Thread references require static lifetime?

While it makes sense intuitively that references passed to spawned threads need to have static lifetimes, I'm unclear about what exactly is making the following code not compile:

use std::sync::Arc;
use std::sync::Mutex;

struct M;

fn do_something(m : Arc<Mutex<&M>>) {
    println!("Ha, do nothing!");
}

fn main() {
    let a = M;
    {
        let c : Arc<Mutex<&M>> = Arc::new(Mutex::new(&a));
        for i in 0..2 {
            let c_clone = c.clone();
            ::std::thread::spawn(move || do_something(c_clone));
        }
    }
}

Compiling this small program gives the following error:

$ rustc -o test test.rs
test.rs:13:55: 13:56 error: `a` does not live long enough
test.rs:13         let c : Arc<Mutex<&M>> = Arc::new(Mutex::new(&a));
                                                             ^
note: reference must be valid for the static lifetime...

It seems to me that the variable a will out-live c_clone, which is what matters in this case...? Hopefully someone can help me understand what I'm missing!

In its essence, the Arc and Mutex wrapping is superfluous: you are passing a reference to something on the local stack. When you spawn a thread with std::thread::spawn, there is nothing linking the lifetimes together; the main thread is quite at liberty to conclude and free anything in it—in this case, including a—before any other threads it spawns even start executing; thus in this case a could refer to freed memory by the time the spawned thread does anything, leaving c_clone as a dangling pointer. This is why the environment of the closure of a spawned thread must be 'static.

Thread references require static lifetime?

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