View Full Version : Zero Point: Solving for it and what it is.
IooqXpooI
Aug10-04, 11:12 AM
Zero Point
What it is:
Zero Point is the point of equivalence of the forces acting on each other of two objects(Gravity and EM). This is where the forces are equal if they were acting on a peice of bread, for instance. If the masses/charges are equal, the Zero Point of the force is directly in the middle, but if it is lopsided, the Zero Point is closer to the lesser mass/charge, due to the inverse square law.
Solving for it:
The basic equations to solve for Zero Point, where:
M is the greater mass.
Q is the greater charge.
q is the lesser charge.
m is the lesser mass(of course, none of this is needed when the masses are equal)
d is the total distance in between the two bodies.
r_1 is the distance of M(or Q) from the Zero Point.
r_2 is the distance of m(or q) from the Zero Point.
Equations:
Gravity-
r_1=\frac{Md - Mr_1}{m}
r_2=\frac{md - mr_2}{M}
Electromagnetism-
r_1=\frac{Qd - Qr_1}{q}
r_2=\frac{qd - qr_2}{Q}
To solve for the Mass(or Charge)-
M=m(\frac{r_1}{r_2})
Q=q(\frac{r_1}{r_2})
m=M(\frac{r_2}{r_1})
q=Q(\frac{r_2}{r_1})
IooqXpooI
Aug10-04, 11:56 AM
C'mon...someone?
IooqXpooI
Aug12-04, 03:59 AM
C'mon! Someone criticize this or tell me why you can't!
Antonio Lao
Aug12-04, 06:16 AM
What you did are just linear combinations of weighted distances similar to adding vectors when they are collinear. And you also take into account that linear coefficients can also be greater than unity which is not normalized. You can get results that are infinite and therefore cannot be part of a solution to your equations.
IooqXpooI
Aug12-04, 12:18 PM
Please, explain it to we who do not have enough knowledge to find your words explaining of your point!
Chronos
Aug12-04, 03:38 PM
I get different answers. Bear in mind that r_1 + r_2 = d.
Using the inverse square law and your notation the ratio of mass [or charge], I get a 'zero point' distance of
M = m\frac{r_1^2}{r_2^2}
Your solution results in
M=m(\frac{d}{r_2} - 1)
Substituting for d
M=m(\frac{r_1 + r_2}{r_2} - 1) which simplifies to
M=m\frac{r_1}{r_2}
Sorry, I just got hung up on this:
acting on a peice of bread, :confused:
Antonio Lao
Aug13-04, 01:51 AM
Please, explain it to we who do not have enough knowledge to find your words explaining of your point!
For normalization of a quantity, the dividend can never be greater than the divisor. This is to say that the quantity is a fraction. And if the quotient is unity then the dividend is equal to the divisor. So to avoid getting infinity in finding a solution when adding fractional numbers, the sum is always unity. When all the fractional numbers (dimensionless quantities such as ratios of masses and distances) do not add to unity then the result is holism. This philosophy says that the whole is greater than the sum of its parts.
Since the total mass of universe cannot be detected as if some mass is missing, the total mass of the universe is a holistic quantity. Since the total size of the universe cannot be determined which is limited by the visible universe, the total distance of the universe is also a holistic quantity.
IooqXpooI
Aug13-04, 10:37 AM
Sorry, I just got hung up on this:
:confused:
I don't know...Just an example-:)
IooqXpooI
Aug13-04, 10:39 AM
For normalization of a quantity, the dividend can never be greater than the divisor. This is to say that the quantity is a fraction. And if the quotient is unity then the dividend is equal to the divisor. So to avoid getting infinity in finding a solution when adding fractional numbers, the sum is always unity. When all the fractional numbers (dimensionless quantities such as ratios of masses and distances) do not add to unity then the result is holism. This philosophy says that the whole is greater than the sum of its parts.
Since the total mass of universe cannot be detected as if some mass is missing, the total mass of the universe is a holistic quantity. Since the total size of the universe cannot be determined which is limited by the visible universe, the total distance of the universe is also a holistic quantity.OK. I am assuming that holistic is wholistic...whatever.
I understand that, though I cannot get that from the original statement.
Antonio Lao
Aug14-04, 07:27 AM
Is there a theory that says anything about the total charge of the universe?
Chronos
Aug14-04, 03:17 PM
Is there a theory that says anything about the total charge of the universe?
Yes. Gauss's Law. Think global charge neutrality. The Laplace equation solution for the total charge of the universe is zero.
Epsilon Pi
Aug14-04, 04:52 PM
Excellent answer! Is it not remarkable that we do not have an equivalent Gauss's Law for the magnetic field, as we cannot have isolated poles? Both the gravitational and the electric field are conservative fields, but the former is a non conservative field, so in a certain sense it is right to say that space is pervaded by a sea of magnetic field, the origin of the so-called Zero-point-energy?, conservative fields on other hand seem to be derived fields.
Regards
EP
Yes. Gauss's Law. Think global charge neutrality. The Laplace equation solution for the total charge of the universe is zero.
