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Author Topic:   Geometry of Spacetime
nwr
Member
Posts: 6408
From: Geneva, Illinois
Joined: 08-08-2005
Member Rating: 5.1


Message 16 of 41 (701535)
06-20-2013 3:28 PM
Reply to: Message 15 by New Cat's Eye
06-20-2013 12:05 PM


The way I understand it, the only way for the time component of your path to be at a true 90 degrees would be if you were not moving.
And the way I see it, is that in space-time there is no such thing as "true 90 degrees" between a time-like direction and a space-like direction. It is all relative to the observers frame.

Fundamentalism - the anti-American, anti-Christian branch of American Christianity

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New Cat's Eye
Inactive Member


Message 17 of 41 (701537)
06-20-2013 4:10 PM
Reply to: Message 16 by nwr
06-20-2013 3:28 PM


And the way I see it, is that in space-time there is no such thing as "true 90 degrees" between a time-like direction and a space-like direction. It is all relative to the observers frame.
Sure. But if you have a particle that is at rest (in whatever reference frame), then aren't the time-like and space-like directions at 90 degrees?

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 Message 16 by nwr, posted 06-20-2013 3:28 PM nwr has replied

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nwr
Member
Posts: 6408
From: Geneva, Illinois
Joined: 08-08-2005
Member Rating: 5.1


Message 18 of 41 (701543)
06-20-2013 7:34 PM
Reply to: Message 17 by New Cat's Eye
06-20-2013 4:10 PM


But if you have a particle that is at rest (in whatever reference frame), then aren't the time-like and space-like directions at 90 degrees?
I'm not sure if that even makes sense.
Take ordinary 2-dimensional space. If we stretch out the x-axis, say rescale it so that what was one unit of length becomes 2 units, then angles change -- assuming that we don't also stretch out the y-axis. So, in some sense, the magnitude of angles is an artifact of how we measure them.
In the case of spatial directions, we normally require rotational symmetry. And if we require rotational symmetry, we cannot stretch out the x-axis without also stretching out the y-axis.
As far as I know, we cannot rotate things from a space-like direction to a time-like direction. So we don't have something like rotational symmetry to normalize our way of measuring. So I think that unavoidably leaves measurements of angles between space-like and time-like directions to be dependent on our arbitrarily chosen standards.

Fundamentalism - the anti-American, anti-Christian branch of American Christianity

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NoNukes
Inactive Member


Message 19 of 41 (701559)
06-21-2013 1:22 AM
Reply to: Message 18 by nwr
06-20-2013 7:34 PM


If we stretch out the x-axis, say rescale it so that what was one unit of length becomes 2 units, then angles change -- assuming that we don't also stretch out the y-axis. So, in some sense, the magnitude of angles is an artifact of how we measure them.
This is true. However, there is a natural set of units that eliminates this problem. If the time and distance units are chosen so that the speed of light equals one, then we have eliminated the issue you describe above.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
I would say here something that was heard from an ecclesiastic of the most eminent degree; ‘That the intention of the Holy Ghost is to teach us how one goes to heaven, not how the heaven goes.’ Galileo Galilei 1615.
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

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Iblis
Member (Idle past 3895 days)
Posts: 663
Joined: 11-17-2005


Message 20 of 41 (701684)
06-23-2013 10:06 PM


Ok, I'm still missing something, probably many things.
Son Goku writes:
Minkowski space, as a plane, just like the normal plane of Euclidean geometry.
I'm taking this to mean that I can keep working in terms of just two components, distance and time. They are sufficient, in the same way that the distance on the ground from me to the bottom of a flagpole, plus its height from the ground, would be sufficient to calculate the true distance from my feet to the top of the pole.
nwr writes:
This, I'm taking to mean, I need to subtract the time instead of adding it. Here let me try it
d - t = x
This seems to be a step in, call it the right dimension, because
this retard writes:
I expect this problem somewhere to make me subtract
but in the whole wrong direction altogether!
subtract the square of distance directly from the square of time.
Look, in my original dealie, distance is 3, the square is 9; time is 4, the square is 16; 9-16=-7, the root of minus 7 is a broken calculator.
This is telling me that it is slower than light travel that is impossible / absurd !!!! my ftl version comes out a perfectly tolerable 2.64etc.
Please tell I just got the terms backward or something. Please?
Son Goku writes:
That first doesn't really belong there does it?
Catholic Scientist writes:
It "takes" time to gain distance.
You sound like you might make sense. Could you give some examples or something, I want to flash on this.
Catholic Scientist writes:
The time coordinate is at a right angle, but the time direction of your path is only at a right angle to your distance when you're at rest. As you increase your velocity, the angle of the time direction becomes more acute. When you start approaching the speed of light, the time direction approaches being parallel to your space direction, and that's how you get length contraction.
I agreed with this completely when I originally posted. The legs of my triangle were my attempt to approximate the view of the "at rest" observer, who sees them at right angles to one another. The hypotenuse and its calculable components was supposed to give the pov of the traveler, who was bending them together by moving.
Now I've got all this "minus" shit and I'm not sure what to do. I'm hoping I can declare something "zero" and just turn my triangle like, upside-down or something.
But I'm not holding my breath ...
nwr writes:
Is that De Broglie? Can you explain him? Perhaps using pictures, or a word problem?
If my audience understood Greek, they wouldn't be New Testament believers, would they.

