Re: The speed of light is c or c+v or c-v depending on the motion of the target
- From: "harry" <harald.vanlintelButNotThis@xxxxxxx>
- Date: Fri, 8 Aug 2008 18:24:49 +0200
Peter Riedt wrote:
On Aug 7, 2:48 pm, "harry" <harald.vanlintelButNotT...@xxxxxxx> wrote:
"Peter Riedt" <rie...@xxxxxxxxxxx> wrote in message
news:6c303558-e5df-42f4-95a0-e4692644f856@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
(c+v+c-v)/2=c
That's true of course. ;-)
The speed of light c is considered to be a constant since at least
Maxwell. However, Michelson and Morley, in the mathematics
underlying their interferometer experiment (MMX) to find evidence
of the ether, used expressions like c+v and c-v. Their idea was
that the formula (c+v +c-v)/2=c applied overall and that the
components of this formula should be applied separately to the
upwind and downwind paths.
So they did. However, it only applies within a single (standard)
coordinate
system:http://www.fourmilab.ch/etexts/einstein/specrel/www/(par.3).
The expressions c-v and c+v are not well received or understood in
this NG.
Indeed, perhaps half of the people here don't understand the above
derivation...
This is because it could lead to the idea that the speed of
light is subject to the speed of the source and this is anathema to
relativity supporters. However, the two expressions have a real
validity. They are mathematical devices to account for the target
moving away or towards the source.
Inexact: both the source and the target may be moving, as indicated
with for example v and w respectively. The expression concerns the
speed of light (also called "closing speed" in modern jargon)
relative to a moving object. Of course, it is also valid for an
object in rest ( c-v AND v=0 => c).
Harry,
the speed of the source is irrelevant. Once the light beam or ray has
left the source it travels independently. Also, let me state that the
speed of the source does not affect the speed of light.
Yes indeed, we agree on that: since the speed of the source is irrelevant it
is inexact to state that we "account for the target moving away or towards
the source".
While the speed of light is always c,
It might be useful if you specify what you mean with "the speed of
light" in this context. Relative to what?
Nobody ever specifies c to be relative to what.
Wrong - in particular Einstein specified this clearly. The fact that this is
commonly omitted is one of the causes of confusion; and the fact that you
didn't answer it makes me wonder...
As c is 300000km/sec
as has been established by many experimeters I believe this limit is
imposed by the properties of the ether.
the distance between source and target will vary with the motion of
objects through space. To compensate for these variations, c-v and
c+v are used. MMX is a good example to explain it. The
interferometer used in MMX is attached to the earth. The earth
moves through space.
Instead of "space" (which is an unknown), for calculations the solar
system is often chosen as reference.
A light beam is sent from a source through the beam splitter of the
interferometer to the mirror at the other end of the parallel arm of
the equipment. The distance between beam splitter and mirror is 11m
as measured in the lab but the earth does not stand still. It and
the mirror move away from the location where the beam splitter was
at the time the light beam passed through it. By the time the light
has caught up with the mirror, it has moved on to a new location,
widening the lab distance of 11m to a distance through space of
11m+. To allow for this extra distance, the speed of light has to
be adjusted to c-v or 300000km/sec-30km/sec giving an effective
299970km/sec.
Apparently you DID use the solar system...
Likewise,
on the return trip, the speed of light has to be adjusted to c+v or
300000km/sec+30km/sec giving an effective 300030km/sec to account
for the fact that the beam splitter is moving towards the light beam
reflected from the mirror. In short, the EFFECTIVE speed of light is
subject to the motion of the target (30km/sec around the sun) but
the ACTUAL speed of light up and down the parallel arm of the
interferometer is always (c+v+c-v)/2 = c. If you understand this
there is no need to worry that the speed of light is not c.
You were so close, and then such a colossal hic-up!
Where did I go wrong?
As I demonstrated just below...
Here's another one:
If you drive 100 km to and fro at 100 km/h without delay, the return
trip will take you 2 hours.
Now, suppose you were slowed down in traffic to 100-10 = 90 km/h on
the first leg, and so you try to compensate for that by driving
100+10 = 110 km/h back; How long will it take in total? What will be
your average speed?
You were so close, and then such a colossal hic-up! Your example
distances to and fro are fixed. The subject however in this thread
involves a moving target, the mirror at the end of the parallel arm of
the interferometer equipment in MMX. It has moved 0.00110011m from the
source by the time the light reaches it.
The distance between the interferometer parts are just as fixed as the
distance between two towns 100 km apart. For you to understand this, we can
observe your car from a boat on the lake which has a speed of -10 km/h: from
that perspective, the speed of your car is 100 km/h on both legs while the
starting point moves. Now please tell the average speed. ;-)
Harald
.
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