Chronos
Aug14-04, 10:53 PM
Excellent answer! Is it not remarkable that we do not have an equivalent Gauss's Law for the magnetic field, as we cannot have isolated poles? Both the gravitational and the electric field are conservative fields, but the former is a non conservative field, so in a certain sense it is right to say that space is pervaded by a sea of magnetic field, the origin of the so-called Zero-point-energy?, conservative fields on other hand seem to be derived fields.
Regards
EP
Spoken like a true EE. I think the magnetic moment eventually submits to the same solution. Try using gauge invariance. As I recall, that was how unification of the EM field with the weak nuclear force was achieved.
Epsilon Pi
Aug15-04, 06:47 AM
But what is a magnetic moment? Its origin is not in an inherent magnetic field, I mean, in the electron? And oh yes, the breaking of parity in the weak nuclear force, but does not the magnetic field and its even parity played a central role on it?
Why should we have waited parity was not broken when the imposed field in the system had an even parity?
Just some questionings about parity
Regards
EP
Spoken like a true EE. I think the magnetic moment eventually submits to the same solution. Try using gauge invariance. As I recall, that was how unification of the EM field with the weak nuclear force was achieved.
Epsilon Pi
Aug15-04, 06:57 AM
Should we consider a generalized Gauss's Law, where antimatter and matter, white gravity bodies and black gravity bodies are included, so we can give reason of what is observed at great astronomical distances and maintain the steady state of the whole gravitational cosmic system?
I mean the total gravitational charge!
Regards
EP
Is there a theory that says anything about the total charge of the universe?
Epsilon Pi
Aug15-04, 08:58 AM
Oh, yes the chemistry of nuclear interactions. My point and concern at this moment is not precisely that chemistry, let the experts on such a field finish their work!
My point has to do with a way of "seeing" the relation between classical concepts and non classical ones:
- can those concepts with which the normal engineer work everyday be put in the same context as those non classical concepts that began with QM?
- Does not EE has worked succesfully for more than 100 years with those non classical concepts by using complex numbers?
- Is it possible to aply that same methodology to QM, to the Lorentz transformation group, to the pendulum and even the gravitational fields?
- can this be done with that methodology used to explain the chemistry of the nuclear interactions, I mean, can all those equations mentioned above be put under that same roof?
Just some questionings
Regards
EP
Try using gauge invariance. As I recall, that was how unification of the EM field with the weak nuclear force was achieved.
Antonio Lao
Aug15-04, 11:35 AM
I mean the total gravitational charge!
Why is the total gravity charge always attractive?
Epsilon Pi
Aug15-04, 11:57 AM
As far as I know there are just two kinds of physical fields, I mean not theoretical fields:
- non conservative fields such as the magnetic field that do not have associated with it a Gauss's Law, having then the chance to have the two kinds of forces:attractive and repulsive: the inherent magnetic center has in it the two polarities
- conservative or derived fields for which Gauss's Law is valid, where we find sort of odd parity; its lines of force go from one center to another center, so just one type of force we have because of this. In the mean while in this kind of field seems to prevail the second law of thermodynamics, it seems not to be the case(?) with the former, as space is surrounded by a sea of a magnetic field, some physicists have been thinking in a Zero-Point-Energy, sort of free energy? But are not both types of field complementary to each other? The latter being used for applications in the mean while the former is the original field?
This is the way I see those things
Best regards
EP
Why is the total gravity charge always attractive?
Antonio Lao
Aug16-04, 08:56 AM
This is the way I see those things
Thanks. I have no question at the moment. But when you say Gauss' law do you mean the existence of a divergence of a vector field while scalar field never can have a divergence?
Epsilon Pi
Aug16-04, 10:54 AM
I mean the divergence of the magnetic field is zero as there are no isolated poles, while both the electric and gravitational fields(conservative fields) have a divergence different from zero, associated with the electric charge or, as it were, the gravitational charge. Gauss's law is recognized as the integral form of one of Maxwell's equation.
Regards
EP
Thanks. I have no question at the moment. But when you say Gauss' law do you mean the existence of a divergence of a vector field while scalar field never can have a divergence?
Antonio Lao
Aug16-04, 11:18 AM
Epsilon Pi, thanks again. Does the divergenceless magnetic field imply that we cannot attribute a speed to the magnetic flux? Or we can only attribute speed of zero and infinity but not any value in between?
Epsilon Pi
Aug16-04, 03:11 PM
It is my pleasure, Antonio, but if we consider that the magnetic field is the real origin of electromagnetic radiation then its speed must not be the speed of light?
Does not the magnetic field imply sort of inherent oscillator?
Regards
EP
PD:
Epsilon Pi, thanks again. Does the divergenceless magnetic field imply that we cannot attribute a speed to the magnetic flux? Or we can only attribute speed of zero and infinity but not any value in between?
Epsilon Pi
Aug16-04, 03:16 PM
Does not the "divergenceless" magnetic field imply sort of inherent energy in that inherent oscillating polarity? sort of Zero-Point-Energy?
