Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

Astro · 2010-08-01 01:48:39

The delta function is not a continuous entity, nor is The Dirac delta is not strictly a function, because any real function that is equal to zero everywhere but a single point must have total integral zero.

However, in the last three days, i pondered a new Dirac Function, one which relied on a different yet i feel rather simple set of derivational clues to make the dirac function not just equal to 0 or 1, but have a range between -1, 0 and 1. I have named this the Triple Delta Function. The mathematical derivation is here:

If we start with $\mathbf{N} \geq 1$ then we note for every $n \geq \mathbf{N}$ we satisfy the condition:

$|x_n - L|\geq \epsilon|$

the integer $\mathbf{N}$ tells you how far you can go as you get closer to $\mathbf{L}$ up to the quantity $\epsilon$ , so it is naturally seen that $\mathbf{N}$ is responsible for the speed of the sequential inequality.

${x_n}n \geq 1$ converges to $\mathbf{L}$ iff for every $\epsilon > 0$ there does exist $n \geq \mathbf{N}$ so:

$|x_n - L|\geq \epsilon|$ is a true statement.

Using this logic however, we can also find that ${(i^2)^n}n \geq 1$ can have a range between $-1$ and $1$ so it oscillates by a mathematical analogue between the imaginary number $-1$ , nothing $0$ and absolute $1$ .

Using ${x_n}n \geq 1=\delta_{nn'}$ as a food for thought, we can now see using this new range in the inequality ${(i^2)^n}n \geq 1$ would lead to a ''function'' which takes into consideration a complex delta function solution.

Of course, the limitation would be $\displaystyle\lim_{x\to \infty}= \mathbf{L}$ thus $\mathbf{L}$ is some number which no longer exhibits a sequence of the number if $x_n=\mathbf{L}$ .

Moving on, i now apply:

$\delta^{nn'n*}= \int_{i^2}^{1} {x_n}= \displaystyle\lim_{n\to n*}$

where $n=-1$ , $n'=0$ and $n*=1$
This last formula actually has more of the appearance of the Kronacker Delta function, so the formula is a tensor product.

Last edited by Astro (2010-08-02 13:12:58)

Astro · 2010-08-01 01:55:11

What is it with the latex here. I can't write out inequalities properly... guh... here we go again.

Astro · 2010-08-01 04:51:32

A way to use the Triple Delta Function.

$-\hbar^2 \frac{\partial}{\partial t} \mu \varepsilon \omega^2 \psi=iaw \hbar^2 (\frac{Mc^{2}}{\hbar}) \psi$

where k²=μεw² and <Ψ|iћ•∂/∂t|Ψ>=ћw known as the energy operator.

Equation 13 would make sense for the inertial mass description since we can invoke the inertial energy description of Mc². All imaginary references are cancelled, so it can be applied to real fields. The equation does not finish there. It has a component left for explanation, and this is the electric and magnetic constants (μ and ε).

They are unified in Maxwells equations as the phase velocity of an electromagnetic wave in a free vacuum με=1/c².

This means that the identity of με on the left hand side of equation 14] can be substituted for 1/c²;

$-\hbar^2 \frac{\partial}{\partial t}\frac{1}{c^{2}} \omega^2 \psi=i \alpha \omega \hbar^2 (\frac{Mc^{2}}{\hbar}) \psi$

And so the equation can now also determine the phase velocity of electromagnetic radiation. This equation also has lined up a possible description of the in phase and out phase response, where the permeability is considered as a function of the frequency, and so, can allow a complex description of the permeability to exist.

