Re: question about photons
- From: The Ghost In The Machine <ewill@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 23 Apr 2005 20:00:05 GMT
In sci.physics, luckyvomit@xxxxxxxxx
<luckyvomit@xxxxxxxxx>
wrote
on 22 Apr 2005 18:33:11 -0700
<1114219991.597289.60410@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
> The current age of the universe is estimated to be 14 billion years.
> How old does the universe 'appear' to a photon of light
>
> thanks
>
Photons are weird beasties. Look at the Lorentz:
x_A = (x_O - v * t_O) / sqrt(1-v^2/c^2)
t_A = (t_O - v * x_O / c^2) / sqrt(1-v^2/c^2)
where v is the velocity along the axis of travel, x is the
position along the axis of travel.
The inverse Lorentz might also be useful:
x_O = (x_A + v * t_A) / sqrt(1-v^2/c^2)
t_O = (t_A + v * x_A / c^2) / sqrt(1-v^2/c^2)
Now, obviously, if v = c, then gamma = 1/sqrt(1-v^2/c^2)
becomes infinite. This causes all sorts of mathematical problems.
The best I can do here is get near c, and then take limits.
If we take O as the observer and A as the photon, we can
try to figure out what t_A is doing, and the best one
can get is to come up with a relationship between
x_A and t_A:
t_A / x_A = (t_O - v * x_O / c^2) / (x_O - v * t_O)
At the limit v = c:
t_A / x_A = (t_O - x_O / c) / (x_O - c * t_O) = 1/c.
However, it gets even weirder. For a photon, at least
as far as we can tell, x_O = c * t_O so t_A / x_A
also becomes the form 0/0.
I have no idea how to properly interpret this result.
--
#191, ewill3@xxxxxxxxxxxxx
It's still legal to go .sigless.
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