If I don't use namespaces, should I care about any of this?

No. Namespaces do not affect any existing code in any way, or any as-yet-to-be-written code that does not contain namespaces. You can write this code if you wish:

<?php
$a = new \stdClass;
?>

This is functionally equivalent to:

<?php
$a = new stdClass;
?>

How do I use internal or global classes in a namespace?

<?php
namespace foo;
$a = new \stdClass;

function test(\ArrayObject $parameter_type_example = null) {}

$a = \DirectoryIterator::CURRENT_AS_FILEINFO;

// extending an internal or global class
class MyException extends \Exception {}
?>

How do I use namespaces classes, functions, or constants in their own namespace?

<?php
namespace foo;

class MyClass {}

// using a class from the current namespace as a parameter type
function test(MyClass $parameter_type_example = null) {}
// another way to use a class from the current namespace as a parameter type
function test(\foo\MyClass $parameter_type_example = null) {}

// extending a class from the current namespace
class Extended extends MyClass {}

// accessing a global function
$a = \globalfunc();

// accessing a global constant
$b = \INI_ALL;
?>

How does a name like \my\name or \name resolve?

Names that begin with a \ always resolve to what they look like, so \my\name is in fact my\name, and \Exception is Exception.

<?php
namespace foo;
$a = new \my\name(); // instantiates "my\name" class
echo \strlen('hi'); // calls function "strlen"
$a = \INI_ALL; // $a is set to the value of constant "INI_ALL"
?>

How does a name like my\name resolve?

Names that contain a backslash but do not begin with a backslash like my\name can be resolved in 2 different ways.

If there is an import statement that aliases another name to my, then the import alias is applied to the my in my\name.

Otherwise, the current namespace name is prepended to my\name.

<?php
namespace foo;
use blah\blah as foo;

$a = new my\name(); // instantiates "foo\my\name" class
foo\bar::name(); // calls static method "name" in class "blah\blah\bar"
my\bar(); // calls function "foo\my\bar"
$a = my\BAR; // sets $a to the value of constant "foo\my\BAR"
?>

How does an unqualified class name like name resolve?

Class names that do not contain a backslash like name can be resolved in 2 different ways.

If there is an import statement that aliases another name to name, then the import alias is applied.

Otherwise, the current namespace name is prepended to name.

<?php
namespace foo;
use blah\blah as foo;

$a = new name(); // instantiates "foo\name" class
foo::name(); // calls static method "name" in class "blah\blah"
?>

How does an unqualified function name or unqualified constant name like name resolve?

Function or constant names that do not contain a backslash like name can be resolved in 2 different ways.

First, the current namespace name is prepended to name.

Finally, if the constant or function name does not exist in the current namespace, a global constant or function name is used if it exists.

<?php
namespace foo;
use blah\blah as foo;

const FOO = 1;

function my() {}
function foo() {}
function sort(&$a)
{
    \sort($a); // calls the global function "sort"
    $a = array_flip($a);
    return $a;
}

my(); // calls "foo\my"
$a = strlen('hi'); // calls global function "strlen" because "foo\strlen" does not exist
$arr = array(1,3,2);
$b = sort($arr); // calls function "foo\sort"
$c = foo(); // calls function "foo\foo" - import is not applied

$a = FOO; // sets $a to value of constant "foo\FOO" - import is not applied
$b = INI_ALL; // sets $b to value of global constant "INI_ALL"
?>

Import names must not conflict with classes defined in the same file.

The following script combinations are legal:

<?php
namespace my\stuff;
class MyClass {}
?>

<?php
namespace another;
class thing {}
?>

<?php
namespace my\stuff;
include 'file1.php';
include 'another.php';

use another\thing as MyClass;
$a = new MyClass; // instantiates class "thing" from namespace another
?>

There is no name conflict, even though the class MyClass exists within the my\stuff namespace, because the MyClass definition is in a separate file. However, the next example causes a fatal error on name conflict because MyClass is defined in the same file as the use statement.

<?php
namespace my\stuff;
use another\thing as MyClass;
class MyClass {} // fatal error: MyClass conflicts with import statement
$a = new MyClass;
?>

Nested namespaces are not allowed.

PHP does not allow nesting namespaces

<?php
namespace my\stuff {
    namespace nested {
        class foo {}
    }
}
?>

<?php
namespace my\stuff\nested {
    class foo {}
}
?>

Dynamic namespace names (quoted identifiers) should escape backslash

It is very important to realize that because the backslash is used as an escape character within strings, it should always be doubled when used inside a string. Otherwise there is a risk of unintended consequences:

<?php
$a = "dangerous\name"; // \n is a newline inside double quoted strings!
$obj = new $a;

$a = 'not\at\all\dangerous'; // no problems here.
$obj = new $a;
?>

Undefined Constants referenced using any backslash die with fatal error

Any undefined constant that is unqualified like FOO will produce a notice explaining that PHP assumed FOO was the value of the constant. Any constant, qualified or fully qualified, that contains a backslash will produce a fatal error if not found.

<?php
namespace bar;
$a = FOO; // produces notice - undefined constants "FOO" assumed "FOO";
$a = \FOO; // fatal error, undefined namespace constant FOO
$a = Bar\FOO; // fatal error, undefined namespace constant bar\Bar\FOO
$a = \Bar\FOO; // fatal error, undefined namespace constant Bar\FOO
?>

Cannot override special constants null, true or false

Any attempt to define a namespaced constant that is a special, built-in constant results in a fatal error

<?php
namespace bar;
const NULL = 0; // fatal error;
const true = 'stupid'; // also fatal error;
// etc.
?>