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# Geometry of Spacetime

Author Topic:   Geometry of Spacetime
Iblis
Member (Idle past 3215 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

a² + b² = c²

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

Iblis
Member (Idle past 3215 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

d² - t² = x²

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.

 Son Goku writes:

That first ² 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

Iblis
Member (Idle past 3215 days)
Posts: 663
Joined: 11-17-2005

 Message 22 of 41 (701768) 06-25-2013 10:01 PM Reply to: Message 21 by NoNukes06-24-2013 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

 Replies to this message: Message 23 by NoNukes, posted 06-25-2013 10:13 PM Iblis has not replied

Iblis
Member (Idle past 3215 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

Iblis
Member (Idle past 3215 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 3215 days)
Posts: 663
Joined: 11-17-2005

 Message 29 of 41 (702947) 07-12-2013 11:24 PM Reply to: Message 27 by NoNukes06-28-2013 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 3215 days)
Posts: 663
Joined: 11-17-2005

 Message 30 of 41 (702952) 07-12-2013 11:32 PM Reply to: Message 11 by Son Goku06-20-2013 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: Message 11 by Son Goku, posted 06-20-2013 7:01 AM Son Goku has not replied

Iblis
Member (Idle past 3215 days)
Posts: 663
Joined: 11-17-2005

 Message 32 of 41 (703021) 07-13-2013 9:58 PM Reply to: Message 31 by NoNukes07-13-2013 11:16 AM

 even bother

I enjoyed studying the example given in the article on Proper Time, I gave the results of my understanding above. Here is a bit of it for reference

wiki writes:

Let there now be another observer B who travels in the x direction from (0,0,0,0) for 5 years of coordinate time at 0.866c to (5 years, 4.33 light-years, 0, 0). Once there, B accelerates, and travels in the other spatial direction for 5 years to (10 years, 0, 0, 0). For each leg of the trip, the proper time is

I plugged in my 3 and 4 and got the same 2.64etc I have been going on about for more than a page. No problem.

I am not finding any similar examples in the article on Proper Length. This is the article from which you have lifted your mistagged definition and one equation, which do not appear to fit together. I am noticing that there are other equations in the article, and some multiplication by c-squared and so on.

But I'm giving you the benefit of the doubt. It would be cruel and senseless of you to pretend there were examples in a place where there aren't any, so I just must not be able to see them for some reason. Please feel free to post them here, thereby alerting the 3 or 4 people in the universe who do not yet know that I am a retard of their duty to throw rocks at me in honor of your pithy wisdom.

I am not swearing out loud in a coffee-house right this very minute.

 This message is a reply to: Message 31 by NoNukes, posted 07-13-2013 11:16 AM NoNukes has replied

 Replies to this message: Message 33 by NoNukes, posted 07-13-2013 11:01 PM Iblis has replied

Iblis
Member (Idle past 3215 days)
Posts: 663
Joined: 11-17-2005

 Message 34 of 41 (703031) 07-13-2013 11:52 PM Reply to: Message 33 by NoNukes07-13-2013 11:01 PM

Did you even bother to look

What discussion? Here's the total dialogue from the talk page

 ... spacelike events in a frame of reference in which the events are simultaneous.Doesn't it mean timelike? If I remember correctly, spacelike events can be in the same spot, but not at the same time. Timelike events, however, can be at the same time, but not in the same place. You don't remember correctly. Timelike events can be related through time. So they can be at the same place but not at the same time. Similartly spacelike events can be related only through space. --EMS

 an improper equation

Here's the actual equation

 L=\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2 - c^2 \Delta t^2},

As given there, yes it's hideous but it probably ought to be

If the c² is just 1=discardable, then my proper length appears to be the same as my proper time only imaginary, ie the root of -7.

Here's another boggle

 Proper length is analogous to proper time. The difference is that proper length is the invariant interval of a spacelike path or pair of spacelike-separated events, while proper time is the invariant interval of a timelike path or pair of timelike-separated events.

Proper time actually turned out to be the relative time experienced by my astronaut when I walked the example. This was cool, I wanted that. So why is it being called invariant and likened to a number that supposedly represents a nonrelative view, ie

 f 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.

Am I just misunderstanding the language? Is this really a calculation of the relative distance? Is there some number I should be plugging in along with my 3²-4² that will make the root of it a real number?

