Is the Speed of Gravity Equal to the Speed of Light?

[pervect]

I don't think it is just a matter of accelerating the charge. It has to be an acceleration that results from an electromagnetic interaction with the charge.


A good example is the EM interaction between an electron and a bending or jiggling magnet in a synchrotron. The electro-magnetic force on the moving electron that is provided by the bending or jiggling magnets in the rest frame of the laboratory would appear to the electron (ie in the moving electron's frame of reference) as a moving field with electric and magnetic components. It would interact with the electric component only. (This has to be the case because in the electron's 'stationary' reference frame, the electron has no magnetic field for the magnet to interact with).

When looking at the interaction between a charge and an external field, you always have the option to use whatever inertial frame of reference you want.

You can either use the intantaneous frame of the charge, or the laboratory frame, whichever is simpler.

In either case you will find that accelerating the charge generates electromagnetic radiation. In many circumstances the motion of the charge is known, but calculating the fields on it is not. The problem of "back reaction" of an accelerating point charges, for instance, is a well known source of infinities in classical EM theory. With quantum theory this issue goes away, as do many of the "point charge" issues, but it's better to avoid the complexities of quantum theory when the problem is basically classical, and this means that it's generally better to analyze the problem in terms of the motion of the charge.

 

[Andrew Mason]

In either case you will find that accelerating the charge generates electromagnetic radiation.

My point, and it may be somewhat heretical, is that it doesn't depend on acceleration at all. It depends on the (time-dependent) interaction of two fields. It is that interaction that causes electromagnetic radiation.

Now such interaction of 'force fields' necessarily involves forces being applied to charges. Since all charges have mass, it necessarily involves acceleration of the charges. But the acceleration of the charges is the result of the field interaction and electromagnetic radiation, not the cause of it. And there is a very simple proof of that.

AM

 

[yogi]

ok guys - it was tongue in cheek - i know all the practical problems of L(di/dt) and the photon drive which always requires a lot of power since your are exhausting a very small mass at a very hi velocity (ideally you would like to exhaust a near infinite mass at near zero velocity). But assume arguendo that even though the force is miniscule - does a finite velocity of field propagation produce a net force for a very long distance between the two magnets that are field coupled by a long mu-metal rod?

 

[Chronos]

ok guys - it was tongue in cheek - i know all the practical problems of L(di/dt) and the photon drive which always requires a lot of power since your are exhausting a very small mass at a very hi velocity (ideally you would like to exhaust a near infinite mass at near zero velocity). But assume arguendo that even though the force is miniscule - does a finite velocity of field propagation produce a net force for a very long distance between the two magnets that are field coupled by a long mu-metal rod?

Irrelevant. If you assume a coupling force, you also assume the burden of 'c' in exchanging forces [information] with the source. As the accelerating body, you will not notice this effect until your stationary counterpart retires. What you are missing is that simultaneity is an illusion under GR rules. Time is not invariant under GR. That is hard to accept. If it is any consolation.. the remote observer still thinks you are the one who is wrong about the time lapse.

 

[Andrew Mason]

ok guys - it was tongue in cheek - i know all the practical problems of L(di/dt) and the photon drive which always requires a lot of power since your are exhausting a very small mass at a very hi velocity (ideally you would like to exhaust a near infinite mass at near zero velocity). But assume arguendo that even though the force is miniscule - does a finite velocity of field propagation produce a net force for a very long distance between the two magnets that are field coupled by a long mu-metal rod?

Not exactly. You have to make the radiation directional. In your example, the EM radiation is omnidirectional, so there is no net force.

AM

 

[yogi]

AM - why is that - the magnetic lines are continuous (no divergence)

 

[Andrew Mason]

AM - why is that - the magnetic lines are continuous (no divergence)

I am not sure what you are asking. Are you asking why there is radiation or why it is in all directions?

It radiates because there is a time dependent magnetic field. The EM wave follows from Maxwell's equations. As far as the direction, why would there be a preferred direction? Since it is a dipole, it will not radiate equally in all directions necessarily, but it will be symmetrical in all directions.

