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Messages - lightarrow

Pages: [1] 2 3 ... 233
1
Physics, Astronomy & Cosmology / Re: What is the ratio between electric and magnetic field in electromagnetic waves?
« on: 01/03/2018 12:56:03 »
Quote from: hamdani yusuf on 01/03/2018 11:37:25
Quote from: lightarrow on 01/03/2018 00:55:04
Quote from: hamdani yusuf on 28/02/2018 12:25:25
What is the ratio between amplitude of electric and magnetic field in electromagnetic waves?
In which unit of measure? In SI is c, as already said: ||E|| = c||B||; in other systems, the rate may not be c; in CGS - GAUSS the rate is 1: ||E|| = ||B||.
And this in the "far field" of the wave, that is far enough from the sources, or things become very complicated

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lightarrow
How can merely change the unit from SI to cgs
Things are different in electrodynamics because the magnetic field B is defined differently, it's not always a mere conversion from a value in a unit of measure to the other value in the other unit, sometimes even the equations are different. But your is a good question. Let's see the effect of this, that is the magnetic force on a moving charge q at velocity v in a field B which, for simplicity, is orthogonal to v.

1) In SI: F = q*v x B where "x" here stays for vectorial product. So, for an em wave in the void in the far field: ||F|| =  q||v||*||B|| = q||v||*||E||/c.

2) In CGS:  F = q*(v/c) x B. So, for an em wave in the far field:
||F|| =  q(||v||/c)*||B|| = q(||v||/c)*||E|| = q||v||*||E||/c

and the two are the same  :)
Quote

In the case of slowly rotating magnet, we can get amplitude of B much larger than amplitude of E when measured nearby.
Yes.
Quote
  But if we measure from adequately far distance, amplitude of B diminish quicker than E until their ratio approach a constant?
Yes. Do you remember what I wrote? Compute the wavelenght λ; how far from the source, let's say at distance r, you should evaluate the fields to be sure that r >> λ?
To compute the wavelenght remember that λ = c/f where f is the frequency. It's a trivial exercise.

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lightarrow

2
Physics, Astronomy & Cosmology / Re: What is the ratio between electric and magnetic field in electromagnetic waves?
« on: 01/03/2018 00:55:04 »
Quote from: hamdani yusuf on 28/02/2018 12:25:25
What is the ratio between amplitude of electric and magnetic field in electromagnetic waves?
In which unit of measure? In SI is c, as already said: ||E|| = c||B||; in other systems, the rate may not be c; in CGS - GAUSS the rate is 1: ||E|| = ||B||.
And this in the "far field" of the wave, that is far enough from the sources, or things become very complicated
Quote
Is it a constant? does it depend on the medium? does it depend on the frequency?
Is it possible to generate pure electric wave or magnetic wave?
You want to know too many things at once  :)
In electrodynamics, things can become very complicated if you want to analyze all the possible situations. In most of the cases you will consider em waves in the void and far from the sources (the distance is many times the wavelenght), then the rate is constant and doesn't depend on frequency. It's not possible to generate a pure electric or a pure magnetic wave: in a wave you necessarily have a time variation of a field and electrodynamics' laws (see Maxwell's equations for example) tells us that every time variation of an electric field generates a magnetic field and every time variation of a magnetic field generates an electric field.

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3
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 25/02/2018 21:13:38 »
Quote from: alancalverd on 25/02/2018 18:08:34
In physics, a field is something that is defined (by a number or a vector) at every point in space. Nothing more, nothing less.
You are talking of the mathematical definition only, here. But in physics a field has a physical meaning too.
Did you read my post where I write that even a static field has energy and a mass too?
Also, I don't get why you say that a field don't propagate.
It's true I haven't attended the forum for a couple of years, but it sounds strange to me that physics has changed so much meanwhile  :)

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4
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 24/02/2018 13:25:26 »
Quote from: Bill S on 23/02/2018 21:58:03
Quote
Conclusion: better not to talk about relativistic mass at all.
I expressed my intended meaning badly.  By "relativistic mass" = inertia, I meant something like: If you meet the term relativistic mass, it is better to think of it as inertia.
Would that be better?
IMHO it would be better "energy" (in the sense "total energy") since relativistic mass is just energy/c2.
Concerning "inertia", not, for what I wrote: you should remember that you have *two* different inertias simultaneously in the same body:
1) tangential inertia = γ3m
2) transversal inertia = γm
where m is the body's (rest/proper/invariant) mass.

