A diff Eq on strings, check out the math.

  • Thread starter Mike2
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  • #1
Mike2
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Check my math!

I've derived a differential equation for strings starting from Stokes Theorems to show that energy is conserved along the world-sheet. These diff eq's involve connection coefficients. And I'm not really sure what it all meants yet. I would appreciate it if some who are more skilled in the art would take a look at this and comment.

The math can be found at:

http://www.sirus.com/users/mjake/diffeq.html [Broken]

Thanks.
 
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Answers and Replies

  • #2
Mike2
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My questions are:
1) Are these equations valid, or have I made a mistake?
2) Are they useful, or just a curiosity?
3) What should be the next step?
 
  • #3
Mike2
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See link in original post.

What does it mean that [tex]b_{\sigma t} = 0[/tex] ? Does this mean that the surface is a geodesic?

And what does it mean that [tex]\Gamma^t_{\sigma t} = 0[/tex] ? Does this mean that there is no force in the direction of time? If F is still an arbitrary field, then does this mean that the string must be travelling in a frame where the time component is zero?
 
  • #4
lethe
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what are you trying to show with this?

what is this F vector field that you define everywhere?

why are you working in R<sup>3</sup>?

normally, in string theory, all vector fields are defined on the worldsheet, instead of on R<sup>3</sup>, which seems very unnatural to me.

halfway through the page, you claim that some quantity is constant, and therefore its derivative with respect to time is zero. which time are you talking about here? am i to infer that this is to be a nonrelativistic theory?

then you go on to claim that you live in R<sup>n</sup>, not R<sup>3</sup>. this is good, i find this much less restricting, but i should mention that the vector calculus you are using throughout this page is only valid in R<sup>3</sup> (like the vector cross product). i m not saying that the results are incorrect, since most of the vector calculus generalizes nicely to any manifold.

you mention that book by frankel. you should read it to learn the language of vectors and forms on manifolds.

you go on and use a bunch of results that rest on the existence of a normal vector to your worldsheet. again, this is only something you have if the worldsheet lives in R<sup>3</sup> although results can be generalized.

finally, at the end, your last equation is seriously messed up. in the last equation, you are adding a scalar to two vectors.

big no-no! the math police will be knocking on your door to take you away soon. and its a good thing too, because if you can prove that any surface in R<sup>n</sup> is flat, just by assuming the existence of an external field, we are all in big trouble.

and by the way, that external field bothers me. what is it? where does it come from?
 
  • #5
lethe
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Originally posted by Mike2
My questions are:
1) Are these equations valid, or have I made a mistake?
i think everything is at least close to correct, except for the very end, where you forget a dot product.


2) Are they useful, or just a curiosity?
unfortunately, as it stands, i don t think this is useful.

3) What should be the next step?

well, i think it contains a mistake, so the first step would be to see if you believe me that there is a mistake, and if so, to fix it.

then, i would strongly recommend you leave vector calculus behind, and embrace differential geometry. you are needlessly restricted by your language and notation.

you can actually find an intro to that stuff somewhere on this forum.

finally, most importantly, if you want this to have any relevance to physics, your model should have a physical interpretation. like, i think you should make more of an effort to make your theory lorentz covariant. and you should come up for some excuse for this F field. it scares me.
 
  • #6
Mike2
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First of all, I want to thank you for responding. I didn't want to get too far off on a tangent if my fundamentals were off. Thanks.


You wrote:
Originally posted by lethe
what are you trying to show with this?

what is this F vector field that you define everywhere?

You might be interested in my webpage at:

http://www.sirus.com/users/mjake/StringTh.html [Broken]

Here I try to explain how physics can be derived from logic. The precepts of logic can be represented graphically, and a coordinate system can be imposed on this set. This is nothing more than the description of a manifold. I'm attempting to show that all events in reality, since they must comply with logic, can be derived from the geometry of events in sample space growing with time. Ambitious, I know. But I don't have a PhD to protect, so I can affort to risk making mistakes along the way.

