## S.R. Defining the present

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### Re: S.R. Defining the present

Even in Ralfativity I have to accept that until a change in velocity is made, the two must be the same age. Age is elapsed time and yet in this discussion I see a valid coordinate time that is different from the elapsed time. I'm trying to understand that. You're saying if I draw 2 more lines my asymmetry in the reciprocity will disappear and I'll see I've created an asymmetry that doesn't really exist.But I think those two lines are there to create a valid elapsed time. That is outside of the discussion. I know how to establish a valid age difference. The question I'm asking here is different:

How can their ages be the same but their clock comparisons for a moment at the 3 ly mark between Bob's network clock and Alce's on-board clock have different times?

The answer is their age difference comes from comparing their on-board proper time clocks and this coordinate time difference comes from comparing 2 other clocks. I'm further saying that time difference forms a concrete foundation on which to base establishment of the age difference once a velocity change is made. This is where ralfativity and relativity diverge. Relativity must consider a complete start to stop spacetime path to establish age difference. I can calculate partial age difference for each change in velocity in real time. I can separate out the permanent age difference for non-stop velocity changes until the spacetime path is complete. In an experiment where no velocity changes are expected before the end of the spacetime path,I can accurately project what the age difference will be before the stop as Alice flies by my remote network clock. Relativity won't make that prediction because it states clock reciprocity between on-board clocks makes that impossible. I agree but when you add in the network clocks it makes the prediction possible. Since no subsequent action will affect that prediction, the age difference is established de facto before the age difference can be truly verified. So then I have to ask, why split hairs? The clock difference might as well be the age difference as it forms an irreversible basis for the final age difference at the end of the spacetime path.

P.S .The STD of two ships flying away or towards the earth at .33c yields a .6c relative velocity between them that does not result in an age difference if they both stop relative to earth and hence each other. I need to STD this because it's a special case where relative velocity and a stop does not result in age difference.
ralfcis
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### Re: S.R. Defining the present

ralfcis » 11 Oct 2017, 13:18 wrote:Even in Ralfativity I have to accept that until a change in velocity is made, the two must be the same age. Age is elapsed time and yet in this discussion I see a valid coordinate time that is different from the elapsed time. I'm trying to understand that. You're saying if I draw 2 more lines my asymmetry in the reciprocity will disappear and I'll see I've created an asymmetry that doesn't really exist.But I think those two lines are there to create a valid elapsed time.

Here is the diagram that I have suggested.

It is completely symmetrical and reciprocal, as one would expect from purely inertial frames. It does not show any propertime differences, just coordinate time differences.

I can calculate partial age difference for each change in velocity in real time. I can separate out the permanent age difference for non-stop velocity changes until the spacetime path is complete. In an experiment where no velocity changes are expected before the end of the spacetime path, I can accurately project what the age difference will be before the stop as Alice flies by my remote network clock.

Not if the scenario is inertial from start to finish. Network clocks indicate coordinate time differences, not propertime differences, unless you can compare Alice's clock with the same network clock twice, as we have discussed before.

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### Re: S.R. Defining the present

I'll describe what is being compared at each BoB and Alice point of 4 and 5 tomorrow. But I only see 1 point where Bob is a blue 5 intersecting with an orange Alice =4. There is no reciprocal equivalent to that in the reverse analysis. You can also get rid of the virtual network clocks because no clock comparison between real clocks can be made and messaged in those instances.
ralfcis
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### Re: S.R. Defining the present

Unfortunately my car has a leaky brake line and I need to fix it before I get to work this morning so I may be delayed in my answer above (I know any delay makes me subject to getting me booted). The subject of this thread is defining the present. I thought of another angle; a time line of events. The order of the messages sent out considering both perspectives simultaneously will determine which messages are sent out simultaneously. I have a feeling that this definition of the present doesn't correspond to any lines of present but probably corresponds to the hyperbolic proper time lines. I don't know yet.
ralfcis
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### Re: S.R. Defining the present

I know how you like my STD's packed with as many lines as possible so this one is everything simultaneous at the 1 year anniversary of Alice's trip from both perspectives. Some lines were erased for better clarity.

