Post by: Dimensional on 14/05/2022 06:08:03
According to a pretty good looking analysis of length contraction and time dilation, the author makes a diagram of a spaceship travelling very fast towards the right side of your screen. At the higher of the 3 clocks/events is the nose of the ship according to an observer on the ground and so is the clock/event below it. For the diagram, here is the link, and just watch from 6:30 to 7:00 .
Now my thought experiment is about what might happen if an object moves really fast in front of the nose of the ship where the lower clock/event is. To make the object appear in front of the nose almost instantaneously, we will say that it came from another spatial dimension z going into your screen. For simplicity sake, let's say this interaction happens at the lower clock/event (where the nose of the ship is).
If an interception between the contracted nose (the lower clock) and the object is possible in the scenario given, it would seem that there would be an interaction with the nose of the ship that will never happen with the nose of the ship in the future (the higher clock).
Post by: Halc on 14/05/2022 06:42:46
At the higher of the 3 clocks/events is the nose of the ship according to an observer on the ground and so is the clock/event below it.This sentence doesn't really parse for me, so hard to figure out what you're trying to say. There seem to be no clocks depicted in the 30 second clip you indicate. The vid around 6:45 shows a spacetime diagram with one event at the rear of the ship and one nose event (the lower one) for the 'ground' frame and another nose event (the upper right one) for the ship frame, each simultaneous with the rear event in their respective frames. Hence the 20m ship is contracted by a sixth in the ground frame. All pretty straight forward SR so far.
Perhaps the 'lower clock
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Now my thought experiment is about what might happen if an object moves really fast in front of the nose of the ship where the lower clock/event is.OK, at some point an object crosses in front of the ship.
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To make the object appear in front of the nose almost instantaneously, we will say that it came from another spatial dimension z going into your screen.Or y, since the diagram only shows the x axis. The vertical axis is t (ground frame).
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If an interception between the contracted nose (the lower clock) and the object is possible in the scenario given, it would seem that there would be an interaction with the nose of the ship that will never happen with the nose of the ship in the future (the higher clock).You're saying the object that crosses in front bumps the nose of the ship as it goes by, but not enough to damage anything. That's an objective event. It must occur in any frame. Perhaps it left a mark. In the ship frame at the time indicated in the video, the bump event you describe has already happened in the past and the mark on the nose is already there. The nose of the ship follows the nose worldline (the dotted green line) and is present at every event along that line. Likewise the tail follows the parallel dotted tail line. Those lines are straight because apparently the ship is not accelerating in this scenario.
Post by: Dimensional on 14/05/2022 18:45:04
At the higher of the 3 clocks/events is the nose of the ship according to an observer on the ground and so is the clock/event below it.This sentence doesn't really parse for me, so hard to figure out what you're trying to say. There seem to be no clocks depicted in the 30 second clip you indicate. The vid around 6:45 shows a spacetime diagram with one event at the rear of the ship and one nose event (the lower one) for the 'ground' frame and another nose event (the upper right one) for the ship frame, each simultaneous with the rear event in their respective frames. Hence the 20m ship is contracted by a sixth in the ground frame. All pretty straight forward SR so far.
Perhaps the 'lower clock
The 3 events are the 3 clocks that I was referring to.
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The bump happens when the ship, in its proper time, is already extended into the future. Its nose is already into the future.QuoteIf an interception between the contracted nose (the lower clock) and the object is possible in the scenario given, it would seem that there would be an interaction with the nose of the ship that will never happen with the nose of the ship in the future (the higher clock).You're saying the object that crosses in front bumps the nose of the ship as it goes by, but not enough to damage anything.
That's an objective event. It must occur in any frame. Perhaps it left a mark. In the ship frame at the time indicated in the video, the bump event you describe has already happened in the past and the mark on the nose is already there. The nose of the ship follows the nose worldline (the dotted green line) and is present at every event along that line. Likewise the tail follows the parallel dotted tail line. Those lines are straight because apparently the ship is not accelerating in this scenario.
