# Theory of LIM

1. Mar 18, 2004

### Antonio Lao

LIM stands for Local Infinitesimal Motion.

LIM is motion of two exclusive space points at the local infinitesimal region of space. The metric can be theorized to be smaller than the Planck length of $$10^{-33} cm$$.

It is known that all fermions possess a magnetic moment. The existence of magnetic moment is a good physical proof that LIM should exist. It is only needed to hypothesize that two space points orbit around each other forming two types of orbit, which are not topologically equivalent. One has an electric charge of +1/6 and the other an electric charge of -1/6. But since electric charge is a property of matter alone, it is more properly to call the electric charge of space points as space charges. When space charges combined to form matter, the space charges are then transformed to electric charges.

Last edited: Mar 18, 2004
2. Mar 18, 2004

### matt grime

'the metric can be theorized to be smaller than planck's constant'? Well, seeing as d(x,x)=0,is that not vacuous. Or are you trying to sy something else?

3. Mar 18, 2004

### Antonio Lao

matt grime,

Matt, it's nice to hear from you again. As a theorist, I cannot say how much smaller than Planck length the metric can be. All I can say is that it must not be zero. It could be inverse of $$10^{34} cm$$. All theories above Planck scale are, I think, well established. I don't want to contradict or to argue with these accepted theories.

I do have a math question for you. When can one be allowed to take only two points of space for analysis without having to consider the closest neighborhood points? Are we dealing with closed or open sets?

Antonio

4. Mar 18, 2004

### matt grime

You appear to have a discrete space, that is in its topology the points are open and closed. You can't mean your metric space is R^n for some n with the usual metric because then there are distances on all scales - it might be that some distances are so small that on those scales the quantum effect is noticable. That doesn't stop there being smaller measurements.

I find it hard to answer your questions because I don't know what you're getting at with them - what analyis, what points, what metric space!

5. Mar 18, 2004

### Antonio Lao

What I am hoping to do is to isolate two points and make something of it. All other neighborhood points are ignored. What I would end up are a bunch of two-point sets that are independent of each other.

My question is: Can we do this by using math?

6. Mar 18, 2004

### matt grime

You really ought to state what the metric space is - that involves the underlying set and the metric on it.

Then can you say what you mean by 'independent'?

7. Mar 19, 2004

### Antonio Lao

I am trying to understand Stoke's Theorem about the curl. I would like to use the concept to make an inverse curl operator. This inverse curl might the metric I am looking for? When this differential operator is applied to a differential force function and take the scalar product of itself or its reflection gives a square of energy.

The LIM that I am trying to define is then the inverse curl.

Can this be done mathematically?

Last edited: Mar 19, 2004
8. Mar 19, 2004

### matt grime

Yes and no.

No because there are plenty of fields with curl zero so you can't do inverses. No because a metric is a map from the space to the real numbers, not a function to vector fields. No because you've not said in which domain you want this to work.

Yes, because you can sidestep two of those. The middle one is fatal, though.

9. Mar 19, 2004

### Antonio Lao

The map looks like this

$$r_1 \times F_1\cdot r_2 \times F_2 \longrightarrow E^2$$

and

$$r\times$$ is the inverse curl operator.

10. Mar 19, 2004

### matt grime

r is the inverse curl (whihc is only defined up to the addition of divergenceless fields or something) but what are r_1, r_2, F_1, F_2, what is the dot, what is E^2, and why is that a metric? The inverse curl ought to give a vector field, a metric gives a scalar (real) number! That is not compatible.

11. Mar 20, 2004

### Antonio Lao

F1, F2 and r1, r2 are vectors.
F1, F2 are the differential force.
r1, r2 are radii of differential path of "almost circular rotation."
E^2 should have unit of square of energy. Is this a scalar in R^2 space?
the dot is the scalar product. The X is the vector product.

12. Mar 20, 2004

### matt grime

A scalar in R^2 space? eh?

E^2 is by definition a scalar as it is the output of the scalar product.

I don't see what this has to do with your initial post, but perhaps that's just because you're not used to writing things for a mathematician to read. I just mean you aren't using words with the strict definitions in mind.

The inverse of curl would have to give a vector output - an element in R^3. A metric gives a scalar output, an element in R. So one cannot be the same as the other as you were initially requiring.

13. Mar 21, 2004

### Antonio Lao

Can you help me make the expression

$$r_1 \times F_1\cdot r_2 \times F_2 \longrightarrow E^2$$

in a definition that a mathematician can read?

14. Mar 21, 2004

### matt grime

That's fine if you explain what r_i and F_i are, though what they have to do with curl I don't see.

It was that you were expecting the inverse of curl (which is not well defined exactly) to be a metric, when one must take values in R the other must take values in R^3.

15. Mar 22, 2004

### Antonio Lao

For a special case:

The absolute magnitudes

$$|F_i| = |F_j|$$
$$|r_i| = |r_j|$$

I don't really need the curl if there is another way of describing double "rotation" in the infinitesimal region of space.

16. Mar 23, 2004

### Antonio Lao

Today I came to realize that the $$r_i$$ can be the wave function $$\psi$$ of quantum mechanics, since they both have the property of length. Is this a misconception?

17. Mar 23, 2004

### Antonio Lao

If $$r_i$$ is the wave function $$\psi$$

and $$F_i$$ is the differential force $$\phi$$

then in a time independent system (force-free):

$$\psi_i \times \phi_i \cdot \psi_j \times \phi_j = E^2$$

18. Mar 23, 2004

### matt grime

In the sense that this implies anything 'with the property of length' could be used, such as my left leg?

19. Mar 23, 2004

### Antonio Lao

The length concept used for ri is the infinitesimal length while your left leg is the integral (the almost infinite sum) of this infitesimal length. Without this infinitesimal length, everything will be in a state of uniformity nothing to measure from or measure to.

This infinitesimal length is one-dimensional (zero-dimensional by experimental verification below the Planck length). But its dynamic states (vibration, rotation, expansion, contraction, etc.) cause it to appear 3-dimensional hence the existence of matter and energy. In this way, does it sound like what the string theorist are saying?

The quantization of length leads to the quantization of matter and energy.

Last edited: Mar 23, 2004
20. Mar 23, 2004

### Antonio Lao

The probability amplitude

$$|\psi|^2$$

is implied in

$$\psi_i \times \phi_i \cdot \psi_j \times \phi_j$$

for i=j.

To see this, the expression of the scalar-vector product must be expanded using Lagrange's identity.