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puzzle in relativity
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Sorcerer1
science forum Guru


Joined: 09 Jun 2006
Posts: 410

PostPosted: Sun Jul 16, 2006 10:36 pm    Post subject: Re: puzzle in relativity Reply with quote

"The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
news:pan.2006.07.16.16.12.35.332064@earthlink.net...
| On Sun, 16 Jul 2006 07:38:25 +0000, Sorcerer wrote:
|
| >
| > "The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
| > news:pan.2006.07.16.01.08.27.22076@earthlink.net...
| > | On Sat, 15 Jul 2006 21:04:30 +0000, Sorcerer wrote:
| > |
| > |
| > | > "The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
| > | > news:pan.2006.07.15.20.33.27.747137@earthlink.net...
| > | > | On Tue, 11 Jul 2006 10:11:09 -0700, cafeinst wrote:
| > | > |
| > | > | > I found this here:
| > | > | >
| > http://www.geocities.com/ResearchTriangle/System/8956/problems/probs.htm
| > | > | >
| > | > | > "Five stuntmen are riding upon a large sled across an icy pond.
In
| > the
| > | > | > center of the pond is a hole exactly the same size as the sled,
| > | > | > and the stuntmen are travelling towards it at near the speed of
| > | > | > light. When they reach the hole, a man standing beside the hole
| > | > | > hits the
| > sled
| > | > | > with a sledgehammer to drive it into the hole. According to the
| > | > | > stuntmen, the hole decreases in length due to the effects of
| > | > | > special relativity, and thus they slide over and do not fall in.
| > | > | > But the man standing beside the hole sees the length of the sled
| > | > | > decrease due to relativity, and so the sled slips easily into
the
| > | > | > hole. Clearly the stuntmen or the hammer man cannot both be
right,
| > | > | > so what happens?"
| > | > |
| > | > | First, the delta-momentum for any speed where relativistic effects
| > | > | are going to be greater than the radius of a molecule is so
| > | > | ridiculously high
| > | >
| > | > Vague. What is the threshold for an effect to change from Newtonian
to
| > | > relativistic?
| > |
| > | All effects are relativistic.
| >
| > Thank you.
| >
| >
| > I did say "greater than the radius of a
| > | molecule". However, your example of moving towards a candle will
| > | definitely slow down one's wristwatch (as observed by someone next to
| > | the candle) -- although the amount of slowdown is ridiculously small
for
| > | someone moving at 3 m/s = 10^-8 c.
| >
| > He walks away from another candle simultanously with walking toward the
| > first.
| > His ridiculous wrist watch needs 6 ridiculous hands, one ridiculous
| > molecule apart,
| > although the separation of the candles is constant and not ridiculous.
|
| The separation of the candles is affected as well. The walker will
| measure, if he bothers to, a distance which will vary from the candles'
| own measurement between themselves.

No he won't, he'll use Doppler radar set in the room and aimed at the other
candle with a digital readout that he can see as he walks. Perhaps you can't
see a digital readout at 20 feet, but I can. Get a bigger digital readout.
[rest snipped]
Androcles
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The Ghost In The Machine1
science forum Guru


Joined: 25 Mar 2005
Posts: 1551

PostPosted: Mon Jul 17, 2006 2:00 am    Post subject: Re: puzzle in relativity Reply with quote

On Sun, 16 Jul 2006 22:36:56 +0000, Sorcerer wrote:

Quote:

