Conference noted::sf

> Say there is a space ship that passes the Earth at a velocity of .99C and an
> Earth-bound tracking station monitors it's progress to a point 1 light away
> from Earth.  How long will it take for the ship to reach that point as
> measured by the ship and the station?

The first problem is to define what you mean by "a point 1 light away from
Earth".  I assume a "light" is a "light-year".  But remember that distance is
as relative as time.  You must specify in what frame of reference the distance
of one light-year is being measured.  Let's assume the Earth is at rest in
an inertial frame of reference (it's not) and use that frame to measure the
distance.  Then let's put a marker buoy at some point one light-year away from
Earth.  Suppose at t[Earth] = t[ship] = 0, the ship passes the Earth, with
v[ship relative to Earth] = .99 c.

Then the ship reaches the buoy at t[Earth] = 1 light-year / .99 c.
Changing "light-year" to "c * years", t[Earth] = 1 c * years / .99 c =
100/99 years.  The light from this event reaches the Earth at 1 + 100/99 years,
or 199/99 years (about 2.01 years).

Changing to the frame of reference in which the ship is at rest, the buoy
is approaching the ship at .99 c.  But since we have changed frames of
reference, we must convert the distance between the buoy and the ship as well.

	d[buoy to ship in ship's frame] =

		d[buoy to ship in Earth's frame] *
		sqrt(1 - [v[difference in velocity between two frames]/c]^2)

		= 1 c * years * sqrt(1 - .99^2)
		~ .141 light-years.

Similarly,

	t[ship reaches buoy in ship's frame] =

		t[ship reaches buoy in Earth's frame] * [same factor]

		= 100/99 * years * sqrt(1-.99^2)
		~ .142 years.

Note that dividing distance to the buoy by time to reach the buoy is equal
to .99c in either frame.  (This will not be a general result; velocities are
usually more complicated vectors than the ones of this case.)


				-- edp

Yes, of course.  That makes sense.  But one last point to clear up:  if the
distance between the Earth and the bouy relative to the ship is only .141 of the
distance relative to the Earth,  so the length of a yard stick relative to the
ship is only .141 of the length of a yard stick relative to the Earth.  So the
ship observer and the Earth observer both measure a light year's worth of yards
to travel. 

(This is worse than trying to figure out the grandfather paradox!)


John B.

Re .2:

If we (on Earth) measured the distance to the buoy with an Earth-meterstick
(a metric version of a yardstick), we would measure off a light-year's worth
of meters.

If the people on the ship measured the distance to the buoy with a
ship-meterstick, they would measure off a light-year's worth of meters.

If we watched the people on the ship measure off the distance with a
ship-meterstick, we would get very dizzy.

Does this help?


				-- edp

Re .2:

I think I messed up something in .1.

If we (on Earth) measured the distance to the buoy with an Earth-meterstick
(a metric version of a yardstick), we would measure off a light-year's worth
of meters.

If the people on the ship measured the distance to the buoy with a
ship-meterstick, they would measure off .141 light-years' worth of meters,
because they see the meterstick as one meter long but see the distance to the
buoy as .141 light-years, because the buoy is moving at .99 c, but the
meterstick is at rest in their frame of reference.

Thus, they figure they will finish the voyage in .141 c * years / .99 c, or
about .142 years.

We figure they will finish the voyage in 1 c * years / .99 c, or about 1.01
years.  But those are our years.  Since we see the people on the ship moving
at .99 c, we see them aging more slowly, at about .141 the rate we are aging.
Thus, they will have aged 1.01 years * .141, or about .142 years by the time
they finish the voyage.



				-- edp

.4 gets it right.

The ship would measure much less than a light year between the Earth
and the buoy.  (.141 to be precise.)  Remember that the earth thinks the
ship has shrunk and the ship think the earth (and the rest of the cosmos)
has shrunk.  Otherwise the people on the ship would realize they had covered
a full light-year in two months and therefor that they are moving; and one
of the basic assumptions of relativity theory is that there is no way to
tell if an inertial references frame is moving.