Post by: Origin on 14/05/2022 20:45:02
The 3 events are the 3 clocks that I was referring to.As far as I could see there were only 2 clocks, one on the ship and one on the ground.
Post by: Origin on 14/05/2022 20:57:17
Now my thought experiment is about what might happen if an object moves really fast in front of the nose of the ship where the lower clock/event is. To make the object appear in front of the nose almost instantaneously, we will say that it came from another spatial dimension z going into your screen. For simplicity sake, let's say this interaction happens at the lower clock/event (where the nose of the ship is).I'm not getting this scenario. Does the intercepting object have to be moving very fast or can the scenario be that the ship hits a pebble.
Post by: Dimensional on 14/05/2022 21:08:25
You should see 3 events/clocks at 6:55.The 3 events are the 3 clocks that I was referring to.As far as I could see there were only 2 clocks, one on the ship and one on the ground.
Post by: Dimensional on 14/05/2022 21:15:15
Just imagine some object moving in front of the ship from another dimension, say dimension z. The object will come from a direction perpendicular to your screen and roll in front of the bottom event/clock, which is also the nose of the ship. There would seem to have to be a collision of some sort that does not happen for the ship in its proper time and length.Now my thought experiment is about what might happen if an object moves really fast in front of the nose of the ship where the lower clock/event is. To make the object appear in front of the nose almost instantaneously, we will say that it came from another spatial dimension z going into your screen. For simplicity sake, let's say this interaction happens at the lower clock/event (where the nose of the ship is).I'm not getting this scenario. Does the intercepting object have to be moving very fast or can the scenario be that the ship hits a pebble.
Post by: Eternal Student on 14/05/2022 22:13:20
It would be easier just to put the static picture on here:
That might help.
video-snip.JPG (226.75 kB . 1815x977 - viewed 2513 times)I think @Halc has already answered the question correctly.
The essence of the idea is that the bump will happen, regardless of which frame of reference you use. However, the spaceman and the planet based observer can disagree about BOTH where and when it happened.
The planet based observer can call out a warning on the radio "... Look out there's about to be a collision at our time t = 10 minutes!" Meanwhile the spaceman will record in his log " A minor collision happened at my time t' = 9 minutes. By the time I received a warning it was too late AND they were wrong about the time".
Best Wishes.
Post by: Dimensional on 14/05/2022 23:01:41
Hi.But in the ship's frame of reference, his nose is advanced further from the origin (0,0) (that they both agree on) in the x position (because the ship is in its proper length) than where the rock rolls into its path. The ship has also advanced further in time from the origin than when the rock rolls into its path.
It would be easier just to put the static picture on here:
That might help.
I think @Halc has already answered the question correctly.
The essence of the idea is that the bump will happen, regardless of which frame of reference you use. However, the spaceman and the planet based observer can disagree about BOTH where and when it happened.
The planet based observer can call out a warning on the radio "... Look out there's about to be a collision at our time t = 10 minutes!" Meanwhile the spaceman will record in his log " A minor collision happened at my time t' = 9 minutes. By the time I received a warning it was too late AND they were wrong about the time".
Best Wishes.
Post by: Eternal Student on 15/05/2022 00:26:48
LATE EDITING: @Halc got a reply in before I finished this. I'm putting this in as it is, just because another way of looking at it might help.
I think I can see some of what you ( @Dimensional ) are trying to say.
If what you said were true then the collision happened at a time before t'=0 for the spaceman. For example the collision could have happened at the spacemans time t' = -1 minute and at a location the spaceman describes as x' = -1 metre.
This assumes the space rocket was always moving at that constant velocity, so the time t' = -1 minute isn't anything weird, it's just 1 minute before the spaceman set his watch to a time t' = 0. Similarly x' = -1 metre just means 1 metre left of the origin.
