Johnny5

Gauss's law tells you that the perpendicular component can only be discontinuos across a surface if there is a surface charge. The parallel component must be continuous.

What if the free charge is zero at the surface? I take it that this means that the perpendicular component must be continuous?

And you are saying the parallel component will be continuous regardless of whether or not there is a surface charge?

What I really want to do is figure out the relationship between the angle of incidence and the angle of reflection, in terms of the permittivities of both mediums. And I want to do that using the boundary conditions. I presume the correct answer is something close to Snell's law.

Mr. Escultura "refuted" trichotomy using the ol' "1 does not equal 0.999..." argument' date=' which we've seen beaten to death here and elsewhere. He argued that it can be both proven that 0.9_ = 1 and 0.9_ < 1, and so trichotomy fails.

http://www.manilatimes.net/national/2004/jun/18/yehey/opinion/20040618opi7.html

 

Needless to say, I'm skeptical, but I'll await further information to see if this gets published in math journals and not merely in a newspaper that Mr. Escultura himself works for.

Anyone who carries out a logical analysis of anything, should be paid attention to, because what if they are right?

I wouldn't call him anything I don't know him, and Andrew Wiles seemed needlessly rude.

There is most definitely some confusion going on as regards

1=.99999999...

But to think you have developed a whole new kind of real number system, is far out there. And it would probably be a help to mr Escultura, to simply show him his error, rather than ridicule him. At least that is my take.

You don't call a scientist who finds an error a kook. He simply is aware that it is not the case that 1=.9999999... and well he is right. They freaking aren't equal.

The LHS and RHS are different kinds of things.

The LHS is a number, and the RHS is an infinite series which doesn't terminate.

PS: Also I don't know if this is true, but years ago a friend of mine told me he watched Andrew Wiles on a PBS broadcast, and when asked about the fact that his "proof" contained an error, Mr. Wiles began crying. My friend found it odd that a mathematician would take it so seriously that he would cry over making a mistake.

No one is perfect all the time... no one.

Unlike some of my recent threads, this one is actually a specific problem I need help with.

I think the answer is snell's law, but its been years since I have solved it.

You can get to Brewster's angle if you keep going.

Here is the problem.

You have an electromagnetic wave incident upon some surface.

There is an incident wave, a transmitted wave, and a reflected wave.

The permittivity constant of the one medium is e₁ and the permittivity constant of the other medium is e₂.

Denote the angle of incidence by q_i, and the transmitted angle by q_t.

Using boundary conditions on the electric field E, find a relationship between the angle of incidence, and the transmitted angle.

Now I have been reading about 5 books on EM, and optics, as far as boundary conditions go.

I remember doing it years ago, and again, none of the books are clear.

I used Gauss' law at the boundary to get the boundary conditions on the E field.

I think Ampere's law gets you the boundary conditions for the B field, but that doesn't matter in this problem, at least i dont think so.

There is a boundary condition for the perpendicular components of the E field, and another for the parallel components.

In the books they just say "electric field is continuous across the boundary."

So here is what i think the answer is, but I'm not sure.

e₁ sin(q_i ) = e₂ sin(q_t )

None of the books cover the vector relationships at the boundary very well.

So any insight would be appreciated.

Thank you

PS: Also, I know that isn't exactly Snell's law, Snell's law is:

n₁ sin(q_i ) = n₂ sin(q_t )

Where n1,n2 are the index of refractions of the mediums.

But there is a relationship between them, the speed of light, and the permittivity.

In general:

n = (em/e₀m₀ )^1/2

Where the numerator is the product of the permittivity and permeability of the medium, and the denominator is for free space.

So equivalently:

n = c(em)^1/2

And letting v denote the speed of light in the medium we also have:

n = c/v

If this is what you thought the issue was then it is a strange thing to say "the square root of -1 leads to the contradiction". Certainly if one assumes i>0' date=' the i^2=-1>0, contradiction, and if i<0, then i*i>0*i, -1>0 again, a contradiction.

[/quote']

 

 

That's the one.

 

Thats how the proof would have to go.

 

if i>0 then problem.

if i= 0 then problem.

if i<0 then problem.

 

Suppose i>0.

By a theorem, i*i>0.

but, i*i=-1, hence -1 can be substituted for i*i, so

-1>0, which is false.

 

Suppose i<0.

 

Product of two negative numbers is positive and a positive number is greater than zero. If (-1)^1/2 is less than zero then i*i>0, and again -1>0, which is false.

 

And of course if i=0 then i*i=0 whence -1=0, which is again false.

 

So you start out with a set U.

 

Then you have a bunch of axioms which hold on the set.

 

You have an undefined binary relation <

 

And the following statement is true:

 

For any x,y elements of U

 

x=y or x<y or x>y

 

Then you encounter i= (-1)^1/2

 

Since the above statement is true, root -1 isn't an element of U.

