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  Strings and their masses

+ 3 like - 0 dislike
1391 views

How do strings present in particles give mass to them? Is it only by vibrating? I have been trying to find the answer but could not find it anywhere, can this question be answered?

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user APARAJITA
asked Aug 23, 2013 in Theoretical Physics by APARAJITA (15 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

2 Answers

+ 3 like - 0 dislike

While it is true that an excited string (hence one with a vibration mode above the ground state) looks like a massive particle from far away, this is not the effect that is supposed to explain the mass of any particle ever seen. This is because the mass of the first excited mode of the string is already huge as far as particle masses go. So in string phenomenology, instead, all particles are modeled by strings in their massless ground state excitation and the actual observed masses are induced, as it should be, by a Higgs effect.

While the excited string states are not supposed to show up at energy scales anywhere close to what is being observed, their presence is still crucial: it is all these heavy particle excitations whose appearance as "virtual particles" in scattering amplitudes serve to make string scattering amplitudes be loop-wise finite, hence renormalized.

See the nLab String Theory FAQ the entry How do strings model massive particles?.

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Urs Schreiber
answered Aug 24, 2013 by Urs Schreiber (6,095 points) [ no revision ]
+ 2 like - 0 dislike

I presume that you are asking about the mass spectrum of string theories.

The mass spectrum of a Classical string theory, or the mass of a string is (due to Special Relativity) given, by:

$$m=\sqrt{-p^\mu p_\mu}=\sqrt N $$

In natural units $c_0=\ell_s=\hbar=1$. Where $N$ is an operator, called the "Number Operator". In Classical string theories, this is continuous. When we quantise the theory, we realise that the new mass spectrum is actually given by:

$$m=\sqrt{N-a} $$

Where $a$ is called the normal ordering constant. Now, $N$ is going to take discrete values, multiples of $\frac12$.

In Bosonic String Theory, $a=1$. In superstring theories, $a$ depends on the sector you' are talking about; it is $0$ in the Neveu-Schwarz sector, and $\frac12$ in the Ramond sector.

Of course, In GSO Projected theories (i.e. the tachyon is removed (yes, even in the RNS (Ramond-Neveu-Schwarz) Superstring, there are tachyons if you don't GSO Project; although this problem is absent in the GS (Green-Schwarz) Superstring)) , a GSO Projection gets rid of certain states and so on, but let's keep things simple right now.

Now, I've only been talking about open strings. What about the closed strings, which are more important, because the open strings are present only in the Type I Superstring theory (and Bosonic, of course (and probably also Type 0A and 0B (not sure))), whereas the closed strings are there in all string theories?

The transition happens to be relatively simple.

You replace $N$ with / $N+\tilde N$ and $a$ with $a+\tilde a$.

EDIT

I also see that in your post, you say "strings in particles". Actually, the particles themselves are strings. And they get their mass as per the vibrational modes $\alpha,\tilde\alpha,d,\tilde{d}$of the string with the Number operator $N$ given by

$$ N = \sum\limits_{n = 1}^\infty {{{\hat \alpha }_{ - n}}\cdot{{\hat \alpha }_n}} + \sum\limits_{r/2 = 1}^\infty {{{\hat d}_{ - r}}\cdot{{\hat d}_r}} $$.

answered Aug 23, 2013 by dimension10 (1,985 points) [ revision history ]
What I actually meant was that, particles gain mass by interacting with Higgs, it also attains mass due to the kinetic energy of gluons and Also due to the vibration of string in them, do all this factor give mass to a single particle?

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user APARAJITA
@APARAJITA: Actually, the vibration of the string (overall given by $N$) is what gives the mass. As for the higgs, it's just that the interaction kind of "gives it this same mass". But note, the Higgs has geometric interpretations in string theory. As for the gluons, it's actually the potential energy, and it's the energy of aan entire composite particle, like a proton, not an individual elementary particle, like the quark, whose mass can be determined by the manipulations in the answer.

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Dimensio1n0
If u don't mind can u explain me the geometric interpretation of Higgs.and also I don't understand how is it potential energy? Isn't it spoked to be kinetic energy ?

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user APARAJITA
@APARAJITA: See this post by Lubos Motll; the last section about "String theory's diverse geometric ways to look at Higgs bosons " , . There are actually 3 different such interpretations.

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Dimensio1n0
@APARAJITA the issue about the geometric interpretation of the higgs in ST could be a good new question ;-)

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Dilaton
But according to dimension10 it has already been answered by lubosmoti, though I have not checked it yet

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user APARAJITA
@APARAJITA: It's actually Lubos Motl's blog post, not an answer. And it's "Motl", not "Moti" : ) .

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Dimensio1n0
Oh ok , then it can be presented as a new question

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user APARAJITA
Wait if particles are string then what about the wave form of particle according to Heisenberg's wave particle duality?

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user APARAJITA
@APARAJITA: That's a very confused statement you make here. Firstly, the wave formulation is due to Schrodinger, not Heisenberg. People usually don't consider the state vector as a wavefunction. And it is certainly not due to the Heisenfbetrg Uncertainty Principle. Secondly, the wave formulation is pretty much dead in real physics nnow (except for historical and intuitive reasons) . People use second quantisation in Quantum Field Theory, i.e. fields. In string theory, we usually use canonical (Heisenberg, state vector) quantisation or suchh. In string field theory, second (co

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Dimensio1n0
@APARAJITA: ntd.) (contd.) quantisation is used I think, not sure. Even if we used the Schroedinger;'s wave formulation , there wouldn't be any clash with particles being strings. Also, note that the Heisengberg Uncertainty principle is modified a bit in string theory.

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Dimensio1n0
First of all I meant schroedinger and second of all see string form of particle is different from wave form , so there has to be a clash , if I'm not wrong

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user APARAJITA
let us continue this discussion in chat

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Dimensio1n0
You said you want to continue conversation on chat

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user APARAJITA

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