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5:00 PM
@barrycarter Yeah, perhaps so. The thing is, people have many different ideas of rudeness.
2 days ago, by David Z
I suspect the poster of that question thinks of good treatment as getting their questions answered.
^^ for example
There is a lot more to be discussed on these topics, but unfortunately we're out of time.
 
@barrycarter it's not cool to be rude, but I don't think anyone hereabouts is deliberately rude.
 
@Danu Bousso is so awesome!
I loved talking to him
I think he sold me; Berkeley it is
:)
 
Though sometimes they can be short - very short! :-)
 
Good chat session everybody. See you at the next one in two weeks!
 
And my soaps just finished downloading
brb gonna do the elliptic bike and watch my soaps
Like a fat mom on a diet
 
5:02 PM
OK, I'm just saying we can have an efficient policy of dealing with them that isn't rude and is standardized.
 
@barrycarter yeah, that's generally a good goal to aim for. Like I said, I have to take off for now but I would like to return to this some other time.
 
Fair enough.
@JohnRennie Shall we continue?
 
BTW I've bookmarked the chat session transcript (let's see if this oneboxes):

Physics chat session 2016-04-05

1 hour ago, 1 hour total – 299 messages, 10 users, 11 stars

Bookmarked 22 secs ago by David Z

in case anyone wants to reference it
 
@JohnRennie OK. Now, always moving twin (AMT) passes NMT at some point. I wish to call that point t=0 timewise. Is that OK?
 
5:06 PM
What point on the diagram I've drawn does the AMT pass through?
 
The blue arrow is the AMT?
 
The blue arrow is the light ray shone from Earth towards the NMT
Gradient is unity for light rays by convention i.e. we actually draw $ct$ on the vertical axis
 
OK, so the AMT's position would be 0.8*t
 
OK, yes that's the gradient - what's the intercept?
 
If I'm understanding the diagram correctly, it would be the origin.
In other words, I declare t=0 when the AMT passes the NMT, which occurs when the 2000 light beam from Earth arrives at NMT time 2010.
 
5:10 PM
I've updated the diagram. Is that what you meant?
 
OK, good
Yes, I think so, let me look at it a bit more
Yes, OK good.
So, at t=0 for NMT and AMT, what does AMT say for Earth's position and time?
I claim it should be (-6,-6)
 
Are we assuming the AMT and NMT synchronise their clocks when they meet, so both agree their meeting point is (0,0)?
 
Correct, yes.
That's what crossing at the origin does, right?
 
x' = \gamma(x - vt)
x = -10 light years
t = -10 years
v = -0.8 light years per year
gamma = 5/3
I get x' = 5/3 ( -10 + 0.8*10) = -10/3
 
And t' = -10/3 ?
 
5:16 PM
t' = gamma (t - vx/c^2) = 5/3 (-10 + 0.8*10) = -10/3
 
OK, so at the moment the twins meet, AMT says "Earth year 2000 occurred 3.3 ly away and 3.3 years ago", is that correct?
 
Yes
 
(thinking)
 
Careful I can smell burning
 
OK, so what current distance and time would the ATM assign the Earth?
 
5:21 PM
Do you mean when AMT passes through the origin what does (t=0, x=-10) transform to?
 
(thinking some more)
Yes, that.
When AMT passes NMT, what time and place does he assign Earth?
Another problem: if Earth year 2000 occurred 3.3 ly away 3.3 years ago, doesn't that mean I'd be on top of Earth at some unrealistic time?
 
Let me just knock up a spreadsheet ...
 
Also, at one point, we had (-10,-10) transforming to (-30,-30). I take it we've abandoned that path.
We're playing Round 3 of Stump @JohnRennie
@JohnRennie Actually, this is all preliminary to the paradox I wish to create. Should we go ahead and discuss that now?
 
Bollocks, I made an error in the sign - I thought 10/3 didn't look correct. Yes, (-10,-10) transforms to (-30,-30) not (-10/3,-10/3).
 
OK, good. That's what I got and @WillO got too.
OK, so Earth 2000 occurred 30 years ago and 30 ly away for AMT.
 
5:31 PM
Yes
 
So, when AMT passes NMT, what coordinates does he assign Earth?
 
I'm just trying to translate that ...
 
It takes AMT 25 years to get 20ly closer to Earth, so he passes NMT at that time. OF course, 25 years AMT time is only 15 years Earth time, so it's Earth 2015 then? (just my guess)
 
I think it means at what value of $x$ does the Earth's trajectory cross the AMT's x axis.
 