 

Then, what is done, is to postulate a superset C.

 

So that U is a subset of C.

 

From this it follows that any element of U is also an element of C, but the converse is false.

 

Clearly the axioms/theorems of the real numbers do not all hold on C.

 

Some axioms hold but not all.

 

What really I mean, is that the binary relation < does not have general meaning across the complex numbers.

 

Let me google this and see what I find. I should have thought of that sooner.

 

Here is the first article I came across: Dr Math

 

 

To much GD nonsense on the web.

 

Actually here's something that seems familiar...

 

The nature of complex numbers

 

I'm going to have a look at that.

 

 

Complex numbers are ordered pairs of real numbers for which multiplication is defined in a special way. Let (a,b) and (c,d) to ordered pairs of real numbers. The product of (a,b) with (c,d) is defined as:

 

(a,b)*(c,d) = (ac-bd,ad+bc).

 

The work at that site all looks familiar. I think it was in my complex variables book.

The only comment I have is if this notional reader knew the proper definitions of a set

And what definition is that?

IN some texts 'set' is undefined.

In others, there is a somewhat goofy definition.

Don't go into modulo yet, really i want to settle this issue about i, for myself.

I remember a proof that i leads to some conflict with ordering.

It was shown to me, probably a decade ago, but i never bothered with it.

But you know, if someone else formulated the argument, it means someone else was thinking about this issue, the one you say i didnt raise.

When I said "square root of minus one on its own" i was being deliberately vague.

I am concerned about ordering of real numbers.

You cannot have two parts of your knowledge that conflict.

Argand plane

You have one axis real, the other imaginary.

So can you order imaginary numbers, as the imaginary axis seems to suggest?

i,2i,3i...

I wish I could remember how that proof went.

Because now i know just a bit more than i did back then.

I avoided the existential point about numbers since it is completely immaterial in the context you stated it. When I said that the square root of minus one may or may not exist it was in the mathematical sense of: the square root of minus 1 does not exist (ie is not an element of) in the set of real numbers, similarly, in Z_4, the integers mod 4 there is no element that squares to give 3, but there is one in Z_2.

Yes i know what sense you used the word Matt, don't worry.

There is a conceptual difference between

existence and 'at least oneness'

In some contexts to exist means to be in the current moment in time.

But for anyone using set theory, the existential quantifier translates best as "there is at least one"

How many men are there?

Right now, the current answer is something like four billion.

How many men are there throughout the whole of time?

The answer to this question was always... at least one.

What issue' date=' Johnny? You haven't raised an issue. What on are you getting at? sqrt(-1) is just some object that may or may not exist within some field.

 

If you were to say, does sqrt(-1)=1 lead to any contradiction (in any field except one of characteristic two) then yes, because you're saying 1=-1. But just (sqrt(-1)? it is not a well formed question. an object can not lead to contradictions until you try to relate it to other things.[/quote']

 

I thought you would never get around to answering my question as to whether or not numbers exist... and you seem to have avoided it. Don't worry I wasn't trying to confuse you, I already know the answer. I wanted to see what you would say.

 

So now, let me read through your paragraph here, to see what you have said.

 

 

This part here is quite good...

 

an object can not lead to contradictions until you try to relate it to other things.

 

Let me see what the hell am I really asking hmm...

 

Well lets say we have a set of consistent axioms, the field axioms.

 

We don't want to do anything to disrupt the set of consistent axioms. They actually have to do with mental operations, that we ourselves perform using symbols.

 

So the last thing in the world we want, is to have a conflicting set of rules. That would be bad.

Do not worry every one, for if the person who didn't know about sets was reading the definition for the first time and had taken the time to read it all properly they'd Know that sets do not have repeated elements and that they are unordered by definition. This whole nonsense about trying to infer something from notation alone is silly - why should they even infer that {a,b,c} is a set?

I actually thought about this just this morning, and decided to do the following in my own work...

Earlier in this thread i said i was toying with the idea that using ; might help the reader infer that all the symbols inside the curly braces denote different elements of the set.

I decided to dispense with that idea completely.

The reason being that {a,b,c,d} is so widely used.

My point remains though. You cannot just pretend there was no point. The issue arose, because I am writing something, and I just want to be clear in my presentation of set theory, to a reader totally unfamiliar to it.

And exactly as Matt says, why should a reader even look at {a,b,c} and infer that the symbols denote a set. He should not. It has to be stated explicitely.

The thing is, I want to write something which is easy to understand, on your first reading.

Now, the most important thing about a set, is how many elements are in it.

So quibbling about whether or not {a,b,c}={a,b} because c=a is not a complete non-issue.