If you are sure to remember that, ok, but when you talk of "inertia" you always have to specify *which of the two*

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5
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 22/02/2018 18:49:35 »
Quote from: Bill S on 22/02/2018 02:09:48
Quote from: Lightarrow
However I don't like to call m "rest" mass because nowadays it's simply called "mass" (the so called "relativistic mass" is an  obsolete term and concept).
How far off the mark would I be if I interpreted this as saying: "rest" mass = total mass;
Ok, unless you give particular interpretations to "total" mass. Better to say "system" mass.
Quote
  "relativistic mass" = inertia? 
Do you like a concept of inertia which is different along different directions? Along the direction of the body's velocity you have a value of inertia, that is: γ3m, where γ is the gamma Lorentz factor:
γ = 1/sqrt[1-(v/c)2];
in the ortogonal direction you have a different value, that is: γm.
Conclusion: better not to talk about relativistic mass at all.
IMHO.

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lightarrow

6
New Theories / Re: Energy =mc^2 or Energy=Hf
« on: 22/02/2018 18:08:55 »
Quote from: sceptic-eng on 20/02/2018 09:01:18
Lightarrow " "both need the electromagnetic current or flux and electrostatic charge voltages for their derivations"".  Let me try to explain.  The hadron collider uses umteen
Sorry, don't know what is "umteen".
Quote

megaWatt hours [V*I*Cos thi*T} of fundamental energy
What do you mean with "fundamental energy" here?
Quote
per second to accelerate or force a particle around a magnetic circle.  YES, but how does that energy vibration
what do you mean with energy "vibration" here?
Quote
transfer into the nucleas of the molecule?? Is it just increasing the magnetic momentum of the particles spin
If you want to use these terms, please at least use them correctly: "the particle's magnetic momentum associated to its spin".
Anyway the answer is not: the magnetic momentum has nothing to do with it; LHC or other similar devices can only accelerate charged particles and even the force which curves them is due to its electric charge:

 F = qE + qv x B

(Lorentz' force).

F = resultant force on the particle;
q = particle's electric charge;
E = electric field applied to the particle by the apparatus;
v = particle's velocity;
B = magnetic field applied to the particle by the apparatus;
In bold the vectorial quantities.
the symbol "x" means "vectorial product".
Quote
or speeding up the proton charge radial velocity within the electron enclosure which remains the same shape [what shape?] without bursting. .
Just meaningless word salad...

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7
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 21/02/2018 23:12:04 »
Quote from: Bill S on 21/02/2018 19:30:11
Could be its silly to ask another question before the previous one has been addressed, but I'm slipping this one in before I forget about it.  Fixation amnesia, whatever that is.  :)
  :)
Quote

Quote from: Lightarrow
Probably I'll amaze you now, but did you know that an electrostatic field has a mass too?
My understanding is that the electrostatic field must carry energy, therefore, according to E=mc2 it must have “mass”.  However, would this equate to rest mass, or would it be better described in terms of inertia?
Good question.
The first you wrote.
As you know, the more general equation is not E=mc2 but is:
E2 = (mc2)2 + (cp)2
where p is the system's total momentum (vector).
If the electric field is stationary, and only in this case, its momentum is zero, so the equation reduces to the much known you wrote.

However I don't like to call m "rest" mass because nowadays it's simply called "mass" (the so called "relativistic mass" is an  obsolete term and concept).

