy and improved constraint satisfaction.

SharedIt: https://rdcu.be/cyhL7

N/A

N/A

# Reviews

### Review #1

• Please describe the contribution of the paper

The paper describes a method to leverage on constraints for semi-supervised training of a deep net to segment images. The crux of the proposed method is an auxiliary loss function, which detects constraint violations in segmentations obtained for images perturbed using the adversarial approach. This loss is stochastic, with gradients computed using the Reinforce rule.

the method is generic and could in theory be applied to any task with structured output and any constraint. In the paper, it is demonstrated to improve the performance of a deep net trained to segment cardiac images from the ACDC data set, when only 3-5% of the training labels are used, with the constraint that each class should form a single connected component.

• Please list the main strengths of the paper; you should write about a novel formulation, an original way to use data, demonstration of clinical feasibility, a novel application, a particularly strong evaluation, or anything else that is a strong aspect of this work. Please provide details, for instance, if a method is novel, explain what aspect is novel and why this is interesting.

1) the idea of using the reinforce rule to minimize the expected number of constraint violations is simple and generic, yet potentially very powerful. 2) the paper is very clear and well written

• Please list the main weaknesses of the paper. Please provide details, for instance, if you think a method is not novel, explain why and provide a reference to prior work.

A) it is a pity that the method is only demonstrated with one constraint. B) … and on one data set. C) The paper does not discuss why the image needs to be perturbed in an adversarial way to run the loss l_cons. It is also not shown experimentally what happens when l_cons is used on the original image. D) I miss a discussion of the limitations of this approach. E) Just for reference, a comparison to another method of semi-supervised segmentation would be nice. This is a minor weakness though, because the advantage of the proposed method is that it accommodates non-differentiable constraints.

• Please rate the clarity and organization of this paper

Excellent

• Please comment on the reproducibility of the paper. Note, that authors have filled out a reproducibility checklist upon submission. Please be aware that authors are not required to meet all criteria on the checklist - for instance, providing code and data is a plus, but not a requirement for acceptance

The description of the method is clear enough for me to attempt to implement it.

• Please provide detailed and constructive comments for the authors. Please also refer to our Reviewer’s guide on what makes a good review: https://miccai2021.org/en/REVIEWER-GUIDELINES.html

I develop here on some weaknesses listed in point 4. ad A) could you suggest other constraints/reward designs for which the method could be used? ad C) what would happen if you evaluated the l_cons loss for r_u=0 and without the VAT loss? Is the adversarial perturbation really necessary in your method? ad D) Are there constraint types for which the loss is ineffective?

Additional question: Q1. one non-obvious weakness is that l_cons only reinforces samples (patches, in this case) that satisfy the constraints. What about penalizing samples that violate the constraints?

Minor comments: i. the role of the subscript i in (7) is not clear until the next section. ii. there is an interrupted sentence in the beginning of 3.2 iii. I cannot find the number of samples used (m) in the paper

Probably accept (7)

• Please justify your recommendation. What were the major factors that led you to your overall score for this paper?

My recommendation is influenced by two main arguments:

• a simple, generic, well described idea, very compelling
• the method was only demonstrated to work with a single constraint type
• What is the ranking of this paper in your review stack?

2

• Number of papers in your stack

5

• Reviewer confidence

Confident but not absolutely certain

### Review #2

• Please describe the contribution of the paper

This paper presents a semi-supervised learning method which incorporate segmentation constraints over virtual adversarial training to produce anatomically-plausible segmentation. The method is validated on the cardiac image segmentation task.

• Please list the main strengths of the paper; you should write about a novel formulation, an original way to use data, demonstration of clinical feasibility, a novel application, a particularly strong evaluation, or anything else that is a strong aspect of this work. Please provide details, for instance, if a method is novel, explain what aspect is novel and why this is interesting.
1. The idea to incorporate segmentation constraint to produce anatomically-plausible segmentation is reasonable.
• Please list the main weaknesses of the paper. Please provide details, for instance, if you think a method is not novel, explain why and provide a reference to prior work.
1. The rationale of proposed method was not clearly elaborated. For example, why reinforcement learning is used in the method, what is the particular advantage.

2. Novelty of proposed method is incremental over VAT.

3. It seems this paper has not been well prepared. (1) No introduction for background in the abstract section. (2) Incomplete sentences/paragraphs in Section 3.2, which is in fact a crucial part in the method.

4. Semi-supervised medical-image segmentation has been a popular topic with a number of methods being proposed recently also in MICCAI 2020. However, the authors didn’t compared with any of the recent SOTA methods.

5. How can the authors demonstrate that there method can produce more anatomically-plausible segmentations? Is there any evidence showing the connections between non-connectivity and anatomically plausibility.

6. The method is only validated on one dataset, which may lead to bias conclusions.

• Please rate the clarity and organization of this paper

Poor

• Please comment on the reproducibility of the paper. Note, that authors have filled out a reproducibility checklist upon submission. Please be aware that authors are not required to meet all criteria on the checklist - for instance, providing code and data is a plus, but not a requirement for acceptance

The authors claim to release the code.

• Please provide detailed and constructive comments for the authors. Please also refer to our Reviewer’s guide on what makes a good review: https://miccai2021.org/en/REVIEWER-GUIDELINES.html
1. In fact this paper is very related to semi-supervised learning, hence keyword “semi supervised” should be included in the title.

2. A figure could be provided to give a clear illustration for the proposed method.

reject (3)

• Please justify your recommendation. What were the major factors that led you to your overall score for this paper?

My rating is mainly from two perspectives:

1. Technical novelty of this paper is incremental and the use of reinforcement is has not been well motivated.
2. Lack comparisons with SOTA and the evaluation is only conducted on one dataset.
• What is the ranking of this paper in your review stack?

4

• Number of papers in your stack

4

• Reviewer confidence

Very confident

### Review #3

• Please describe the contribution of the paper

This paper proposed a semi-supervised method with an additional non-differentiable constrain attached to the VAT framework, to enhance model robustness and accuracy. They have validated their performance on cardiac image segmentation.

• Please list the main strengths of the paper; you should write about a novel formulation, an original way to use data, demonstration of clinical feasibility, a novel application, a particularly strong evaluation, or anything else that is a strong aspect of this work. Please provide details, for instance, if a method is novel, explain what aspect is novel and why this is interesting.

++ The problem they tackled, i.e., how to exploit the shape priors (e.g., the anatomical constraints) in semi-supervised learning, is interesting.

++ A new usage of reinforcement learning, i.e., using reinforcement learning to solve the non-differentiable loss term.

• Please list the main weaknesses of the paper. Please provide details, for instance, if you think a method is not novel, explain why and provide a reference to prior work.

For methods, there are several ambiguous explanations, which make it hard to follow. – In the 2nd term in Eq.3, how the constrain violation term could lead to better anatomical shapes? – For someone who is not familiar with reinforcement learning, I do not quite understand why using reinforcement learning could solve the non-differential loss? Since you used reinforcement learning (RL), what do other commonly-used RL terms including environment and status, refer to in the context? – It would be greatly helpful if there is a brief figure to illustrate your framework and insights. – There may have a typo in Eq.4. It should be $x_{u}+r_{u}$, instead of $x+r_{u}$ ,right?

For experiments, I noticed that the improvements over VAT are not significant, i.e., <1% in Dice.

• Please rate the clarity and organization of this paper

Satisfactory

• Please comment on the reproducibility of the paper. Note, that authors have filled out a reproducibility checklist upon submission. Please be aware that authors are not required to meet all criteria on the checklist - for instance, providing code and data is a plus, but not a requirement for acceptance

Since several components are coarsely described with no code available, I am not positive about it.

• Please provide detailed and constructive comments for the authors. Please also refer to our Reviewer’s guide on what makes a good review: https://miccai2021.org/en/REVIEWER-GUIDELINES.html

It would be great if the authors could elaborate more on several details in methods

1. Regarding the 2nd term in Eq.3, (1) How the constrain violation term would help the network to learn more complex or better anatomical shapes? (2) Why using reinforcement learning could solve the non-differential loss?

2. Regarding Eq. 6, how to sample the discrete segmentation mask? What exactly the constrain mean to direct the reward function to be 1 or 0?

3. In the 3rd paragraph in Section 3.2, you mentioned you utilized a constrain to impose the segmentation regions to have a single connected component. I was wondering if it could generalize well to other applications. For cardiac segmentation on organ structures, it should be fine as the segmented regions like LV, Myo and RV, indeed should be one single piece. However, for some lesions, like the myocardial infarction regions, since they are naturally scattered, is the proposed constrain also appliable here?

