Author Topic: Woodward's effect  (Read 803002 times)

Offline JohnFornaro

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Re: Woodward's effect
« Reply #80 on: 02/10/2013 02:01 pm »
Again, it seems more like Woodward is trying to fit the universe to do what he wants it to do instead of trying to figure out how it /actually/ works and only then exploiting it.

Again, I applaud your amazing capability of reading people's minds.
It goes along with the non-locality of the Woodward Effect. ;)

Hardy har har.   I gotta say, that's pretty good!
Sometimes I just flat out don't get it.

Offline Patchouli

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Re: Woodward's effect
« Reply #81 on: 02/10/2013 02:13 pm »
General Relativity isn't compatible with the Mach principle, despite what Woodward claims. GR is intrinsically /local/. The existence of gravity waves actually supports this.

General Relativity also is not compatible with most theories of quantum physics nor can it describe what happens inside a black hole.
When you try the math things break down pretty badly.

« Last Edit: 02/10/2013 02:15 pm by Patchouli »

Offline Robotbeat

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Re: Woodward's effect
« Reply #82 on: 02/10/2013 02:55 pm »
Nothing can really explain what goes on inside a black hole. But we are shielded by the event horizon, which also makes experiment with it not possible.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Offline JohnFornaro

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Re: Woodward's effect
« Reply #83 on: 02/10/2013 03:40 pm »
Not sure whether this has been posted here yet, but Heidi Fearn's presentation is up:
http://physics.fullerton.edu/~jimw/ASPW2012.pdf

[finger raised.]

On P. 6:  phi=GM/R.

So uhhhh, what's the radius of the universe?  Since it is expanding, what is the force constant that the following equations seem to be demonstrating?

IOW, is the M-E a constant, or is it getting larger as time goes on?

You all skip so much of the math.  It would be nice to start at the beginning.
Sometimes I just flat out don't get it.

Offline grondilu

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Re: Woodward's effect
« Reply #84 on: 02/10/2013 04:02 pm »
On P. 6:  phi=GM/R.

So uhhhh, what's the radius of the universe?  Since it is expanding, what is the force constant that the following equations seem to be demonstrating?

This joins my initial question in this thread.  It really depends on your cosmological model, and the necessary simplifications you put in it.

As mentioned a bit later, I'm not sure it matters much.  What matters is that you can get the speed out of the integral and thus find out the static field phi, whatever its expression on the radius and mass of the universe actually is.

In his paper, Sciama doesn't even bother expressing phi in terms of M and R, anyway.   He writes A = phi/c v pretty much directly.
« Last Edit: 02/10/2013 04:52 pm by grondilu »

Offline Robotbeat

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Re: Woodward's effect
« Reply #85 on: 02/10/2013 05:36 pm »
Not sure whether this has been posted here yet, but Heidi Fearn's presentation is up:
http://physics.fullerton.edu/~jimw/ASPW2012.pdf
Uses Comic Sans. Blasphemy!
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Offline cuddihy

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Re: Woodward's effect
« Reply #86 on: 02/10/2013 06:19 pm »
Not sure whether this has been posted here yet, but Heidi Fearn's presentation is up:
http://physics.fullerton.edu/~jimw/ASPW2012.pdf

[finger raised.]

On P. 6:  phi=GM/R.

So uhhhh, what's the radius of the universe?  Since it is expanding, what is the force constant that the following equations seem to be demonstrating?

IOW, is the M-E a constant, or is it getting larger as time goes on?

You all skip so much of the math.  It would be nice to start at the beginning.

R= radius of the observable universe. "Observable" means that, since all forces and information travel at the speed of light max, the maximum observable radius is (age of universe)*(speed of light).

Yes, it changes over time just as in an expanding universe G changes and has changed over time, albeit very slowly.

Offline grondilu

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Re: Woodward's effect
« Reply #87 on: 02/10/2013 06:27 pm »

In its 2004 paper, "Tests of Mach’s Principle with a Mechanical Oscillator", John Cramer makes an interesting point about Woodward being probably wrong about his method for testing his effect.

« Woodward and his students [3,4] have attempted
to observe the predicted effect by producing an
unbalanced force (say, to the right) that is expected to
arise when the inertia varying test mass is accelerated
to the right when it has low inertia and to the left when
it has high inertia. They report having observed small
unbalanced forces, near the limits of their sensitivity,
which are about five orders of magnitude smaller than
the predicted effect.

