# Why don't strings have a huge Planck mass?

1. Sep 19, 2009

### johne1618

If a string has a Planck length then why doesn't it have a huge Planck mass?

Perhaps it does but its negative gravitational self-energy counteracts this huge positive mass/energy.

Maybe a closed string can be thought of as a Planck energy electromagnetic wave trapped by its own gravitational field so that it is forced to orbit itself in a Planck scale loop. The positive rotational energy of the wave would be exactly balanced by the negative gravitational field so that the total rest mass/energy of the object would be zero. Maybe this would be a model of a graviton.

Last edited: Sep 19, 2009
2. Sep 20, 2009

### tom.stoer

The zeroth oscillations of string are (nearly) massless. All other vibrational modes have higher masses and should be comparable to Planck mass.

3. Sep 20, 2009

### johne1618

But, as I understand it, any string is under a huge tension of c^4/G. If it is a Planck length long then Energy=Tension*Distance implies that it has a Planck mass just from its tension alone regardless of how it is oscillating.

I think this tensional energy is in fact negative and purely classical gravitational in origin rather than quantum as can be seen from the fact that the expression for the tension involves G but not hbar. It is the quantum uncertainty in the string's momentum which provides the positive energy which balances this negative gravitational energy.

Last edited: Sep 21, 2009
4. Sep 20, 2009

### tom.stoer

In the quantization of the bosonic string several consistency conditions emerge lead to a state space similar to the harmonic oscillator. After normal ordering of the Hamiltonian the "1/2" of the ground state energy disappears from the spectrum, so the lowest energy level is something like "string tension times zero" which is of course zero.

See e.g. http://xxx.lanl.gov/PS_cache/hep-th/pdf/9411/9411028v1.pdf

5. Sep 20, 2009

### johne1618

Thanks. I'll have a look at the Polchinski workshop paper but it looks too advanced for me to follow.

So you are saying that the lowest *oscillatory* energy level is zero.

Fair enough.

But I still think there is a story to be told about the balance between the positive energy in the string due to its uncertainty in momentum and the negative energy contained in its tension.

Last edited: Sep 20, 2009
6. Sep 20, 2009

### tom.stoer

Normal ordering is used in quantum field theory to have an empty vacuum state. An annihilation operator should annihilate the vacuum - that's the very reason for the name. The commutator generates the "1/2". This is typically dropped to have a vacuum with zero energy. There is no rigorous proof, it's just a physical argument to set the vacuum energy scale to zero.

7. Sep 20, 2009

### johne1618

Sorry I've just edited the thread under you.

I accept that the lowest *oscillatory* energy level is zero.

But as I said above:

I still think there is a story to be told about the balance between the positive energy in the string due to its uncertainty in momentum and the negative energy contained in its tension.

8. Feb 22, 2010

### presura

I am also looking for such a story... What are the contribution of tensional energy (streching the string) and vibrational energy. What is the zero energy in 26 dimensions of the first excited state? Is it simply zero? Than what about the tensional energy, is it not present? Or, in other words, take the masless particule in 26 dimensions, how comes the the total energy (vibrational and tensional) of the first excited state is zero? I expected only the virbational energy to be zero, I did not expect the tensional energy... Any help here?