If (because I can imagine it will be coming) the rocket was actually assumed to be stationary to start with and then quickly accelerated before holding a steady velocity throughout the rest of the motion..... Then the diagram presented is insufficient. All those worldlines are straight lines not curves, so no acceleration is shown. You need a different diagram. With the right diagram, the spaceman will find the nose of his rocket has existed at the collision event. To say that another way, at some time and place the collision did occur for the spaceman.
Another minor note or comment: I'm not sure why the video showed only the time axis sloping and implied the spaceman would naturally choose to continue using the same x-axis as the planet observer. Presumably the video presenter was trying to help and keep everything simple. The planet observer's x axis would seem like an awkward blend of some space and time to the spaceman. It's far better to imagine that both the spacemans axis will start to tilt away from the planet observers axis and effectively close up like a pair of scissors.
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Feinsteinrelativelyeasy.com%2Fimages%2Fspecialrelativity%2FLorentz_Minkowski14.png&hash=23ae4ecac2fbd22fb53d8cc73168327f)
Best Wishes.
Post by: Dimensional on 15/05/2022 19:10:04
That just means that in the ship frame, the interaction with the object took place before the rear of the ship reached the origin event. The nose is present at every event along the right dotted worldline, so all those event happen in every frame. Ditto with the other dotted line.Yes, I understand that that is what is suppose to happen in GR. In GR the universe is like a 4 dimensional static structure, so I can't just inject new world lines if the block is already there. In other words, my question does not even get off the ground if we just assume that GR is correct. So, I am trying to understand the details of how my scenario can't happen.
Post by: Dimensional on 15/05/2022 19:31:12
Hi.At exactly 6:43, he has the other part of the Minkowski diagram.
LATE EDITING: @Halc got a reply in before I finished this. I'm putting this in as it is, just because another way of looking at it might help.
I think I can see some of what you ( @Dimensional ) are trying to say.
If what you said were true then the collision happened at a time before t'=0 for the spaceman. For example the collision could have happened at the spacemans time t' = -1 minute and at a location the spaceman describes as x' = -1 metre.
This assumes the space rocket was always moving at that constant velocity, so the time t' = -1 minute isn't anything weird, it's just 1 minute before the spaceman set his watch to a time t' = 0. Similarly x' = -1 metre just means 1 metre left of the origin.
If (because I can imagine it will be coming) the rocket was actually assumed to be stationary to start with and then quickly accelerated before holding a steady velocity throughout the rest of the motion..... Then the diagram presented is insufficient. All those worldlines are straight lines not curves, so no acceleration is shown. You need a different diagram. With the right diagram, the spaceman will find the nose of his rocket has existed at the collision event. To say that another way, at some time and place the collision did occur for the spaceman.
Another minor note or comment: I'm not sure why the video showed only the time axis sloping and implied the spaceman would naturally choose to continue using the same x-axis as the planet observer. Presumably the video presenter was trying to help and keep everything simple. The planet observer's x axis would seem like an awkward blend of some space and time to the spaceman. It's far better to imagine that both the spacemans axis will start to tilt away from the planet observers axis and effectively close up like a pair of scissors.
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Feinsteinrelativelyeasy.com%2Fimages%2Fspecialrelativity%2FLorentz_Minkowski14.png&hash=23ae4ecac2fbd22fb53d8cc73168327f)
Best Wishes.
I have a question that will probably clear things up a lot for me. But first, let's call the event where the spaceman and the Earth people agree (0,0). Relative to (0,0), where in space and time is the nose of the rocket for the spaceman when the rock rolls in front of the ship? To keep it simple, I don't think we have to give numbers, just put if it is =, < or > than 0. And let's assume that the rock rolls in front of the ship at the same event as shown at the nose of the ship on the ground.
The answer to this question should help me a lot.
