# Motion in space

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1. Jun 3, 2015

### Stephanus

Dear PF Forum
I want to know about these questions that are still bothering me,
Does the universe have prefered frame of refference?
Motion is relative, why 1 clock experiences time dilation while the other doesn't?

V = $\sqrt\frac{3}{4} ≈ 86.6\%$
If V define ratio of speed of light, then Lorentz factor is 2.

There are 4 probes.
Clocks are synchronized and reset. Time = T0
Their distance is 100 lys according to Pic 01

It's been 100 years since the clocks are synchronized and reset.
They have been sending digital signals.
The signals contain respective time and status, the time when they fire their rocket.

Then BB and B, preprogrammed, fire their rockets for 1 second, catapult them for V and stop.

A would have said. "Wow, B is traveling 86.6%c, but BB still stays". Is that right?
Because there's no way that A will know that BB has fired its rocket.

And after 10 years in A clock, will A still consider BB still at rest?
Because A only receives BB time and status, which is the rocket has never been fired, yet.
Does A still see that BB hasn't been contracted yet, and sees that B is already contracted?
Does A read BB clock showing T0+10 years?
And one more thing, tell me if I calculate correctly.
After 10 years in A clock, B keep sending signals
A will read B clock... at
L = V * Tb,
L = Distance traveled by B to send signal so that the signal will be received by A in 10 years.
Ta = L, Time taken by Light to reach A from L
Ta + Tb = 10
L + L/V = 10
L=10V/(1+V)
L = 4.641 Ly
Tb = 5.359 Ly
Lorentz factor is 2, so the clock in B reads 2.6795 years.
Is that right, that in 10 years, A will read the signal that B is T0+102.6795 years?

Back to T100
Would A see Pic03 the instant if rocket B and BB are fired and shot at V?

B is contracted while BB isn't, is this the right picture?

What if rocket A (and AA) is fired?
Would A see Pic04 the instant A (and AA) is shot at V?

Both BB, B and the distance is contracted by 50%

Does this why twins paradox occur.
Because in both cases the distance between A-AA and B-BB decreases but there is asymmetry?
But I don't ask when B reaches AA or A reaches BB, it's much more complicated I think with length contraction, where BB can reach A before B reach AA. I'll ask once I get this picture.

2. Jun 3, 2015

### pervect

Staff Emeritus
I'm not sure why these questions are in a strike-out font, or why you're asking them again. I guess what we said earlier didn't register :(. This does not make me hopeful that what we say will register this time, either, but I'll try.

To recap:

1) No, the universe doesn't have a preferred frame of reference. To clarify : experimental efforts to detect such a preferred frame of reference have to date all failed, and special relativity (conceived of in part beccaue of these failures of experiment to detect any such preferred frame) does not allow one.

2) The twin paradox is not an actual paradox. Rather, it's supposed to be an educational tool. Unfortunately, some people never figure it out :(

3) Motion is relative, and both clocks experience time dilation. I would recommend the PF thread https://www.physicsforums.com/threa...on-implies-relativity-of-simultaneity.805210/ "symmetrical time dilation implies the relativity of simultaneity" which describes how this is possible. I am tempted to repeat the arguments here, but I don't think it's asking too much to click on a link that answers your questions.

3. Jun 3, 2015

### Staff: Mentor

A cannot see (in the literal sense of receiving light signals) BB moving until light signals emitted by BB after he starts moving reach A. Does that answer your question?

Again, A cannot see (in the literal sense of receiving light signals) BB moving until light signals emitted by BB after he starts moving reach A. Does that answer your question?

A does not see (in the literal sense of receiving light signals) either B or BB as contracted. If B is large enough that light signals from different parts of B take significantly different times to reach A, then A will see (in the literal sense of receiving light signals) B as rotated (google "Penrose-Terrell rotation"). Similarly for BB, when light signals from BB emitted after he starts moving reach A.

A sees (in the literal sense of receiving light signals) BB's clock showing whatever reading was on BB's clock when the light signals were emitted.

As above, A sees (in the literal sense of receiving light signals) B's clock showing whatever reading was on B's clock when the light signals were emitted. I haven't verified your calculation, but if that was the intent of your calculation, then you have the right method.

A does not see (in the literal sense of receiving light signals) the distance from BB to B. A only sees (in the literal sense of receiving light signals) light signals coming from B and BB, with B's and BB's clocks showing certain readings. A can deduce the distance from BB to B, at some instant of A's time, only by using additional information: what A knows of BB's and B's state of motion, and whatever simultaneity convention A is using.

See above on "seeing" length contraction.

See above on "seeing" distances.

No.

4. Jun 3, 2015

### Stephanus

Thank you very much Pervect for your effort.
The reason why those lines were struck is: Actually that's what I want to know, why there's twins paradox.
But to know about twins paradox, I have to know what is inertial frame of reference, why the universe has no frame of reference.
And to understand twins paradox, I have to learn what is Lorentz transformation/contraction
All those questions are too much for me.
So I just settled with motion in space. That's why I struck those lines.

5. Jun 3, 2015

### Stephanus

Thank you very much PeterDonis for your help.

6. Jun 3, 2015

### Stephanus

Yes and yes and yes

I'll contemplate.

Ok.

So every thing A knows about BB (and about everything else in the universe) depends ONLY after light signals emitted by BB reach A.
I've seen you type this phrase very often.

Thanks for your confirmation. Okay..., now even if I haven't reached my target yet, but at least I'm no the right track.

Okay guys, thanks for you invaluable helps.

