mouse@mcgill-vision.UUCP (der Mouse) (11/08/86)
In article <1388@trwrb.UUCP>, galins@trwrb.UUCP (Joseph E. Galins) writes: > If the elevator is accelerating, then after a long, long time its > speed will reach the speed of light then pass it. Not if current theories are correct. > I know that the problem here has to do with the fact that F=mA is > false when the speed approches 'c'. Well, this depends on how you interpret the symbols. As I recall, if you interpret m as relativistic mass then it is true. > In fact F approches infinity near 'c'. Really? Where does the energy to exert such a force come from? > So with a constant (or even increasing but finite) force, wouldn't > the acceleration necessarly slow down as the rider approched 'c' and > hence notice that he was in an elevator? Yes, EXCEPT that as the rider views it, the acceleration is constant. Time dialation, remember. It's the outside observer that sees acceleration as slowing down. > In other words, with a constant acceleration eventually you would be > going at a speed of 'c' with no more acceleration therefore losing > the 'gravity' feeling. (1) You can't reach c. (2) The rider still feels the acceleration as proportional to the force; it's just the observer with respect to whom the elevator is moving near c (the "stationary" observer) that sees the acceleration slack off. > Then is it impossible to simulate gravity via acceleration for a > 'long' period of time? Not so. The only difference between "gravity as created by a mass" and "gravity as created by acceleration" (I use the quotes because gravity *is* acceleration, but I don't see any simple, brief, and more accurate way of putting it) is the structure of the field (ie, parallel versus converging lines of force....). Note another difference. If the elevator rider measures his velocity with respect to "the universe", whatever that means, he will find it to be steadily increasing. The guy on the planet will not. For "the universe", picky theorists may substitute, say, the background black-body radiation (a suspiciously absolute frame of reference, if you ask me!). der Mouse USA: {ihnp4,decvax,akgua,utzoo,etc}!utcsri!mcgill-vision!mouse think!mosart!mcgill-vision!mouse Europe: mcvax!decvax!utcsri!mcgill-vision!mouse ARPAnet: think!mosart!mcgill-vision!mouse@harvard.harvard.edu Aren't you glad you don't shave with Occam's Razor?
news@cit-vax.Caltech.Edu (Usenet netnews) (11/13/86)
Organization : California Instititute of Technology Keywords: Background radiation From: myers@hobiecat.Caltech.Edu (Bob Myers) Path: hobiecat!myers In article <546@mcgill-vision.UUCP> (der Mouse) writes: >Note another difference. If the elevator rider measures his velocity >with respect to "the universe", whatever that means, he will find it to >be steadily increasing. The guy on the planet will not. > >For "the universe", picky theorists may substitute, say, the background >black-body radiation (a suspiciously absolute frame of reference, if >you ask me!). Hey, here's a fairly interesting question! What happens to the background black-body radiation when you start moving faster? Obviously, it gets doppler shifted. What does the distribution of radiation with wavelength look like? Is it still a black-body curve? All sorts of interesting questions with this one! -Bob Myers
gwyn@brl-smoke.ARPA (Doug Gwyn ) (11/13/86)
In article <1167@cit-vax.Caltech.Edu> myers@hobiecat.UUCP (Bob Myers) writes: >doppler shifted. What does the distribution of radiation with wavelength >look like? Is it still a black-body curve? This is harder to compute than it may first appear. I seem to recall that the answer is, no, isotropic black-body radiation does NOT look like isotropic black-body radiation with respect to a moving frame.