The vector field F that I refer to is the gradient of an underlying probability density function. Surely, the world-sheet sweeps out more samples (of at least space time) with time. And this probability density function describes the density of the samples that are being added with time. What are these sample of whose probability density function is being used? I don't know. It may not matter. Whatever it is, the same mathematics applies.


why are you working in R3?

Yes, yes. I do intend to generalize using the notation of differential geometry and tensors on manifolds, etc. But I want to make sure I was headed in the right direction using something I could at least visualize.




normally, in string theory, all vector fields are defined on the worldsheet, instead of on R3, which seems very unnatural to me.

Are you saying that manifolds of higher dimensions do not have a tangent space? I'm not quite sure what you are saying.


halfway through the page, you claim that some quantity is constant, and therefore its derivative with respect to time is zero. which time are you talking about here? am i to infer that this is to be a nonrelativistic theory?

If you understood the part about the line integral be conserve along the world sheet, then isn't that the definition of being conserved, that their time derivative is zero?



finally, at the end, your last equation is seriously messed up. in the last equation, you are adding a scalar to two vectors.
.


You mean I went from:

[tex]
\,\frac{{d\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over F} }}{{dt}}\,\,\, \cdot \,\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over x} _\sigma + \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over F} \cdot (\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over x} _\sigma \Gamma _{\sigma t}^\sigma + \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over x} _t \Gamma _{\sigma t}^t + b_{\sigma t} \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over N} )
[/tex]


To:

[tex]
\,(\frac{{d\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over F} }}{{dt}} + \Gamma _{\sigma t}^\sigma \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over F} {\rm{)}}\,\, \cdot \,\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over x} _\sigma + \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over x} _t \Gamma _{\sigma t}^t + b_{\sigma t} \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over N} {\rm{ = 0}}
[/tex]

When it should be:

[tex]
\,(\frac{{d\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over F} }}{{dt}} + \Gamma _{\sigma t}^\sigma \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over F} {\rm{)}}\,\, \cdot \,\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over x} _\sigma + \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over F} \,\, \cdot \,\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over x} _t \Gamma _{\sigma t}^t + \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over F} \,\, \cdot \,\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over N} b_{\sigma t} {\rm{ = 0}}
[/tex]

You're right! Thanks! If it weren't so pains taking to write out these formulas in html, I wouldn't be so distracted. And this forum's LaTex interpreter is still flakey, not regenerating gifs after editing and resubmitting.

Still if those vectors are linearly independent then vectors in those directions are also independent so that the results are the same. Or is it that if those vectors are independent AND F is not zero, then the results are the same.
 
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  • #7
lethe
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Originally posted by Mike2
First of all, I want to thank you for responding. I didn't want to get too far off on a tangent if my fundamentals were off. Thanks.
no sweat.


Here I try to explain how physics can be derived from logic. The precepts of logic can be represented graphically, and a coordinate system can be imposed on this set. This is nothing more than the description of a manifold. I'm attempting to show that all events in reality, since they must comply with logic, can be derived from the geometry of events in sample space growing with time. Ambitious, I know. But I don't have a PhD to protect, so I can affort to risk making mistakes along the way.
i m not sure what this means "to derive string theory from logic alone". every derivation follows the same format: start with some physical assumptions, use logic to see what must follow from those assumptions, arrive at a result, hopefully something which can be measured in the lab.

logic is the only way any reasoning is ever done, but without some physical assumptions, you do not have a physical theory.


The vector field F that I refer to is the gradient of an underlying probability density function.
so here is one of your physical assumptions. that there is some probability density function.

alright, i still don t like it very much, but i will accept it for now.
Surely, the world-sheet sweeps out more samples (of at least space time) with time. And this probability density function describes the density of the samples that are being added with time. What are these sample of whose probability density function is being used? I don't know. It may not matter. Whatever it is, the same mathematics applies.
didn t follow this at all. what is a "sample", in this context?


