So, as suggested here, I open this thread about Woodward's effect.I try to understand his paper called Recent Results of an Investigation of Mach Effect Thrusters, and there are already some details I don't get in the first equations.Here is a mathbin: http://mathbin.net/154127I'm not sure this 3/2 factor really matters or what but already it nags me. Funny thing is that if the test particle was "outside" of the universe, then sure, one could use the gauss theorem or stuff like that and the universe would provide a potentiel just as if it was ponctual. But if I assume the test particle is in the middle of a universe, then there is this 3/2 factor. That's weird.

I thought I'd make my first post here.

Start with Sciama 1953. Which, of course, as is well known, was the best year of the twentieth century.Quote from: JohnFornaro on 02/09/2009 06:57 PMI thought I'd make my first post here.

Quote from: JohnFornaro on 02/20/2013 02:30 PMStart with Sciama 1953. Which, of course, as is well known, was the best year of the twentieth century.Quote from: JohnFornaro on 02/09/2009 06:57 PMI thought I'd make my first post here.I'm not starting with something Woodward says his theory doesn't depend on.

Math?

If I'm going to evaluate something, I'm going to need to know what the /actual/ basis for the theory is. And, of course, this "slipperiness" of appealing to Sciama and then saying you don't depend on it is a sign you're trying to pull a fast one.

...What is your angle here? I'm rather patient, but this is getting lame. You clearly don't have a handle on the math, but appear to be too proud to admit it....

Math without context is useless. You can build a system of consistent math with no connection with the real world, and although it may make you /seem/ impressive, it means diddly squat for physics. We're trying to establish what equations Woodward depends on, not ancillary stuff.So, supporters of Woodward's Effect, provide a single document which we can examine.

Quote from: JohnFornaro on 02/20/2013 04:27 PM...What is your angle here? I'm rather patient, but this is getting lame. You clearly don't have a handle on the math, but appear to be too proud to admit it....A challenge. I'm in physics grad school, the math isn't the limiting factor it's the time to sift through it all.Where is the "seminal paper" on this effect? We need /one/ thing which we can examine.

Quote from: Robotbeat on 02/20/2013 04:31 PMQuote from: JohnFornaro on 02/20/2013 04:27 PM...What is your angle here? I'm rather patient, but this is getting lame. You clearly don't have a handle on the math, but appear to be too proud to admit it....A challenge. I'm in physics grad school, the math isn't the limiting factor it's the time to sift through it all.Where is the "seminal paper" on this effect? We need /one/ thing which we can examine.I have provided the one seminal paper.

Quote from: JohnFornaro on 02/20/2013 04:32 PMI have provided the one seminal paper.No, you did not. You gave me a Sciama paper, which Woodward explicitly says his "theory" doesn't depend on.

I have provided the one seminal paper.

...Sciama 1953 gives the context, consistent with the real world. You keep dodging for no apparent purpose other than pride.

Over a century has passed since Ernst Mach conjectured that the cause of inertia should somehow be causally related to the presence of the vast bulk of the matter (his "fixed stars") in the universe. Einstein translated this conjecture into "Mach’s principle" (his words) and attempted to incorporate a version of it into general relativity theory (GRT) by introducing the "cosmological constant" term into his field equations for gravity. Einstein ultimately abandoned his attempts to incorporate Mach’s principle into GRT. But in the early 1950s Dennis Sciama revived interest in the "origin of inertia"” ....

Math without context is useless.

I don't want context...

Now find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!

Give me math to prove or disprove and the physical context (by this I mean assumptions, not papers by different authors that is not dependent on the problem in question).

Quote from: LegendCJS on 02/20/2013 04:32 PMNow find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!That is what Woodward is claiming: He has found a periodic fluctuation of mass.I don't know if you can do this or not.

