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View Full Version : Calling all smart math people......

UncleBeer
10-11-1999, 01:56 PM
http://www.ravenna.com/blackhole.cgi?6

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"The problem with the world is that everyone is a few drinks behind." - Humphrey Bogart

Satan
10-11-1999, 01:59 PM
x = 0

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Yer pal,
Satan

Pickman's Model
10-11-1999, 03:53 PM
Hmm, I'm disappointed......I thought if I left another blank post dangling out there, I'd get a lot more comments. Apparently not.

Ah, well. Trolling aside, see if any of you can explain this to me in terms that make sense to a mathematical moron. I take this material from page 295 of Goodbye, Darkness, by William Manchester; (Boston: Little, Brown, and Company, 1979):

____________________
/ 2 2 2 2 2
ds= V c dt ---dx ---dy ---dz

I don't have a good program for this, so if it looks screwy, it's ds equals the square root of c squared times dt squared minus dx squared minus dy squared minus dz squared.

Supposedly this is something called "Minkowski's Clock Paradox", and I have never been able to find anybody who can explain to me what it means. I asked several different math profs in school, and only one of them came up with a partial answer; he said he thought it had something to do with the relationship of distance to time on the surface of a moving sphere. If that's the case, then I am assuming that "d" means "distance", "s" means "sphere" or "surface", "t" means "time", and "x", "y", and "z" are variables of some description. I have no clue as to what "c" might be. So, what does it mean, all you brilliant people out there who are smarter than I am? I await your explanations.

tracer
10-11-1999, 03:58 PM
Pickman's model wrote:

ds equals the square root of c squared times dt squared minus dx squared minus dy squared minus dz squared.

Lemme help with the notation:

ds = SQRT(c2 * dt2 - dx 2 - dy2 - dz2)

(The secret for the exponents is to use the following HTML sequence: &lt;SUP&gt;2&lt;/SUP&gt; . )

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The truth, as always, is more complicated than that.

tracer
10-11-1999, 04:01 PM

The various "d"s do not stand for distance. They are the differential symbol. "dt" means an infinitessimally small change in t, "dx" means an infinitessimally small change in x, etc.. Furthermore, x y and z are all spatial coordinates, and t is time. c is of course the speed of light. It's a problem from Relativity.

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The truth, as always, is more complicated than that.

Pickman's Model
10-11-1999, 04:39 PM
Thanks, Tracer; however, I'm still foggy on this. Okay, so it's relativity, but what does it mean? Or is it one of those things that pertains more to philosophy than math, like, if I skip a stone across a pond, does that mean that skipping it is what the future was? Or, can you put your foot into a running stream and pull it back out of the same stream, or is it a different stream, since time, distance, and water have all changed since putting your foot in to begin with? I don't follow.

Pickman's Model
10-12-1999, 12:31 AM

Dorf OnLine
10-12-1999, 12:34 AM
That's a toughie...but I'll work on it.

Keeves
10-12-1999, 12:36 AM
42

Earl Snake-Hips Tucker
10-12-1999, 12:37 AM
Yes, pick the other door.

Polycarp
10-12-1999, 12:38 AM
Funny, I thought you said "Hieronymous Bosch."

tracer
10-12-1999, 02:12 PM
I'm not entirely sure myself, as I've never wrestled with Minkowski's Clock Paradox (unless this is the same thing as the Twin Paradox with a different name).

However, I do know that in Relativity, time is usually treated as one more component in a 4-dimensional space-time vector. I.e. instead of saying something is at position (x, y, z) at time t, Relativitists would say that something is at a spacetime "position" of (x, y, z, ct). t is multiplied by c, here, to make the units come out the same, among other reasons.

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Quick-N-Dirty Aviation: Trading altitude for airspeed since 1992.

Pickman's Model
10-12-1999, 04:57 PM
Okay, so if "d" represents a very small change in the variable that follows it, then "s" obviously must stand for "speed", I would guess. Ergo:

the change in speed equals the square root of:
the speed of light squared
multiplied by
the change in time squared
minus
the change in "x" squared
minus
the change in "y" squared
minus
the change in "z" squared.

I am assuming that all these differential designations are multiplied by the variable that follows them, as in dt2 equals d times t2, right?

So in other words, this would pertain to a moving sphere; say, planet Earth, and x,y, and z could be New York, Cairo, and Rangoon, for example?

(I'm sorry if I'm appearing to be terribly dense, but you have to remember, with my math skills, I have trouble balancing a checkbook.) :)

Pickman's Model
10-12-1999, 05:10 PM
And the speed of light is still figured at 186,300 miles per second (299,792.8 kilometers per second), is it not?

