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# Geometry of Spacetime

Author Topic:   Geometry of Spacetime
Iblis
Member (Idle past 3440 days)
Posts: 663
Joined: 11-17-2005

 Message 1 of 41 (701430) 06-18-2013 10:11 PM

Hi, I guess I need some help.
I want to show time as the 4th dimension using the Pythagorean Thingie. I'm pretty sure it's doable, cavediver, Son Goku and others have gone at it partway more than once. But I need more detail.
Here's the dealie. If I have two roads square to one another, such that I can travel 3 miles to the intersection, turn right, and travel 4 miles to the destination
then if I go "as the crow flies" instead, it will only be 5 miles. This is because
a + b = c
that is, 3 x 3 = 9, 4 x 4 = 16, 9 + 16 = 25, 25 = 5 x 5. Yeah?
So, if time is in fact a 4th dimension at right angles to space, then if I travel 3 light years in 4 years (averaging 75% of C) then my total experience should be 5 years-and-light years altogether. If I were to treat space as the constant, 3 light years, then my experienced time would be only 2 years, a 50% time dilation. But I know that's wrong, space gets compressed too. Just for a lark, treat time as a constant, 4 years, then I would only experience 1 light year of travel, 33% space dilation. Also wrong, of course.
I'm guessing, sort of feeling my way along, and I come up with the notion that the proportions of the squares can guide me. Perhaps my perceived distance can be 9 / 25 of 5 = 1.8 light years; and my perceived time can be 16 / 25 of 5 = 3.2 years.
This seems reasonable enough, but I'm sure I'm still missing something. I can't see any reasons why I couldn't switch space and time around, and get 3.2 light years in 1.8 years. Symmetrical?
I expect this problem somewhere to make me subtract the square of distance directly from the square of time. With my reversed figure, 4 light years in 3 years, this would be 9 - 16 = -7
Then when it makes me get the square root of -7 my calculator could explode and so forth. Yeah?
But I don't see anything in this figure that really makes me do that. And I need it. I neeed my audience to be able to see, very clearly and simply why ftl travel is impossible / absurd. What am I missing? Please someone make me smarter than I am.
Evolution of the physics model, Pythagoras to La Maitre. Cosmology, please.
Edited by Iblis, : god does NOT play horseshoes with the wtfverse
Edited by Iblis, : does he?
Edited by Iblis, : praps it's more like mumblety peg

 Replies to this message: Message 3 by nwr, posted 06-18-2013 11:57 PM Iblis has not replied Message 7 by New Cat's Eye, posted 06-19-2013 11:09 AM Iblis has not replied Message 11 by Son Goku, posted 06-20-2013 7:01 AM Iblis has replied

Posts: 3966
Joined: 09-26-2002

 Message 2 of 41 (701432) 06-18-2013 11:13 PM

Thread Copied from Proposed New Topics Forum
Thread copied here from the Geometry of Spacetime thread in the Proposed New Topics forum.

nwr
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Posts: 6303
From: Geneva, Illinois
Joined: 08-08-2005
Member Rating: 5.0

 (1)
 Message 3 of 41 (701434) 06-18-2013 11:57 PM Reply to: Message 1 by Iblis06-18-2013 10:11 PM

a + b = c
With the Pythagorean formula, we measure distance in two dimensions as $\color{white} \sqrt{x^2 + y^2}$. The equivalent for three dimensions is to measure distance as $\color{white} \sqrt{x^2 + y^2 +z^2}$
Once we add in a time dimension, that becomes $\color{white} \sqrt{x^2 + y^2 +z^2 -t^2}$
This assumes that we have scaled things so that $\color{white} c$, the velocity of light, is 1.
That's pretty much the metric that is the basis for relativity.
Edited by nwr, : No reason given.

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

 This message is a reply to: Message 1 by Iblis, posted 06-18-2013 10:11 PM Iblis has not replied

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Coyote
Member (Idle past 1650 days)
Posts: 6117
Joined: 01-12-2008

 Message 4 of 41 (701435) 06-19-2013 12:02 AM Reply to: Message 3 by nwr06-18-2013 11:57 PM

This may seem like a dumb question, but I am math challenged. (I took three semesters of calculus, but they were all Calculus 101.)
Why do you + x, y, and z but —t? Why not +t?
And how do you know?