7. Jun 3, 2015

### Mentz114

The reason for the twins 'paradox' is because every clock measures its own path through spacetime. No clock or thing experiences universal time.

That is the way things appear to work.

8. Jun 4, 2015

### Stephanus

That's the questions that I would like to know in the end. But now, I have to understand motion in space

9. Jun 4, 2015

### Mentz114

Non-accelerated motion in space is easy - it does not exist.

If someone is in a closed box and feels no weight - how can they tell if they and the box are moving ?

10. Jun 4, 2015

### harrylin

I have the impression that your philosophy is at odds with the ones I know of, and possibly it's incompatible with SR. Can you explain how clocks in space can "record" time differently according to the philosophy that their motions do not exist?

11. Jun 4, 2015

### Mentz114

We may be using different definitions of 'motion in space' since I'm not sure what the OP means.

Using the word 'motion' without the 'relative' qualifier has no meaning except if there is absolute motion. I wanted to discourage that idea.

I think your philosophy is unimpugned.

12. Jun 4, 2015

### Stephanus

13. Jun 4, 2015

### pervect

Staff Emeritus
I'm interpreting this as the sticking point for you is the Lorentz transform. Why is this so difficult?

The Lorentz transform is based on two simpler notions, the notion of addition and the notion of proportionality.

Addition is simple enough - it's just addition. I'd feel funny trying to explain how to add. And I really think that the concept is well understood by most (I could be fooling myself, I suppose).

Proportionality is a bit trickier but not a lot trickier. It's just multiplication, it says that you multiply one number by a constant (or maybe a parameter) in order to get another number. I really think it's likely that most PF posters understand proportionality too, I've seen many lay posters successfully use the concept, and use arguments based on it.

It seems to me that a fair number of posters just give up (I'm guessing you're one of them) when confronted with combining the two concept - one has an equation that involves the sum of two terms (addition), each term f which is a multiplication of a variable by a paramter (a proportionality). And I don't understand why this is perceived as such an obstacle.

In math, if we choose units such that c=1, we write:

$x' = \gamma x - \beta \gamma t \quad t' = \gamma t - \beta \gamma x$

14. Jun 4, 2015

### Stephanus

Okay,,, if C = 1, than we can get rid of C from the equation as long as it's a multiplication not addition such as in $\beta$
Is it the time dilation and length contraction formula?
But.., whose x, whose t are we talking here?
One participant will see the other participant is contracting, the other will see the same.
A will see B is contracting, B will see A is contracting.
But when they meet, the clocks are not the same.

15. Jun 4, 2015

### Mentz114

No one can se a Lorentz transformation. It is an abstract formula that connects the rulers and clocks of 2 coordinate systems in inertial motion wrt each other.

What you will see if another something approaches you is blue shift in the light they send towards you. From this you can say that their clocks are running faster.
If they are receding from you then you get the opposite - their clocks are running slower. That is the physical facts.

Have another look at the diagrams ghwells made earlier.

That is because all clocks show their own spacetime interval. Forget time dilation/length contraction. Clocks and Doppler is where the physics is.

16. Jun 4, 2015

### Stephanus

Ahh, Doppler.
Nugatory and PeterDonis have said Doppler in my previous thread and I ignore it.

17. Jun 5, 2015

### harrylin

But what is your idea, by which you want to replace an idea that you dislike? Once more: according to you, motion relative to what causes a difference in recorded "time"? Do you suggest that motion relative to a construct of our mind (an inertial "reference frame") can cause one clock to get behind on another clock?

18. Jun 5, 2015

### A.T.

Physics is about predicting what the clocks will show, and the concept of inertial reference frames is useful for that. Philosophizing about "causation" on the other hand, is often futile.

19. Jun 5, 2015

### harrylin

Sure; I merely ask Mentz114 to clarify the philosophy that he is promoting here to Stephanus.

20. Jun 5, 2015

### Mentz114

I have no idea what you are talking about.

21. Jun 5, 2015

### 1977ub

The clocks are not the same because one traveler has moved in multiple inertial frames and one has not. The apparent slowness of one person's clock as measured by another during a particular leg of the journey turns out not to be important.

Two people A & B are moving roughly toward one another and pass close to each other. When they do, they synchronize their clocks, and then they are coasting apart. What could ever cause their clocks to come together again? Either A or B will have to turn on the jets and change to a different inertial frame to catch up with the other again in the future. Whichever person does that will feel the acceleration as they change frames, and it is this person's clock which will read less time when the clocks meet up again.

22. Jun 6, 2015

### Mentz114

That may be true for the case where one twin remains inertial. But how do you tell the ages of the twins if the both go on a trip in opposite directions and then meet up ?

23. Jun 6, 2015

### 1977ub

If they "go in opposite directions" they are both inertial during that initial period. There is no way to say which is the "stay at home" twin. Whichever one changes direction, stopping the separation, and starts moving toward the other will have aged less when they meet again. That person's trip takes place in 2 inertial frames.

24. Jun 6, 2015

### Mentz114

But in my setup they both stop and turn round and meet.

The answer is - whichever twin has the shortest path through spacetime will have aged less.

25. Jun 6, 2015

### Staff: Mentor

This is not exactly wrong, but it is terribly misleading. Clocks moving on different paths through spacetime generally register different amounts of time because the lengths of the paths are generally different - it's like finding that cars driven on different routes through space generally end up recording different mileages on their odometers. Inertial frames are pretty much a red herring here, as the amount of time elapsed on the clock on its path through spacetime is the same in all frames, whether inertial or not.