ethan@utastro.UUCP (Ethan Vishniac) (11/13/86)
In article <5363@brl-smoke.ARPA>, gwyn@brl-smoke.ARPA (Doug Gwyn ) writes: > In article <1167@cit-vax.Caltech.Edu> myers@hobiecat.UUCP (Bob Myers) writes: > >doppler shifted. What does the distribution of radiation with wavelength > >look like? Is it still a black-body curve? > > This is harder to compute than it may first appear. > I seem to recall that the answer is, no, isotropic > black-body radiation does NOT look like isotropic > black-body radiation with respect to a moving frame. The radiation field still looks like a blackbody *in any given direction* but a moving observer will see an angle dependent temperature. -- "More Astronomy Ethan Vishniac Less Sodomy" {charm,ut-sally,ut-ngp,noao}!utastro!ethan - from a poster seen ethan@astro.AS.UTEXAS.EDU at an airport Department of Astronomy University of Texas
markb@sdcrdcf.UUCP (Mark Biggar) (11/14/86)
In article <1167@cit-vax.Caltech.Edu> myers@hobiecat.UUCP (Bob Myers) writes: >In article <546@mcgill-vision.UUCP> (der Mouse) writes: >>For "the universe", picky theorists may substitute, say, the background >>black-body radiation (a suspiciously absolute frame of reference, if >>you ask me!). >Hey, here's a fairly interesting question! What happens to the background >black-body radiation when you start moving faster? Obviously, it gets >doppler shifted. What does the distribution of radiation with wavelength >look like? Is it still a black-body curve? The background black-body radiation almost defines a privileged reference frame. You can almost define Absolute Rest as meaning there is no observed doppler shift in the background radiation. In fact, there is an observed doppler shift in the background radiation as seen from earth. This has been used to get a fairly good estimate of the real velocity of the earth, sun, milky way, etc. with respect to the rest of the universe (the so called fixed stars). Mark Biggar {allegra,burdvax,cbosgd,hplabs,ihnp4,akgua,sdcsvax}!sdcrdcf!markb
ran@ho95e.UUCP (RANeinast) (11/14/86)
> Hey, here's a fairly interesting question! What happens to the background > black-body radiation when you start moving faster? Obviously, it gets > doppler shifted. What does the distribution of radiation with wavelength > look like? Is it still a black-body curve? > All sorts of interesting questions with this one! > -Bob Myers A Doppler shifted blackbody curve (actually, it can be shifted by any "geometric" means, such as gravitational redshift) is another blackbody curve, but with a new temperature shifted exactly as the frequencies are shifted. This means that if you look ahead of you, you will see a blackbody curve for a higher temperature, and if you look behind, you will see a blackbody curve for a lower temperature. This directional anisotropy (although quite small) has been detected in the background 3 degree radiation, and is generally interpreted as the cumulative (from motion about the sun, center of galaxy, galactic group) speed of the earth relative to the original big bang. -- ". . . and shun the frumious Bandersnatch." Robert Neinast (ihnp4!ho95c!ran) AT&T-Bell Labs
matt@oddjob.UUCP (Matt Crawford) (11/17/86)
It's odd that nobody has trotted out this follow-on question... What does the background radiation look like to an *accelerating* observer? In light [sic] of the answer, consider Courtney Footman's puzzle of a week back. _____________________________________________________ Matt University crawford@anl-mcs.arpa Crawford of Chicago ihnp4!oddjob!matt (Still recruiting for SO-SO: Scientists, Occultists, Skeptics and Optimists. Membership 0004 and growing!)
