Propellantless Field Propulsion and application

[Dr_Fierro]

"Paul,

Your correspondent is looking for a flaw in the argument as most would: in this case, taking an approximation which might suppress something that leads to a cancelation of the effect that is suspected.  He's right, of course, that in general the derivative of gamma is not zero.  It is easy to calculate.  You get gamma cubed times the dot product of the three velocity and three acceleration divided by c^2.  If the three velocity is not equal to zero (that is, if you are in some frame other than that of instantaneous rest of the test particle in this case), then dgamma/dt will likely be non-zero.  But in the frame of instantaneous rest, the three velocity is zero -- and so too is dgamma/dt.

If you stop and think about this for a few moments, it should be plain that dm0/dt =/= 0 is a real physical effect.  You should not be able to transform it away by choice of particular coordinates.  Especially since the Lorentz transformations do not depend on acceleration.  There is no fundamental error in Equation (A4).

Best,

Jim"

Star-Drive:

Jim's reply to SiriusGrey's objection is indeed correct, since his derivation in "Flux Caps & Origin of Inertia -
Appendix A" proceeds in the particle's rest frame. However, it seems the derivation is fatally flawed in that the
external four-force F is taken as the time rate change of the particle four-momentum (without the constant rest
mass constraint), instead of the correct expression F = (rest mass)*(four-acceleration), which is valid irrespective
of the rest mass being constant or variable. Using the latter expression, there is nothing left for obtaining the
transient mass terms, unless I am missing further subtleties of Jim's work.

[Sith]

No, you don't.  For atmospheric travel you'd want a simple application of thrust, rather than trying to create negative mass, and M-E thrusters (if they work, and work as well as March et al. hope) would be perfectly suited for that.  No need for conventional engines.

Let's assume that there is a zero or negative mass propulsion drive made by us several decades from now. Let's also assume that the craft travels in the atmosphere accidentally. What shield will be used against air resistance?

[cuddihy]

I would suspect Aluminum+ the intelligence not to reenter at high relative speeds higher than Mach 3 or so

[cuddihy]

However, it seems the derivation is fatally flawed in that the
external four-force F is taken as the time rate change of the particle four-momentum (without the constant rest
mass constraint), instead of the correct expression F = (rest mass)*(four-acceleration), which is valid irrespective
of the rest mass being constant or variable.

It's great to see some physics red meat on this thread after a desert of Months perhaps longer.

Dr Fierro, can you explain more why F=m0•a(4) =/= F=dp/dt,
especially in the context of not assuming at the start mass fluctuations are impossible prima fascia?

(pls excuse the clunky iPhone notation)

[JohnFornaro]

..iPhone notation..

Kinda what we used in hi-school.  What the problem is?   Oh.  Style.

[Star-Drive]

"Paul,

Your correspondent is looking for a flaw in the argument as most would: in this case, taking an approximation which might suppress something that leads to a cancelation of the effect that is suspected.  He's right, of course, that in general the derivative of gamma is not zero.  It is easy to calculate.  You get gamma cubed times the dot product of the three velocity and three acceleration divided by c^2.  If the three velocity is not equal to zero (that is, if you are in some frame other than that of instantaneous rest of the test particle in this case), then dgamma/dt will likely be non-zero.  But in the frame of instantaneous rest, the three velocity is zero -- and so too is dgamma/dt.

If you stop and think about this for a few moments, it should be plain that dm0/dt =/= 0 is a real physical effect.  You should not be able to transform it away by choice of particular coordinates.  Especially since the Lorentz transformations do not depend on acceleration.  There is no fundamental error in Equation (A4).

Best,

Jim"

Star-Drive:

Jim's reply to SiriusGrey's objection is indeed correct, since his derivation in "Flux Caps & Origin of Inertia - Appendix A" proceeds in the particle's rest frame. However, it seems the derivation is fatally flawed in that the external four-force F is taken as the time rate change of the particle four-momentum (without the constant rest mass constraint), instead of the correct expression F = (rest mass)*(four-acceleration), which is valid irrespective of the rest mass being constant or variable. Using the latter expression, there is nothing left for obtaining the transient mass terms, unless I am missing further subtleties of Jim's work.