Cardio done :V
So much poweeeer
 
5:35 PM
i.e. in the AMT's frame when t = 0 what is the value of $x$ for the earth.
 
Yes, I agree. What is that number?
Let's make it easier (pause for laughter). When AMT passes NMT, he is 6 light years from Earth, are we in agreement on that?
 
The point (t=-10, x=8) transforms into (t'=0,x'=-6)
 
Hello!
 
@JohnRennie So the Earth is 8 ly away when AMT passes NMT (in AMT's frame)?
 
@FenderLesPaul Careful though---what if he's not going to be your supervisor?
 
5:38 PM
So when the AMT's closk reads zero Earth is indeed six light years away
 
I don't know do I finish reading Resnick Halliday.
 
@JohnRennie processing that...
 
@JohnRennie This has still not been resolved? :P
 
@JohnRennie OK, so when the AMT clock reads 0, he assigns the Earth (-6,-6)?
 
I thought I understood SR till Barry got started on me :-)
 
5:39 PM
@Danu I broke @JohnRennie -- how many rep points do I get?
 
@barrycarter no, he doesn't assign the Earth anything. In the AMT's frame the Earth's trajectory is a straight line that passes through the point (0, -6).
 
@JohnRennie Rephrasing: there exists a time when AMT passes NMT. AMT assigns this as t=0. At that exact instant, he also looks at Earth, adjusts for light travel time and says: "The Earth is at position x, and time t". What are x and t?
 
What the AMT would say is that when the Earth is at the point (0, -6) a clock on the Earth would read t=+8 years
Assuming the Earth and the NMT had agreed that the zero of time is 2010.
 
Ummm, if the Earth is at the point (0, anything), doesn't that mean it's right on top of me?
 
(t,x) remember
 
5:44 PM
Oh, shoot.
 
I know it's silly but we normally write the time component first.
 
So Earth is 6 ly (I know, I've flipped on that a few times myself) when I pass NMT, correct?
 
It makes sense when we move to different dimensions, e.g. string theory, because the time component is always $v^0$.
 
And AMT regards Earth time as 2010, just the same as NMT time?
Wait, why t=+8 years? If the Earth is at (0, -6), wouldn't that be the year 2010?
 
The time on AMT clock agrees with the time on NMT's clock when they meet, but only when they meet. However at this moment the Earth clock doesn't agree with AMT's clock but does agree with NMT's clock.
That's because the time axes of the AMT and NMT are rotated wrt each other
 
5:48 PM
OK, but at t=0, AMT assigns the Earth the position -6, right?
 
Yes
 
And NMT assigns the Earth position -10 (always), right?
 
Yes
 
hmmm, thinking
So, we really can't compare the Earth 2010 that AMT is seeing to the Earth 2000 that the NMT is seeing?
 
Remember that simultaneity breaks down in SR. Events that are simultaneous in one frame are not simultaneous in a different frame.
 
5:50 PM
OK, but here's the paradox. NMT, stirred by a fraternal love for this brother, abandons his sedentary ways and accelerates to 0.8c. Now, he must see the exact same thing his brother sees, right?
 
So you mean the Earth jumps from x=-10 to x=-6, and it's clock jumps from t=0 to t=8.
Now didn't I write that down somewhere ..
Oh yes, that would be in my answer to your question.
 
Something like that. I'm saying that AMT says Earth is 6 ly away and is at the year 2015, right?
Wait, do I mean 2010?
 
This fiddling with the year is confusing me. Let's just agree 2010 is t = zero.
 
Agreed.
So, AMT says "Earth is 6ly away and the time on Earth is ... t=0"?
 
When AMT passes NMT Earth is at x = -6 and the clock on Earth shows t = +8
 
5:54 PM
OK, I think I can still break things with that setup. When you say Earth clock shows, you're ignoring light travel time, right?
 
Yes
 
If you included light travel time, what AMT actually sees is Earth at t=+2, right?
 
Didn't I answer that in my first answer to your question?
 
I don't remember, sorry. But I'm trying to derive a contradiction, so play along :)
What you said was that accelerating didn't affect what the eyeballs would see by much.
Maybe 40K seconds (or you later revised it).
 
0
A: Rapid acceleration does what to inertial frame clocks?

John RennieThe question is easily answered by drawing a spacetime diagram in the Earth's rest frame: I've scaled the time axis so that a light ray travels at 45º. You start at $x=10$ light years so the light ray emitted from the Earth at $t=-10$ years reaches you just as you start. The red line shows yo...

