Tachyon expectation values [Archive]

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Red Bull

Oct25-04, 07:09 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Dear Forum,\n\nI have a specific question on tachyon expectation values in bosonic string\ntheory. In a flat 26d background one has something like:\n\n-\\partial^2 T(x) - 4T(x)/(\\alpha\')=0 (*)\n\n((9.9.2) in Polchinski with V=0). So this is the first two orders in an\nalpha\' expansion, right?\n\n[Moderator\'s note: Yes, except that the mass term -4T/ \\alpha\' is the\nleading term, while you wrote it as the second term. LM]\n\nAt the lowest order, the only solution is T(x)=0. Now at the next order,\nJP states the solution:\n\nT(x)=exp(q.x) for q^2=-4/(\\alpha\') (**)\n\nClearly this is a solution of (*), but what confuses me is that it solves\n(*) by having the first term in the alpha\' expansion cancelling exactly\nthe next order term, rather than cancellation order by order in \\alpha\'.\n\n[Moderator\'s note: If you just look at different powers of \\alpha\' in\nvarious expressions, does not it solve all the confusion? In string\ntheory, one has many examples in which "quantum" effects cancel against\n"classical" effects. For example, the Green-Schwarz mechanism is the\ndiscovery in 1984 how one-loop hexagon anomalies are canceled against\n*tree* diagrams in 10-dimensional N=1 supergravity coupled to\nsuper-Yang-Mills. Also, the Fischler-Susskind mechanism is about a\ncancellation between different topologies of the worldsheet between\none another - one is forced to add certain counterterms to a genus g\nsurface to cancel the contributions of singular genus (g+1) surfaces,\nat least in bosonic string theory. The two examples I mentioned are\nabout apparently different powers of g, but if you look carefully\nat these powers, of course that they match, much like they match in\nyour case (for example, the vertices of tree diagrams in the GS\nmechanism come with extra powers). The cancellation between seemingly\ntree and seemingly quantum effects in string theory is what makes\nits quantization *necessary* - string theory is a purely quantum\ntheory. LM]\n\nI think a problem with this is that the _next_ order in the expansion will\nbe something like (\\alpha\')\\partial^4 T(x), and if you plug a solution\nlike (**) into this then this third order term will not in any sense be\nsmall compared to the first two. Surely you want your solution of (*) to\nbe a good approximation to a solution to all orders in alpha\'? And this\ncan happen only if the the higher order terms are negligible when you plug\nin your putative solution - or am I missing something obvious?\n\n[Moderator\'s note: Yes, I do think that you are missing something. For\nexample, there can be no term like \\partial^4 T(x) in your equation.\nThis would, indeed, modify the tachyon mass. But we know that the tachyon\nmass is *exactly* -4 / \\alpha\' for closed string tachyons @ g=0. This is an\nexact result to all orders in \\alpha\', although - of course - it ignores\ninteractions i.e. terms with higher powers of g. You can easily see\nthat the vertex operator for the tachyon with the right, "naive" momentum\nis exactly a tensor field. There are no other terms in the effective\nfield theoretical equation of motion for the tachyon at g=0. Such terms\nwould have higher powers of T. It may be a good idea if you were trying\nto calculate the exact results as functions of \\alpha\'. There is no point\nof doing the \\alpha\' expansions if you can solve the problem exactly.\nIf you solve it exactly, you will see that such problems can\'t occur.\nIncidentally, in the open string case, string field theory can give you\na nice nontrivial potential for the tachyon. But such a potential already\ncontains all *tree level* diagrams, including the interacting ones.\nHere you seem to be talking about g=0 string theory, and g=0 string\ntheory in flat space is simple and solvable, and has no corrections.\nWe only need to use \\alpha\' expansions for nonlinear theories (nonlinear\nsigma model) that we cannot solve directly. But the flat space in d=26\nat g=0 is trivial, and the equation (*) is exact. At higher orders in g,\nthe modifying terms have higher powers of T, too.\nEveryone is welcome to write a better answer. Best, Lubos LM]\n\nthanks\nd70yxj\n\n\n\n\n______________________ ____________\nDo you Yahoo!?\nYahoo! Mail - You care about security. So do we.\nhttp://promotions.yahoo.com/new_mail\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Dear Forum,

I have a specific question on tachyon expectation values in bosonic string
theory. In a flat 26d background one has something like:

-\partial^2 T(x) - 4T(x)/(\alpha')=0 (*)

((9.9.2) in Polchinski with V=0). So this is the first two orders in an
\alpha' expansion, right?