AM

 

[yogi]

AM - well - I am not sure myself of where we are. Let me pose this question instead --as it has always bothered me - let us assume an electro magnet that we energize by turning on a switch. The current rises exponentially until it is limited by the coil resistance (we can limit the current to a fixed value at some point so the magnetic field will be constant thereafter). Once the local field max value is reached, what happens to the field that is extending outwardly in space presumably at the velocity of light - no more energy is being added to the field (although we may continue to dissipate energy in the resistor) so the magnitude of the local field will stablized - but what is happening to the continuous magnetic field lines - do they continue to grow until they reach the limit of the universe even though no more energy is being added to the system?

The difference between the propagation of a photon and a field lies in the fact that the energy is contained in the photon itself - once it passes a certain place it carries the energy and momentum with it - but in a continuous field - the entire space behind the field front remains intact - with a certain energy density. From whence then commeth the needed energy to propagate the magnetic field

 

[Andrew Mason]

Let me pose this question instead --as it has always bothered me - let us assume an electro magnet that we energize by turning on a switch. The current rises exponentially until it is limited by the coil resistance (we can limit the current to a fixed value at some point so the magnetic field will be constant thereafter).

I am not sure about that. The inductive reactance limits current flow. In a pure inductor (zero resistance), the current lags the applied voltage by $\pi/2$ (alternating current) so there is practically 0 energy ($V \cdot \vec I = VIcos\theta = 0)$ consumed in alternately setting up and collapsing a magnetic field. The only loss is to em radiation.

Once the local field max value is reached, what happens to the field that is extending outwardly in space presumably at the velocity of light - no more energy is being added to the field (although we may continue to dissipate energy in the resistor) so the magnitude of the local field will stablized - but what is happening to the continuous magnetic field lines - do they continue to grow until they reach the limit of the universe even though no more energy is being added to the system?

The expanding field is an electromagnetic wave. The energy of the wave front is supplied by the part that precedes it.

The difference between the propagation of a photon and a field lies in the fact that the energy is contained in the photon itself - once it passes a certain place it carries the energy and momentum with it - but in a continuous field - the entire space behind the field front remains intact - with a certain energy density.;/quote] I don't think there really is a difference between the propagation of an EM wave and propagation of a field. A time dependent magnetic field propagates as an EM wave. A photon corresponding to this increasing magnetic field has an enormous wavelength because the frequency approaches 0. So the energy of such a photon approaches 0. The energy in the magnetic field rapidly diminishes with distance, as well due to dipole geometry.

From whence then commeth the needed energy to propagate the magnetic field

The increasing magnetic field has to propagate as an em wave not as a magnetic field. The energy comes from the electrical energy in the original current.

These are interesting questions. There is a lot of depth to EM theory. But Maxwell's equations always seem to provide the answer.

AM

 

[yogi]

But once the field is established we have an effective permanent magnet - the field occupies a continually increasing volume if it makes its presence known at velocity c - radio waves and photons fall off inverse square because they are spread over the surface of an expanding sphere - but a field is an energy per unit volume affair - the radiation equations won't apply,

I don't see how it can spread as an em wave. When you interrupt the current that produces the field, the total energy w/i the entire volume occupied by the field collapses producing a voltage spike. If we had left the current on for one hour the extent of the field would be 60 x 3 x 10^8 meters in radius - so we would have to wait one hour to recover all the field energy? Obviously this over simplified model of magnetic interaction is wrong - but...

 

[Andrew Mason]

But once the field is established we have an effective permanent magnet - the field occupies a continually increasing volume if it makes its presence known at velocity c - radio waves and photons fall off inverse square because they are spread over the surface of an expanding sphere - but a field is an energy per unit volume affair - the radiation equations won't apply,

They will if that field changes with time. Since magnetic field lines all pass through the core, the energy of the entire field is contained within the core. Just as one can think of the energy of an electric field as being contained entirely within the field between charges (there is no energy contained in the field 'outside' the charge distribution) all magnetic energy of a static magnetic field is contained within the core of the magnet.