Dorf OnLine
10-12-1999, 06:18 PM
Dude, try a search on the web. I got 26 hits on metacrawler with 'minkowski clock paradox'. This one seems kinda readable; http://www.weburbia.com/physics/twin_spacetime.html

Trouble with your checkbook? Do the math profs help with those questions as well? :)

tracer
10-12-1999, 06:54 PM
Pickman's Model wrote:

So in other words, this would pertain to a moving sphere; say, planet Earth, and x,y, and z could be New York, Cairo, and Rangoon, for example?

Uh ... no. x, y, and z are the three rectangular spatial coordinates for any ONE point in space. I.e. x could be the distance north of Cairo, y could be the distance east of Cairo, and z could be the distance above Cairo. Cairo would, by this definition, have (x,y,z) coordinates of (0,0,0). A plane flying directly over Cairo at an altitude of 5,000 meters would have (x,y,z) coordinates of (0,0,5000m). A man standing 1000 meters to the north and 2000 meters to the west of Cairo would have (x,y,z) coordinates of (1000m,-2000m,0).

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Quick-N-Dirty Aviation: Trading altitude for airspeed since 1992.

Omniscient
10-12-1999, 08:45 PM
You are speaking of the "Twins Paradox", heres a link...read. I will not attempt to explain it since my grasp is marginal at best and I tok a semester course on it.

Omniscient
10-12-1999, 09:15 PM
By the way, s doesn't stand for speed, speed is almost always v for velocity a vector, rarely the scalar speed. V is more commonly defined as "dx dy dz" as in change in x, y and z over time.

Your equation is not correct either. dx2 etc. is against convention. It unclear if your squaring the derivative or showing a second partial derivative.

Pickman's Model
10-12-1999, 10:08 PM
Ah, HA! So my conception was way out in left field. It must have been the "sphere" idea that threw me off......it seems like all this has to do is with the relationship of time to distance (like, does it travel in a straight line, or does it curve?), and nothing to do with speed or the shape of a particular surface.

Thanks, Tracer, Dorf, and Omniscient, for your explanations and links. I read both links, by the way; I could grasp some of it, but once I got into the heavier stuff, it was the same sensation I used to get in Algebra class: I could feel the circuit breakers in my brain begin to snap off, one by one, until the formulas magically changed into gibberish right in front of me bleedin' eyes. Mathematical comprehension for me is sort of like pouring water into a glass: you can only fit do much in, and after that, you can keep pouring if you want to, but all it'll do is just run off the top. I'm SO glad I was a History major.

In any event, I know more than I did, and I do appreciate the help. Thanks again guys!

Atrael
10-13-1999, 08:49 AM
Now I have to ask Pickman, where the hell did you stumble across this?..and what made you so interested in it?..I mean I've come across all sorts of higher Physics questions, and I just skip right over them. They make almost no sense to me, and I personally believe that most of them are hoeee anyway.(yes, that's the technical term for B.S.). I'm all for expanding my horizon, but damn you went waaaayyyy beyond the call of duty there.....

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I haven't lost my mind, I have a tape backup around somewhere.

Pickman's Model
10-13-1999, 10:34 AM
LOL. It came from William Manchester's book, Goodbye, Darkness, which is a combination memior and historical account of the war in the Pacific in World War II.

Manchester himself was a Marine, and he was telling about the times he and the guys in his company would be huddled down in holes during a lull in the fighting, and somebody would introduce some groovy concept to talk about. Many of them were draftees and college-educated (there was no deferment for college students in WW II the way there was in Vietnam), and one guy, named Wally Moon, was a budding physicist who'd been at MIT. He was the one who introduced Minkowski. Manchester wrote the equation down in his diary, but had never been able to understand it. He said there were four Phi Beta Kappas in his section, but when Wally started going into the 4th and 5th dimensions, he started to lose his audience. Another guy named Knocko told him, "Bubba, this is too deep for Dixie. The day they introduce an entrenching tool into Alabama, it'll spark an industrial revolution." I'd have to agree with that; not the part about Alabama, but about it being too deep for Dixie. Too deep for me, too.

Anyway, I was intrigued by this funny-looking equation, that's all. I wanted to know what it meant, what it was about. None of my math profs in college could tell me, like I said; so I just thought I'd throw it out here and see what came up. Interesting, but as I said, I believe I'll stick to History. The equations I saw in those two links above looked about as comprehensible as Klingon written in kanji to me.

As an aside, if anybody is interested, Goodbye, Darkness is one of the best books I ever read, bar none. I'd recommend it to anybody. It's out of print in hardcover, but you can find it in a Dell paperback edition at any good bookstore. Buy it and read it---it is excellent, in every sense of the word. Something there for everyone, I think.

tracer
10-13-1999, 02:43 PM
Omniscient wrote:

Your equation is not correct either. dx2 etc. is against convention. It unclear if your squaring the derivative or showing a second partial derivative.

That may be, but the first URL provided in this thread (by Dorf OnLine) used exactly this notation at one point. Without explaining it, I might add. I was a mite confused by it, m'self.

