Author Topic: Woodward's Effect - MATH ONLY  (Read 23933 times)

Offline QuantumG

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Woodward's Effect - MATH ONLY
« on: 02/15/2013 05:21 AM »
Original comment of previous thread:

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/154127

I'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.

Please leave your hopes and dreams for propellantless propulsion in the other thread. Similarly, we don't care if you think it's not going to work, or if you think it is against the laws of physics, God, dog or hamster. Leave it in the other thread or yell it at the neighbors.

In here, math, discussion of math and preferably some links to other people explaining the math.

Okay?
I hear those things are awfully loud. It glides as softly as a cloud. What's it called? Monowhale!

Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #1 on: 02/15/2013 05:27 AM »
We need to make this falsifiable... Ie, don't rely on Sciama then say in the next breath you don't rely on Sciama.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #2 on: 02/15/2013 06:02 AM »
Question:  Can Woodward's effect be described in a common language (i.e. axiomatic theory) to GR (e.g. abstract differential geometry)?

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #3 on: 02/19/2013 01:07 AM »
Some backup reading:
« Last Edit: 02/19/2013 01:16 AM by JohnFornaro »
Sometimes I just flat out don't get it.

Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #4 on: 02/19/2013 03:44 AM »
About a week's worth of reading.
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Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #5 on: 02/20/2013 02:30 PM »
Start with Sciama 1953.  Which, of course, as is well known, was the best year of the twentieth century.

I thought I'd make my first post here.
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Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #6 on: 02/20/2013 02:33 PM »
Start with Sciama 1953.  Which, of course, as is well known, was the best year of the twentieth century.

I thought I'd make my first post here.
I'm not starting with something Woodward says his theory doesn't depend on.
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 - MATH ONLY
« Reply #7 on: 02/20/2013 03:17 PM »
Start with Sciama 1953.  Which, of course, as is well known, was the best year of the twentieth century.

I thought I'd make my first post here.
I'm not starting with something Woodward says his theory doesn't depend on.

You are free to say that, but Woodward and Heidi Fearn mention this document specifically as a starting point.
Sometimes I just flat out don't get it.

Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #8 on: 02/20/2013 04:00 PM »
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.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Offline Cinder

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Re: Woodward's Effect - MATH ONLY
« Reply #9 on: 02/20/2013 04:16 PM »
Math?
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Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #10 on: 02/20/2013 04:18 PM »
Math?
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.

Random math to satisfy Cinder:
« Last Edit: 02/20/2013 04:21 PM by Robotbeat »
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 - MATH ONLY
« Reply #11 on: 02/20/2013 04:27 PM »
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.

I'm not saying that, and they're not saying that.  It's the first thing woodward and Fearn mention. 

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.

I can't do the math either. 
Sometimes I just flat out don't get it.

Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #12 on: 02/20/2013 04:31 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. ;) Well, I'm not going to bite until I know it's actually going to do some good. If you can just throw up yet another paper by saying "ah HA! Well, Woodward Effect doesn't depend on *blahblah*, it still works based on *blah*," I'm not going to waste my time.

I'm in physics grad school, the math isn't the limiting factor it's the time to sift through it all (and really, given enough time, you too could learn the math, starting with all the stuff here: http://www.khanacademy.org/math/calculus and then buying a few upper level textbooks and working through all the examples).

Where is the "seminal paper" on this effect? We need /one/ thing which we can examine. This is a challenge to supporters of the Woodward Effect.
« Last Edit: 02/20/2013 04:35 PM by Robotbeat »
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Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #13 on: 02/20/2013 04:31 PM »
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.

I'm not a supporter, but I provided the single document upon which Woodward began his theorization.  So, you basically have a problem with me?  Not the paper?

Sciama 1953 gives the context, consistent with the real world.  You keep dodging for no apparent purpose other than pride.
Sometimes I just flat out don't get it.

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #14 on: 02/20/2013 04:32 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.
Sometimes I just flat out don't get it.

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #15 on: 02/20/2013 04:32 PM »
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 momentum

momentum = mass*velocity

if 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.)
« Last Edit: 02/20/2013 04:33 PM by LegendCJS »
Remember: if we want this whole space thing to work out we have to optimize for cost!

Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #16 on: 02/20/2013 04:36 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.
No, you did not. You gave me a Sciama paper, which Woodward explicitly says his "theory" doesn't depend on.

And really, I don't think you're a Woodward supporter.
« Last Edit: 02/20/2013 04:39 PM by Robotbeat »
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Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #17 on: 02/20/2013 04:41 PM »

I 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.

Where did he say that?
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Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #18 on: 02/20/2013 04:44 PM »
...

Sciama 1953 gives the context, consistent with the real world.  You keep dodging for no apparent purpose other than pride.
I don't want context, I want some solid paper I can examine, to either prove, disprove, or show is unknowable. Sciama's theory isn't entirely accepted (i.e. it's not "textbook"), but Sciama also doesn't suppose it's possible to do the Woodward effect and Woodward claims not to rely on Sciama, so I don't see the point when we're talking about slinging equations.
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 - MATH ONLY
« Reply #19 on: 02/20/2013 04:45 PM »
From:

0015-9018/04/1000-1475/0 © 2004 Springer Science+Business Media, Inc

Quote
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"” ....
Sometimes I just flat out don't get it.

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #20 on: 02/20/2013 04:48 PM »
Math without context is useless.

I don't want context...

The farce is strong in this one...
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Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #21 on: 02/20/2013 04:48 PM »
So freaking what? It's not widely accepted and taken as nearly axiomatic ("textbook") the way GR is now.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Re: Woodward's Effect - MATH ONLY
« Reply #22 on: 02/20/2013 04:51 PM »
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!

That is exactly what Woodward is claiming:  He has found a periodic fluctuation of mass.

What's more, he claims that his experiment shows how much it fluctuates.  Not as much as he might want or predict, mind you.  Mind everybody, actually.

I don't know if you can do this or not.
« Last Edit: 02/20/2013 05:00 PM by JohnFornaro »
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Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #23 on: 02/20/2013 04:52 PM »
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).

I think you're just trolling, now.
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 - MATH ONLY
« Reply #24 on: 02/20/2013 04:55 PM »
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).

I know where, and have shown where to start looking for the math, but I cannot give the math.   Neither can you.  No blame to you from me on that.
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Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #25 on: 02/20/2013 04:58 PM »
Burden on proof is on supporters to come up with something that can be examined. I'm going to sit here, lazy, until that is done.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

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Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #26 on: 02/20/2013 04:58 PM »
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!

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/dt

And plot F.  Is it periodic?  Then no net force exists.  Game over.  Thread Dead, stop wasting Chris' bandwidth.
« Last Edit: 02/20/2013 05:11 PM by LegendCJS »
Remember: if we want this whole space thing to work out we have to optimize for cost!

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #27 on: 02/20/2013 05:14 PM »
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!

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


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.
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Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #28 on: 02/20/2013 05:18 PM »
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!

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/dt

And 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.
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.

" ***he adds electricity****" seriously? Irrelevant.

They want to build a thruster.  They say the thrust comes form the push heavy/ pull light cycle.  Where is the electricity in this besides making the mass fluctuate?  Even if its making the push and pull happen (i.e. giving you your V(t)) that obviously averages to zero, no excel exercise needed there, and it doesn't change anything I said.
« Last Edit: 02/20/2013 05:30 PM by LegendCJS »
Remember: if we want this whole space thing to work out we have to optimize for cost!

Offline GeeGee

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Re: Woodward's Effect - MATH ONLY
« Reply #29 on: 02/20/2013 06:01 PM »
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 momentum

momentum = mass*velocity

if 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.

Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #30 on: 02/20/2013 06:49 PM »

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.


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?

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #31 on: 02/20/2013 06:58 PM »
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 momentum

momentum = mass*velocity

if 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.
Remember: if we want this whole space thing to work out we have to optimize for cost!

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #32 on: 02/20/2013 07:01 PM »

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.


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.
« Last Edit: 02/20/2013 07:04 PM by LegendCJS »
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Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #33 on: 02/20/2013 07:30 PM »
...if both mass and velocity are functions of time then the chain rule applies.

F = v*dm/dt + m*dv/dt.

My understanding is very basic; if any little thing is wrong, I get lost. 
The product rule is a special case of the chain rule and is what is applicable here.  I think.