The t and t' axes are labeled bob blue/alice red, stationary/moving. The first point we hit is bob stationary at t=.8. Why is this even significant? This intersects the virtual orange line of present network clock from Alice at t'=1. As I said in the previous post, no single spaceship is going to drag around a sync'd clock network stationary relative to the moving ship to check Bob's relative motion. But relativity allows one to postulate that if Alice's ship did have its own clock network attached to it, Bob's on-board clock would read .8 while Alice's network and on-board clocks would read 1.

The only physical way to check this is to send a light signal from Bob when he is .8 (light blue line) and subtract the light signal delay multiplied by the doppler ratio (1/2 at .6c) from the point Alice receives the message. So the light signal delay is 1.2/2 = .6 and that is subtracted from 1.6, when Alice receives the signal, and it confirms that Alice will be 1 when Bob was .8. It's a very weird definition of the present and I don't see any real purpose for it but relativity uses it extensively with the relativity of simultaneity to solve the twin paradox. I don't.

There are several other simultaneous events between Bob and Alice at Alice t'=1. The hyperbolic black line joins all the on-board clocks = 1 at any relative velocity between Bob and Alice. So this is also another definition of the present joining Alice's and Bob's on-board clocks to signify that at any constant relative motion, they both age at the same rate at the same on-board clock time.

Let's do the two pink lines next as they are more difficult to understand. They are both light signals from Alice to Bob when Alice reaches 1 on her on-board clock. Whether she's moving or stationary, the time on her on-board clock is the same. The shorter pink line is from Alice moving to Bob stationary and it takes .75 yrs to reach Bob. The longer pink line is Alice stationary to Bob moving and it takes 1.5 yrs to reach Bob but don't forget to multiply that by the doppler ratio to get .75yrs, the same as the shorter pink line. I don't know the physical meaning of this. Is this a purely mathematical construct or does it reflect on the constancy of c somehow?

The darker blue line intersecting Alice's virtual orange dashed network clock line of present and Alice's on-board clock at t'=1 is Bob's blue network clock line of present. Alice's pink light signal from this point contains the information that her on-board clock reads t'=1 at the same instant Bob's remote network clock at the same point in space reads 1.25. It takes .75 yrs for the info to reach Bob at t=2. Subtracting the delay reveals Bob was indeed 1.25 at the simultaneous moment Alice was 1.

The longer pink line contains the same message even though it's not immediately apparent. Bob's moving network clock is the same as his stationary network clock except the line is slanted. It intersects Alice's on-board stationary clock at t=1. Alice sends this info via the long pink signal where it reaches Bob at 2 and it takes 1.5/2 = .75 yrs to get there with the doppler ratio correction. The horizontal dashed orange where the long pink signal starts is Alice's virtual network clock like the other slanted orange dashed line. So right here is the elusive symmetry I could not see before. I will redraw Jorrie's diagram to highlight it.

So there it is, how true reciprocal symmetry looks like. I was wrong about the details (symmetry is weird but not broken) but the diagram still shows the only true clock comparison between Bob and Alice during motion from either perspective is Bob is 5 and Alice is 4 at a common present. Any other comparison involves comparing future or past clocks to a present clock which, as I said, is not real but virtual. Discuss amongst yourselves for now, I've run out of time.
ralfcis
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### Re: S.R. Defining the present

ralfcis » 13 Oct 2017, 18:13 wrote:As I said in the previous post, no single spaceship is going to drag around a sync'd clock network stationary relative to the moving ship to check Bob's relative motion.

But you do agree that there is nothing stopping each inertial frame from having such a set of physical, synchronized clocks? Since both frames are 'stationary', there is no 'dragging around' of clocks. Each frame just put them there and they stay there.

