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From: Tony M on 23 Apr 2010 14:47
On Apr 23, 1:44 pm, PD <thedraperfam...(a)gmail.com> wrote:
> On Apr 23, 12:23 pm, Tony M <marc...(a)gmail.com> wrote:
>
> > As PD above, please consider c = 3E8 m/s.
>
> And since I did all those calculations, what more do you need?

I'll tell you later. Thank you for taking the time to do that.
From: xxein on 23 Apr 2010 22:27
On Apr 23, 8:49 am, Tony M <marc...(a)gmail.com> wrote:
> A spaceship with a mass of 1000kg is initially at rest relative to an
> external observer. The spaceship emits one photon with an energy of
> 1kg x c^2 as measured by the observer. We define a closed system
> including the spaceship and the photon. Please provide the values for:
> -invariant mass of the system
> -total energy of the system
> -velocity of the center of mass, and momentum of the system as
> measured by the observer
> -velocity of spaceship as measured by the observer
> -rest mass of the spaceship (kg, minimum 4 decimals)

xxein: To start with, there is nothing but a relative mass for an at
rest observer wrt the spaceship. There is no invariant mass because
mass is defined as m=c^2/E. If c is a constant in physics for all
frames, then why is there a Doppler effect?

Total energy is also measured relativistically for the same reason.

What velocity of the center of mass? You said that the observer is at
rest wrt it.

What velocity can a comoving observer measure? 0 v.

Rest mass of the spaceship? 1000 kg (relativistically). The same as
all the garbage you smoked to write this.

From: BURT on 24 Apr 2010 00:45
On Apr 23, 7:27 pm, xxein <xx...(a)comcast.net> wrote:
> On Apr 23, 8:49 am, Tony M <marc...(a)gmail.com> wrote:
>
> > A spaceship with a mass of 1000kg is initially at rest relative to an
> > external observer. The spaceship emits one photon with an energy of
> > 1kg x c^2 as measured by the observer. We define a closed system
> > including the spaceship and the photon. Please provide the values for:
> > -invariant mass of the system
> > -total energy of the system
> > -velocity of the center of mass, and momentum of the system as
> > measured by the observer
> > -velocity of spaceship as measured by the observer
> > -rest mass of the spaceship (kg, minimum 4 decimals)
>
> xxein:  To start with, there is nothing but a relative mass for an at
> rest observer wrt the spaceship.  There is no invariant mass because
> mass is defined as m=c^2/E.  If c is a constant in physics for all
> frames, then why is there a Doppler effect?
>
> Total energy is also measured relativistically for the same reason.
>
> What velocity of the center of mass?  You said that the observer is at
> rest wrt it.
>
> What velocity can a comoving observer measure? 0 v.
>
> Rest mass of the spaceship?  1000 kg (relativistically).  The same as
> all the garbage you smoked to write this.


Energy of motion is in an absolute space frame. The math is Gamma for
energy or mass speed through space.

Mitch Raemsch
From: Tony M on 24 Apr 2010 12:42
On Apr 23, 2:47 pm, Tony M <marc...(a)gmail.com> wrote:
> On Apr 23, 1:44 pm, PD <thedraperfam...(a)gmail.com> wrote:
>
> > On Apr 23, 12:23 pm, Tony M <marc...(a)gmail.com> wrote:
>
> > > As PD above, please consider c = 3E8 m/s.
>
> > And since I did all those calculations, what more do you need?
>
> I'll tell you later. Thank you for taking the time to do that.


That's it, no more takers?!

I was hoping that everyone would have an opinion, as usual. People
debating the Theory of Relativity should find this exercise trivial;
it should take you little time and effort to post the answer. For me
this was high-school physics, I would rate its difficulty between
clueless and amateur.
And the whole point of it was to get numerical answers. As you can all
see, when discussing principals, notions and theory we get as many
opinions and interpretations as we have writers. Numbers are either
right or wrong, they add up or they don't. I think they leave little
room for interpretation and BS.
This exercise was intended first as a survey, to test the consistency
of the answers, and second, I would have considered it as a form of
introduction (like stating ones name), so me and others, would know
whom we're talking to.
I have tried to make the exercise as clear as I could at the time, so
that most people would understand what's given and what's asked for. I
think it's clear enough and whoever doesn't get it would likely not be
able to provide the answers either. So, like they say at weddings,
“speak now or forever hold you peace”!
From: Tony M on 24 Apr 2010 12:46
On Apr 23, 2:47 pm, Tony M <marc...(a)gmail.com> wrote:
> On Apr 23, 1:44 pm, PD <thedraperfam...(a)gmail.com> wrote:
>
> > On Apr 23, 12:23 pm, Tony M <marc...(a)gmail.com> wrote:
>
> > > As PD above, please consider c = 3E8 m/s.
>
> > And since I did all those calculations, what more do you need?
>
> I'll tell you later. Thank you for taking the time to do that.

That's it, no more takers?!

I was hoping that everyone would have an opinion, as usual. People
debating the Theory of Relativity should find this exercise trivial;
it should take you little time and effort to post the answer. For me
this was high-school physics, I would rate its difficulty between
clueless and amateur.

And the whole point of it was to get numerical answers. As you can all
see, when discussing principals, notions and theory we get as many
opinions and interpretations as we have writers. Numbers are either
right or wrong, they add up or they don't. I think they leave little
room for interpretation and BS.

This exercise was intended first as a survey, to test the consistency
of the answers, and second, I would have considered it as a form of
introduction (like stating ones name), so me and others, would know
whom we're talking to.

I have tried to make the exercise as clear as I could at the time, so
that most people would understand what's given and what's asked for. I
think it's clear enough and whoever doesn't get it would likely not be
able to provide the answers either. So, like they say at weddings,
“speak now or forever hold your peace”!
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