In this specific case, we can take into consideration that spin can also have a value of zero, and now that we are allowed a complex description of the permeability, so plugging in the Triple Delta Function can describe the angular frequency between the imaginary, real and zero integer values:

$i\hbar \frac{\partial}{\partial t} \int_{i^2}^{1} \frac{1}{c^{2}} \omega^2 \psi=i \alpha \omega \delta^{nn'n*} \pm \hbar(\frac{Mc^{2}}{\hbar})\psi$

Last edited by Astro (2010-08-01 04:52:44)

AlphaNumeric · 2010-08-01 11:56:33

I couldn't resist registering to comment on what a massive waste of your time you're doing here Reiku. You're still posting meaningless LaTeX expressions I see. Never mind coming up with physicially meaningless equations you're not even able to come up with mathematically valid expressions but then that's your typical behaviour, you try to copy-edit things from work others have done and in doing so you demonstrate you don't understand what you're editing because you produce gibberish.

Really, what do you think you're accomplishing by wasting years by making stuff up which is just gibberish? I really don't get it. You said on that forum you mod that I judge you too quickly but I don't. You've been wasting your time for years, you aren't misunderstood or unfairly judged. I know you want to delude yourself that deep down everyone thinks you're a genius but that simply isn't the case. The funny thing is if you'd spent the last 3 or 4 years actually doing proper mathematics and physics you'd now be in a position to be able to understand the Dirac delta function, measures, tensor products and quantum mechanics. Instead you took the easier option of just copy-editing work you didn't understand and unfortunately it amounts to nothing. How sad.

Astro · 2010-08-02 09:26:41

LOL!

OMG - the condescending God of sciforums has honored me by his appearance. How lucky for me?

But yet again - you have failed in many remarks above, as they are inconsistent or simply not true. None of the work above is anything i have copied from anyone. And neither are they physically-meaningless, nor do i believe the written work is incorrect - if it is, and indeed, you did not escape your own ego to advance an account here, perhaps you could elaborate on the inconsistencies, rather than raising a voice.

You should know by now physics is not about being right - it's about being proven wrong. Please, do so.

Astro · 2010-08-02 09:34:19

''Really, what do you think you're accomplishing by wasting years by making stuff up which is just gibberish? I really don't get it.''

I believe whether the equations are gibberish is indeed within the eye of the beholder.

''You've been wasting your time for years, you aren't misunderstood or unfairly judged.''

Really? And who are you the judge of anyone? Certainly not me sir. I live a very happy life, and someone like you is incapable of deterring me from ''wasting years'' on writing physics - an interesting disparaging remark, one that is disproportional to a true scientist, whose goal should be to entice the investigation of the physical sciences. And what makes you think if i am proven wrong, that i will be ashamed of it? Einstein himself as a good analogy, made several mathematical mistakes in his first electromagnetic paper of relativity. He also made many quantum mistakes as well. Being wrong, is not a crime against science.

''The funny thing is if you'd spent the last 3 or 4 years actually doing proper mathematics and physics you'd now be in a position to be able to understand the Dirac delta function, measures, tensor products and quantum mechanics.''

What do you know what i have been studying the past 3 years? Have you got a hidden camera in the corner of my house, or should i start wrapping my head with tinfoil?

'' Instead you took the easier option of just copy-editing work you didn't understand and unfortunately it amounts to nothing. How sad.''

This is a random set of inconsistent lies.

Astro · 2010-08-02 12:55:38

To proove this is simply not 'copying' and then editing - there are more things i have considered. Here is a challenge for you. Having a PhD, it should be relatively simple for you.

Working from the last equation, we no longer require the integral to be set from -1 to 1. This is a hint.

taking the equation i created $\bar{e} (\mu D) \psi + \bar{e} E\mathbb{l} = \frac{1}{c^2} \bar{e} E \mathbb{l} e^{i \omega t} + \frac{1}{2} \hbar \omega$

where D is displacement and E(l) is a kinetic energy term from the electromagnetic equation $\bar{e} E\mathbb{l}=1/2M_0v^2$ for an ionized particle, thus the fancy (l) coefficient is the distance involved with the travelling system. I want you to prove that the equation is correct for a non-zero value of $E \mathbb{l}$ . If you can, you might be able to see how these equations can be useful, just as any equation is, so long as it can be proven correct - how useful they are, where they are useful and when they are useful, of course, is a different yet complicated set of questions.