4. It would be greatly helpful if there is a brief figure to illustrate your framework and insights.

5. In experiments, I am curious about the performance of VAT + shape constrain in [28].

probably reject (4)

• Please justify your recommendation. What were the major factors that led you to your overall score for this paper?
1. The explanations on the key component, i.e., the constrain violation, is ambiguous and hard to follow.
2. The performance gains are less significant, i.e., <1% in Dice.
• What is the ranking of this paper in your review stack?

4

• Number of papers in your stack

5

• Reviewer confidence

Somewhat confident

# Primary Meta-Review

• Please provide your assessment of this work, taking into account all reviews. Summarize the key strengths and weaknesses of the paper and justify your recommendation. In case you deviate from the reviewers’ recommendations, explain in detail the reasons why. In case of an invitation for rebuttal, clarify which points are important to address in the rebuttal.

Strengths

• The idea of using reinforcement learning to include non-differentiable constraint to produce anatomically-plausible segmentation is highly compelling.
• The paper is very clear and well written.

Weaknesses

• The method is only validated on one dataset.
• The authors did not compare with any of the recent SOTA methods.
• Novelty of proposed method is incremental over VAT, wrt to all metrics.

Regarding the remark of Rev#2 « Novelty of proposed method is incremental over VAT. i.e., <1% in Dice. »: I completely agree. Thus I advise the authors not to « boldify » the top 1 result, unless statistical significance has been demonstrated, cf for example 77.40 and 77.85 are likely to be statistically similar given the std deviation. However let me acknowledge that the increment in terms of Dice is not what the authors put forward in the text, as they mainly comment on the non-conn metric. That said, I believe even the experimentations with the VAT/Entropy min bring interesting insight as to how to adversarial examples to enforce constraint in segmentation.

Despite some justified limits noted by the reviewers, I think the paper has some merits, from a methodological perspective, since proposing a framework that allow to incorporate non-differentiable constraints in segmentation network is a hot topic and of great importance. Also the paradigm of reinforcement learning has received scarce attention for image segmentation models. I believe the authors can bring clarifications to some key questions.

The questions to address in the rebuttal are inspired by all reviewer’s concerns: 1) It would be good to highlight the benefits of using reinforcement learning in this context, for reader unfamiliar with this topic. In this regard, please also explicit « Why using reinforcement learning could solve the non-differential loss? »  A figure would be definitely helpful to help readers unfamiliar with RL. 2) Please explain why method can produce more anatomically-plausible segmentations 3) Please discuss the limitations of this approach. 4) Please discuss why the image needs to be perturbed in an adversarial way to run the loss l_cons. It is also not shown experimentally what happens when l_cons is used on the original image.

Other important points, as noted by Rev#2:

• This paper is related to semi-supervised learning, hence keyword “semi supervised” should be included in the title: the superiority of the method outside of a semi-supervised setting has not been shown, so
• Please remove the unfinished sentence of section 3.2 « In this paper, we consider a well-known and broadly »

In addition of course, if space is left, the authors can also correct any misunderstanding and address additional points raised by the reviewers.

• What is the ranking of this paper in your stack? Use a number between 1 (best paper in your stack) and n (worst paper in your stack of n papers).

8

# Author Feedback

• Novelty of our method (R2)

As underlined by the Meta and R1, we propose a generic and efficient approach for adding any segmentation constraint on top of any segmentation network. For R2’s comment, we argue that our method is not an extension of VAT, but rather uses adversarial training to generate examples violating given constraints without having prior knowledge of these constraints. Our goal is not directly improving the segmentation Dice but instead reducing anatomically-impossible predictions by the network. Experiments show our method to yield a statistically greater constraint satisfaction (non-conn) than VAT and Entropy minimization.

• Benefits of RL (M,R2,R3)

The link between RL and our method is two-fold. 1) As the policy network in RL, the segmentation network predicts label probabilities at pixels. However, the agent’s action in RL and segmentation labels in our work are discrete. 2) Like the reward function in RL, the segmentation constraints in our work can be added on a generic learning strategy, without requiring prior knowledge on that constraint or a differentiable loss function. We achieve these goals via the Reinforce algorithm that is commonly used in policy gradient RL methods.

• Anatomically-plausible segmentation of method (M,R2,R3)

Previous work as [3] showed that a high Dice does not always imply a good segmentation perceived by clinicians. For instance, a segmented region with 90% Dice may still contain holes or disconnected regions that are anatomically impossible. Unlike approaches focusing solely on accuracy measures like Dice, our method also considers complex topological constraints like connectivity which cannot be easily modeled in a loss function. We use adversarial training to concentrate the learning on examples violating the constraints, so the network can learn to avoid making such incorrect predictions on new examples. We demonstrate our method on a hard-to-model constraint with a high applicability: connectivity. Results show our method to give less disconnected pixels compared to other tested approaches (paired t-test p < 0.05).

• Need for adversarial perturbation (R1,R3)

The constraint loss based on the Reinforce algorithm produces a non-zero gradient only if the constraint is violated. Since the constraint can be complex and is not known beforehand, there is no guarantee that it will be violated during training hence that the network will learn to avoid this violation. In our method, we use adversarial training to focus on examples with a non-zero gradient for the constraint loss, thus speeding-up the learning. Meanwhile, the standard VAT term increases the network’s robustness to noise. To answer R1’s question, we tested our method without adversarial training (i.e., r_u=0) and found it gave a similar Dice but a higher HD and constraint violation (non-conn).

• Limitations of our method (M,R1)

Limitations include: 1) the Reinforce algorithm requires sampling several segmentations from the predicted probabilities otherwise the optimization may be unstable; 2) although our method allows adding any constraint, evaluating some constraints during training may be computationally expensive.

• Other datasets and SOTA (R1,R2)

Our method can be added as a plug-in on top of any segmentation approach. We further demonstrate its advantage by adding it on two SOTA approaches, Co-training (Co-T) and Mean Teacher (MT), and testing it on the PROMISE12 dataset (5% labeled):

   | Dice | HD | Non-conn Ent. min | 50.21 (1.93) | 13.52 (3.95) | 30.33 (6.46) VAT | 50.96 (3.42) | 13.44 (3.57) | 31.96 (2.23) Co-T | 48.92 (2.74) | 9.72 (0.92) | 33.09 (4.06) MT | 67.97 (1.88) | 11.72 (0.76) | 11.33 (3.10) CaVAT | 57.45 (1.23) | 15.60 (2.25) | 21.53 (1.31) Co-T + CaVAT | 63.01 (1.62) | 4.21 (0.72) | 16.10 (3.49) MT + CaVAT | 70.87 (1.70) | 10.17 (0.60) | 9.67 (2.24)

• Paper title (R2)

We will modify the title to include the term semi-supervised.

# Post-rebuttal Meta-Reviews

## Meta-review # 1 (Primary)

• Please provide your assessment of the paper taking all information into account, including rebuttal. Highlight the key strengths and weaknesses of the paper, clarify how you reconciled contrasting review comments and scores, indicate if concerns were successfully addressed in the rebuttal, and provide a clear justification of your decision. If you disagree with some of the (meta)reviewer statements, you can indicate so in your meta-review. Please make sure that the authors, program chairs, and the public can understand the reason for your decision.

This paper presents novel and powerful ideas to constrain image segmentation, as acknowledged by the reviewers. There were issues on the motivation, the impact and the limitations of the method. To my opinion, the authors have addresses these issues in their rebuttal, hence this paper deserves acceptance for its interest regarding methodology.

• After you have reviewed the rebuttal, please provide your final rating based on all reviews and the authors’ rebuttal.

Accept

• What is the rank of this paper among all your rebuttal papers? Use a number between 1 (best paper in your stack) and n (worst paper in your stack of n papers).

3

## Meta-review #2

• Please provide your assessment of the paper taking all information into account, including rebuttal. Highlight the key strengths and weaknesses of the paper, clarify how you reconciled contrasting review comments and scores, indicate if concerns were successfully addressed in the rebuttal, and provide a clear justification of your decision. If you disagree with some of the (meta)reviewer statements, you can indicate so in your meta-review. Please make sure that the authors, program chairs, and the public can understand the reason for your decision.

I believe the paper is quite borderline, with an interesting RL based application to shape constraints for segmentation, but with substantial drawbacks: clarity of exposition, limitation in experiments (few baselines), limitations in quality of results, etc.