[Moderator's note: Yes, except that the mass term -4T/ \alpha' is the
leading term, while you wrote it as the second term. LM]

At the lowest order, the only solution is T(x)=0. Now at the next order,
JP states the solution:

T(x)=\exp(q[/itex].x) for [itex]q^2=-4/(\alpha') (**)

Clearly this is a solution of (*), but what confuses me is that it solves
(*) by having the first term in the \alpha' expansion cancelling exactly
the next order term, rather than cancellation order by order in \alpha'.

[Moderator's note: If you just look at different powers of \alpha' in
various expressions, does not it solve all the confusion? In string
theory, one has many examples in which "quantum" effects cancel against
"classical" effects. For example, the Green-Schwarz mechanism is the
discovery in 1984 how one-loop hexagon anomalies are canceled against
*tree* diagrams in 10-dimensional N=1 supergravity coupled to
super-Yang-Mills. Also, the Fischler-Susskind mechanism is about a
cancellation between different topologies of the worldsheet between
one another - one is forced to add certain counterterms to a genus g
surface to cancel the contributions of singular genus (g+1) surfaces,
at least in bosonic string theory. The two examples I mentioned are
about apparently different powers of g, but if you look carefully
at these powers, of course that they match, much like they match in
your case (for example, the vertices of tree diagrams in the GS
mechanism come with extra powers). The cancellation between seemingly
tree and seemingly quantum effects in string theory is what makes
its quantization *necessary* - string theory is a purely quantum
theory. LM]

I think a problem with this is that the _next_ order in the expansion will
be something like (\alpha')\partial^4 T(x), and if you plug a solution
like (**) into this then this third order term will not in any sense be
small compared to the first two. Surely you want your solution of (*) to
be a good approximation to a solution to all orders in \alpha'? And this
can happen only if the the higher order terms are negligible when you plug
in your putative solution - or am I missing something obvious?

[Moderator's note: Yes, I do think that you are missing something. For
example, there can be no term like \partial^4 T(x) in your equation.
This would, indeed, modify the tachyon mass. But we know that the tachyon
mass is *exactly* -4 / \alpha' for closed string tachyons @ g=0. This is an
exact result to all orders in \alpha', although - of course - it ignores
interactions i.e. terms with higher powers of g. You can easily see
that the vertex operator for the tachyon with the right, "naive" momentum
is exactly a tensor field. There are no other terms in the effective
field theoretical equation of motion for the tachyon at g=0. Such terms
would have higher powers of T. It may be a good idea if you were trying
to calculate the exact results as functions of \alpha'. There is no point
of doing the \alpha' expansions if you can solve the problem exactly.
If you solve it exactly, you will see that such problems can't occur.
Incidentally, in the open string case, string field theory can give you
a nice nontrivial potential for the tachyon. But such a potential already
contains all *tree level* diagrams, including the interacting ones.
Here you seem to be talking about g=0 string theory, and g=0 string
theory in flat space is simple and solvable, and has no corrections.
We only need to use \alpha' expansions for nonlinear theories (nonlinear
\sigma model) that we cannot solve directly. But the flat space in d=26at g=0 is trivial, and the equation (*) is exact. At higher orders in g,
the modifying terms have higher powers of T, too.
Everyone is welcome to write a better answer. Best, Lubos LM]