I don't see how it can spread as an em wave. When you interrupt the current that produces the field, the total energy w/i the entire volume occupied by the field collapses producing a voltage spike.

Not unless you have an infinitely large conducting loop. Only the flux contained within the area of the conducting loop induces the voltage spike (Faraday's law):

$$\oint_S E\cdot ds = - \frac{d\phi_B}{dt} = -\frac{\partial}{\partial t}\int_S B\cdot dA$$

The field outside the conducting loop doesn't produce any energy at all. It is all in the core. The field lines are closed so any field line extending to infinity passes through the core.

If we had left the current on for one hour the extent of the field would be 60 x 3 x 10^8 meters in radius - so we would have to wait one hour to recover all the field energy?

Actually 3600 x 3 x 10^8 m. But that is only if you have a current loop that big.

AM

 

[yogi]

Correct about the 3600 - I was thinking minutes instead of hours. But about the location of the field energy, seems we are on different pages in different books - the energy density of a magnetic field is inversly proportional to the 4th power of the distance from the pole, or for a single moving charge q having velocity v it is
[(q^2)(v^2) sin^2 theta]/[8(pi)(c^2)(r^4)] where theta is the angle between the velocty vector and the point where the energy density is being measured. Same is true of energy density for an electric charge and for the energy density of a gravitational field (always a inverse 4th power factor).

 

[Andrew Mason]

Correct about the 3600 - I was thinking minutes instead of hours. But about the location of the field energy, seems we are on different pages in different books - the energy density of a magnetic field is inversly proportional to the 4th power of the distance from the pole, or for a single moving charge q having velocity v it is
[(q^2)(v^2) sin^2 theta]/[8(pi)(c^2)(r^4)] where theta is the angle between the velocty vector and the point where the energy density is being measured. Same is true of energy density for an electric charge and for the energy density of a gravitational field (always a inverse 4th power factor).

There is a lot of confusion created by textbook writers regarding energy density of a magnetic field. (Just in case anyone thinks we are completely off the original topic of this thread, there are also potentially important analogies to gravitation and the possible speed of propagation of gravity).

Feynman talks about the ambiguous location of the field energy in II-27 of his Lectures. See esp. 27-4 'The Ambiguity of the Field Energy'. Quote: "It is interesting that there seems to be no unique way to resolve the indefiniteness in the location of the field energy".

The energy of the magnetic field of a magnet or current loop can be thought of in various ways: ie. in terms of the energy of another magnet or moving charge placed in that field, or the self energy of the field (the net electrical energy spent in creating the field).

In the case of another magnet, it is the work done by the electrical current in that magnet against the magnetic force from the first that creates the energy in the field (e.g current in a motor armature coil placed between the field magnets) . So it is not the field of the first that provides the energy. In the case of a moving charge or conductor, the energy is entirely derived from the kinetic energy of the charge or conductor, not the field (eg a generator armature coil moving through the magnetic field of the field magnets). It is only the self energy of the magnetic field that has any real bearing on the energy of the magnetic field.

Now, the self energy of the magnetic field of electromagnet A (emA), say - a function of the energy used to set up the field of emA- can be (almost) completely transferred to another coil if that coil encloses the core of emA.

If the other coil is some distance away from the core and encloses only a certain proportion x of the total field lines from emA, only that proportion of the self energy of the field will be transferred. Does that mean that the energy was in the area of the field enclosed by the remote coil? That may be one way of looking at it. But accessing that energy depends upon the current in the coil of emA changing with time. The energy cannot be obtained any other way. So does it make sense to talk about that region of space having magnetic energy? Or should we talk about the flow of energy from the coil of emA to the remote coil which occurs when the current in emA changes. In the latter case, there is only the energy of the core and the transfer of that energy into the surrounding space when the energy changes.