"The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
news:pan.2006.07.16.16.12.35.332064@earthlink.net...
| On Sun, 16 Jul 2006 07:38:25 +0000, Sorcerer wrote:
|
|
| > "The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
| > news:pan.2006.07.16.01.08.27.22076@earthlink.net...
| > | On Sat, 15 Jul 2006 21:04:30 +0000, Sorcerer wrote:
| > |
| > |
| > | > "The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
| > | > news:pan.2006.07.15.20.33.27.747137@earthlink.net...
| > | > | On Tue, 11 Jul 2006 10:11:09 -0700, cafeinst wrote:
| > | > |
| > | > | > I found this here:
| > | > |
| > http://www.geocities.com/ResearchTriangle/System/8956/problems/probs.htm
| > | > |
| > | > | > "Five stuntmen are riding upon a large sled across an icy
| > | > | > pond.
In
| > the
| > | > | > center of the pond is a hole exactly the same size as the
| > | > | > sled, and the stuntmen are travelling towards it at near the
| > | > | > speed of light. When they reach the hole, a man standing
| > | > | > beside the hole hits the
| > sled
| > | > | > with a sledgehammer to drive it into the hole. According to
| > | > | > the stuntmen, the hole decreases in length due to the effects
| > | > | > of special relativity, and thus they slide over and do not
| > | > | > fall in. But the man standing beside the hole sees the length
| > | > | > of the sled decrease due to relativity, and so the sled slips
| > | > | > easily into
the
| > | > | > hole. Clearly the stuntmen or the hammer man cannot both be
right,
| > | > | > so what happens?"
| > | > |
| > | > | First, the delta-momentum for any speed where relativistic
| > | > | effects are going to be greater than the radius of a molecule is
| > | > | so ridiculously high
| > |
| > | > Vague. What is the threshold for an effect to change from
| > | > Newtonian
to
| > | > relativistic?
| > |
| > | All effects are relativistic.
|
| > Thank you.
|
|
| > I did say "greater than the radius of a
| > | molecule". However, your example of moving towards a candle will
| > | definitely slow down one's wristwatch (as observed by someone next
| > | to the candle) -- although the amount of slowdown is ridiculously
| > | small
for
| > | someone moving at 3 m/s = 10^-8 c.
|
| > He walks away from another candle simultanously with walking toward
| > the first.
| > His ridiculous wrist watch needs 6 ridiculous hands, one ridiculous
| > molecule apart,
| > although the separation of the candles is constant and not
| > ridiculous.
|
| The separation of the candles is affected as well. The walker will
| measure, if he bothers to, a distance which will vary from the candles'
| own measurement between themselves.

No he won't, he'll use Doppler radar set in the room and aimed at the
other candle with a digital readout that he can see as he walks. Perhaps
you can't see a digital readout at 20 feet, but I can. Get a bigger
digital readout. [rest snipped]

So you're postulating 1 normal readout at candle 2, 1 reversed readout at
candle 1, 1 mirror (to look at the reversed readout at candle 1 in back of
the walker [*]), and two radar guns, mounted back to back, somewhere on
the walker.

Interesting solution, but it won't help you. The guns will of course
be doing the actual measurement (the candles don't really have a good
method of determining when the guns were fired), and the readouts simply
get the signal from the gun after the round trip, as a display device.

Lessee...I'm going to have to grind this out. I'll assume the walker has
traversed 2 seconds from candle 1 *in the ballroom frame* [+]; things get
a little complicated otherwise. It's bad enough in this frame.

Also, since c = 1, 3 m = 1 * 10^-8 lightsecond. ()_B refers to the
ballroom (rest or candle) frame; ()_W to the walker's. For simplicity the
origins are placed at the point of firing.

c = 1 light-second/second [+]
v = 10^-8
g = 1/sqrt(1-v^2/c^2) = 1/sqrt(1-10^-16)
Candle 1: (-2 * 10^-8,0)_B = (-g * (2 * 10^-Cool, +g * (2 * 10^-16))_W
Candle 2: (+8 * 10^-8,0)_B = (g * (8 * 10^-Cool, -g * (8 * 10^-16))_W
Walker: (0,0)_B = (0,0)_W

The guns fire simultaneously, in the walker's frame, but the candles
won't see it that way. The candles have mirrors of their own, of course
-- to reflect the radar radio wave. In the ballroom frame, the wavefront
is dopplered slightly (redshifted for candle 1, blueshifted for 2) but
will reach the candles at the following times.

Candle 1: (-2*10^-8, 2*10^-8)_B
=(-2.00000002*10^-8*g,2.00000002*10^-8*g)_W

Candle 2: (8*10^-8, 8*10^-8)_B
= (7.99999992*10^-8*g, 7.99999992*10^-8*g)_W

So the radar guns will measure times of 4.00000004*10^-8 g
from the walker to candle 1, and 15.99999984*10^-8*g
from the walker to candle 2. Total measured lightpath distance:
g*19.99999988*10^-8 .