Replies to this message:
 Message 21 by NoNukes, posted 06-24-2013 3:19 AM Iblis has replied
 Message 39 by Son Goku, posted 07-19-2013 1:04 PM Iblis has replied

  
NoNukes
Inactive Member


Message 21 of 41 (701687)
06-24-2013 3:19 AM
Reply to: Message 20 by Iblis
06-23-2013 10:06 PM


Look, in my original dealie, distance is 3, the square is 9; time is 4, the square is 16; 9-16=-7, the root of minus 7 is a broken calculator.
Assuming units where c=1, you can express the proper time as follows:
Where distances are changing at a constant rate with respect to time, then you can use this expression:
This is the form that I have seen used to show how the twin paradox works. By the way, just because an expression generates an error on your calculator does not mean that the expression is improper.
Edited by NoNukes, : No reason given.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
I would say here something that was heard from an ecclesiastic of the most eminent degree; ‘That the intention of the Holy Ghost is to teach us how one goes to heaven, not how the heaven goes.’ Galileo Galilei 1615.
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

This message is a reply to:
 Message 20 by Iblis, posted 06-23-2013 10:06 PM Iblis has replied

Replies to this message:
 Message 22 by Iblis, posted 06-25-2013 10:01 PM NoNukes has replied

  
Iblis
Member (Idle past 3895 days)
Posts: 663
Joined: 11-17-2005


Message 22 of 41 (701768)
06-25-2013 10:01 PM
Reply to: Message 21 by NoNukes
06-24-2013 3:19 AM


NoNukes writes:
just because an expression generates an error on your calculator
Yeah yeah, this is just my expressive way of saying / showing the idea that the objection to ftl travel is supposed to be, the "imaginary" numbers involved.
Ok, this looks like, a step in the actual right direction; in the sense that we now appear to be subtracting the space from the time. This will give us i's in the right place, I suspect. Why is this different from what we were seeing before, as from Son Goku for example?
Also, what do those d's signify? Is there something I'm supposed to be multiplying everything by? Or do they just mean distance, and if so, why is there one on the t?
And even moreso, what do those deltas signify?

This message is a reply to:
 Message 21 by NoNukes, posted 06-24-2013 3:19 AM NoNukes has replied

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NoNukes
Inactive Member


Message 23 of 41 (701769)
06-25-2013 10:13 PM
Reply to: Message 22 by Iblis
06-25-2013 10:01 PM


This will give us i's in the right place, I suspect. Why is this different from what we were seeing before, as from Son Goku for example?
You can develop the equations with either negative time contributions and positive space contributions or vice versa.
The "d"s in the first equation indicate that dt, dx, dy, dz are differential values. The integral is a line integral over the trajectory of a particle.
In the equation with the deltas, deltas mean 'change'. The second equation can be be used where the change in time, and x, y, z coordinates in a case where those quantities each vary linearly with time from a starting point to the ending point.
Your questions suggest that you've got quite a bit of studying including learning a tiny bit of math before you are going to understand special relativity. Try reading the wikipedia article I linked as a starting point.
Edited by NoNukes, : No reason given.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
I would say here something that was heard from an ecclesiastic of the most eminent degree; ‘That the intention of the Holy Ghost is to teach us how one goes to heaven, not how the heaven goes.’ Galileo Galilei 1615.
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

This message is a reply to:
 Message 22 by Iblis, posted 06-25-2013 10:01 PM Iblis has not replied

  
Iblis
Member (Idle past 3895 days)
Posts: 663
Joined: 11-17-2005


Message 24 of 41 (701938)
06-28-2013 12:14 AM


Anybody?
You can develop the equations with either negative time contributions and positive space contributions or vice versa.
WTMFF?!?
. . .
Ok, is there in fact anybody anybody on this site with the insight and patience to help me understand how this minus stuff for duration works and why we are acting like this
nwr writes:
and this
are in some sense interchangeable, or descriptive of the same situation, or, whatever it is they are that allows them both to be here.
. . .
To reiterate, I expect imaginary numbers when my d is larger than my t, and ordinary decimals when my t is larger than my d. Yeah?

Replies to this message:
 Message 25 by nwr, posted 06-28-2013 9:03 AM Iblis has not replied
 Message 26 by New Cat's Eye, posted 06-28-2013 9:48 AM Iblis has not replied
 Message 27 by NoNukes, posted 06-28-2013 10:47 AM Iblis has replied

  
nwr
Member
Posts: 6408
From: Geneva, Illinois
Joined: 08-08-2005
Member Rating: 5.1


Message 25 of 41 (701947)
06-28-2013 9:03 AM
Reply to: Message 24 by Iblis
06-28-2013 12:14 AM


Re: Anybody?
Some of this is basic calculus, and some of it comes from the mathematical model that Einstein used in special relativity.
Einstein's intuition told him that the velocity of light should be the same for all observers, and the Michelson-Morley experiment seemed to confirm that. But that was incompatible with the traditional Newtonian/Euclidean view of space, which saw time and distance as independent. So the problem was one of finding a new metric which connected time and space in such a way that the velocity of light could be the same for all observers.
It worked very well. It accurately predicted motion in particle accelerators. One could deduce which seemed to accurately account for the energy seen in radioactive materials, and which was confirmed by nuclear physics.
And yes, many people at that time thought it counter-intuitive. But science is a pragmatic enterprise, and it is hard to beat "it works very well."