Why is this article so shitty compared to the other one, and what in the world makes you think it's sufficient?

Maybe I will just post the whole thing and ask for votes on the shittiness quotient™

Edited by Iblis, :

Edited by Iblis, :

 This message is a reply to: Message 33 by NoNukes, posted 07-13-2013 11:01 PM NoNukes has replied

 Replies to this message: Message 35 by NoNukes, posted 07-14-2013 12:00 AM Iblis has replied

Iblis
Member (Idle past 3215 days)
Posts: 663
Joined: 11-17-2005

 Message 36 of 41 (703033) 07-14-2013 12:13 AM Reply to: Message 35 by NoNukes07-14-2013 12:00 AM

Re: Did you even bother to look
 the meaning of proper length

1) In special relativity, the proper length of an object moving at velocity close to that of the speed of light is its length as measured by an observer in the rest frame of the object.
Will: Dude, that 30cm ruler travelling at the speed of light looks to be only about 28.5cm.
Nathaniel: Yeah, that's because it's proper lenth is 30cm

2) A massive shit.
Nathaniel: Dude, I just crimped out a proper length.
Will: Yeah, thats a proper brown length.

 This message is a reply to: Message 35 by NoNukes, posted 07-14-2013 12:00 AM NoNukes has replied

 Replies to this message: Message 37 by NoNukes, posted 07-14-2013 1:06 AM Iblis has not replied

Iblis
Member (Idle past 3215 days)
Posts: 663
Joined: 11-17-2005

 Message 38 of 41 (703332) 07-18-2013 11:41 PM

reality bites
 NoNukes writes:The observed length of the object varies

Yep, got it. That's when I posted the urbandictionary above, which helped make it clear to me.

Proper Length has nothing to do with the distance traveled, it is the vehicle that is getting dilated in a sort of "doppler effect" of the dimension in line with trajectory. The only relation to my original question is the symmetry with proper time for elucidation, and perhaps that it was, or at least could put me closer to, what Son Goku was talking about. Yes?

Back to the actual question, I'm totally headfucked about the distance part of relativity, aren't I? I e there's no dilation of the distance traveled from the viewpoint of my astronaut, he perceives himself as having traveled 3 light years. He just thinks he did it in 2.64etc years, at warp factor 1 point something.

I think I will tell my crowd of Believers that it is 2.35ish anyway. It will be ok! They will Believe it, that's what they do ...

 not fun anymore

Yes, that's fine, you're off the hook. No need to answer this even, why I made it General.

Thanks for helping !!!

Iblis
Member (Idle past 3215 days)
Posts: 663
Joined: 11-17-2005

 Message 41 of 41 (703444) 07-21-2013 10:47 PM Reply to: Message 39 by Son Goku07-19-2013 1:04 PM

Son Goku ftw
 actually makes no difference

Thanks brother, this is helping a lot. I'm sure you see what I am doing here, but I'm going to continue because it really is making me smarter. Or call it, less stupid

 imaginary with respect to each other

So would it be correct to say that this business we hear widely in the media and in physics classes taught by coaches, about how ftl is "impossible" because of the square root of negative 1, is just a pedagogical canard?

Or would it be better to say that since the theory uses Minkowski space, and Minkowski subtracts instead of adding like Pythagoras, and the theory has been proooven right, and we see plenty of cases of traveling slower than light, then it is ftl that must be imaginary?

Or am I going to far here? I know that time dilation due to gravitational influences has been supported, but I can't find any references this minute that show data for a real-life test of dilation due to acceleration.

 in calculations of physical quantities the imaginary number drops out

Are you just saying that, we don't have any real-world ftl events to measure? Or is there deeper stuff here I'm not flashing on presently?

I'm still missing something in this "proper length" stuff. The relief of realizing it had less to do with my question than one might think has passed and I'm left a curious monkey going huh. I can't seem to make a real world example cause I'm not flashing on what I should use for the t quantity there. Is the c stuck in to make sure I realize it is speed I should plug, not 0 = "simultaneous" ? An example might help, showing how fast my yardstick is traveling vs how long it looks.

Still looking ...

 This message is a reply to: Message 39 by Son Goku, posted 07-19-2013 1:04 PM Son Goku has not replied

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