# A diff Eq on strings, check out the math.

Originally posted by lethe
you want a complete proof? stokes theorem says $$\mbox{\int_Cd\omega = \int_{\partial C} \omega}$$. so if $$\mbox{\quad C}$$
is a circle, then $$\mbox{\partial C}$$ is zero, regardless of whether $$\mbox{\omega}$$ vanishes or not.

"if C is a circle, then $$\mbox{\partial C}$$ is zero" - you say? We can use Stoke's Theorem on an open surface with a closed curve as a boundary. We cannot use Stoke's Theorem on a closed surface without boundary. Likewise, we cannot apply Stoke's Theorem to a closed curve with no boundary. A boundary on an open curve would just be its end points. Then Stoke's Theorem in that case would simply reduce to the fundamental theorem of calculus. But sorry, you cannot apply Stoke's Theorem when there is no boundary. So your argument does not seem to say anything about the integral or the integrand.

like another counter example? the electric field in the vicinity of a point charge is nonzero. this is a conservative field, so any loop integral is zero.
Neither does this seem to say anything about the time derivative of a conserved integral. I said conserved integral, not conservative field. A conservative field means you get back what you put in when you end where you started. But I don't see how that has anything to do with a time dervative of an integral that is a constant which is not necessarily zero.

Originally posted by Mike2
We can use Stoke's Theorem on an open surface with a closed curve as a boundary. We cannot use Stoke's Theorem on a closed surface without boundary.
i suggest you learn stokes theorem a little better. you have a copy of Frankel, right? check it out, he has the details....

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Originally posted by lethe
i suggest you learn stokes theorem a little better. you have a copy of Frankel, right? check it out, he has the details....
Yes, I'll have to read it again with just that in mind. But you know Frankel, a little here and a little there; it's like reading a bible code. Have you ever seen the notation where $$\mbox{\partial C}$$=0 in any book somewhere. I don't remember seeing this. However, on second thought, I wonder what happens as the boundary does approach zero (though not zero), and the curve approaches being closed, then the integral over the boundary goes to zero, and you might be right. Thank you.

Where would that leave me? I would always have to specify the closed line integral of the time derivative is zero?

Originally posted by Mike2
Have you ever seen the notation where $$\mbox{\partial C}$$=0 in any book somewhere.
yes

I don't remember seeing this. However, on second thought, I wonder what happens as the boundary does approach zero (though not zero), and the curve approaches being closed, then the integral over the boundary goes to zero, and you might be right. Thank you.
my pleasure.

Where would that leave me? I would always have to specify the closed line integral of the time derivative is zero?
i think it leaves you in this position (although i admit that i haven t checked too carefully): you cannot conclude that anything is zero just because a closed line integral of that thing is zero for an arbitrary closed line.

i think it leaves you in this position (although i admit that i haven t checked too carefully): you cannot conclude that anything is zero just because a closed line integral of that thing is zero for an arbitrary closed line.
Just a minute. I've thought about it again. Let me know if I am understanding you correctly. You're saying that

$$\oint\limits_C {\frac{d}{{dt}}(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} \cdot d\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over l} {\rm{)}}} \, = \,\,\frac{d}{{dt}}\,\oint\limits_C {\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} \cdot d\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over l} } \, = \,\,0$$

because $$$\partial C$$$ would be zero when the integral after the time derivative on the right is cast in the form

$$$\int\limits_C {d\omega } \,\, = \,\,\int\limits_{\partial C} \omega$$$

with $$$d\omega \,\, = \,\,\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} \, \cdot \,d\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over l}$$$.

Is this right?

But that would make Stoke's Theorem in the form

$$$\int {\int\limits_S {(\nabla \times \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} )\, \cdot \,d\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over S} \,\, = \, \oint\limits_{\partial S = C} {\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} \, \cdot \,d} } } \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over l}$$$

be indentically zero for any F whatsoever simply because of the form on the right is zero as you say. Then there would be no Stoke's theorem to begin with. But since there is a Stoke's Theorem,

$$$\oint\limits_C {\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} \, \cdot \,d} \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over l}$$$

is not identically zero for any F whatsoever. So the fact that the integral is zero for any C whatsoever must mean that

$$$\frac{d}{{dt}}\,(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} \, \cdot \,d\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over l} {\rm{)}}$$$

is identically zero. Is this correct? Or did I lose your point?

