Aethelwulf

I've noticed a small mistake... don't know if anyone is following this carefully enough to have noticed.

 

The equation

 

[math]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

Is written wrong. It should be

 

[math]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

The reason why, is imagine we have what we began with

 

[math]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}[/math]

 

Take the dot product of the unit vector on both sides gives

 

[math]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

Now taking the curl of F is

 

[math]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

So I dropped out a curl on the right. And because I dropped out this curl, it makes the new Larmor equation invalid and the reason why is because the extra nabla operator would bring down a 1/length dimension not accounted for.

 

 

 

Oops! The rest is fine though, just a small hiccup in my last derivation.

 

 

In light of this equation however

 

[math]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

I could rewrite the Larmor energy as

 

[math]\Delta H = \frac{2\mu}{\hbar Mc^2 e}(\vec{F}_{ij} \cdot \hat{n}) L \cdot S[/math]

 

But I am sure many agree that is not very interesting.

 

 

 

Going back to the mistake... (and this is really starting to make my brain cells burst), taking the curl of a force does gives F/length, however, the unit vector cancels these out and what I have again is the force, ... very circular... since this would be true, then we know what the force is anyway...

 

[math]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n} [/math]

 

Now, just a moment ago I found the Larmor energy written as

 

[math]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/math]

 

Notice the 1/r term which is not in my original case. If that where true, and we plug in my force example again

 

[math]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}}\hat{n} ( L \cdot S)[/math]

 

(If I am doing this right) the unit vector would cancel out with the radius term and what would be left with is

 

[math]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/math]

 

Now if that wasn't confusing I don't know what is.

 

I've either had it right from the beginning, or I have made a tiny mistake which is making a huge impact on my understanding of my own equation. Any help would be gladly appreciated.

 

 

 

I found a solution to my problem (I think). The Larmor equation is

 

[math]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/math]

 

What I kept deriving was:

 

[math]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

What I really needed was the original derivation

 

[math]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}[/math]

 

Then taking the curl of F gives

 

[math]\nabla \times \vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

Which removes the unit vector because nabla again is 1/length.

 

So what I think I have ended up with was the right derivation for the modified Larmor energy except for a dot product made on the unit vector after all, it would be

 

[math]\Delta H = \frac{2\mu}{\hbar Mc^2 e} (\nabla \times \vec{F}_{ij}) L \cdot S[/math]

 

 

What a mess my brain got into. Assuming this is correct mind you.

 

I found a solution to my problem (I think). The Larmor equation is

 

[math]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/math]

 

What I kept deriving was:

 

[math]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

What I really needed was the original derivation

 

[math]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}[/math]

 

Then taking the curl of F gives

 

[math]\nabla \times \vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

Which removes the unit vector because nabla again is 1/length.

 

So what I think I have ended up with was the right derivation for the modified Larmor energy except for a dot product made on the unit vector after all, it would be

 

[math]\Delta H = \frac{2\mu}{\hbar Mc^2 e} (\nabla \times \vec{F}_{ij}) L \cdot S[/math]

 

 

What a mess my brain got into. Assuming this is correct mind you.

It is a very complex question that requires sound knowledge of physics we are only starting to understand. We need a proper theory that includes quantum gravity to tackle the initial singularity.

 

The more I think about it the notions of before, causality, time and so on are probably not so clear near the singularity. I very much suspect that these notions may not really be appropriate. But I am speculating here.

 

Short answer we do not understand physics very near and at the singularity.

 

It's true what you say.

 

When people say, ''physics breaks down at the singularity'' many of them are referring to the uncertainty principle when you try and squeeze particle into a more confined region - which is totally forbidden by quantum mechanics. However, I believe that if it only exists for a very short period of time, who gives a damn about the violation? It's almost poetic to think that the universe arose from an uncertainty.

 

Of course... here is another problem for you to chew on.