The only physical way to check this is to send a light signal from Bob when he is .8 (light blue line) and subtract the light signal delay multiplied by the doppler ratio (1/2 at .6c) from the point Alice receives the message. So the light signal delay is 1.2/2 = .6 and that is subtracted from 1.6, when Alice receives the signal, and it confirms that Alice will be 1 when Bob was .8. It's a very weird definition of the present and I don't see any real purpose for it but relativity uses it extensively with the relativity of simultaneity to solve the twin paradox. I don't.

Me neither. This is an extremely roundabout way to make a simple observation. Alice simply puts friend Charlene at a distance -0.6 in her frame and Charlene reads Bob's clock when she passes him.

Let's do the two pink lines next as they are more difficult to understand. They are both light signals from Alice to Bob when Alice reaches 1 on her on-board clock. ... I don't know the physical meaning of this. Is this a purely mathematical construct or does it reflect on the constancy of c somehow?

A quite unnecessary complication to put into the mix, but it is part of the hyperbolic structure of spacetime that follows from the postulate that all inertial frames are equivalent. The isotropy of the speed of light follows from that.

So there it is, how true reciprocal symmetry looks like. I was wrong about the details (symmetry is weird but not broken) but the diagram still shows the only true clock comparison between Bob and Alice during motion from either perspective is Bob is 5 and Alice is 4 at a common present.

It would have been enlightening if you have done it completely symmetrically, because that night have rescued you from the false statement that followed. There is no common present for Alice and Bob, unless you restrict it to where they pass each other. At other times, they have their own private 'the presents'. I'm sure you know hat I mean.

BurtJordaan
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### Re: S.R. Defining the present

Since both frames are 'stationary', there is no 'dragging around' of clocks.

But as soon as you invoke motion between them then one of the networks of stationary clocks is either dragged past the other (Alice moving) or Alice has to remain stationary with her and all her network clock engines ablazing stationary to some point outside the universe while the universe ambles past. Either way Alice's engines are on.

But the practical considerations of physically establishing Alice's network clock aren't even important. I just learned the asymmetry never stemmed from the reciprocity, it was inherent from the fact that one of the network clocks constructs some weird present made up of one of the network clocks having an on-board present at one end of the network clock line of present and the other on-board clock either in the past (motion separating) or future (motion approaching).

Relativity applies a bandaid called relativity of simultaneity to mask this problem. RoS is nothing but the difference between Bob's stationary network clock that is in the same present as Alice's on-board clock and this weird present between Alice's virtual network clock and Bob's on-board clock which is in the past (or future if approaching). RoS is a bank of time that increases as they separate but is granted as a lump sum payment of time that gets used up by the guy approaching. When Alice separating hands off to Charlie approaching, there is more than a handoff of her clock reading. There is a reckoning and balancing between their RoS time banks.

Me neither.

But you do. The equations that you use are based on comparing only 2 clocks, RoS and Einstein's clock sync method all put in the mix. Those are the ingredients and initial assumptions I don't need and still manage to come up with the correct answer. If you want me to believe in those things, you need to come out of your comfort zone and show me where my interpretation is not correct taking on my thought patterns. You've said many times you don't wish to do this so instead of a discussion I get booted.

There is no common present for Alice and Bob

Pure semantics. The title of this thread is defining the present. But you just brush that all aside and use a definition of the present that you say is a term not even used in relativity without any rebuttal to all the different kinds of "present" I was able to determine. The STD is a diagram of everything that occurs simultaneously between Bob and Alice and you just go right back to your 2 clock comparison method when I keep proving there are 4 (3 actually since I discount Alice's network clock as being real).