But typically, it is thought that mass is constant, therefore dm/dt=0; hence F=m(dv/dt).  Of course, a=dv/dt.  So what am I missing here?

If mass could fluctuate, you could then say:

F = v*dm/dt + m*dv/dt. (Which is also F = dp/dt.)

So I kinda get where you're coming from, but I don't know how to solve it with a fluctuating mass.  I can't get mass on one side of the equation.  Can I divide by dt/dm?  dt/dv?

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.
« Last Edit: 02/20/2013 07:33 PM by JohnFornaro »
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Offline Cinder

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Re: Woodward's Effect - MATH ONLY
« Reply #34 on: 02/20/2013 08:03 PM »
 This is my last post in this thread.


Random math to satisfy Cinder:
Playing with words.  The topic is not random math but "Woodward" math.

"Lazy" is correct.  You sabotage this topic for the sake of your opinion.  You cannot pretend QED till it is.  The burden of proof is not on Woodward & co because they have yet to make as absolute an assertion WRT their proposed model(s) as you have on this forum.

This kind of attitude is why science is not totally politics-proof. 
« Last Edit: 02/20/2013 08:03 PM by Cinder »
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Offline Robotbeat

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Re: Woodward's Effect - MATH ONLY
« Reply #35 on: 02/20/2013 08:05 PM »
What am I trying to prove or disprove? Give me one (Woodward) paper.
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 - MATH ONLY
« Reply #36 on: 02/20/2013 08:16 PM »
Ok, but clearly, we all understand that you didn't say you'd even read it, much less explain it:
Sometimes I just flat out don't get it.

Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #37 on: 02/20/2013 08:34 PM »

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.

If we're talking relativistic velocities, we need to consider momentum p as a 4-vector.  Changes to a 4-momentum can only be made via a 'Lorentz' boost.  http://en.wikipedia.org/wiki/Four-momentum

If you do that then "conventional wisdom" is that you don't consider the inertia itself to change under relativistic velocities.

Woodward might say something different... I've only just finished Sciama '53...

Offline 93143

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Re: Woodward's Effect - MATH ONLY
« Reply #38 on: 02/20/2013 08:50 PM »
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.

Ever heard of Galilean invariance?

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #39 on: 02/20/2013 08:57 PM »
Yes, and the evidence is that leaking something while moving at a particular velocity doesn't produce a net force proportional to that velocity.

Wait a sec.  Are you saying that SLS leaks?  And that's why it can't get 70 tons to LEO on its own?

What are we paying these people to do?
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Re: Woodward's Effect - MATH ONLY
« Reply #40 on: 02/20/2013 09:02 PM »

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.


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

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #41 on: 02/20/2013 09:14 PM »

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.


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

Indeed, start loosing particles (i.e. mass) and F=ma goes out the window.  F = dp/dt still applies.  No disagreement from me.
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Offline Chris Bergin

Re: Woodward's Effect - MATH ONLY
« Reply #42 on: 02/20/2013 09:43 PM »
This thread's a bit over my head, but I want everyone to remember not to shout other people's opinions down, or be distruptive. It's a big forum, always fine to have other threads on your own opinions.

Offline GeeGee

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Re: Woodward's Effect - MATH ONLY
« Reply #43 on: 02/20/2013 10:56 PM »
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 momentum

momentum = mass*velocity

if 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.

Uh, what? I just told you where your complaint is addressed. The OP said "In here, math, discussion of math and preferably some links to other people explaining the math." So do you really want to be proven wrong, or are you just here to sling mud?

Online KelvinZero

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Re: Woodward's Effect - MATH ONLY
« Reply #44 on: 02/20/2013 11:04 PM »
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 momentum

momentum = mass*velocity

if 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.

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #45 on: 02/21/2013 03:04 PM »
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 momentum

momentum = mass*velocity

if 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.

That doesn't change the outcome.  Please take the time to do the numerical experiment if you don't trust me.  Its only 3 steps.

Take this medicine ball being passed back and forth.  Open excel or matlab or whatever program you like.  Pick a periodic function for its position.  Any function you like, but if you don't have a favorite a good old sinusoid will do.  Now differentiate it to get velocity v(t), and again to get acceleration a(t).  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.

