View Full Version : Why don't strings have a huge Planck mass?
johne1618
Sep19-09, 10:51 AM
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.
tom.stoer
Sep20-09, 04:05 AM
The zeroth oscillations of string are (nearly) massless. All other vibrational modes have higher masses and should be comparable to Planck mass.
johne1618
Sep20-09, 08:02 AM
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.
tom.stoer
Sep20-09, 08:38 AM
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
johne1618
Sep20-09, 12:33 PM
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.
tom.stoer
Sep20-09, 12:55 PM
Why choose "normal ordering" of the creation and annihilation operators rather than any other scheme of ordering?
I wish I could get more of a handle on this stuff but it is too technical for me!
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.
johne1618
Sep20-09, 01:09 PM
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.
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