Quote from: JohnFornaro on 02/20/2013 04:51 PMQuote from: LegendCJS on 02/20/2013 04:32 PMNow find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!That is what Woodward is claiming: He has found a periodic fluctuation of mass.I don't know if you can do this or not.I sense you missed the point. For this proof take Woodward at his word. Allow for periodic mass fluctuations. It doesn't matter. You can not get a net force out of the effect EVEN IF ITS TRUE

Quote from: LegendCJS on 02/20/2013 04:58 PMQuote from: JohnFornaro on 02/20/2013 04:51 PMQuote from: LegendCJS on 02/20/2013 04:32 PMNow find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!That is what Woodward is claiming: He has found a periodic fluctuation of mass.I don't know if you can do this or not.I sense you missed the point. For this proof take Woodward at his word. Allow for periodic mass fluctuations. It doesn't matter. You can not get a net force out of the effect EVEN IF ITS TRUE I encourage everyone to try it. Open Matlab or your favorite spread sheet. Create a periodic function called M(t), and one called V(t) . Be creative, get crazy, try to make the fantasy come true.Then calculate F as shown:F = v*dm/dt + m*dv/dtAnd plot F. Is it periodic? Then no net force exists. Game over. Thread Dead, stop wasting Chris' bandwidth.He adds electricity to provide the motive force, AIUI. So I don't think the simplification you provided is complete. You gotta have an energy supply to change the direction of the momentum.

Quote from: JohnFornaro on 02/20/2013 04:51 PMQuote from: LegendCJS on 02/20/2013 04:32 PMNow find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!That is what Woodward is claiming: He has found a periodic fluctuation of mass.I don't know if you can do this or not.I sense you missed the point. For this proof take Woodward at his word. Allow for periodic mass fluctuations. It doesn't matter. You can not get a net force out of the effect EVEN IF ITS TRUE I encourage everyone to try it. Open Matlab or your favorite spread sheet. Create a periodic function called M(t), and one called V(t) . Be creative, get crazy, try to make the fantasy come true.Then calculate F as shown:F = v*dm/dt + m*dv/dtAnd plot F. Is it periodic? Then no net force exists. Game over. Thread Dead, stop wasting Chris' bandwidth.

Woodwards effect or not, mass fluctuations or not, the math to show how you can not get a force out of any push heavy/pull light scheme is pretty simple, and can be worked by anyone who has seen the chain rule in intro calculus.With words:Force = change in momentummomentum = mass*velocityif both mass and velocity are functions of time then the chain rule applies.F = v*dm/dt + m*dv/dt.Now find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!(happy to be proven wrong.)

I didn't provide this simplification, Newton did. p is momentum. F = dp/dt is the proper form of Newtons second law. F = ma is only true if m is a constant.

Quote from: LegendCJS on 02/20/2013 04:32 PMWoodwards effect or not, mass fluctuations or not, the math to show how you can not get a force out of any push heavy/pull light scheme is pretty simple, and can be worked by anyone who has seen the chain rule in intro calculus.With words:Force = change in momentummomentum = mass*velocityif both mass and velocity are functions of time then the chain rule applies.F = v*dm/dt + m*dv/dt.Now find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!(happy to be proven wrong.)I'm not sure, but I think you're referring to the vdm/dt term argument, which is addressed in the following papers on page 1 of this thread: "Origin of inertia JF Woodward 2004" in Appendix B , "Refutation 02 ORNL of Woodward" and "Refutation 03 ORNL Woodward of ORNL" on page 7.

Quote from: LegendCJS on 02/20/2013 05:18 PMI didn't provide this simplification, Newton did. p is momentum. F = dp/dt is the proper form of Newtons second law. F = ma is only true if m is a constant.This is a physics question not a maths one but hey: what is the experimental evidence that F = dp/dt is more correct than F = ma? Without observing mass fluctuations, how can we tell the difference?It may be that nature is somewhere in between no?

...if both mass and velocity are functions of time then the chain rule applies.F = v*dm/dt + m*dv/dt.

Random math to satisfy Cinder:

We do know that relativity wise, mass does fluctuate with velocity; the faster the velocity, the greater the mass. So that's the "push heavy" part.What I gather from Woodward's experiment is that the nuclei of PZT thing resonate, or vibrate or move back and forth under the varying electromagnetic field of the capacitor, which has an A/C current of "x" Hertz.My intuition tells me that the dern nuclei vibrate back and forth at "x" megahertz, and so what if they do change mass? On the next cycle, they change mass in the other direction. The thing just sits there and vibrates.

Because you can apply F = dp/dt to a leaking water balloon or any other system whose mass is changing in a conventional way and get all the experimental evidence you need, silly!

Yes, and the evidence is that leaking something while moving at a particular velocity doesn't produce a net force proportional to that velocity.