You are done with step 1 of 3.

Woodward and you are saying that the mass alternates periodically. I.e. m(t) is periodic.  Pick any function you want for m(t) as long as it is periodic.  Make sure it is heavy when you want it to be and light when you want it to be.  Please do it.  Make it satisfy your condition: "was it heavy or light at the precise moment that it was given a shove."  You know when it was given a shove because you can refer to v(t) and a(t).  Have you picked an m(t) that does this yet? yes? good. Step 2 of 3 complete.

Now plug these functions into this formula to finish step 3:

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

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) 

But there is no use arguing with wordy posts and analogies about something that is this easily verifiable or falsifiable.

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.
« Last Edit: 02/21/2013 03:14 PM by LegendCJS »
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Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #46 on: 02/21/2013 04:40 PM »
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.

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.

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

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?
Sometimes I just flat out don't get it.

Offline GeeGee

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Re: Woodward's Effect - MATH ONLY
« Reply #47 on: 02/21/2013 05:29 PM »


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?

If you look back to the bottom of page 2, you'll see that I specified the papers in which the vdm/dt term argument (which LegendJCS is appealing to) is addressed.

Offline MikeAtkinson

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Re: Woodward's Effect - MATH ONLY
« Reply #48 on: 02/21/2013 05:53 PM »

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.

When person A throws the ball he is accelerating it (with lets assume constant) acceleration a1 (for a time t1), lets also assume that when he catches the ball he decelerates it with a constant deceleration which is also a1. Lets assume person B catches and throws the ball with an acceleration a2 (for a time t2), which is in the opposite direction to a1. So the acceleration graph will have a period of positive acceleration a1, followed by zero acceleration followed by a period of negative acceleration a2.

To impart the same velocity change to the ball we need a1 * t1 = - a2 * t2 = 1.5 m/s.

In the medicine ball example the ball does not change mass during the catch and throw, so we can use F = ma. Integrating we get a net sum of zero assuming m1 == m2.

In the medicine ball example the mass change occurs between the throws and catches. This is where Woodward comes in, you have a momentum change (a change of mass, but not velocity).  Using F = v*dm/dt + m*dv/dt. V is constant, so m*dv/dt = 0, using classical dynamics the forces from the mass changes will exactly cancel out the missmatch in forces on the throwers A and B.

Woodward somehow thinks that he v*dm/dv term does not exist and so the missmatch in forces on A and B are not cancelled.

To be fair to Woodward it seems he has both the mass and velocity varying at the same time, rather than separated as in the medicine ball example, I'm still trying to follow his assumptions and math, they are rather complicated, so they obscure the situation.


Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #49 on: 02/21/2013 06:01 PM »
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.

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.

Going with the medicine ball analogy for now.

You left out a vital point of the description.  When person A has the ball the ball is 0.9kg.  He throws the ball at 0.75 m/s.  Somewhere in the middle the ball changes mass to 1.1kg.  Let us say this change takes place at the midpoint of the distance between person A and B. When this change happens the ball must conserve kenetic energy.  So because it is getting more mass it must slow down. The ball that was 0.9kg and moving at 0.75m/s is now a ball at 1.1kg and moving at 0.6136 m/s when person B catches it, not 0.75 m/s.

Person B tosses the 1.1kg ball back to person A at 0.75 m/s.  When the ball reaches half way it sheds mass to become 0.9kg again, but it has to speed up, so it is now moving at 0.9167 m/s.

The cycle repeats.

For person A they gain the momentum of mass time change in velocity for their catch and throw: 0.9*(0.75+0.9167) = 1.5

Person B does the same: 1.1*(0.75+0.6136) = 1.5. in the other direction.

Net Force = 0.

Guess what 1kg*(0.75 + 0.75)  equals....
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Offline MikeAtkinson

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Re: Woodward's Effect - MATH ONLY
« Reply #50 on: 02/21/2013 06:10 PM »
Yes, I used conservation of momentum, you used conservation of energy, in both cases Woodard says they are broken locally, with the excess momentum and energy being transferred the rest of the universe.