When the AMT passes the NMT what the AMT would see is the light that left the Earth at t = -10.
 
5:58 PM
And would thus conclude t=-4 currently?
 
I don't follow that argument
 
See what I'm saying? You can't have t=+8 and t=-4 at the same time.
OK, AMT is 6 ly from Earth and sees Earth at t=-10 including light travel time, right?
 
Oh wait. So, if that light took 6 years to get here.... wait, is that 6 years AMT time?
 
The light wan't emitted when the Earth was six light years away - it hasn't been travelling for six years
 
6:02 PM
OK, let's go back a step. If you're in a reference frame, how do you assign times to remote objects? I thought the answer was: what you see right now plus light travel time (as you compute it). Not true?
 
I suspect the problem is your obsession with the idea of assigning times to remote objects
 
But that's what relativity does, right?
 
In the AMT's frame the Earth's trajectory is a line.
 
OK, that makes sense. Earth distance is 6-0.8*t or something
 
The distance between any two points on that line is calculated using the metric.
And that distance is what we call the proper time (or sometimes proper distance)
The proper time between any two points on a trajectory is equal to the elapsed time shown on a clock comoving with the travelling object.
 
6:06 PM
OK, I think I'm starting to see.
 
So if we can agree on a clock synchronisation at some point (typically a meeting point) we can calculate elapsed time along the trajectory in either direction from our clock synchronisation point.
 
So, if I'm moving at 0.8 c wrt Earth and I'm 6 ly away, and see (with my eyes) Earth is at t=0, what time do I say those light rays left Earth to strike my eyeballs now?
 
Does that mean at what spacetime point in my coordinates was the light ray emitted, or what was the time shown by clocks on Earth when the light ray was emitted. The two are different.
 
What time was shown by Earth clocks.
 
t = -10
 
6:10 PM
OK, because I'm 10 ly away from Earth in Earth's frame?
 
No hang on, I'm getting mixed up.
 
You know, when Captain Kirk confused bots, they had the decency to blow up...
 
Let me go back to my spacetime diagram ...
 
It's a good thing @JohnDuffield isn't here, he would love this.
 
Hush you fool
You'll get us all banned
 
6:17 PM
Oh no, I just summoned him, didn't I?
 
No, t = -10 is correct. But at the time Earth was at x' = -30.
The light was emitted when the Earth's clock showed -10 and the AMT's clock showed -30.
 
OK, let's try it this way. At the moment AMT passes NMT, can we agree that the same light from Earth is reaching them?
 
Yes
 
@JohnRennie So they both see a big sign saying "t=0" from Earth. Now, we ask both: when the light rays that just hit your eyes left Earth, where was Earth, and what time was it on Earth
 
In the AMT's frame the light was emitted when the the Earth was at x' = -30 and t' = -30. At that time the Earth's clock read -10. The light reached the AMT when the Earth was at x' = -6 and at that moment the Earth's clock read +8 years.
The ratio 18/30 is just the time dilation factor of 0.6
What we should really do is draw the spacetime diagram in the AMT's rest frame and label up the Earth's trajectory with the proper times. However I'm not doing that now.
 
6:24 PM
(still thinking)
 
@MAFIA36790 I said I might.
 
So, AMT says: when the Earth was 30 ly away from me 30 years ago, they emitted a signal reading t=0 ?
 
A signal reading t = -10
 
OK, right.
AMT says: at that time, NMT was 24 ly away?
 
30 light years away. The signal was emitted at (t' = -30, x' = -30)
 
6:28 PM
That's the Earth, though, right? NMT is 10ly away from Earth (in their stat frame).
 
The NMT and Earth agree on their times, so at the moment the light ray was emitted the simultaneous point on the NMT's trajectory was (t = -10, x = 0). We need to transform that point into the AMT's frame.
And the result is (t' = -13.333, x' = -16.666)
 
OK, but I don't think that should be difficult? When AMT says t=-30..... ok.
 
Note that the two spacetime points are simultaneous in the Earth/NMT frame but not in the AMT frame.
 
OK, so AMT says "that event occurred 30 ly away 30 years ago" (ie, Earth saying t=-10). good so far?
 
Earth saying t = -10
 
6:33 PM
Sorry, fixed.
But yes?
 
Yes
 
And I'm seeing it now, since that light just got to me at my t=0
 
Yes
 
At t=0, AMT is 6 ly from Earth, right?
 
