John Berry wrote:
On 3/8/07, *Stephen A. Lawrence* <[EMAIL PROTECTED]
<mailto:[EMAIL PROTECTED]>> wrote:
So how about you try working through the mathematics of the
contradictions you think you've found in relativity, and post the
results here?
I mean, work them through using the Lorentz transforms. I'll be happy
to argue them with you, if you'll actually work through the math rather
than just blowing off the calculations and calling it all "bunk".
Here's the thing though.
I'm not attacking the equations as self contradictory, I'm bypassing all
of that by pointing out that there is no way for them to be correct
because it is well known that SR functions based on the idea that you
can not assess the rate of time in another frame without distortion.
I don't understand what you mean by this.
In the SR model the time associated with any particular frame of
reference is a scalar field. The "rate of time" in that frame is the
gradient 1-form of the scalar field. The rate of time in that frame, as
measured by any particular observer, is that 1-form applied to the
observer's 4-velocity.
The problem is that the scalar fields which represent the "time"
coordinate for observers in relative motion are not identical.
The reason that instantaneous communication is said to break down SR is
not that it is instantaneous but that it presupposes no distortion by
effects such as the Doppler effect.
I don't understand what you mean by that.
Here's the deal on instantaneous communication: Given any two
spacelike-separated events, E1 and E2, it's possible to find 3 frames of
reference, "a", "b", and "c", such that:
-- in frame "a" E1 occurs before E2
-- in frame "b" E1 and E2 are simultaneous
-- in frame "c" E2 occurs before E1
We don't care about frame "b" here, but frames "a" and "c" cause real
trouble with instantaneous communicators.
If we have instantaneous communication which works in _any_ frame of
reference, via, say, a hyperwarp communicator carried by an observer in
that frame, then we can do this:
-- In frame "a" an observer, "O1", at event E1, "hyperwarps"
information about E1 to another observer, "O2", located at the point in
space in frame "a" where E2 will later occur, "upstream" of event E2.
This is possible because E1 occurs _before_ E2 in frame "a".
-- In frame "c" an observer, "O3", who is just at this moment zipping
past observer "O2", copies the information from "O2" and hyperwarps it
to a fourth observer, "O4", who is at the spacial location in frame "c"
where event E1 _will_ _occur_. This is possible because E2 occurs
_before_ E1 in frame "c".
-- "O4" has received information about E1 before it happens, and can
now take action to keep event E1 from ever taking place. This is a
contradiction.
If we assume reality doesn't actually allow for contradictions, then
either simultaneity is _not_ relative or instantaneous communication
between arbitrary observers is _not_ possible.
If we restrict instantaneous communication to a single, preferred
inertial frame, then we avoid the contradictions outlined above. As you
said, that's not "special relativity" any more; it's got a preferred
rest frame. But none the less, if we believe that simultaneity is
relative, and we don't think reality allows for contradictory
situations, then if instantaneous communication is really possible, it
must presumably behave that way.
As numerous sci-fi writers have observed, if general time travel is
possible, then we will most likely _never_ discover it. The problem is
that once it's discovered, it's possible for someone to go back in time
and prevent the discovery. Hard to calculate the odds, but given enough
-- uh -- _time_, it might very well be inevitable that such a thing
would happen -- and once the discovery was prevented, the result would
be "stable", as nobody could go back and change things again. In other
words, the only "stable" universe is one in which nobody knows how to
travel back in time, whether or not it's hypothetically "possible".
The methods I have given allow observers in 2 different frames to
observe each others rate of time and agree fully.
Not exactly; the methods you gave allow each observer to measure the
other's rate of time passage at some point in the past. And as to
whether they "agree", that depends on what you mean by the word.
In any case that doesn't allow for any instantaneous transfer of
information. If you disagree please post an actual example, using
numbers, in which information is transfered between frames faster than C.
This can be achieved by observing travel towards and away and working
out the Doppler effect as I have laid out*,
Of course.
or by communication during a
flyby, if the flyby is incredibly distant then even a flyby at .99c
could last a while, if very close it may be fleeting but could pretty
much fit the bill for being practically instantaneous too.
The simplest is to allow them to come very close, and each watches the
other's clock, and sees how fast it's ticking.
Each will see the other's clock ticking slower than his.
You need to actually work out the exact details of this, and the paths
the information follows, if you think there's a contradiction. You
cannot get past this one without working through the math, because the
devil is in the details.
