Dave,
To me it appears you are making two assumptions, which is OK but should be
qualified
1). Space is empty and not full of energetic particles/cosmic rays that
will penetrate and decay you and ruin your trip
2) Time actually exists and is not really just a rate of decay, ie on Earth
we all decay over 80 or so quantum orbits around the Sun, some of us faster
than others, depending upon vacuum density, which varies because we are not
in a smooth quantum vacuum field in space or on Earth.
Einstein said time was an illusion which I believe to be true.
On Tuesday, November 19, 2013, David Roberson wrote:
> I read the responses and find the answers to have varying degrees of
> relevance. So far we have not followed closely what an outside observer
> measures. At this point, I agree with SR and GR that other observers
> looking at the spaceship will see that it moves at a velocity that does not
> exceed c.
>
> But, the rocket man on board does not have to look outside to figure out
> how much he has accelerated. Unless his computer is defective, he smoothly
> reaches c and will exceed it as he maintains constant acceleration given
> enough time. The beginning of his trip can be experienced by anyone today
> that goes on board a normal rocket. There is no magic in that case. And,
> since velocity is relative, once he reaches say 10% of the speed of light
> according to his calculations, he can stop the engine. This can be
> repeated indefinitely into the future if he wishes and at no point would he
> consider his ship as being different except for having lost mass due to
> exhausting some to generate thrust.
>
> This operation is totally consistent with the rules of SR and GR as far
> as I can determine. The spaceman can even measure his ship's mass by using
> his accelerometer while monitoring the mass of the exhaust leaving his ship
> at a relative velocity of c. He will not see anything unusual about these
> calculations at any speed according to his local reference frame.
>
> The muon calculations support this situation quite well when we choose a
> viewpoint riding along with the particle. Again, other external observers
> moving at different velocities than us do not agree with our muon
> accessment. But, none of them agree with each other either so that is not
> surprising. :-) Recall the galactic red shift? Some of the far off
> observers would see our particle as moving slower than light or perhaps
> zero relative to their velocity. Many of these guys would think that the
> spaceship was moving away from them when it began its acceleration and
> after enough time actually come to rest according to their observations.
>
> Perhaps we should look at the rocket ship from other perspectives soon.
>
> Dave
>
>
> -----Original Message-----
> From: Eric Walker <eric.wal...@gmail.com <javascript:_e({}, 'cvml',
> 'eric.wal...@gmail.com');>>
> To: vortex-l <vortex-l@eskimo.com <javascript:_e({}, 'cvml',
> 'vortex-l@eskimo.com');>>
> Sent: Tue, Nov 19, 2013 10:09 am
> Subject: Re: [Vo]:Local Calculated Velocity of Space Ship
>
> Dave, I've asked our question on phsyics.stackexchange.com -- here's
> what has come back so far:
>
> http://physics.stackexchange.com/q/87047/6713
>
> On Mon, Nov 18, 2013 at 10:44 PM, David Roberson
> <dlrober...@aol.com<javascript:_e({}, 'cvml', 'dlrober...@aol.com');>
> > wrote:
>
> As I have mentioned on occasions, I see plenty of evidence that both
>> forms of relativity are strongly supported by the behavior of such machines
>> as the LHC.
>>
>
> See @dmckee's comment to Suzan Cioc's answer.
>
>
>> One of the implications of SR is that each observer should experience his
>> own local time and motions as being completely normal regardless of any
>> relative motion with respect to other observers.
>>
>
> I believe this only applies when no acceleration is involved. Once one
> of the parties steps on the gas pedal, you have a situation where "symmetry
> is broken," and the considerations change.
>
> Eric
>
>
