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.

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

..iPhone notation..

"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.

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

Quote from: 93143 on 11/30/2010 07:22 amNo, 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?

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?

Quote from: cuddihy on 12/01/2010 11:52 pmDr 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."

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.

Quote from: Star-Drive on 12/03/2010 09:27 amProvided 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 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.

Quote from: Dr_Fierro on 12/03/2010 11:01 am"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.

"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."

Quote from: Sith on 12/05/2010 11:25 ammlorrey 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.