 

In the beginning, we ascribe time as t=0 at time zero - then t=1. At t=1, the very first moment of existence, we are led to believe that the universe suddenly appeared as a single ''dot'' - a tiny point without dimensions or space. This is where the first inconsistency or oxymoron arises if you like. If there was no space, there was certainly no time, so how can we talk about ''first instances'' and t=1? In relativity, time is part of the manifold, but in the beginning there was no space - so... how can we really talk about time as a fundamental object?

So, are you saying in your paper, its not meant to be the operator, because it was an error when it was error in the translation?

Science never did separate from philosophy. They are inseperable.

 

But it's a small point. For Hegel and me it's a science of logic, but we'd have to sit down and clarify exactly what we mean by science to decide, and it doesn't really matter.

 

Still, as metaphysics is prior to physics, then on your view we can say that physics rests on non-scientific foundations, and I don't think you would be happy with that idea.

 

I suppose we could call it all natural philosophy.

 

EDIT: It's interesting though. I just checked that link you gave to a definition and to my mind it would cover philosophy (or philosophy done in a certain way). 'Analysis' would be the key word. But I'm happy to call metaphysics a discipline and have done with it.

 

EDIT 2: Extract from that article - "The definition of science given here differs with the 'official' definitions given by the Science Council and the American Physical Society (APS), and with the definitions given in scientific and philosophical textbooks."

 

 

I don't believe one can say that

 

''Still, as metaphysics is prior to physics, then on your view we can say that physics rests on non-scientific foundations, and I don't think you would be happy with that idea. ''

 

In science, real science, you have to test to see if an idea is falsifiable. If the framework is not scientific or falsifiable, how can it be a science? Indeed, how can anything like that be linked to a real science?

 

 

Mandatory XKCD (webcomic).

 

 

And this is the link to the XKCD webcomic.

 

 

Ah, but physics is largely mathematical - it's hard to separate the two.

This next derivation, as I continue this work, will be concerned with the quantization condition on the mass of a particle (the gravitational charge). This was discussed by Motz in his paper, which is referenced in post 1.

 

We begin with the equation

 

[math]G = \frac{rc^2}{2Gt^2}[/math]

 

which can be found in [1]. Taking the square root of both side and rearranging yields and then multiplying through by [math]\sqrt{M}[/math],

 

[math]\sqrt{G}M = \sqrt{\frac{Mr^3}{2t^2}}[/math]

 

 

Which is the gravitational charge. As shown in the beginning of this work, I equated the gravitational charge to a gravitational energy multiplied by the schwartschild radius, [math]\sqrt{G}M = \sqrt{E_gr_s}[/math]

 

we simply substitute now

 

[math]\sqrt{\frac{Mr_{s}^{3}}{2t^2}} = \sqrt{E_g r_s}[/math]

 

and let [math]r = r_s[/math] on the left hand side of the equation. Now multiply through [math]2t^2[/math]

 

[math]\sqrt{Mr_{s}^{3}} = \sqrt{E_g r_s 2t^2}[/math]

 

Seeing the gravitational energy for it's relativistic equality [math]Mc^2[/math] and then cancelling out some of our terms on both sides of the equation yields

 

[math]\sqrt{r_{s}^{2}} = \sqrt{c^2 2t^2}[/math]

 

We can fudge this result for

 

[math]r_{s} = \sqrt{c^2 2t^2}[/math]

 

since there is no such thing as a negative radius. Notice this derivation is the mathematical equivalent of the Minkowski relation, where here we are swapping the radius for the affine length [math]x=ct[/math]

 

Using the relationship [math]c = \omega r_s[/math] (which I could derive on request) we have

 

[math]r_{s} = \sqrt{2t^2 \omega^2 r_{s}^{2}}[/math]

 

We may take the cross product for the frequency and the radius

 

[math]r_{s} = \sqrt{2t^2 (\omega^2 \times r_{s}^{2})}[/math]

 

Multipling energy on both sides gives the square of the gravitational charge again

 

[math]E_g r_{s} = Mc^2\sqrt{2t^2 (\omega^2 \times r_{s}^{2})}[/math]

 

Square everything we have

 

[math]G^2M^4 = M^2c^4 2t^2 (\omega^2 \times r_{s}^{2})[/math]

 