Dr_Fierro:

Dr. Woodward's response is below and attached:

"Paul,

Well, if the assertion about the four force were correct, there wouldn't be any predicted effects.  But it is not correct.  It seems to me obvious that the restmass of an object can be changing, and that this must have dynamical consequences.  But perhaps I have lived with this too long.  If the correct expression for the four force is the derivative with respect to proper time of the four momentum -- as it is -- then transient fluctuations in the rest masses of things do have dynamical consequences.

In any event, to counter the latest assetion, I long ago included in presentations of the derivation the words of Wolfgang Rindler in the second edition of his text on special relativity.  I attach the PPT slide of his words.  I don't think Mach effects will be so easily dismissed as by the claim that the definition of the four force doesn't allow them.  :-)

Best,

Jim"


From StarDrive:

BTW, you also have to remember that Dr. Woodward, his graduate student Tom Mahood, and myself have all experimentally demonstrated the existence of the effects of these mass fluctuations to one degree or another.  They are real and engineerable.  It's about time we got on with the task of doing so.

Paul March
Friendswood, TX

Update:

"Rindler's point is that the restmasses of objects in collision transiently change.  And when that happens dm0/dt is not zero and the four force has a non-vanishing time-like part.  That's where Mach effects come from.

Best,

Jim"

[Cinder]

I ask with no implied meaning: If the effects are proven real, why does Woodward (IIRC) refuse funding?

[Star-Drive]

I ask with no implied meaning: If the effects are proven real, why does Woodward (IIRC) refuse funding?

Cinder:

First off, what is satisfactory proof of principle data for me, may not be satisfactory proof of principle data for you, academics or investors.  We all have our own and sometimes very different tolerance levels for uncertainty.  Secondly, because Dr. Woodward is now in his late sixties fighting lung cancer, he doesn't want to deal with the complicating issues surrounding accepting investment funding.  There are always strings attached to such funding and Jim doesn't want to deal with those restrictions on how he performs his research or what he does with the time that is left to him on Earth.

[aceshigh]

No, you don't.  For atmospheric travel you'd want a simple application of thrust, rather than trying to create negative mass, and M-E thrusters (if they work, and work as well as March et al. hope) would be perfectly suited for that.  No need for conventional engines.

Let's assume that there is a zero or negative mass propulsion drive made by us several decades from now. Let's also assume that the craft travels in the atmosphere accidentally. What shield will be used against air resistance?

wrong question. I rather ask what "shield" can Earth use against a relativistic spacecraft impact.

in fact, if we are going to have craft travelling at relativistic speeds, we better also develop faster than light craft at once, so we can have probes to detect relativistic objects and warn us IN TIME to coordinate defenses.

[Cinder]

ME thrusters would change the picture by so much, that such a question as above wouldn't be asked on the premise of today's state of things.  E.G.  With ME tech the world can probably afford satellites easily enough that such a threat as relativistic projectiles could have some network of sentries dedicated to detecting them and coordinating interception, even if just by commanding some structure to move itself into the relativistic projectile's path to mitigate damage down on Earth.

[Dr_Fierro]

Dr Ferrer, can you explain more why F=m0•a(4) =/= F=dp/dt,
especially in the context of not assuming at the start mass fluctuations are impossible prima fascia?

cuddihy:

By using F=m0*dV/dt, where t stands for the proper time and V for the four-velocity, Eq. (A4)
of Woodward's paper becomes: F=-(m0*dV(0)/dt, f). Now, when dividing by m0, the new Eq. (A6)
is (F/m0)=-(dV(0)/dt, f/m0). Remember that in the rest frame dV(0)/dt=0; anyway, by taking the
four-divergence the new Eq. (A9) becomes: -(1/c)*d˛V(0)/dt˛-div(f/m0)=4*pi*G*rho0. Besides
some technicalities you can see that no time derivatives of the rest mass appears in the new
Eq. (A9), leaving no room for transient source terms as proposed by Woodward.
Regards.

