I Why don't strings have a Planck mass?

jcap

Summary
The intrinsic Planck mass of a string might be largely cancelled by its negative gravitational binding energy so that its net mass is small.
I understand that strings have a size of roughly the Planck length $l_P$ of $10^{-35}$ m.

If that is the case then one would expect that their mass would be roughly the Planck mass which is an enormous $10^{19}$ GeV.

(Strings that have small spins, like standard model particles, are about $l_P$ in length see https://physics.stackexchange.com/a/315166/22307)

In order to model the particles of the standard model their effective mass must be much smaller than the Planck mass.

Is the intrinsic Planck mass of a string largely cancelled by its negative gravitational binding energy so that its net mass is small?

For example assume that the gravitational binding energy of a string is roughly equal to its intrinsic mass energy then we have

$$\frac{GM^2}{R}\sim Mc^2$$

For a quantum object the uncertainty principle gives us the relationship

$$Mc\ R \sim \hbar$$

(I'm assuming a particle model such that its effective rest mass $M$ is entirely due to its internal momentum $P=Mc$ i.e. a zero rest mass particle confined to move around at the speed of light inside a box of size $R$)

Thus we find that

$$R \sim \sqrt{\frac{\hbar G}{c^3}} \sim l_P$$

Therefore, due to negative gravitational binding energy, Planck length strings are effectively massless. Thus they can reasonably model low-spin standard model particles which are very light compared to the Planck mass.

Related Beyond the Standard Model News on Phys.org

mitchell porter

First, something of a technical detail - very important for actual string theory, but a detail with respect to your argument:

In the standard model, in the absence of the Higgs field, all the elementary particles are actually massless. The Higgs field interacts with itself in such a way that its potential energy is minimized at a nonzero field value, and the Higgs field also has quantum numbers (hypercharge and weak isospin) which cause it to interact with the electroweak bosons and all the fermions, in such a way as to give them an effective mass.

String phenomenology has to work in the same way. That is, one needs string states which are not just light, but actually massless, and then a counterpart of the Higgs field that will work the same way as in the standard model.

But OK, whether the string is exactly massless or not, the natural mass scale here is the Planck mass. And you are right that light strings appear in string theory, as a result of cancellations between Planck-scale energies.

However, at least in the string theory of the past fifty years, these cancellations do not involve gravitational binding energy in the way you suggest. It may well be possible to define some other notion of string which works in that way.

But in the orthodox string theory, the cancellations involve the quantum energy levels of the string's vibratory modes. Interactions, gravitational or otherwise, play no part in those calculations. If I can find a satisfactory account of how it works, I will link to it.

jcap

Thanks for your answer! I need to do more homework to find out about the mass cancellations in standard string theory that you suggest.

Demystifier

2018 Award
The lowest mass excitations of strings in a flat background have zero mass because the mass operator is defined with a normal ordering of string-position operators.

jcap

Isn't normal ordering just a "trick" to define the Hamiltonian so that there is no zero point energy? I understand that experiments show that there is such a thing as zero point energy in systems such as cooled atoms for example.

see https://physics.stackexchange.com/a/433824/22307

mitchell porter

The lowest mass excitations of strings in a flat background have zero mass because the mass operator is defined with a normal ordering of string-position operators.
I believe that statement is wrong - that it is an attempt to simplify the logic of string theory, which ends up actually misrepresenting it.

You can motivate the form of the mass operator, by expressing the momentum-energy of a classical string, in terms of its oscillation modes. When you go to the quantum theory, you have to care about commutation relations among the level-raising and level-lowering operators for the different oscillation modes. There is a parameter in those commutation relations, the central charge c, which is initially undetermined.

When you construct the quantum theory, you find that the central charge and the dimension of space-time have to take specific values. There are several different-looking ways to make the quantum theory, and the way that the correct values for those two quantities emerge, also looks different in each case. One path to the correct values, does involve the zero-point energy of the string modes, and I'm guessing that this is what @Demystifier is talking about.