Last edited by Astro (2010-08-04 10:33:45)

Astro · 2010-08-02 16:21:48

You do realize AN, that you followed me from physorg.com to sciforums, to now here. Are you somehow infatuated with me?

If not, then i truely do hope you are more than just words, because the above can be solved/proven.

AlphaNumeric · 2010-08-02 22:18:25

Astro wrote:
But yet again - you have failed in many remarks above, as they are inconsistent or simply not true. None of the work above is anything i have copied from anyone.

Well given you haven't ever done anything to do with convergence or quantum mechanics or functional analysis it stands to reason you have copied the equations, with more changes, as you have no actual knowledge of the relevant areas. You do it all the time, its nothing new.

Astro wrote:
And neither are they physically-meaningless, nor do i believe the written work is incorrect

Not only are they physically wrong but they are mathematically wrong, you can't even write down coherent algebra.

Astro wrote:
- if it is, and indeed, you did not escape your own ego to advance an account here, perhaps you could elaborate on the inconsistencies, rather than raising a voice.

Why do you always do this, saying "So where's my errors then?", as you know full well you will be called on them. For instance :

Astro wrote:
If we start with $\mathbf{N} \geq 1$ then we note for every $n \geq \mathbf{N}$ we satisfy the condition:

$|x_n - L|\geq \epsilon|$

How can you 'note' a result when you haven't defined anything and x_n is without context. If you bothered to read the place you lifted that from you'd note that x_n was given a specific meaning, such as a sequence with some specific properties. You haven't provided any explanation. And you put in a spurious half mod size after the epsilon, making the expression as algebraicly unsound.

Astro wrote:
the integer $\mathbf{N}$ tells you how far you can go as you get closer to $\mathbf{L}$ up to the quantity $\epsilon$

No, it gives the lower bound for the term in the sequence {x_n} such that the value of that term and all terms which follow it in the sequence are within a distance (as defined by the norm) of epsilon from a given point, namely L. You've done little more than incorrectly try to state the definition of a limit, yet you word your sentences as if you've shown some result.

Astro wrote:
so it is naturally seen that $\mathbf{N}$ is responsible for the speed of the sequential inequality.

'Speed' has nothing to do with it. N tells you how far down the sequence you must go to be sure to be closer to the limit than a specified tolerance, epsilon. This is basic mathematical analysis, 1st year stuff.

${x_n}n \geq 1$ converges to $\mathbf{L}$ iff for every $\epsilon > 0$ there does exist $n \geq \mathbf{N}$ so:

$|x_n - L|\geq \epsilon|$ is a true statement.

No, it isn't a true statement. It is a true statement if and only if such an N exists. You state the conditions and requirements for such a sequence to exist, thus acknowledging sometimes such an N doesn't exist and yet you then state that x_n, which you haven't given any constraints or properties to, definitely satisfies that. Any diverging sequence being an example where no such L and N exist.

Astro wrote:
But yet again - you have failed in many remarks above, as they are inconsistent or simply not true. None of the work above is anything i have copied from anyone.

Well given you haven't ever done anything to do with convergence or quantum mechanics or functional analysis it stands to reason you have copied the equations, with more changes, as you have no actual knowledge of the relevant areas. You do it all the time, its nothing new.

Astro wrote:
And neither are they physically-meaningless, nor do i believe the written work is incorrect

Not only are they physically wrong but they are mathematically wrong, you can't even write down coherent algebra.

Astro wrote:
- if it is, and indeed, you did not escape your own ego to advance an account here, perhaps you could elaborate on the inconsistencies, rather than raising a voice.

