将对象实例化为具有命名空间的类

我不知道该怎么解释，但我会尽力的。好的，我有这三个文件:

• Theme.php

  path: /shared/models/Theme.php
  class: Theme
  namespace: namespace models;
  

• Custom.php

  path: /themes/default/Custom.php
  class: Custom
  namespace: this class does not use namespace
  

• Settings.php

  path: /controllers/Settings.php
  class: Settings
  namespace: this class does not use namespace
  

在我的Settings.php中看起来像:

<?php
class Settings
{
    public function apply()
    {
        $theme = new 'models'Theme();
        $theme->customize(); // in this method the error is triggered
    }
}

现在看下面的Theme类:

<?php
namespace models;
class Theme
{
    public function customize()
    {
        $ext = "/themes/default/Custom.php";
        if (file_exists($ext))
        {
            include $ext;
            if (class_exists('Custom'))
            {            
                $custom = new Custom(); 
                //Here, $custom var in null, why???
            }
        }
    }
}

Message: require(/shared/models/Custom.php) [function.require]: failed to open stream: No such file or directory
Line Number: 64

为什么解释器试图从另一个目录加载Custom类，而不是用$ext var指定?