It seems to me that the Primary MR values the RL application to anatomical constraints and would like to see the paper accepted. I mostly agree, and believe the rebuttal helps clarify some of the issues, including adding new results on a new dataset with new baselines.

I am learning towards a conditional accept, given that the authors include everything they added to the rebuttal with a careful analysis, even in cases where their method does not improve statistically on VAT. The main reason for acceptance is the uniqueness of the method-task combination, which makes it worthwhile to discuss at MICCAI.

• After you have reviewed the rebuttal, please provide your final rating based on all reviews and the authors’ rebuttal.

Accept

• What is the rank of this paper among all your rebuttal papers? Use a number between 1 (best paper in your stack) and n (worst paper in your stack of n papers).

12

## Meta-review #3

• Please provide your assessment of the paper taking all information into account, including rebuttal. Highlight the key strengths and weaknesses of the paper, clarify how you reconciled contrasting review comments and scores, indicate if concerns were successfully addressed in the rebuttal, and provide a clear justification of your decision. If you disagree with some of the (meta)reviewer statements, you can indicate so in your meta-review. Please make sure that the authors, program chairs, and the public can understand the reason for your decision.

The author added a context-aware virtual adversarial training module into the semi-supervised segmentation pipeline, aiming to preserve the anatomical structure better. The problem to tackle is interesting and well motivated. The experimental results is also promising. The AC agrees that the dice score should not be the only metric to focus on. However, there are several limitations of current versions: more explanation of RL is needed, the module is only tested on one dataset. A lot of edits are needed to address these concerns. Recommend to reject based on current version.

• After you have reviewed the rebuttal, please provide your final rating based on all reviews and the authors’ rebuttal.

Reject

• What is the rank of this paper among all your rebuttal papers? Use a number between 1 (best paper in your stack) and n (worst paper in your stack of n papers).

15


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# An Exceptionally Technical Discussion of AESToE

 P: 40 moveon said "... the only algebras that contain both bosonic and fermionic generators are superalgebras ...". No, that is not true. As Pierre Ramond in hep-th/0112261 said "... exceptional algebras relate tensor and spinor representations of their orthogonal subgroups ...", and the exceptional algebra 248-dim E8 contains 120 generators corresponding to the tensor/bosonic part 120-dim adjoint Spin(16) and 128 generators corresponding to the spinor/fermionic part 128-dim half-spinor Spin(16) and their commutation relations do close into E8. However, Pierre Ramond went on to say in that paper: "... Spin_Statistics requires them [ the adjoint/bosonic and half-spinor/fermionic ] to be treated differently ...", so any model you build with E8 must somehow treat them differently. For example, you might just construct a Lagrangian into which you put the 128 half-spinor fermionic generators into a fermion term and 8 of the 120 bosonic generators into a spacetime base manifold term and 120-8 = 112 of the 120 bosonic generators into a gauge boson curvature term. Then you might have disagreement as to how natural (or ad hoc) is such an assignment of parts of E8 to terms in a Lagrangian, but all should agree that you have "treat[ed] them differently" as required by Spin-Statistics. However, in Garrett's 13-grading decomposition of the 240 root vectors of E8 5 + 6 + 15 + 20 + 30 + 30 + 28 + 30 + 30 + 20 + 15 + 6 + 5 some of the graded parts contain both bosonic terms and fermionic terms, for example the central 28 has both circles (bosons) and triangles (leptons and quarks), which has led Thomas Larsson to complain (on Cosmic Variance): "... both fermions and bosons belong to the same E8 multiplet. This is surely plain wrong. ...". I think that the point of Thomas Larsson is that the model must treat the fermions and bosons differently to satisfy Spin-Statistics so the fermionic generators must be put into some part of the model where the bosonic generators are not put so if you decompose the generators into multiplets some of which contain both fermionic and bosonic generators (as in Garrett's 13-grading decomposition) then you are not respecting your multiplets when you, from a given multiplet, put some of them into a fermionic part of the model and some of them into a bosonic part of the model. This is not merely an objection of ad hoc assignments of generators to parts of the model, it is an objection that the assignments do not respect the chosen decomposition into multiplets. Tony Smith PS - It is possible to choose a decomposition that does keep the bosonic and fermionic generators separate, the simplest being 64 + 120 + 64 where the 120 is bosonic and the 64+64 = 128 is fermionic.
P: 9
 Quote by Tony Smith moveon said "... the only algebras that contain both bosonic and fermionic generators are superalgebras ...". No, that is not true. As Pierre Ramond in hep-th/0112261 said "... exceptional algebras relate tensor and spinor representations of their orthogonal subgroups ...", and the exceptional algebra 248-dim E8 contains 120 generators corresponding to the tensor/bosonic part 120-dim adjoint Spin(16) and 128 generators corresponding to the spinor/fermionic part 128-dim half-spinor Spin(16) and their commutation relations do close into E8.

Oh yes, this is of course very well known since ages. But those tensor and spinor rep generators are all bosonic, and close into the usual E8 commutator relations. My point is, apparently still not appreciated, that if some of the generators are made fermionic (as it happens for superalgebras), then they cannot produce the E8 commutation relations (and jacobi identities etc) any more. The opposite seems to be claimed here all over, so I'd like to see, how. Please prove this by writing them down!

And if the E8 commutation relations are not there, there is no E8 to talk about. There is "somewhat" more to E8 than a drawing of the projection of its polytope...
 P: 19 This from Tony's website might be good to look at (I'm sure Tony can say more if needed): http://www.valdostamuseum.org/hamsmi...eStdModel.html
 P: 102 One needs to distinguish between spin and statistics. There are two types of statistics: fermions, which anticommute and obey Pauli's exclusion principle, and bosons, which commute. There are also two types of spin: spinors, which have half-integer spin, and tensors and vectors, which have integer spin. The spin-statistics theorem asserts that physical fermions always have half-integer spin and physical bosons have integer spin. But this is non-trivial and surprisingly difficult to prove. In contrast, BRST ghosts are fermions with integer spin, and therefore unphysical. Physical and unphysical fermions are not the same. What is quite easy to prove is that statistics is conserved, i.e. [boson, boson] = boson [boson, fermion] = fermion {fermion, fermion} = boson. People like Lee, Peter and Bee know this, of course, and it must be obvious that putting both bosons and fermions into the same E8 multiplet violates this fundamental principle. That they don't emphasize this simple fact but instead complain about manners is something that I find surprising and quite disappointing.
 P: 40 Here is what I hope is a concrete example of what I think that Thomas Larsson is saying (please feel free to correct my errors): If you were to (not what Garrett did) make a physics model by decomposing E8 according to its e17 5-grading: g(-2) = 14-dim physically being spacetime transformations g(-1) = 64-dim physically being fermion antiparticles g(0) = so(7,7)+R = 92-dim physically being gauge bosons g(+1) = 64-dim physically being fermion particles g(+2) = 14-dim physically being spacetime transformations then that would be consistent with spin-statistics because the products fermion(-1) times fermion(+1) would be gauge bosons(-1+1=0) the products of gauge bosons(0) times gauge bosons(0) would be gauge bosons(0+0=0) the products of gauge bosons(0) times fermions(-1) would be fermions(0-1=-1) the products of gauge bosons(0) times fermions(+1) would be fermions(0+1=+1) etc The point is that if you have fermions and bosons mixed up together in the same part of the graded decomposition, you do not get good spin-statistics, but it is possible to decompose in a way that you do get good spin-statistics and that is something that should be taken into account in model-building. Tony Smith PS - Sorry for burying stuff like fermion(-1) times fermion(-1) giving spacetime(-2) into an "etc" (sort of like spinor x spinor = vector) but in this comment I am just trying to make a point and not build a complete model here.
 P: 360 sambacisse, The issue is a bit more complicated than it appears because of how the real representations are mixed together in exceptional groups into complex representation spaces, relying on an inherent complex structure. This sort of thing is described halfway through John Baez's TWF253 for the case of E6. When describing so(3,1) reps in terms of sl(2,c) this is further complicated, and when swapping in conjugated anti-fermions it's more complicated still -- because one has to be clear in each step which complex structure one is conjugating with respect to. I thought I had this figured out several years ago, but I don't like to make statements about complicated things without having slowly worked through them in detail. So I've stayed out of the arguments. Of course, I can say that the worst case scenario is that one might have to use a complex E8. moveon, Tony addressed this a bit, and I'll try to summarize the specific case in the paper. The E8 Lie algebra may be naturally decomposed into a D4+D4 subalgebra, and everything else. In terms of the number of elements, this decomposition is: (28+28)+64+64+64 which I don't consider a "grading," but it relates to gradings. The important thing is the Lie brackets. If we label the D4+D4 elements "bosons," and the rest "fermions," the brackets are as Thomas Larsson has helpfully described. Now, if the E8 symmetry is broken such that the "fermion" part of the Lie algebra is pure gauge, then that part of the connection may be replaced by Lie algebra valued Grassmann fields. We end up with a D4+D4 valued connection 1-form field, $$\underline{H}_1+\underline{H}_2$$, and three other fields, the first of which is the first generation fermions, $$\Psi$$, which are Grassmann valued E8 Lie algebra elements. Because of the structure of E8, the Lie brackets between these give the fundamental action: $$[\underline{H}_1+\underline{H}_2,\Psi] = \underline{H}_1 \Psi - \Psi \underline{H}_2$$ The brackets between two $$\Psi$$'s are in D4+D4, but these terms vanish in the action. Notice that there is no symmetry here relating the fermions to bosons. That symmetry was destroyed when we broke the E8 symmetry by adding the terms we did to the action. I did that by hand in my paper, and Lee talks about how that can happen dynamically in his. There is a cute trick in the BRST literature whereby these objects can be formally added in a generalized connection: $$\underline{H}_1+\underline{H}_2+\Psi$$ Since I like cute math tricks, I used it -- allowing all fields to be written as parts of this "superconnection," with the dynamics coming from its generalized curvature.
 P: 40 So, it seems to me that: 1 - Garrett has shown that his physical identifications of E8 generators are consistent with spin-statistics; 2 - Garrett is not claiming that any BRST ghost-fermions-with-integer-spin are physical, but is just using one of the technical "math tricks" from BRST literature in order to construct his "superconnection" containing both gauge boson curvature terms and curvature terms derived from spinor/fermions; 3 - Garrett has explicitly broken full E8 symmetry so that it is irrelevant whether or not Garrett's physics stuff (whether it is Pati-Salam or not) fits inside E8, so that Jacques Distler's arguments about it not fitting inside E8 are irrelevant. 4 - However, just as Jacques Distler's comments were useful in seeing that E8(8) might be more useful than E8(-24), it may be that his comments about Pati-Salam vs. the Standard Model might also be useful indicators that Garrett's model should perhaps be put directly in terms of the minimal Standard Model than in terms of Pati-Salam. Tony Smith PS - If I had to guess, I would guess that Garrett used Pati-Salam because he thought that it was an established particle physics model, and its use would make his E8 model more acceptable to conventional physicists. Since it has turned out otherwise, maybe just using the plain vanilla minimal Standard Model plus MacDowell-Mansouri gravity might be a way to go. PPS - It is unfortunate that a "food-fight" atmosphere has obscured much of the sensible physics in discussions on some parts of the web, and I would like to say that I very much appreciate the moderate (in more meanings than one) atmosphere here on Physics Forums. Such moderation-in-climate does not come about without moderation-in-the-other-sense, and that takes effort, which I appreciate very much.
P: 9
Garret,