A bit off. [%]

It gets worse. If the walker is 5 seconds in (as determined by the
master ballroom clock), we get:

Firing:
Candle 1: (-5 * 10^-8,0)_B = (-g * (5 * 10^-Cool, +g * (5 * 10^-16))_W
Candle 2: (+5 * 10^-8,0)_B = (g * (5 * 10^-Cool, -g * (5 * 10^-16))_W

Bounceback:
Candle 1: (-5 * 10^-8,5*10^-8)_B
= (-5.00000005 * 10^-8 * g, 5.00000005 * 10^-8 * g)_W
Candle 2: (+5 * 10^-8,5*10^-8)_B
(-4.99999995 * 10^-8 * g, 4.99999995 * 10^-8 * g)_W

Total measured lightpath distance is now g*20.00000000*10^-8. Not only is
the measured distance multiplied by g (making it too long to begin with),
but it's *changing* depending on where the walker is. No doubt you think
this is totally cuckoo, and I for one don't know whether this particular
effect has been verified. I'm hoping I'm at least calculating it
correctly.

BTW, the distance is more like 98 feet. But no problem; make the display
as big as you like; many Doppler radar units are posted around school
zones around here, and can be read from a considerable distance, with an
implied hint to "PLEASE...SLOW DOWN FOR THE KIDDIES!".

Quote:
Androcles

[*] If you prefer, 2 mirrors and a normal readout at candle 1 would also
work. The 2 mirrors would be mounted at a 90 degree angle in that case.

[+] I'll assume that the ballroom floor has little clocks in it,
synchronized to the master ballroom clock. The master ballroom clock will
be light-delayed from the perspective of each of these clocks; the delay
depends on the distance, as measured in the ballroom frame, from the floor
clock to the ballroom clock. This is the best SR can do regarding
"absolute time".

[%] The lightpath distance is twice the distance between candles, as the
radar signals are roundtripping.

--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
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Edward Green
science forum addict


Joined: 21 May 2005
Posts: 95

PostPosted: Mon Jul 17, 2006 3:02 am    Post subject: Re: puzzle in relativity Reply with quote

cafeinst@msn.com wrote:

Quote:
I found this here:
http://www.geocities.com/ResearchTriangle/System/8956/problems/probs.htm

"Five stuntmen are riding upon a large sled across an icy pond. In the
center of the pond is a hole exactly the same size as the sled, and the
stuntmen are travelling towards it at near the speed of light. When
they reach the hole, a man standing beside the hole hits the sled with
a sledgehammer to drive it into the hole. According to the stuntmen,
the hole decreases in length due to the effects of special relativity,
and thus they slide over and do not fall in. But the man standing
beside the hole sees the length of the sled decrease due to relativity,
and so the sled slips easily into the hole. Clearly the stuntmen or the
hammer man cannot both be right, so what happens?"

There must be something wrong with me -- I can't get worked up over
this silly paradox. I remember what it looks like when an airplane
travelling a mere 500+ miles/hr hits a building. What's going to
happen to your stupid sled is going to make what happened to that
airplane look like a landing you could walk away from.

Of course it's important that there are five stuntmen; if there were
six stuntmen, or five stuntmen and a performing dwarf, things are going
to turn out a little differently.
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Sorcerer1
science forum Guru


Joined: 09 Jun 2006
Posts: 410

PostPosted: Mon Jul 17, 2006 7:41 am    Post subject: Re: puzzle in relativity Reply with quote

