Ralfativity pt 3: Relative Velocity and the Myth of the Preferred Frame

The speed of empty space reprinted so you don't need to look for it.

Let's say you launch a tube of water at .1c with a laser inside and outside the tube and a time comparator at the end to measure the speed difference between the two beams of light. The speed of light through water happens to be .75c so the combined velocity of the tube will bring the speed of light within the tube wrt the outer laser to .8c using the velocity combo law.

But the tube has a small leak and as it evacuates, the speed within the tube gets closer and closer to the outside c. When the tube is evacuated, both the light within and without the tube are at c. The vacuum sealed in the tube is the same as the vacuum outside the tube which means space can't move or be moved. The sealed tube is not moving the vacuum within it.

The following STD shows light signals communicating between Bob and Alice. The ones from Bob are in yellow and from Alice, pink.

Since Bob and Alice started together at t=0 in the instantaneous present (they were co-located), all Alice has to say in her pink message to Bob st t'=4 is that she has aged 4 yrs. Age is the passage of time through time measured by her clock. (Her time component through space is 1 yr as discussed above.) When Bob gets the message at t=8, he can calculate from the message contents that it took 3 of his years to reach him and that he was 5 in his delayed present when Alice sent it.

Alice simultaneously gets a yellow message sent from Bob that he is 2 from her perspective in her delayed present of t'=4. She can calculate from the message contents that it took 2.4 of her years for the light to travel the 3 ly from Bob. This is c' =Yc=x/t' which is the mixed perspective of the common distance using Alice's clock. From Bob's perspective c=x/t=3/3=1c. From Alice's perspective c=c'/Y=1.25/1.25=1c so no conflict with relativity.

A few of you might be screaming that this is a trick to hide length contraction in some fancy math footwork. Oh how quickly we forget the previous post.

Alice's perspective of Bob at t'=4 in Alice years is t=3.2 in Bob years. So from Alice's perspective of Bob, Bob's light signal has moved 1.2 ly in 1.2 of his years which equals c. (Similarly Bob's relative velocity through space from Alice's perspective is 1.92/3.2= .6c.) No tricks here, c is c from all perspectives without needing length contraction.

So what have we determined thus far. From Bob's perspective when Bob's age is t=5, Alice's age is t'=4 plus a year hidden in the distance between them so they're both technically 5. From Alice's perspective, when Alice's age t'=4, Bob's age is t=3.2 plus .8yrs hidden in the distance between them so they're both 4. From the .33c green line perspective, Bob and Alice see each other as 4 which is how old they would see each other along the purple line of instantaneous present.

The reason the reading at Bob t=5 seems out of whack with the rest is the reciprocal time dilation at Alice t'=4 and Bob t=3.2 is actually Bob t=4 and Alice t'=3.2, not Bob t=5 and Alice t'=4.

Oops I made a mistake c'=c/R not c/Y.

Since STD depictions are all about relative velocity between two participants, the results must be the same no matter what common background frame is being used. Since all depictions yield the same results, any depiction chosen is as valid as any other and is not the dreaded preferred frame.

The following Loedel and Minkowski (without length contraction) STD's look very different from each other but actually yield the same results when properly understood. The first step is to understand the difference between stationary and moving depictions and Bob's and Alice's relative perspectives. To do so strips away any dependence on the background frame and circumvents limitations of depiction.

The Loedel isn't rotated to create the Minkowski, it's pushed over and the lines of simultaneity will shift. The units labelling the velocity lines will also shift and require re-labelling.

The distances will be compared between Bob and Alice separating from each other at .33c relative to earth and Alice separating at .6c from Bob depicted as stationary. Earth will not be depicted as stationary with Bob in this STD, that is another depiction with the green and blue lines overlapping.

We'll just go through and check the numbers and illustrate the techniques to get at those numbers.

1. Loedel Bob's perspective of Alice at t=2.5 is she has travelled .67-(-.83)=1.5ly. Remember there is no length contraction in ralfativity, the x-axis is shared and invariant. Bob's velocity is 1.5/2.5 = .6c. Alice's v' from Bob's perspective using Alice's time and invariant space is 1.5/2 =.75c. Same results for the Minkowski except the distance travelled is 1.5-0=1.5ly.

2. Loedel Bob's perspective of Alice at t=2 is she has travelled ..53-(-.67)=1.2ly. Bob's velocity is 1.2/2 = .6c. Alice's v' from Bob's perspective is 1.2/1.6 =.75c. Same results for the Minkowski except the distance travelled is 1.2-0=1.2ly.

3. Loedel Alice's perspective of Bob at t'=2 (using ralfativity without having to redraw Alice as stationary) is the red line from t'=2 to t=1.6 then the blue line to t'=1.28. The corresponding distance he travels is .422-(-.533)=.96ly. So his v=.96/1.6 = .6. Same results for the Minkowski except the distance travelled is .96-0=.96ly. Alice at t'=2 sees Bob move only .96ly while Bob sees Alice at t'=2 move 1.5ly. Why? because there are 2 different perspectives of Alice at t'=2 which have nothing to do with which one you depict as moving or stationary.