Naked Science Forum

Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Kean1 on 09/11/2019 13:13:18

Title: Can you help me understand "speed"?
Post by: Kean1 on 09/11/2019 13:13:18
(this is my first post on here so be gentle with me!)
Quoted from latest episode "eggs, eyes etc..., in answer to the question: "How fast a rocket can humans safely travel in?"
"So the Earth rotates at 600 miles an hour. So we're currently on the earth going 600 miles an hour, because we're on the earth we're also going 600 miles an hour. The Earth is speeding around the sun at 70,000 miles per hour. So we are also speeding around the sun at 70,000 miles an hour. The sun is speeding through the galaxy at 450,000 miles an hour, which means we're also spinning at 450,000 miles an hour."

I'm probably dim, but how can I understand "speed" in this statement? Are the references always implicit? For example, in the first statement "we're on earth going 600 mph", should I implicitly understand that the "reference" (if that's even the correct word!) is the centre of the earth, and this jumps to the sun in the second statement "The Earth is speeding around the sun at 70,000 miles per hour". In other words, in astronomy, when speeds are mentioned, should I just assume that I should understand the quoted speed relative to an "obvious" reference.

The reason I'm asking the question is because I simply don't "get" relativity, time dilation etc...e.g. in the classic example of my "twin" charging off in a rocket at near the speed of light aging differently to me, am I not simply "aging" at the same rate relative to my twin, because relative to him/her, I'm charging off in the opposite direction at an equal and opposite speed, therefore cancelling out any warping of time?
Hope that makes (some) sense!
Title: Re: Can you help me understand "speed"?
Post by: Halc on 09/11/2019 13:31:40
Quoted from latest episode "eggs, eyes etc..., in answer to the question: "How fast a rocket can humans safely travel in?"
"So the Earth rotates at 600 miles an hour.
Maybe where they are (Montreal?). At the equator, it is over 1000 mph.  25000 mile circumference in 24 hours, no?

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So we're currently on the earth going 600 miles an hour, because we're on the earth we're also going 600 miles an hour.
We're going 600 mph relative to the center of Earth.  Speed is always relative, so not specifying the relation is wrong.  I'm not moving 600 mph relative to my mailbox. I'm nearly stationary relative to that.

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The Earth is speeding around the sun at 70,000 miles per hour. So we are also speeding around the sun at 70,000 miles an hour. The sun is speeding through the galaxy at 450,000 miles an hour, which means we're also spinning at 450,000 miles an hour."
Just so, relative to the sun and the galaxy respectively.

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I'm probably dim, but how can I understand "speed" in this statement? Are the references always implicit? For example, in the first statement "we're on earth going 600 mph", should I implicitly understand that the "reference" (if that's even the correct word!) is the centre of the earth, and this jumps to the sun in the second statement "The Earth is speeding around the sun at 70,000 miles per hour". In other words, in astronomy, when speeds are mentioned, should I just assume that I should understand the quoted speed relative to an "obvious" reference.
You got it.  Speed is always relative to something, and language usually lets you get away with letting it be implicit, but the reference is always there.  The speedometer on my car shows speed relative to the road under it, and doesn't bother to say that explicitly.
Velocity (a vector) is also a relation.  So I might be moving at speed 600 mph relative to Earth, and some guy on the planet opposite me also is moving at the same 600 mph speed, but our velocity is very different since we're moving in different directions.  Speed is a scalar, and velocity is a vector.

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The reason I'm asking the question is because I simply don't "get" relativity, time dilation etc...e.g. in the classic example of my "twin" charging off in a rocket at near the speed of light aging differently to me, am I not simply "aging" at the same rate relative to my twin, because relative to him/her, I'm charging off in the opposite direction at an equal and opposite speed, therefore cancelling out any warping of time?
Under relativity, it is the other twin that is moving relative to either twin, so in the current inertial frame of either twin, the other twin is aging slower.  Not intuitive, but that's how it works.
When the 'travelling' twin turns around, his inertial frame changes, and in that frame the twin back home simultaneous with the turnaround event is much older than his age in the outbound frame.  That's the relativity of simultaneity in action, and is why one twin is older than the other when they finally meet again.
Title: Re: Can you help me understand "speed"?
Post by: jeffreyH on 09/11/2019 14:04:52
You are attracted to the earth by gravity. This is important because it puts you in an accelerating frame. But you say, "I'm standing still. How can I be accelerating?" If there was a hole under your feet you would start falling. So the ground is supporting you.