Quote from: Celebrimbor on 02/20/2013 06:49 PMQuote from: LegendCJS on 02/20/2013 05:18 PMI didn't provide this simplification, Newton did. p is momentum. F = dp/dt is the proper form of Newtons second law. F = ma is only true if m is a constant.This is a physics question not a maths one but hey: what is the experimental evidence that F = dp/dt is more correct than F = ma? Without observing mass fluctuations, how can we tell the difference?It may be that nature is somewhere in between no?Because you can apply F = dp/dt to a leaking water balloon or any other system whose mass is changing in a conventional way and get all the experimental evidence you need, silly! Its also in every physics/ mechanics textbook ever written for an audience that is familiar with calculus.

Quote from: LegendCJS on 02/20/2013 07:01 PMQuote from: Celebrimbor on 02/20/2013 06:49 PMQuote from: LegendCJS on 02/20/2013 05:18 PMI didn't provide this simplification, Newton did. p is momentum. F = dp/dt is the proper form of Newtons second law. F = ma is only true if m is a constant.This is a physics question not a maths one but hey: what is the experimental evidence that F = dp/dt is more correct than F = ma? Without observing mass fluctuations, how can we tell the difference?It may be that nature is somewhere in between no?Because you can apply F = dp/dt to a leaking water balloon or any other system whose mass is changing in a conventional way and get all the experimental evidence you need, silly! Its also in every physics/ mechanics textbook ever written for an audience that is familiar with calculus.Before I flood my kitchen, I checked good old Wikipedia...The second law is only valid for a fixed set of particles. If you loose some of them from your "balloon" then the law does not apply.http://en.wikipedia.org/wiki/Newtons_second_law#Newton.27s_second_law

Quote from: GeeGee on 02/20/2013 06:01 PMQuote from: LegendCJS on 02/20/2013 04:32 PMWoodwards effect or not, mass fluctuations or not, the math to show how you can not get a force out of any push heavy/pull light scheme is pretty simple, and can be worked by anyone who has seen the chain rule in intro calculus.With words:Force = change in momentummomentum = mass*velocityif both mass and velocity are functions of time then the chain rule applies.F = v*dm/dt + m*dv/dt.Now find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!(happy to be proven wrong.)I'm not sure, but I think you're referring to the vdm/dt term argument, which is addressed in the following papers on page 1 of this thread: "Origin of inertia JF Woodward 2004" in Appendix B , "Refutation 02 ORNL of Woodward" and "Refutation 03 ORNL Woodward of ORNL" on page 7.What can those possibly say besides "I get to ignore Newtons Laws because I'm special" anyway?*this response deliberately flippant because I want to provoke someone into doing the reading and summarizing it for me because the burdon of proof is on them/ the supporters and I do not have the time or will to dig into it myself.

Quote from: LegendCJS on 02/20/2013 04:32 PMWoodwards effect or not, mass fluctuations or not, the math to show how you can not get a force out of any push heavy/pull light scheme is pretty simple, and can be worked by anyone who has seen the chain rule in intro calculus.With words:Force = change in momentummomentum = mass*velocityif both mass and velocity are functions of time then the chain rule applies.F = v*dm/dt + m*dv/dt.Now find a periodic function of mass and a periodic function of velocity of your choice that gives you a non periodic Force? YOU CAN NOT DO IT!(happy to be proven wrong.)Hi, is this the point I brought up earlier? A heavy mass going one way and a light mass going the other doesnt really define what happens. The important bit is, was it heavy or light at the precise moment that it was given a shove in the other direction? If it were both heavy and light during this shove, dependent on its current velocity (moving right = heavy, moving left = light) then everything would cancel.If however it were heavy (or light) for the entire shove it is easy to get an unbalanced force.Think of two people passing a medicine ball between them. If the ball becomes heavy before it reaches the left hand person and light before it reaches the right, then the left hand person will be pushed left faster than the right hand person is pushed right. If they are connected by a rope the total system would get a net push left.

Sums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics.

Changing the subject somewhat: If Woodward's mass fluctuation theory fails on such an elementary analysis as yours purports to be, how come he's presenting papers still, like at the Summer 2012 (Joint Propulsion Conf. AIAA)? If his theory is so obviously wrong, how come nobody else is publicly deriding it with absolute certainty? Is there something wrong with the peer review process?

Can you put it in writing? And use the medicine ball analogy. It's a word problem, with what I think are all the necessary assumed numbers:We have person A and B to the left and right, respectively. They are separated by distance d, 1.5 m. We have the medicine ball M, with an at rest mass of 1kg. Person A throws it to person B with a velocity of .75m/sec. Person B catches it, but it has a mass of 1.1kg. Person B throws it back to A at .75 m/sec, and it has a mass of 0.9kg. Repeat. Ignore the effects of air friction and gravity.