thanks
d70yxj

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Lubos Motl

Oct25-04, 08:43 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Mon, 25 Oct 2004, Red Bull wrote:\n\n> -\\partial^2 T(x) - 4T(x)/(\\alpha\')=0 (*)\n\nOne more comment. You see that Polchinski, above (9.9.2), calls it\n"linearized tachyon field equation". The word "linearized" means that it\nis the truncation to terms that are linear in T and/or its derivatives -\nbut it contains *all of them*. It is the *exact* linearized equation. All\nother terms that may modify this equation are non-linear, i.e. they are\nproducts of at least two copies of T (or one T and at least on other\nfield).\n\nYou may also consider the propagation of the tachyons on slightly curved\nbackgrounds. Indeed, then the analogous equation could have terms of\nhigher order in \\alpha\', depending on the geometry. However, this \\alpha\'\nwould be combined into \\alpha\' / a^2 where a is roughly the curvature\nradius of the geometry. Indeed, you could then use your argument,\ncomparing these terms to the basic terms, and you would conclude that the\nperturbative expansion may break down - but it only breaks down if \\alpha\'\n/ a^2 is of order one or larger. This \\alpha\' / a^2 would be your real\ndimensionless expansion parameter, and if it is small, then the terms with\nhigher powers of \\alpha\' are guaranteed to be smaller. If it is large and\nthe curvature is "string scale or bigger", then the perturbative \\alpha\'\nexpansion breaks down, as you correctly say.\n\nAnother general thing: if two terms cancel, then they must be of the same\norder in \\alpha\'. ;-) This may be just a matter of terminology, but if two\nthings depend differently on \\alpha\' parameterically, they cannot cancel.\nIn order to get the right powers of \\alpha\', you must calculate the\ncontributions to the dimension from all factors. For example, quantum\nmechanics guarantees that the total dimension of the operator\nexp(i.k.X(z)) is -\\alpha\'.k^2/2, even though it is classically\ndimensionless.\n\nIf you\'re using a consistent perturbative expansion, then - sort of by\ndefinition - you are indeed comparing the terms with the same powers,\norder by order, separately - that\'s true by definition. This, in a sense,\nproves that your original presentation how the tachyon equation is\nsatisfied was not really any perturbative expansion. Both terms were of\nthe same order - they had to be because they canceled. ;-) Something\ncalled the "virial theorem" implies that the different terms in these\nquadratic actions are always of the same order. Both of them are, in the\nproper counting, at leading order in \\alpha\'. The free tachyon with\nmomentum p, p^2=4/ \\alpha\', is the zeroth order approximation - and you\nmay consider corrections, but the linear corrections only appear in curved\nspace, and are suppressed by \\alpha\' / a^2.\n____________________________________________ __________________________________\nE-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/\neFax: +1-801/454-1858 work: +1-617/384-9488 home: +1-617/868-4487 (call)\nWebs: http://schwinger.harvard.edu/~motl/ http://motls.blogspot.com/\n^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Mon, 25 Oct 2004, Red Bull wrote:

> -\partial^2 T(x) - 4T(x)/(\alpha')=0 (*)

One more comment. You see that Polchinski, above (9.9.2), calls it
"linearized tachyon field equation". The word "linearized" means that it
is the truncation to terms that are linear in T and/or its derivatives -
but it contains *all of them*. It is the *exact* linearized equation. All
other terms that may modify this equation are non-linear, i.e. they are
products of at least two copies of T (or one T and at least on other
field).

You may also consider the propagation of the tachyons on slightly curved
backgrounds. Indeed, then the analogous equation could have terms of
higher order in \alpha', depending on the geometry. However, this \alpha'
would be combined into \alpha' / a^2 where a is roughly the curvature
radius of the geometry. Indeed, you could then use your argument,
comparing these terms to the basic terms, and you would conclude that the
perturbative expansion may break down - but it only breaks down if \alpha'/ a^2 is of order one or larger. This \alpha' / a^2 would be your real
dimensionless expansion parameter, and if it is small, then the terms with
higher powers of \alpha' are guaranteed to be smaller. If it is large and
the curvature is "string scale or bigger", then the perturbative \alpha'
expansion breaks down, as you correctly say.

Another general thing: if two terms cancel, then they must be of the same
order in \alpha'. ;-) This may be just a matter of terminology, but if two
things depend differently on \alpha' parameterically, they cannot cancel.
In order to get the right powers of \alpha', you must calculate the
contributions to the dimension from all factors. For example, quantum
mechanics guarantees that the total dimension of the operator
\exp(i.k.X(z)) is -\alpha'.k^2/2, even though it is classically
dimensionless.