By the by, if the earth and ship try to settle their differences about who
has actually been shrunk, they will be unable to.  The law of simultanaety
(which says there isn't any such thing) says that if the ship tries, for
example, to mark the earth at bow and stern as it flies by, the earth will
claim that the mark at the stern was made before the mark at the bow, thereby
stretching the marks out unfairly.  

	HRB

Ok you special relativity freaks, try this one on....

Observer 1 is sitting on top of a barn which has the front door open and
the back door closed. The barn is 9.5 meters long as measured in the frame
of observer 1.

Observer 2 is running toward the barn carrying a 10 meter (as measured in
his frame) pole. Observer 1 sees observer 1 coming and measures the pole
as being 9 meters long.

Question: since the barn is obviously longer (to observer 1) than the pole
carried by observer 2, it is possible the following sequence to happen:
1. observer 2 runs into the barn, 2. the front door closes, 3. the back
door opens and 4. observer 2 runs out the back. In other words, the pole
clearly fits within the barn. But there's a problem... from observer 2's
frame of reference, he's carrying a 10 meter pole and the barn is less than
9 meters long....   Resolve the paradox.

	Jerry

ps. this is a classic of special relativity and the aha that I got 20 years
ago is still ringing in my head.

What does Observer2 measure the length of the barn as?????

Nigel

Re .7:

.9 * 9.5 meters = 8.55 meters.


				-- edp

re .7 and .8, yup that's correct. Now for the hard part... Extra points to
anyone who lists the math giving the answer.

	Jerry

Since the topic is special relativity, here is one of my favorite effects:

	Consider a rectangle with vertices A, B, C, and D:

			C	A
					---> v
			D	B




			    O

	An observer at the position marked by O sees the rectangle traveling
	at a velocity whose direction is indicated by the arrow.  To this
	observer, the rectangle appears to be rotated:

			       A
			C
				 B
			  D

	Explain this effect.  Does the angle ABD appear to be a right angle?


				-- edp

Re .6:


Obviously, the two observers disagree about which happened first: 


Observer 1 (on top of the barn)        Observer 2 (carrying the pole)

1.  Front door closes                  1.  Back door opens
2.  Back door opens                    2.  Front door closes


I say obviously because I haven't the foggiest notion why...


John B.

Re .6:

Wouldn't you know it:  I wrote down all the equations and left them at home
this morning.  Oh well, here goes:

Without loss of generality, let's place the origin of observer one's inertial
frame at the west (front) door of the barn when the west (back) end of the
pole passes it (so the front end of the pole is already in the barn).  For
convenience, let's place the origin of observer two's inertial frame at the
same location.  In both frames, the x axis is directed from west to east.
The y and z axes can be ignored.

Let f = sqrt(1 - (v/c)^2).  Let l and b be the length of the pole and the
length of the barn, respectively, in observer one's frame.  Let l' and b' be
the same lengths in observer two's frame.  Thus,

	f = .9,
	l = 9 m = 10m * f = l' * f, and
	b = 9.5 m = 8.55m / f = b' / f.

There are two events we are concerned with.  The first is the west end of the
pole passing the west end of the barn.  In the first inertial frame, this
occurs when x = 0 and t = 0.  The equations to transform these to coordinates
in the second frame are:

	x' = (x - vt)/f and
	t' = (t - vx / c^2)/f.

Note that v is the velocity of the second frame with respect to the first.
Substituting 0 for x and 0 for t, we get:

	x' = (0 - 0v)/f = 0 and
	t' = (0 - 0v / c^2)/f = 0.

Thus, the west end of the pole passes the west end of the barn at the origin
of the second frame.

