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  4. Can you help me understand "speed"?
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Can you help me understand "speed"?

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Online yor_on

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Re: Can you help me understand "speed"?
« Reply #20 on: 31/03/2020 18:38:27 »
If the universe would rotate it also would allow a Gödel universe with closed timelike curves. https://web.archive.org/web/20070701033428/http://www.ettnet.se/~egils/essay/essay.html

But to get something to rotate you seem, at least to me, to need a frame of reference relative it can be defined to rotate? If you check the essay you will find something entirely different about the idea there. It somehow avoid that question.

to me it falls down to whether you by defining it so that all 'laboratories' in the universe, using the same techniques described in the essay, find a equal 'rotation' taking place, in what way do you define the universe as a whole 'objective frame' not to rotate, or if you like 'rotate'?
=

What I mean is that it to me by necessity must introduce a 'objective' universe, relative a 'outside', which this 'rotation' takes place. Maybe it's possible to avoid by keeping the solution 'local' at all 'points' but it then gives me a very complicated universe methinks.To me it connects to Mach's principle https://en.wikipedia.org/wiki/Mach%27s_principle
==

Actually this whole idea may be possible, locally defined, which I find rather frustrating as that there is where I come from myself. And I don't believe in time travels. So to get it right we better build one of those laboratories and see what we will see :)
« Last Edit: 31/03/2020 18:59:28 by yor_on »
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Re: Can you help me understand "speed"?
« Reply #21 on: 31/03/2020 19:13:18 »
Quote from: Bill S on 31/03/2020 16:51:56
Quote from: Halc
That said, I cannot think of a single example of a non-rotating object except for fundamental particles that lack a property of angular moment.

Possibly the entire Universe?
I've always found that interesting contradictions arise when the universe is treated as an object.  If anything, the universe is a spacetime structure, and moving (rotating or otherwise) spacetime is a contradictory concept.
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Re: Can you help me understand "speed"?
« Reply #22 on: 31/03/2020 19:29:30 »
If you take your local clock Bill it always tick in one direction. In Gödels universe all (local) clocks also must tick in a same direction to have a relevance for the one we are in. So to get back 'in time' you then need to introduce another frame of reference with a clock that ticks 'backwards, relative yours. You then also need to define that frame of reference as being in the same universe (and possible to reach by you). I don't think that is possible unless we define it so that 'time' and 'clocks' are non existent. That's not what we see, we see frames of reference in where we can define some other frame to have stopped ticking (relative our own clock) , like a event horizon, but I don't know of any frame in where you can find a clock going backwards.
=

syntax
« Last Edit: 31/03/2020 19:33:36 by yor_on »
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Re: Can you help me understand "speed"?
« Reply #23 on: 31/03/2020 19:38:19 »
Yeah, that's true Halc. But if we keep it strictly local there are no contradictions. The contradictions comes when we compare our local definitions to another 'frame of reference'. Every local frame has some principles that joins them, one of them being 'c' and the arrow of time.
=

Maybe you can call it limits. One limit being how far you can stretch a 'objective universe'. To stretch it into Gödels you must introduce a timeless universe. Because the idea of time travel is that you can go back to a SpaceTime position you already passed temporarily. And as the essay points out, it kills causality, which then also mean that everything we build science on must be wrong, and it isn't. It makes any idea of a repeatable experiment questionable.

So even if locally possible it must fail as soon as you introduce a SpaceTime position you already passed temporarily, as I think then. Unless you connect it to 'multi verses' which to me is a another questionable idea.

( But which would make for a pretty cool Science Fiction :)
« Last Edit: 31/03/2020 23:25:44 by yor_on »
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Re: Can you help me understand "speed"?
« Reply #24 on: 31/03/2020 20:11:26 »
A example of it. Presuming you can split 'time' into Plank scale you now can use that to 'go back' to a same SpaceTime position, for let's say a normal life span, around 75-85 for a European male human. We can all meet up and populate the same instant, on the same earth.. It will be a incredible lot of me, meeting me, there :)

And I pity my girlfriend, as she meets us...
=

( Actually this becomes a definition of a 'infinity' presuming that all of 'me', every Planck time, goes back to the same SpaceTime position. We won't even be able to fit, and that poor girl? With all of it 'happening instantly' as far as she is concerned..)