Yes, yes. I do intend to generalize using the notation of differential geometry and tensors on manifolds, etc. But I want to make sure I was headed in the right direction using something I could at least visualize.
fair enough. like i said, using the language of euclidean R3 will be a hindrance to you eventually, but for now, i suppose this is OK.


Are you saying that manifolds of higher dimensions do not have a tangent space? I'm not quite sure what you are saying.
that is not at all what i am saying. what i am saying is that the fundamental fields in string theory, as it is usually formulated, are vector fields on the worldsheet (that is, they are vector (or spinor) valued functions of &sigma; and &tau;, not of t, x, y, and z. those fields appear only as low energy effective field theories from the string spectrum.

you, on the other hand, are starting with this field F that is defined on the spacetime, not on the world sheet.


If you understood the part about the line integral be conserve along the world sheet, then isn't that the definition of being conserved, that their time derivative is zero?
yeah, that s fine, i just wish it were covariant.



You mean I went from:
.......
You're right! Thanks! If it weren't so pains taking to write out these formulas in html, I wouldn't be so distracted. And this forum's LaTex interpreter is still flakey, not regenerating gifs after editing and resubmitting.

Still if those vectors are linearly independent then vectors in those directions are also independent so that the results are the same. Or is it that if those vectors are independent AND F is not zero, then the results are the same.

no! the fact that the sum of the inner product of some vectors is zero cannot be used to infer that the sum of the vectors must be zero. not even if the vectors are linearly independent. a scalar equation contains much less information than a vector equation.

for example, if F points in the 4th spatial dimension, and x_sigma, x_tau, and N point in the i, j, k directions, the scalar product will be zero, even if none of the coefficients is zero.

and this is a good thing! if you had somehow proved that the existence of F makes the Christoffel symbols vanish, then it would be impossible to have nonflat surfaces! which is crazy! i would have to report you to the math police for torture.
 
  • #8
Mike2
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Originally posted by lethe
no sweat.



i m not sure what this means "to derive string theory from logic alone". every derivation follows the same format: start with some physical assumptions, use logic to see what must follow from those assumptions, arrive at a result, hopefully something which can be measured in the lab.

logic is the only way any reasoning is ever done, but without some physical assumptions, you do not have a physical theory.

Actually, the probability density function is necessary to describe some kinds of logic in this graphical representation.



so here is one of your physical assumptions. that there is some probability density function.

alright, i still don t like it very much, but i will accept it for now.

I think my "physical assumption" was to introduce time. Events changing with time give the string as a boundary. Yes, this 2D world-sheet does not enclose any volume in more than a 3D world. But the 2D sheet itself is a manifold that grows with time.


didn t follow this at all. what is a "sample", in this context?

Like I said, it doesn't really matter what the samples are samples of. The math is still the same. Relying only on the math and forgetting what kind of objects fills the space is common in the study of probability spaces.



that is not at all what i am saying. what i am saying is that the fundamental fields in string theory, as it is usually formulated, are vector fields on the worldsheet (that is, they are vector (or spinor) valued functions of &sigma; and &tau;, not of t, x, y, and z. those fields appear only as low energy effective field theories from the string spectrum.

you, on the other hand, are starting with this field F that is defined on the spacetime, not on the world sheet.


If the world sheet can go anywhere in space, then isn't this saying that the field exists everywhere? It may be that the only thing giving this field any definate meaning or value is the boundary conditition imposed by the world-sheet. What good is it if you have a diff eq without boundary conditions?

Or, since everything is evaluated on the world sheet, how could one tell if they exist off the world sheet?





no! the fact that the sum of the inner product of some vectors is zero cannot be used to infer that the sum of the vectors must be zero. not even if the vectors are linearly independent. a scalar equation contains much less information than a vector equation.

for example, if F points in the 4th spatial dimension, and x_sigma, x_tau, and N point in the i, j, k directions, the scalar product will be zero, even if none of the coefficients is zero.

and this is a good thing! if you had somehow proved that the existence of F makes the Christoffel symbols vanish, then it would be impossible to have nonflat surfaces! which is crazy! i would have to report you to the math police for torture.

