Re: Double-slit experiment

On Jan 13, 12:52 pm, Sam Wormley <sworml...@xxxxxxxxx> wrote:
Strich.9 wrote:
On Jan 12, 6:07 pm, Sam Wormley <sworml...@xxxxxxxxx> wrote:
Strich.9 wrote:
On Jan 12, 4:58 pm, Sam Wormley <sworml...@xxxxxxxxx> wrote:
Strich.9 wrote:
On Jan 12, 4:23 pm, Sam Wormley <sworml...@xxxxxxxxx> wrote:
Strich.9 wrote:
The famous double slit experiment proves how correct quantum theory
is.  Any doubter can easily perform the double slit experiment and
confirm QUANTUM theory for himself.  Sadly, the same cannot be said
for relativity.  In fact, anybody with a good background in math can
easily prove special relativity to be wrong.
   Actually that's not true--relativity is self consistent
   mathematically and empirically correct.
How about the...   TWIN PARADOX?  (Note how one twin is exempt from
time dilation)
   You are not the first person to misunderstand "The twin paradox"
   which is not a paradox at all. Pick one perspective and stick
   with it... no paradox.
CORRECTION: You mean UNDERSTAND...
Note: The standard twin paradox is explained by SR/GR only by DENYing
or EXEMPTing the other twin from time dilation...
   You are not the first person who doesn't really want to understand
   the twin paradox. We cannot deny you your pleasure.- Hide quoted text -

- Show quoted text -

You are one of many idiots who do not understand the fallacy of
exempting one twin from time dilation...

   See if this will help you:http://www.physicsforums.com/showthread.php?t=205447


Why don't you ask Taylor and Wheeler if they will put this crap in
their textbook!?!



There is no twin paradox - mathematical proof

For the twin paradox to be considered a true paradox the framing of the scenario must be
stringent, that is to say we cannot permit assumptions to be ignored. Therefore I must
start with a short description of the twin paradox followed by identification of the
inherent assumptions.

 From Einstein Light (http://www.phys.unsw.edu.au/einstein..._paradox.htm):
Jane and Joe are twins. Jane travels in a straight line at a relativistic speed v to some
distant location. She then decelerates and returns. Her twin brother Joe stays at home on
Earth. The situation is shown in the diagram, which is not to scale (see link above).

Joe observes that Jane's on-board clocks (including her biological one), which run at
Jane's proper time, run slowly on both outbound and return leg. He therefore concludes
that she will be younger than he will be when she returns. On the outward leg, Jane
observes Joe's clock to run slowly, and she observes that it ticks slowly on the return
run. So will Jane conclude that Joe will have aged less? And if she does, who is correct?
According to the proponents of the paradox, there is a symmetry between the two observers,
so, just plugging in the equations of relativity, each will predict that the other is
younger. This cannot be simultaneously true for both so, if the argument is correct,
relativity is wrong.
(The author goes on to explain that asymmetry resolves the paradox, this is an not
entirely satisfactory explanation.)

There are a few assumptions, which are perhaps only obvious when one takes time to search
for them.

"(S)ome distant location" appears sufficiently vague as to avoid creating problems but an
inherent assumption is that this location shares the same frame as Joe.

By placing Joe on Earth we hide the other assumption, which is that we also share the same
frame as Joe.

The at-a-distance observations of each other's clock potentially hides an assumption of
instantaneous transmission of information. I doubt the author intended that, but it must
be remembered that information is not transmitted instantaneously.

"(Jane) decelerates and returns" is distracting. As the author correctly points out this
is a point of asymmetry. But a similar scenario (to be shown shortly) while show that it
doesn't matter which decelerates and changes direction - Jane or the entire universe. It
is generally assumed that the period during Jane changes direction is insignificant enough
to ignore.

Finally, "Jane travels in a straight line at a relativistic speed v" begs the question
"relativistic speed v relative to what?" The assumed answer is "relative to both Joe and
the distant location" (and we the readers). This is a direct consequence of the assumption
that Joe and the distant location share the same frame (and that we also share that frame).

Let me provide an analogous scenario.

Joe floats in a space suit with two clocks.

Jane sits at one end of an extremely long structure with another two clocks. At the other
end of the structure is a beacon. According to Jane, the structure has a length of L. Joe
knows this.

Jane and Joe pass each other twice, at relativistic velocities of v and -v. Joe and the
beacon pass each other twice, also at relativistic velocities of v and -v (Jane and the
beacon are fixed to the same structure and hence share the same frame).

Four events are noteworthy:

1. Joe and Jane are collocated as they pass for the first time. Their clocks begin
measuring time elapsed.

2. Joe and the beacon are collocated as they pass for the first time. Joe's clocks are
paused and the beacon sends a message to Jane's clocks to pause.