 This message is a reply to: Message 3 by nwr, posted 06-18-2013 11:57 PM nwr has replied

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nwr
Member
Posts: 6303
From: Geneva, Illinois
Joined: 08-08-2005
Member Rating: 5.0

 Message 5 of 41 (701438) 06-19-2013 12:50 AM Reply to: Message 4 by Coyote06-19-2013 12:02 AM

Why do you + x, y, and z but —t? Why not +t?
It's connected to the fact that there's a minus in the wave equation:
$\color{white} \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} - \frac{\partial^2 f}{\partial t^2} = 0$
Light (electromagnetic fields) satisfy the wave equation.
And how do you know?
I'll admit that my relativity is a bit rusty.

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

 This message is a reply to: Message 4 by Coyote, posted 06-19-2013 12:02 AM Coyote has not replied

NoNukes
Inactive Member

 Message 6 of 41 (701440) 06-19-2013 2:08 AM Reply to: Message 4 by Coyote06-19-2013 12:02 AM

Why do you + x, y, and z but —t? Why not +t?
You mean -i*t rather than -t since the coordinates are squared in the given equation.
The glib answer would be that the universe just does not work that way. This is not a calculus question per se, but a special relativity question.
Proper Time, Coordinate Systems, Lorentz Transformations | Internet Encyclopedia of Philosophy
quote:
STR predicts that motion of a system through space is directly compensated by a decrease in real internal processes, or proper time rates. Thus, a clock will run fastest when it is stationary. If we move it about in space, its rate of internal processes will decrease, and it will run slower than an identical type of stationary clock. The relationship is precisely specified by the most profound equation of STR, usually called the metric equation (or line metric equation).
(1) [sorry I'm too lazy to latex the equation]
This applies to the trajectory of any physical system. The quantities involved are:
Dt is the amount of proper time elapsed between two points on the trajectory.Dt is the amount of real time elapsed between two points on the trajectory.
Dr is the amount of motion through space between two points on the trajectory.
c is the speed of light, and depends on the units we choose for space and time.

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 4 by Coyote, posted 06-19-2013 12:02 AM Coyote has not replied

New Cat's Eye
Inactive Member

 Message 7 of 41 (701446) 06-19-2013 11:09 AM Reply to: Message 1 by Iblis06-18-2013 10:11 PM

if time is in fact a 4th dimension at right angles to space
I don't think it is. Here's my current conception of it, though I may be wrong:
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.
Stand with your arms pointing straight out to your side at 90 degrees and look straight forward. You're going to be moving forward in the direction you're looking, but while your standing still you're arms are pointing in the perpendicular direction. The amount that your arms are pointing forward towards the direction you are looking (traveling) is how much time contributes to your position, which is zero right now at rest. As you begin moving faster in the forward direction, move your arms towards the direction you are facing. This is an increase in the time component of the distance. More of your path includes the time direction the faster you are going. When you reach the speed of light, your arms are pointing straight forward and time component makes up all of your distance (the space component is contracted). A photon doesn't really experience passing through space, its more like the spot it is going to is contracted to its current position.
But all that is just off the cuff so I'm sure I screwed something up. Still, I think that's the jist of it.

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New Cat's Eye
Inactive Member

 Message 8 of 41 (701447) 06-19-2013 11:11 AM Reply to: Message 4 by Coyote06-19-2013 12:02 AM

Why do you + x, y, and z but —t? Why not +t?
It "takes" time to gain distance.

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NoNukes
Inactive Member

 Message 9 of 41 (701473) 06-19-2013 6:44 PM Reply to: Message 7 by New Cat's Eye06-19-2013 11:09 AM

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.
What does it mean to be 'at rest'?

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 7 by New Cat's Eye, posted 06-19-2013 11:09 AM New Cat's Eye has replied

 Replies to this message: Message 10 by Taq, posted 06-19-2013 9:11 PM NoNukes has replied Message 13 by New Cat's Eye, posted 06-20-2013 10:47 AM NoNukes has replied

Taq
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Posts: 9683
Joined: 03-06-2009
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 Message 10 of 41 (701475) 06-19-2013 9:11 PM Reply to: Message 9 by NoNukes06-19-2013 6:44 PM

What does it mean to be 'at rest'?
It means relative velocity. All frames of reference are equally valid, so at rest means not moving in relation to your indicated frame of reference.
As an example, if you saw someone sitting in front of a tree you would probably say that the sitting person is not moving with respect to the tree. However, they are moving with respect to the Sun as the Earth hurls around it, and the Sun itself is moving about the center of the Milky Way. None of these frames of reference is THE frame of reference, btw.
Edited by Taq, : No reason given.
Edited by Taq, : No reason given.