Unfortunately, this scheme for observing the
inertia variation appears to be at odds with the
relativistically invariant form of Newton's 2nd law of
motion:

F = dp/dt = m dv/dt + v dm/dt

Since the inertial mass m of the test body is
expected to vary with time, the last term of equation (1)
cannot be ignored. It is not surprising, in view of
Newton's 3rd law of motion, that for any sinusoidal
variations of the mass around a central value, the force
contribution from the v dm/dt term is found to
precisely cancel the supposed "unbalanced force"
arising from the m dv/dt term, leading to a time-
averaged net force of zero on the overall system.
From this simple calculation, it appears that
unbalanced force searches are not good tests of the
proposed effect. »


I wonder if Woodward has taken this into account since then.

Offline cuddihy

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Re: Woodward's effect
« Reply #88 on: 02/10/2013 06:28 pm »
I don't understand your focus on the weight or apparent weight of the active mass.  Nothing terribly interesting happening there, the amount of mass actually experiencing the effect is very small, and it's a transient. It's the apparent change in inertia that really matters, because with the push that is what provides the useful force.

You say you don't understand the focus on weight and then you say that what matters is  change of inertia.  Well, according to general relativity, isn't there an exact correspondence between gravitational mass (aka. weight) and inertial mass (the tendency to resist to an external force)?

Just because there's a correspondence doesn't mean it's the same thing. In fact it's a key point for Mach Principle that whereas gravitational mass is a local effect that is only observed in the immediate vicinity of a massive object, inertial mass is an explicitly non-local phenomenon that local objects have a negligible effect on, because compared to the mass of the rest of the universe the mass of a massive object is so small. I won't have time to dig the discussion out today (kid bday party), but it was explicitly calculated by someone in Woodward's circle, it's buried in one of those linked papers on UCF's site.

Offline grondilu

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Re: Woodward's effect
« Reply #89 on: 02/10/2013 07:16 pm »
Just because there's a correspondence doesn't mean it's the same thing.

I've always thought that's precisely what Einstein meant, though.

Edit.  Also, I wrote "correspondence" but the correct term is "equivalence", which is less ambiguous.

Quote
In fact it's a key point for Mach Principle that whereas gravitational mass is a local effect that is only observed in the immediate vicinity of a massive object, inertial mass is an explicitly non-local phenomenon that local objects have a negligible effect on, because compared to the mass of the rest of the universe the mass of a massive object is so small.

Something tells me there is something wrong in this reasoning.  I think it's because you consider mass (either gravitational or inertial) as being an effect of some sort.  I think Mach was talking about inertial forces resulting from the action of distant stars, not the inertial mass.

Mass, either gravitational or inertial, is a form of energy.  E=mc2  does not come in two flavors.
« Last Edit: 02/10/2013 09:22 pm by grondilu »

Offline cuddihy

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Re: Woodward's effect
« Reply #90 on: 02/11/2013 12:48 am »
Just because there's a correspondence doesn't mean it's the same thing.

I've always thought that's precisely what Einstein meant, though.

Edit.  Also, I wrote "correspondence" but the correct term is "equivalence", which is less ambiguous.

Quote
In fact it's a key point for Mach Principle that whereas gravitational mass is a local effect that is only observed in the immediate vicinity of a massive object, inertial mass is an explicitly non-local phenomenon that local objects have a negligible effect on, because compared to the mass of the rest of the universe the mass of a massive object is so small.

Something tells me there is something wrong in this reasoning.  I think it's because you consider mass (either gravitational or inertial) as being an effect of some sort.  I think Mach was talking about inertial forces resulting from the action of distant stars, not the inertial mass.

Mass, either gravitational or inertial, is a form of energy.  E=mc2  does not come in two flavors.


As long as we're being precise, relativity means that you can't tell the difference between the two via normal means, not that they're the same. If the Woodward Effect works, we've got the exception that proves the distinction.

You're right, I was referring to the "observed action on" each, not the mass itself. Mass is mass. You can tell apart weight from inertial mass only in the case that observed inertial mass is variant...which is what Woodward suggests.

Woodward posits a transient reduction in the observed inertia. Transients related to gravitation would be insignificant due to, again, the scale of inertial back-reaction forces vs. the small scale of local gravitation.
« Last Edit: 02/11/2013 12:50 am by cuddihy »

Offline grondilu

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Re: Woodward's effect
« Reply #91 on: 02/11/2013 01:10 am »
You can tell apart weight from inertial mass only in the case that observed inertial mass is variant...which is what Woodward suggests.

Woodwards predicts a variation of inertial mass, and thus weight.   Really I don't think it is possible to distinct inertia from gravity, as I understand it they really are the same thing according to the equivalence principle.

Gravity is the inertial force you feel when the ground prevents you from following an inertial reference frame.