Why do you always do this, saying "So where's my errors then?", as you know full well you will be called on them. For instance :

Astro wrote:
If we start with $\mathbf{N} \geq 1$ then we note for every $n \geq \mathbf{N}$ we satisfy the condition:

$|x_n - L|\geq \epsilon|$

How can you 'note' a result when you haven't defined anything and x_n is without context. If you bothered to read the place you lifted that from you'd note that x_n was given a specific meaning, such as a sequence with some specific properties. You haven't provided any explanation. And you put in a spurious half mod size after the epsilon, making the expression as algebraicly unsound.

Astro wrote:
the integer $\mathbf{N}$ tells you how far you can go as you get closer to $\mathbf{L}$ up to the quantity $\epsilon$

No, it gives the lower bound for the term in the sequence {x_n} such that the value of that term and all terms which follow it in the sequence are within a distance (as defined by the norm) of epsilon from a given point, namely L. You've done little more than incorrectly try to state the definition of a limit, yet you word your sentences as if you've shown some result.

Astro wrote:
so it is naturally seen that $\mathbf{N}$ is responsible for the speed of the sequential inequality.

'Speed' has nothing to do with it. N tells you how far down the sequence you must go to be sure to be closer to the limit than a specified tolerance, epsilon. This is basic mathematical analysis, 1st year stuff.

Using this logic however, we can also find that ${(i^2)^n}n \geq 1$ can have a range between $-1$ and $1$ so it oscillates by a mathematical analogue between the imaginary number $-1$ , nothing $0$ and absolute $1$ .

Wrong. You've just defined the condition for a sequence to converge and then you pick a sequence which doesn't converge. You failed to realise you're not stating a result but a definition and you failed to realise your example doesn't satisfy your definition!

If you think I'm wrong on this then please provide an L and N whereby for all n>N |x_n - L| < 0.1.

Furthermore the values of the sequence don't reproduce the effects/properties of the Dirac function.

Using ${x_n}n \geq 1=\delta_{nn'}$ as a food for thought

The tex'd ${x_n}n \geq 1=\delta_{nn'}$ is sloppy notation, as you put an inequality associated to the indices into the expression. You don't specify n' and thus x_n is ill defined. The Kronecker delta isn't a sequence and has nothing to do with convergence.

we can now see using this new range in the inequality ${(i^2)^n}n \geq 1$ would lead to a ''function'' which takes into consideration a complex delta function solution.

You haven't done anything to do with the Dirac delta function. You haven't generalised it, you haven't come up with something which does the same job, you haven't had anything to do with complex numbers, as you squared i to give -1, the choice of using i^2 rather than -1 is superfluous. I'll come back to this in a second.

Of course, the limitation would be $\displaystyle\lim_{x\to \infty}= \mathbf{L}$ thus $\mathbf{L}$ is some number which no longer exhibits a sequence of the number if $x_n=\mathbf{L}$ .

You take a limit x -> infinity without saying what is being limited. Even allowing for the fact you might be meaning x_n you don't take the limit by varying x, you do it by varying n, ie $L = \displaystyle\lim_{n\to \infty}x_{n}$ . But your function $x_{n} = (-1)^n$ doesn't have a limit, it doesn't converge.

$\delta^{nn'n*}= \int_{i^2}^{1} {x_n}= \displaystyle\lim_{n\to n*}$

where $n=-1$ , $n'=0$ and $n*=1$

And now its clear why you said i^2 instead of -1, you have partly copied this from somewhere else where the integral range was -1 to +1 and to make it seem more complicated you've changed some -1's to i^2. The change is pointless, as complex numbers have nothing to do with what you're talking about and you thus demonstrate you both copy from other places and that you don't understand what you copy. You did it on PhysOrg after you copied a post of BenTheMan from Sciforums and changed 'H' to the Hamiltonian, not realising H was a short hand for 'Higgs'. Furthermore you do the same as the previous mistake, taking a limit but not stating what you're taking the limit of. And you've just pulled a triple index delta out of nowhere without defining it.

This last formula actually has more of the appearance of the Kronacker Delta function, so the formula is a tensor product.