The case of the energy of an electric charge is similar. There is an energy between two charges or as a result of a charge distribution. Electrical energy can only be defined in terms of two charges and their separation. (The electrical potential energy of a charge Q at point x is defined as the path integral of the electrical force on a unit charge moving from infinity to point x.) Charge is always conserved so it can never be turned on or off. So it is meaningless to talk about the space surrounding a single charge as having a 'self-energy' density. The energy is in the field between charges. The energy does not extend to an infinite volume of space around the charge in the absence of another charge in that space.

AM

 

[yogi]

I would argue conversely that all energy is in the field - take a charged sphere and integrate over the surface outward to infinity - you get a value that corresponds to the energy required to charge the sphere. Also - if you look at some other places in Feynman's lectures you will see that he makes the case for the proposition that the energy required in charging a parallel plate capacitor does not come from the circuit current flowing into the plates , but from field energy flux converging inwardly to take up the volume between the plates.

 

[Andrew Mason]

I would argue conversely that all energy is in the field - take a charged sphere and integrate over the surface outward to infinity - you get a value that corresponds to the energy required to charge the sphere. Also - if you look at some other places in Feynman's lectures you will see that he makes the case for the proposition that the energy required in charging a parallel plate capacitor does not come from the circuit current flowing into the plates , but from field energy flux converging inwardly to take up the volume between the plates.

There are various ways of looking at electromagnetic phenomena. The fact that they may provide the same mathematical result doesn't mean that they all represent physical reality. The concept of 'lines of force' for example provide a useful geometrical model for describing and predicting the behaviour of magnetic fields. But it doesn't mean that such lines exist.

Since mass and energy are equivalent, and if, as you suggest, the energy of the field of an electric charge or electromagnet pervades all of space, then the mass associated with that energy must pervade all of space. If that is the case, the field has 'rest mass' and an associated gravitational field. Perhaps someone can think of an experiment to determine if such rest mass or gravitational field actually exists.

AM

 

[ZapperZ]

There are various ways of looking at electromagnetic phenomena. The fact that they may provide the same mathematical result doesn't mean that they all represent physical reality. The concept of 'lines of force' for example provide a useful geometrical model for describing and predicting the behaviour of magnetic fields. But it doesn't mean that such lines exist.

Since mass and energy are equivalent, and if, as you suggest, the energy of the field of an electric charge or electromagnet pervades all of space, then the mass associated with that energy must pervade all of space. If that is the case, the field has 'rest mass' and an associated gravitational field. Perhaps someone can think of an experiment to determine if such rest mass or gravitational field actually exists.

AM

Not necessarily!

What if I can decouple the rate at which charge carriers move and the rate of mass flow? Then I can prove that an E&M field need not be always tied to any "mass" (rest or effective, or otherwise). In other words, what if I can violate the Wiedemann-Franz law?

Well, I can! There have been at least 2 recent experimental results showing clear signatures of mass-charge (and spin) separation.[1,2] These unambiguously show the coupling of mass from charge. One can only conclude that EM fields need not contain in it any need for any kind of "mass".

Zz.

1. R.W. Hill et al., Nature v.414, p.711 (2001).
2. T. Lorenz et al., Nature v.418, p.614 (2002).

 

[Andrew Mason]

Well, I can! There have been at least 2 recent experimental results showing clear signatures of mass-charge (and spin) separation.[1,2] These unambiguously show the coupling of mass from charge. One can only conclude that EM fields need not contain in it any need for any kind of "mass".

Does that not mean that the field itself does not contain self-energy?

See also "Does an electric charge curve space time?". https://www.physicsforums.com/showthread.php?t=43998

AM

 

[ZapperZ]

Does that not mean that the field itself does not contain self-energy?

See also "Does an electric charge curve space time?". https://www.physicsforums.com/showthread.php?t=43998

AM

Self-energy isn't necessarily "mass".

My point here is simply to show that you CAN decouple EM interactions from gravitational interactions. So EM fields need not carry any "mass" or require the presence of one, which is what you claimed.

Zz.