Quote from: LegendCJS on 02/21/2013 03:04 PMSums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics. I appreciate your taking the time for these explanations. Sadly, I still need tutoring.Can you put it in writing? And use the medicine ball analogy. It's a word problem, with what I think are all the necessary assumed numbers:We have person A and B to the left and right, respectively. They are separated by distance d, 1.5 m. We have the medicine ball M, with an at rest mass of 1kg. Person A throws it to person B with a velocity of .75m/sec. Person B catches it, but it has a mass of 1.1kg. Person B throws it back to A at .75 m/sec, and it has a mass of 0.9kg. Repeat. Ignore the effects of air friction and gravity.I don't see how you turn this description into a periodic function. And this just assumes that the medicine ball changes mass, which I understand you to be saying doesn't matter, even if it were to be true. Reword the description, if I haven't descibed what Kelvin suggests.What you seem to be describing is a tuning fork, which just vibrates back and forth, and does not change its momentum with respect to an external frame of reference. Ignoring the friction of the material of the tuning fork, it would vibrate forever, bu not go anywhere. If the momentum is to be changed in a preferential direction, energy should be injected into the equation somehow, presumably causing this imbalance in the mass of the medicine ball. Or an imbalance in the masses of the two forks in the tuning fork. I don't think that your equation is complete.

Quote from: LegendCJS on 02/21/2013 03:04 PMSums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics. I appreciate your taking the time for these explanations. Sadly, I still need tutoring.Can you put it in writing?

and call me crazy if F turns out not to be periodic (and please share your details is it isn't, believe me I want "star trek" style space drives as much as any dreamy eyed science fiction fan)

If you want a function representing discrete shove events start with a(t) and make it a square wave, and integrate to get v(t) and position.

So because it is getting more mass it must slow down.

Quote from: LegendCJS on 02/21/2013 06:01 PMSo because it is getting more mass it must slow down.I guess you missed it the first time...Ever heard of Galilean invariance?You have to be very careful with that v*dm/dt term.Or, to put it another way - getting more mass from where?

Quote from: JohnFornaro on 02/21/2013 04:40 PMQuote from: LegendCJS on 02/21/2013 03:04 PMSums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics. I appreciate your taking the time for these explanations. Sadly, I still need tutoring.Can you put it in writing? Well since this is a MATH ONLY thread, lets not necessarily appeal to intuition about physics.The claim is that "products of periodic functions are periodic". Well, lets find out (without appealing to infinite examples in Matlab or the authority of high school math):What is a periodic function? Lets say it is a continuous, smooth function of some parameter, t, with the property that,a(t) = a(t+T) (with T>0; we also assume T to be the smallest value and we call it the period).And now lets double down on the periodic functions. No need for them to have the same period, so we havea: a(t)=a(t+Ta); and b: b(t)=b(t+Tb)Now definef: f(t)=a(t)b(t).Is f(t) periodic? Lets try shifting f(t) along by a whole number multiple of Ta:f(t + nTa) = a(t+nTa)b(t+nTa) = a(t)b(t+nTa). This will only be equal to f(t) if nTa happens to be some whole number multiple of Tb.Oh... so the product of two periodic functions is periodic if and only if the ratio of the two periods is rational.

You continue to miss the point.The v in v*dm/dt is not frame invariant. This throws a huge wrench in your explanation.Take the medicine ball's frame of reference. Why should gaining more mass suddenly add momentum in a particular direction where before it had none?

We are on the same page.

Quote from: JohnFornaro on 02/21/2013 04:40 PMQuote from: LegendCJS on 02/21/2013 03:04 PMSums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics. I appreciate your taking the time for these explanations. Sadly, I still need tutoring. ... Well since this is a MATH ONLY thread, lets not necessarily appeal to intuition about physics.The claim is that "products of periodic functions are periodic". Well, lets find out (without appealing to infinite examples in Matlab or the authority of high school math) ...

Quote from: LegendCJS on 02/21/2013 03:04 PMSums of periodic functions either cancel out or are periodic. Derivatives (where they exist) of periodic functions are periodic, and products of periodic functions are periodic. This is just high school math being applied to a formula from college freshmen (at an engineering school) physics. I appreciate your taking the time for these explanations. Sadly, I still need tutoring. ...