The second event is the east end of the pole passing the east end of the barn.
In the first inertial frame, this takes place at x = b (the location of the
east end of the barn).  At t = 0, the east end of the pole is at x = l, so it
must travel to x = b.  This takes a length of time equal to (b-l)/v, where v
is the velocity of the pole (conveniently the same as the v in the equations
above).  Thus, the second event takes place at x = b and t = (b-l)/v.
Substituting these values into the transformation equations, we get:

	x' = (b - v(b-l)/v)/f and
	t' = ((b-l)/v - vb / c^2)/f.

Simplifying the first equation yields:

	x' = (b - (b-l))/f = l/f = l'.

Multiplying the right side of the second equation by v/v yields:

	t' = v((b-l)/v - vb / c^2)/vf
	   = ((b-l) - b v^2 / c^2)/vf
	   = (b(1 - (v/c)^2) - l)/vf
	   = (b*f^2 - l)/vf
	   = (bf - l/f)/v
	   = (b'-l')/v
	   = (l'-b')/(-v).

What do these equations mean?  Note that x' = l', so the east end of the pole
passes the east end of the barn at a distance l' from the origin of the
second frame.  Since the origin of the second frame started at the event where
the west end of the pole passed the west end of the barn, the origin was at
the west end of the pole.  Since this frame travels at the same speed as the
pole, the origin remains at the west end of the pole.  Thus, the east end of
the pole is always the length of the pole away from the west end, or l'.
Thus, l' is the correct location for any event involving the east end of the
pole, such as its meeting the east end of the barn.

For the time coordinate, note that the east end of the barn is at b' at t = 0.
It must travel to l' for the second event (the barn is traveling with a
velocity of -v in this frame).  The time required for this is (l'-b')/(-v),
which is the indicated value of t'.  This value is negative, but that is not
a paradox.  The two observers do not agree on the order of events.  Note that
distance in space between the two events is greater than the distance light
can travel in the difference in time between the two events (about 1.15 meters
in the first frame), so there are no problems with cause and effect -- nothing
could travel fast enough from one event to the other to cause something to
happen at that event, in either frame.

I do not know much about general relativity, but I would like to.  Does anybody
have anything interesting to say about it?


				-- edp

Have a look at _Turtle_Geometry_.  The last chapter describes implementation
of a relativistic turtle, complete with gravitational effects.  It is the
only intuitive explanation I have ever seen for general relativity.

	HRB

re .11 and .12. Well done! The importance of Special Relativity was in its
precise description of how we measure time (sequence of events) and distance
(indirectly as a function of time). This paradox is the best illustration of
both that I know of. Extra points in .12 for showing the distance effects so
precisely. The interrelated nature of the measurement of distance and time
lead Einstein into the entire space-time concept which is one of the corners
of General Relativity. It's interesting to treat this paradox from the
equations for General Relativity (which I saw done once but can't reproduce).
In this derivation, you distort the space-time coordinates of one frame to
another and achieve the same result.

I wish I could help you with General Relativity, but I know some of the
conceptual stuff but never really delved into the math of it. And without the
math, it's all hand waving and parlor chat.

	Jerry

ps. Now I have to dust off my math and do your distorted rectangle...

I should mention that the rectangle is not drawn to scale.  Also, the illusion
might not hold up to good measurements; it is (I think) an illusion rather than
an actual rotation.

Incidentally, doesn't the Uncertainty Principle tell us a 10 meter pole can
fit in a 9.5 meter barn for a very small period of time, even though it is at
rest?


				-- edp

edp: try "GRAVITATION" by Wheeler et al.
	for the full particulars on the book, give me a call or
	ENET and I can supply the info;	it's only 1500 or
	2000 pages long. Lots of illustration (both diagrams and
	text) and an interesting split for those wanting the math
	and those who don't dare. Last time I checked it was a 
	paltry $50.00 or so (paperback). BBiigg paperback !

re #1.  The tau eqn seemed strange... {(1-v^2)/c^2} so have read
	1-(v^2/c^2). That may have had an impact on your numbers;
	I did not do a check.

Re .16:

> re #1.  The tau eqn seemed strange... {(1-v^2)/c^2} so have read
> 	1-(v^2/c^2). That may have had an impact on your numbers;
> 	I did not do a check.