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lightarrow   

8
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 21/02/2018 15:57:28 »
Quote from: Bill S on 20/02/2018 20:19:45
Quote from: lightarrow
The fact it needs a source doesn't mean it can't exists after being generated by it. You switch on a laser device: it generates a light beam; you switch off the device: the light beam keeps going towards its target.
Sorry, this makes no sense to me. 

“You switch on a laser device: it generates a light beam”.  The device is the source of the beam.
“you switch off the device: the light beam keeps going..”.  Of course, but the device is still the source of the beam!
But the device is switched off. If the electric current or other source of energy is not there any longer, you can destroy the device, vaporize it, and nothing changes: no field is generated.

The actual source of the field are the electric current's charges in the specific potential inside the light emitting diode (if it's a laser device of this kind). When you switches it off it's just as a piece of inert stone, for what concerns the laser beam, that is, the field.
Quote
Quote
The ancient romans created the Adrian wall; the wall is still there, but the ancient romans don't exists anymore.
Surely, the ancient Romans remain the “source” of Hadrian’s Wall; even if they are no longer with us.   
In this case the ancient romans are the electric's current charges: once switched off, the laser beam remains (if you shooted it in the void, travelling in outer space) but the electric current which generated the field is gone.

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lightarrow

9
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 21/02/2018 15:42:22 »
Quote from: geordief on 19/02/2018 17:34:39
i
Quote from: lightarrow on 19/02/2018 16:05:55
Quote from: geordief on 18/02/2018 23:40:43
Quote from: jeffreyH on 18/02/2018 22:32:55
The energy of a field exists. Otherwise we wouldn't have any forces. It isn't tangible. That does not mean that it is simply an abstract concept.
the field is not just model and  fit for scrap if superseded?
Probably I'll amaze you now, but did you know that an electrostatic field has a mass too?

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lightarrow
Can I rationalize that by saying (correctly I hope) that the source  of this effect (the source =the charges)  is not local but spread out in a wave form?

That might be word salad  as..... (well you can probably guess :-[  )
No, the fact sources are spread out or not doesn't change the properties of the electrostatic/electromagnetic field outside of them; in case it changes the electrostatic's field energy inside the charge. This is an old problem, studied by Abraham - Lorentz, Dirac, Feynman and others and partially solved by Quantun Electrodynamics; look for electron's electromagnetic mass and the 4/3 problem, e. g.  https://en.m.wikipedia.org/wiki/Electromagnetic_mass
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10
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 19/02/2018 16:05:55 »
Quote from: geordief on 18/02/2018 23:40:43
Quote from: jeffreyH on 18/02/2018 22:32:55
The energy of a field exists. Otherwise we wouldn't have any forces. It isn't tangible. That does not mean that it is simply an abstract concept.
the field is not just model and  fit for scrap if superseded?
Probably I'll amaze you now, but did you know that an electrostatic field has a mass too?

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lightarrow

11
New Theories / Re: Energy =mc^2 or Energy=Hf
« on: 19/02/2018 15:56:12 »
Quote from: sceptic-eng on 19/02/2018 07:33:53
Physics and cosmology are at odds as both the above statements are correct but contradictory. The reason for the difficulty is that both need the electromagnetic current or flux and electrostatic charge voltages for their derivations but these balancing electric forces are not included in the base equations above are they?
 Newton never said that gravity was the only accelerating force in space and Einstein never said that electric forces did not exist in space.
 We want the truth about balancing the universe not mumbo-jumbo theories about dark energy/matter or multi-universes. Why not just balance the galaxies using the electromagnetic forces that must be there. ACSINUK
Sorry but I'll be a bit critic about you statements.

First, equations in Physics means nothing if decontextualized: which is the system/s on which you are applying those equations? Wich is the environment with which it exchanges energy and/or others?

Second, physics and cosmology can't be in contrast since cosmological models are made using current physics.

Third, the sentence "both need the electromagnetic current or flux and electrostatic charge voltages for their derivations" means absolutely nothing said in this way. If it means something, you have to specify exactly and in detail what you are talking about (maybe you are referring to something you have discussed recently with someone in this forum, which  I'm not aware of, I don't know).