I may be wrong though, as I'm really no expert in GR.  But long time ago that was one thing I remembered from a vulgarization text about it.
« Last Edit: 02/11/2013 01:14 am by grondilu »

Offline JohnFornaro

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Re: Woodward's effect
« Reply #92 on: 02/11/2013 01:24 am »
Quote from: cuddihy
Just because there's a correspondence doesn't mean it's the same thing.

Quote from: grondilu
I've always thought that's precisely what Einstein meant, though.

Edit.  Also, I wrote "correspondence" but the correct term is "equivalence", which is less ambiguous.

Which is fine and good, but answers not the math of Sciama at all. 

I recall the correct term being "equivalence" as well.  Even so, ten dollars equals, say, fifteen francs.  (what do I know about money)  In this example, the two sums are said to be "equivalent", even tho they are denominated as different things.

There is some difference between mass and inertia, but in terms of "dollars and francs", they have the same value.  Sciama's theory purports to describe the "dollar" and the "franc", that is, mass and inertia.

Sciama, the dear reader will remember, suggested in 1953 that "The principle of equivalence is a consequence of [his inertial theory, not an initial axiom."  This is a part of his "tentative theory to account for the inertial properties of matter".

So going back to P. 6, phi=GM/R, in that ASPW2012 article, since I recognize and understand this equation:

G, M, and R are changing over time.  So must phi, unless it is asserted that G, M, and R change in such a fashion that phi remains constant.  R is the "Hubble" sphere, defined as (ct), the speed of light times the elapsed time since the Big Bang.  But that radius is the observable universe, not the radius of the universe as it exists.

No evidence exists to suggest that the boundary of the observable universe constitutes a boundary on the universe as a whole, nor do any of the mainstream cosmological models propose that the universe has any physical boundary in the first place ...

It appears that there are parts of this universe which are not observable to us.  It is thought that "dark energy" is contributing to the increasing acceleration of the radius of the universe.  The struggle for me is partly this:  Let's say we have a universe which started from one Big Bang, and during an early period of FTL expansion, a good bit of the universe got away from our light cone of observation.  But how could causality stop at the radius of observability? Doesn't make sense to us simpletons.  I prefer to think that the entire universe is fraught with causality of a sort that is different from the speed of light observability.

If this line of reasoning has any merit, then what happens "there" can affect what happens "here", and vice versa.  That is, there is some kind of "action at a distance".  What inertia is has to do with this "action at a distance".  According to Mach's Principle:

Quote from: that ASPW2012 article
The inertial mass of a body is determined by the distribution and flow of mass-energy in the universe.

The mass of the universe is not static; it is moving.  Even so, if your "brain" were big enough, you could conceive of a "fixed frame" of the universe, which is what I think Mach is surmising.  After all, the universe, if that's all there is, cannot be moving with respect to anything else; all of its constituent atoms can be moving with respect to one another, however.  Which leads me to believe that the center of mass of the universe is not static, but is moving around the universe in such a way as to constantly express that it is the center of mass.

With the universe getting bigger, somehow the scalar term of the "fixed frame" is getting bigger too, but apparently that doesn't matter.  As to the pragmatic effect of the fixed inertial frame being different from "absolute space", I don't get that either, but hey.  For all practical purposes, the two terms could be interchangeable.  Besides, Sciama contends that you don't need "absolute space" anyhow, so an understanding of the difference may not be necessary for the layman!

Anyhow... Heidi Fearn, in 2012, asserts that Sciama was right in 1953, and that phi=c^^2.

But first, the "gravelectric" field needs to be proven correct.  Intuitively, the M-E device does convert electricity into forward momentum; if it goes "up" then it defies gravity.  So it seems to me that his equation relates electricity and gravity.

*************************************

Don't ask me for the math.  I'm just the "idea" guy.

« Last Edit: 02/11/2013 01:29 am by JohnFornaro »
Sometimes I just flat out don't get it.

Offline grondilu

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Re: Woodward's effect
« Reply #93 on: 02/11/2013 02:02 am »
The struggle for me is partly this:  Let's say we have a universe which started from one Big Bang, and during an early period of FTL expansion, a good bit of the universe got away from our light cone of observation.  But how could causality stop at the radius of observability? Doesn't make sense to us simpletons.  I prefer to think that the entire universe is fraught with causality of a sort that is different from the speed of light observability.

I think the limit of observability is actually the last scattering surface.  Beyond it, no light can be seen because it would come from a time in the universe when it was so hot that light was coupled with matter and could not escape.  Just at the limit what you see is basically the cosmic background radiation:  it's the first light the universe ever emitted.

