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  Energy of a string

+ 3 like - 0 dislike

What is the correct definition of the energy of a string ?

I suddenly get confused with the definition of the energy of a string. Considering, for instance, a bosonic open string in the light-cone gauge, We have $H = p^-=p_+$, the hamiltonian, and we have $p^0$, which appears in the expansion of the string in light-cone gauge $X^o(\sigma, \tau) = x^0 + \frac{p^0}{p^+}\tau + ...$.

a) In Zwiebach's First course in string theory (1), there is a calculus of the entropy of a string, which begins (Chapter $16.2$ p $354$) by a function partition for a non relativistic string with fixed points (the quantum violin string). For a given energy $E = N \hbar w_0$, where $N$ is the level, the entropy is computed (this is direcly linked to the number of partitions p(n)). More precisely, an expression for the entropy, $\sim T$, and the energy ,$\sim T^2$, are obtained, se we get a $S \sim \sqrt{E}$, and plugin the $E \sim N$ expression, we get a $S \sim \sqrt{N}$.

"Suddenly", in chapter $16.3$ page $361$, it is said that "we now return to relativist strings that carry no spatial momentum, and this happens if the open string endpoints end on a D-0 brane, so the energy levels are given by the rest mass of its quantum states". Of course, we have $m^2 \sim N$ (for large $N$), so $m\sim \sqrt{N}$, so if the energy ($p^0 ?$) is identified with $m$, we have $S \sim E$ Now, if we take the hamiltonian $H$, for large $N$, we have $H \sim N$, so $S \sim \sqrt{H}$

So, this confused me, because I have $2$ law expressions for the entropy, depending on the definition of the energy.

b) An other problem, is saying that the energy of the string is proportionnal to its length. If we look at $H$, we have $H=p^- \sim \frac{1}{l}$ (with zero trans verse momenta $p^i$), while $p^+ \sim \frac{l}{\alpha'}$. So, may I write that the correct energy is $p^o \sim (p^+ + p^-)?$

This post imported from StackExchange Physics at 2014-03-12 15:51 (UCT), posted by SE-user Trimok
asked Aug 17, 2013 in Theoretical Physics by Trimok (955 points) [ no revision ]
I really don't see the problem here in (a) ? Ok, the entropy is different for relativistic strings, what's wrong with that ?

This post imported from StackExchange Physics at 2014-03-12 15:51 (UCT), posted by SE-user Dimensio1n0
@Dimension10 : If you read carefully, there are 2 expressions $S \sim m$ and $S \sim \sqrt{H}$

This post imported from StackExchange Physics at 2014-03-12 15:51 (UCT), posted by SE-user Trimok
Oh, the 2 expressions in the "Suddenly" paragraph itself... +1.

This post imported from StackExchange Physics at 2014-03-12 15:51 (UCT), posted by SE-user Dimensio1n0

1 Answer

+ 6 like - 0 dislike

In general, the energy is a quantity conserved due to the translational invariance in time or, equivalently, the generator of these translations in time. So depending on what we mean by time, we also have different meanings of the word energy.

In the usual treatment of the spacetime, the energy $E=p^0$ is the generator of translations in the usual time $t=x^0$. In the spacetime light cone gauge, the energy is often meant to be $E= p^-$ and generate translations in $x^+$, the light-cone-gauge counterpart of "time" that defines the slices in such a gauge.

A single string has mass going like $\sqrt{N}$ so this is also its energy $p^0$ in the rest frame where $\vec p=0$. However, in the light cone gauge, $$p^- = \frac{(p^i)^2+m^2}{2p^+} $$ note that this is just the usual $p_\mu p^\mu = m^2$ solved for $p^-$ in the light cone gauge, so that the $m^2$ in the numerator goes like $N$ (the total excitation of the string) rather than $\sqrt{N}$.

Also, the energy may mean the world sheet energy, the generator of translations in the world sheet temporal coordinate $\tau$. Then, for an excited string, the energy goes like $H\sim L_0+\tilde L_0\sim N$ much like in the light cone gauge. However, there are different coefficients in these two interpretations of the energy and one has to treat the nonzero modes and ghosts differently.

I am confident that all the standard textbooks make it clear from the context which "energy" they are talking about.

Concerning the second problem, what was clearly meant was the "long/classical string" and its energy $p^0$ in the rest frame of the whole string. Such a long string has the contribution to its rest mass $\Delta m^2 = (T L)^2$ where $T$ is the string tension (the linear energy density) and $L$ is the length. There are also contributions to $m^2$ going like $N/ \alpha'$ etc. from the string excitations.

The world sheet energy or the light-cone-gauge energy have contributions from long wound strings that go like $L^2$ (because they are linear functions of $m^2$, not $m$).

When you wrote $p^0=(p^++p^-)/\sqrt{2}$, you missed the usual $1/\sqrt{2}$ normalization but this relationship holds in general (definition of the light cone gauge). Yes, the energy that was scaling linearly in the length (and tension) was $p^0$. However, you don't gain anything by rewriting $p^0$ in terms of the light cone $p^\pm$ because the energy that is proportional to the length of the string has to be measured in the rest frame of the long string in the spacetime where $\vec p=0$ which also implies $p^+=p^-$. When you impose this equality between $p^+$ and $p^-$ and the assumption about the right squared mass of the long string, you will also derive $p^0\sim TL$. But there's no point in setting $p^+=p^-$ etc. because the formula for $p^-$ in the light cone gauge always explicitly contains $1/p^+$ and the formula for $p^-$ may always be seen – once you multiply it by $p^+$ and subtract $(p^i)^2$ from the result – to be equivalent for the formula for $m^2$. So in the light cone gauge, to derive the "linear density of the string", you must still know that $m^2$ goes like $(TL)^2$, so you effectively assume (or precalculate) what you want to derive, anyway, $m_0\sim TL$.

This post imported from StackExchange Physics at 2014-03-12 15:51 (UCT), posted by SE-user Luboš Motl
answered Aug 17, 2013 by Luboš Motl (10,278 points) [ no revision ]

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