Post by: Halc on 15/05/2022 23:15:49
In other words, my question does not even get off the ground if we just assume that GR is correct.First of all, SR is sufficient for this case since there is no gravity involved. If you're assuming that your assumptions somehow contradict relativity theory, then maybe it's time to rethink your assumptions. Relativity theory does not in any way forbid an object passing in front of another object.
At exactly 6:43, he has the other part of the Minkowski diagram.6:42 actually is when the nose of the ship is at the event where the rock crosses in front.
atRockEvent.JPG (34.78 kB . 430x267 - viewed 2438 times)The Minkowski diagram still shows the frame of the 'ground'. If in the frame of the ship, the ship would be stationary and not be progressing to the right like they depict. But the animation shows the ship at events that are simultaneous in the ship frame.
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Relative to (0,0), where in space and time is the nose of the rocket for the spaceman when the rock rolls in front of the ship?The event of the rock rolling in front is not present at the origin event, so the spatial location and time of that rock event is frame dependent (per relativity of simultaneity). The physical event is objective, but the abstract coordinates assigned to that event are coordinate system (frame) dependent.
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To keep it simple, I don't think we have to give numbers, just put if it is =, < or > than 0.As depicted in the animation, the rock rolling event is after the origin event in the ground frame, and before the origin event in the ship frame. You can see in the picture attached that the rear of the ship (the only part that will be present at the origin event) has yet to reach the origin and is in fact off the edge of the screen still.
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And let's assume that the rock rolls in front of the ship at the same event as shown at the nose of the ship on the ground.There is no ground in the picture. There's a frame where the hypothetical ground is statationary, but no ground appears anywhere. It is presumed to be a long way off, and this ship is just going by the planet at 0.55c
Post by: Eternal Student on 15/05/2022 23:21:03
Here's the bit of the spacetime diagram we need.
[ Invalid Attachment ]
The Black axis show the time, t, and the space, x, axis used by the planetary based observer.
The Red axis, shows the time, t' and space x' axis that the spaceman will use.
The green line is the worldline of the nose of the rocket.
The brown circle shows the rock collision event, it's just a single point on the green worldline but I've made it bigger because there isn't anyway we can see just a single dot.
Identify what you need to in that diagram before I make a mess of it and draw on some reference grid lines so that we can read off the (x,t) values that describe the rock collision event for the planet based observer and also the (x',t') values which describe the same event in the spacemans co-ordinates.
Here's some gridlines for the (red) spaceman axis. Slide along those gridlines and the rock collision event is at (5, -1) for the spaceman. It happens at x' = 5 and t' = -1.
[ Invalid Attachment ]
Now let's clear those gridlines out of the way and put gridlines on for the planetary based observer.
[ Invalid Attachment ]
The rock collision event occurred at (x,t) = ( 5, 1), or x = 5, and t = 1 for the planetary based observer.
I hope that makes sense.... the exact placements might have been off slightly (I only sketched the diagrams not calculated them with any precision etc) but hopefully you can see how to use these diagrams and make sense of the stuff.
Best Wishes.
Post by: Dimensional on 16/05/2022 00:43:22
(I read everything that you put, but I think this part right here is at the heart of the matter)At exactly 6:43, he has the other part of the Minkowski diagram.6:42 actually is when the nose of the ship is at the event where the rock crosses in front.
In my thought experiment, I want the rock to move in front of the ship when the ship's back fin is at the origin (0,0). The origin is where we can all agree on where the back fin is located.
Post by: Eternal Student on 16/05/2022 01:03:30
I think you edited that screenshot. It looks like you've put a blob where the rock event is and then run a thin blue line back to the spaceman's time axis.
You can't run that line perfectly horizontally to read off the spaceman's time value. It has to run parallel to the spaceman's space axis. See the diagrams I presented just earlier.
Best Wishes.
Post by: Dimensional on 16/05/2022 01:10:34
Thank you very much for your diagrams; I really appreciate them. They clearly show what is happening.
