What is your EM crankosity?
in [Relativity]
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From: Jim Black on 27 Sep 2009 18:12 On Wed, 23 Sep 2009 14:32:54 -0700 (PDT), Benj wrote: > Let's now jump to #9 For Bill. Relativity implies that clocks slow as > they approach the speed of light. So the question comes: "do > electromagnetic clocks in moving reference frames slow because of > retardation without even considering relativity?" The calculations > have been done by Oleg Jefimenko in his book "Electromagnetic > Retardation and the Theory of Relativity" in Chapter 10 "the rate of > moving clocks". What he finds after calculating the slowing rate for a > number of electromagnetic oscillators (clocks) is that some indeed do > slow at the Einsteinian rate. Some do not. Some do but the amount of > slowing depends upon orientation. He notes that his calculations show > that electromagnetic clock slowing is not a relativistic effect at > all, but is a dynamic effect where slowing in general may not be > proportional to gamma. His conclusion therefore is that there is no > such thing as "time dilation". And Jefimenko notes that while > calculation of the electromagentic "clocks" give implications, we are > not provided with ANY information as to how one would calculate the > slowing of biological clocks as say found in the "Twin Paradox". But > relativity concerns aside, the main point here (as proved by the > calculations of the various oscillators) is that various > electromagnetic "clocks" do NOT all slow by the same amount when > viewed in a moving frame. OK? I'm not aware of any purely electromagnetic clocks in the strictest sense; they all have some mechanical parts, such as wires and the electrons in them. If you model those with Newtonian mechanics, of course it's possible to get an answer different from special relativity. So I would say that "leading to the conclusion" is the only wrong part of #9. Relativistic time dilation affects not just electromagnetic clocks, but any self-contained clock that does not rely on outside parts which may not be set in motion along with the clock. Examples of clocks failing the "no outside parts" condition are pendulum clocks and WWV receivers. On the other questions: For #5, Susan's answer is best; the statement is only approximate; there is a small error term. Statement #4, on the other hand, is only valid as an approximation if E is not too much larger or smaller than cB. Otherwise the B field from the transformed E field can swamp the original B field, or vice versa. #7 is too sloppily posed for me to call true or false, but I will say that the actual length of the line segment in the observer's frame is different from the actual length of the line segment in its own frame. Unless the motion is perpendicular to the line segment, of course. I'm interested in seeing your reasons for calling it false; then there might be something to discuss. #8 is wrong because of "cosmic rays." I agree with your reasons for the others being wrong. #10 is also wrong on the uniqueness claim. -- Jim E. Black (domain in headers) How to filter out stupid arguments in 40tude Dialog: !markread,ignore From "Name" +"<email address>" [X] Watch/Ignore works on subthreads
From: Jonah Thomas on 27 Sep 2009 18:31 Jim Black <fmlast3(a)organization.edu> wrote: > Benj wrote: > > > Where the problem comes in, is I said that it is a CHANGING MAGNETIC > > FIELD that creates this induced E field. This is flat out wrong. > > Faraday's law may be a true relationship, but it's NOT a causal > > relationship. The actual true mechanism is that a changing current > > (charges) creates BOTH a magnetic field AND the induced electric > > field that can produce the induced current in wires etc. Both the > > magnetic field AND the electric fields travel away from the changing > > source current at the speed of light and hence are simultaneous and > > therefore cannot "cause" one another. > > I agree that the changing magnetic field does not cause the "induced" > electric field. But you can interpret Faraday's Law as a causal > relationship, but it has to be the other way around. That is, an E > with nonzero curl causes a changing B field. And a B field with a > curl that does not equal mu0 J causes a changing E field. Here's how I like to think of it. I can't guarantee that this approach will be useful to anyone else. Ah, notation. I want to just type, and so I will use the notation I use for myself. When it doesn't cause confusion, I will use t to mean the partial derivative wrt time, and D to mean the del operator, the vector of partial derivatives wrt distance. p to mean the scalar term of the 4-potential, and A to mean the vector term of the 4-potential. Then what happens if we take all the partial derivatives on the 4-potential? (t,D)(p,A)=( tp-D.A , Dp + tA + DxA ) So Dp+tA is precisely the electric field, and DxA is the magnetic field, and tp-D.A is a gauge. Our 12 partial derivatives for the vector term amount to the sum of the electric and magnetic fields. If you change to another inertial frame then you get the same result but the contribution of electric and magnetic terms is different. it's the same field, just viewed differently, as partial derivatives on different variables. When you look at it this way, electric and magnetic fields aren't causing each other, they're just descriptions of different parts of the total field, and when you change the way you describe one of them then you need to change the other one to match or you'll be describing some other field. Maxwell's equations then turn into a set of constraints on how the 4-potential can change. You can do anything you want provided it fits into the constraints (and provided you can figure out how to make it happen). I certainly don't want to say that the way you look at it is wrong, and I don't claim the approach I sketched out will be useful to you. I hope it may be of some use.