It is just as important to explain this kind of thing to the reader, as it is to explain to the reader that...

{a,b,c} denotes a set. (As Matt pointed out)

So here is what I finally settled on...

I will write using the following approach...

"In any discussion in which {a,b} denotes a set, unless explicitely stated otherwise, the symbols a,b denote different elements of the set."

The italicized portion is all that is required to avoid confusion, teach, and use currently accepted notation in the way it was intended.

If anyone has any helpful comments, they are welcome.

Regards

Does the question even have any meaning? I'd say not. How about: does Table Mountain lead to any contradiction?

That is one way to avoid the issue altogether.

Do numbers exist?

Well?

in the sense of mv' date=' it would be. in qm would it be? i think it would, because the difference in time would make a period longer/shorter which would change the wavelength thereby changing the momentum.

 

in your frame of reference, do you have momentum? in classical physics, i would say no, but in qm, i would say yes.

 

if momentum is relative, does that make energy relative? wait, thats a stupid question: m=m/(1-(v^2)/(c^2)).[/quote']

 

It appears you basically have the idea.

 

I equate classical and quantum mechanical momentum, so I would say that momentum is relative in either case.

 

Regards

The title is self explanatory.

I would like to hear different people's answer to this.

I am aware of this equation' date=' thank you for helping... [/quote']

 

I try.

 

Sometimes it works, sometimes not.

 

Regards

 

PS: if you are taking a class in classical mechanics, then you are well on your way.

 

But again, to ground yourself I suggest these three formulas:

 

 

E = hf

c=f l

p=h/l

I have a hard time grasping the concept of the propabilty of a particle acting as a wave.

 

Can someone explain any better please?

Understanding quantum mechanics without knowing the whole history of physics which preceded it' date=' is difficult but I won't say impossible.

However, the usage of probability theory is going to be confusing no matter what you do.

Firstly, to even understand what ultimately you would have to, requires you to know some probability theory, so that you can make intelligent statements.

Quantum mechanics has led many physicists to strange statements like many universes, parallel universes, just many strange sentences.

All of that is a result of them trying to interpret the mathematics using their natural language.

Its not that they don't understand the mathematics, it's that they do.

But something weird is happening between the paper, and their thoughts.

If you want to understand anything at all, you need something as a basis.

For quantum mechanics, I recommend you start with just one simple formula:

E = hf

You would do better to start with that formula, than worry about probability theory right away. Probability theory didn't enter until the 1920's.

What I am telling you is this.

Your desire is to understand quantum mechanics, not probability theory.

Quantum mechanics is based upon an experimental discovery.

Actually several discoveries, which happened apparently within just a few years.

But no matter what, start off with understanding just that one equation.

Learn its history, learn the meaning of the symbols. Then proceed to make things more and more complex.

E is the energy of a photon.

h is Planck's constant of nature.

f is the frequency of a photon.

Ask some physicists to explain f to you, and you will get different answers.

Ask them what is waving with frequency f too, and again you will get different answers.

c = speed of light = 299792458 meters per second = f l

So

E = hf

c = f l

the f in both formulas denotes the same thing.

Those equations are your starting point.

The first one originates aroung 1900.

Then Niels Bohr, around 1913, came up with a model for a hydrogen atom.

He only had three postulates, and his theory adequately explained the wavelengths of the photons emitted by glowing hydrogen gas.

Hey what happened to all my work????

It's too much so I'm not going to repeat it.

No' date=' I'm not saying that. [x,y'] is the notation for the commutator. It means:

 

[x,y]=xy-yx.

 

Notation for commutator, yeah.

So in the case where the quantities x,y commute then xy=yx whence it follows that [x,y]=0.

When you write [A, A⁺] does A⁺ denote the complex conjugate of A?

When the latex is working again, I'll show you what to do.

Hey peoples... i have a quick question...

 

if i want to find the moment of inertia from a the gradient of a graph.

Moment of inertia of what shaped object?

And this graph, it was what vs what exactly? What were the units of the x axis, and what were the units of the y axis.

There is a formula for the tension of a string, and it has a mass density dependence, but I dont remember the formula for it offhand.

Regards

There is nothing wrong with that. You can live your life quite happily that way. Science isn't everything.

If someone could show me a solid reason to study manifolds, I would pick it up as fast as I could. But really, I would rather study Lagrangian dynamics first, and everytime I start to read about it, it's worse than having a tooth pulled.

Somewhere in the universe is the center of mass of the universe, it is a place it is somewhere you can go there and move your head around and look in any direction.

You can set up a three dimensional rectangular coordinate system with its origin at that place, and you can use that reference frame to define the position of every single object in the universe.

Since you dont need to do anything else to discuss the positions of things, you can use the simple Geometry used by Euclid to understand motion.

I see no point in trying to curve space, and study manifolds and what not.