If you're using a consistent perturbative expansion, then - sort of by
definition - you are indeed comparing the terms with the same powers,
order by order, separately - that's true by definition. This, in a sense,
proves that your original presentation how the tachyon equation is
satisfied was not really any perturbative expansion. Both terms were of
the same order - they had to be because they canceled. ;-) Something
called the "virial theorem" implies that the different terms in these
quadratic actions are always of the same order. Both of them are, in the
proper counting, at leading order in \alpha'. The free tachyon with
momentum p, p^2=4/ \alpha', is the zeroth order approximation - and you
may consider corrections, but the linear corrections only appear in curved
space, and are suppressed by \alpha' / a^2.
__{_______________________________________________ _____________________________}
E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
eFax: +1-801/454-1858 work: +1-617/384-9488 home: +1-617/868-4487 (call)
Webs: http://schwinger.harvard.edu/~motl/ http://motls.blogspot.com/
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Red Bull

Oct25-04, 05:21 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Lubos Motl wrote: [Moderator\'s note: replies of others will be\nappreciated, I don\'t have enough time for\na better answer. LM]\n\n> Here you seem to be talking about g=0 string theory, and g=0 string\n> theory in flat space is simple and solvable, and has no corrections.\n> We only need to use \\alpha\' expansions for nonlinear theories (nonlinear\n> sigma model) that we cannot solve directly. But the flat space in d=26\n> at g=0 is trivial, and the equation (*) is exact.\n\nLubos, thanks for your replies; I think I see now what I missed, but let\nme check I am interpreting things correctly. So one can see easily from\nthe g=0 string spectrum that there is a tachyon of mass 4/alpha\', and I\nguess the linearized effective action in Polchinski can be reproduced\nentirely, using just this piece of information. As you say it must be\nexact.\n\nOn the other hand, I was (incorrectly, I think) assuming that those two\nterms in the linearized equation (*) were in fact the first two terms in a\nbeta function, calculated a la Friedan - in which case there would be no\nobvious reason not to have terms looking like \\partial^4 T (at least I\nthink not, but maybe this is still wrong). If I do make such a calculation\nI presume there will indeed be linear terms in T at all orders in alpha\',\nbut that conversely the first two terms won\'t look exactly like (*) - is\nthis at least in principle correct?\n\nOne other thing on the non-linear terms in T; do these appear in a beta\nfunction calculation? It looks like only linear terms in T will show up\nwhen you expand T around a background field in the canonical way. The\nmetric seems to be a special case - the nonlinear beta function appears\nbecause terms involving the Riemann tensor appear when one canonically\nexpands any spacetime tensor field around some background value. I may\nagain be missing something, but is there a reference where these nonlinear\nterms in T are calculated?\n\n> Another general thing: if two terms cancel, then they must be of the same\n> order in \\alpha\'. ;-) This may be just a matter of terminology, but if two\n> things depend differently on \\alpha\' parameterically, they cannot cancel.\n> In order to get the right powers of \\alpha\', you must calculate the\n> contributions to the dimension from all factors.\n\nYep, OK, I think this is a matter of terminology, and I sort of knew what\nI said might not really express what I meant. It\'s probably not important,\nbut let me explain\n\nthough in case I am still missing something. Suppose I have an expansion\nin alpha\' that doesn\'t truncate at any particular order, but just goes on,\nand also is linear in the field T:\n\n\\sum_n \\alpha\'^n H_n(T)=0\n\nwhere the Hs are differential operators (or whatever).\n\nSuppose then I have a solution T_0 at lowest order, H_0(T_0)=0. If I want\nto adjust this to be a solution at the next order, I guess it\'s natural to\nconsider T_1=T_0+\\alpha\'(\\delta T_0). Then the lowest order in alpha\'\nstill cancels but the perturbation at the lowest order cancels with T_0\nplugged into H_1:\n\nH_1(T_0)+H_0(\\deltaT_0)=0.\n\nwhile all the pieces left over are O(\\alpha\'^2) and negligible at this\norder. This is what I meant by cancelling order by order, not sure if it\nis clearer now. For example if I have a metric that solves Einstein\'s\nequations, but want to find a perturb it to be a solution to Einstein plus\n\\alpha\' corrections, I think one probably looks for a new metric which is\nthe original one plus a correction of order alpha\', would that make sense?\n\nConversely, in this kind of situation a solution like (**) which directly\ncancels the first term with the next order up would have higher order\nterms not negligible (unless they are identically zero i.e. the alpha\'\nexpansion truncates and is exact at some order, as in (*)). Hope this\nmakes sense but thanks again for putting me right on (*).\n\n__________________________________\nDo you Yahoo!?\nYahoo! Mail Address AutoComplete - You start. We finish.\nhttp://promotions.yahoo.com/new_mail\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Lubos Motl wrote: [Moderator's note: replies of others will be
appreciated, I don't have enough time for
a better answer. LM]