Fourth, you talk of "these balancing electric forces are not included in the base equations above" but the "equations above" give the energy, not the force.

Fifth, I have totally no idea of what you mean with "We want the truth about balancing the universe".

Sixth, which are the: "Why not just balance the galaxies using the electromagnetic forces that must be there"?

On which popular book did you find these (incorrect) things?

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lightarrow

12
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 18/02/2018 22:50:23 »
Quote from: Bill S on 18/02/2018 19:39:29
Quote from: Chiral
As I understand it, a field needs no source. A field just is. It's a sort of accounting or mapping tool that we can use to describe the universe. Every field exists everywhere, even if it has zero amplitude at specific places, or even everywhere.
You are saying, quite specifically, that a field (e.g. the electron field?)
"electron field" is something else, let's call it with a better name: electric (or electromagnetic) field generated by the electron.
Quote

 is the pre-existing entity – it needs no source.
The fact it needs a source doesn't mean it can't exists after being generated by it. You switch on a laser device: it generates a light beam; you switch off the device: the light beam keeps going towards its target.
Quote
If this is so, what meaning does it have to talk of the electron as the source of the field?
The ancient romans created the Adrian wall; the wall is still there, but the ancient romans don't exists anymore   :)

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13
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 17/02/2018 18:40:30 »
Quote from: opportunity on 17/02/2018 11:08:45
With energy annihilation, as for instance to resolve Dirac's postulate,
What do you mean with "Dirac's postulate"?
Quote
  does field annihilation propose a source?
Sorry I can't understand what you mean here too.

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lightarrow

14
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 17/02/2018 18:34:15 »
Quote from: geordief on 16/02/2018 22:29:19
Secondly I should have described my potential sensors as "conductive" rather than "charged".
Then that's not a problem too: a piece of dielectric material which has a charge separation (let's say a plastic or glass stick with a charge + at one end and a charge - at the other) or a simple molecular dipole (e.g. a water molecule) will rotate in an electric field which is not parallel to its axis.
Quote
Apologies :-(
No problems at all...  :)

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lightarrow

15
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 17/02/2018 15:35:59 »
Quote from: jeffreyH on 17/02/2018 13:44:30
@lightarrow hadn't noticed you around for a while. Welcome back.
Thank you.
Yes, I have been away for a couple of years, too busy.

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lightarrow

16
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 17/02/2018 15:27:04 »
Quote from: opportunity on 16/02/2018 12:09:07
Would that theoretical construction you're proposing aim to alter the reference
It's not an "alteration": I have just specified (simple) initial conditions; the OP didn't specify them.
Quote
in view to make it more understandable?
The fact I used simple initial conditions is certainly in view of that.   
Quote
Sounds wonky but is theory here taking over fact?
Well, if we are still talking of a universe with ONLY two charges and nothing else, it's not simply theory, it's plain phylosophy   :)

But if we want to talk about something with some physical meaning and admit the existence of a little more than two charges only, i.e. at least a detector and a physicist  :) then the above simplification is acceptable.

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lightarrow

17
Chemistry / Re: How can pH be adjusted by using CaCO3 - calcium carbonate?
« on: 16/02/2018 17:23:54 »
Quote from: dgt20 on 14/02/2018 22:02:31
What other simple chemical experiments are there used to alter pH of grapes?
If you add MgCO3 or Mg(OH)2 instead of CaCO3, it helps you feel less muscular fatigue: athlets take sources of Mg2+ for that purpose  :) I don't know which will be the wine  final taste, however...

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lightarrow

18
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 16/02/2018 17:09:43 »
Quote from: geordief on 16/02/2018 11:23:32
Quote from: lightarrow on 16/02/2018 08:31:13
The couple of charges, let's assume for simplicity they can be considered initially stationary in an inertial frame (difficult to define in a universe where only those charges exists, however  ;)) generate an electrostatic field in all space, which have its precise energy. The two charges are sources of this field.

