🚀 go-pugleaf

In article <32892F...@extro.ucc.su.oz.au>,
JeffBD  <jbd...@extro.ucc.su.oz.au> wrote:
>Jon A. Maxwell wrote:
>> 
>> Scott Hess wrote:
>> 
>> ] You _cannot_ go faster than the speed of light while remaining
>> ] in our universe.  Almost by definition, if you are going faster
>> ] than the speed light travels in our universe, then you are
>> ] _not_ in our universe.
>> 
>> Scott, just curious if you have two bodies coming at each other
>> really fast, why will gravity not accelerate them to the speed of
>> light?  Is it because the faster they get the less mass they
>> have, so less gravity?  What if one or more of the bodies has
>> mass enough to be a black hole?

The faster something goes (relative to some observer), the more mass
it will have (relative to the same observer).  The object itself (be
it a space ship with a person inside, or a computer run drone) will
not detect any mass change in itself because it is stationary relative
to itself (but it will see a mass gain in the observer mentioned
before)... cool, eh?

>> 
>> ] The problem is that the amount of acceleration to add a given
>> ] amount of velocity increases exponentially to infinity.
>> 
>> Doesn't gravity also increase in an exponential manner?
>> 
>> --
>>      thur  Mail Address: LordA...@vt.edu or jmax...@vt.edu
>>   n  r
>>   a JAMax "Not even SATAN himself can find anything wrong with
>>   h o   w  my Ultrix setup.  Ultrix: A mature operating system."
>>   tan lle                     --JAM
>
>Two bodies are both moving at 3/4 the speed of light towards each other.
>Doesn't this mean that relative to one of the bodies, the other body is
>travelling at 1.5 times the speed of light? If this is possible, doesn't
>this apparently violate the so-called constant of C? If not, why not? 
>

No... 
I heard once upon a time this was called the doppler affect (for
relativistic mechanics, not sound), but someone told me I was
incorrect, so I'm not sure what it's called.  Anyway...

I believe the formula is
v = (v1 + v2)/(1 + v1*v2)
(v1 and v2 are ratios of the speed of light.. like .75, not absolute
numbers like 225000000 m/s)
you said they were both going 3/4 the speed of light, so that means
v1 = v2 = .75
v = (.75 + .75)/(1 + .75*.75)
  = 1.5/(1 + .5625) = .96c

So one body would see the other body moving at .96c
BTW, you'll notice if you plug in v1 = v2 = 1 (both velocities are the
speed of light), the formula spits out 1, meaning if you're in a car
going the speed of light (yeah, magic), and you turn on your headlights,
the beam would shine away from you at the speed of light.  Also, if you
turned on your taillights, the beam would again shine away from you at
the speed of light.. nifty, eh? :)

For a more thorough explanation of how this works and why, go read
sci.physics.relativity.  You might also check rec.arts.startrek.tech,
where this discussion is thrown around frequently.  FAQ's are posted
there... mind you, this is *NOT* light reading (no pun intended).

-- 
Glenn Lamb - mum...@netcom.com -- Finger mum...@netcom.com for PGP Key.
PGPprint = E3 0F DE CC 94 72 D1 1A  2D 2E A9 08 6B A0 CD 82
Reply address set to mum...@no-spam.netcom.com to fudge USENET scanners.
Be sure to send replies to mum...@netcom.com