A few weeks ago, I started reading books on string theory. One thing that really seemed confusing or contradictory was that string theory explains that the energy of a superstring gives mass to the particle. Meaning, the more energy the string holds, the more it vibrates the more mass it gives to the particle.

It seems like if a photon doesn't have any mass then the string wouldn't have any energy and therefore wouldn't even exist. My question is, how can a photon exist as a particle according to string theory if it has $0$ mass, which I think implies the superstring has no energy?

  • $\begingroup$ I don't know how string theory specifically handles this but a photon does have energy even though it doesn't have mass. From a standard models perspective, I believe the energy that is manifest as mass has to do with the interaction with the Higgs field. $\endgroup$ Dec 12 '13 at 19:43

It is incorrect to say that the energy of a string directly gives us the mass of the particle. While it is true that more the oscillations on the string, higher the mass, the relation between the oscillations and the mass it not that of a simple proportionality.

What's really happening is that the string has some energy $E$ (due to oscillations on it) and a momentum $\vec{p}$ (since it is moving in space), which is not affected by the oscillations. The mass of the particle that such a string relates to is given by the relativistic dispersion relation $$ m = \sqrt{\frac{E^2}{c^4} - \frac{p^2}{c^2} } $$ where $p^2 = \vec{p}\cdot\vec{p}$.

Now as you can see, it is possible to choose oscillations on the string such that $E^2 = p^2 c^2$ which gives us a zero mass particle. Indeed, this is precisely how string theory recovers massless particles. Now, whether said massless particle is a photon or not depends on some other internal properties of the string itself.

ASIDE: In bosonic string theory, it is possible to have a string with NO EXCITATIONS, in which case $E=0$. One then recovers a particle with imaginary mass called the tachyon. Existence of such tachyons is problematic in any theory, as it leads to instabilities. This problem is solved in superstrings, where certain constraints does not allow one to have strings with no excitations.

  • $\begingroup$ Does this mean that a generic vacuum state for string theory has no zero-mass particles? $\endgroup$ Dec 12 '13 at 20:55
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    $\begingroup$ @peter the zero point energy is typically a linear function of dimension, $D$. We know particular states must be massless, because of their Little group. We must, therefore pick a critical dimension such that the state is massless. $\endgroup$
    – innisfree
    Dec 12 '13 at 21:29
  • $\begingroup$ I don't get it. How do you get E and p to act like independent parameters, and "choose" E to have some value, regardless of p? I would think that E would contain a part that was due to the energy of the oscillations, plus another part that was due to the string's motion through space. What you're describing doesn't seem to make any sense at all compared to how special relativity normally works. $\endgroup$
    – user4552
    Nov 8 '19 at 1:31
  • $\begingroup$ @BenCrowell The momentum $p$ fixes the kinetic energy part of $E$ arising from linear motion of the string. One can then add oscillations to the string while keeping its linear momentum the same and in doing so increase the total energy $E$ of the string keeping $p$ fixed. One then uses the dispersion relation shown above to compute the invariant mass of this configuration. It is in this way that a single string in string theory gives rise to many different particles. By the phrase "choose E", I simply meant "choose the oscillations on the string". $\endgroup$
    – Prahar
    Nov 9 '19 at 14:50
  • $\begingroup$ @BenCrowell - There is, of course, a quantization condition that puts constraints on what type of oscillations the string is allowed to have which therefore translates to a quantization of the energy $E$ as well (not all of it, the kinetic energy $E$ is still continuous, but the vibratory energy is quantized). $\endgroup$
    – Prahar
    Nov 9 '19 at 14:54

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