Logic and math do not sway believers who do not even need discussions to arrive at the truth they already know. Unfortunately not one member of your panel of elders, who apparently also don't want any form of discussion despite the readership, ever chime in with their complaints in public. This is why I need my thread to stay on the physics forum, in case a physicist with an open mind is willing to correct me on my terms or agree with me. He won't be on the personal theories forum because that's just the wild west of conspiracy theories.

Anyway I'll clean up the first STD in the next post and add a second that will include Bob's signals. Then the purest symmetrical scenario of Bob and Alice leaving earth (the reference frame) at .33c each.
ralfcis
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### Re: S.R. Defining the present

ralfcis » October 14th, 2017, 12:19 pm wrote:Either way Alice's engines are on.

Surely not? She is inertial, and will remain inertial while her engines are off (ignoring any slight gravitational influences). A body remains at rest or in constant motion until a force acts upon it.

And I don't see how the practicalities of establishing a clock network are relevant. It is theoretically possible for Alice and Bob to each have their own clock network.
Positor
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### Re: S.R. Defining the present

She had to get moving somehow. Can the universe fire up some engines to get moving? No so Alice's ship is like a needle on a record and the universe is the record but the arm pivot is the stationary point Alice must be stopped relative to in order for the universe to whiz past her. Again this is impractical. Alice is moving, it's only from her perspective that the Earth is moving away from her. So if Alice is moving, she must be dragging an entire clock network with her. Again none of these details are important to the point I'm making so I'll agree with whatever you have to say about the details. It's just an obstructionist distraction from the main discussion.

I also said the impracticalities are a consideration. Physics main application is on things that would physically happen. Do you think anyone would waste any time creating a moving clock network to verify relativity when it absolutely doesn't advance any physical application. Maybe you can explain how a clock reading from the past gets on a line of present with a clock reading from the future. I misspelled impracticalities, would you rather discuss that instead?

The main thing is a present involving past and future as concurrent is not a theory I choose to believe if there is a simpler explanation that has practical results. The main discussion for this part of the thread is I'm saying Bob=5 and Alice =4 cannot be reversed so it is concrete. Relativity says it's not because reciprocity can make that time some other valid value until the concrete gets set by a stop. I'm also saying there's a preferred frame due to the disqualification of one of the 4 network clocks. Just stating the obvious that relativity says no to that doesn't negate my argument, a logical counter-argument does. But in order for that to happen, someone needs to understand what I'm saying and not quote scripture.
ralfcis
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### Re: S.R. Defining the present

With the moderating issues dealt with here, I just want to put a few hanging technical issues straight, for whoever may still be interested in the physics here.

Ralf wrote:She had to get moving somehow. Can the universe fire up some engines to get moving?

We were discussing two inertial-only observers for the complete 'test argument', as per your Minkowski diagrams. Whether Alice took off with a huge explosion that gets her to move away from Bob, makes no difference to this scenario. If you wanted a period of acceleration, it can be calculated, but then it is a different scenario, with different results.

Ralf wrote:The main discussion for this part of the thread is I'm saying Bob=5 and Alice =4 cannot be reversed so it is concrete. Relativity says it's not because reciprocity can make that time some other valid value until the concrete gets set by a stop.

Relativity says that all inertial frames have spacetime paths that are equivalent until you change one or more frame's state to non-inertial. This makes your "Bob=5 and Alice =4" a coordinate dependent observation and will have different values for different inertial observers. Reciprocity is something that logically follows from the equivalence of inertial frames.

Another way to put it, is that inertial frames follow equivalent paths through spacetime, provided that they stay inertial. As soon a you change an inertial clock's spatial path by accelerating it, its spatial path length between two events gets longer, which in relativity translates to shorter elapsed time, not 'slower time'. This is due to the constancy of the timelike spacetime interval: $\Delta s^2 = \Delta t^2 - \Delta x^2$ between two events.

Whoever wants to understand relativity, this is a basic concept that must be mastered. Putting up all sorts of thought experiments and trying to solve them without recognizing this basic premise, is doomed to confusion and frustration.

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