james@reality1.UUCP (james) (11/18/86)
IN article <546@mcgill-vision.UUCP>, mouse@mcgill-vision.UUCP (der Mouse) wrote: > In article <1388@trwrb.UUCP>, galins@trwrb.UUCP (Joseph E. Galins) writes: > > So with a constant (or even increasing but finite) force, wouldn't > > the acceleration necessarly slow down as the rider approched 'c' and > > hence notice that he was in an elevator? > > Yes, EXCEPT that as the rider views it, the acceleration is constant. > Time dialation, remember. It's the outside observer that sees > acceleration as slowing down. Hmmm... I take it you mean that if the rider measures acceleration in terms of the gravity the rider feels, then it is constant. But if the rider measures acceleration in terms of the observer that was left behind? I guess it all depends on how you want to define acceleration maybe? After all, since the rider can't see the observer recede faster than the speed of light, eventually the change in velocity between the rider and observer would have to fall off. > Note another difference. If the elevator rider measures his velocity > with respect to "the universe", whatever that means, he will find it to > be steadily increasing. The guy on the planet will not. Huh? Are you suggesting that the observer could view the rider as having no acceleration even while the rider viewed acceleration? I would think the guy on the planet would still measure the _velocity_ as increasing, but the acceleration as decreasing... > For "the universe", picky theorists may substitute, say, the background > black-body radiation (a suspiciously absolute frame of reference, if > you ask me!). Ah, but absolute relative to what? :-) One other thing that confuses me about all of this. The rider takes off from the planet at relativistic speeds, and the observer on the planet sees the rider "suffer" time dilation: the rider appears to move more slowly through time. I think I see where this comes from. But what does the rider see of the planet? Since the planet is also receding relativisticly from the rider, the rider observes the planet suffering the same fate: time dilation by the same factor as the planetary observer measured. Now, when the rider arrives at the destination, the rider must conclude that the rider's clocks are *ahead* of the planet's, since the rider clearly observed the planet's clocks "slow down". Similarly the planetary observer must also conclude that the rider's clocks are *behind* the planetary clocks. Clearly I'm missing something rather fundemental, but it seems to me that I have to be able to view the situation from both the rider's standpoint and the planet's and get the same answers (that's what relativity is about in this example?). When the rider returns to the planet, every cutsey story says the rider is much "younger" than the contemporaries left behind because of time dilation. Yet somehow I much account for the fact that the rider observed the rider's contemporaries age slower than the rider as the rider left the planet... Not very well put, but I think the dilema is stated. My high school physics isn't letting me in on the secret... -- James R. Van Artsdalen ...!ut-ngp!utastro!osi3b2!james "Live Free or Die"
KFL@MX.LCS.MIT.EDU (11/22/86)
From: "Keith F. Lynch" <KFL@MX.LCS.MIT.EDU>
From: ucbcad!ames!nike!ll-xn!mit-eddie!husc6!ut-sally!ut-ngp!melpad!reality1!james@UCB-Vax.arpa (james)
Hmmm... I take it you mean that if the rider measures acceleration in
terms of the gravity the rider feels, then it is constant. But if the
rider measures acceleration in terms of the observer that was left behind?
The observer who was left behind may be:
1) Left behind in empty space, in which case he would observe the
elevator accelerating away from him, which it really would be. The
observer would be weightless because he is not accelerating and
there are no large masses nearby.
2) Left behind in an elevator shaft on Earth, in which case he would
observe the elevator accelerating away from him. Actually, HE is
accelerating away from IT, but he can't know that without peeking.
The observer would be weightless (until he reaches the bottom of the
shaft!) because he in unsupported.
Once again, there is no way to tell which is the case without looking
"outside".
I guess it all depends on how you want to define acceleration maybe?
After all, since the rider can't see the observer recede faster than
the speed of light, eventually the change in velocity between the rider
and observer would have to fall off.
This relativistic decrease in acceleration would be observed in either
case, so it can't be used to distinguish between them.
One other thing that confuses me about all of this. The rider takes off
from the planet at relativistic speeds, and the observer on the planet
sees the rider "suffer" time dilation: the rider appears to move more
slowly through time. ... But what does the rider see of the planet?
Since the planet is also receding relativisticly from the rider, the
rider observes the planet suffering the same fate: time dilation by the
same factor as the planetary observer measured.
Now, when the rider arrives at the destination, the rider must conclude
that the rider's clocks are *ahead* of the planet's, since the rider
clearly observed the planet's clocks "slow down". Similarly the planetary
observer must also conclude that the rider's clocks are *behind* the
planetary clocks. ... it seems to me that I have to be able to view the
situation from both the rider's standpoint and the planet's and get the
same answers (that's what relativity is about in this example?).