And you round it off with a pointless dropping of buzzwords, as you haven't done anything to define a tensor product expression and even allowing for that you haven't defined what precisely the delta is.

Astro wrote:
You should know by now physics is not about being right - it's about being proven wrong. Please, do so.

Physics is about justifying why you your work has viability, the "Prove me wrong" attitude is not how science works.

Astro wrote:
You should know by now physics is not about being right - it's about being proven wrong. Please, do so.

Physics is about justifying why you your work has viability, the "Prove me wrong" attitude is not how science works.

Astro wrote:
I believe whether the equations are gibberish is indeed within the eye of the beholder.

So 1+1=3 is 'in the eye of the beholder' then? Unfortunately for you there is such a thing as right and wrong in mathematics and no matter how you copy-edit things 1+1 is not 3.

Astro wrote:
Really? And who are you the judge of anyone? Certainly not me sir.

I'm just calling a spade a spade.

Astro wrote:
And what makes you think if i am proven wrong, that i will be ashamed of it?

I didn't say shame. But there's something definitely amiss if you have to pretend for so long, to so many people who see through you that you're doing physics when you know full well you're not.

Astro wrote:
Einstein himself as a good analogy, made several mathematical mistakes in his first electromagnetic paper of relativity. He also made many quantum mistakes as well. Being wrong, is not a crime against science.

Einstein did actual research, which everyone gets wrong from time to time. Einstein didn't go to his local physics library and copy down some equations from a book, relabel a few things and then go door to door trying to convince people he's done physics. Einstein actually put his physics where his mouth was, producing clear, predictive viable models which passed peer review. I know you want to delude yourself into a comparison with Einstein but you are nothing like him.

Astro wrote:
What do you know what i have been studying the past 3 years? Have you got a hidden camera in the corner of my house, or should i start wrapping my head with tinfoil?

You and I crossed paths years ago and in that time you've posted massive numbers of threads like this, all of them garbage. You've lied about the education you've got, claiming to be doing Riemannian geometry pre-university. You couldn't even multiply out two brackets properly. Whatever you've been 'studying' it hasn't been a science.

Astro wrote:
This is a random set of inconsistent lies.

Your posts say otherwise. And it wouldn't be the first time, now would it?

Astro wrote:
Here is a challenge for you. Having a PhD, it should be relatively simple for you.

What do I need to prove? As you just said, I have a PhD in physics. I don't claim all knowledge but I can do the things I claim to capable of doing. You're now trying to shift the burden of proof away from yourself, trying to avoid justifying your own position and 'work'. Your last equation in the psot further up is something you've botched together, so there's no reason for me to try and turn your crap into something else you've made up. If you've got work which is right then why aren't you submitting it to journals?

Astro wrote:
You do realize AN, that you followed me from physorg.com to sciforums, to now here. Are you somehow infatuated with me?

I know you want to believe the world revolves around you but it doesn't. I've known you post on other forums for a while, I simply was bored and thought that because you're such an easy target I'd point and laugh at you. I think you deserve ridicule because you clear learn absolutely nothing from the many many times you've had your ignorance slammed in your face by anyone who didn't sleep through secondary school science class.

Astro wrote:
If not, then i truely do hope you are more than just words, because the above can be solved/proven.

Wonderful straw man, you make (I have a dirty mouth and should be ashamed) and then demand someone else justified it. Then, when they don't because you've failed to clearly define WTF you're talking about, you'll declare "Oh look, Mr PhD can't do it!" as if its my fault you can't think coherently. If you've got some proof of something, submit to a journal. You claimed last year on SciForums you'd got something accepted, whatever happened to that? Oh yeah, you hadn't.