Rearrange

 

[math]\frac{G^2M^4}{2M^2c^4 t^2} = (\omega^2 \times r_{s}^{2})[/math]

 

This is just

 

[math]\frac{G^2M^4}{2\hbar^2} = (\omega^2 \times r^2)[/math]

 

Halving this all out we have finally the quantization condition on the mass with the exception that the right handside is expressed in terms of a cross product between the angular frequency and the radius

 

[math]\frac{GM^2}{\hbar} = (\omega \times r_s)[/math]

 

Here is a link which explains the cross product between the angular frequency and the radius. http://www.physics.s.../360notes12.pdf

 

[1] http://arxiv.org/ftp/arxiv/papers/0810/0810.1629.pdf

So, the universe is composed of gravitational charge field. We have not known the field property quite well yet. LHC results will tell whether it is true or not.

 

The Higgs Field was such a field - it provided a similar charge to particles. The Higgs mechanism occurs whenever a charged field has a vacuum expectation value.

 

The only slight difference, is that in Motz' work, we don't know what the field is that he defines as the field which causes the charge. I've likened mass to the idea that an electron experiences an electric charge by moving in an EM-field, so the question is why a particle with mass moving in a gravitational field would not be the same? There is obviously going to be unique dynamics however, such as a type of coupling to the field, but there are similar couplings in a Higgs Field.

Right, so I have had time to read it.

 

Generally, it is saying the metric possesses a timelike killing vector, which is actually makes the metric static - or t-symmetric in your case example.

 

I find it hard to believe still, a metric as such does not gives up energy... as you yourself established, the SC metric is not an accurate representation of reality. My case in the OP would highlight that. So it seems, there is definitely a problem with its accuracy - perhaps the way it is treated mathematically is the problem.

We do not know the detail process about giving mass to a object. But the getting mass process seems to related to getting gravity. General relativity has no problem to say like this.

 

Well Lloyd Motz a cosmologist who did work on quantum gravity actually defined mass as a gravitational charge.

 

http://www.gravityresearchfoundation.org/pdf/awarded/1971/motz.pdf

 

Charge is just the coefficients of the Lie Algebra.

 

I wrote something up on it not long ago if you care to read it http://www.scienceforums.net/topic/66985-a-forgotten-theory-of-mass/

Most people simply don't understand much physics.

Can it be meaningfully said to have won over everything?

 

I think it's fair to say, we don't know everything about biology or chemistry either... in fact, we don't know everything about any science. We do however, know what physics wants to deal with and that is the fundamentals.

 

Personally, I don't think of philosophy a science. It perhaps once begged to answer scientifically-rooted questions, but the scientific methodology overtook this with experimental physics.

 

 

Light does cause gravity, but it is very weak. Whilst the photon does not contain a gravitational mass, it can cause the curvature of space because any type of energy can do this.

 

Large masses and even small particles draw each other in because they distort the spacetime fabric/continuum.

No. I didn't want anything. I was merely making a challenge. I think I decided not to post the result in open forum. If you'd like to see the answer please send me a PM and I'll send it to you.

 

Thanks for the link. When I get a printer I'll prinit it out and read it. I suspect that is wrong though. I can't concentrate reading involved documents online. Sitting too long causes pain and the pain distracts me.

 

By the way, you gave the value of the energy density for a magnetic field, not the mass density. Please take note that if the answer was just mass density = energy density/c[sup2[/sup] then I'd never have asked. That would be trivial and not worth discussing about.

 

The answer is also in this paper that I wrote - http://arxiv.org/abs/0709.0687

 

Can you define [math]\Box[/math] this more clearly, you say Minkowski defined it as the mechanical mass, but I have never seen this before - I believe this symbol is more generally used to define the d'Alembertian.

That seems a little harsh I've read through Lynds papers myself and they seem interesting, he is published after all (http://cdsweb.cern.c...rd/622019?ln=en) but anyway that might be straying off topic. What do you think makes this one look rubbish, I've only skimmed over it myself.

 

No self-serving paper nowadays would even dream of citing Peter Lynds.