 

[Andrew Mason]

Self-energy isn't necessarily "mass".

My point here is simply to show that you CAN decouple EM interactions from gravitational interactions. So EM fields need not carry any "mass" or require the presence of one, which is what you claimed.

Well I didn't exactly claim that they did. I said they would have to have mass if 'self energy' was contained in the field. But I said that the field itself contains no self energy.

The 'self energy' is simply a tool that helps us to apply electrodynamics. Like lines of force. It doesn't equate to physical reality. Your point about the 'decoupling' of mass from EM fields shows that this must be correct. Unless, of course, one invents a new concept of energy that does not obey E=mc². I prefer not to go that route.

AM

 

[Andrew Mason]

I would argue conversely that all energy is in the field - take a charged sphere and integrate over the surface outward to infinity - you get a value that corresponds to the energy required to charge the sphere.

The question is: what are you integrating? You are summing the work done by a charge dq = $\sigma dA$ in moving from the surface of the sphere to infinity. I agree that when you disassemble the sphere that way that is the energy that is released. Why does that have to be contained in the field outside the spherebefore it is disassembled like that? Why is that energy not simply a part of the 'mass' of the sphere?

I suggest that there should be an experimental way of determining whether the charged sphere has that additional mass. If it does, the energy resides in the field within the charge distribution (ie. between the charges on the surface of the sphere) not in the field that extends to infinity.[/QUOTE]

AM

 

[ZapperZ]

Well I didn't exactly claim that they did. I said they would have to have mass if 'self energy' was contained in the field. But I said that the field itself contains no self energy.

The 'self energy' is simply a tool that helps us to apply electrodynamics. Like lines of force. It doesn't equate to physical reality. Your point about the 'decoupling' of mass from EM fields shows that this must be correct. Unless, of course, one invents a new concept of energy that does not obey E=mc². I prefer not to go that route.

AM

You said

Since mass and energy are equivalent, and if, as you suggest, the energy of the field of an electric charge or electromagnet pervades all of space, then the mass associated with that energy must pervade all of space. If that is the case, the field has 'rest mass' and an associated gravitational field. Perhaps someone can think of an experiment to determine if such rest mass or gravitational field actually exists.

The decoupling of charge and mass means that there isn't necessary any "mass" associated with "that energy". I could have just EM field without having to talk about ANY mass.

You also need to be careful when you talk about "self-energy". The self-energy out of classical E&M differs in nature from the self-energy within QFT. While the self-energy in classical E&M tends to be "subtracted" out when one tries to find the energy density in a unit volume, for example, the self-energy in QFT arises out of higher-order perturbation expansion. The single-particle Green's function, for example, has the real and imaginary part of the self energy that contain within them all of the higher-order interactions. You cannot just simply assign these to be the "mass" of the field.

Zz.

 

[yogi]

AM - you can also look at it is the energy required to bring the charges to the surface of the sphere -

Most of these types of discussions get murky because we tend to think of charges and masses as discrete chunks of something - but the more we probe the more we find only space and relationships between things - there never seems to be any basic entity that we can describe as a particle in the sense we are familiar with - in the last analysis. are we not dealing with some form of spatial stress at every level - some distortion of space and time - i.e., a field.

 

[Andrew Mason]

Most of these types of discussions get murky because we tend to think of charges and masses as discrete chunks of something

Their appearance as chunks of something may be murky, but their discreteness is not. Especially charge. Charge is quite remarkable because it is universally invariant - in classical electrodynamics, relativity, and quantum mechanics. No other quantity seems to have this property.

are we not dealing with some form of spatial stress at every level - some distortion of space and time - i.e., a field.

Ok. And if matter can be represented by a field such field must have energy (eg. the nuclear 'field' between quarks in a proton or neutron;the em field between electrons and protons;). But not every field is an energy field. I don't see the basis for saying a bare electric field/magnetic field/gravitational field represents self energy. It represents energy only if there is another charge/current/mass in the field.