Offline MikeAtkinson

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Re: Woodward's Effect - MATH ONLY
« Reply #51 on: 02/21/2013 06:14 PM »
That is where Woodward's analogy of sand falling off a train breaks down. The mass of the system (train + falling sand) does not change, so there is no v*dm/dv term, until the sand reaches the ground - then the system either has to include the earth or the force decelerating the sand needs to be taken into account.

Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #52 on: 02/21/2013 06:50 PM »
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.

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 have
a: a(t)=a(t+Ta); and b: b(t)=b(t+Tb)

Now define

f: 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. 

Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #53 on: 02/21/2013 07:16 PM »

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) 


Now that we know that F can be aperiodic.  Why should this allow space drives?

The thing that we need to know is whether the average net force has positive magnitude when averaged over many cycles.  A lack of periodicity necessary but not sufficient for this to be true.

Offline MP99

Re: Woodward's Effect - MATH ONLY
« Reply #54 on: 02/21/2013 07:21 PM »
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.

ISTM that "square wave" *requires* your input periodic functions to be symmetrical around zero? (Please correct me if not.) If so, are you sure that doesn't guarantee the result you state?

Take the attached image (the headline illustration at http://en.wikipedia.org/wiki/Periodic_function) as being my a(t) function, with zero at the "x" line. I don't need to do anything more than integrate that to end up with a position which trends upwards. It's completely unphysical for the system we're considering, but is more like the pulsed engine on a V-1 rocket.

ISTM a constant mass with an a(t) *rectangular* function with positive & negative segments of equal duration but peaks of +2 and -1 will get a net force over the cycle. (Vary mass from 0.999 to 1.001 if you like.) Same for a rectangular wave from +1 to -1 where the positive part of the curve is twice the duration of the negative part.

Again, I'm not saying this represents the system, so can you just list all the requirements for the mass & position curves to represent the system as described by it's advocates?

Also, I'm not clear quite what the position curve is supposed to represent? It's clearly not the absolute position of an unconstrained propulsive element, because a requirement that it be a periodic system would *require* it to have no net movement. So, I'm thinking that it would represent an absolute position in a system which is bolted to an "infinite" mass like the Earth. Then the net of F over a cycle is the thrust an untethered system would experience.

cheers, Martin
« Last Edit: 02/21/2013 07:25 PM by MP99 »

Offline 93143

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Re: Woodward's Effect - MATH ONLY
« Reply #55 on: 02/21/2013 07:36 PM »
So 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?
« Last Edit: 02/21/2013 07:48 PM by 93143 »

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #56 on: 02/21/2013 07:45 PM »
So 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?
My point isn't to argue that mass fluctuations are true or not.  It is to say that it doesn't matter, you can't use them to make a propulsive force.  Let a magic fairy give the medicine ball more mass or take it away for all I care.
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Offline 93143

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Re: Woodward's Effect - MATH ONLY
« Reply #57 on: 02/21/2013 07:49 PM »
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?
« Last Edit: 02/21/2013 07:50 PM by 93143 »

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #58 on: 02/21/2013 07:58 PM »
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.

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 have
a: a(t)=a(t+Ta); and b: b(t)=b(t+Tb)

Now define

f: 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. 

Remember the whole framing is "push heavy, pull light" thrusting.  The periods of the pushing and pulling oscillation and the periods of the mass fluctuations had better be the same or at least some integer ratio or after a little time you will be pushing light and pulling heavy and there would be no hope of getting anywhere.

You might end up displacing your thruster some distance form its starting position (in a frictionless world) but the instant you turned it off and all the moving parts came to rest the thruster would be back where it started.

The product of two periodic function, if they have different frequencies (inverse periods) will be a function composed of a component at the sum of the two frequencies and a component at the difference of the two frequencies.  If the ratio of periods isn't rational the resulting function's period will be infinite.
« Last Edit: 02/21/2013 08:17 PM by LegendCJS »
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Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #59 on: 02/21/2013 08:05 PM »
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?

I never said it does.  I agree. It can not.  We are on the same page.  Preaching to the choir. We are citing the very same thing to show why thrusting form any Woodwards effect will not work. Check my numbers from my example.  Or if you are not replying to me don't make your posts ambiguous about who you are replying too.