In the meantime, Jim has confirmed these results, albeit not accepting F=m0*dV/dt as the correct
definition of the four-force.

P.S.: The thread's dynamics is a little bit too fast for my taste and my allowable spare time.
I will do my best to keep with your pace.

[Star-Drive]

Dr Ferrer, can you explain more why F=m0•a(4) =/= F=dp/dt,
especially in the context of not assuming at the start mass fluctuations are impossible prima fascia?

cuddihy:

By using F=m0*dV/dt, where t stands for the proper time and V for the four-velocity, Eq. (A4) of Woodward's paper becomes: F=-(m0*dV(0)/dt, f). Now, when dividing by m0, the new Eq. (A6) is (F/m0)=-(dV(0)/dt, f/m0).  Remember that in the rest frame dV(0)/dt=0; anyway, by taking the four-divergence the new Eq. (A9) becomes: -(1/c)*d˛V(0)/dt˛-div(f/m0)=4*pi*G*rho0.  Besides some technicalities you can see that no time derivatives of the rest mass appears in the new Eq. (A9), leaving no room for transient source terms as proposed by Woodward.

Regards.

In the meantime, Jim has confirmed these results, albeit not accepting F=m0*dV/dt as the correct definition of the four-force.

P.S.: The thread's dynamics is a little bit too fast for my taste and my allowable spare time.  I will do my best to keep with your pace.

Provided below is Dr. Woodward's latest reponse to Dr. Fierro's above comments to cuddyhi.  There appears to be some sort of misunderstanding developing here surrounding the definition of dv/dt in an instantaneous rest frame having to be zero, or not.  Fierro indicates it has to be zero apparently by definition, but Rindler and Woodward say no it does not have to be zero.  Hmmm...


"Sorry, not only is his definition of the four force wrong, he claims that dv/dt in the instantaneous rest frame is zero.  That is simply wrong.  This person has decided that the derivation must be wrong, and is making stuff up to get that result.  That is not good physics.  And the definition of the four force is the derivative with respect to proper time of the four momentum."


BTW, here is Wolfgang Rindler's Bio:

http://en.wikipedia.org/wiki/Wolfgang_Rindler
http://en.wikipedia.org/wiki/Rindler_coordinates

[Dr_Fierro]

Provided below is Dr. Woodward's latest reponse to Dr. Fierro's above comments to cuddyhi.  There appears to be some sort of misunderstanding developing here surrounding the definition of dv/dt in an instantaneous rest frame having to be zero, or not.  Fierro indicates it has to be zero apparently by definition, but Rindler and Woodward say no it does not have to be zero.  Hmmm...


"Sorry, not only is his definition of the four force wrong, he claims that dv/dt in the instantaneous rest frame is zero.  That is simply wrong.  This person has decided that the derivation must be wrong, and is making stuff up to get that result.  That is not good physics.  And the definition of the four force is the derivative with respect to proper time of the four momentum."

To all:

I am sorry for not being clear enough about notation. V(0) stands for the timelike component of the four-velocity, i.e., V(0)=c/alpha, with alpha=sqrt(1-v˛/c˛) right?. dV(0)/dt is thus the timelike component of the four acceleration. As Dr. Woodward already demonstrated in some earlier post, in the particle's rest frame that component is exactly equal to zero.

When time allows, I will post my (and others') objections to Jim's preferred definition of the four-force.

[mlorrey]

ME thrusters would change the picture by so much, that such a question as above wouldn't be asked on the premise of today's state of things.  E.G.  With ME tech the world can probably afford satellites easily enough that such a threat as relativistic projectiles could have some network of sentries dedicated to detecting them and coordinating interception, even if just by commanding some structure to move itself into the relativistic projectile's path to mitigate damage down on Earth.

Sith's question isn't about shields against relativistic impactors, but shields for the ship against what he sees as air resistance being a problem. A replacement for a solid TPS.