OK so let me translate this in my language.. your superconnection does not take values in the Lie algebra of E8 as some generators are fermionic (they square to zero, eg).
Therefore the curvature, or field strength does not take values in all of E8, but in D4+D4 only. The full commutation relations of E8 are therefore not non-trivially realized. So in what sense then does E8 play a role? It seems that the purpose of your E8 is to organize, as a bookkeeping device, the fermionic part of the spectrum in terms of the coset E8/(D4+D4), as far as their quantum numbers are concerned.

This is linked to the "breaking" of E8. There are different notions of a symmetry being broken. Usually in particle physics a symmetry is spontaneously broken, which means it is "still there" albeit non-linearly realized. It reflects itself in terms of Ward identities of the low energy effective theory.
There is an energy scale above which the symmetry is restored and the theory is in an "unbroken phase". So one may speak of an "underlying" symmetry.

In contrast, you write a theory where there is no E8 symmetry to begin with (ie, its commutation relations are not fully realized) and there is no energy scale above which it is restored. So calling it "breaking" may be misleading...it is just not there. It is a bit like saying the standard model has monster group symmetry, although most of it is broken.

 Quote by Tony Smith So, it seems to me that: 1 - Garrett has shown that his physical identifications of E8 generators are consistent with spin-statistics; ..... 3 - Garrett has explicitly broken full E8 symmetry so that it is irrelevant whether or not Garrett's physics stuff (whether it is Pati-Salam or not) fits inside E8, so that Jacques Distler's arguments about it not fitting inside E8 are irrelevant.

To 1- .... they are not the generators of E8. They are the generators of some superalgebra whose bosonic piece is D4+D4.

To 2- .... it seems to me that the claim was that that the standard model spectrum can be organized in terms of E8/(D4+D4) (rather, of the relevant non-compact real forms). That has been shown by Distler not to be the case.