I thought you were the math police.:smile:

Yes, I caught that mistake after rewriting it again. That eq is not the sum of linear independent vectors. So I have no right to say their coefficients are zero. These inner products can be interpreted as projecting everything parallel to F. So I cannot say each term is zero. That's what I get trying to think after a 9 hour day at work while watching TV at the same time. eeessshhhh.

Thanks for the help
 
  • #9
lethe
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Originally posted by Mike2
Actually, the probability density function is necessary to describe some kinds of logic in this graphical representation.
i have no idea what that means.


I think my "physical assumption" was to introduce time.
perhaps it was. but if so, it was a much milder assumption. most mathematical systems can be parametrized by a real number, which we might as well call time. this is not so bold.

on the other hand, most physical models start with a vacuum, and introduce fields on spacetime only to represent something physical, like the gravitational, electromagnetic, or fermionic field.

here you have a field, but i have no idea what physical quantity it represents.

Events changing with time give the string as a boundary. Yes, this 2D world-sheet does not enclose any volume in more than a 3D world. But the 2D sheet itself is a manifold that grows with time.

grows with time? what does that mean? its area increases? why?

Like I said, it doesn't really matter what the samples are samples of. The math is still the same. Relying only on the math and forgetting what kind of objects fills the space is common in the study of probability spaces.
it is a common practice in all of modern mathematics to forget what the mathematical objects physically represent. they only rely on mathematical formalism

however, i thought we were doing physics, not mathematics. in physics, every term in your equation should represent something physical, and you should be able to tell me what. i have yet to figure out what this probability field represents physically.



If the world sheet can go anywhere in space, then isn't this saying that the field exists everywhere?
No, not at all! just because the worldsheet can go everywhere, doesn t mean that it does! it certainly doesn t fill all spacetime, so there will always be points in spacetime where a field on the worldsheet is undefined.

It may be that the only thing giving this field any definate meaning or value is the boundary conditition imposed by the world-sheet. What good is it if you have a diff eq without boundary conditions?
not much good.

Or, since everything is evaluated on the world sheet, how could one tell if they exist off the world sheet?
if something is only defined on the worldsheet, it doesn t have any meaning to ask if it exists off the worldsheet.

I thought you were the math police.:smile:
i m on sebatical this semester.

Yes, I caught that mistake after rewriting it again. That eq is not the sum of linear independent vectors. So I have no right to say their coefficients are zero. These inner products can be interpreted as projecting everything parallel to F. So I cannot say each term is zero. That's what I get trying to think after a 9 hour day at work while watching TV at the same time. eeessshhhh.

Thanks for the help

yeah, you seem to have a good grasp of vector calculus, and just made a silly mistake.

but i don t think any of this has anything to do with string theory.
 
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  • #10
Mike2
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Originally posted by lethe
i have no idea what that means.

Surely you are aware that a probability density function is required in the study of probability spaces from which we get inductive logic, right?


on the other hand, most physical models start with a vacuum, and introduce fields on spacetime only to represent something physical, like the gravitational, electromagnetic, or fermionic field.

here you have a field, but i have no idea what physical quantity it represents.

That's precisely what I am trying to avoid. If I postulate the existence of some physical quantity as an axiom of my system, then this commits the logical error of petitio principii, or a.k.a. begging the question. It is like trying to explain life on earth by suggesting that aliens from outer space fertilized the earth. The question is then begged, "well, then where did the aliens come from?"

If we are ever to achieve a TOE, then that explaination cannot start with something physical since that is what you are trying to explain. However, if it can be shown that logic dictates certain kinds of interactions, and reality is logical, then we must admit that these interaction do take place in reality. And we should antisipate their manifestation. The only alternative is to assert that reality contains inconsistencies.


grows with time? what does that mean? its area increases? why?