Once you are falling freely you are then in a frame indistinguishable from an inertial or non accelerating frame. But you say "How can I be in an inertial frame? I am accelerating down into a hole."

Welcome to relativity!
Title: Re: Can you help me understand "speed"?
Post by: PmbPhy on 09/11/2019 16:05:16
(this is my first post on here so be gentle with me!)
Quoted from latest episode "eggs, eyes etc..., in answer to the question: "How fast a rocket can humans safely travel in?"
There is no limit to how fast human in free-fall can move other than the speed of light, which can be shown that nothing can move faster than that. You've been thinking about acceleration and then quoting speed. If the ship is free fall then it can move at any speed less than the speed of light.
Title: Re: Can you help me understand "speed"?
Post by: yor_on on 10/11/2019 15:19:09
Nice to see you Pmb, and interesting comment that I totally agree too..

As for  'speeds' you always will need to define something from where you measure it. In relativity there is no defined universal platform of a null speed. And all stars planets etc 'uniform motion' then becomes equivalent, as defined locally. By that I mean you f.ex doing a experiment on earth trying to see if it moves, being inside a 'black box' without external information. That also leads us to the equivalence principle in where a uniform constant acceleration becomes inseparable from the gravity you find yourself to have inside that box.
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Although there is a difference, thinking of it.

Presume you set up a experiment in where you place a light emitter in the middle of your box. Then place two detectors, one at the 'front' of the box, the other at the 'end'. If you're in a gr.....

damn, it won't matter :) The light red and blue shifts both in your gravitational 'field', as well as it will do in a accelerating box. so there will be no difference measured, if we ignore the way earth f.ex spins and  'wobble away'. Will blame this on being away from physics for a while.

What you could use the experiment for is to define the 'gravity' at where the box is situated, but that's all. Excluding outside information it won't tell you a thing about any possible 'speed'. Although if you knew the mass, wouldn't that make a difference?

No not really, if we instead let the box 'float' outside a gravitational potential (aka in 'deep space') you won't see any red and blue shifts, no matter what you would define your speed to be versus some origin.

Ideally that is, because the mass of the box might interfere somewhat with your experiment. Discounting that there will be no proofs for a motion inside that box. This can also be called being in a same 'frame of reference' which just means that you the box and your experiment are 'motionless' relative each other (in this case).

But 'frames of reference' are tricky.
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spelling and syntax
Title: Re: Can you help me understand "speed"?
Post by: Kean1 on 12/11/2019 07:53:15
Thanks all for your replies.
Halc, you wrote (many thanks!):
"Under relativity, it is the other twin that is moving relative to either twin, so in the current inertial frame of either twin, the other twin is aging slower.  Not intuitive, but that's how it works.
When the 'travelling' twin turns around, his inertial frame changes, and in that frame the twin back home simultaneous with the turnaround event is much older than his age in the outbound frame.  That's the relativity of simultaneity in action, and is why one twin is older than the other when they finally meet again."

Really not intuitive... "So in the current inertial frame of either twin, the other twin is aging slower".
What I just can't get past is this idea that time appears to be slowing down for both twins, it just depends on whether you are measuring from one twin's position or the other's...