I would thus advise to look for superalgebras instead of E8. There exist even exceptional ones; they have been classified by Katz, and a useful ref is hep-th/9607161. Choosing one with D4+D4 as its bosonic piece (and a suitable real form) may be more successful. Also, superalgebras are consistent with Coleman-Mandula (that's why supergravity works).
 P: 751 moveon, thanks for that; a very very illuminating comment.
P: 19
 Quote by moveon In contrast, you write a theory where there is no E8 symmetry to begin with (ie, its commutation relations are not fully realized) and there is no energy scale above which it is restored. So calling it "breaking" may be misleading...it is just not there. It is a bit like saying the standard model has monster group symmetry, although most of it is broken. To 1- .... they are not the generators of E8. They are the generators of some superalgebra whose bosonic piece is D4+D4. To 2- .... it seems to me that the claim was that that the standard model spectrum can be organized in terms of E8/(D4+D4) (rather, of the relevant non-compact real forms). That has been shown by Distler not to be the case.
The full E8 symmetry would seem to be E8/D8, I personally am more familiar with E8/E7xSU(2) and so on down the A-D-E series but maybe one can do something with E8/D8. The D4+D4 part seems after symmetry breaking so one should not expect any E8/(D4+D4) physics.
 P: 40 moveon "... advise[s] to look for superalgebras instead of E8. There exist even exceptional ones; they have been classified by Katz, and a useful ref is hep-th/9607161. Choosing one with D4+D4 as its bosonic piece (and a suitable real form) may be more successful. ...". hep-th9607161 is indeed a nice reference. Thanks for it. However (please correct me where I am wrong) when I look at it for exceptional Lie superalgebras, I see only three: F(4) which is 40-dimensional; G(3) which is 31-dimensional; and D(2,1;a) which is 17-dimensional, so none of them are large enough to contain 28+28=56-dimensional D4+D4. From Table III on page 13, it seems that the only one with a Dm bosonic part is D(m,n) which has bosonic part Dm (+) Cn which the describe on page 37 as being "... osp(2m|2n) ...[ which ]... has as even [ bosonic ] part the Lie algebra so(2m) (+) sp(2n) ...". osp(2m|2n) is the basis for supergravity and, in his book Supersymmetry (Cambridge 1986 at page 113), Peter G. O. Freund says "... In extended supergravity of type N the largest internal nonabelian gauge group is O(N), corresponding to a gauged osp(N|4) ... The largest nonabelian gauge symmetry is O(8) ...". So, since the sp(4) in Freund's notation, which is sp(2) in some other notations accounts for gravity and therefore for one of the D4, you have the O(8) for the other D4, so it seems to me that N=8 supergravity is the only superalgebra based model that could reasonably be seen as fitting something like Garrett's D4 + D4 model-making scheme. As Freund discusses in some detail in chapter 23, N = 8 supergravity and concludes "... all this makes the ultimate absence of a compelling and realistic spectrum all the more frustrating. ...". In chapter 26, Freund discusses the related 11-dimensonal supergravity, but as far as I know there has been no satisfactory realistic 11-dim supergravity or N=8 supergravity model. Therefore, to work with D4 + D4 it seems to me that you must abandon superalgebras because they either do not have it or have not been shown to work (despite much effort), and that ordinary exceptional Lie algebras, which have both bosonic and spinor parts, are a useful place to look for building models, and that Garrett has done a good job of seeing how the root vector generators of E8 can be assigned physically realistic roles in constructing a useful physics model, and therefore is worth a substantial amount of research effort (comparable to that spent so far on supergravity). Tony Smith
 P: 360 moveon, Your translation is interesting, but all fields in the paper are valued in the Lie algebra of E8. I'm not yet certain that the first generation doesn't work in real E8, because of the unusual complex structure employed -- but even if it doesn't work, complex E8 would.
 Sci Advisor HW Helper P: 1,204 Gosh, it seems to me that in QFT signs are arbitrary and in any observable, fermions always appear in pairs. In that sense, what you really need is to have your fermions square to zero and your bosons not. Zero is not a valid quantum state. To get the equivalent, all you really have to do is make the square of a fermion be "not a valid quantum state", it doesn't actually have to be zero. What fermions do to each other when you permute them is not a physical observable. Quantum mechanics is a probability theory. To get a probability in QFT you begin with an amplitude, which is a complex number, computed as $$\langle 0 | stuff | 0\rangle$$. Then you take the squared magnitude, that is, you multiply your amplitude by its complex conjugate: $$\langle 0 | stuff | 0\rangle \langle 0 | stuff^* |0\rangle$$ Now suppose you commute two creation operators in "stuff" and get a minus sign. That minus sign is cancelled by the minus sign that you get when you commute the same two observables in its Hermitian conjugate. No change to the observable whether the result of the commutation is +1 or -1. So suppose you start with a bosonic QFT and you have a boson $$\psi$$ that you want to give "fermion statistics" to. Add a term to the Hamiltonian of $$\kappa\;\psi\psi$$. Let $$\kappa \to \infty$$ to prevent it from being energetically possible. The result is a mixed fermion / boson theory by symmetry breaking. To put the above argument in QM form, consider the ancient physics test problem, "what happens to an electron if you rotate it by 360 degrees?" Every physicist knows the answer: "it gets multiplied by -1". But that is only true in the spinor representation. In the density matrix representation of a quantum state, spinors appear in pairs and the result of rotating them is to change the density matrix representation by -1 x -1 = 1, or not at all. The act of rotating a fermionic wave function by 360 degrees is related to the act of switching the order of creation operators as is discussed in many QFT textbooks. To put this in into the operator language, let Q be an operator, we wish to compute the average value of Q for a quantum state produced by the application of say four creation operators on the vacuum to make a four particle state. Label the four particles "k,n,u,j". So the 4-particle state is $$k^*n^*u^*j^*|0\rangle$$. Then the average of the operator Q over this quantum state is: $$\langle 0 | j\; u\; n\; k\; | Q |\;k^*\;n^*\;u^*\;j^*| 0\rangle$$ Suppose you've got the above worked out for k, n, u, and j fermion creation and annihilation operators. You might write Q in terms of these creation and annihilation operators, but when you're done writing it, you will have some ordering and you won't have to rearrange them. Now you can consider the same theory, but with the commutation relations of the k, n, u, j changed (but the operator Q left alone). The ensemble average will be the same as there will be no further need to commute the creation and annihilation operators. You get what you get. And if you want to change the order of the k, n, u, j, then you will be doing it twice and a sign change will cancel. Another case is when the quantum state is a superposition. For example, consider $$j^*\;u^*\; - u^*\;j^*\;|0\rangle$$. If j and u are bosons the result is just zero, no more to say. For fermions, you get twice your choice, of ordering. Choose one of the orderings and relabel your fermions as bosons. No problems. Problems happen when you try to modify your operators (built from creation and annihilation operators with assumed commutation relations) at the same time as you modify the rules you use for how your creation and annihilation operators operate on the vacuum state. But if you do that you will be making a circular argument if you use that to say that the choice of commutation relations is an observable -- what you've done is modified the observable, not the quantum state itself.
 P: 587 So this is perhaps a step down in technicalness from the discussion of the last few pages, but this is something I have been wondering for awhile and have only just figured out how to ask correctly: Something that I keep running across in discussions of symmetry groups is the distinction between local or internal symmetries, and global or spacetime symmetries. In general the idea seems to be that local symmetries, things like quantum phase invariance, apply at a point (or at least to a single structure?); global symmetries, like poincare invariance, apply to "everything". Are the symmetries of Garrett's E8 construction local, or spacetime symmetries? E8 here contains both things which are usually given as examples of spacetime symmetries, like the Lorentz group, and also things which are usually given as examples of local symmetries, like electroweak SU(2)xU(1). Meanwhile, E8 is here used as a "gauge group", and for some reason I have gotten the impression that all "gauge" symmetries are local symmetries. Are all the E8 symmetries local? Or do they somehow incorporate a mix of local and spacetime symmetries? And if all of the symmetries in the E8 theory are local, then are there assumed to still be any "background" global/spacetime symmetries which exist apart from the symmetries E8 describes?
 P: 360 Hello Coin, All symmetries in this construction are local. The so(3,1) is a local symmetry of the frame, which is a local map from spacetime tangent vectors to a local rest frame, consistent with the equivalence principle. Now, when there are solutions, which give some spacetime, this may or may not have global symmetries.
 P: n/a Dear Garrett, Have you contacted CERN directly to be sure they include the predictions of your E8 model in the particle collision data they will capture when they start the LHC experiments soon in 2008 ? As I understand the situation, only a small fraction of the LHC collision data will be captured and stored, the rest is lost forever (CERN does not have enough computer memory storage). What data they do capture is what is predicted by current Standard Model, perhaps some new physics stuff--but, are you 100 % sure they will capture data that can be used to test (e.g., falsify) the predictions of your E8 model ?
 P: 360 Hello Rade, This E8 theory isn't developed well enough to produce such predictions with sufficient confidence. There are pretty clearly twenty or so new particles predicted, but until the problems with the theory are worked out, their properties are kind of up in the air. But, there are plenty of LHC observations which wouldn't be compatible with this theory, so it does have some predictive power in that sense. In any case, the theory needs to be developed further before specific predictions can be made with any confidence. There's a long way to go.
P: n/a
 Quote by garrett Hello Rade, This E8 theory isn't developed well enough to produce such predictions with sufficient confidence. There are pretty clearly twenty or so new particles predicted, but until the problems with the theory are worked out, their properties are kind of up in the air. But, there are plenty of LHC observations which wouldn't be compatible with this theory, so it does have some predictive power in that sense. In any case, the theory needs to be developed further before specific predictions can be made with any confidence. There's a long way to go.
Thank you very much for your clarification. It just seems that it would be a such a great lost to science if the "properties" (by this I mean the LHC collision patterns) of these 20 new particles predicted by your version of E8 are the types of patterns that CERN will never capture and store. Could you please provide some input on nature of the:
plenty of LHC observations which wouldn't be compatible with this [E8] theory

 Related Discussions Beyond the Standard Model 208 General Discussion 20 General Discussion 4 General Physics 0 Medical Sciences 22


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# 2.7: Example Problems

## 2.7.1 Plane Symmetry.

### (1) A Uniformly Charged Plane.

Consider a plane which is infinite in extent and uniformly charged with a density of σ Coulombs/m2 ; the normal to the plane lies in the z-direction, Figure (2.7.6). The charge plane is located at z=0. It is clear from symmetry that the electric field can have only a component normal to the plane since for every charge element at x,y there is an exactly similar charge element at -x,-y such that the transverse field components at the point of observation cancel. All of the charges on a ring of radius r produce the same z-component of electric field at the point of observation, P; they may therefore be lumped together to obtain

$\left.d E_{p}\right|_{z}=\frac{d q}{4 \pi \epsilon_{0}} \frac{\cos (\theta)}{R^{2}}=\frac{\sigma 2 \pi r d r}{4 \pi \epsilon_{0}} \frac{z}{R^{3}}, \nonumber$

therefore

$E_{z}=\left(\frac{\sigma}{2 \epsilon_{0}}\right) z \int_{0}^{\infty} \frac{r d r}{\left(z^{2}+r^{2}\right)^{3 / 2}}. \nonumber$

Upon carrying out the integration one obtains an electric field that is independent of z and, for positive z, has the value $$\sigma / 2 \epsilon_{0}$$:

$E_{z}=\frac{\sigma}{2 \epsilon_{0}}. \label{2.25}$

Note that for points to the left of the charge plane the electric field points along -z: ie. $$E_{z}=-\sigma / 2 \epsilon_{0}$$. There is a discontinuity of magnitude $$\sigma / \epsilon_{0}$$ in the z-component of the electric field at the charge plane.