 

He's was published because the board who looked over his work obvious can't discern a good scientific debate. This point was raised by many scientists at the time of his.... fascinatingly bizarre leap to fame. They said his paper was .... ok, but certainly wasn't ground-breaking and most where surprised it managed to get published in that specific journal.

 

Anyway, that's makes this work rubbish. But, I will take a better look later.

Why gravity is proportional to mass? Graviton?

 

It's most likely that a graviton doesn't exist.

 

Anyway, they are proportional because, you can't have gravity without the presence of mass - mass is the presence of gravity, is the present of distortions, is the presence of spacetime curvature and acceleration.

 

The acceleration understanding comes of Einsteins elevator experiment and the weak equivalence principle. The curvature appears from Einstein's Field Equations which explains how matter warps spacetime and how spacetime tells matter how to move in the presence of gravity.

It looks like rubbish. It even cites Peter Lynd, who is a college drop out with little behind his papers than philosophical rambling... personally, Peter Lynd is an oddity who should never have became famous.

 

And I say that politely.

But anti-neutrino comes from the neutron decay process.

Matter makes neutrino (for example, fusion reaction), but it has no interaction with matter.

Do we have anything wrong?

 

Nothing is wrong. Decay processes does not mean that it interacts with matter, only that neutrino's are a by-product of certain decay processes.

Neutrino can pass through the Earth. Are neutrinos not affected by Earth underground material? But neutron can not penetrate through the Earth. What is the difference? Stability?

 

Nuetrino's are barely effected by anything because they don't couple to ordinary materials very well. Neutrino's pass through your body every second with ease.

Ok, I have decided to derive some more relationships for this thread. I came to realize that this could be represented a new way.

 

 

In my thread, I explain how (at least) one problem with unification of all physics is that when you wind the clock back in the universe to the big bang, you come to a point with zero spacetime - this condition, the point where everything is believed to have come from called the big bang is a problem for physics and has been known for a while, because the more you squeeze particle into a single confined position (or point), you inevitably violate the uncertainty principle. One of the consequences of doing this I explained, was that the force to do this would be tremendously large, and I derived a force equation to help explain this phenomenon:

 

[math]F_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \mu(\hat{n} \cdot \sigma_{ij})^2 = \frac{\partial V(r_{ij})}{\partial r_{ij}} \mathbf{I}[/math]

 

Which took advantage of a spin network, to also explain an uncertainty of the spacetime. -- keep in mind that [math]\hat{n}^{2}_{i} = 1[/math]

 

[math]\nabla \times \vec{F}_{ij} = \begin{vmatrix} \hat{n}_1 & \hat{n}_2 & \hat{n}_3 \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \hat{F_x} & \hat{F}_y & 0 \end{vmatrix}[/math]

 

(Incidently, the lowercase ij denotes particle's 1 and particles 2 in the [math]F_{ij}[/math] term and shouldn't be mistaken for the unit vectors, however, if they were, they work out on the same column as would be found in the determinant matrix.

 

This gives

 

[math]\nabla \times \vec{F}_{ij} = \frac{\partial F_y}{\partial z} \hat{n}_1 + \frac{\partial F_x}{\partial z} \hat{n}_2 + (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}) \hat{n}_3[/math]

 

Now, this just gives

 

[math]\nabla \times \vec{F}_{ij} = (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}) \hat{n}_3[/math]

 

and

 

[math]\nabla \times \vec{F}_{ij} \cdot \hat{n} = (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}) [/math]

 

A bit exhaustive to write out so I hope it came out ok.

 

Anyway, the force equation in my original work I have in front of me had the force equation written as

 

[math]F_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}[/math]

 

We can apply the derivation above to it and it yields

 

[math]\nabla \times \vec{F}_{ij}\cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math]

 

Because as mentioned before, [math]\hat{n} \cdot \hat{n} = \hat{n}^2 = 1[/math]

 

I have noticed in the cumbersome book-keeping of latex, I have written the Larmor energy wrong: There should be no second order derivatives. That's quite an ugly typo that has been carried on through the equations. Is there any way it can be changed for me?