What's the error in the derivation?

 

I said the so-called "linearized" Sagnac effect was essentially a Michelson inteferometer. The Sagnac effect is inherently a rotational effect' date=' so saying you want to look at it in one dimension, i.e. linearize it, results in an oxymoron.

 

If light behaves as you seem to think - that the overall speed is c+v or c-v, when the source moves at v with respect to some preferred frame (the ether), then you should end up with a fringe shift in the Michelson interferometer that is dictated by v.[/quote']

 

I was trying to understand the derivation, I had a book, they were using S and S`, and giving a presentation. And I already had formulated a clear argument that SR self contradicts. I went and found a book, covering MM, the reason I was even looking for the book in the first place, is because armed with the logic, I figured naturally the MM derivation contains an error.

 

I read and read... and stopped reading the moment i saw the error.

 

I still have the book, but as I recall, the authors switched frames, and mixed up frame measurements somehow.

 

In other words...

 

 

They defined something like this say:

 

 

A = B/C

 

 

but B was measured in frame S

 

and C was measured in frame S`

 

the derivation contained an error, as I've said.

The one thing that I perhaps should have stated (skipped my mind earlier) is that

 

A = {a' date='b,c} => |A| = 3 if a, b and c are all distinct elements.[/quote']

 

Well see I don't want the 'if' left open.

 

I need notation which is unambiguous, and that is the point.

 

If that notation leaves open the possibility that some of the symbols being looked at inside the curly braces could denote the same element of the set, then I need a new notation which clearly says to the reasoning agent, there are 7 elements in this set... immediately, and you do not need to reason otherwise about it.

 

I was toying with this:

 

{a;b;c;d;e;f;g}

 

 

So if an author wants to leave the possibility open, as to what the symbols inside the braces denote, they can use

 

{a,b,c}

 

and if the author wants to close off the possibility, he can use the semicolon instead of the comma.

 

Obviously, this would only make sense in my work, but that's all its supposed to do.

 

But of course, the whole point of my question was this...

 

If it is already standard practice that when you see {x,y,z,p,d,q} that you can instantly infer (just by looking) that there are six elements in the set, then i needn't bother with the semicolon thing.

Erm... not that I'm aware of. Everything I have seen or read has always infered that if A = {a,b,c} then |A| = 3.

Ok, thanks. That makes life easier from a reasoning standpoint of course, it's just I am writing something, and I was covering this issue.

It's a reasoning issue actually. if i took the time to explain it to you you would follow, but i just needed the question answered.

Well perhaps for your own edification...

The title of the work is

"Treatise On The Transmission Of Knowledge"

Suppose that someone doesnt know anything about set theory at all.

They then see the following symbolism being used by the author, to explain set theory...

{a,b,c,d}

It is natural for them to initially wonder about several things...

One of which is, "does order matter"

Symbolically if order mattered, it would have been handled this way,

(a,b,c,d), by some particular author.

But see they don't know anything about that notation yet.

Now, assume they are accustomed to using the = symbol, and know the reflexive, transitive, and symmetric properties of equality.

They are well accustomed to having to draw the conclusion that such and such = something symbolically different, yet equivalent in some sense.

So they don't, a priori know if they must also do this with the set theoretic notation which uses the curly brackets.

Hence, if when that notation is used by some author, it is impossible for different symbols to denote the same element of the set, that author must supply the learner with that information, otherwise be remiss.

If you don't get it by now, don't worry about it.

You told me what I wanted to know.

I googled on "roster method" but got no clear answer to something that should be stated immediately in anyone's logical presentation of set theory.

Kind regards

PS: One more thing.

I am covering something called "tasking"

which has to do with demands made upon external reasoning agents, mental tasks that they must perform, in order to learn something from you.

When teaching, you seek to minimize the tasks they must carry out, while maximizing the rate of knowledge transfer.

but then this kind of thing is being covered in my treatise.

Also Dave, forgive me if I overanalyze. It happens by accident.

Johnny I agree that space is Euclidean, but an Euclidean 3-manifold can have 18 different topologies, please read the paper

I have to be honest with you... the pictures were cool but they prove space is simple 3-D Euclidean. Just look at the ones with the straight lines going off into the distance. The lines are straight.

I have no idea what even one of the 18 "topologies" is after looking.

Well basically because I was only looking to find an error as fast as possible.

What in the heck was I supposed to see?

I have read probably over 100 different books on set theory over the past 15 years.

I have a question, it's about notation and that's all it is about.

Suppose that someone uses the following notation for a set:

{a,b,c}

Must I infer that the set has three elements in it OR

have they left open the possibility that the set contains one element, or possibly even two elements?

I could adjust my logical structure to accomodate anything, but the question is only about standard usage of the notation above.

Thank you