### (2) The Potential Function for a Uniformly Charged Plane.

It is interesting to calculate the potential function from a direct application of Equation (2.2.4) for the potential generated by a charge distribution. Referring to Figure (2.7.6) one can calculate the contribution to the potential at the observation point P:

$d V_{p}=\frac{d q}{4 \pi \epsilon_{0}} \frac{1}{R}=\frac{\sigma 2 \pi r d r}{4 \pi \epsilon_{0}} \frac{1}{\sqrt{r^{2}+z^{2}}}\nonumber$

and therefore for z>0

$V_{p}=\frac{\sigma}{2 \epsilon_{0}} \int_{0}^{\infty} \frac{r d r}{\sqrt{r^{2}+z^{2}}}=\frac{\sigma}{2 \epsilon_{0}}[\sqrt{\infty}-z]. \nonumber$

This is an example of an integration that does not converge. In order to obtain a proper integral one can use a trick. Suppose that the charged plane was not infinite in extent, but that it was in the form of a large, but finite, disc having a radius D. There would then be no problem; one would have

$V_{p}=\frac{D \sigma}{2 \epsilon_{0}}-\frac{z \sigma}{2 \epsilon_{0}}=\text { constant }-\frac{z \sigma}{2 \epsilon_{0}}.\nonumber$

One is always free to subtract a constant from the potential without changing the value of the corresponding electric field distribution. One can simply subtract the constant from the above equation to obtain

$V_{p}=-\frac{z \sigma}{2 \epsilon_{0}}, \quad \text { for } z>0. \nonumber$

Notice that Vp does not change sign for z less than zero because the integration involves the square root of z2 . Thus the complete expression for the potential function, valid for all values of z, is given by

$V_{p}=-\frac{\sigma}{2 \epsilon_{0}}|z|, \label{2.26}$

see Figure (2.7.7). In Equation (\ref{2.26}) the zero for the potential function has been chosen so that the potential is zero on the plane. The potential function is continuous as the field point P moves through the plane from left to right in Figure (2.7.7): the electric field component normal to the plane undergoes a discontinuity.

### (3) The Field of a Uniformly Charged Plane Using Gauss’ Theorem.

We could have deduced that $$\vec E$$ is independent of position, except at the charge plane, directly from the fourth Maxwell equation $$\operatorname{div}(\overrightarrow{\mathrm{E}})=\left(1 / \epsilon_{0}\right)\left(\rho_{f}-\operatorname{div}(\overrightarrow{\mathrm{P}})\right)$$. In this application there is no electric dipole density since $$\vec P$$ = 0 everywhere. Consequently, also $$\operatorname{div}(\overrightarrow{\mathrm{P}})=0$$ everywhere.

As we have seen in (1) above using symmetry arguments $$\vec E$$ has only a z-component; moreover, this component, Ez, cannot depend upon x,y if the charge distribution is uniform and infinite in extent: any shift of the infinite charge distribution in the x-y plane can not be detected by a fixed observer. Therefore, in this case,

$\operatorname{div}(\overrightarrow{\mathrm{E}})=\frac{\partial E_{z}}{\partial z}=0 \nonumber$

except at z=0 since the charge density is zero everywhere except at z=0. Thus Ez = constant everywhere except at z=0.

Moreover, it is obvious from Coulomb’s law and symmetry that the magnitude of the electric field for z < 0 must have the same value as the magnitude of the electric field for z > 0, although the direction of the field switches 180 degrees upon passage through the charge plane. For this very symmetric case one can obtain the magnitude of the electric field directly from Gauss’ Theorem.

$\int \int \int_{V} d V d i v(\overrightarrow{\mathrm{E}})=\int \int_{S} d S(\overrightarrow{\mathrm{E}} \cdot \hat{\mathbf{n}})=\frac{Q}{\epsilon_{0}}, \nonumber$

where V is the volume bounded by the closed surface S and Q is the total charge contained in the volume. Apply this theorem to a parallelepiped having its edges oriented along the co-ordinate axes as shown in Figure (2.7.8). The box shown in the figure contains an amount of charge σAz Coulombs since the area of the side of the box whose normal lies along z is just Az. Now calculate the surface integral of the electric field over the surface of the

parallelepiped. This integral is easy to carry out because the electric field is constant in magnitude and in direction. Over four sides of the box shown in Figure (2.7.8) the direction of the electric field is parallel with the surface, thus perpendicular to the surface normals, and the scalar product of the electric field and the normal to the surface, $$\hat{\mathbf{n}}$$, vanishes so that these sides contribute nothing to the surface integral. Over the two ends of the box the electric field is parallel with the surface normal so that the scalar product $$\vec E$$· $$\hat{\mathbf{n}}$$ just becomes an ordinary product, and the contribution to the surface integral from each end is E0Az: notice that at each end the electric field is parallel with the direction of the outward normal because the field reverses direction from one side of the charge sheet to the other. The surface integral in Gauss’ theorem is given by

$\int \int_{S} d S(\overrightarrow{\mathrm{E}} \cdot \hat{\mathbf{n}})=2 E_{0} A_{z}. \nonumber$

But the total charge contained in the box is σAz, so that the electric field must have the magnitude $$E_{0}=\sigma /\left(2 \epsilon_{0}\right)$$ in agreement with Equation (\ref{2.25}).

### (4) An Electric Double Layer.

Consider two oppositely charged uniform charge sheets separated by a distance of 2d meters, as illustrated in Figure (2.7.9). The electric field generated by each charge sheet is uniform, independent of z, and directed normal to the planes of charge. In the regions outside the charged planes,ie. z > d or z < d, the electric field is zero because the fields generated by the oppositely charged planes have opposite directions and therefore cancel. Between the two charged planes the fields due to the two planes have the same orientation and therefore add to produce the total field $$E_{z}=-\sigma / \epsilon_{0}$$. A plot of Ez vs z exhibits discontinuities at z= d, Figure (2.7.10). The potential function outside the double layer is constant corresponding to zero electric field. However, the potential on the right hand side of the double layer is different from the potential on the left hand side. The step in the potential is related to the strength of the double layer:

$\Delta V=\frac{2 \sigma d}{\epsilon_{0}}=P_{d} / \epsilon_{0}, \nonumber$

where Pd is the dipole moment per unit area in Coulombs/meter.

### (5) A Uniformly Polarized Slab.

Consider a slab that is uniformly polarized along the z-axis as shown in Figure (2.7.11). The strength of the polarization density is P0, and there are no free charges anywhere. One can define a bound charge density from the relation ρb = −div($$\vec P$$). This bound charge density generates an electric field just as surely as does the free charge density, ρf . Inside the slab of Figure (2.7.11) the polarization density is $$\overrightarrow{\mathrm{P}}=\mathrm{P}_{0} \hat{\mathbf{u}}_{z}$$; thus ∂Pz/∂z is zero inside the slab and therefore div($$\vec P$$), and hence ρb, is zero. Outside the slab $$\vec P$$ = 0 so that here also div($$\vec P$$) = 0 and therefore ρb = 0. One might be misled by the fact that the bound charge density vanishes both inside and outside the slab into thinking that the bound charge density is zero everywhere. However, the bound charge density does not vanish on the slab surfaces. The derivative ∂Pz/∂z is singular at z=0 and at z=d ie. on the faces of the slab. This singularity is integrable: if one integrates the derivative from z = −$$\epsilon$$ to

z = +$$\epsilon$$, for example, the result is

$\int_{-\epsilon}^{+\epsilon}\left(\frac{\partial \mathrm{P}_{z}}{\partial z}\right) d z=\mathrm{P}_{z}(+\epsilon)-\mathrm{P}_{z}(-\epsilon)=\mathrm{P}_{0}, \nonumber$

because Pz(−$$\epsilon$$) = 0 and Pz(+$$\epsilon$$) = P0. Notice that the value of the integral is independent of the small interval ǫ, and, in particular, it remains finite even in the limit as $$\epsilon$$ → 0. The integrand has the character of a Dirac δ − function: $$\partial \mathrm{P}_{z} / \partial z=\mathrm{P}_{0} \delta(z)$$, where the function δ(z) is a strongly peaked function having zero width but infinite amplitude in such a way that it’s integral is just equal to unity. But this means that the charge density on the surface at z=0, ρb = −∂Pz/∂z, z, is a very sharply peaked integrable function of z: it is in fact a surface charge density of strength −P0 Coulombs/meter2 . Similarly, there will be a surface charge density of strength +P0 Coulombs/meter2 on the surface at z=d. The electric field distribution produced by a uniformly polarized slab in which the polarization density lies parallel with the slab normal is exactly the same as that which would be produced by two uniformly charged planes carrying charge densities of σ = ±P0, see Figure (2.7.9). Outside the uniformly polarized slab the electric field is zero. Inside the slab there is a uniform electric field, $$\mathrm{E}_{z}=-\mathrm{P}_{0} / \epsilon_{0}$$, whose direction is opposite to the direction of the polarization density; it is called a depolarizing field because it tends to act so as to reduce the polarization density. The potential outside the slab will be constant, both on the left and on the right, but the potential on the right will be larger than that on the left by the amount $$\Delta V=\mathrm{P}_{0} d / \epsilon_{0}$$ Volts. It is interesting to examine the auxiliary field $$\overrightarrow{\mathrm{D}}=\epsilon_{0} \overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{P}}$$. Outside the slab both $$\vec E$$ and $$\vec P$$ are zero so that $$\vec D$$ must also be zero. Inside the slab $$\vec D$$ has only a z-component, and

