[meteorite-list] Speed-of-light question

From: Sterling K. Webb <sterling_k_webb_at_meteoritecentral.com>
Date: Wed, 26 Aug 2009 03:01:10 -0500
Message-ID: <78F97420209945D0AD7691D5BF79C826_at_ATARIENGINE2>

Since our inertial frame is HERE,
the universe is expanding away from
us and WE are standing still, just like
in good ol' medieval cosmology. We are,
by our own definition, a rest frame.
Therefore, no motion = no dialation of
time.

Around these parts, one second is
one second long, bub!

Now, if you're in a distant galaxy, far,
far away, and we're red-shifted --- well,
that's another matter.


Sterling K. Webb
----------------------------------------------------
----- Original Message -----
From: "Mark Ford" <mark.ford at ssl.gb.com>
To: <meteorite-list at meteoritecentral.com>
Sent: Wednesday, August 26, 2009 2:54 AM
Subject: Re: [meteorite-list] Speed-of-light question


>
> 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 04:01:10 AM PDT