AM

 

[yogi]

When we consider a vector field, we assign a magnitude and direction to each point. The field exists whether or not a test particle is inserted. To find the force, we apply (usually multiply) the local value of the field by the value of the test charge - I would agree that this does not prove the existence of an energy density at the point - but there is a gradient in all divergent or convergent fields, and when the test charge is moved along the gradient, there is a change in the potential -

Seems we have drifted a long way from the question as to the velocity of the G field.

Regarding the electron charge and its discreteness - I would agree that all electrons are alike and that "e" is a temporally invarient constant of the universe

I might add, if you are not already aware of the fact, the electron charge can be used to derive a set of units analogous to Planck units - but they will have different values. In other words - you use e instead of h as one of the fundamental constants and you get different values for the so called fundamental units of mass, time and length. Stoney did this prior to Planck.

 

[Andrew Mason]

When we consider a vector field, we assign a magnitude and direction to each point. The field exists whether or not a test particle is inserted. To find the force, we apply (usually multiply) the local value of the field by the value of the test charge - I would agree that this does not prove the existence of an energy density at the point - but there is a gradient in all divergent or convergent fields, and when the test charge is moved along the gradient, there is a change in the potential -

No question. My point, and really my only point, is that the field does not represent energy UNLESS there is another charge placed in it.

Seems we have drifted a long way from the question as to the velocity of the G field.

I'm not sure about that. In order to understand a gravitational field propagating in space one has to understand the nature of propagation of any field. If the field does not represent energy in itself, 'nothing' is propagated simply by changing position.

Nothing should prohibit the field from 'progagating' faster than c in this one limited sense: if the source of the field (charge, current or mass) is moving at uniform speed relative to another frame of reference, the field at a distance d from the source changes instantaneously in step with the movement of the source.

This is unremarkable in itself until the source slows its motion. Then it is a question whether within a time t<d/c the field at d is the potential that would have existed if the source had not slowed. If it is not (and I suggest it is not) the question is "how can this occur without violating the principle that c cannot be exceeded".

The concept of the 'retarded potential' is used (the potential at d is really the potential due to the position of the charge at time t = d/c earlier) and it is said that the field from the slowed source simply propagates at speed c after it slows. It is said that gravity has to obey this same principle.

I say, there is no need to assume that the electromagnetic field or a gravitational field propagates at all due to the slowing of the source. The whole thing can be explained by relativity. One has to abandon the view that EM radiation results from the charge accelerating.

AM

 

[yogi]

AM - But even though the field does not represent energy per se until you place a second charge - it does exist as some sort of potential -

I would like to ponder your point further - but let me see if I understand it - are you saying that a moving charge or mass conveys its present position instantly to all parts of the universe - but It is not a propagation of energy - but it is a force field that is revealed only when a test particle is inserted. But what if the test particle(s) is/are already present - e.g., all the other masses in the universe for example. Are they acted upon instantly vis a vis the present location of the moving source? Also, even though the field is not energy, does it not convey information in contravention of SR's prohibition against FTL signaling?

 

[Andrew Mason]

I would like to ponder your point further - but let me see if I understand it - are you saying that a moving charge or mass conveys its present position instantly to all parts of the universe - but It is not a propagation of energy - but it is a force field that is revealed only when a test particle is inserted. But what if the test particle(s) is/are already present - e.g., all the other masses in the universe for example. Are they acted upon instantly vis a vis the present location of the moving source?

Yes, but the key is the meaning of the term 'instantly'. Since the situation of a moving charge and fixed observer is exactly equivalent to a rest charge and a moving observer, the moving observer will observe the field to correspond to the position of the charge that the observer measures at the same instant that he measures the field.

Also, even though the field is not energy, does it not convey information information in contravention of SR's prohibition against FTL signaling?

No. It only conveys information when the field changes.

AM

 

[yogi]

AM - Doesn't it convey information instantly when the source of the field changes by moving relative to the rest of the universe?

 

[Andrew Mason]

AM - Doesn't it convey information instantly when the source of the field changes by moving relative to the rest of the universe?