I'll spare you the ordeal of "banging your head against a brick wall" trying to clarify this further and I'll just have to dig deeper (down the rabbit hole!) and try harder to get my head around it.
Perhaps, if it's not too much of a bother, and you understand my confusion, do you happen to know of any links to explanations that might help me...just idle curiosity as I don't NEED to know this. The trouble I've had is that everything I've read/watched on Youtube that describes, for e.g., the twin thought experiment, always takes as a given that the "reference point" is the twin/you "here on earth", let's say. But if I assume that "I" am the other twin, the one zooming off in the space ship how can I demonstrate/reason that it is "I" that am moving and not the twin on earth that is zooming away from me at an equal and opposite velocity??
Title: Re: Can you help me understand "speed"?
Post by: Colin2B on 12/11/2019 10:02:59
But if I assume that "I" am the other twin, the one zooming off in the space ship how can I demonstrate/reason that it is "I" that am moving and not the twin on earth that is zooming away from me at an equal and opposite velocity??
Relatively you can’t. But let’s be realistic about this, you are the one travelling in the ‘twins scenario’ and Earth isn’t going to suddenly come and find you in your frame.  However, if you waited hanging in space for the Earth to complete its orbit and return to you, people on earth would have aged less than you as the final measurement is made in your frame.
Title: Re: Can you help me understand "speed"?
Post by: Halc on 12/11/2019 12:56:47
Really not intuitive... "So in the current inertial frame of either twin, the other twin is aging slower".
That is basic time dilation.  Time 'runs' at full speed if you're stationary, and all objects are stationary in their own frames.  This is consistent with Galilean relativity: "Physics is the same in any frame".  Time dilation means that time slows down for moving things, and in any frame, it is everything not stationary in that frame that is moving.  So relative to the twin in the ship, the Earth twin is the one moving.  Earth is just another ship after all, just a larger one than usual.

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What I just can't get past is this idea that time appears to be slowing down for both twins, it just depends on whether you are measuring from one twin's position or the other's...
Be careful about 'appears to be'. Time is computed to be slower in these other frames. The twins are not in each other's presence, and hence neither has a direct way to measure the other.  In fact, if a clock is approaching you fast, it will 'appear' to be running faster, but that's mostly due to Doppler effect, and is why light from approaching galaxies is blue shifted despite being dilated a bit slower.
Say there are twins X and Y moving apart moving fast enough for 2x dilation.
In the frame of one twin X, his clock reads 5 and the clock of the other twin simultaneously (in X's frame) reads say 8.  These are arbitrary numbers, let's just say the value of the minute hand.
Two minutes later (in X's frame), X's clock reads 7 and Y's clock simultaneously (in X's frame) reads 9.
Note that I put 'in X's frame' next to each mention of simultaneous since simultaneity is relative, and two events simultaneous in one frame are not simultaneous in another.  The word is ambiguous without this reference. This (relativity of simultaneity) is the most important part because this explains why one twin comes home objectively younger than the other.

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Perhaps, if it's not too much of a bother, and you understand my confusion, do you happen to know of any links to explanations that might help me...just idle curiosity as I don't NEED to know this.
https://en.wikipedia.org/wiki/Relativity_of_simultaneity
That is probably one of the best places to start. They have some nice diagrams/gifs, and there's a description of the classic train thought experiment from which RoS is demonstrated from the postulate of frame independent light speed. Wiki is great for this kind of thing since it is full of hyperlinks you can click on for detail on any part that into which you want to go deeper.

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The trouble I've had is that everything I've read/watched on Youtube that describes, for e.g., the twin thought experiment, always takes as a given that the "reference point" is the twin/you "here on earth"
OK, I know that. For that very reason, I always choose the frame of the guy in the ship or the guy in the train, which really helps one discard a lot of the assumptions about absolute motion.  With a westbound train it is pretty much true.  A really fast train going west is actually more stationary (accelerates less) and has Earth rotating underneath it.  Few trains move at fast enough speeds to actually cancel rotation, but jets sometimes do.