Originally posted by Mike2

But that would make Stoke's Theorem in the form

$$$\int {\int\limits_S {(\nabla \times \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} )\, \cdot \,d\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over S} \,\, = \, \oint\limits_{\partial S = C} {\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over F} \, \cdot \,d} } } \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over l}$$$

be indentically zero for any F whatsoever simply because of the form on the right is zero as you say. Then there would be no Stoke's theorem to begin with. But since there is a Stoke's Theorem,
i don t see what s wrong with stokes theorem being identically zero. this equation is identically zero, and that s OK. it does not mean that "there is no Stokes theorem" as you say. in fact, this is evidence that stokes theorem is correct. the left hand side of this equation is zero because F is a gradient, and the curl of a gradient is zero, and the right hand side is zero because the boundary of a boundary is zero.

but so what?

Originally posted by lethe
the left hand side of this equation is zero because F is a gradient, and the curl of a gradient is zero, and the right hand side is zero because the boundary of a boundary is zero.

but so what?
Well, if the right hand side is zero simply because it is the boundary of a boundary, then it would be zero irregardless of whether F is a scalar or a vector valued function. And it is this that would make Stoke's theorem of no effect.

Originally posted by Mike2
Well, if the right hand side is zero simply because it is the boundary of a boundary, then it would be zero irregardless of whether F is a scalar or a vector valued function. And it is this that would make Stoke's theorem of no effect.
this stuff is equivalent to the conservation of energy: if you integrate the force field around a closed loop, you get the work done around that loop. it is always zero. this is a nontrivial statement. it follows from either one of two dual statements: the derivative of a derivative is zero, or the boundary of a boundary is zero.

you should check frankel, he has the details.

Originally posted by lethe
this stuff is equivalent to the conservation of energy: if you integrate the force field around a closed loop, you get the work done around that loop. it is always zero. this is a nontrivial statement. it follows from either one of two dual statements: the derivative of a derivative is zero, or the boundary of a boundary is zero.
Yes, we do have both situations in Stoke's theorem in the form of a surface integral to a line integral. The line integral is closed (thus no boundary), and the integrand is an exact differential. But this is not the most general form of Stoke's theorem. The general theorem has an integral over an open region of an exact differential converting to an integral over an ALWAYS CLOSED boundary of that open region of NOT NECESSARILY an exact differential. If the general theorem is not to equate to zero in the general form, then integrating over a closed boundary need not result in zero. For they ALWAYS convert to integration over a CLOSED boundary of one less dimension.

you should check frankel, he has the details.
What pages.

Originally posted by Mike2
Yes, we do have both situations in Stoke's theorem in the form of a surface integral to a line integral. The line integral is closed (thus no boundary), and the integrand is an exact differential. But this is not the most general form of Stoke's theorem.
no, of course it is not the most general form of stokes theorem. but it is still a situation in which stokes theorem applies. In this case, which applies to your derivation, stokes theorem says the following: the closed line integral of a derivative is zero, regardless of whether the integrand is nonzero.

in your derivation, you infer from the fact that the closed line integral is zero that the integrand itself is zero. this is false, as demonstrated by this application of stokes theorem.

The general theorem has an integral over an open region of an exact differential converting to an integral over an ALWAYS CLOSED boundary of that open region of NOT NECESSARILY an exact differential. If the general theorem is not to equate to zero in the general form, then integrating over a closed boundary need not result in zero. For they ALWAYS convert to integration over a CLOSED boundary of one less dimension.
yeah, that s the gist of stokes theorem. it applies to your derivation.

let me show you:
you have
$$\frac{d}{dt}\oint_C\mathbf{F}\cdot d\mathbf{l}$$
and you assume that
$$\mathbf{F}=\nabla\rho$$
so you have the integral of an exact form over a closed cycle. this is zero, by stokes, and no information about the integrand can be inferred.

i have said this before: i don t even have to bother trying to apply stokes theorem, because i can already tell your equation is wrong. if it were correct, there would be no such thing as electric potential.

What pages.
pg 111 and subsequent.