The answer must be 'no', of course. But just how to analyse this is difficult. Relativistic effects must be taken into account. There appears to be more than one 'correct' approach (ie. the use of advanced, retarded or classical instantaneous potentials seems to produce the same solution).

Consider an observer O at the origin in his frame of reference and moving at speed v relative to charge Q. O is measuring electric potential from Q continuously (using a test charge q<<Q). At time=0, O measures the position of Q to be d. I suggest that he measures the potential to be kQ/d.

Then, at time=t in his frame, O measures the potential of Q. But at a slightly earlier time = t1<t where t-t1<d/c as measured in O's frame, Q experienced a sudden acceleration in the direction of O's motion and never makes it as close to O as d-vt. What potential does Q measure at time=t? Is it E=kQ/(d-vt)? I think the answer is: 'yes'.

Feynman spent a great deal of time and effort on this kind of electrodynamic question, as did John Wheeler. I get the sense from reading Feynman that the 'correct answer' and physical explanation was still a matter of debate (at least it seems it was 40 years ago). Perhaps someone will be able to provide us with a more up to date perspective. My physics on this is 30 years out of date, I am afraid.

AM

 

[yogi]

AM - Good to find another out of date physicists - I also am of Feynman-Wheeler vintage - I particularly like their approach to problems - they always sought a physical analogy rather than abstraction. As far as the issue of the speed at which fields make their presence known, I don't think the experiments are conclusive one way or the other. Nor do I think it is good to blindly accept Einstien's prejudice re the ultimate velocity at which information might be conveyed, although I would still regard Einstein as the greatest contributor to Science since Newton. My first wife could talk so fast I am sure she must have violated at least some prohibition against FTL communication.

Regards

Yogi

 

[Andrew Mason]

AM - Good to find another out of date physicists - I also am of Feynman-Wheeler vintage - I particularly like their approach to problems - they always sought a physical analogy rather than abstraction. As far as the issue of the speed at which fields make their presence known, I don't think the experiments are conclusive one way or the other.

The fact may be that one can get experimentally equivalent results for the field of a charge using theories based on retarded, advanced or "half retrarded/half advanced" (I think this means the instantaneous values) potentials, which seems to be what Feynman thought.

I have the highest regard for Feynman (who doesn't?). For me, his ability to take a different approach to something - which was always equivalent to the way others looked at it - was what distinguished him from most others. His fascination with the principle of 'least action' is a good example. It underlies his novel approach to quantum theory, QED (involving the sum over probabilities and decoherence). But it also is another way of looking at Newtonian mechanics and general relativity. I find that I rarely fully understand Feynman and I probably misunderstand a lot of what he says. But he is sure interesting to read, if only for the occasional glimmer.

Nor do I think it is good to blindly accept Einstien's prejudice re the ultimate velocity at which information might be conveyed, although I would still regard Einstein as the greatest contributor to Science since Newton.

I don't think it is fair to say that Einstein had a prejudice re: c as the ultimate velocity. His belief was based on evidence. Prejudice usually refers to a conclusion one reaches without facts. If one agrees that the speed of light is independent of its source (which is based on evidence), there is no other conclusion that one can reach. So saying that Einstein had a prejudice re: the ultimate velocity (c) is equivalent to saying that he had a prejudice that the speed of light is independent of the motion of its source.

AM

 

[yogi]

AM - I would say that very few doubt that the speed of light is independent of the source - that is always the situation with wave phenomena - but Einstein took it further by asserting that the receiver (the observer in motion) would always measure light to have a velocity c as well - that was/is the bold step - and it was an assertion that was not required by the experiments - MMx had provided good evidence that the over and back velocity would be measured as c - but there are no experiments that have conclusively proved that the round trip velocity will be measured as c - it is taken as constant for one way measurements - in the derivation of the transforms - but it is not even to this day confired by experiment - although GPS provides some pretty solid data that it is - at least in the Earth centered reference frame.