On 07.12.2016 13:47, Thomas 'PointedEars' Lahn wrote:
> Paul B. Andersen wrote:
>
>> […] can you find any error in my following statement?
>
> Without having read your paper, based on what I already understand about
> special relativity, I can find an inconsistency in it and in previous
> statements that might be caused by choice of words only.
>
>>
>> The Li-ion clock is measured to run slow in the lab frame,
Let the "Li-ion clock" be an observer with a clock hereafter called
'the moving clock'. Let's call the time of the moving clock t'.
Given an inertial frame of reference which we will call the 'lab frame'.
We will call the coordinate time of this frame t.
A number of synchronized clocks in this frame will show
the 'coordinate time' t.
Let the 'moving clock' move the speed 0.866c, gamma =2.
(the units may be nano-seconds and light-nano-seconds, or similar)
Event E0: moving clock at position x=0
---------------------------------------
t': 0 ->v (moving clock)
t: 0 0
|------|--------> x (lab frame) t0' = 0 t0 = 0 x0 = 0
x: 0 0.866
Event E1: moving clock at position x=0.866
------------------------------------------
t': 0.5 ->v (moving clock)
t: 1 1
|------|--------> x (lab frame) t1' = 0.5 t1 = 1 x1=0.866
x: 0 0.866
The moving clock will measure the time (t1'-t0') = 0.5
between the events E0 and E1. This is a proper time
because it is measured by one clock which is present
at both the events, which means that the spatial interval
in the rest frame of the clock is zero.
So the invariant spacetime interval between E0 and E1 is 0.5.
In the lab frame the temporal component of the interval
between the events is (t1-t0) = 1 while the spatial
component of the interval is (x1-x0) = 0.866.
The invariant interval is sqrt(1^2 - 0.866^2) = 0.5
The 'rate of the moving clock' is the ratio between
the proper time (interval) between the events (0.5),
and the projection of this interval in the lab frame (1).
dt'/dt = (t1'-t0')/(t1-t0) = 0.5
This is commonly expressed as:
"The rate of the moving clock as measured in
the lab frame is 0.5"
or:
"The moving clock runs slow as measured in the lab frame."
These statement must not be taken literally, though.
Proper clocks always run at their proper rate, they never
speed up or slow down.
But the rate of the moving clock appear to be slower
in the lab frame because of the projection mentioned above.
>
> ACK.
>
>> and if the Li-ion observe the coordinate time of the lab
>
> JFTR: That is not possible.
>
>> as it passes by the (imaginary) co-ordinate clocks in the lab frame, it
>> would observe the co-ordinate time to run fast.
Let's observe the scenario from the moving clock's point of view:
Event E0: moving clock at position x=0
---------------------------------------
t': 0 (moving clock)
t: 0 ?
v<- |------|--------> x (lab frame)
x: 0 ?
t0' = 0 t0 = 0 x0 = 0
Event E1: moving clock at position x=0.866
------------------------------------------
t': 0.5 (moving clock)
t: ? 1
v<- |------|-------> x (lab frame)
x: ? 0.866
t1' = 0.5 t1 = 1 x1=0.866
The moving clock can only observe the time (and position)
of the coordinate clock it is adjacent to and finds:
dt/dt' = (t1-t0)/(t1'-t0') = 2
So the moving clock can conclude that by reading off
a sequence of different coordinate clocks, the coordinate time
of the lab frame appears two run fast compared to his clock.
====================================================
But he can conclude nothing about the rate of any of
the clocks in the lab frame.
====================================================
In fact, the moving clock's observations are exactly
the same as the observations made by the experimenter:
dt/dt' = 2, so dt'/dt = 0.5.
So the moving clock can conclude that it must run slow
as measured in the lab frame.
>> But _a specific stationary clock in the lab frame_
>> runs slow as measured in the rest-frame of the Li-ion.
So how can the moving clock measure the rate of the clock at
x=0 in the lab frame? The answer is that to do the measurement
we must have at least one other comoving synced clock.
So we introduce a second clock at the position x' = -0.866
t: 0
v<- |--------> x (lab frame)
x: 0
t': 0 0
|------|---------> x' (moving clock frame)
x': -0.866 0
t: 0.5
v<- |--------> x (lab frame)
x: 0
t': 1 1
|------|---------> x' (moving clock frame)
x': -0.866 0
The rate of the clock at x = 0 in the lab frame can be measured
by comparing its reading with the clock at x'=0 (the 'moving clock')
when they are adjacent, and then comparing it to the clock
at x' =-1 when they are adjacent.
The clock at x=0 runs at the rate dt/dt' = 0.5 as measured in
the rest frame of the 'moving clock'.
Mutual time dilation.
> I do not understand this statement. Are there two times in the lab frame,
> only one running according to clocks?
I hope you understand it now.
>
>> That's mutual time dilation.
>
> That means to me that each observer is observing the other’s clock running
> slower than theirs.
Indeed.
> How can then “the co-ordinate time […] run fast”?
It can't.
But if a moving observer reads a sequence of coordinate
clocks as he passes them, he will find that the numbers
he reads off the series of clocks increases faster than
the number he reads off his own clock, so to him,
the coordinate time of the lab frame appears to run
faster than his own clock.
But this is an artefact of his own movement, and
the clocks he passes are obviously not affected
in any way by his motion. They all run an normal rate.
--
Paul
https://paulba.no/