- #36
harrylin
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I would have appreciated it if you clarified some of your claims. But never mind, Stephanus can enquire more if he likes.Mentz114 said:I have no idea what you are talking about.
I would have appreciated it if you clarified some of your claims. But never mind, Stephanus can enquire more if he likes.Mentz114 said:I have no idea what you are talking about.
Stephanus said:Can we go on?
I have a new question here.
View attachment 84791
There are four probes, each has a clock.
A1, A2, B1, and B2.
The clocks are synchronized.
And then B1-B2 moves to A2 direction.
##v = 0.6c \implies \gamma = 1.25##
B travels for 20 years wrt B's clock then stops.
So B is in the same frame again as A
B1 will meet A2, B2 will meet A1.
Q1: Is this right?
B is preprogrammed to stop when their clock read 20 years. Because for B2, there's no way B2 knows that B1 has stopped.
The instant B accelerates, B1 will see this.
View attachment 84792
Thank you.Mentz114 said:If you could only show these set ups as spacetime diagrams, the answers would be clearly seen
I'll try to explain how to make an ST plot.
If something moves along the x direction starting at x=0 with velocity v=0.25 we can work out its position at different times
t=0, x=0
t=1, x=v
t=2, x=2v
t=3, x=3v
and so on. If the(x,t) pairs are plotted on a graph with the x-axis horizontal and t-axis vertical, they lie on a straight line inclined to the right.
Repeat this for something coming the other way starting at x=8 with v=-0.3
t=0, c=8
t=1, x=8-0.3
t=2, x=8-(2*0.3)
t=3, x=8-(3*0.3)
and so on.
This is a straight line inclined to the left. The point where the lines cross is where they meet
From this diagram we can deduce the answers to many questions - eg what will their clocks show when they meet ?
I urge you strongly to plot this graph - and show it here so I can tell if it is correct.
If you can understand this concept you will make a great step forward.
Mentz114 said:The point where the lines cross is where they meet
From this diagram we can deduce the answers to many questions - eg what will their clocks show when they meet ?
I urge you strongly to plot this graph - and show it here so I can tell if it is correct.
If you can understand this concept you will make a great step forward.
Stephanus said:Sorry, drawing the graphs by no mean an easy task.
Okay...
So they meet at ##x = 7\frac{1}{7}## and their clocks show, say ##2\frac{6}{7}##
I don't know if this is how we translate the equation in English.
And tell me something from this graph.
Does the direction always to the top. They can go left and right, but always to the top?
View attachment 84811
Ahhh, the other way around.Mentz114 said:No. The red line is wrong. The endpoint is more like (t=14, x=6)
Wait...Mentz114 said:Excellent. That is correct.
Remember that the things are moving only along the x-axis. The vertical axis is 'motion in time'.
Now imagine someone who is not moving in this diagram. Their worldline is a vertical line. Can you see that is logical because they have no motion in the x-direction but their clock is always ticking. The times and distances on this chart are in the frame of that observer.
Now to work out the clock times for those worldlines ( yes, you have drawn worldlines) we use ##\tau^2 = dt^2-dx^2## to work out the hypotenuses of the two triangles ( see the pic). Could you do that ? It is just arithmetic.
What do you mean by dt?Mentz114 said:we use ##\tau^2 = dt^2-dx^2## to work out the hypotenuses of the two triangles ( see the pic).
It just hit me. Are the red and blue line angles limited between 450 and 1350?Mentz114 said:[..]to work out the hypotenuses of the two triangles ( see the pic).
##\sqrt{AD^2 - CD^2}##, are you sure? that will put ##\sqrt{<0}##Mentz114 said:Sorry my notation is not clear.
Please look at the picture herewith. Can you see the lines AC, AB and so on ?
The proper time for the red worldline is ##\sqrt{AD^2-CD^2}##
Come on ! You are replying too quickly ! Think first !Stephanus said:##\sqrt{AD^2 - CD^2}##, are you sure? that will put ##\sqrt{<0}##
Mea culpa, I tought AD - AC, sorry.Mentz114 said:Come on ! You are replying too quickly ! Think first !
It is ##\sqrt{(14.54)^2 - (3.64)^2}.##
Yes, the our calculations matches. Now I'm using calculator.Mentz114 said:I don't know. I can't follow what you've written and I'm not going to finish your calculations.
The answers are
red worldline ##\tau = \sqrt{(14.54)^2 - (3.64)^2} = 14.08 ##
other worldline ##\tau = \sqrt{(14.54)^2 - (8-3.64)^2} = 13.87##
stationary worldline ##\tau = 14.54##
Wait...Mentz114 said:Can you work out the time on the blue worldline when a light beam from the origin hits it ?