I can't find what you are referring to.


				-- edp

re .16 and .17 - the error in the Lorentz Fitzgerald equation is in the
original posting, 228.0, not 228.1.  That's why you couldn't find it.

Re .18:

I did not use the equations given in .0.


				-- edp

From:	PALUMBO        27-AUG-1985 09:02  
To:	EDECK
Subj:	Special relativity


Ed,

That problem was easy, but.....

A.  What happens when you lock the back door of the barn?  Does the
    pole fit inside?

B.  And, when the back door is locked, what happens to the back end of
    the pole?

Shelley

I told her there was no back end; it was a monopole.

"All things are relative...my relatives are all things..."
(Any better answers?)

Re .20:

Neither of the events (the back door closing or the front door opening) can be
said to have occurred before the other.  Nor would it be correct to say either
the pole was entirely inside the barn or it was always partially outside the
barn.  This is because the events are too far apart in space for either to have
had any effect on the other, so there is no before-after relationship.  And
since the back and the front of the barn are too far apart, we cannot talk
about the entire barn at once; there is no "split second" in which the entire
barn was in some state.  We must consider each event separately.

Consider a river which splits into an east branch and a west branch.  The east
branch travels 10 miles before rejoining the west branch, which has traveled 15
miles at that point.  If there is a point, A, on the east branch five miles
downriver from the split and a point, B, on the west branch 12 miles downriver
from the split, which point is farther downriver?  Well, point B is 12 miles
downriver from the split, so it might be farther downriver.  But it is only
three miles upriver from the rejoining, so it is less upriver than point A.
Wouldn't being less upriver be more downriver?  The problem is that "downriver"
is not well-defined.  Similarly, in the barn-pole problem, "after" is not
well-defined; time does not flow in a single stream. 


				-- edp                          

    There is a discussion somewhere in SF about FTL travel and its physical
    interpretation, but I can't find it right now, so I'm putting this
    here, since this note is about time dialation. The following was
    written for a similar discussion in STAR_TREK.
    
    sm
    
                  <<< THEBAY::DRC0:[NOTES$LIBRARY]STAR_TREK.NOTE;1 >>>
                                 -< Star Trek >-
================================================================================
Note 40.88                        Time travel!                          88 of 88
CACHE::MARSHALL "beware the fractal dragon"          35 lines  22-AUG-1986 06:18
                           -< all done with mirrors >-
--------------------------------------------------------------------------------

    finally decided to dig out my Special Relativity book and see what
    he (A.P.French) has to say about FTL.
    
    	For suppose an event P(x,t) could cause an event Q(x+dx,t+dt)
    	through the agency of a signal having a velocity u > c. Then 
    	the time interval between the events as viewed in some other
    	frame moving with velocity v would be given by:
    
    		dt' = g(dt - v*dx/c^2)
    		    = g*dt(1- u*v/c^2)

        		       {	    /--------------/ }
    			       {where g = \/ (1 - v^2/c^2)   }
    
    	Thus if u>c, we could find a range of values of v ( < c) such
    	that dt' and dt were of opposite sign. All inertial observers
    	with velocities greater than (c^2)/u could conclude from their
    	observations that event P was caused by Q, rather than the other
    	way round.... it would make the laws of physics appear different
    	to different observers - at least as long as we have a basis
    	for knowing the direction in which time is advancing, as judged,
    	for example, by our own ageing.
    			            (SPECIAL RELATIVITY, A.P.French. p118)
                                    
    Whew. What he is saying is that suppose we have two ships, one that
    can move FTL, the other just very fast. Now send the FTL ship to
    Alpha Centauri at 2*c. Their will be some speeds for the second
    ship, where it *appears* that the other one arrives at A.C. before
    it left Earth.
    
    But I say, so what. Appearances are always deceiving. No *physical*
    paradoxes are created, only perceptual paradoxes. But magicians
    do that all the time.
    
    sm