Pickman's Model: Believe it or not, the Twin Paradox can be explained completely without having to use ANY formulas. It's like this:

1) Special relativity says that if you're standing still, and you look at someone else going past you at 86.6% of the speed of light, you will see his clock moving twice as slow as yours, even though he will see his clock running at normal speed.

2) Special relativity ALSO says that if you're moving at 86.6% of the speed of light, and you look at someone else standing still as you pass him, you will see HIS clock moving twice as slow as yours, even though he will see his clock running at normal speed.

So, if two identical twins are separated at age 20, one of whom stays on Earth for 40 years, and the other of whom takes off in a rocket ship and whizzes around the cosmos at 86.6% of the speed of light and then returns to Earth 40 years later -- which one will have aged 40 years, and which one will have only aged 20 years? Remember, each sees the other's clock as moving half as fast as their own while the rocket-bound twin is in flight.

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Quick-N-Dirty Aviation: Trading altitude for airspeed since 1992.

10-22-1999, 07:41 PM

This appears to be an implicit differential equation for the instantaneous magnitude of a space-time interval. (s would be the magnitude of the space-time interval).

I don't see anything wrong with the notation. It doesn't appear to say anything specifically about the twins paradox, but I could see how it might be used to talk about it.

The only reason that I could see a need for this equation would be to describe the effects of acceleration within special relativity (which would require the complete neglection of gravity). Getting to the meat of the twins paradox does involve understanding how acceleration plays in.

Usually in special relativity we talk about uniform linear motion. The magnitude of a space-time interval in that case can simply be determined as

sqrt((x2-x1)2 + (y2-y1)2 + (z2-z1)2 - c2(t2-t1)2)

Hrmm. There's a difference in signs, isn't there? I'll hafta look into that. Not sure why the differential equation has reverse signs.

10-22-1999, 08:11 PM
Just read the twins paradox link. The equation that PM provided is identical to the equation they give on that page:
<img src=http://www.weburbia.com/physics/eq-tausq.gif>
They just look slightly different, because they are talking in units based on c = 1.

Still not sure as to why the signs are different from the version I knew.

The Ryan
10-23-1999, 01:44 AM
Although I have read extensively into both matrhematics and physics, I do not consider myself an abolute authority, so keep that in mind when reading the followiing explanation:
In three dimensional curves, s usually stands for arc length. That is, if a particle goes from a point on a curve that has an s value of say 5 to a point with an s value of 6, the particle has gone one unit length. Using the pythogorian thereom, (s1-s2)^2~= (x1-x2)^2+(y1-y2)^2+(z1-z2)^2 if s is very small. Mathematically, ds represents the concept of the change in s if the change is very small. So |ds|=|s1-s2| and ds=sprt of a bunch of stuff.
Now, if we go to four dimensions, things get more complicated. It's easy to imagine s in three dimensions to be amount of time that the particle has been moving. After 5 seconds, it has such and such an x coordinate, such and such a y coordinate, etc. But in four dimensions, time becomes a coordinate in which the particle "moves". That is, the particles "traces" a path through not only height, width, and lenth, but also time.
So what does s represent? I can't really explain that; the best I can say is that s is the amount time that the particle experiences. So if in a particular frame of reference, a particle spends t seconds moving through d distance (d^2= x^2+y^2+z^2; if your substitute that expression for d^2 in the following equation you'll end up with the equation of the day), the amount of time that the particle experiences is sprt(t^2-d^2). (Assuming that time and distance are represented in the "same" units).
If you know anything about relativity, this should make at least a modicum of sense to you; the more time a person in one frame of reference experiences, the more everyone experiences. However, the more someone else moves, the faster they must be moving, and therefore the less time they experience. If, in your frame of reference, someone is moving at the speed of light, then d and t will be equal (because the conversion factor between them is defined to be the speed of light) so s^2=0; the person will experience no time.
Another way of looking at this is that space is imaginary time; not imaginary in the sense that it doesn't exist but imaginary in the mathematical sense of being a multiple of the square root of negative one. Space squared is negative time squared and vice versa.
Another interesting aspect of this equation is that given a particle that is expoeriencing no forces other than gravity, and given two points (in four dimensional space) that you know that the particle visits, you can take all the possible paths between the points and calculate the s value for each path. Whichever path has the greatest s value will be the one that particle takes.
For instance, suppose you have to leave Los Angeles at 8 AM and arrive at New York at exactly 10 PM. Now, in case are not aware, the mass of the earth distorts space-time, making people deep in its gravity well (that is, close to its surface) experience slightly less time that those that are far away from the surface. So if you want to spend the most time in your cross country flight as possible (have some last minute work to get done or something) it might be a good idea to get away from the gravity well. But if you go too far away, and then come back, you'll have to accelerate to a high velocity to make it out and back in the time alloted. Going at a high speed means that you experience less time. So you want a balance between getting away from the earth's gravity well and not going to fast. It turns out that the best possible curve is basically a parabola, that is, exactly the path that you would take if you left the earth with a certain velocity, did not accelerate, and just followed a ballistic (that is, unpowered) trajectory to New York. What looks to us like curved paths are actually the longest possible paths, are therefore, in a sense, the "straighest" paths. That is, they are geodesics on the space-time manifolds, to throw out a bunch of fancy terms.
As for the sign difference, that's just a matter of notation. If you take t^2 to be positive, then that's "time distance". A positive time distance between event A and and event B ("event" refers to a point in four dimensional space) represents the most time that it could have taken a signal to go from event A to event B. If the time distance is positive, then in every frame of reference event A will appear to occur before event B. The amount of time will be relative to reference frame, but the order is absolute. If, however, one takes space be positive, then one gets the space distance. If two events have a positive space distance, then it means that there is more space between the two events than can be crossed by any particle moving at a speed equal to or less than light speed. Different reference frames will disagree about which event occurred "first" since the events are "too far apart" to get an absolute order.
I hope this has shed at least some light (no pun intended) onto the issue. If you have specific points you're still confused about, or clarifications to make, please let me know. I'm sure that there are more elegant ways of explaining this stuff; in fact Feynman probably has written one somewhere.