Regards
EP
Epsilon Pi, thanks again. Does the divergenceless magnetic field imply that we cannot attribute a speed to the magnetic flux? Or we can only attribute speed of zero and infinity but not any value in between?
Antonio Lao
Aug16-04, 03:54 PM
Does not the magnetic field imply sort of inherent oscillator?
What do you think is the speed or frequency of the oscillation? This seems to be a rotational motion (spin or constant angular acceleration of some sorts) as opposed to a translational linear motion.
Could this angular acceleration be the motion of 1D space when it is subjected to an extremely strong force at a local infinitesimal region of spacetime?
The magnitude of this force is the ratio of Planck energy over Planck length.
Epsilon Pi
Aug17-04, 06:54 AM
Yes, it is a rotational motion as is inferred from the mathematical model I have in mind, I mean Euler relation, and the frequency of oscillation will be that one given by Plank's formula, isn't ?
Had not been the main problem with "black bodies", the fact they used sort of dipolar electric model, not an inherent polarity model, as that one of an inherent magnetic field?
Regards
EP
What do you think is the speed or frequency of the oscillation? This seems to be a rotational motion (spin or constant angular acceleration of some sorts) as opposed to a translational linear motion.
Antonio Lao
Aug17-04, 07:01 AM
Epsilon Pi,
Thanks. Looking up something. Will reply asap.
Antonio Lao
Aug17-04, 10:05 AM
According to the book by Linus Pauling and E. Bright Wilson, Jr "Introduction to Quantum Mechanics: with Applications to Chemistry" on page 72, the zero-point energy is given by
E_0 = \frac{1}{2} h \nu_0
where \nu_0 is the fundamental frequency of the oscillator.
My question is what is the value of this fundamental frequency?
Antonio Lao
Aug17-04, 10:16 AM
The angular frequency is defined as
\omega = 2 \pi \nu_0 = \frac{2 \pi}{T}
where T is the period of the wave motion.
The circumference of a circle is 2 \pi r , therefore the above definition of angular frequency implies a unitless radius of 1.
Antonio Lao
Aug17-04, 10:27 AM
For all dimensional waves, the wavelength can be defined as four times the unitless radius. The angular frequency can be calculated to give a value greater than light speed
\omega = \frac{ \pi c}{2}
Antonio Lao
Aug17-04, 10:42 AM
Furthermore, if the invariance given by
\vec{a} \cdot \vec{r} = c^2
is inserted into the simple formula for angular frequency
\vec{a} \cdot \vec{r} = \left( \frac{2 \omega}{\pi} \right)^2
Epsilon Pi
Aug17-04, 02:06 PM
Are you not violating the uncertainty principle with a unitless radius of 1? At those levels you cannot suppose a particle with a definite radius, can you?
Regards
EP
therefore the above definition of angular frequency implies a unitless radius of 1.
Epsilon Pi
Aug17-04, 02:12 PM
Remember it is not a wave, it is not a particle, it is an energetic system, whose state cannot be determined completely
Regards
RP
The circumference of a circle is 2 \pi r , therefore the above definition of angular frequency implies a unitless radius of 1.
Antonio Lao
Aug17-04, 03:01 PM
I think, in physics, the unitless radius is replace by the phase angle of a wave. a right triangle of sides 3-4-5 always has the same complementary angles regardless of how the sides are scaled as long as the ratio remains 3-4-5. The invariance of the phase angle is a scaling transformation invariance applicable to all similar right triangles at any given scale. And all trigonometric functions is the ratio of two sides of a particular right triangle. In a sense, it is applying the Pythagorean theorem over and over again. This theorem is the basis for the definition of a length and a distance and any higher dimensional metric.
Epsilon Pi
Aug17-04, 03:13 PM
If I have in mind, at the background Euler relation, in a certain sense I can follow you, but then in it, phase angle is quite different from the radius or amplitude of that wave.
Regards
EP
I think, in physics, the unitless radius is replace by the phase angle of a wave...In a sense, it is applying the Pythagorean theorem over and over again. This theorem is the basis for the definition of a length and a distance and any higher dimensional metric.
Antonio Lao
Aug17-04, 03:13 PM
The existence of a minimum triangular surface area of 1/2 corresponding to a unit square does not depend on the curvature of spacetime where the surface is embedded and clearly demonstrated in an Euclidean geometry using the parallel axiom.
Antonio Lao
Aug17-04, 04:04 PM
The Euler's identities are given by
e^{+ i \theta} = cos \theta + i sin \theta
and
e^{- i \theta} = cos \theta - i sin \theta
A complex number z is given by
z = r \left( cos \theta + i sin \theta \right)
So Euler's identities is the same as when r=1 and
z = e^{ i \theta}
Antonio Lao
Aug17-04, 04:10 PM
But the complex number is defined as z = x + iy. This is like adding apples to oranges where x is an apple and iy is an orange. Physically, I still failed to understand this but mathematically, I guess, anything is possible.
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