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But if I assume that "I" am the other twin, the one zooming off in the space ship how can I demonstrate/reason that it is "I" that am moving and not the twin on earth that is zooming away from me at an equal and opposite velocity??
Colin's answer to this is correct. There is no way to detect absolute motion, or more precisely, no local test.  The two twins moving apart at high speed is entirely symmetrical, and thus each ages slower in the frame of the other.  But one of them significantly changes reference frames, and does so a long distance from the other.  That's what makes the difference when they are reunited.  Remember RoS?
Say the rocket takes 5 years (ship time) each way, with 2x time dilation.  At the turnaround point in the outgoing frame, the clock on the stationary ship reads 5 and simultaneous (in that outgoing frame), the clock of the distant (moving away) twin reads 2.5 years.  Now the first twin turns around and is now stationary in a very different inbound frame.  In that frame, his clock still reads 5 (he hasn't gone anywhere yet), but the clock of the distant (moving towards us) twin reads 17.5 years.  During the remaining part of the exercise which the stationary twin waits for Earth to come to him, he ages 5 more years (10 total), and the traveling guy ages 2.5 more (20 total).  The Earth guy has aged twice that of the 'stationary' twin.
RoS is illustrated there.  Relative to two different frames (inbound and outbound), the turnaround event is simultaneous with two very different events (times 2.5 and 17.5) back on Earth.
Title: Re: Can you help me understand "speed"?
Post by: Origin on 14/11/2019 14:18:12
How fast a rocket can humans safely travel in?"
Any speed won't hurt a person.  As a result of tremendous speed other things like radiation could kill you, but the speed itself is not an issue.  If you are on a plane going at, say 600 mph you would not even feel like you are moving, the same would be true if you were travelling at 10,000,000 kph.
Title: Re: Can you help me understand "speed"?
Post by: Bill S on 17/11/2019 16:34:09
Quote from: Kean1
Perhaps, if it's not too much of a bother, and you understand my confusion, do you happen to know of any links to explanations that might help me...just idle curiosity as I don't NEED to know this.

  http://www.owl232.net/papers/twinparadox.pdf

I found Michael Huemer’s explanation very helpful. 
MTW; never doubt the value of curiosity, it might try the patience of others, but it’s seldom idle. :)
Title: Re: Can you help me understand "speed"?
Post by: LB7 on 17/11/2019 20:55:25
Any speed won't hurt a person.  As a result of tremendous speed other things like radiation could kill you, but the speed itself is not an issue.  If you are on a plane going at, say 600 mph you would not even feel like you are moving, the same would be true if you were travelling at 10,000,000 kph.
True in translation, not in rotation like a circular path. The links between the atoms are weak and don't like centrifugal forces. Does a pure translation can exist in the Universe ?
Title: Re: Can you help me understand "speed"?
Post by: Bored chemist on 17/11/2019 21:05:00
Any speed won't hurt a person.  As a result of tremendous speed other things like radiation could kill you, but the speed itself is not an issue.  If you are on a plane going at, say 600 mph you would not even feel like you are moving, the same would be true if you were travelling at 10,000,000 kph.
True in translation, not in rotation like a circular path. The links between the atoms are weak and don't like centrifugal forces. Does a pure translation can exist in the Universe ?

One of the interesting properties of light is rectilinear propagation.
Title: Re: Can you help me understand "speed"?
Post by: LB7 on 17/11/2019 21:25:25
One of the interesting properties of light is rectilinear propagation.
Except when the light is close to a black hole, so like there is matter in the Universe, the trajectory of light is a pure translation ? I mean, EXACTLY a translation ?
Title: Re: Can you help me understand "speed"?
Post by: hamdani yusuf on 20/11/2019 04:15:31
Cannot parse this. What do you mean by 'translation'?
I think it's about linear motion, which has straight line trajectory.
Title: Re: Can you help me understand "speed"?
Post by: yor_on on 23/11/2019 17:03:46
One of the interesting properties of light is rectilinear propagation.
Except when the light is close to a black hole, so like there is matter in the Universe, the trajectory of light is a pure translation ? I mean, EXACTLY a translation ?