One way to kill a universe.
And a relation
« Last Edit: 31/03/2020 20:35:45 by yor_on »
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Re: Can you help me understand "speed"?
« Reply #25 on: 31/03/2020 23:54:11 »
Back to basics for a moment. Forget accelerations, gravitation, rotation and stuff. Let a radioactive nucleus expel two electrons A and B in opposite directions, each travelling in a straight line at 0.6c relative to the residual nucleus. It's a rare event but by no means impossible. Momentum is conserved. What is the speed of B relative to A?   
« Last Edit: 31/03/2020 23:56:15 by alancalverd »
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Re: Can you help me understand "speed"?
« Reply #26 on: 01/04/2020 01:51:10 »
Quote from: alancalverd on 31/03/2020 23:54:11
Let a radioactive nucleus expel two electrons A and B in opposite directions, each travelling in a straight line at 0.6c relative to the residual nucleus. It's a rare event but by no means impossible. Momentum is conserved. What is the speed of B relative to A?
About .882c
Conserved momentum has little to do with that.
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Re: Can you help me understand "speed"?
« Reply #27 on: 01/04/2020 17:01:07 »
Quote from: alancalverd on 31/03/2020 23:54:11
Back to basics for a moment. Forget accelerations, gravitation, rotation and stuff. Let a radioactive nucleus expel two electrons A and B in opposite directions, each travelling in a straight line at 0.6c relative to the residual nucleus. It's a rare event but by no means impossible. Momentum is conserved. What is the speed of B relative to A?   
As measured from what frame of reference?   
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Re: Can you help me understand "speed"?
« Reply #28 on: 01/04/2020 17:21:02 »
Quote from: Janus on 01/04/2020 17:01:07
Quote from: alancalverd on 31/03/2020 23:54:11
Back to basics for a moment. Forget accelerations, gravitation, rotation and stuff. Let a radioactive nucleus expel two electrons A and B in opposite directions, each travelling in a straight line at 0.6c relative to the residual nucleus. It's a rare event but by no means impossible. Momentum is conserved. What is the speed of B relative to A?   
As measured from what frame of reference?
Both A and B. If it were one but not the other that would be a genuine paradoxical contraction.

From the frame of the nucleus it's 1.2c. 2c is actually the maximum speed allowed in the universe. 1c is just for objects in motion relative to the observer, and then it only applies to linear velocity, not angular.

So if two object are moving relative to the observer in a straight line in different directions the relative velocity of each will be subject to the velocity addition formula from sr but not when you then take those to work out their velocity relative to each other in your frame. They'll of course be moving at <c relative to each other in their own frames.
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Offline Janus

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Re: Can you help me understand "speed"?
« Reply #29 on: 01/04/2020 19:47:10 »
Quote from: A-wal on 01/04/2020 17:21:02
Quote from: Janus on 01/04/2020 17:01:07
Quote from: alancalverd on 31/03/2020 23:54:11
Back to basics for a moment. Forget accelerations, gravitation, rotation and stuff. Let a radioactive nucleus expel two electrons A and B in opposite directions, each travelling in a straight line at 0.6c relative to the residual nucleus. It's a rare event but by no means impossible. Momentum is conserved. What is the speed of B relative to A?   
As measured from what frame of reference?
Both A and B. If it were one but not the other that would be a genuine paradoxical contraction.

From the frame of the nucleus it's 1.2c. 2c is actually the maximum speed allowed in the universe. 1c is just for objects in motion relative to the observer, and then it only applies to linear velocity, not angular.