The flaw is the assumption that they ever get together again. The
rule about clocks slowing down is strictly true only for unaccelerated
observers. And if two unaccelerated observers start in the same place,
they will never meet eachother again.
I know it's kind of hard to imagine that both are correct in
considering the other one's clock to be running slower that their own.
The best way to see it is to draw two perpendicular lines on a piece of
paper, one line representing time, the other line representing the
direction in space in which the two observers are seperating. The lines
you have drawn are the space and time axes as seen by one observer. To
represent the space and time axes seen by the other observer, draw two
other lines, which are not perpendicular, but which angle inwards from
the lines you have by the same amount. Of course the other observer
"sees" those lines as perpendicular, and the original two as angled
inwards.
One of my favorite thought experiments is to imagine a large set of
synchronized clocks, all at rest relative to eachother, set up at
arbitrary grid points in empty space. Someone sitting on one of the
clocks observes a clock 186,282 miles away and observes it striking
the hour one second late, which he correctly explains as the speed of
light delay. But now a second similar set of synchronized clocks
comes hurtling through this set. Does the observer see these clocks
as being synchronized with eachother? He does not. He sees the
trailing clocks as lagging behind, and the leading clocks as being
set ahead. But an observer at rest with respect to the second set
of clocks will see that his set are in synch and the FIRST set are not.
Which observer is right? They both are.
A similar paradox is the garage door paradox. This relies on the
concept that things in rapid motion are shortened in the direction of
motion. The driver of a very fast 10 foot long car sees his car as
still being ten feet long, but sees a garage, normally 10 feet long,
as being only 5 feet long. The garage operator sees his garage as
being ten feet long, but sees the fast car as being foreshortened to
only five feet long. The paradox is that the garage operator thinks
that he can have both garage doors, front and back, (briefly) closed
while the car is passing through. But the driver is equally convinced
that he cannot pass through the garage unless both doors are open.
Think about it. I won't answer it for now. Remember, the rules are:
1) Motion in a straight line is relative, i.e. no experiment can show
which state of uniform motion is at rest (from which it is concluded
that the concept of "at rest" is bogus). This comes from Newtonian
physics.
2) The speed of light in a vacuum is the same to all observers. This
can be deduced from Maxwell's equations (1861) and was confirmed
directly by Michelson and Morley in 1881.
Special Relativity is the simplest theory with these as axioms. It
can all be derived by high school algebra. The fact that it wasn't for
over 40 years shows how reluctant people are to question things that
have been long taken for granted, such as the constancy of space and of
time.
By the way, it can also be proven that if these axioms are true, that
sending information faster than light is not possible without reversing
causality (i.e. time travel). SF authors would do well to accept time
travel if they postulate FTL travel, unless they instead explain which
one of the two axioms turned out to be wrong despite overwhelming
evidence. A plausible case could be made that the cosmic background
noise defines an absolute frame of reference. I suspect that Doug
Gwyn's vehement rejection of the cosmic background explanation for the
noise might be at least partly due to his heavy emotional investment in
relativity theory, and due to a perhaps subconscious perception that
this cosmic background presents a threat to that investement. Then
again perhaps my eager acceptance of the cosmic background explanation
for the noise is for the same reason, since I *WANT* FTL travel to be
possible but can't bring myself to accept reverse causality.
Of course facts are facts, and are not influenced by our hopes or
fears or misconceptions.
When the rider returns to the planet, every cutsey story says the rider
is much "younger" than the contemporaries left behind because of time
dilation.
This involves general relativity, not special relativity. General
relativity expalins the interactions of ACCELERATING and GRAVITATING
objects, while special relativity only applies to objects moving in
straight lines at constant speeds in the absense of gravity.
These cutesy stories are correct, according to general relativity.
In addition to the above two axioms, general relativity has a third:
3) Acceleration is indistinguishable from gravity.
