David--

The guy need only account for the loss of mass energy by adding the amount of 
energy transferred to the negative energy sea. 

Of course, if he does not consider a negative energy sea exists, he cannot 
properly account.  He is stuck with an observation that makes no sense to him. 

His reaction less drive converted what was originally linear momentum of real 
particles to the intrinsic property of angular momentum energy,  which he does  
not accounted for in measuring the the rest mass of real particles.  The rest 
mass of his ship has decreased from his counting of particles, the angular 
momentum of the universe has been transferred to the negative sea--the Dirac 
sea.  

Bob
  ----- Original Message ----- 
  From: David Roberson 
  To: [email protected] 
  Sent: Monday, November 24, 2014 8:23 AM
  Subject: Re: [Vo]:They call me a moron. A reply.


  Yes, he can determine that he has changed velocity by looking outside the 
ship at other objects.  That is why I proposed the recent posting where he 
returns to the original location and velocity.  That procedure counters the 
thought that a final velocity change can obscure any problems due to usage of 
the reactionless drive.  Special Relativity is generally considered capable of 
countering the natural feeling that a particular velocity is important in 
space, but with zero velocity change there is no need to play that card.

  The guy must reconcile where the mass of his ship has gone after using the 
reactionless drive.

  Dave







  -----Original Message-----
  From: Bob Cook <[email protected]>
  To: vortex-l <[email protected]>
  Sent: Mon, Nov 24, 2014 10:38 am
  Subject: Re: [Vo]:They call me a moron. A reply.



  David--

  You stated:

  <<<After the drive is shut down the ship stops accelerating and comes to rest 
in space. Even though the new velocity is different than the old one before the 
drive operates, a guy onboard the ship can not determine that he is moving. >>>

  Yes he can determine he is moving.  All he needs to do is look out the window 
and see that he  is moving relative to objects that were fixed before he 
started his travel and are assumed to have remained fixed.  

  Bob
    ----- Original Message ----- 
    From: David Roberson 
    To: [email protected] 
    Sent: Sunday, November 23, 2014 9:21 PM
    Subject: Re: [Vo]:They call me a moron. A reply.


    The fact that energy can be extracted from the battery to drive the 
microwave source is certainly possible.  No one will ague against that point.  
The problem is that this energy can be depleted without having anything to show 
for its loss.  If taken to the extreme most of the ship can be converted into 
energy by some nuclear process to supply power for the drive mechanism.

    After the drive is shut down the ship stops accelerating and comes to rest 
in space.  Even though the new velocity is different than the old one before 
the drive operates, a guy onboard the ship can not determine that he is moving. 
 He will not have any kinetic energy relative to himself.  He sees that his 
ships mass has depleted but has nothing to show where it went.  With a normal 
drive the guy can see the exhaust that is moving relative to him which contains 
all of the converted energy.

    Dave







    -----Original Message-----
    From: Eric Walker <[email protected]>
    To: vortex-l <[email protected]>
    Sent: Mon, Nov 24, 2014 12:02 am
    Subject: Re: [Vo]:They call me a moron. A reply.


    On Sun, Nov 23, 2014 at 8:26 PM, David Roberson <[email protected]> wrote:


      I encourage anyone out there with knowledge about how to overcome the 
obvious problems to offer their input.


    One thought here -- the "reactionless drive" that I am aware of being in 
the recent news is the EmDrive.  That one involves the generation of microwaves 
and their reflection in a cavity.  It's not clear whether anyone other than 
Nasa and the inventor believe that it works as advertised.  But if it does, 
note that energy must be expended to generate the microwaves, e.g., by a 
battery, to which the usual E=mc^2 conversion will apply.


    Eric

Reply via email to