$\mathrm{D}_{z}=\epsilon_{0} \mathrm{E}_{z}+\mathrm{P}_{z}=-\mathrm{P}_{0}+\mathrm{P}_{0} \equiv 0. \nonumber$

The normal component of $$\vec D$$ is continuous across the slab surfaces. In general, in the absence of any surface free charge density the normal component of $$\vec D$$ must be continuous across the interface.

Now consider a uniformly polarized slab in which the polarization density vector lies in the plane of the slab as shown in Figure (2.7.12). In this case the polarization has only an x-component, and that x-component, Px, is a function only of z. This means that $$\operatorname{div}(\overrightarrow{\mathrm{P}})=\partial \mathrm{P}_{x} / \partial x=0$$ everywhere. There is no bound charge density anywhere, and there is no free charge density, by supposition, so that there are no sources for the electric field. A uniformly

polarized slab in which the polarization lies in the slab plane generates no macroscopic electric field.

It is a general rule that a spatial variation of the polarization density, $$\vec P$$, produces an effective volume charge density $$\rho_{b}=-\operatorname{div}(\overrightarrow{\mathrm{P}})$$. In addition to this effective volume charge density, any discontinuity in the normal component of $$\vec P$$ across a surface produces an effective surface charge density given by

$\sigma_{b}=-\left(\mathrm{P}_{2 n}-\mathrm{P}_{1 n}\right). \nonumber$

These bound volume and surface charge densities must be used, along with the free charge distributions, to calculate the electric field and potential distributions.

### 2.7.2 A Spherically Symmetric Charge Distribution.

Consider a charge distribution that is spherically symmetric but one in which the charge distribution ρ(r) may have an arbitrary dependence upon the co-ordinate r, see Figure (2.7.13). In this case the electric field can have only a radial component by symmetry. The magnitude of this radial component, Er, cannot depend on position on the surface of a spherical surface centered on

the center of symmetry of the charges because any rotation of the distribution around the center of symmetry leaves the charge distribution unaltered. It follows that the surface integral of the electric field over the surface of a sphere of radius R is just $$4 \pi R^{2} \mathrm{E}_{r}) so that from Gauss’ theorem $\mathrm{E}_{r}=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{1}{R^{2}}\right) \int_{0}^{R} d r \rho(r) 4 \pi r^{2}. \nonumber$ Thus for the electric field outside the charge distribution one has $4 \pi R^{2} \mathrm{E}_{r}=\frac{1}{\epsilon_{0}} \int_{S p h e r e} \rho(r) d V o l=\frac{Q}{\epsilon_{0}}. \label{2.27}$ Here Q is the total charge contained in the distribution. The electric field outside a spherically symmetric charge distribution looks like the field of a point charge! The potential function generated by any symmetric charge distribution must be independent of the spherical polar co-ordinates θ, φ because Eθ and Eφ are both zero. The potential function can be calculated from ∂V/∂r = −Er. Note that the potential must be a continuous function of r even if the charge density distribution, ρ(r), contains discontinuities. The potential function is continuous everywhere except on a surface containing an electrical double layer. ## 2.7.3 Cylindrical Symmetry. ### (1) A uniformly Charged Line. Let charges be distributed uniformly along the z-axis with a charge density of ρL Coulombs/meter as shown in Figure (2.7.14). It is easy to see, using Coulomb’s law, that the electric field generated by this distribution can have only a radial component; the transverse components generated by equivalent charge elements symmetrically disposed around the origin at +z and at -z cancel each other out. Furthermore, the radial electric field strength,Er, cannot depend upon the angle θ because the line charge exhibits rotational symmetry; ie. the charge distribution remains unaltered by a rotation through any angle around the z-axis. Er also cannot depend upon position along z since the line is taken to be infinitely long. Apply Gauss’ theorem to a cylindrical surface centered on the line charge, and 1 meter long and having a radius of r meters. The charge contained within this cylinder is ρL Coulombs. The surface integral of the electric field is easy to carry out because the electric field is parallel with the surface normal at every point on the cylinder surface; on the end surfaces the electric field contributes nothing to the surface integral because it lies in the surface, i.e. \(\overrightarrow{\mathrm{E}} \cdot \hat{\mathbf{u}}_{n}=0$$, where $$\hat{\mathbf{u}}_{n}$$ is the unit normal to the surface. It follows from Gauss’ Theorem that

$2 \pi r \mathrm{E}_{r}=\frac{\rho_{L}}{\epsilon_{0}}, \nonumber$

and therefore

$\mathrm{E}_{r}=\frac{\rho_{L}}{2 \pi \epsilon_{0}}\left(\frac{1}{r}\right) \quad \operatorname{Volts} / m. \nonumber$

Since the electric field has only a radial component it follows that the corresponding potential function, V, cannot depend upon the co-ordinates z,θ. But Er = −∂V/∂r, so that one can use

$\mathrm{V}(r)=-\left(\frac{\rho_{L}}{2 \pi \epsilon_{0}}\right) \ln (r)\nonumber$

as the potential function for a uniformly charged line. The logarithmic variation of V(r) with r can also be deduced from a direct application of Poisson’s equation, Equation (2.2.5). In this application the free charge density ρf = 0 everywhere except at the origin, r=0. There is assumed to be no electric dipole distribution anywhere so that $$\vec P$$ = 0 and ρb = −div($$\vec P$$) = 0 everywhere. Thus ∇2V = 0. But in cylindrical polar co-ordinates, and in the absence of any variation of V with angle or with displacement along z

$\nabla^{2} \mathrm{V}=\frac{\partial}{\partial r}\left(r \frac{\partial \mathrm{V}}{\partial r}\right)=0. \nonumber$

It follows from this that r∂V/∂r = constant = a, and therefore V = a ln(r) + constant in agreement with the above expression for V(r): remember that one can add or subtract a constant from the potential function without changing the electric field calculated from it.

### (2) A Line of Dipoles.

Consider a line of dipoles uniformly distributed along the z-axis with a density of PL Coulombs; the dipoles are supposed to be oriented along the x-direction

as shown in Figure (2.7.15). The line of dipoles can be modelled by two line charges separated by a very small distance d, as shown in Figure (2.7.15). The resulting potential function at P, the point of observation, written in cylindrical polar co-ordinates can be calculated as follows using the potential function for a uniformly charged line:

$\mathrm{V}_{P}=-\frac{\rho_{L}}{2 \pi \epsilon_{0}} \ln \left(r_{+}\right)+\frac{\rho_{L}}{2 \pi \epsilon_{0}} \ln \left(r_{-}\right)=\frac{\rho_{L}}{2 \pi \epsilon_{0}} \ln \left(r_{-} / r_{+}\right). \nonumber$

But $$r_{-} \cong r_{+}+d \cos (\theta)$$ so that

$\mathrm{V}_{P}=\frac{\rho_{L}}{2 \pi \epsilon_{0}} \ln \left(1+\frac{d \cos (\theta)}{r_{+}}\right)\nonumber$

and

$\lim _{d \rightarrow 0} \mathrm{V}_{P}=\frac{\rho_{L} d}{2 \pi \epsilon_{0}} \frac{\cos (\theta)}{r}=\frac{1}{2 \pi \epsilon_{0}} \frac{\overrightarrow{\mathrm{P}}_{L} \cdot \vec{r}}{r^{2}}, \nonumber$

where PL is the line density of dipoles. The electric field components are

$\mathrm{E}_{r}=-\frac{\partial \mathrm{V}_{P}}{\partial r}=\frac{\mathrm{P}_{L}}{2 \pi \epsilon_{0}} \frac{\cos (\theta)}{r^{2}},\nonumber$

$\mathrm{E}_{\theta}=-\frac{1}{r} \frac{\partial \mathrm{V}_{P}}{\partial \theta}=\frac{\mathrm{P}_{L}}{2 \pi \epsilon_{0}} \frac{\sin (\theta)}{r^{2}},\nonumber$

$\mathrm{E}_{z}=0. \nonumber$

A cross-section through the electric field distribution in a plane normal to the line of dipoles looks similar to the field distribution around a point dipole. The electric field at θ = 0 is directed along the +x direction and so is the field at θ = $$\pi$$: both fields have the strength $$\mathrm{P}_{L} /\left(2 \pi \epsilon_{0} r^{2}\right)$$. The electric fields at θ = $$\pi$$/2, 3$$\pi$$/2 are directed along the -x direction, and have the strength $$\mathrm{E}_{\theta}=\mathrm{P}_{L} /\left(2 \pi \epsilon_{0} r^{2}\right)$$.