It really wouldn't hurt you to read my website at:

http://www.sirus.com/users/mjake/StringTh.html#consider [Broken]

Time gives us more samples of something. Every instant of time is a different sample of everything. Haven't you looked at something and then looked again to make sure? So time gives us more samples parameterized by some coordinate system, a growing manifold. But you might ask why time should be so necessarilly logical. Well, first there is nothing, then there is something. Whatever it is, it is represented as a growing manifold, since it is an object parameterizable with coordinates. Something from nothing implies a growth with time.

The probability density function is just a generic description of a distribution of objects, whatever they may be. It doesn't matter. Everything may ultimately be desribed as ether existing or not. If it exists, then it has a density throughout space. First you have whether it exists or not. Then you have how much of it is distributed through space and time. This much applies to everything. There is no escaping it. What I am attempting is the logically necessary laws of physics, no experiment is required. It MUST be.


it is a common practice in all of modern mathematics to forget what the mathematical objects physically represent. they only rely on mathematical formalism

however, i thought we were doing physics, not mathematics. in physics, every term in your equation should represent something physical, and you should be able to tell me what. i have yet to figure out what this probability field represents physically.

Our math can ONLY be a representation of what is real, not the reality of the physical objects themselves. The probability density function is a distribution of whatever might be physical.




No, not at all! just because the worldsheet can go everywhere, doesn t mean that it does! it certainly doesn t fill all spacetime, so there will always be points in spacetime where a field on the worldsheet is undefined.

The Feynman path formulation employs every imaginable path covering all of space. So this is equal to assuming a field throughout all of space and not just on one particular world-sheet.
 
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  • #11
lethe
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Originally posted by Mike2
Surely you are aware that a probability density function is required in the study of probability spaces from which we get inductive logic, right?

no, i am not aware that a probability density function is required to study probability spaces. in fact, i don t know what probability spaces are.


That's precisely what I am trying to avoid. If I postulate the existence of some physical quantity as an axiom of my system, then this commits the logical error of petitio principii, or a.k.a. begging the question. It is like trying to explain life on earth by suggesting that aliens from outer space fertilized the earth. The question is then begged, "well, then where did the aliens come from?"
there are an infinite number of mathematically consistent models. without some physical input, you have no way to tie it to reality.

here is an example: einsteins theory of relativity in a vacuum is a consistent model. it can never be applied to systems with matter in them. in order to describe systems with matter, i have to hypothesize some matter field. it is logically impossible to think that you can derive a physical model without starting with some physical assumptions.

If we are ever to achieve a TOE, then that explaination cannot start with something physical since that is what you are trying to explain. .....
i think you misunderstand what physics is, and what it is capable of.


It really wouldn't hurt you to read my website at:

http://www.sirus.com/users/mjake/StringTh.html#consider [Broken]
meh. not enough interest.

Time gives us more samples ....
i still have no idea what this means.

... What I am attempting is the logically necessary laws of physics, no experiment is required. It MUST be.

good luck with that. obviously, i believe this approach is fundamentally misguided.


Our math can ONLY be a representation of what is real, not the reality of the physical objects themselves. The probability density function is a distribution of whatever might be physical.
the difference between reality and a model of reality is a semantic issue, in my opinion, or at best a metaphysical one.


The Feynman path formulation employs every imaginable path covering all of space. So this is equal to assuming a field throughout all of space and not just on one particular world-sheet.
you should learn the Feynman path integral formulation a little better.

the path integral formulation evaluates the action over all possible worldsheets, with a field defined locally only on the worldsheet, not in spacetime.
 
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  • #12
Mike2
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Originally posted by lethe

there are an infinite number of mathematically consistent models. without some physical input, you have no way to tie it to reality.

