A function $a$ defined on $T^*\mathbb{R}^n$ is called a classical symbol of order $m$ provided it is smooth and there exists $\tilde{a}\in C^\infty(\mathbb{R}^n\times S^{n-1}\times\mathbb{R}_+)$ so that for all $(x,\xi)\in T^*\mathbb{R}^n$ with $|\xi|>1$, $$a(x,\xi)=|\xi|^m \tilde{a}\left(x,\frac{\xi}{|\xi|}, \frac{1}{|\xi|}\right).$$
I find this definition confusing, as I don't seem why the order is well-defined. For example, say that $$a(x,\xi)=|\xi|^2.$$ I'd like to say that the order is $2$. But, we can write $$a(x,\xi)=|\xi|^2\cdot \tilde{a}\left(x,\frac{\xi}{|\xi|}, \frac{1}{|\xi|}\right),$$ where $$\tilde{a}\left(x,\frac{\xi}{|\xi|}, \frac{1}{|\xi|}\right)=1,$$ or $$a(x,\xi)=|\xi|^3\tilde{a}\left(x,\frac{\xi}{|\xi|}, \frac{1}{|\xi|}\right),$$
where $$\tilde{a}\left(x,\frac{\xi}{|\xi|}, \frac{1}{|\xi|}\right)=\frac{1}{|\xi|},$$ or even
$$a(x,\xi)=|\xi|\tilde{a}\left(x,\frac{\xi}{|\xi|}, \frac{1}{|\xi|}\right),$$
where $$\tilde{a}\left(x,\frac{\xi}{|\xi|}, \frac{1}{|\xi|}\right)=\frac{1}{|\xi|^{-1}}={|\xi|}.$$
All of the conditions seem to be satisfied, so how is the order well-defined?
If this is a silly question, feel free to leave a comment, and I'll delete it.