Which is where I came in, so I'll climb down off my soapbox now.
...Keiththroopw@dg_rtp.UUCP (11/22/86)
> james@reality1.UUCP (james) >> mouse@mcgill-vision.UUCP (der Mouse) >>> galins@trwrb.UUCP (Joseph E. Galins) >>> So with a constant (or even increasing but finite) force, wouldn't >>> the acceleration necessarly slow down as the rider approched 'c' and >>> hence notice that he was in an elevator? >> Yes, EXCEPT that as the rider views it, the acceleration is constant. > But if the rider > measures acceleration in terms of the observer that was left behind? How would you suggest the rider do this without looking outside? >> Note another difference. If the elevator rider measures his velocity >> with respect to "the universe", whatever that means, he will find it to >> be steadily increasing. The guy on the planet will not. Well, one thing it means is looking outside the elevator. > One other thing that confuses me about all of this. > [description of classic twin "paradox", omitted] > My high school physics isn't letting me in on the secret... My college physics DOES let me in on the secret... but just how many times should I post the same thing over and over? Well, maybe just once more, since this aspect of it hasn't come up in this particular newsgroup for a while (though you'd already know the answer if you were paying close attention to the "instant message channel" exchange). This is a repeat of a talk.philosophy.misc fragment, with a small correction pointed out to me by ucbvax!brahms!weemba. We have a stay-at-home A, and a passing traveler, B. A says something like "First, take the event on B's timeline that I consider simultaneous with 'now' on my timeline. Second, take another event on B's timeline that I consider simultaneous with a point N seconds ago on my timeline. B will think that M seconds have passed between these two events, and M will be less than N." Fair enough. But change the B's to A's in the above statement, and it must be *just* *as* *true* when B says it. How can this be? Well, note that the point on A's timeline M seconds ago (according to A) doesn't *HAVE* to be simultaneous with the point on B's timeline of M seconds ago (according to B). How does it all work? See the diagram below. Events are 1, 2, 3, 4 and 5. Worldlines are A and B. A and B cross at 3, so event 3 is on both A's and B's timeline. | . . . . | . . . . * | . . . . * ^ | . . .. * | | A ****** 1 2 ************ 3 ************ space | .. * .. | . * . . | .. * . . | . 4* . . | 5 . . . | B * . . . . +------------------------------------------------- time --> Note that in this diagram, events located along the vertical dotted lines are considered to be all simultaneous by A, and events located along the slanted dotted lines are considered to be simultaneous by B. This is a space-time geometric cosequence of the generally-given-as-algebraic "laws" of special relativity. So what have we got? Well, the time interval 1-3 is what A considers N seconds. A considers 1 and 4 to be simultaneous. The interval 5-3 is what B considers N seconds. B considers 5 and 2 simultaneous. We can see that A considers 2-3 to be less than N seconds long, B considers 4-3 to be less than N seconds long. So, lo and behold, A thinks that B is slower, and B thinks that A is slower. And all because space and time are (to a limited extent) interchangeable, just as various spatial dimensions are. Fancy that! Note that all these relationships stay the same from any point of view, whether A's, B's, or somebody else going a different constant speed. Of course, none of these others will agree that 1-3 or 5-3 is N seconds long, but so what? Even A doesn't think that 5-3 is N seconds (A thinks it is longer!), nor does B think that 1-3 is N seconds (again, B thinks it is longer!). Ok. Fine. But how does all this hoo-hah relate to the twin "paradox"? After all, either A or B has to be "really" slower, right? Wrong. As B travels away from A, B considers events on slanted lines to be "simultaneous". When B accelerates to stop at some distance from A, B has changed reference frames, and now considers events on vertical lines to be "simultaneous". Thus, for any stretch of the trip with no acceleration (and with B and A moving relative to each