[meteorite-list] Speed-of-light question

From: Mark Ford <mark.ford_at_meteoritecentral.com>
Date: Wed, 26 Aug 2009 08:54:17 +0100
Message-ID: <29A9DB45B84970458190D7D39BD42C497246C6_at_gamma.ssl.atw>

Reminds me of a question I was asked a while back - what's the average
time dilation of all mass/particles in the universe, due to the
expansion rate of the universe - i.e how much younger is the universe
now than it 'should be' if it was static?

(I had to think about that one!)

I guess technically since time was created at T=0 then the answer is
simply 'now'!?

Mark



-----Original Message-----
From: meteorite-list-bounces at meteoritecentral.com
[mailto:meteorite-list-bounces at meteoritecentral.com] On Behalf Of Rob
Matson
Sent: 26 August 2009 08:28
To: Mexicodoug; Meteorite-list at meteoritecentral.com
Subject: [meteorite-list] Speed-of-light question

Hi All,

Doug was first with the correct answer: 1/sqrt(2) * speed of light
or a little more than 70% of the speed of light. I figured it
might come down to a race between Doug and Sterling. ;-)

Here's an alternative way of looking at the problem which will
give you the correct answer almost immediately. The trick is to
assume that *ALL* objects travel at the same "velocity" in
4-dimensional space-time, and for convenience we'll call this
velocity "c". For simplicity, assume linear motion along just
one spatial axis -- let's just call it the X-axis and make it
horizontal. Now add a perpendicular axis (traditionally the
Y-axis) but instead we're going to call it the T-axis (the
velocity component in the time-axis direction):

  ^
  |
  |
T |
  |
  +--------->
       X

A vector representing the velocity of any object will have a
length of c. Any object traveling at the speed of light (e.g.
a photon) is represented by the vector of length c parallel
to the X-axis; in other words, time stands still for this
object. And any object at rest gets represented by a vector
of length c parallel to the T-axis; all the "motion" is in
the direction of time.

For our problem, we're looking for the vector that has equal
velocity components in both the X-axis and T-axis (X=T).
Obviously this is a 45-degree angle clockwise from +T (or
counterclockwise from +X). So the component of the 4-D velocity
that is in the spatial direction is C*COS(45), while the
component of the 4-D velocity that is in the time direction
is C*SIN(45). Voila!

When you accelerate from a stand-still, your 4D velocity vector
rotates away from vertical and toward horizontal (by a
miniscule amount). Using the simple system above, you can easily
figure out the required velocity in order to cover 2 light-years
distance in one year, 4 light-years in one year, etc.

--Rob

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Received on Wed 26 Aug 2009 03:54:17 AM PDT