After the charges have annihilated,  that field disappears too (not immediately of course, the information travels at c).  Its energy will be conserved in the form of em radiation which adds to the 2 gammas (infact it could be included in the gammas itself).

So the actual field generated by the charges disappears with them.

Not immediately, however. Maybe the question should be referred to this field which disappears at speed c from the charges centre.

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lightarrow


Any mechanism whereby the field could be measured ? 
With another charge  :)
When you study in a (good)  physics text what is an electrostatic field, you find a phrase like: "... the force on a test charge q per unit of that charge, providing q is so small not to perturb the charge distribution which is source of that force". 

But if the (initial) charge distribution is made only of a couple e+ e- and the test charge is close to them, the measurement's error will be high. Anyway it shouldn't be a big problem, if we only want to reveal detect this field.
Quote

Is there such a  (theoretical) thing as a measurement that is not made at the macro level?
I don't think so, at least at the current state of knowledge/technology; but again it doesn't seem to me a big problem: a measurement's apparatus can amplify a microscopic system like a single particle.
Quote
Might it help if there were more than two charges (not practically but theoretically)  and  can an electron field
What do you mean with "electron field"? I ask because it exists in QFT, but I'm not sure you are talking about this here, maybe you mean "electric field generated by the charges".
Quote
  be detected or measured by  sensors that are not themselves electrically charged ?
If with "not electrically charged" you mean that the system is overall neutral but it can have charges spatially separated (as in a capacitor or an electret or a simple dipole) then the answer is yes: the field can be detected and measured: your dipole will feel a torque if immersed in an external field which is not parallel to its axis.

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19
Physics, Astronomy & Cosmology / Re: Does a field require a source?
« on: 16/02/2018 08:31:13 »
Quote from: jeffreyH on 14/02/2018 23:11:30
Say we have a universe that contains only one electron and one positron. They collide and annihilate producing two gamma rays. Since the sources of the fields are gone how are the gamma rays propagating? Does the electromagnetic field, or any other field, require a source?
The couple of charges, let's assume for simplicity they can be considered initially stationary in an inertial frame (difficult to define in a universe where only those charges exists, however  ;)) generate an electrostatic field in all space, which have its precise energy. The two charges are sources of this field.

After the charges have annihilated,  that field disappears too (not immediately of course, the information travels at c).  Its energy will be conserved in the form of em radiation which adds to the 2 gammas (infact it could be included in the gammas itself).

So the actual field generated by the charges disappears with them.

Not immediately, however. Maybe the question should be referred to this field which disappears at speed c from the charges centre.

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lightarrow

20
Physics, Astronomy & Cosmology / Re: Can this spinning magnet generate electricty?
« on: 23/07/2016 12:35:52 »
Quote from: Atomic-S on 19/07/2016 07:07:47
Consider a ring (toroid) composed of magnetizable material, that is converted into a permanent magnet having the lines of flux running circularly within the material about the toroid's  axis. The lines of flux are everywhere at right angles to the axis,  circulating around it within the toroid.  Because they never enter or exit the toroid's surface, the toroid has no magnetic poles and no external field. 

Now imagine mounting this toroid on a nonmagnetic axle that passes through the center but is at right angles to the axis, so that the axle also meets the toroid itself at opposite places.   We shall assume that the axle is affixed to the toroid as by epoxy, but does not actually penetrate it, so that it has no effect on the toroid's internal magnetic field.   For mechanical reasons, it also extends beyond the toroid on each side so the thing can be mounted in bearings with a drive mechanism. Now imagine the toroid spinning upon this axle. 

Question:  What effect if any will be experienced by electrical conductors brought into the near vicinity of the spinning toroid?
For a magnetizable material I don't think you can say there is no external magnetic field outside the stationary toroid. Maybe you can make it negligible in some way, e.g. covering the toroid's surface with mu-metal?
https://en.wikipedia.org/wiki/Mu-metal
If it worked, would this be the same for your experiment or you actually don't want any other kind of device/material added?

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