But I'm not aware of any derivation which has the simple form he describes. For example, in one derivation, you look at all the "first level" states that can be created by the oscillator operators, which should form a multiplet connected by rotation in n-dimensional space, and you find that there aren't enough of these states to form a rotation multiplet for a massive state, but just the right number if it is a massless state; and this is what allows you to deduce the space-time dimension.

It could be that I have gotten lost in the complexities of these rival derivations, and that at some deeper level, the reason that @Demystifier gives is the fundamental explanation - but I don't think so. I think he has jumped to the conclusion that the determination of central charge and dimension in string theory, is just another version of "normal ordering removes the vacuum energy", applied to the conformal field theory of the string worldsheet; and I think this is incorrect. But I am ready to be shown otherwise.

jcap

Improved Argument

The Energy $E$ of a fundamental string due to its length $L$ goes like
$$E\sim TL$$
where string tension $T$ is given by
$$T \sim \frac{1}{l_P^2}$$
(Using natural units $\hbar=c=1$ with Planck length $l_P$)

(e.g. see http://www.damtp.cam.ac.uk/user/tong/string/string.pdf Eqn. (1.16) etc)

Now the gravitational binding energy $E_G$ of the string goes like
$$E_G \sim -\frac{G E^2}{L}$$
Since $G = l_p^2$ then we have $G \sim 1/T$.

Thus the gravitational binding energy $E_G$ also goes like
$$E_G \sim -\frac{E^2}{TL} \sim -TL$$

It seems that 1-d strings might be unique in that their rest mass energy can be cancelled by their gravitational binding energy to produce a net rest mass of zero.

In general for an object with dimension $n$ we have $E \sim L^n$ and $E_G \sim -E^2/L \sim -L^{2n-1}$ so that the rest mass of the object can only be cancelled by the gravitational binding energy if $n=1$.

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Demystifier

2018 Award
I believe that statement is wrong - that it is an attempt to simplify the logic of string theory, which ends up actually misrepresenting it.

You can motivate the form of the mass operator, by expressing the momentum-energy of a classical string, in terms of its oscillation modes. When you go to the quantum theory, you have to care about commutation relations among the level-raising and level-lowering operators for the different oscillation modes. There is a parameter in those commutation relations, the central charge c, which is initially undetermined.

When you construct the quantum theory, you find that the central charge and the dimension of space-time have to take specific values. There are several different-looking ways to make the quantum theory, and the way that the correct values for those two quantities emerge, also looks different in each case. One path to the correct values, does involve the zero-point energy of the string modes, and I'm guessing that this is what @Demystifier is talking about.

But I'm not aware of any derivation which has the simple form he describes. For example, in one derivation, you look at all the "first level" states that can be created by the oscillator operators, which should form a multiplet connected by rotation in n-dimensional space, and you find that there aren't enough of these states to form a rotation multiplet for a massive state, but just the right number if it is a massless state; and this is what allows you to deduce the space-time dimension.

It could be that I have gotten lost in the complexities of these rival derivations, and that at some deeper level, the reason that @Demystifier gives is the fundamental explanation - but I don't think so. I think he has jumped to the conclusion that the determination of central charge and dimension in string theory, is just another version of "normal ordering removes the vacuum energy", applied to the conformal field theory of the string worldsheet; and I think this is incorrect. But I am ready to be shown otherwise.
My statement was somewhat sloppy. The correct statement is that the mass squared operator is the normal-ordered one plus a divergent constant. The divergence can be tamed by zeta-function regularization, which leads to the (in)famous "equality"
$$1+2+3+...=-\frac{1}{12}$$
after which one gets the correct mass spectrum. For details please see Zwiebach, A First Course in String Theory, Chapter 12. If you use 1st edition the relevant equations are (12.98)-(12.108), or if you use 2nd edition the relevant equations are (12.101)-(12.111).

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"Why don't strings have a Planck mass?"

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