Re: Does photons really travel?
- From: The Ghost In The Machine <ewill@xxxxxxxxxxxxxxxxxxxxxxx>
- Date: Fri, 9 May 2008 08:16:16 -0700
In sci.physics.relativity, Smooth John
<yoshioory@xxxxxxxxxx>
wrote
on Tue, 6 May 2008 14:53:18 -0700 (PDT)
<703f99c7-720b-42d5-b446-12edf9fca3aa@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
On May 6, 6:19 am, The Ghost In The Machine
<ew...@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
In sci.physics.relativity, Smooth John
<yoshio...@xxxxxxxxxx>
wrote
on Mon, 5 May 2008 12:40:55 -0700 (PDT)
<5d7b2167-a326-4860-abbc-c4a68a1d5...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
From a photon point of view, it never travel. For it, now is all the
time.
Debatable. Let's assume that a photon starts at (x,t) = (0,0),
from the still observer's point of view. Since we can't quite use the
Lorentz we need to get creative regarding limits, but we do know
that
x' = (x-vt)/sqrt(1-v^2/c^2)
Why (x-vt) and not (x+vt), explain
(x,t) is world coordinates; (x',t') are coordinates in the photon's
system. This isn't exactly accurate (as a photon is an expanding
sphere) but if we assume a point particle moving along the positive X
axis, at 1 second (x,t) = (c,1) and (x',t') = (0,?).
What is your location, the light departs from you
or arrives to you?
I'm surfing the light wave, dude. ;-)
t' = (t-vx/c^2)/sqrt(1-v^2/c^2)
similar here
for any massive particle.
Since for a photon x^2 = c^2t^2, we can substitute:
x'^2 = (x-vt)^2 / (1 - v^2/c^2)
= (x^2 - v^2t^2 - 2xvt) / (1 - v^2/c^2)
= (c^2t^2 - v^2t^2 - 2xvt) / (1 - v^2/c^2)
t'^2 = (t-vx/c^2)^2 / (1 - v^2/c^2)
= (t^2 - v^2x^2/c^4 - 2xv/c^2) / (1 - v^2/c^2)
= (t^2 - v^2t^2/c^2 - 2xv/c^2) / (1 - v^2/c^2)
= x'^2/c^2
Last time I checked
(x-vt)^2 was (x^2 + v^2t^2 - 2xvt)
Why you do mistakes?
Why do you? This one is indeed an error; the equations
should be
x'^2 = (x-vt)^2 / (1 - v^2/c^2)
= (x^2 + v^2t^2 - 2xvt) / (1 - v^2/c^2)
= (c^2t^2 + v^2t^2 - 2xvt) / (1 - v^2/c^2)
t'^2 = (t-vx/c^2)^2 / (1 - v^2/c^2)
= (t^2 + v^2x^2/c^4 - 2xv/c^2) / (1 - v^2/c^2)
= (t^2 + v^2t^2/c^2 - 2xv/c^2) / (1 - v^2/c^2)
= x'^2/c^2
Turns out not to make any difference.
This is fine for any massive particle, but since the
denominator becomes 0 at the limit, the best I can do
there is note that the numerator must be 0 as well,
which basically means it will be created and destroyed
in an instant (since the photon is fixed at x'=0 in its
coordinate-space).
So, in a way, you're right, it never travels. We'll never
know anyway; we weigh too much, and even the best diets
won't help... ;-)
Paradoxically, when for photons is now all the time, since the
beginning
Big Bang, it still takes billions of years for the rest of us.
This is likely to be impossible.
I don't see a problem here. Could you clarify?
I dont know if I can. If for light is "now" all the time,
implies that light never ages from own point of view.
Therefore the detected light and radiation from BigBang is
100% fresh.
However, the new light generated after the BigBang has same
age as the primordially BigBang light. Even when light comes
later also in the future into existence, from our point of view, light
has the same age. Therefore from lights point of view
light is the same light for ever
If you understand this, I dont in any case
The "tired light" hypothesis is occasionally trotted
out here. It doesn't quite work.
--
#191, ewi...@xxxxxxxxxxxxx
Linux sucks efficiently, but Windows just blows around
a lot of hot air and vapor.
** Posted fromhttp://www.teranews.com**
--
#191, ewill3@xxxxxxxxxxxxx
Useless C/C++ Programming Idea #992381111:
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