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## Adobe Photoshop Elements

Adobe Photoshop Elements is a free image editing program included with your copy of Adobe Photoshop. While it is often touted as an easy and inexpensive way to edit photos, I don’t recommend it for anything serious, but it’s a good budget alternative.


* Less expensive
* Free
* Quick to use
* Has many of the features of a professional version


* Little learning curve, and not as robust
* Can take hours and hours to get used to

Adobe Photoshop Elements is easy to get started with and often comes with some basic templates to get you started.

The following sites have tutorials to get you started:

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The latest version of Photoshop Elements, Photoshop CC 2018 has plenty of exciting features that most users will use: advanced tools and layers, an innovative, easy-to-use workspace, content-aware enhancements, and more.

As a professional photographer or digital designer, you probably do not know Photoshop Elements well. In this post, you’ll get a brief introduction to the Photoshop Elements Editor: what it is and what you can do with it.

Editor Overview

In this part of the Photoshop Elements article series, we will explore the basic functions and tools of the Photoshop Elements Editor.

With a few clicks, you can create new photos, edit photos, or retouch photos in the Photoshop Elements Editor. You can do this by either using the predefined actions or customizing these actions using the built-in actions panel.

Creating New Photos with the Photoshop Elements Editor

It is quite difficult to say the truth about how easy or difficult it is to create new photos with Photoshop Elements. There are two things that you can do.

The first is to use a predefined action or a custom action. We will go into each of these options in detail, so don’t worry if you feel like you don’t know anything about Photoshop Actions. You can learn everything you need to know about the Photoshop Actions Editor in part 3 of this series.

The second method is to use the Photoshop Elements Editor’s built-in Photo Editor. By default, all of the actions on the New Photo panel are set to “Custom,” which allows you to create a new photo from scratch.

Some actions even come with predefined templates so you don’t need to do anything.

Step 1. Open the Photo Editor

First, select Photoshop Elements from the main menu. Then, open the Photo Editor by selecting New > Photo.

Step 2. Choose the Photo

In this step, you can create a new photo from scratch or import a photo from the album.

Step 3. Create a new photo

After you select Photo from the menu, you will notice that the Photo Editor on the right has two icons:

The first icon will take you to the Photo Editor. You can just choose a layout you want (we will take a look at this icon more in step 5).

The second icon will allow you to select from many predefined templates. You can also edit these

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Is there an intuition for why the chiral anomaly precisely cancels the classical electromagnetic Chern-Simons action?

I’m reading an article by Bernard and Leutwyler on the chiral anomaly. It contains the following statement that seems to be a bit wrong (or at least incomplete):

for a theory with $U(1)$ gauge symmetry the classical action is $S = \int d^3 x \mathcal{L}_{EM}$,

where the Lagrangian density in Lorentz gauge is given by
$$\mathcal{L}_{EM} = -\frac{1}{4}F^{\mu
u} – \frac{m^2}{4}A^\mu A_\mu = \frac{1}{2}(\partial \vec{A})^2 + \frac{m^2}{2} \vec{A}^2$$
and it is evaluated in $x^0 = t, x^1 = x, x^2 = y, x^3 = z$.
A similar statement can be found here in 4.3.1 where the authors claim that

The classical equations of motion are $\frac{\partial}{\partial x^\mu} F^{\mu
u} = m^2 A^

which is also not true.
To the best of my understanding, if a classical field $A(x)$ has mass $m$, then we can expect that the equations of motion are $\frac{\partial}{\partial t} F(t,x) = m^2 A(t,x)$. This is indeed true for the classical EM field $\vec{A}$, but it does not imply that the classical action is $\int d^4 x F(t,x)$ (it only implies that the classical field $\vec{A}$ has the same mass $m$).
How can we ensure that the classical EM action is indeed $\int d^4 x F(t,x)$ instead of $\int d^4 x \mathcal{L}_{EM}$?


The classical action for the EM field is indeed given by $\int d^4x \mathcal{L}_{EM}$, where the Lagrangian density is

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define( [

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