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OK. So what you are saying is that the closed integral of an exact form is zero independent of the what the integrand is, as long as it is an exact form, right? If the integrand was not an exact form, then the closed boundary would not matter, or if the integrand was exact but the integral was not closed that would not make it zero either. But when you have both a closed boundary and an exact form, then it is zero irregardless of the specifics of the exact form. Is this right?

Originally posted by Mike2
OK. So what you are saying is that the closed integral of an exact form is zero independent of the what the integrand is, as long as it is an exact form, right? If the integrand was not an exact form, then the closed boundary would not matter, or if the integrand was exact but the integral was not closed that would not make it zero either. But when you have both a closed boundary and an exact form, then it is zero irregardless of the specifics of the exact form. Is this right?
can you please stop saying "irregardless"? it is not a word.

and yes, the integral of an exact form over a closed cycle is zero (i feel like i have said this already)

I concede your point that just because a closed loop integral is zero regardless of the size of the closed loop doesn't mean that the integrand is zero. The closed loop integral of conserved fields is a counter example where the integral is zero regardless of the loop, but the integrand is not zero. Thank you.

I guess my problem is understanding Stoke's theorem in itself. For it seems that any open integral of the differential of a form converts to a closed integral of the form. But then either it is impossible to integrate the form, or it also is a differential that can be integrated, and this over no boundary, so that everything always equates to zero.

It would seem that when I used Stoke's theorem to show a conserved closed integral from one side of a world sheet to the other, that all I really have is zero equals zero. If the curl of a gradient is zero, than ANY closed loop integral in that field is zero, right?

I suppose the only diff eq I have is the curl of the gradient equals zero with periodic boundary conditions over the closed strings.

Originally posted by Mike2
.... regardless....
thank you
Thank you.
you re welcome

I guess my problem is understanding Stoke's theorem in itself. For it seems that any open integral of the differential of a form converts to a closed integral of the form. But then either it is impossible to integrate the form, or it also is a differential that can be integrated, and this over no boundary, so that everything always equates to zero.
if the form is not exact, and the region of integration is not a boundary, then stokes theorem can tell you nothing about the value of the integral, and in general, it need not be zero.

stokes theorem only applies when the integral involves an exact form or a boundary. and in general, this integral need not be zero.

but when the integral involves both, then it must be zero.

Originally posted by lethe
thank you

you re welcome

if the form is not exact, and the region of integration is not a boundary, then stokes theorem can tell you nothing about the value of the integral, and in general, it need not be zero.

stokes theorem only applies when the integral involves an exact form or a boundary. and in general, this integral need not be zero.

but when the integral involves both, then it must be zero.
I guess I'm asking since when is the integrand of a solvable integral not "exact" as the word is used in Stoke's theorem. Doesn't "exact differential form" exactly mean that it is integrateable? Since integration is the inverse of differentiation, how can you integrate anything else but a differential?

If that is the case, then Stoke's theorem would always lead to an integral over a closed boundary. Isn't $$\partial\ C$$ always itself without a boundary, in other words, the boundary of $$\partial\ C$$ is always zero?

So if we apply Stoke's theorem more than once you always end up with a closed integral whose boundary is zero of an integrand that is always an exact differential (=integrateable?), then doesn't it always equate to zero? I don't know how you can escape the second application of Stoke's theorem being over a boundary of zero since the first application give a closed boundary. But I might be mistaken about an exact differential form being the same a some integrand that is integrateable.

Originally posted by Mike2
I guess I'm asking since when is the integrand of a solvable integral not "exact"
the integrand is exact when it is the derivative of something else. when it is not, then it is not exact.

for example, the form

$$\mathbf{B}\cdot d\mathbf{l}$$
is not exact, for the magnetic field, say, in the vicinity of a wire carrying a current. and therefore stokes theorem is not trivial for an integral of this expression. it need not be zero. in particular, if you integrate this expression over a curve that is not closed (has boundary), then stokes theorem does not apply at all.

on the other hand, the form
$$\mathbf{E}\cdot d\mathbf{l}$$
is exact, and so the integral of this around a closed curve is always zero, even if the curve is itself not the boundary of something else.

as the word is used in Stoke's theorem. Doesn't "exact differential form" exactly mean that it is integrateable?
no. every smooth integrand is integrable, whether or not it is closed or exact.