3. Joe and the beacon are collocated as they pass for the second time. Joe's clocks
restart measuring time elapsed and the beacon sends a message to Jane's clocks to resume
measuring time elapsed.

4. Joe and Jane are collocated as they pass for the second time. Their clocks stop
measuring time elapsed and Joe and Jane exchange one of their clocks. Neither consults the
other as they each attempt to work out what the other's clock will read.

Observe that I do not say who reverses direction. For the purposes of the mind experiment,
we can say that both Joe and Jane were anaethetised while one of them reversed direction,
but neither knows which of them has now changed direction relative to the third observer
(the reader).

There is an asymmetry in this scenario, but Jane and Joe cannot determine on whose part
that asymmetry lies.

Jane's calculations:

Joe is in motion relative to Jane. Jane calculates that Joe must take a period of 2L/v to
travel between events 1 and 2 and events 3 and 4. She further calculates that because Joe
is in motion, his clocks will run slow and will show a time elapsed of γ.2L/v where γ =
sqrt (1-v^2/c^2). She can check her clock and see that the first period elapsed (between
event 1 and event 2) was L/v + L/c and the second period elapsed (between event 3 and
event 4) was L/v-L/c for a total of 2L/v.


Lookie. No distance contraction on Joe's part as seen by Jane. Joe
is exempted from length contraction. Note also that there that is no
time dilation on the observer's co-inertial clock (Jane) while there
is on the trans-inertial clock (Joe)

Joe's calculations:

Jane is in motion relative to Joe. Joe therefore calculates that Jane's structure is
foreshorted by a factor of γ. Therefore the time elapsed while the entirety of the
structure passes twice will be γ.2L/v. Sure enough, he checks his clock and sees that this
is the case.


Here time dilation of the observer's co-inertial clock (Joe) is
introduced incorrectly. Note that Joe's clock is not in motion
relative to Joe.

Working out what Jane's clock will read is a little more complex. Joe knows that not only
is Jane's structure foreshortened, but her clocks will also run slow by a factor of γ.

The first period elapsed can therefore be calculated as follows (noting that Jane's
relative motion is in the same direction as the message from the beacon to Jane's clocks):

t1=γ.(γ.L/v + γ.L/(c-v))=γ^2.(L/v + L/(c-v))
=γ^2.(L/v.(c^2-v^2)/(c^2-v^2) + L(c+v)/(c^2-v^2))
=γ^2.(c^2.L/v - Lv + Lc + Lv)/(c^2-v^2)
=γ^2.(c^2.L/v + Lc)/(c^2-v^2)


Lookie, time dilation to the second power!!! This is the part you can
submit to Taylor and Wheeler. This is the part that turns relativity
physics from mere fallacy to absolute comical.


but since γ^2 = 1 - v^2/c^2 = (c^2 - V^2)/c^2,

t1=(c^2 - V^2)/c^2 . (c^2.L/v + Lc)/(c^2-v^2)
=(c^2.L/v + Lc)/c^2
=L/v + L/c

The same process can be used to calculate that the second period elapsed is L/v-L/c. The
total time elapsed on Jane's clock, as calculated by Joe, will be 2L/v - precisely the
same as calculated by Jane.

There is no disagreement and there is no paradox, merely a poorly frame scenario.

cheers,

neopolitan- Hide quoted text -

- Show quoted text -

Anymore attempts??? Keep em coming.

Strich, IQ 200, exposing the logical flaws of any interpretation or
reinterpretation of the twin paradox.
.

Relevant Pages

  • Re: Double-slit experiment
    ... easily prove special relativity to be wrong. ... Jane travels in a straight line at a relativistic speed v to some distant location. ... Her twin brother Joe stays at home on Earth. ... Joe observes that Jane's on-board clocks, which run at Jane's proper time, run slowly on both outbound and return leg. ...
    (sci.physics)
  • Re: Intercepting and Misdirecting email, and it be done with exchange?
    ... to the intended recipient, but to end up in her mailbox instead. ... mailbox or a folder on the ES as if it were undeliverable or badmail? ... If Jane wants to see everything that's going on, ... Joe gets fired. ...
    (microsoft.public.exchange.admin)
  • Re: Intercepting and Misdirecting email, and it be done with exchange?
    ... If Jane wants to see everything that's going on, ... Joe gets fired. ... intended mailbox, and then delivered to her instead. ... Can't think of a way to block the user from sending emails whilst at ...
    (microsoft.public.exchange.admin)
  • Re: Cannot get a grip on house values on west coast
    ... Where do Joe and Jane Blue Collar live in CA? ... That's $72,000/year in mortgage ... but would not be able to afford to buy the same house today. ...
    (misc.consumers.house)
  • Re: GMT Sunday 29th March 09
    ... wtwjgc (Joe) wrote: ... And we did have all those clocks *spring* ... That's no excuse! ...
    (uk.people.silversurfers)