10-23-1999, 02:51 AM
Although I can sell you exactly one pound of candy for \$12.35, made up of candy priced at \$11.47/pound and candy priced at \$14.72/pound (I hope you don't mind; I had to break some of the pieces to get it to come out right), I lay no claim to being a smart math person. Especially after reading this thread. Feel free to flame me out of existence for having the temerity to show myself here.

Still, I pride myself on a certain knack for finding out stuff, even after I've hit what some might call a brick wall. Perhaps going around the wall can be tried.

Atrael, thanks for asking abuot how Pickman's Model got interested in this stuff; it's given me an idea.

Officer Model (and forgive me if I've inadvertently demoted you from Inspector or something), WWII was not that long ago. Some of the people who fought in it might even still be alive. Does Mr. Manchestr's book mention whether Wally Moon survived the war? Is there anything to be gained by attempting to find Mr. Moon and asking him?

C K Dexter Haven
10-23-1999, 06:41 AM
There's no second derivative involved, this is the standard distance formula.

In two dimensions, it's Pythagoras, the distance between the two points is:
h^2 = x^2 + y^2

If we use cartesian coordinats, then the distance along the x-axis is x2 - x1 (those should be subscripts), which we represent by dx (meaning a change in the x-variable.)

Thus, in 3-dimensions, the distance between two points is given by:

SQRT [ (dx)^2 + (dy)^2 + (dz)^2 ]

Einstein is the one who postulated that time is simply a fourth dimension, and that you can incorporate time into the formula by:
(a) multiplying it by c (speed of light) to get it to the same units (feet or meters or whatever); and
(b) put a negative sign in front of it to reflect that time is different from the space coordinates.

Thus, the distance between two objects in four-dimensional space-time is given by:

SQRT[ (dx)^2 + (dy)^2 + (dz)^2 - (c * dt)^2 ]

That's pretty much what you got there, Pickman -- it's Einstein's way of reflecting the distance between two points (not on sphere, but in cartesian coordinates) when you know their location in the xyz-space and you know their separation by time t.

h = SQRT[ (dx)^2 + (dy)^2 ]

10-23-1999, 03:49 PM
I'm a little sick today, so hopefully it won't be the cough syrup talking...

The Ryan, I think I agree with the general gist of what you are saying.

Describing s as arc-length is right on. The arc is normally referred to as a worldline. I referred to it as the magnitude of a spacetime interval, because the length of the worldline is the scalar element in the vector that is normally called a "space-time interval". I guess we could also call ds an infinitessimal wordline segment.

I can't really explain that; the best I can say is that s is the amount time that the particle experiences. -- The Ryan
You lose me here a bit. One of the things that is special about a space-time interval is that it is "invariant", i.e. the interval is the same for all observers at all velocities.

As for the sign difference, that's just a matter of notation. If you take t^2 to be positive, then that's "time distance". -- The Ryan

Doh! Yes, of course. It is the difference between describing a "timelike" interval vs. a "spacelike" interval.

You briefly mentioned imaginary time, and I feel like fleshing that out, since Hawking likes to talk about it.

Given:
c = speed of light
i = square root of -1
t = time
Then the value "cit" gives us the quantity that Hawking refers to as imaginary time.

If we say that T=cit, then we can write

sqrt((x2-x1)2 + (y2-y1)2 + (z2-z1)2 + (T2-T1)2)

And thus we can describe spacetime with homogeneous dimensions.