Since integration is the inverse of differentiation, how can you integrate anything else but a differential?
that s right. you cannot integrate anything other than a differential form. this does not imply that the differential must be exact or closed.

If that is the case, then Stoke's theorem would always lead to an integral over a closed boundary. Isn't $$\partial\ C$$ always itself without a boundary, in other words, the boundary of $$\partial\ C$$ is always zero?
yes. $$\partial C$$ is a boundary, and it is closed (without boundary), because $$\partial^2=0$$

So if we apply Stoke's theorem more than once you always end up with a closed integral whose boundary is zero of an integrand that is always an exact differential (=integrateable?)
only when the integrand is exact, can you apply stokes theorem this way, and it is only in this case that the integration becomes trivial.

, then doesn't it always equate to zero? I don't know how you can escape the second application of Stoke's theorem being over a boundary of zero since the first application give a closed boundary. But I might be mistaken about an exact differential form being the same a some integrand that is integrateable.
yes, you are mistaken. good thing too, or else we would all look pretty foolish for using a theorem that is trivial.

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Yes, there is some info on exact differential forms at:

http://mathworld.wolfram.com/ExactDifferential.html

It is interesting that they define exact in terms of path independence.

I'm wondering is there a similar definition for exact differential forms of higher dimensionality, such as p(x,y)dxdy+... ? Are they independent of the boundary they are integrated over?

You may remember that I put up a web page stating that the closed line integral is conserved along the world-sheet. I used Stoke's theorem. But now on second thought, isn't it true that any closed line integral can be equated to a surface integral? Now, since the integrand is the gradient of scalar function, then ANY open surface integral of the curl of that gradient of a scalar would be zero. Thus, any closed line integral of the gradient of a scalar is always zero. For we would not have to define the open surface we would integrate over if we were to equate the line integral to a surface integral. It wouldn't matter becaue any such surface integral would always be zero.

My point? So what if zero is conserved. We could not distinquish between no interaction from the situation of many interactions because both cases would result in zero. Oh well, back to the drawing board.

Originally posted by Mike2
Yes, there is some info on exact differential forms at:

http://mathworld.wolfram.com/ExactDifferential.html

It is interesting that they define exact in terms of path independence.

I'm wondering is there a similar definition for exact differential forms of higher dimensionality, such as p(x,y)dxdy+... ? Are they independent of the boundary they are integrated over?
their definition is a little limited: it only applies to 1-forms.

here is my definition: a differential form a is exact iff a=db where b is some other differential form.

this definition is equivalent to theirs, in the case that a is a 1-form. i use stokes theorem to prove the equivalence.

you could describe them as forms that are "path independent", although you will have to change the word "path" to something else, since you don t integrate a higher order differential over a path, but a higher dimensional surface.

frankel probably has a definition as well, and it probably agrees with mine, not with mathworld.

Originally posted by Mike2
I used Stoke's theorem. But now on second thought, isn't it true that any closed line integral can be equated to a surface integral?
only if the closed loop is the boundary of some surface. this is not necessarily always satisfied. in some spaces with nontrivial topology, there exist closed loops that are not themselves the boundary of some surface.

but in R3, this is always satisfied: a closed loop is always the boundary of some surface. R3 has trivial topology.

Now, since the integrand is the gradient of scalar function,
why is the integrand a gradient? i thought you said any integral. now you are restricting yourself to integrals of exact forms. why?

this is the assumption that makes all your calculations turn out to be trivially zero. not all differentials are exact, and not all integrations are trivial. see above example.

then ANY open surface integral of the curl of that gradient of a scalar would be zero. Thus, any closed line integral of the gradient of a scalar is always zero.
yes, of course. but this is a trivial statement, because the curl of a gradient is always zero.

For we would not have to define the open surface we would integrate over if we were to equate the line integral to a surface integral. It wouldn't matter becaue any such surface integral would always be zero.

My point? So what if zero is conserved.
yes, "so what?" indeed.
We could not distinquish between no interaction from the situation of many interactions because both cases would result in zero. Oh well, back to the drawing board.
i would recommend "back to the classroom" instead. master as much of frankel as you can possibly stomach. there is quite a lot of good mathematics in there.