"The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
news:pan.2006.07.17.01.31.03.395068@earthlink.net...
| On Sun, 16 Jul 2006 22:36:56 +0000, Sorcerer wrote:
|
| >
| > "The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
| > news:pan.2006.07.16.16.12.35.332064@earthlink.net...
| > | On Sun, 16 Jul 2006 07:38:25 +0000, Sorcerer wrote:
| > |
| > |
| > | > "The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
| > | > news:pan.2006.07.16.01.08.27.22076@earthlink.net...
| > | > | On Sat, 15 Jul 2006 21:04:30 +0000, Sorcerer wrote:
| > | > |
| > | > |
| > | > | > "The Ghost In The Machine" <ewill3@earthlink.net> wrote in
message
| > | > | > news:pan.2006.07.15.20.33.27.747137@earthlink.net...
| > | > | > | On Tue, 11 Jul 2006 10:11:09 -0700, cafeinst wrote:
| > | > | > |
| > | > | > | > I found this here:
| > | > | > | >
| > | >
http://www.geocities.com/ResearchTriangle/System/8956/problems/probs.htm
| > | > | > | >
| > | > | > | > "Five stuntmen are riding upon a large sled across an icy
| > | > | > | > pond.
| > In
| > | > the
| > | > | > | > center of the pond is a hole exactly the same size as the
| > | > | > | > sled, and the stuntmen are travelling towards it at near the
| > | > | > | > speed of light. When they reach the hole, a man standing
| > | > | > | > beside the hole hits the
| > | > sled
| > | > | > | > with a sledgehammer to drive it into the hole. According to
| > | > | > | > the stuntmen, the hole decreases in length due to the
effects
| > | > | > | > of special relativity, and thus they slide over and do not
| > | > | > | > fall in. But the man standing beside the hole sees the
length
| > | > | > | > of the sled decrease due to relativity, and so the sled
slips
| > | > | > | > easily into
| > the
| > | > | > | > hole. Clearly the stuntmen or the hammer man cannot both be
| > right,
| > | > | > | > so what happens?"
| > | > | > |
| > | > | > | First, the delta-momentum for any speed where relativistic
| > | > | > | effects are going to be greater than the radius of a molecule
is
| > | > | > | so ridiculously high
| > | > | >
| > | > | > Vague. What is the threshold for an effect to change from
| > | > | > Newtonian
| > to
| > | > | > relativistic?
| > | > |
| > | > | All effects are relativistic.
| > | >
| > | > Thank you.
| > | >
| > | >
| > | > I did say "greater than the radius of a
| > | > | molecule". However, your example of moving towards a candle will
| > | > | definitely slow down one's wristwatch (as observed by someone next
| > | > | to the candle) -- although the amount of slowdown is ridiculously
| > | > | small
| > for
| > | > | someone moving at 3 m/s = 10^-8 c.
| > | >
| > | > He walks away from another candle simultanously with walking toward
| > | > the first.
| > | > His ridiculous wrist watch needs 6 ridiculous hands, one ridiculous
| > | > molecule apart,
| > | > although the separation of the candles is constant and not
| > | > ridiculous.
| > |
| > | The separation of the candles is affected as well. The walker will
| > | measure, if he bothers to, a distance which will vary from the
candles'
| > | own measurement between themselves.
| >
| > No he won't, he'll use Doppler radar set in the room and aimed at the
| > other candle with a digital readout that he can see as he walks. Perhaps
| > you can't see a digital readout at 20 feet, but I can. Get a bigger
| > digital readout. [rest snipped]
|
| So you're postulating 1 normal readout at candle 2, 1 reversed readout at
| candle 1, 1 mirror (to look at the reversed readout at candle 1 in back of
| the walker [*]), and two radar guns, mounted back to back, somewhere on
| the walker.
|
| Interesting solution, but it won't help you. The guns will of course
| be doing the actual measurement (the candles don't really have a good
| method of determining when the guns were fired), and the readouts simply
| get the signal from the gun after the round trip, as a display device.
|
| Lessee...I'm going to have to grind this out. I'll assume the walker has
| traversed 2 seconds from candle 1 *in the ballroom frame* [+]; things get