 

it shouldn't be

 

[math]\Delta H_L = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} (\hat{n}\cdot \sigma_{ij}) \begin{pmatrix} \alpha \\ \beta \end{pmatrix}[/math]

 

It should be

 

[math]\Delta H_L = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V (r_{ij})}{\partial r_{ij}} (\hat{n}\cdot \sigma_{ij}) \begin{pmatrix} \alpha \\ \beta \end{pmatrix}[/math]

 

These ''squared'' results came from... well I was supposed to just square the unit vector and spin terms but it seems I got a bit carried away.

 

A new form of the Larmor Equation

 

And, so previously we derived a new form of the Larmor Equation:

 

[math]\Delta H_L = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V (r_{ij})}{\partial r_{ij}} (\hat{n}\cdot \sigma_{ij}) \begin{pmatrix} \alpha \\ \beta \end{pmatrix}[/math] (1)

 

But we also derived

 

[math]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/math] (2)

 

This means we can come to a completely new form of the equation again, by plugging in 2. with eq 1.,

 

[math]\Delta H_L = \frac{2\mu}{\hbar Mc^2 e} (\nabla \times \vec{F}_{ij} \cdot \hat{n}) \frac{1 + cos \theta}{2}[/math]

 

(we have just exchanged the matrix form for the more direct form of the angle between the spin vectors [math]\frac{1 + cos \theta}{2}[/math] ) Compare this with the original version of the Larmor Energy and you can see why it can be done

 

[math]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial U r}{\partial r} L \cdot S[/math]

 

Of course, it's probably not too surprising that we can involve the curl of the field in such a way in these equations, from the Larmor energy point of view... after all, the Larmor Precession is all about the magnetization energy of a field and the way it rotates. In this sense, we are using the curl so we must be calculating infinitesimal rotations of the field in three dimensional space.

 

And the angle between the two spin vectors is just a natural consequence of the mathematics used, but doesn't seem to be a coincidence in it appearing since in the original Larmor energy, it had [math](L \cdot S)[/math] which can be defined as the angle between two spins [math]|L||S|cos \theta[/math].

Cheers.

 

yeah, I didn't know that either... was thinking more along the lines of row-echelon.

 

 

 

I've had no replies to this topic... does no like my idea :/

Can anyone make clear what the sense is, if any, in saying that a photon and a neutrino, two elementary particles, are incomparable? Here I am especially inquiring about their existence as measurable entities.

 

 

Incomparable? Do you mean, is there anything about a neutrino that can be similar to a photon? Actually there is... something called the Weyl Limit.

 

I'll write out the maths if you want, but its not nice for anyone whose not adapt to quantum mechanics... The neutrino is found to be almost massless... It has therefore a very very very miniscule mass. You can take the Dirac Equation and rewrite it as a set of decoupled equations. They are only coupled up to a limit, and that is when the mass equals zero [math]M=0[/math].

 

Now, from what I can understand of the Weyl Limit, is that the nuetrino mass is so small that it may as well just act like a massless particle. So in your case, a photon and a neutrino has a little something in common, at least from the perspective of the limit just mentioned.

 

 

 

 

But as other posters have shown, there are plenty differences between the two, measurably.

There is no such thing as a flux, passage or flow to time.

 

 

I can assure you of this.

All I asked for was a more in-depth explanation and you have now provided it. I will certainly follow the link... I could be very wrong in my approach which is why I questioned it.

 

You do mentioned the radiation a few times:

 

''The SC metric is not necessarily an accurate representation of reality if you're taking into account small details like radiation. It is useful, for example, to model the Sun with the SC metric. It's obviously not a 100% accurate model since the Sun is not perfectly symmetrical, it's giving off radiation, it's rotating a bit, etc. Nonetheless, the SC metric is a good approximation.''

This was the kind of idea I had in mind... except for a non-rotating body.

I guess which science pawns what is a matter of personal belief. Personally, physics is the science... the only science rather capable of explaining the origins of any other science.

Where is this thread in the forum, I can't find it anywhere?