为什么我们调用 start() 方法时会执行 run() 方法，为什么我们不能直接调用 run()方法？

如果我们直接调用 run() 方法 是不会创建新线程的。
start() 源代码：

/**
     * Causes this thread to begin execution; the Java Virtual Machine
     * calls the <code>run</code> method of this thread.
     * <p>线程被执行，JVM调用run方法
     * The result is that two threads are running concurrently: the
     * current thread (which returns from the call to the
     * <code>start</code> method) and the other thread (which executes its
     * <code>run</code> method).
     * <p>
     * It is never legal to start a thread more than once.多次调用start方法启动一个线程是非法的
     * In particular, a thread may not be restarted once it has completed
     * execution.
     *
     * @exception  IllegalThreadStateException  if the thread was already已经启动的线程再次start，异常
     *               started.
     * @see        #run()
     * @see        #stop()
     */
    public synchronized void start() {
        /**
         * This method is not invoked for the main method thread or "system"
         * group threads created/set up by the VM. Any new functionality added
         * to this method in the future may have to also be added to the VM.
         *
         * A zero status value corresponds to state "NEW".
         */
        if (threadStatus != 0)//状态校验  0：NEW 新建状态
            throw new IllegalThreadStateException();

        /* Notify the group that this thread is about to be started
         * so that it can be added to the group's list of threads
         * and the group's unstarted count can be decremented. */
        group.add(this);//添加进线程组

        boolean started = false;
        try {
            start0();//调用native方法执行线程run方法
            started = true;
        } finally {
            try {
                if (!started) {
                    group.threadStartFailed(this);//启动失败，从线程组中移除当前前程。
                }
            } catch (Throwable ignore) {
                /* do nothing. If start0 threw a Throwable then
                  it will be passed up the call stack */
            }
        }
    }

    private native void start0();

通过源码我们可以看到，在start()中调用了native方法 start0()，在start0()中会创建一个新的线程调用run()。 所以，如果我们直接调用 run() 方法 是不会创建新线程的。