If you base your theory on observation and measurement, these are only contingencies and interpretations, which could be wrong. Are you sure you measured right? Maybe you need to look more closely. Perhaps the truth is in the fine details. The trial and error methods is used only because you are not sure of the contingencies you use as the basis of your argument or theory.

You can invent a theory based on the existence of unicorns. For example, where unicorns go, fairies are sure to follow. But most have not seen a unicorn. And others may have only thought they saw a unicorn.

Yes, there are an infinite number of contingent systems that are consistent. Each of these is a list of propositions that have a different interpretation than some other contingent system. But I am not supposing that we start with some contingent suppositions. Consistency is proven by abstracting these contengencies to the level of general propositions and showing that none contradicts the others. At this level of abstract logic, there is only one consistent system; we call it logic. What I am proposing is that we do not leave this level. If we don't then our conclusions must be right and apply to everything in reality.



here is an example: einsteins theory of relativity in a vacuum is a consistent model. it can never be applied to systems with matter in them. in order to describe systems with matter, i have to hypothesize some matter field. it is logically impossible to think that you can derive a physical model without starting with some physical assumptions.

Then there must be something incomplete about his theory. In otherwords the theory fails because it doesn't apply to every proposition or "event" whatsoever, but its validity is contingent on which set of propositions you use.



quote:
--------------------------------------------------------------------------------
Time gives us more samples ....
--------------------------------------------------------------------------------

i still have no idea what this means.

What? The propogation of a string is described as a world sheet. This world sheet is described as a 2D manifold. It's area increase with time. It is a manifold that "grows" with time. I'm not proposing anything else.

The "sample space" I speak of is nothing more that the overall manifold of space-time that you speak of. Unions and intersections are included in both. The "open events" I speak of is nothing more than the world-sheet that you speak of. The "probability density function" is probably nothing more than the "background" potential that you speak of. There is only a semantic difference. It is due to the path of study I took to get here. Whereas physicists merely postulate strings, world-sheets, and manifolds, the approach I take has a much more appealing feature of being a parameterization of logic. The postulated view only begs the question of its postulates. But my view is the natural extention of logic.

Do I really need to justify the existence of this "background potential" of a "probability density function"? Surely, I can propose a function that describes whether something exists at a given point or not. If anything exists at all in space-time (including space-time), then that is a legitamate function. You can't deny it without denying any and all of existence itself. For you are saying that there is no way to describe that something exists there. So whatever may exist, whether particle or field, it is fair to describe it with a function that says it exists there. Then if there exists more here than there, then that function may be interpreted as a density function. And then if we speak of a relative density with respect to a region normalized to 1, then you are speaking of what is in effect a probability density function.


you should learn the Feynman path integral formulation a little better.

the path integral formulation evaluates the action over all possible worldsheets, with a field defined locally only on the worldsheet, not in spacetime.


Still each point in space time would have a sheet going through it with a field on it. So what's the difference? You still end up with a field at each point is space-time.
 
  • #13
lethe
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Originally posted by Mike2
So what's the difference?

the difference is the domain of definition, which is quite a significant difference.
 
  • #14
ObsessiveMathsFreak
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I think there's one problem.

When you say that the circulation around any loop, and hence the rate of change of circulation about any loop, is zero you then say that it follows that the rate of change of the entire function is everywhere zero.

This does not nessesarily follow.

For instance it is not the case for gravity.
 
  • #15
lethe
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Originally posted by ObsessiveMathsFreak
I think there's one problem.

When you say that the circulation around any loop, and hence the rate of change of circulation about any loop, is zero you then say that it follows that the rate of change of the entire function is everywhere zero.

This does not nessesarily follow.

yes, i didn t notice this before, but Obsessive is correct.
 
  • #16
Mike2
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Do you mean where I bring the derivative inside the integral? Or do you mean when I say the integral is zero for any curve, therefore the integrand is zero?
 
  • #17
lethe
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the latter.
 