other), both B and A think that the other has slowed down. But during B's acceleration, B skips large parts of A's history, because B's line through space-time indicating what is the "present moment" rotates, and sweeps along a large part of A's history. The same occurs when B accelerates to return home. The difference in the twin's timelines is that A didn't change inertial frames, and B did. Simple as that. -- Our tool is diagram and explanation... uh, two, our TWO tools are diagrams and explanataions, and appeal to authority... uh THREE our THREE tools are diagrams, explanations, appeal to authority, and... AMONG our tools ARE: diagrams, explanations, appeal to authority, and a fanatical devotion to Physics! --- (NOBODY expects the RELATIVISTIC INQUISITION!) -- Wayne Throop <the-known-world>!mcnc!rti-sel!dg_rtp!throopw
ccplumb@watnot.UUCP (11/23/86)
In article <255@sri-arpa.ARPA> KFL@MX.LCS.MIT.EDU writes: >1) Motion in a straight line is relative, i.e. no experiment can show > which state of uniform motion is at rest (from which it is concluded > that the concept of "at rest" is bogus). This comes from Newtonian > physics. > >2) The speed of light in a vacuum is the same to all observers. This > can be deduced from Maxwell's equations (1861) and was confirmed > directly by Michelson and Morley in 1881. > > It can also be proven that if these axioms are true, that sending >information faster than light is not possible without reversing >causality (i.e. time travel). SF authors would do well to accept time >travel if they postulate FTL travel, unless they instead explain which >one of the two axioms turned out to be wrong despite overwhelming >evidence. A plausible case could be made that the cosmic background >noise defines an absolute frame of reference. >[...] >I *WANT* FTL travel to be possible but can't bring myself to accept >reverse causality. Of course facts are facts, and are not influenced >by our hopes or fears or misconceptions. I agree - I want FTL travel too, but physics won't let me have it... Given the choice you offer above, I'd take 1, and violate 2. Perhaps this is due to not having studied Maxwell's equations, but I have hope that someone will find a way of contorting space-time to violate it. With 10 dimensions to play with, it might be possible. Axiom #1 seems too fundamental to play with, but perhaps someone would like to propose its sacrifice. -Colin Plumb (ccplumb@watnot.UUCP) P.S. Someone once figured out that a super-dense, rapidly spinning cylinder could act as a time machine. I don't know if the proof was invalid, but this seems to offer another opening for FTL - accept reverse causality! Stephen Hawking's been giving causality a beating, anyway. Zippy says: OMNIVERSAL AWARENESS?? Oh, YEH!! First you need 4 GALLONS of JELL-O and a BIG WRENCH!!... I think you drop th'WRENCH in the JELL-O as if it was a FLAVOR, or an INGREDIENT... ...or...I...um...WHERE'S the WASHING MACHINES?
zdenek@heathcliff.columbia.edu (Zdenek Radouch) (11/24/86)
In article <706@dg_rtp.UUCP> throopw@dg_rtp.UUCP (Wayne Throop) writes: >> [somebody about twin paradox] >> My high school physics isn't letting me in on the secret... > >My college physics DOES let me in on the secret... but just how many >times should I post the same thing over and over? Well, maybe just once >more... I don't think the problem with understanding relativistic physics is as simple as you put it. To be more precise I don't think there are too many people around who DO KNOW how to use space-time diagrams and at the same time CAN'T EXPLAIN the twin paradox. My guess is that there are two big groups with the following comments: 1. Why the hell does he repeat this over ond over? 2. Why the hell does he repeat this over ond over? The first group is annoyed because they don't learn anything new, whereas the second group is annoyed because they don't learn anything new, too. Oh, I almost forgot to say that the first group understands the problem and the second doesn't. So I am not sure about your "one more explanation". zdenek P.S. Could anybody who didn't understand the twin paradox and does understand it now as a result of Wayne's article please raise hand and say I'm wrong? zdenek@cs.columbia.edu or ...!seismo!columbia!cs!zdenek