Yes
 
6:36 PM
Hmmm, thinking.
 
In the AMT frame the light has been travelling for 30 years by the time AMT saw it, meaning it has travelled 30 light years.
In that time the Earth's clock has ticked forward by 30/gamma = 18 years.
 
Sure, I'm OK with that. The problem I'm having is seeing what happens now that the AMT is 6 ly away.
I'm convinced that he must also be seeing other light beams emitted from Earth.
 
Making the time on Earth at the moment AMT saw the light +8 years.
Which is where we came in
 
Which makes sense since AMT was 18 ly away Earth distance.
 
Yes
 
6:39 PM
OK, AMT can now ask "since I'm 6 ly from Earth, I will see Earth at some time and distance 6 years from now." Is that correct?
IE, eyeball striking.
 
In six years time AMT will see a light ray carrying an image of the Earth clock showing +8
 
Ah, OK! So when I cross NMT, I say Earth's coords are t=8, x=-6 ?
 
Yes. Were I the sardonic sort I'd point out that I mentioned that about an hour ago :-)
 
@JohnRennie Fortunately, you're not :P
And the NMT says the coords are t=-10, x=-10 ?
Oooh, wait.
 
Well, to be precise the Earth's proper time is t = +8 when x' = -6
 
6:43 PM
That's where simultaneity comes in.
 
The coordinates are just t' = 0, x' = -6
 
You've won this round @JohnRennie. But I shall return. I still think there's an inconsistency here somewhere.
 
@barrycarter the NMT says the Earth's coordinates are t=0, x=-10. However the light ray the NMT just received shows the clock reading -10.
 
I'm going to draw some correct x-vs-t diagrams and include light travel time... next time, the advantage shall be MINE!
You're not the only one who can draw, @JohnRennie ... you're not the only one (fading out as I flee in my spaceship)
 
Spacetime diagrams are essential for brains not designed to be relativistic. Have I mentioned that relativity is a geometrical theory?
 
6:46 PM
Mathematicians don't need diagrams. We use matrices instead.
See, I haven't had any trouble with SR until I actually tried to understand the physical interpretation.
 
However, between us we have managed to stop 0celo7 and Slereah arguing about the maths of GR for over an hour!!!
 
HA HA HA!
 
Which is an unparalleled achievement :-)
 
I have no problem plugging and chugging numbers to get answers.... but when I started to think about what those answers actually mean in the real world... blargh.
 
You mathematicians invented differential geometry ...
 
6:48 PM
Yes, but we never think about the physical interpretation of anything we do.
We work hard to create mathematics that has no practical use.
Thus, my plan for an underground stackexchange reputation exchange.
 
Even after more years than I care to think about I still find myself confused about what's actually happening with SR. If you want to go forward we need to draw the spacetime diagram in the AMT's frame.
However that's a job for another day.
 
@JohnRennie I'm going to draw a diagram my way and see what happens.
 
Fair enough. Stick it on Google draw so we can share it.
 
I think eyeball time and distance are too often neglected in relativity.
@JohnRennie I'll be sure to stick it somewhere you can get to it.
 
I'm off now, summon me if you want to pick up where we left off.
 
6:54 PM
(takes the ring) Thank you. How can we ever repay you?
I'm taking a break from this stuff myself for a bit.
 
7:33 PM
@barrycarter : I wouldn't love it I'm afraid Barry. I'm afraid to say I think you're making to much of a meal of this SR stuff. I've answered your questions, it's simpler than you think. By the way, you have to say my name three times to summon me.
 
7:50 PM
@JohnDuffield oooh, just like that other guy.
@JohnDuffield I thought finding an inconsistency in the currently accepted theory of relativity would elate you. Unforunately, the evil Rennie has won this round.
I am now torn between doing actual productive work and following this black hole of relativity.
 
@barrycarter : elate me? I'm an amateur "relativist". I root for relativity.
 
@JohnDuffield (before we get too far, we may want to "get a room"). Yes, but your interpretation of it varies from the generally accepted interpretation.
 
@barrycarter : not as much as you think.
 
@0celo7 I think you defined the pre-image of a subset through f (not sure).
Given: Set A, Set B, a - elements of A, b - elements of B. $f(A) \subseteq B$
$f(A) = \{a \in A|f(a)\}$
for some a
where $f(a) = b$
is that a good definition of the image of set $A$ through/under function $f$?
 

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