### 2.7.4 A Uniformly Polarized Ellipsoidal Body.

Consider a uniformly polarized body of arbitrary shape that is immersed in vacuum. The bound volume charge density associated with a uniform polarization density is zero since div($$\vec P$$) = 0. The bound surface charge density is not zero because there is a discontinuity in the normal component of $$\vec P$$ on the surface of the body; this surface charge density is given by $$\sigma_{b}=\overrightarrow{\mathrm{P}}_{0} \cdot \hat{\mathbf{n}}$$ where $$\hat{\mathbf{n}}$$ is the unit vector normal to the surface of the body. Notice that this surface charge density varies from place to place on the surface and that it

changes sign when one considers opposite points on the surface, see Figures (2.7.16 and 2.7.17). In fact, it is a consequence of charge conservation that the sum over all surface charges on a body that carries no free charges must be zero. The surface bound charge density distribution can be used to calculate the electric field and potential function everywhere in space. An analytical calculation is usually out of the question, and the problem must usually be solved by means of a numerical summation. One could, in principle, calculate the electric field components by means of Coulomb’s law, but it is usually more convenient to work with the integral for the potential function, Equation (2.2.6). For the particular case in which the uniformly polarized body has an ellipsoidal shape the calculation of the potential function and the electric field can be carried out analytically. See, for example, J.A. Stratton, Electromagnetic Theory, McGraw-Hill, N.Y., 1941, section 3.25. The surprising result is that the electric field inside a uniformly polarized ellipsoid is uniform. Usually this internal electric field is not parallel with the direction of the polarization density. The internal electric field and the polarization are parallel only if the polarization vector is directed along a principal axis of the ellipsoid defined by the equation for its surface

$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}+\left(\frac{z}{c}\right)^{2}=1. \label{2.28}$

Here a,b,c are the three semi-axes that define the ellipsoid. Outside the ellipsoid the electric field is not uniform: it resembles a distorted dipole field and falls off at large distances like 1/r3 . In the principle axis co-ordinate system defined by Equation (\ref{2.28}) the electric field components inside the ellipsoid are simply related to the polarization components:

$\mathrm{E}_{x} =-\mathrm{N}_{x} \frac{\mathrm{P}_{x}}{\epsilon_{0}}, \label{2.29}$
$\mathrm{E}_{y} =-\mathrm{N}_{y} \frac{\mathrm{P}_{y}}{\epsilon_{0}}, \nonumber$
$\mathrm{E}_{z} =-\mathrm{N}_{z} \frac{\mathrm{P}_{z}}{\epsilon_{0}}. \nonumber$

The depolarization coefficients Nx, Ny, and Nz are pure numbers that depend upon the parameters of the ellipsoid (a,b,c of Equation (\ref{2.28})). They can be calculated using elliptic integrals. The depolarization coefficients obey a sum rule:

$\mathrm{N}_{x}+\mathrm{N}_{y}+\mathrm{N}_{z}=1. \label{2.30}$

This sum rule makes it very easy to deduce values for the depolarization coefficients for very symmetrical bodies. Consider the following examples:

(a) Sphere. By symmetry Nx = Ny = Nz. Therefore from the sum rule each must be equal to 1/3

(b) Cylinder. This is the limiting case in which one dimension of the ellipsoid, the z dimension say, becomes very long. The two transverse depolarizing factors must be equal by symmetry. On the other hand, if the cylinder is polarized along its length any surface bound charge density can only be associated with the ends; but the ends are infinitely far away and consequently any charges on them produce an infinitely small electric field. This means that a cylinder that is uniformly polarized along its length will produce no electric field. It can be concluded that for such a cylinder Nz = 0. It follows from the sum rule that if Nx = Ny then each must be equal to 1/2.

(c) A Thin Flat Disc. Think of an ellipsoid of revolution which is thin along the z-direction but which has a very large radius R in the xy plane. By symmetry Nx = Ny . If the edges of the disc are very far from its center, ie. R → ∞, then the electric field near the center of the disc due to surface charges on the disc edges must become vanishingly small ( remember that

the field due to an element of charge decreases like 1/R2 ). It follows that Nx = Ny → 0 in the limit as R → ∞. From the sum rule Equation (\ref{2.30}) one can conclude that Nz → 1. The field inside an infinite disc that is polarized parallel with its axis of symmetry has the value $$\mathrm{E}_{z}=-\mathrm{P}_{z} / \epsilon_{0}$$ in agreement with the result previously deduced for a transversely polarized slab, section (5) above, and Figure (2.7.11).

Two commonly encountered special cases of ellipsoids of revolution are shown in Figures (2.7.18 and 2.7.19). In each case one need only specify the depolarizing coefficient for the axis of revolution, the z-axis. The other two depolarizing factors are equal and can be calculated from the sum rule Equation (\ref{2.30}). Case(a) is a pancake shaped ellipsoid, Figure (2.7.18). For this case

$\mathrm{N}_{z}=\frac{\mathrm{R}^{2} \mathrm{d}}{\left(\mathrm{R}^{2}-\mathrm{d}^{2}\right)^{3 / 2}}\left(\frac{\sqrt{\mathrm{R}^{2}-\mathrm{d}^{2}}}{\mathrm{d}}-\arctan \left(\frac{\sqrt{\mathrm{R}^{2}-\mathrm{d}^{2}}}{\mathrm{d}}\right)\right), \quad \text { where }(\mathrm{d} / \mathrm{R})<1. \label{2.31}$

In the limit d/R → 0 the depolarizing factor is given approximately by

$\mathrm{N}_{z} \cong 1-\frac{\pi}{2}\left(\frac{\mathrm{d}}{\mathrm{R}}\right). \label{2.32}$

Case(b) is a cigar shaped ellipsoid, Figure (2.7.19). For this case

$\mathrm{N}_{z}=\frac{\left(1-\psi^{2}\right)}{\psi^{3}}\left(\frac{1}{2} \ln \left(\frac{1+\psi}{1-\psi}\right)-\psi\right), \text { where } \psi=\sqrt{1-(\mathrm{R} / \mathrm{d})^{2}}. \label{2.33}$

In the limit as the cigar becomes very long, (d/R) → ∞ the demagnetizing coefficient can be expressed as

$\mathrm{N}_{z} \cong\left(\frac{\mathrm{R}}{\mathrm{d}}\right)^{2}\left(\ln \left[\frac{2 \mathrm{d}}{\mathrm{R}}\right]-1\right).\label{2.34}$

For a general ellipsoid it can be shown that

$\begin{array}{l} \mathrm{N}_{x}=\left(\frac{a b c}{2}\right) \int_{0}^{\infty} \frac{d s}{\left(s+a^{2}\right) R_{s}} \\ \mathrm{N}_{y}=\left(\frac{a b c}{2}\right) \int_{0}^{\infty} \frac{d s}{\left(s+b^{2}\right) R_{s}} \\ \mathrm{N}_{z}=\left(\frac{a b c}{2}\right) \int_{0}^{\infty} \frac{d s}{\left(s+c^{2}\right) R_{s}} \end{array}$

where $$R_{s}=\left(\left[s+a^{2}\right]\left[s+b^{2}\right]\left[s+c^{2}\right]\right)^{1 / 2}$$.

(See J.A. Stratton, Electromagnetic Theory, McGraw-Hill, N.Y., 1941. Section 3.27.)


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