You said "then i truely do hope you are more than just words", yet at the end of the day my physics and mathematics work is more than just words, I have done some peer reviewed work and I can actually do physics, applying it to the real world (hell, its my job) while on the other hand you've stuck making stuff up on forums, desperately lying to people you don't know who don't believe you, all in the hopes of convincing others (and thus yourself) you are not just wasting your time. Yes, I registered here specifically to reply to you but so what, you're entertainment to me and I have something to show for my time doing mathematics and physics away from the computer, you have nothing.

Astro · 2010-08-03 13:10:01

Oh dear alphanumeric... did my equations at the end stump you?

I will read your post, answer every retort, but i am very disappointed you could not solve my posed question.

Astro · 2010-08-03 18:56:18

In fact, i've pretty much explained in the otherb thread, i do not just make things up and hope for the best. I vision what i want, then use the already existing formulas to help manifest some description of them, even if it is conceptually-based. I'm still waiting for you to prove the question i posed. Will it take much longer, or should i show you how you can prove it?

Astro · 2010-08-04 10:34:57

Ok, i've given you long enough.

If there is an energy present, then the displacement has to be set to zero. I told you it was pretty simple.

$\bar{e} E\mathbb{l}= \frac{1}{2}Mv^2 + \int_{0}^{2 \pi r} D dt$

Thus the part $\bar{e} (\mu D) \psi$ would be equal to zero, but hence, the equation

$\bar{e} (\mu D) \psi +\bar{e} E \mathbb{l} = \frac{1}{c^2} \bar{e} E \mathbb{l} e^{i \omega t} + \frac{1}{2} \hbar \omega$

Has an added dimension of kinetic energy, hence the object is travelling in a circle, but it's energy is easily non-zero.

Last edited by Astro (2010-08-04 10:35:26)

AlphaNumeric · 2010-08-13 12:30:36

Astro wrote:
Oh dear alphanumeric... did my equations at the end stump you?

Tell me, do you believe the things you post or do you know you're obviously a liar?

Astro wrote:
I will read your post, answer every retort, but i am very disappointed you could not solve my posed question.

Yes, my PhD (yes, I have it all completed now) counts for nothing compared to your deliberately gibberish questions. Tell me, if I linked to a 1st year university homework problem sheet would you try to answer them? I doubt it.

Or shall I just link to that thread on PhysOrg where you challenged me to a 'physics off' and you couldn't answer a single question, other than the attempt you made to pass off someone else's work as your own? Or how about all the threads on SciForums where you get called on your ignorance? I can provide plenty of examples of my working understanding, as well you know but I imagine you're playing to an imaginary audience.

Astro wrote:
I'm still waiting for you to prove the question i posed.

Ah, your excuse. You make up something then refuse to justify your own position until someone answers your question which you know you've just made up.

Like I said, how about we consider an impartial 3rd party for questions, so that I can't claim you're just making stuff up?

Astro wrote:
Ok, i've given you long enough.

Wow, you waited a while morning! Do you really think I was just sitting there pressing refresh waiting for you to post? No one thinks you're that important. Even for your entertainment value.

Astro wrote:
Has an added dimension of kinetic energy,

They aren't dimensions. Well done on misusing terminology.

Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

#1 2010-08-01 01:48:39

Triple Delta Function

#2 2010-08-01 01:55:11

Re: Triple Delta Function

#3 2010-08-01 04:51:32

Re: Triple Delta Function

#4 2010-08-01 11:56:33

Re: Triple Delta Function

#5 2010-08-02 09:26:41

Re: Triple Delta Function

#6 2010-08-02 09:34:19

Re: Triple Delta Function

#7 2010-08-02 12:55:38

Re: Triple Delta Function

#8 2010-08-02 16:21:48

Re: Triple Delta Function

#9 2010-08-02 22:18:25

Re: Triple Delta Function

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#10 2010-08-03 13:10:01

Re: Triple Delta Function

#11 2010-08-03 18:56:18

Re: Triple Delta Function

#12 2010-08-04 10:34:57

Re: Triple Delta Function

#13 2010-08-13 12:30:36

Re: Triple Delta Function

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