I'll repeat:

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.
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Offline 93143

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Re: Woodward's Effect - MATH ONLY
« Reply #60 on: 02/21/2013 08:07 PM »
We 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.
« Last Edit: 02/21/2013 08:13 PM by 93143 »

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #61 on: 02/21/2013 08:09 PM »
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.

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) ...

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

Legend didn't mention "intuition" at all, for one thing.  Furthermore, from a grammatical standpoint, the way I read his comment, the meaning of his remark about hi-school math is not an "appeal", it is his anecdotal remembrance of his personal experience.  Sure, he offers the gratuitous side that either his hi-school was better than mine, or else that I shoulda paid better attention, but hey.  Hopefully, his remembrances will be recognized as immaterial to the math part of this thread, and won't be continued.

Back to the math tutoring...

Legend describes a stable system, not a system where energy is being converted into some combination of entropy and forward momentum.  As I see it.
Sometimes I just flat out don't get it.

Offline LegendCJS

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Re: Woodward's Effect - MATH ONLY
« Reply #62 on: 02/21/2013 08:41 PM »
We 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...

« Last Edit: 02/21/2013 08:44 PM by LegendCJS »
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Offline 93143

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Re: Woodward's Effect - MATH ONLY
« Reply #63 on: 02/21/2013 08:52 PM »
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...

You made that up to support your contention.  (Never mind the fact that assuming the rest of the universe is stationary with respect to the two persons is kinda silly, but that's beside the point.)

The scenario you propose isn't at all obvious from the physics (well, it's trivially obvious, but it's not obviously true); please note that we are now dealing not with Newtonian mechanics but with the details of Woodward's theory of inertia.

Note also that inertial effects are independent of velocity, as is Woodward's mass-fluctuation equation...

...

Anyway, now you understand why it matters where the extra mass comes from...

...

As an aside, note that the second mass-fluctuation term in Woodward's equation, while small, is not a zero-average term...
« Last Edit: 02/21/2013 09:15 PM by 93143 »

Online KelvinZero

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Re: Woodward's Effect - MATH ONLY
« Reply #64 on: 02/21/2013 10:07 PM »

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.
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.

Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #65 on: 02/22/2013 07:06 AM »

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.
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.

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.

Now, Lorentz invariance is a very good starting point for a theory.  If you close your eyes and try to build a theory of nature that is maximally symmetric, you might include Lorentz invariance because it says laws of physics are the same at any velocity (even relativistic velocities).

However, the astronimical observations do not appear to be Lorentz invariant.  The entire observable universe (by and large) appears to have an absolute velocity.  If it didn't then we would see galaxies flying in all directions at any old velocity.  We don't.  If we account for the Hubble expansion, galaxies are (pretty much) sitting still with respect to one another.  Why?  Because at some point in the Early Universe, the entire universe (including the CMB and even parts that are not observable) was causally connected.  The Lorentz symmerty was broken on a cosmological scale - a preferred velocity was selected.

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?  Yes, but again the entire universe is assumed to have been causally connected very early on. 

Could there have been some kind of periodic fluctuation back then that leads to periodic mass fluctations now?  I suppose so, but would this not be affecting all inertia all the time?

Offline Celebrimbor

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Re: Woodward's Effect - MATH ONLY
« Reply #66 on: 02/22/2013 07:19 AM »

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


How very dare you.  I have no intention of appealing to anyone! ;)
« Last Edit: 02/22/2013 07:19 AM by Celebrimbor »

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #67 on: 02/22/2013 12:59 PM »
How very dare you. 

That is a wording that I shall immediately adopt.
Sometimes I just flat out don't get it.

Offline 93143

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Re: Woodward's Effect - MATH ONLY
« Reply #68 on: 02/24/2013 07:25 AM »
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.

Uh, that depends strongly on your underlying assumptions...

Like I said, it matters "where" the mass fluctuation comes from/goes...

Regarding Galilean vs. Lorentz, the discussion at that point was exclusively Newtonian, and I felt that dragging Lorentz into it would just muddy the waters.

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

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.

Quote
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.

No, that's not what he does.  Lorentz symmetry is with reference to velocity (and orientation).  Sciama's math shows a preferred acceleration.  Totally different, and completely consistent with GR and Lorentz covariance; in fact it explains something they had to assume (the idea of an "inertial frame").

