# What can make a string massless?

1. Jul 9, 2011

### kcajrenreb

What makes some strings have mass, and others none? (eg. graviton vs. electron)

2. Jul 9, 2011

### mathman

I am completely ignorant about string theory. However I am under the impression that the theory has not been developed to the point where anyone can answer your question.

3. Jul 9, 2011

### Kevin_Axion

It's determined by the energy in the vibrations, the higher the frequency, the higher the mass.

4. Jul 9, 2011

### kcajrenreb

So a massless string has no vibration at all?

5. Jul 10, 2011

### kcajrenreb

Kevin_Axion, you didn't really answer my question, how can a string become massless?

6. Jul 10, 2011

### DaveC426913

A string does not "become" massless. Mass is one (of several) resultant properties of a string's vibrations.

i.e.: Some strings vibrate in a fashion such that they manifest a property of mass, whereas some vibrate so that they do not manifest a property of mass.

7. Jul 10, 2011

### Jakeman002

http://superstringtheory.com/basics/basic5a.html

this should answer your questions. The basic principle is how the certain modes of the one dimensional string behave under the group symmetries such as E(8)xE(8) which allows us to identify how the different modes of the string behave as particles i.e being a fermion or a boson. The main answer basically about the mass is the group being used which is most likely these days G2's exceptional 8 x exceptional 8 built from the octonian generators.

8. Jul 10, 2011

### kcajrenreb

Okay, I see now. Thanks guys.

9. Jul 11, 2011

### Demystifier

Not really. A string with no vibration at all actually has a NEGATIVE mass squared. This perhaps counterintuitive fact can only be understood through quantum mechanics of strings. In supesymmetric string theory such a no-vibration mode of string is unphysical, so the next lowest (first physical) mode of vibration is massless.

10. Jul 11, 2011

### kcajrenreb

Yeah, a string with negative mass squared is a tachyon, right? And supersymmetry was made to avoid that problem, correct?

11. Jul 11, 2011

### Kevin_Axion

Sorry for the delay, I was waiting for a response from a string theorist I was talking to:

"Sorry for the delayed reply, I have been trying to think of the best way to explain it. What you end up getting is what is known as a tower of excitations, simply put

$m^2 = (\frac{n\pi}{L})^2$

Where L is a length related to compactification, ie its a small number and n=0,+/- 1,+/-2... so that its the n=0 mode that gives you a massless particle in string theory. This isn't anything fundamental though.
For the next bit you will need tex the world add on for your browser.
In general we require gauge bosons to be massless, regardless if we are doing String Theory, GR, or any other field theory. Consider E&M, its Lagrangian density is

$\mathcal{L}=F ^{\mu\nu} F_{\mu\nu}$

Where F is the field strength tensor

$F_{\mu\nu}= \partial_\mu A_\nu-\partial_\nu A_\mu$

After some algebra you get

$F_{\mu\nu} F^{\nu\mu}=\partial_\mu A_\nu \partial ^\mu A ^\nu -\partial_\nu A_\mu \partial ^\mu A ^\nu$

Schematically this is just

$F ^2 = (\nabla A) ^2$

Which looks like part of the lagrangian that leads to the Klein Gordon equation. The crucial part that it is missing is the m2 A2 term. What happens if we put that in by hand? Its not good. A is the Gauge field, its the photon when we quantize the theory. Since its the gauge field it must obey gauge symmetry, for abelian fields this means that the lagrangian must be invariant under the transformation

$A_\mu\rightarrow A_\mu -\partial_\mu f$

where f is an arbitrary function. When we add a mass term to our original lagrangian it loses gauge invariance, therefore all gauge fields, by definition, have to be massless. The higgs mechanism allows a way around this for low energies, but that is another discussion entirely."

Last edited: Jul 11, 2011
12. Jul 12, 2011

Right.