Originally posted by lethe

yes, "so what?" indeed.
Yes, what I'm trying to get at is whether this concept of a conserved integral is trival or not. I started with Stoke's theorem using a surface that could then be split into two separate closed curves with a cylindrical surface between them. And this integral turned out to be conserved (one side equals the other) because Stoke's theorem says that the surface integral and thus the line integral was zero. If the whole line integral is zero, then the two separate halves must be equal. But then I thought that each of the separate closed curve integrals would be trivially zero because they could just as easily serve as the closed boundary of ANY surface integral of the curl of a gradient of a scalar. This type of conservation would be useless because you could never know how many interactions it took to get to the end, a million times zero is equal to one times zero. I've already taken this much off of my website.

i would recommend "back to the classroom" instead. master as much of frankel as you can possibly stomach. there is quite a lot of good mathematics in there.
If I had the time and money... perhaps. But there is such a plethera of math that I'd want to limit search to what is relevant. I wouldn't bother at all except that the theorists seem to be putting a lot of effort into begging the question with their curve fitting methods when they should be trying to make physics a parameterization of logic. You're not going to finish physics by postulating the existence of something you can't explain.

Originally posted by Mike2
Yes, what I'm trying to get at is whether this concept of a conserved integral is trival or not.
there are many conserved integrals that are not at all trivial. here is one:

$$\begin{gather*} Q=\sideset{}{}\int_\text{spatial slice} j^0_\text{EM}d\sigma\\ \frac{dQ}{dt}=0 \end{gather*}$$

.... because Stoke's theorem says that the surface integral and thus the line integral was zero. If the whole line integral is zero, then the two separate halves must be equal. But then I thought that each of the separate closed curve integrals would be trivially zero because they could just as easily serve as the closed boundary of ANY surface integral of the curl of a gradient of a scalar. This type of conservation would be useless because you could never know how many interactions it took to get to the end, a million times zero is equal to one times zero.
yes, your "conservation law" was trivial, and i would not call it a conservation law at all. i hope it is clear why, at this point.

I wouldn't bother at all except that the theorists seem to be putting a lot of effort into begging the question with their curve fitting methods
what curve fitting methods? i don t think curve fitting comes up in frankel at all.

when they should be trying to make physics a parameterization of logic.
how do you "parametrize logic"? logic is a language. i can t even imagine how to make sense of this notion.

You're not going to finish physics by postulating the existence of something you can't explain.
i have no idea what you mean here.

look, if you know better than professional physicists how to "fix" the "curve fitting" methods that modern physicists use, and incorporate logic properly, by all means, write up your proposal. but if your calculations are going to make use of a theorem like Stokes theorem, you had better make damn sure that you know how to use stokes theorem, and for such things, education is key. this is why i suggest reading frankel. i can recommend many other books too, if frankel doesn t meet your needs.

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Originally posted by lethe
look, if you know better than professional physicists how to "fix" the "curve fitting" methods that modern physicists use, and incorporate logic properly, by all means, write up your proposal. but if your calculations are going to make use of a theorem like Stokes theorem, you had better make damn sure that you know how to use stokes theorem, and for such things, education is key. this is why i suggest reading frankel. i can recommend many other books too, if frankel doesn t meet your needs.
Isn't that how they started using "string theory" to begin with, the data seem to fit a function that also describe strings? No, they don't seem to be starting from logic and working their way to physics (Top down). But they are looking for the best equation (curve) that fit the data and imposing a physical interpretation on the elements of the equation (Bottom up). I think that we should not lose sight of either and put some effort into Top down perspectives.

The trouble with studying all the available math and physics is that you can get lost in the forest of too much information. I've read some of the books mostly for informational purposes to see what is available. But I don't use it every day, and I didn't do the exercises. So I am not pretending to know everything or that I've have a complete theory yet. What I have is a strong intuitional sense that I am starting to describe. And before I get too far, I ask for comments from those who do have more experience is the mathematical details. Thank you.

Originally posted by Mike2
Isn't that how they started using "string theory" to begin with, the data seem to fit a function that also describe strings?
OK, i admit that you are correct. string theory was first discovered by finding a function to fit the Regge slope.

but that is not at all what string theory looks like today. since there are no experiments to check string theory against, all research done these days in string theory is purely mathematical. the Regge slope has nothing to do with strings, according to current theory. it was just a coincidence.