One of the genuine powers of the result for the mass-shell condition from the last post is that it enables us to calculate the mass of a string (with arbitrary motion) in the classical theory. It is a fairly straightforward procedure. We start by recalling the mass-energy relation,

[ M^{2} = -p^2 ]

We of course know the right-hand side. We can write,

[ M^{2} = -p^2 = 2p^{+}p^{-} – p^{I}p^{I} ]

[= frac{1}{alpha^{prime}}L_{0}^{perp} – p^{I}p^{I} ]

Now all we need to do is substitute for $L_{0}^{perp}$ and tidy things up,

[ M^{2} = frac{1}{alpha^{prime}}(frac{1}{2}sum_{p in mathbb{Z}} alpha_{p}^{I} alpha_{-p}^{I}) – p^{I}p^{I} ]

[= frac{1}{alpha^{prime}} (frac{1}{2}alpha_{0}^{I}alpha_{0}^{I} + sum_{p geq 1} (alpha_{p}^{I})^{*}(alpha_{p}^{I}) – p^{I}p^{I} ]

[ = frac{1}{alpha^{prime}} sum_{p=1}^{infty} (alpha_{p}^{I})^{*}(alpha_{p}^{I}) geq 0 ]

This is a terrific result in the classical theory. But, ultimately, it is problematic (Wray, 2009) because in physics, and especially in string theory, we want massless particles! The Standard Model is all massless until the introduction of the Higgs. So, obviously, we need massless strings. To that end, notice this expression for the mass of a string can only be massless if all $alpha$ terms are zero. But if all $alpha$ terms are zero, the string behaviour is rendered trivial.

The good news is that in the quantum theory we will see how this formula can be altered in just such a way that it will allow for a lot of massless states.

References

Joseph Polchinski. (2005). “*String Theory: An Introduction to the Bosonic String*“, Vol. 1.

Kevin Wray. (2009). “*An Introduction to String Theory*“.

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