So if two object are moving relative to the observer in a straight line in different directions the relative velocity of each will be subject to the velocity addition formula from sr but not when you then take those to work out their velocity relative to each other in your frame. They'll of course be moving at <c relative to each other in their own frames.
My question was aimed at alancalverd, as he does not specify the frame for the answer.  Halc gave the answer as measured by A or B,  and as you mentioned, the separation speed between A and B is 1.2c in the frame of the nucleus. 
The point being that the answer depends on the reference frame.
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Re: Can you help me understand "speed"?
« Reply #30 on: 01/04/2020 21:27:46 »
Quote from: Janus on 01/04/2020 19:47:10
My question was aimed at alancalverd
It missed, sorry. :)
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Re: Can you help me understand "speed"?
« Reply #31 on: 03/04/2020 13:17:02 »
Quote from: Halc on 09/11/2019 13:31:40
Speed is always relative to something
So is distance. Length is just distance between one point on an object boundary and another point on the same object boundary along an axis.
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Re: Can you help me understand "speed"?
« Reply #32 on: 03/04/2020 15:08:24 »
Quote from: hamdani yusuf on 03/04/2020 13:17:02
Quote from: Halc on 09/11/2019 13:31:40
Speed is always relative to something
So is distance. Length is just distance between one point on an object boundary and another point on the same object boundary along an axis.
Under relativity, length is just distance between a pair of events, which is different than distance between a pair of spatial points.  When I measure a length-contracted object in some frame where it is moving, I'm comparing two different events than had I taken the same length measurement in the frame of the object.
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Re: Can you help me understand "speed"?
« Reply #33 on: 03/04/2020 15:49:11 »
A mirror is placed 1 light hour away from a powerful laser pointer. Right when the laser is turned on, Alice move from laser pointer towards the mirror at 1 m/s. Bob stays where the laser pointer is.
Who will see the reflected laser first?
What's the speed of light before and after reflection when measured by Alice? Is it still the same as distance divided by travel time?
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Re: Can you help me understand "speed"?
« Reply #34 on: 03/04/2020 17:38:41 »
Quote from: hamdani yusuf on 03/04/2020 15:49:11
A mirror is placed 1 light hour away from a powerful laser pointer. Right when the laser is turned on, Alice move from laser pointer towards the mirror at 1 m/s. Bob stays where the laser pointer is.
Who will see the reflected laser first?
What's the speed of light before and after reflection when measured by Alice? Is it still the same as distance divided by travel time?
To make this simple, we will assume Alice is already moving towards the mirror at 1 m/s relative to the pointer as she passes it and it is turned on.
Because she is moving at 1m/sec relative to both pointer and mirror, she would not measure the distance between them as being 1 light hr, but rather 0.99999999999999999443674971973191 light hr.  as measured relative to herself, the laser will travel at c, towards the the mirror, while the mirror moves towards her at 1m/sec.  It will take 0.99999999666435905358172976595532 hrs for the laser and mirror to meet.  During which time the distance between Alice and the mirror would decrease by 3599.9999879916925928942271574392 meters.  The laser will then return at c from this distance to Alice.  At this point, I'm going to quit giving numbers, because it is just too cumbersome to keep working with all these decimal places.
Instead, I'm going to change the scenario to use a velocity for Alice that is easier to work with.
Thus Alice now moves at 0.6c. relative to laser pointer and mirror.
Now:
As Alice passes the pointer, she will measure the distance to the mirror as being 0.8 light hrs, and the mirror will be approaching at 0.6c, while the light from the laser will be speeding towards it at c relative to Alice.  Thus, according to Alice, the light will take (0.8 lh/(1c+.6c) = 0.5 light hr to meet up with the mirror.  During which time, the distance between Alice and the Mirror will have decrease by 0.5 hr * 0.6c = 0.3 lh to 0.5 light hr.   The refelcted light, traveling at c will take 0.5 hr to get back to Alice.  Total time for Alice for round trip of laser  1 light hr.
Now, if we work it out from Bob's view:
The light travels at c towards the Mirror, arrives at the mirror in one hour and returns 1 hr later, total trip time 2 hr.
Alice, is moving towards the mirror at 0.6 c, so, in the time it take for the light to reach the Mirror, Alice will have moved to be 0.6 light hrs closer to the mirror to be 0.4 lh away from it, and will meet with the returning light in another (0.4 lh/(1c+0.c) = 0.25 hr,  total time between firing of laser and it meeting up again with Alice is 1.25 hrs.  However, since Alice is moving at 0.6 c relative to Bob, He will measure her clock as time dilated by a factor of 0.8 and only tick off 1.25hr * 0.8 = 1 hr.  The same amount of time that Alice says here clock ticked off.
If we go back to Alice's view, we can also work out how much time she would say ticks off on Bob's clock between Firing the laser and the light returning to Bob.
As noted above, Alice measures 1 hr for the light to meet up with her. During which time, she would measure Bob's clock as being time dilated and ticking off 0.8 hr.  Now in that hr, Bob, with a relative velocity of 0.6c, has moved 0.6 lh away from Alice.  The light passing Alice on it way to him has to "chase after" the receding Bob at c.  This takes 0.6 lh (1-0.6c) = 1.5 hours by Alice's clock. During which time Bob's clock accumulates 1.5 hr *.8 = 1.2 hr.  Added to the 0.8 hr already accumulated equals 2 hrs total time accumulated by Bob's clock according to Alice.  The same as what Bob's recorded.
Both Alice and Bob agree that the reflected Laser hits Alice first. ( though Alice would say that when the Light hits her, Bob's clock read 0.8 hr, and Bob would say that when the light hit Alice his clock read 1.25 hrs, And Alice would say that when the light returns to Bob, her clock reads 2.5 hrs, while Bob would say that her clock reads 1.6 hrs.
 