  • #18
Mike2
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Originally posted by lethe
the latter.

Think about it. Take as an example an ordinary integral of f(x) over some interval. If it happens that this integral is zero no matter what interval you integrate over, it must be that f(x) is zero identically. Do you agree with this? When is this not the case?

What I showed is that the line integral of the gradient of a scalar over a closed curve is conserved no matter what world-sheet, or world-tube connects initial and final closed curves. It doesn't matter what size or shape the closed curve took. The integral is still conserved no matter how it propogates. The derivative of a constant is zero. Since it is legitimate to take the derivative inside the integral since the variable of integration is independent of the variable of differentiation, the integral of the derivative is zero no matter what the interval of integration. So it certainly seems as though this implies that the integrand itself is zero. Does this help?
 
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  • #19
lethe
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Originally posted by Mike2
Does this help?

i thought you were the one asking for help. i do not need help. you are in error. do you need help to see your error?
 
  • #20
Mike2
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Originally posted by lethe
i thought you were the one asking for help. i do not need help. you are in error. do you need help to see your error?

At this point, I am open to corrections. If you've found an error, then by all means, state what it is. What exactly is my error?
 
  • #21
lethe
653
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Originally posted by Mike2
At this point, I am open to corrections. If you've found an error, then by all means, state what it is. What exactly is my error?

you state that because the integration is zero over all possible closed loops, the integrand must be zero. this is false. i can see this in a couple of ways.

first, let me look at your explanatory example of a real valued function. it is true that if the integral of a function along any path from a to b is 0, then the function must be identically 0, but only if the points a and b are not the same point.

if they are the same point, then the integration must give me zero for a broad class of nonzero integrands:

on the real line, for example, you see
[tex]
\oint_C f(x) dx = \int_a^b f(x) dx + \int_b^a f(x) dx=0
[/tex]
which is identically zero, for any &fnof;

this argument can be generalized to [tex]\mbox{$\mathbb{R}^3$}[/tex] and above.

alternatively, you can make a physical argument. in electrostatics, for example, the electric field is the gradient of electric potential. therefore (static) electric field is a conservative field, and its integration around any loop must be zero.

if your argument were correct, then i could conclude from this that the integrand [tex]\mbox{$\mathbf{E}\cdot d\mathbf{s}$}[/tex] would have to be zero everywhere. in other words, the potential difference between any 2 points is zero.

clearly, this is not true.

OK, does that help?
 
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  • #22
Mike2
1,313
0
Originally posted by lethe
you state that because the integration is zero over all possible closed loops, the integrand must be zero. this is false. i can see this in a couple of ways.

first, let me look at your explanatory example of a real valued function. it is true that if the integral of a function along any path from a to b is 0, then the function must be identically 0, but only if the points a and b are not the same point.

if they are the same point, then the integration must give me zero for a broad class of nonzero integrands:

on the real line, for example, you see
[tex]
\oint_C f(x)\! dx = \int_a^b f(x)\! dx + \int_b^a f(x)\! dx=0
[/tex]
which is identically zero, for any &fnof;

this argument can be generalized to [tex]\mbox{$\mathbb{R}^3$}[/tex] and above.

I don't think so, because on the real line you necessarily have to retrace your steps to get back to "a". In higher dimensions you don't have to retrace your steps to get back to "a". Isn't the closed curve parameterized by the real line? So if my argument works for the real line, then it should be valid for anything parameterized by the real line.


alternatively, you can make a physical argument. in electrostatics, for example, the electric field is the gradient of electric potential. therefore (static) electric field is a conservative field, and its integration around any loop must be zero.

if your argument were correct, then i could conclude from this that the integrand [tex]\mbox{$\mathbf{E}\cdot d\mathbf{s}$}[/tex] would have to be zero everywhere. in other words, the potential difference between any 2 points is zero.

clearly, this is not true.

