geistkiesel:
Thanks for the clarification. When you wrote,"An inertial frame A accelerates a transducer probe A’ ", I took that to mean that the probe was accelerated in some coordinate system A. But it appears that what you mean is that both A and A' can be considered as accelerating together.
No. A’ accelerates from A in order to get a sufficient distance apart such that the measurements may be made. This is the only reason for having the A’ probe available in the first place.
geistkiesel:
I assume you also mean that A and A' maintain a constant separation at all times in the frame of A. Thus, the distance d2 is the fixed distance between A and A', provided that d2 is measured in the frame of A and A'.
It all sounds good to me.
geistkiesel:
The travel time of a light pulse from A to A' and back, is therefore 2 (d2)/c, as measured in the frame of A or A' (which are the same inertial frame). i.e.:
t1 - to = (d2)/c
t2 - t1 = (d2)/c
Look at the schematic. A and A’ are continually moving in one direction the distance from A’ back to A is constantly shrinking and expanding proportional to the relative directions of frame and photon. As the pulse leaves A and heads toward A’, A and A’ always move the +X direction, the light direction change in a regular fashion. If you take Figure 1 and repeat the light pulse sequence, each round trip will be the same duration.
Your statement here made me sit up and take notice.
Lets start with a simple model and set ct = d1 + VAt. This says the light in leaving A must travel a distance d1, plus the distance VAt, the distance the frame moves during the time the light has moved a distance d1 (see the schematic – Figure 1).
geistkiesel:
So, when the light reflects A has your claim that delta tn +1 - tn is constant is therefore correct.
Now, you define:
"D = d1 – d2 (where D is the distance A travels during t2 - t0)"
Which reference frame was D measured in? And which frame was d1 measured in? Which was d2 measured in?
All measurements in the experiment are in the AA' frame when Va = Va'.
It is problematic to measure D in the reference frame of A or A', since A and A' both accelerated prior to time to.
But now both have the same velocity and therefore the same clock tick rates. Check the schematic below James R. D results from the measurement of t0, t1 and t2, where at least two pulses are used in order to confidently generate a known t1. As all math is performed on the A frame d1 = t1 - t0 and d2 = t2 - t1. Therefore, d1 - d2 = D, the distance the frame travels during t2 - t0. Getting ahead slightly, D = Va(t2 - t0) = (t1 – t0) - (t2 - t1) where the first and second expressions on the right are d1 and d2 respectively.
We can perhaps avoid this problem by using a third reference frame - one that sees A and A' accelerate from rest to speed Va. Call that frame, frame X.
Let us suppose that the various distances D, d1 and d2 are measured in frame X, then.
In frame X, it is still true that
t1 - to = (d2)/c
t2 - t1 = (d2)/c
(where we now understand the times to be measured using frame X's clocks, and not A or A' 's clocks).
The following was stated by James R in the previous 5 lines above.
“Let us suppose that the various distances D, d1 and d2 are measured in frame X, then.”
[I didn’t see it at first, but the second expression above should read, t2 - t1 = (d1)/c, you have “d2”, clearly a typo(?)]
What you are suggesting would add unnecessary complications and, without more than an instinct, I do not see how an inertial frame X, albeit in the same frame of reference as A, could make the same measurements with a close approximation to the resolution (accuracy) as produced on the A frame, taking into account the necessity of the light motion being intimately tied to the frame motion, or utilizing an added reference frame that I see as “problematic. Perhaps you see something I missed or that I was unable to detect in your post.
It is critical that all measurements be conducted on the A frame. This frame emits the pulse along the axis of motion and on the pulse is reflected back to the A frame. The outward pulse is a measure of d1, that is, d1 = t1 – t0. The in bound reflected pulse is a measure of d2 = t2 – t1. Here it is important to understand that the measurement of t2 completes the measurement protocol.
Similarly, I am a loss to see how the previous motion of the A frame has any rationally arrived at effect on the results and am not quite following your claim that
“It is problematic to measure D in the reference frame of A or A', since A and A' both accelerated prior to time t0.”
You must explain this to me.
Up until the A’ was launched, the motion histories of A and A’ are identical as A’ began the journey as cargo on the A frame.
I wanted to relate this thread as occurring exclusively in "deep space" where A and its cargo, including the A' transponder probe, exist as a solitary unit. Remember the measurement is of unaccelerated translatory motion, which to my 1st edition of “Handbook of Astronautical Engineering”, Chapter 11 ‘Relativistic Rocket Mechanics”, the SRT postulate is stated as, “It is impossible to measure or detect unaccelerated translatory motion in deep space”. I intended to maintain the thread with a minimum of actors, physical, parameters if you will, which to my thinking would require restricting the experiment to one inertial frame measuring its own motion. Your reasoning for including a third frame I find difficult to follow, perhaps this is because my post was not understood as intended.
Now you say:
"D = d1 – d2 (where D is the distance A travels during t2 - t0)"
As I see it D = d1 - d2 where D is the distance A travels BEFORE t0. The distance A travels in time t2 - t0 is: VA(t2 – t0), but this is not equal to D.
VA(t2 – t0), is most certainly equal to D. Check the schematic as D, as described there, only appears after t0.
Let us suppose that the various distances D, d1 and d2 are measured in frame X, then.
D cannot be the distance A travels before t0, because no measurements of any kind were made beforet0; no pulses occurred before t0; Your modification using the X frame drastically alters the experimental arrangement, and together with the statement that D was established before t0, make no sense, to me. This experiment is like a mini-Big Bang that is nothing happens before t0.
Slow down James R. As stated d1 and d2 are the measurements that begin with t0, include t1 and finally end with t2. d1 = t1 - t0 and d2 = t2 - t1.