> Here you seem to be talking about g=0 string theory, and g=0 string
> theory in flat space is simple and solvable, and has no corrections.
> We only need to use \alpha' expansions for nonlinear theories (nonlinear
> \sigma model) that we cannot solve directly. But the flat space in d=26
> at g=0 is trivial, and the equation (*) is exact.

Lubos, thanks for your replies; I think I see now what I missed, but let
me check I am interpreting things correctly. So one can see easily from
the g=0 string spectrum that there is a tachyon of mass 4/\alpha', and I
guess the linearized effective action in Polchinski can be reproduced
entirely, using just this piece of information. As you say it must be
exact.

On the other hand, I was (incorrectly, I think) assuming that those two
terms in the linearized equation (*) were in fact the first two terms in a
\beta function, calculated a la Friedan - in which case there would be no
obvious reason not to have terms looking like \partial^4 T (at least I
think not, but maybe this is still wrong). If I do make such a calculation
I presume there will indeed be linear terms in T at all orders in \alpha',
but that conversely the first two terms won't look exactly like (*) - is
this at least in principle correct?

One other thing on the non-linear terms in T; do these appear in a \beta
function calculation? It looks like only linear terms in T will show up
when you expand T around a background field in the canonical way. The
metric seems to be a special case - the nonlinear \beta function appears
because terms involving the Riemann tensor appear when one canonically
expands any spacetime tensor field around some background value. I may
again be missing something, but is there a reference where these nonlinear
terms in T are calculated?

> Another general thing: if two terms cancel, then they must be of the same
> order in \alpha'. ;-) This may be just a matter of terminology, but if two
> things depend differently on \alpha' parameterically, they cannot cancel.
> In order to get the right powers of \alpha', you must calculate the
> contributions to the dimension from all factors.

Yep, OK, I think this is a matter of terminology, and I sort of knew what
I said might not really express what I meant. It's probably not important,
but let me explain

though in case I am still missing something. Suppose I have an expansion
in \alpha' that doesn't truncate at any particular order, but just goes on,
and also is linear in the field T:

\sum_n \alpha'^n H_n(T)=0

where the Hs are differential operators (or whatever).

Suppose then I have a solution T_0 at lowest order, H_0(T_0)=0. If I want
to adjust this to be a solution at the next order, I guess it's natural to
consider T_1=T_0+\alpha'(\delta T_0). Then the lowest order in \alpha'
still cancels but the perturbation at the lowest order cancels with T_0
plugged into H_1:H_1(T_0)+H_0(\deltaT_0)=0.

while all the pieces left over are O(\alpha'^2) and negligible at this
order. This is what I meant by cancelling order by order, not sure if it
is clearer now. For example if I have a metric that solves Einstein's
equations, but want to find a perturb it to be a solution to Einstein plus
\alpha' corrections, I think one probably looks for a new metric which is
the original one plus a correction of order \alpha', would that make sense?

Conversely, in this kind of situation a solution like (**) which directly
cancels the first term with the next order up would have higher order
terms not negligible (unless they are identically zero i.e. the \alpha'
expansion truncates and is exact at some order, as in (*)). Hope this
makes sense but thanks again for putting me right on (*).

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Red Bull

Oct29-04, 06:39 AM