Quote from: LegendCJS on 02/21/2013 08:05 PMWe are on the same page.No, we are not.Take the medicine ball's frame of reference. Why should gaining more mass suddenly add momentum velocity in a particular direction where before it had none?Galilean invariance. If you can't respect it, you're using Newtonian physics wrong.

Well this magic mass is supposed to come from the "rest of the universe", which has 0 velocity in its own reference frame, or 0.75 m/s velocity in the reference frame of the medicine ball. That is why there is a preferential effect on the motion of the ball in the direction of the place from which the mass came from (the rest of the universe). The additional 0.2kg of mass being added to the medicine ball (which is stationary in its own reference frame) is moving at 0.75 m/s towards person A when it appears, thus slowing the ball on its journey to person B...

A 0.9kg ball moving at 0.75 m/s has momentum of: m*v = 0.6750.All of the sudden it gains mass (not that I believe this can happen either) of 0.2kg to become 1.1kg. But it didn't/can't gain any momentum in any direction. So momentum has to be the same as the 0.9kg ball. That is why the speed decreased to 0.6136 m/s, so the p = m*v product doesn't change.

Quote from: LegendCJS on 02/21/2013 08:05 PMA 0.9kg ball moving at 0.75 m/s has momentum of: m*v = 0.6750.All of the sudden it gains mass (not that I believe this can happen either) of 0.2kg to become 1.1kg. But it didn't/can't gain any momentum in any direction. So momentum has to be the same as the 0.9kg ball. That is why the speed decreased to 0.6136 m/s, so the p = m*v product doesn't change.Im a bit late back to the conversation, just wanted to say this is where I am having the problem also. I had assumed that when the ball magically gained mass it kept its current velocity because any other interpretation seems inconsistent to me.I'll keep out of the debate because I don't think I have any arguments beyond the ones others have introduced.

I appreciate your math comments, Celebrimbor, but you too need to be careful with appeals.

How very dare you.

If you subscribe to Galilean invariance, or more currently Lorentz invariance, then it is equally inconsistent to say that the ball stays moving at the same velocity in any reference frame including the one of A and B.

However, the astronimical observations do not appear to be Lorentz invariant.

Sciama shows that this selection of an "absolute reference frame" provides a possible mechanism by which the distant matter of the universe introduces local inertia.

Does this require a causal connection to the unobservable universe?

Lorentz symmetry describes a fairly fundamental principle of physics that is unrelated to the fact that the velocity spread of observable matter is relatively narrow. Like you said, this is to be expected based on the Big Bang theory; it doesn't mean the laws of physics themselves are dependent on velocity.

R was taken by Sciama as the radius of the "Hubble sphere", that is, the product of the speed of light and the age of the universe. A more accurate calculation would have employed the "particle horizon", the sphere centered on Earth within which signals traveling at the speed of light can reach Earth. The particle horizon encompasses considerably more material than the Hubble spere. Sciama also neglected the expansion of the universe.Thes issues notwithstanding, Sciama's work triggered an at times intense debate about the origin of inertia...

What throws me is that Sciama can use faulty assumptions, yet his work built on those assumptions is still thought to be accurate?

Woodwards new book arrived early this week. It starts out with a "dreamy" outline of stargates and stuff, then switches into heavy math, where he starts out with the basics, but moves quickly from m=E/c^2 (p.29)to:G_{mu v} = R_{mu v}- 1/2g_{mu v} = -(8 pi G/c^4)t_{muv} (p.30)with no explanation getting the reader from calculus to relativity.

Quote from: JohnFornaro on 03/07/2013 08:38 PMWhat throws me is that Sciama can use faulty assumptions, yet his work built on those assumptions is still thought to be accurate?Not accurate. Correct (or so we hope). There's a difference. If you actually calculate out his result, it's nowhere near reality, because ... it's a mathematical demonstration of an effect that behaves qualitatively like inertia, caused by distant gravitating matter in a model universe.The idea is not built on the assumptions. The assumptions are used in the demonstration of the idea, but they can be changed without destroying the idea.If you use his idea with accurate assumptions and the full relativistic equivalent of his solution, using modern cosmological data, you supposedly get F = 0.92ma, which is pretty good...

Are you looking for the detailed math calcs?

If you use his idea with accurate assumptions and the full relativistic equivalent of his solution, using modern cosmological data, you supposedly get F = 0.92ma, which is pretty good...