I hope that makes sense.... the exact placements might have been off slightly (I only sketched the diagrams not calculated them with any precision etc) but hopefully you can see how to use these diagrams and make sense of the stuff.
Best Wishes.
Is your origin (0,0) at the very back of the back fin of the ship like in the video?
Post by: Dimensional on 16/05/2022 01:13:05
Hi.I understand. I didn't do that. I think that was part of the video.
I think you edited that screenshot. It looks like you've put a blob where the rock event is and then run a thin blue line back to the spaceman's time axis.
You can't run that line perfectly horizontally to read off the spaceman's time value. It has to run parallel to the spaceman's space axis. See the diagrams I presented just earlier.
Best Wishes.
Post by: Eternal Student on 16/05/2022 02:51:37
Is your origin (0,0) at the very back of the back fin of the ship like in the video?Yes. The origin is intended to be exactly where they placed it in the original video (although I only sketched it, I didn't get a ruler and compass).
The planet based observer says the back of the fin is at x= 0 when t =0. Spaceman says the back of the fin is at x'=0 when t' =0.
- - - - - - - - - -
Just to emphasize one issue, although in my diagrams it looks like the x and x' co-ordinates of the rock collision event are both +5, they aren't actually exactly the same. That's just that the diagram is only a sketch and I haven't placed all the gridlines exactly the same space apart etc. I just want to dispel the notion that there was any reason why they had to agree on the spatial location of the event... there isn't.
I've run the precise calculation with these figures (they are roughly what was used in the video).
Set the velocity of the rocket = half the speed of light.
Use units for measuring time and distance so that the speed of light, c = 1 in those units (Just to be clear that's not going to be seconds and metres. It's just conventional to set c = 1).
Set the rock collision event to co-ordinates (x, t) = ( +5.00 , +1.00 ) as was shown in my diagram for the planetary observer.
This becomes (x', t') = ( +5.20 , -1.73) in the spacemans co-ordinate system.
So, with these figures, the spaceman and planet based observer disagree on the both the location and time of the collision event.
Best Wishes.
Post by: Dimensional on 16/05/2022 05:03:49
Hi.I need to think about this more.Is your origin (0,0) at the very back of the back fin of the ship like in the video?Yes. The origin is intended to be exactly where they placed it in the original video (although I only sketched it, I didn't get a ruler and compass).
The planet based observer says the back of the fin is at x= 0 when t =0. Spaceman says the back of the fin is at x'=0 when t' =0.
Post by: Dimensional on 24/05/2022 23:15:26
Hi.Yeah this makes sense to me. Thank a lot.Is your origin (0,0) at the very back of the back fin of the ship like in the video?Yes. The origin is intended to be exactly where they placed it in the original video (although I only sketched it, I didn't get a ruler and compass).
The planet based observer says the back of the fin is at x= 0 when t =0. Spaceman says the back of the fin is at x'=0 when t' =0.
- - - - - - - - - -
Just to emphasize one issue, although in my diagrams it looks like the x and x' co-ordinates of the rock collision event are both +5, they aren't actually exactly the same. That's just that the diagram is only a sketch and I haven't placed all the gridlines exactly the same space apart etc. I just want to dispel the notion that there was any reason why they had to agree on the spatial location of the event... there isn't.
I've run the precise calculation with these figures (they are roughly what was used in the video).
Set the velocity of the rocket = half the speed of light.
Use units for measuring time and distance so that the speed of light, c = 1 in those units (Just to be clear that's not going to be seconds and metres. It's just conventional to set c = 1).
Set the rock collision event to co-ordinates (x, t) = ( +5.00 , +1.00 ) as was shown in my diagram for the planetary observer.
This becomes (x', t') = ( +5.20 , -1.73) in the spacemans co-ordinate system.
So, with these figures, the spaceman and planet based observer disagree on the both the location and time of the collision event.
Best Wishes.