From: Jim Black on 27 Sep 2009 20:39 On Sun, 27 Sep 2009 18:31:18 -0400, Jonah Thomas wrote: > Here's how I like to think of it. I can't guarantee that this approach > will be useful to anyone else. > > Ah, notation. I want to just type, and so I will use the notation I use > for myself. When it doesn't cause confusion, I will use > t to mean the partial derivative wrt time, and > D to mean the del operator, the vector of partial derivatives wrt > distance. > p to mean the scalar term of the 4-potential, and > A to mean the vector term of the 4-potential. > > Then what happens if we take all the partial derivatives on the > 4-potential? > > (t,D)(p,A)=( tp-D.A , Dp + tA + DxA ) So that others can follow, it should be stated that you're using quaternions here, with (t,D) representing the quaternion with scalar part t and vector part D. If I'm guessing your meaning right. > So Dp+tA is precisely the electric field, The electric field is -Dp-tA. > and DxA is the magnetic field, > and tp-D.A is a gauge. I don't understand what you mean by this statement. A gauge is a particular choice of p and A given E and B. How does this relate? > Our 12 partial derivatives for the vector term > amount to the sum of the electric and magnetic fields. I get a difference. > If you change to another inertial frame then you get the same result That the equation is valid in any frame is true but not very useful, especially since I don't know what the significance of this expression is. > but > the contribution of electric and magnetic terms is different. it's the > same field, just viewed differently, as partial derivatives on different > variables. That E and B are components of the same field is a fairly standard idea, but I don't see how it or anything afterwards follows from the above. Maybe I'll try to work through the chain of events in the LC circuit using the potentials instead of the fields, though. That could be interesting. -- Jim E. Black (domain in headers) How to filter out stupid arguments in 40tude Dialog: !markread,ignore From "Name" +"<email address>" [X] Watch/Ignore works on subthreads
From: RichD on 27 Sep 2009 21:00 On Sep 25, Benj <bjac...(a)iwaynet.net> wrote: > > > Benj knows quite a bit about science and more than you, > > > going by the evidence so far. > > > Star Trek physics? > > Shut up or I'll stuff tribbles in your mailbox. Lines we never heard: "Bring the ship to a halt, Mr. Chekov." "Relative to what, Kapitan?" -- Rich
From: Jonah Thomas on 27 Sep 2009 21:52
Jim Black <fmlast3(a)organization.edu> wrote: > Jonah Thomas wrote: > > > Here's how I like to think of it. I can't guarantee that this > > approach will be useful to anyone else. > > > > Ah, notation. I want to just type, and so I will use the notation I > > use for myself. When it doesn't cause confusion, I will use > > t to mean the partial derivative wrt time, and > > D to mean the del operator, the vector of partial derivatives > > wrt > > distance. > > p to mean the scalar term of the 4-potential, and > > A to mean the vector term of the 4-potential. > > > > Then what happens if we take all the partial derivatives on the > > 4-potential? > > > > (t,D)(p,A)=( tp-D.A , Dp + tA + DxA ) > > So that others can follow, it should be stated that you're using > quaternions here, with (t,D) representing the quaternion with scalar > part t and vector part D. If I'm guessing your meaning right. Yes, of course. > > So Dp+tA is precisely the electric field, > > The electric field is -Dp-tA. Ooops. I like to use the negative of the electric field because it makes my notation easier. OK, (t,-D)(p,-A)=( tp-D.A , -Dp - tA + DxA ) There. > > and DxA is the magnetic field, > > and tp-D.A is a gauge. > > I don't understand what you mean by this statement. A gauge is a > particular choice of p and A given E and B. How does this relate? Sorry, it's one possible gauge choice, you set it to zero. It's like the lorenz gauge with a sign change. For each gauge you can go through some complications in doing (t,D)(p,A) to get the gauge along with the proper vector part. Like, you could get the lorenz gauge with (t,-D)(p,0) - (0,A)(t,D) = (tp+D.A , -Dp -tA + DxA) This other gauge is the one that's simplest to get. > > Our 12 partial derivatives for the vector term > > amount to the sum of the electric and magnetic fields. > > I get a difference. > > > If you change to another inertial frame then you get the same result > > That the equation is valid in any frame is true but not very useful, > especially since I don't know what the significance of this expression > is. > > > but > > the contribution of electric and magnetic terms is different. it's > > the same field, just viewed differently, as partial derivatives on > > different variables. > > That E and B are components of the same field is a fairly standard > idea, but I don't see how it or anything afterwards follows from the > above. > > Maybe I'll try to work through the chain of events in the LC circuit > using the potentials instead of the fields, though. That could be > interesting. I hope it gives you something interesting. I'm not in a position to make any guarantees. |