Fundamentalism - the anti-American, anti-Christian branch of American Christianity

This message is a reply to:
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New Cat's Eye
Inactive Member


Message 26 of 41 (701948)
06-28-2013 9:48 AM
Reply to: Message 24 by Iblis
06-28-2013 12:14 AM


Re: Anybody?
Have you studied calculus?

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 Message 24 by Iblis, posted 06-28-2013 12:14 AM Iblis has not replied

  
NoNukes
Inactive Member


Message 27 of 41 (701958)
06-28-2013 10:47 AM
Reply to: Message 24 by Iblis
06-28-2013 12:14 AM


Re: Anybody?
why we are acting like this...are in some sense interchangeable, or descriptive of the same situation, or, whatever it is they are that allows them both to be here.
I have some patience, but to date you don't seem willing to make much of an effort on your own.
Let's discuss what the equations are mean. The equation for proper time gives the time that an observer traveling a trajectory would measure. Observers not following that same trajectory could measure different times. This is explained in a fairly detailed manner with a couple of example calculations at the Wikipedia article on proper time located here.
We might also discuss the "proper length" between events rather than proper time. That formulation leads to those equations with the negative signs.
Proper length - Wikipedia
[quote]In special relativity, the proper length between two spacelike-separated events is the distance between the two events, as measured in an inertial frame of reference in which the events are simultaneous. So if the two events occur at opposite ends of an object, the proper length of the object is the length of the object as measured by an observer which is at rest relative to the object.[\quote]

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
I would say here something that was heard from an ecclesiastic of the most eminent degree; ‘That the intention of the Holy Ghost is to teach us how one goes to heaven, not how the heaven goes.’ Galileo Galilei 1615.
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

This message is a reply to:
 Message 24 by Iblis, posted 06-28-2013 12:14 AM Iblis has replied

Replies to this message:
 Message 29 by Iblis, posted 07-12-2013 11:24 PM NoNukes has replied

  
Iblis
Member (Idle past 3895 days)
Posts: 663
Joined: 11-17-2005


Message 28 of 41 (702457)
07-06-2013 11:51 PM


Sorry, had to table this question for a bit until I could get a lot more generic Excedrin super-cheap.
NN writes:
The equation for proper time gives the time that an observer traveling a trajectory would measure.
I'm starting to grok this. In my example, 3 light years in 4 years = 75% of c, the time experienced by my astronaut is only 2.64ish years. And he can't travel faster than light because he gets the imaginary numbers only if d is bigger than t. This is good, even if it's disappointing because I don't yet have a good self-evident easy-to-explain reason why I'm subtracting.
CS writes:
It "takes" time to gain distance.
So again, I really need more of that kind of stuff.
NN writes:
"proper length"
I'm really not getting this. By the definition there, wouldn't the "proper length" in my example just be the 3? The equation is making it look like it would be 2.64etc * I, what am I missing?
And how do I figure out the "improper" length, ie the distance experienced by my astronaut?
NN writes:
you don't seem willing
If you only knew, brother.

My big triangle was a lot easier to explain. The relative distance wouldn't by any chance be 2.35ish would it? That would be Swell.
Edited by Iblis, : TALKING triangles
Edited by Iblis, : wtfever

  
Iblis
Member (Idle past 3895 days)
Posts: 663
Joined: 11-17-2005


Message 29 of 41 (702947)
07-12-2013 11:24 PM
Reply to: Message 27 by NoNukes
06-28-2013 10:47 AM


Re: Anybody?
in which the events are simultaneous.
If the events were simultaneous, what would the t represent?

This message is a reply to:
 Message 27 by NoNukes, posted 06-28-2013 10:47 AM NoNukes has replied

Replies to this message:
 Message 31 by NoNukes, posted 07-13-2013 11:16 AM Iblis has replied

  
Iblis
Member (Idle past 3895 days)
Posts: 663
Joined: 11-17-2005


Message 30 of 41 (702952)
07-12-2013 11:32 PM
Reply to: Message 11 by Son Goku
06-20-2013 7:01 AM


Re: Geometry
Those paths can be drawn on both Euclidean space and Minkowski space.
These guys are telling me that your equation is for calculating "proper length", which from the description I got seems like it should just be the 3 light years in my example. I'm beginning to understand how to calculate the relative time experienced by my astronaut, which I'm understanding to be the reasonable 2.64 and change I get from the square root of (4 squared minus 3 squared). But I could probably flash on this if I could calculate the relative distance.

This message is a reply to:
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