Sciama's first paper on the subject (1953) is kinda cool.  He starts with a simple vector approximation and derives Newton's laws of motion and gravitation, plus the centrifugal and Coriolis effects, based purely on the mass potential of the observable universe.  He also suggests a method of estimating the cosmic mass density based on the value of the gravitational constant and the age of the universe; more recent calculations in reverse have recovered F=ma within about 8% based on current cosmological data.

Woodward's math, like Sciama's, is all Lorentz invariant.

Quote
Does this require a causal connection to the unobservable universe?

No.  Sciama's calculations were retarded and cut off at the observability horizon Hubble radius (not the same thing, but the former would have been more correct; regardless, the point stands).  Woodward continually refers to the observable universe, not merely the universe, as being the source of inertia.
« Last Edit: 03/07/2013 10:23 PM by 93143 »

Online KelvinZero

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Re: Woodward's Effect - MATH ONLY
« Reply #69 on: 02/24/2013 01:21 PM »
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.

Btw, there is probably also an anthropic reason. If there were no smoothness to the velocities we saw in our vicinity we probably couldnt survive to evolve intelligence and observe them. Even if pockets of smooth velocity only appeared by shear mind boggling coincidence in infinitesimally remote corners of the universe we shouldn't be surprised to observe ourselves in the middle of one.

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #70 on: 03/07/2013 08:38 PM »
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:

Gmu v = Rmu v- 1/2gmu v = -(8 pi G/c^4)tmuv (p.30)

with no explanation getting the reader from calculus to relativity.

By page 35, Woodward points out:

Quote
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? 

Just an update.
« Last Edit: 03/07/2013 08:38 PM by JohnFornaro »
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Offline 93143

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Re: Woodward's Effect - MATH ONLY
« Reply #71 on: 03/07/2013 10:05 PM »
What 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 his simplified math and inaccurate assumptions sink it (he even said it was just a simplified model, and that an improved treatment would follow in another paper).  But the idea is suggestive; 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...
« Last Edit: 03/07/2013 11:44 PM by 93143 »

Offline cuddihy

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Re: Woodward's Effect - MATH ONLY
« Reply #72 on: 03/08/2013 12:24 AM »
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:

Gmu v = Rmu v- 1/2gmu v = -(8 pi G/c^4)tmuv (p.30)

with no explanation getting the reader from calculus to relativity.


What does this mean? Are you looking for the detailed math calcs? Because those are in the papers he included, most notably "Killing Time."

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #73 on: 03/08/2013 01:18 PM »
What 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...

Well, thanks for the grammatical nit, which I agree with.  At some point, accuracy and correctness are related.  If the observations of the universe are faulty and incorrect, one can only go but so far in discovering the physical rules which govern it.

There is a story told that mankind is extremely lucky to have had a moon which orbits Earth very obviously, with phases, eclipses, and so forth.  Perhaps ancient woman would have noticed that the planets moved against the fixed background of the stars, perhaps not.  Point being that this huge nearby celestial object stimulated our early interest in the heavens, and that our scientific advancement would have happened at a much slower pace, if at all, without Luna.

A larger point being that those epicycle models of the solar system were almost made to work, pre-Copernicus.  Wrong assumptions, almost hammered into a coherent theory.

Anyhow...

Quote from: Cuddihy
Are you looking for the detailed math calcs?

Well yeah, because so far, they are not included. I'm only on p.37 or so.  The "backup" material skips an awful lot of the intermediate steps, assuming that those who are already relativistic math initiates are already completely familiar with the material and can already accept Woodward's line of reasoning.

If Woodward's math and subsequent experimental regimen already presented an airtight case for stargates, then it's surprising that the idea is met with so much resistance from the physics and math "establishment".

This leads me to believe that his argument, math, and logic may not be correct.
Sometimes I just flat out don't get it.

Offline JohnFornaro

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Re: Woodward's Effect - MATH ONLY
« Reply #74 on: 03/08/2013 01:22 PM »
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...

That is an accurate reporting of his claim.

I'm sure I speak for a number of interested people in asking you to work out for us the "full relativistic equivalent of his solution", so that we could see how the "magic" is accomplished.

I'll buy ya a beer.
Sometimes I just flat out don't get it.

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