I don't see the derivative of your line integral being used in your argument. If you say that the integration around ANY loop is zero, then how can it be that the time derivative of this closed line integral is not zero? Or are you just testing my language skills?
 
  • #23
lethe
653
0
Originally posted by Mike2
I don't think so, because on the real line you necessarily have to retrace your steps to get back to "a". In higher dimensions you don't have to retrace your steps to get back to "a". Isn't the closed curve parameterized by the real line? So if my argument works for the real line, then it should be valid for anything parameterized by the real line.
my argument is independent of parametrization. i can parametrize my curve by the real line, if you like.

it is also independent of the fact that the path "retraces itself". i like this particular example, because it makes it really obvious why the integration of an exact form must be zero along any closed curve: to go around a closed form, you have no choice but to retrace your steps. for every step you take along the postive z-axis, you have to come back down, if you are going to end where you started. the same for all axes. the argument is general.



I don't see the derivative of your line integral being used in your argument. If you say that the integration around ANY loop is zero, then how can it be that the time derivative of this closed line integral is not zero? Or are you just testing my language skills?

you argue the following: since the integral of something is zero for any closed loop, the integrand must be zero

this is false, independently of whether the integrand is the time derivative of something else.
 
  • #24
Mike2
1,313
0
Originally posted by lethe

you argue the following: since the integral of something is zero for any closed loop, the integrand must be zero

this is false, independently of whether the integrand is the time derivative of something else.

I would appreciate a more complete proof. You agreed with me in the case that if the integral over any interval on the real line (in one direction only) was zero that the integrand is zero. And you've agreed that it is legitimate to parameterize the closed curve with the real line. So I don't see your objection. Do you have a better counter-example?
 
  • #25
lethe
653
0
Originally posted by Mike2
I would appreciate a more complete proof. You agreed with me in the case that if the integral over any interval on the real line (in one direction only) was zero that the integrand is zero. And you've agreed that it is legitimate to parameterize the closed curve with the real line. So I don't see your objection. Do you have a better counter-example?

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

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.

the integration is independent of the parametrization, so it does not matter that i can use the real line to parametrize the integration.
 
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  • #26
Mike2
1,313
0
Originally posted by lethe
you want a complete proof? stokes theorem says [tex]\mbox{$\int_Cd\omega = \int_{\partial C} \omega$}[/tex]. so if [tex]\mbox{$\quad C$}[/tex]
is a circle, then [tex]\mbox{$\partial C$}[/tex] is zero, regardless of whether [tex]\mbox{$\omega$}[/tex] vanishes or not.


"if C is a circle, then [tex]\mbox{$\partial C$}[/tex] 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.
 
  • #27
lethe
653
0
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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  • #28
Mike2
1,313
0
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 [tex]\mbox{$\partial C$}[/tex]=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?
 
  • #29
lethe
653
0
Originally posted by Mike2
Have you ever seen the notation where [tex]\mbox{$\partial C$}[/tex]=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.
 
  • #30
Mike2
1,313
0
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

[tex]
\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
[/tex]

because [tex]\[\partial C\][/tex] would be zero when the integral after the time derivative on the right is cast in the form

[tex]\[
\int\limits_C {d\omega } \,\, = \,\,\int\limits_{\partial C} \omega
\][/tex]

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

Is this right?

But that would make Stoke's Theorem in the form

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

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,

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

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

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

is identically zero. Is this correct? Or did I lose your point?
 
  • #31
lethe
653
0
Originally posted by Mike2

But that would make Stoke's Theorem in the form

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

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?
 
  • #32
Mike2
1,313
0
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.
 
  • #33
lethe
653
0
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.
 
  • #34
Mike2
1,313
0
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.
 
  • #35
lethe
653
0
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
[tex]
\frac{d}{dt}\oint_C\mathbf{F}\cdot d\mathbf{l}
[/tex]
and you assume that
[tex]
\mathbf{F}=\nabla\rho
[/tex]
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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