Beyond the last scattering surface, there is what I think is called the cosmological horizon. It is the point to which the expansion makes space itself go away from us at light speed.  Beyond that, things still exist but whatever happens can never affect us in anyway.  There can indeed be lots of stuffs there, and it can expand to thousands of times the size of the observable universe, but it is indeed causally separated.


Quote
With the universe getting bigger, somehow the scalar term of the "fixed frame" is getting bigger too, but apparently that doesn't matter.

Indeed it doesn't matter, because in the end what you use is the literal expression phi, not GM/R.   I don't know why Woodward mentions it.  Sciama does not.
« Last Edit: 02/11/2013 02:03 am by grondilu »

Offline JohnFornaro

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Re: Woodward's effect
« Reply #94 on: 02/11/2013 12:54 pm »
I think the limit of observability is actually the last scattering surface. ... Beyond the last scattering surface, there is what I think is called the cosmological horizon.

Yes, and the oracle explains that fairly well to the layman.  But still, there was one cause, the Big Bang.  Now there are multiple "causalities", since this part of the u. is too far from that part of the u.  And what about the stuff on that "horizon"?  Objects there can "see" beyond the horizon, as well as into the "interior", where we are.

If there is to be a "fixed frame", it must contain the whole universe, regardless of the speed of light.  Otherwise, it would just be an arbitrary frame, dependent on the observer's location.  You know:  "Uhhhh... the stars look pretty 'fixed' from this location..."

If the "fixed frame" can be said to exist, then there must be "action at a distance", and the idea that inertia is the inevitable result of this "action" can start to be considered.

Quote from: JF
With the universe getting bigger, somehow the scalar term of the "fixed frame" is getting bigger too, but apparently that doesn't matter.

Quote from: Grondilu
Indeed it doesn't matter, because in the end what you use is the literal expression phi, not GM/R.   I don't know why Woodward mentions it.  Sciama does not.

Unfortunately, this does not explain the matter at all.  It is fundamental to the argument that phi=c^^2.  They must mention this for a reason; the term is not thrown into the line of argument without reason.
Sometimes I just flat out don't get it.

Offline grondilu

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Re: Woodward's effect
« Reply #95 on: 02/11/2013 01:03 pm »
Unfortunately, this does not explain the matter at all.  It is fundamental to the argument that phi=c^^2.

I did not get this either.  phi = GM/R is not fundamental to the argument that phi=c^2 since Sciama does not use it.  Sciama writes about a page and half to justify it.  It's page 38 to 40.  Not much maths, but essentially cosmological and relativistic considerations.  I don't get it all.

Offline mrmandias

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Re: Woodward's effect
« Reply #96 on: 02/11/2013 01:05 pm »

This is on the same level as mind-reading (which I was accused of earlier) and telekinesis, and it should be given the same level of skepticism.

That's absurd.  I was taking you seriously until you got histrionic. 

You can't defeat silly claims with silly claims.  Double down on the silly claim if you want, but you'll only be doubling down on a patent falsehood.

Offline mrmandias

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Re: Woodward's effect
« Reply #97 on: 02/11/2013 01:13 pm »
Cuddihy, Grondilu, Fornaro, et al.,
I'm getting a lot out of this thread.  Thanks.

Offline JohnFornaro

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Re: Woodward's effect
« Reply #98 on: 02/11/2013 01:57 pm »
Unfortunately, this does not explain the matter at all.  It is fundamental to the argument that phi=c^^2.

I did not get this either.  phi = GM/R is not fundamental to the argument that phi=c^2 since Sciama does not use it.  Sciama writes about a page and half to justify it.  It's page 38 to 40.  Not much maths, but essentially cosmological and relativistic considerations.  I don't get it all.

Huh?  Are you referring to Sciama 1953?, where page 38 begins with:

"since the change of rho with time is very small..."

You call those pages "Not much math?"  Sheesh.
« Last Edit: 02/11/2013 02:00 pm by JohnFornaro »
Sometimes I just flat out don't get it.

Offline JohnFornaro

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Re: Woodward's effect
« Reply #99 on: 02/11/2013 01:58 pm »

This is on the same level as mind-reading (which I was accused of earlier) and telekinesis, and it should be given the same level of skepticism.

That's absurd.  I was taking you seriously until you got histrionic. 

You can't defeat silly claims with silly claims.  Double down on the silly claim if you want, but you'll only be doubling down on a patent falsehood.

This is absolutely true.  Chris will throw math at people pretty readily.  Here he does not.
Sometimes I just flat out don't get it.

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