There is nothing of interest here prior to t0, nor after t2. I considered clock perturbations due to acceleration of the A' frame, but up to the time A' was launched A and A' shared an identical history. Now, once A' was launched I considered some perturbation to A' tick rate (acceleration effects, some might refer this to SRT time dilation). However, I discarded the possibility that the A' suffered some permanent tick rate changes and presumed both frame clocks tick the same, they are, after all the same frame of reference. But then perhaps the time-of-day indicated on both frame clocks are not the same after A’ surged ahead. There is no unambiguous way of asking A’ what the time of day (tod) is indicated on his clocks as the spatial separation precludes any confident 'time hacking' of the A and A’ clocks. Mulling over this problem brought me to the solution of using the delta-tn as indicated which brought a soft wind of evaporation to the enigma as t1 is not a simple ad hoc substitution for observed data.
You next say:
"D= (t1 – t2) – (t2 – t1) = Va (t2 – t0) = 2t1 – (t2 – t0)
Va = 2t1/(t2 – t0)– 1"
This is still a muddle, as far as I can tell. Can you please explain what you're saying here?
The first term in your post above after the “=” is incorrect. Either I made a typo or you did. The expression should read (see the underlined t0), D = (t1 – t0) – (t2 – t1) = Va (t2 – t0) = 2t1 – (t2 – t0)
Va (t2 – t0) = 2t1 – (t2 – t0)
And dividing through by the t2 – t0 term we arrive at,
Va = 2t1/(t2 – t0) – 1
[See below for an interesting and different method of calculation.]
From t0 on and until t1, the time of flight of the out going pulse, the A frame moved a distance Va(t1 - t0) during the time the light moved a distance d1. The light moves a bunch, the A frame but a relatively small distance. But we do not know what t1 is until the pulse is returned to A and to the measurement of t2. The return signal from A' includes t1 embedded in the signal. Now we have a t0, and an ambiguous t1 and we are not confident
that t1 is usable until a second identical pulse is emitted with the identical protocol as the previous pulse.
Now we useful data are available for some arithmetic. The frame is moving in the direction of the emitted pulse (and toward A' and always along the X axis). When the light traveled from A to A' the A frame traveled a distance Va(t1 - t0), but the value of VA is not known from this alone, obviously. Remember, velocity, V, is defined as, in general,
V = (Xn+1 - Xn)/(Tn+1 - Tn)
At this point d1 may not be calculated before d2 is measured, as this then and only then, is the t1 data available as t1 is embedded in the signal received by A at t2. Simpler said no calculations are permitted until the completion of the round trip travels of the pulse. When t2 is measured the d1 calculation is performed using appropriate data, the only data available, d1 = t1 - t0 for c = 1. Nothing is known of d1 until the return of the light from A' to A, which is received by A at t2 with the additional information that the time t1 is embedded in the signal ala a transponder signal (So I repeat myself occasionally).
Here is the crux. As the light moves a known distance during the time of flight from A to A' (which is available when the signal returns at t2) the A frame moves some distance, unknown up to here. After reflection from A', the light continues its travels to the A frame where/when the light and the A frame meet. This received pulse signals the end of the total distance the light travels during which time the frame is moving steadily forward. The light motion began at t0, where now the values of (X0,t0) are known, or are they? Actually X0 is not known because we did not measure an actual X0, but this is of no concern as we are not measuring X0 and X1 individually, directly, or otherwise. Rather
we are measuring the difference of X1 - X0, the distance the frame travels during t2 - t0.
The schematic may be of some use here.
1. As the pulse moves a distance d1, the frame moves a distance Va(t1 - t0), Va being unknown at this point.
2. The pulse then travels the distance d2, (A' back to A) during t2 - t1, while the frame continues its constant velocity. The A frame will continue moving, hypothetically speaking, until the light motion ceases.
3. When the pulse is received at A at t2 the measurement of the continuing frame motion ceases, or has ceased. There are three parameters observed in the experiment’s data base, t0, t1, and t2 with a minimum of six data points, two for each of the three times measured.
The actual direction of the pulse motion is irrelevant, the crucial parameter here is the time-of-flight of the pulse along d1 and d2 as all distances can be determined, for example, by reflections between mirrors ala Michelson-Morley. Any sufficient number of arbitrary reflections will suffice as we are measuring non-vectored distances which should be only be sufficiently large to insure an acceptable experimental error in measurement, which can be as fine tuned by extending the time of flight until reaching some distance where gains in accuracy can no longer be determined – the point where the return on investment in accuracy has reached a point of diminished returns that approach zero. Why subtract d1 - d2? We use the difference to determine the ΔX for our upcoming velocity calculation. Using Figure 1 we see that if the light simply continued along an un-reflected path that D would not change, but in this case we would not be able to determine its value. When the motion of the frame and reflected pulse meet (t2 is the indicated point in Figure 1), we then have confidence in the calculated distance D, which is the delta-X in the velocity expression, as opposed to being the numbers describing the individual X0 and X1. D is the distance the frame moves during t2 – t0.
To now answer your specific question, we begin by equating D = (d1 - d2), when expressed mathematically (for c = 1) becomes,
D = (t1 – t0) – (t2 – t1) = VA(t2 – t0).
The first () term is d1, the second, d2, and D expressed as VA(t2 – t0) where here the delta-t covers the entire time of flight of the pulse.
VA = (t1 – t0) – (t2 – t1)/(t2 – t0).
= (t1 – t0 – t2 + t1)/(t2 – t0)
= [2t1 – (t2 – t0)]/(t2 − t0)
= (2t1)/(t2 - t0) – (t2 − t0)/(t2 − t0)
= (2t1)/(t2 - t0) – 1
There are a lot of interesting twists and turns up to now and I trust some of your ‘muddle’ has been clarified. I am looking forward to further comments.