Here's a very brief summary of the scoop: If "A" wants to measure "B"'s
clock rate, A must observe B's clock at _two_ _different_ _times_. (You
can't get a rate of change with a single measurement.) At those two
different times, B is in two different locations in A's frame of
reference. So, we have ONE location in B's frame, whose time is being
checked at TWO locations in A's frame.
That is an asymmetry, and that is the heart of the matter. This
operation, no matter how you imagine A performs the measurements, ends
up applying the gradient of A's time coordinate to B's velocity vector
and comparing the result with B's proper clock.
Conversely if B measures A's clock rate, he's applying the gradient of
B's time coordinate to A's velocity vector and comparing the result to
A's proper clock.
They're using two different gradient functions, which are at an angle
relative to each other; hence they do not get the obvious (reciprocal)
answers.
To put it another way, if "@" is the partial derivative sign and "d" is
the ordinary total derivative sign, then, along any particular path, we
DO have the following:
dt/dtau = 1/(dtau/dt)
However, we do _NOT_, in general, have the following:
@t/@tau = 1/(@tau/@t)
The reason, in short, is that the partial derivative implicitly
determines the path on which the derivative is taken, and the paths for
@t/@tau and @tau/@t are not the same.
For example, take the 1-dimensional Lorentz transforms for time and space:
t' = g * (t - v*x)
x' = g * (x - v*t)
Inverting that, we get
t = g * (t' + v*x')
x = g * (x' + v*t')
This just simple linear algebra, for the case where x and t can be
described as functions of x' and t', _or_ x' and t' can be described as
functions of x and t.
Now let's look at some derivatives, with the assumption that "B", in the
primed frame, passed location 0 in both frames at time 0 in both frames,
and is moving at speed "v" along A's "x" axis. So, we also have
x = v*t
OK? Now let's take the total derivative along the path B follows, and
the partial derivative (which doesn't depend on the path):
dt'/dt = d/dt (g*(t-v*x))
= g * (1 - v^2)
= 1/g
@t'/@t = @/@t (g*(t - v*x))
= g
Note well: The partial and total derivatives of B's time with respect
to A's time are _INVERSES_.
Now, we'll finally get to the point: If A measures B's clock rate by
watching B's clock, he's checking B's time at two points along the path
B follows through A's frame. That's the _total_ _derivative_ of B's
time with respect to A's time along that path, and it's 1/g. If B is
moving at 0.866C, A sees B's clock ticking at 1/2 the rate of A's clock.
However, when we use A's _SINGLE_ clock to measure the time at TWO
DIFFERENT POINTS in B's frame of reference, then we are taking the rate
of change in B's time as A's time changes, _BUT_ with _NO_ change in A's
spacial coordinate. That is the partial derivative of B's time with
respect to A's time. In other words, it's the rate at which B's time
changes, while holding B's "x" coordinate _fixed_. That is, as we saw
above, "g". When measured this way, B's time goes _faster_ than A's
time; this is sometimes called "time contraction", and it's the inverse
of "time dilation", and it is, in fact, the measurement B makes when he
observes A's clock at two different moments in order to see how fast A's
time is passing: that's the total derivative of B's time along the path
which goes straight down A's time axis.
Sigh ... I bet this didn't help in the least, did it?
The 3 different ways are either: 1:Use a computer to work out the level
of Doppler distortion,2: Communicate to the other frame it's apparent
time rate as you observe it and visa versa, if both is you get the same
score then there is no time dilation only Doppler effects.
3: And finally have 2 positions A & B in the same reference frame
measure the apparent rate of time of a vehicle moving from one to the
other, the Doppler effect is positive for A and negative for B so they
can be added together to remove the Doppler component.
Basically any way you slice it, there is no way to stop an accurate
observation of the time rate another frame is experiencing which means
that if any time dilation is present it is observable and agreed on by
both parties, it is absolute.
Sure, more or less. But to measure a rate you always need to measure
the value at two points. And, as I pointed out above, the way you
choose those two points is critical. For any _particular_ set of two
observations A and B will agree. However, B can't use the same set of
two observations to measure the rate at which A's clock is ticking, as A
uses to measure the rate at which B's clock is ticking, exactly because
the clocks are in motion relative to each other. A fixed point in the
space of one frame or the other must be chosen, and that choice of fixed
point determines the result.