| a little complicated otherwise. It's bad enough in this frame.
|
| Also, since c = 1, 3 m = 1 * 10^-8 lightsecond. ()_B refers to the
| ballroom (rest or candle) frame; ()_W to the walker's. For simplicity the
| origins are placed at the point of firing.
|
| c = 1 light-second/second [+]
| v = 10^-8
| g = 1/sqrt(1-v^2/c^2) = 1/sqrt(1-10^-16)
| Candle 1: (-2 * 10^-8,0)_B = (-g * (2 * 10^-Cool, +g * (2 * 10^-16))_W
| Candle 2: (+8 * 10^-8,0)_B = (g * (8 * 10^-Cool, -g * (8 * 10^-16))_W
| Walker: (0,0)_B = (0,0)_W
|
| The guns fire simultaneously, in the walker's frame, but the candles
| won't see it that way. The candles have mirrors of their own, of course
| -- to reflect the radar radio wave. In the ballroom frame, the wavefront
| is dopplered slightly (redshifted for candle 1, blueshifted for 2) but
| will reach the candles at the following times.
|
| Candle 1: (-2*10^-8, 2*10^-8)_B
| =(-2.00000002*10^-8*g,2.00000002*10^-8*g)_W
|
| Candle 2: (8*10^-8, 8*10^-8)_B
| = (7.99999992*10^-8*g, 7.99999992*10^-8*g)_W
|
| So the radar guns will measure times of 4.00000004*10^-8 g
| from the walker to candle 1, and 15.99999984*10^-8*g
| from the walker to candle 2. Total measured lightpath distance:
| g*19.99999988*10^-8 .
|
| A bit off. [%]
|
| It gets worse. If the walker is 5 seconds in (as determined by the
| master ballroom clock), we get:
|
| Firing:
| Candle 1: (-5 * 10^-8,0)_B = (-g * (5 * 10^-Cool, +g * (5 * 10^-16))_W
| Candle 2: (+5 * 10^-8,0)_B = (g * (5 * 10^-Cool, -g * (5 * 10^-16))_W
|
| Bounceback:
| Candle 1: (-5 * 10^-8,5*10^-8)_B
| = (-5.00000005 * 10^-8 * g, 5.00000005 * 10^-8 * g)_W
| Candle 2: (+5 * 10^-8,5*10^-8)_B
| (-4.99999995 * 10^-8 * g, 4.99999995 * 10^-8 * g)_W
|
| Total measured lightpath distance is now g*20.00000000*10^-8. Not only is
| the measured distance multiplied by g (making it too long to begin with),
| but it's *changing* depending on where the walker is. No doubt you think
| this is totally cuckoo, and I for one don't know whether this particular
| effect has been verified. I'm hoping I'm at least calculating it
| correctly.
|
| BTW, the distance is more like 98 feet. But no problem; make the display
| as big as you like; many Doppler radar units are posted around school
| zones around here, and can be read from a considerable distance, with an
| implied hint to "PLEASE...SLOW DOWN FOR THE KIDDIES!".
|
| > Androcles
|
| [*] If you prefer, 2 mirrors and a normal readout at candle 1 would also
| work. The 2 mirrors would be mounted at a 90 degree angle in that case.
|
| [+] I'll assume that the ballroom floor has little clocks in it,
| synchronized to the master ballroom clock. The master ballroom clock will
| be light-delayed from the perspective of each of these clocks; the delay
| depends on the distance, as measured in the ballroom frame, from the floor
| clock to the ballroom clock. This is the best SR can do regarding
| "absolute time".
|
| [%] The lightpath distance is twice the distance between candles, as the
| radar signals are roundtripping.
|
| --
| #191, ewill3@earthlink.net
| It's still legal to go .sigless.



You continue to handwave something you call 'g' which you have yet to
derive.
It won't help you, since the derivation of g is from this very experiment.
The numbers to choose for simplicity (courtesy McCullough)
are v = 3, c = 5, room = 80.

That's an awful lot of writing to say that objectively two candles are
(and remain) a fixed distance apart, independent of the subjective
impressions of someone walking from one to the other. I didn't read it all.
It won't help you either, if Einstein wants to define time his way,
"we establish by definition that the ``time'' required by light to travel
from A to B equals the ``time'' it requires to travel from B to A"
he can, but I don't believe him and I call him a fuckin' lunatic since
it is known from the real Sagnac experiment that his thought experiment
is a load of s**t, just as aether was a load of imaginary s**t destroyed by
MMX.
Santa Claus, on the other hand, is real.

Androcles.
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