Hits it.Mentz114 said:Can you work out the time on the blue worldline when a light beam from the origin hits it ?
Stephanus said:No, not perpendicular. But 450 from vertical lines. That's how x = t
Tell me. Which one?
Thanks. But I already calculated before your alert popped up. Is it time dilation in PF forum?Mentz114 said:This one. The WL of the light is as you say ##x=t## and the thing is on ##x=8-vt## so you can get the intersection on the appraoching WL.
Then make a triangle and work out a ##t^2-x^2## and you got it.
No, no, no. Don't be sorry. I liked that.Mentz114 said:Solving ##x=t## with ##x=8-vt## gets ##t=8-vt## so ##t=8/(1+v)=6.15##
and ##x=6.15##.
I'm sorry to keep you awake.
If by origin you mean (0,0), then ##8/1.3 = 6.15##Mentz114 said:Solving ##x=t## with ##x=8-vt## gets ##t=8-vt## so ##t=8/(1+v)=6.15##
and ##x=6.15##.
I'm sorry to keep you awake.
The light that comes from the origin.Mentz114 said:I drew the light ray on the graph. It is thin yellow line.
Back again to Twins Paradox.Ibix said:Events can exist anywhere, and they don't "travel". Things that are traveling appear as sloped lines in your graph. Events are points on your graph. More physically an event is something like turning on your rockets. That happens at a point in space at an instant in time, which can't be said to be moving.
That said, you are correct that events in the magenta region cannot be reached from the origin without exceeding the speed of light. An example would be a car crash in London and a simultaneous (in some frame) car crash in New York. If one was at the origin the other was in the magenta region.
Correct. Although I'm not sure how you intend to accelerate a supernova to 0.1c.Stephanus said:Back again to Twins Paradox.
Supposed there's a Type 1a Supernova, 1000 lys away. We know that it's 1000 lys because of its brightness. And suddenly it moves toward us at 0.1 c.
What would we see? Nothing changes, is this right?
The supernova would be blue-shifted, yes. You don't need to say "towards us" in this context.Stephanus said:Suddenly, 1000 years later, we'll see that the star is blue shifted toward us or at least it changes. Is this right? And keeps getting bigger and bigger.
Not really. We will see the supernova evolving rapidly due to the decreasing lightspeed delay. But if we correct for the changing delay we will find that the supernova is evolving more slowly than a non-moving supernova, and this is due to time dilation.Stephanus said:And also 1000 years later we'll experience that time in the supernova runs faster
I know, I know, supernova only lasts a few weeks or years.
Yes, although again there is an effect from changing light travel time and an effect from time dilation. So we would, again, see that the supernova was running fast, but could subtract out the light travel time and deduce that time was running slow at the supernova.Stephanus said:Now, what if WE move toward the supernova. We'll see it blue shifted right away. Is this right? We'll experience doppler effect right away and we'll experience that time in the supernova runs faster.
We will always experience time for us at one second per second. That's more or less what time is.Stephanus said:In both cases, we'll experience that times slows for us, but at different date. Time dilation and length contraction work both ways.
But if we move, we'll experience those effect RIGHT AWAY.
No. When an object's worldline reaches you, it has crashed into you. When an event's light cone reaches you, it is possible to see that event. You can't see it before because the light didn't have time to reach you yet.Stephanus said:But if we at the 'rest frame' and the other object is moving toward us. We'll notice that time/length,etc from that object also dilated/contracted WHEN the object's WORLDLINE reaches us?
Is that so?
You aren't really talking about simultaneity here. You are talking about when people perceive events, which is a separate topic.Stephanus said:Is that why there's Twins Paradox? Simultaneity of events?
Come on..., I was just trying to say that even though an object moves toward us, we'll never know if it HAS MOVED toward us until it worldline "crashes" us. Or I should have said, if it's a probe with laser signal emitted to us, we'll never know that the probe is moving toward us, until the last signal emitted when the probe still at rest reaches us. But if we move toward the supernova/probe, we'll realize at once that that thing is getting closer. Especially a type 1a supernova, as you certainly know it's a standard candle.Ibix said:Correct. Although I'm not sure how you intend to accelerate a supernova to 0.1c.
Yes, yes. Thanks. It greatly improves my logic. Thank you. We could have flown away from Andromeda 100 km/s, still its light is blueshifted.Ibix said:The supernova would be blue-shifted, yes. You don't need to say "towards us" in this context.
You're taking the hypotenuse route.Ibix said:Let me ask you a question. I drive from A to B in a straight line, and my odometer reads 100 miles. You take a triangular route from A to B via C, and your odometer reads 120 miles. But we started in the same place and finished in the same place. How would you explain the difference in our odometer readings to me?