The misunderstanding of course is the idea that you need to go particularly fast in atmosphere with ME thrusters. The fact is you don't, any more than a space elevator needs to go hypersonic. An ME thruster driven vehicle can ascend through the atmosphere at even subsonic velocities, maintain that speed through the atmosphere, out of LEO regions, and directly into escape trajectories (because, after all, even an escape velocity of 25,000 mph on earth or in LEO will deplete to near 0 at the top of the gravity well due to gravity losses, what matters is the total energy applied). It would be less efficient to do it this way, but it totally eliminates the need for hypersonic TPS materials, so overall your vehicle will be lighter.

If you are running, say, a 1.1 meter diameter superconducting polywell as your power generator with a minimalist shadow shield , out of depleted uranium and your hydrogen tank, you could power a rather minimal sized vessel, basically a pressurized cabin for the needs of the crew, the cargo compartment, the hydrogen coolant/fuel tank, the thrusters, and the fusion generator and radiator wings/fins/landing struts. This is the point at which you get to start entertaining science fictiony ship designs.

Stuff even Bussard didn't consider.

[Cinder]

That's how I read Sith too, the first time I read it and before Aceshigh took a different perspective, and I agree.  I think it's more plausible that the early days of METs would look more like blimps without the blimps' massive volume, rather than everything suddenly being made hyper-sonic.

I mean.. If you've got METs, wouldn't that enable station keeping of sorts of a space elevator if you stuck METs along the ribbon?  Basically a vertical railroad.  Anyway, literally the whole picture of transportation changes.  Orbital/space access becomes just a subset.

[Sith]

mlorrey interpreted me right. As you all know air is a fluid with it's own viscosity. I was talking about drag at high velocity. How to avoid it, etc. Maybe we need more than just shape geometry. Maybe we need some advanced laser cutter that would work as an air spike.

[Star-Drive]

Provided below is Dr. Woodward's latest reponse to Dr. Fierro's above comments to cuddyhi.  There appears to be some sort of misunderstanding developing here surrounding the definition of dv/dt in an instantaneous rest frame having to be zero, or not.  Fierro indicates it has to be zero apparently by definition, but Rindler and Woodward say no it does not have to be zero.  Hmmm...


"Sorry, not only is his definition of the four force wrong, he claims that dv/dt in the instantaneous rest frame is zero.  That is simply wrong.  This person has decided that the derivation must be wrong, and is making stuff up to get that result.  That is not good physics.  And the definition of the four force is the derivative with respect to proper time of the four momentum."

To all:

I am sorry for not being clear enough about notation. V(0) stands for the timelike component of the four-velocity, i.e., V(0)=c/alpha, with alpha=sqrt(1-v˛/c˛) right?. dV(0)/dt is thus the timelike component of the four acceleration. As Dr. Woodward already demonstrated in some earlier post, in the particle's rest frame that component is exactly equal to zero.

When time allows, I will post my (and others') objections to Jim's preferred definition of the four-force.

Dr. Fierro:

While you and your associates are considering your response to Dr. Woodward’s preferred definition of the four-force, you might also review a book review that Dr. Woodward did for a Brian Greene book entitled “The Fabric of the Cosmos” that covers some of Dr. Woodward positions on 4D spacetime analysis.  See: http://en.wikipedia.org/wiki/Brian_Greene

I’m also including an excerpt from Brian Greene’s book on how one is to calculate the radius of the causally connected universe for your reference.  In the meantime below is Dr. Woodward latest comments on this forum thread topic.