So the answer to what the speed of the light is according to Alice both before and after refection, it is c ( relative to Alice.) just like it is c relative to Bob as measured by Bob's.   It is also distance divided by travel time, keeping in mind that Bob and Alice do not measure either distance or time the same.
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Re: Can you help me understand "speed"?
« Reply #35 on: 04/04/2020 05:12:00 »
Quote from: Janus on 03/04/2020 17:38:41
Both Alice and Bob agree that the reflected Laser hits Alice first.
Is there any reference frame which sees that reflected light hits Bob first due to relativity of simultaneity?
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Re: Can you help me understand "speed"?
« Reply #36 on: 04/04/2020 06:34:42 »
Quote from: hamdani yusuf on 03/04/2020 15:49:11
Who will see the reflected laser first?
Always Alice, in any frame.  She's closer when it comes back. Janus goes so far as to compute how much sooner, which is a frame dependent thing.
Quote
What's the speed of light before and after reflection when measured by Alice? Is it still the same as distance divided by travel time?
Under SR, speed of light is constant in any frame, measured by anybody.  Yes, distance over time, by definition.
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Re: Can you help me understand "speed"?
« Reply #37 on: 04/04/2020 16:54:07 »
Quote from: hamdani yusuf on 04/04/2020 05:12:00
Quote from: Janus on 03/04/2020 17:38:41
Both Alice and Bob agree that the reflected Laser hits Alice first.
Is there any reference frame which sees that reflected light hits Bob first due to relativity of simultaneity?
No.  Relativity of simultaneity cannot reverse the causality of events.  In this case, Alice is always between the mirror and Bob. and the the reflected light always leaves Bob, hits the mirror and returns to Bob, in that order. Ergo, the light must pass alice first on its way back to Bob.
Everyone in all  frames agree that the light leaves Bob as Alice passes,  both Bob's and Alice's clocks read 0, that Alice's clock reads 1 hr when the reflected light hits it and Bob's clock reads 2 hrs when the light reaches it.
All frames will also agree that Alice's clock reads 1 hr before Bob's clock reads 2 hr.

Below shows the Space-time diagrams for these events in as measured from different frame of reference.

* Image11.gif (23.41 kB . 1536x1106 - viewed 3901 times)
Top left is "Bob(blue line) and mirror(red line) at rest", with Alice (green line) moving at 0.6c  The yellow lines are the light moving at c.
In these diagrams simultaneous events are directly horizontal to each other.
as we see, the returning light hits Alice before Bob.
Next over is "Alice at rest", with Bob and The mirror moving at 0.6 c .
Again, the reflected Laser hits Alice first on its way to Bob.

Next is a frame of reference has a relative velocity of 0.6c with respect to Bob and the mirror, but in the opposite direction from Alice.  Note while the vertical separation (representing time difference) is less for the period between the light hitting Alice and hitting Bob, it still hits Alice first.
Lastly, we have a frame with a relative velocity of ~0.882 c relative to Bob, in the same direction as the last frame.
The time gap between The light hitting Alice and Bob has shrunk even more, but it still hits Alice first.
we can keep increasing the velocity of this frame relative to Bob closer and closer and closer to c, and while the time gap between the light hitting Alice and hitting Bob will continue to decrease, it will reverse the order of these events (or even shrink the time gap to zero).
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Re: Can you help me understand "speed"?
« Reply #38 on: 06/04/2020 08:39:19 »
Quote from: Janus on 03/04/2020 17:38:41
Now, if we work it out from Bob's view:
The light travels at c towards the Mirror, arrives at the mirror in one hour and returns 1 hr later, total trip time 2 hr.
Alice, is moving towards the mirror at 0.6 c, so, in the time it take for the light to reach the Mirror, Alice will have moved to be 0.6 light hrs closer to the mirror to be 0.4 lh away from it, and will meet with the returning light in another (0.4 lh/(1c+0.6c) = 0.25 hr,  total time between firing of laser and it meeting up again with Alice is 1.25 hrs.
I can calculate in Bob's frame when and where Alice will see the reflected light without using relativistic velocity addition.
x=where Alice see reflected light
t=when Alice see reflected light
from Alice's travel x=0.6ct
from light's travel (2-x)=ct
1.2-0.6x=x
x=1.2/1.6=0.75 light hours
t=x/0.6c = 1.25 hours

In Alice's frame, it would be translated to
1/γ=0.8
x'=0.75*0.8 = 0.6 light hours
t'=1.25*0.8 = 1 hours

Do I get the correct numbers?
« Last Edit: 06/04/2020 11:02:24 by hamdani yusuf »
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Re: Can you help me understand "speed"?
« Reply #39 on: 06/04/2020 12:12:46 »
Quote from: hamdani yusuf on 06/04/2020 08:39:19
In Alice's frame, it would be translated to
1/γ=0.8
x'=0.75*0.8 = 0.6 light hours
t'=1.25*0.8 = 1 hours

Do I get the correct numbers?
I agree on the 1 hour, but the mirror then must have reflected the light after 30 minutes, and thus the (moving) mirror is 0.5 light hours away when it reflects the signal.
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