Re: The Present.


"Henri Wilson" <HW@..> wrote in message
news:2vro92tt7mfu9nev7684cdcoqoi5fisgpp@xxxxxxxxxx
On Fri, 23 Jun 2006 16:50:52 -0400, "Len Gaasenbeek" <gaasbeek@xxxxxxxxxx>
wrote:

THE PRESENT

I think we all agree that we live in the present, since it is impossible
for
any of us to live in the past or live in the future. So what exactly is
the
present?

The present is the point in time where the past ends and the future
begins.
Whereas the past and the future are both infinitely long, the present is
of
infinitely short duration. This is the case because we can envision what
the 'past present' was like at any point of time before the 'current
present
'. Similarly, we can imagine what the 'future present' might be like at
any
point of time beyond the 'current present'.

Now if person A is holding hands with person B, it is obvious that the
present is the same point in time for both A and B. It is also clear
that
both A and B are aging at the same rate, as per Greenwich Mean Time used
as
a basis to calculate the present time throughout the world (or universe).

That is to say, if the Greenwich time is lets say 2 o-clock p.m., it is
the
same time all over the world, i.e. 2 p.m. Even if A lives in Toronto and
B
lives at a planet one light-hour away, they both find themselves at the
same
point in time at any given point in time, that is to say they both live
in
the same present and age at the same rate.

Now if A looks at B's clock, it will appear to run exactly one hour slow
to
his clock. Yet when B looks at A's clock, it also will appear to run one
hour slow to his clock. This is the case because it takes A's clock
image
one hour to reach B and B's clock image one hour to reach A, because it
takes the light image of the clocks that long to travel the distance
between
A and B.

It follows from the above that an 'observed image' is not reality as it
is
always older than its present 'actual image', by the length of time it
took
for the image to reach the observer.

For example, in the case when A lives one light-hour away from B, A may
observe B to be perfectly healthy, whereas in reality someone murdered
him h
alf an hour ago.

In the case where A and B travel away from each other at a constant
velocity, A will always observe B where he used to be at some time in the
past, while travelling at an observed speed which is slower than his
actual
speed. Similarly B will observe A as he looked some time ago and where
he
was located some time ago.

The above concept can become confusing when we speak of the time at which
the observer makes the observation (which is the present), as compared to
'the observed time' or age of the observed person.

For example, if A lives one light-hour away from B, A will observe B as
he
looked one hour ago. Now the time at which A makes the observation could
also be called 'the observed time', as compared to the observed time of
B's
image as seen by A. (And vice versa)

So to be clear we really should differentiate between 'the observer's
time'
which is the 'real' or 'actual' time, and 'the observed time' of the
observed object, which is some time in the past.

OK so far Len ...I think most of us here are familiar with the difference
between REAL and APPARENT observations.

Next let us suppose that at time zero, B starts to move away from A at a
speed of 1/4 c. What will A see?

At any future point in time, A will observe B at an 'observed distance'
where B used to be a quarter of an hour ago. Similarly B will be
traveling
at an 'observed speed' which is less than his 'actual speed', and B will
appear to be of an 'observed age' which is less than his real or 'actual
age



Or to put it in mathematical form:
The observed distance traveled = (tvc) / (v+c)
The observed speed = (vc) / (v+c)
The observed age = (tc) / (v+c)
The observed aging rate = c / (v+c)
Where:
t = the elapsed time since the observed object left the observer.
v = the actual speed of the observed object.
c = the speed of light.

Didn't you know that the light from B was traveling to A at c-v....or
only
3/4c?
..........................................................
To Henry,

The above statement is incorrect if A and B are in the same EFOR.
Sure the light leaves B towards A at 3/4 c but it immediately falls in step
with other light it meets, causing it to travel towards A at c.

If B's planet is larger than A's planet (the earth), it will create a moving
EFOR
which will cause B's light to travel towards A at 3/4 c as you suggest. But
as
B's light approaches the earth's EFOR it will once more travel at close to
c.

In this case the average speed at which B's light travels towards earth can
be
calculated as per formula [20] in my second paper 'Frames of Reference'.

Len.
.............................................................

For further details see the second of my 'Selected Papers' titled:
'Frames of Reference' at web-site: http://www2.rideau.net/gaasbeek

Since it is generally accepted that electromagnetic radiation travels at
c
in a vacuum, the gravitational attraction between two moving bodies is
also
a function of their observed, rather than their actual location.

In other words, two bodies that pass each other will experience an
attractive gravitational force which depends on where each body observes
the other to be, rather than where they actually are at a given point in
time. Not only will the magnitude of the attraction be different than
expected, but its direction will different too.

This, among other things, explains why some space-crafts are experiencing
an
unexplained change in their velocity, the further away they move from
earth.

Enjoy, Len. June 23, 2006.







HW.
www.users.bigpond.com/hewn/index.htm

Appropriate message snipping is considerate and painless.



.

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