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Son Goku
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Posts: 1226
From: Ireland
Joined: 07-16-2005

 (2)
 Message 11 of 41 (701489) 06-20-2013 7:01 AM Reply to: Message 1 by Iblis06-18-2013 10:11 PM

Geometry
But I don't see anything in this figure that really makes me do that. And I need it. I neeed my audience to be able to see, very clearly and simply why ftl travel is impossible / absurd. What am I missing? Please someone make me smarter than I am.
Typically we represent the spacetime of special relativity, called Minkowski space, as a plane, just like the normal plane of Euclidean geometry.
This is because they are both flat spaces, so you can carry over intuition from one to the other and the easiest way to encode this intuition is by drawing them the same way, i.e. as a plane.
Now on a 2D plane, if dx and dy are the distances between two points on the x and y axis, then the distance between both of the points is: $\color{white} s = \sqrt{dx^2 + dy^2}$ for 3D space: $\color{white} s = \sqrt{dx^2 + dy^2 + dz^2}$ for 4D space: $\color{white} s = \sqrt{dx^2 + dy^2 + dz^2 + dt^2}$
Of course you could naturally ask the question, what if I put a minus sign in front of one of the terms, I'd get: $\color{white} s = \sqrt{dx^2 + dy^2 + dz^2 - dt^2}$
So there are two possible four dimensional spaces here. The first one, with the + sign is called 4D Euclidean space and the second one with the - sign is called 4D Minkowski space.
They behave quite differently, but they are both valid spaces mathematically. It just turns out that our universe is the second one, not the first.
You can compare a universe with the first type of distance rule (basically the universe of Aristotles physics) to the real world and it fails to match the behaviour of the real world, unless things are moving very slowly.
So the first thing is, you can't get this minus sign from the triangle. The triangle is just a path through the space or more accurately the composition of three paths. Those paths can be drawn on both Euclidean space and Minkowski space.
In other words, the picture only represents a triangle in a flat space, whether that space is Minkowskian or Euclidean is an extra detail you have to supply.
So, to my mind, you could write down pythagoras' theorem, calculate distances and then ask "why" must that be the rule that triangles obey. The truth is that it isn't the only logical possibility. Then you could suggest the new one, with the minus sign, which is our actual universe.
Edited by Son Goku, : No reason given.
Edited by Son Goku, : No reason given.

 This message is a reply to: Message 1 by Iblis, posted 06-18-2013 10:11 PM Iblis has replied

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NoNukes
Inactive Member

 Message 12 of 41 (701507) 06-20-2013 10:19 AM Reply to: Message 10 by Taq06-19-2013 9:11 PM

It means relative velocity. All frames of reference are equally valid, so at rest means not moving in relation to your indicated frame of reference.
Thanks. I was really checking to see what Catholic Scientist meant by 'at rest'?

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 10 by Taq, posted 06-19-2013 9:11 PM Taq has not replied

New Cat's Eye
Inactive Member

 Message 13 of 41 (701510) 06-20-2013 10:47 AM Reply to: Message 9 by NoNukes06-19-2013 6:44 PM

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.
What does it mean to be 'at rest'?
The impossible state of not-moving. Since everything is moving, there's always going to be some contribution to your path from the time direction. That is, it would never be at a true right angle.

 This message is a reply to: Message 9 by NoNukes, posted 06-19-2013 6:44 PM NoNukes has replied

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NoNukes
Inactive Member

 Message 14 of 41 (701517) 06-20-2013 11:51 AM Reply to: Message 13 by New Cat's Eye06-20-2013 10:47 AM

The impossible state of not-moving. Since everything is moving, there's always going to be some contribution to your path from the time direction. That is, it would never be at a true right angle.
Taq has anticipated where my question is going. As I understand your answer, you seem to believe that there is some rest state that is impossible to achieve. Instead the situation is that some objects are at rest with respect to some frame, possibly an inertial frame. However there is no special inertial frame which defines motion at rest.
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 13 by New Cat's Eye, posted 06-20-2013 10:47 AM New Cat's Eye has replied

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New Cat's Eye
Inactive Member

 Message 15 of 41 (701520) 06-20-2013 12:05 PM Reply to: Message 14 by NoNukes06-20-2013 11:51 AM

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. I suppose you could achieve that with the proper inertial frame, but that didn't seem to be what the OP was asking.

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