Not sure what you mean by translation, but light always take the shortest path between points. That means that light does not 'bend' It just follows the shortest 'path' through SpaceTime according to what I understand. And around a black hole space is not flat but 'curved'.
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It's like Halc wrote above. And comparing a rotational motion to a geodesic is like comparing a apple to a orange. A rotation (carousel) is GR and accelerations, a geodesic is SR and 'relative/uniform motion', so BC is perfectly correct. As a example you can think of earth rotating the sun. That's not a 'rotation' ala GR, instead it's the 'straightest path' through a 'curved' SpaceTime. And the proof for that is the absence of effects of a acceleration from it, on any of us on earth. It's a 'relative uniform motion' following a geodesic, with only 'proper mass' acting and being acted upon.

(Well, earth do spin around its axis too which could be considered a form of acceleration so it's not as simple as I would like it, and then you have the equivalence principle in where earths gravity becomes a 'acceleration' all on its own)

But if we ignore those two for now the average speed of earth relative its 'orbit' around the sun is 66,627 mph (107,226 km/h).. If we think of the system earth-sun as a merry go round with its axis being the sun I think we should notice a 'force' acting on us.
Title: Re: Can you help me understand "speed"?
Post by: Kean1 on 31/03/2020 07:05:49
Rather a long time coming but many thanks for all your replies...the "Corona time warp" now means I have time to try and investigate and understand what you've written!
Title: Re: Can you help me understand "speed"?
Post by: yor_on on 31/03/2020 13:30:09
Kean, you can think of 'speeds' as a result of what frame of reference you use.

In relativity (SR) a speed is translatable to being still, by choosing the appropriate frame of reference. And the difference between a 'speed' and a acceleration is always local, defined by your weight. All 'speeds', including 'gravitational accelerations' are without 'weight' (when tested by using a scale). So for you being inside a black box all speeds disappear. You need another frame of reference for defining it.
Title: Re: Can you help me understand "speed"?
Post by: 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?
Title: Re: Can you help me understand "speed"?
Post by: yor_on 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'?
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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
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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 :)
Title: Re: Can you help me understand "speed"?
Post by: yor_on 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.
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syntax
Title: Re: Can you help me understand "speed"?
Post by: yor_on 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.
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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 :)
Title: Re: Can you help me understand "speed"?
Post by: yor_on 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...
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( 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
Title: Re: Can you help me understand "speed"?
Post by: 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?   
Title: Re: Can you help me understand "speed"?
Post by: Janus on 01/04/2020 17:01:07
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?   
Title: Re: Can you help me understand "speed"?
Post by: A-wal on 01/04/2020 17:21:02
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.
Title: Re: Can you help me understand "speed"?
Post by: Janus on 01/04/2020 19:47:10
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.
Title: Re: Can you help me understand "speed"?
Post by: A-wal on 01/04/2020 21:27:46
My question was aimed at alancalverd
It missed, sorry. :)
Title: Re: Can you help me understand "speed"?
Post by: hamdani yusuf on 03/04/2020 13:17:02
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.
Title: Re: Can you help me understand "speed"?
Post by: 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?
Title: Re: Can you help me understand "speed"?
Post by: Janus on 03/04/2020 17:38:41
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.
Title: Re: Can you help me understand "speed"?
Post by: hamdani yusuf on 04/04/2020 05:12:00
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?
Title: Re: Can you help me understand "speed"?
Post by: Janus on 04/04/2020 16:54:07
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.
 [ Invalid Attachment ]
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).
Title: Re: Can you help me understand "speed"?
Post by: hamdani yusuf on 06/04/2020 08:39:19
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?
Title: Re: Can you help me understand "speed"?
Post by: hamdani yusuf on 06/04/2020 22:44:01
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.
We can refer to Janus' diagram for crosschecking the result.
Is there anything wrong?
Title: Re: Can you help me understand "speed"?
Post by: hamdani yusuf on 09/04/2020 08:40:58
The picture below shows the same situation, but I've added a mirror that's stationary in Alice's frame 1 light hour behind to make the situation more symmetrical. When Alice and Bob meets, they shine light to both mirrors.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=78081.0;attach=30470;image)
Left picture is from Bob's frame while right picture is from Alice's frame
green=Alice's position
blue=Bob's position
red=Bob's mirror
purple=Alice's mirror