“Paul,

When one talks about frames of reference in mechanics generally, unless otherwise specified, it is assumed that an inertial frame of reference is being talked about -- especially when one specifies that the velocity of some object therein is taken to be zero (think Newton's first law).  Instantaneous rest frames are routinely used to deal with acceleration in special relativity for precisely the reasons I used them in the derivation.  Folks not too familiar with special relativity, I suppose, can easily become confused by this technical "trick".  But those with more than a nodding familiarity with SRT should know immediately what is being talked about.  In the now 20 years since Mach effects were first raised, and 15 years since I got the derivation sorted out correctly, only once has any physicist raised issues about this aspect of the derivation.  The one who did, a world class general relativist (hired by the ORBs to try to discredit me), when I pointed out that d(cm)/dt = cdm/dt =/= 0 in general, immediately dropped his objection.  In his case, it was an inadvertent oversight.  In the present cases, I hope that is true too.  For if it isn't, it means you are dealing with people who don't know what they are talking about.

Best,

Jim

P.S.  The relativist's critique, which he sent me before proceeding, had a very elegant derivation of an interesting point.  I encouraged him to publish it.  He did so a year or two later.  :-)  It really was quite elegant."

[Cinder]

mlorrey interpreted me right. As you all know air is a fluid with it's own viscosity. I was talking about drag at high velocity. How to avoid it, etc. Maybe we need more than just shape geometry. Maybe we need some advanced laser cutter that would work as an air spike.

I don't see why METs would force anything to fly faster.  If anything, they'd allow substantial savings thanks to all the low speed lift they enable due to not completely depending on aerodynamic lift.  High speeds not being a necessary consequence of ME propulsion, the aerodynamics involved are the same "ordinary" aerodynamic concerns already worked out up to date.

[cuddihy]

"Sorry, not only is his definition of the four force wrong, he claims that dv/dt in the instantaneous rest frame is zero.  That is simply wrong.  This person has decided that the derivation must be wrong, and is making stuff up to get that result.  That is not good physics.  And the definition of the four force is the derivative with respect to proper time of the four momentum."

<snippy snip>

To all:

I am sorry for not being clear enough about notation. V(0) stands for the timelike component of the four-velocity, i.e., V(0)=c/alpha, with alpha=sqrt(1-v˛/c˛) right?. dV(0)/dt is thus the timelike component of the four acceleration. As Dr. Woodward already demonstrated in some earlier post, in the particle's rest frame that component is exactly equal to zero.

<snip>

“Paul,

When one talks about frames of reference in mechanics generally, unless otherwise specified, it is assumed that an inertial frame of reference is being talked about
[/quote]

Hmm...are we talking past each other? This seems like a simpler objection than that to me.

Dr. Fierro says: F=dP/dt is wrong, but that F= Mo*dV/dt is correct. This seemed an extraordinary statement to me because they ought to be identical statements, so I was looking for an explanation.

I was thinking if you start from Mo*V (the instantaneous definition of momentum) rather than than P (momentum), and then you differentiate WRT time to get F, you have started with the assumption that Mo does not change. Of course that eliminates any change in Mo. You've pulled it out of the integral! It seemed circular to me.

So I was hoping Dr. Fierro would explain why it is wrong to start from P rather than Mo*V, but perhaps I am not asking clearly because he appears to actually be starting, prima facie, from the assumption that V is the proper place to start rather than P.

Obviously I'm no physicist, so I was hoping if the choice of P or V could be explained, I would understand the objection better as well as get a better insight into Woodward's derivation...

[mlorrey]

mlorrey interpreted me right. As you all know air is a fluid with it's own viscosity. I was talking about drag at high velocity. How to avoid it, etc. Maybe we need more than just shape geometry. Maybe we need some advanced laser cutter that would work as an air spike.

I don't see why METs would force anything to fly faster.  If anything, they'd allow substantial savings thanks to all the low speed lift they enable due to not completely depending on aerodynamic lift.  High speeds not being a necessary consequence of ME propulsion, the aerodynamics involved are the same "ordinary" aerodynamic concerns already worked out up to date.

Quite right, as I detailed in my response to Sith. There is no need for hypersonic speeds in atmosphere for either launch or reentry once you have ME thrusters of sufficient performance to achieve T/W of >1. You could essentially turn a normal airliner into a lunar passenger ship, replacing the turbine engines with ME thrusters and sticking a nuke (fission or polywell fusion, either or) in the cargo bay.