Length contraction merry-go-round

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Scott H - 15 Jul 2008 01:13 GMT

From The Principle of Relativity, p. 116:

"We suppose that the circumference and diameter of this circle have been
measured with a unit measure infinitely small compared with the radius, and
that we have the quotient of the two results. If this experiment were
performed with a measuring-rod applied at rest relatively to the Galilean
system K, the quotient would be pi. With a measuring-rod at rest relatively
to K' [in uniform rotation], the quotient would be greater than pi."

I'm having a bit of trouble understanding this.

Suppose that in K, I measure the circumference of a fast merry-go-round to
be 10 rod-lengths, and suppose that in K' I get a larger number due to
length contraction: 15 rod-lengths. What would happen if I marked off 10
intervals with a marker? Shouldn't the length between one mark and the next
contract like a rod? How could that happen without new marks appearing?

How can the same not be said of any measuring-rod arbitrarily small?

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Androcles - 15 Jul 2008 02:13 GMT

| From The Principle of Relativity, p. 116:
|
[quoted text clipped - 14 lines]
|
| How can the same not be said of any measuring-rod arbitrarily small?

Well, see, all you need do is suppose.
"We suppose that the circumference and diameter of this circle..."
We suppose that Einstein was sane...
Suppose that in K, I measure the circumference of a fast merry-go-round to
be 10 rod-lengths...
Suppose Santa Claus comes down chimneys. How come his red
suit isn't black with soot?
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?

Hint: He was f.cking crazy.

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Spaceman - 15 Jul 2008 02:51 GMT

> From The Principle of Relativity, p. 116:
>
[quoted text clipped - 7 lines]
>
> I'm having a bit of trouble understanding this.

That is because most of relativity is a bunch of bullshit.

Signature

James M Driscoll Jr
Spaceman

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Sue... - 15 Jul 2008 06:51 GMT

> From The Principle of Relativity, p. 116:
>
[quoted text clipped - 14 lines]
>
> How can the same not be said of any measuring-rod arbitrarily small?

What does "uniform motion" and "uniform rotation" mean to you?

If light does not move like a massive particle, can you
justify the application of length contraction to
spatial displacements?

See:

"Relativity Principle"
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html

"Time Dependent Maxwell's Equations"
http://farside.ph.utexas.edu/teaching/em/lectures/node50.html

Sue...

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Jim Black - 15 Jul 2008 07:37 GMT

> From The Principle of Relativity, p. 116:
>
[quoted text clipped - 12 lines]
> intervals with a marker? Shouldn't the length between one mark and the next
> contract like a rod? How could that happen without new marks appearing?

Do you mean what would happen if you started with the merry-go-round at
rest, marked it, and then started it turning? If so, the merry-go-round
will distort or break, no matter how strong it is, from geometrical
considerations alone.

> How can the same not be said of any measuring-rod arbitrarily small?

Not sure what you mean here.

Signature

Jim E. Black (domain in headers)
How to filter out stupid arguments in 40tude Dialog:
!markread,ignore From "Name" +"<email address>"
[X] Watch/Ignore works on subthreads

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Scott H - 15 Jul 2008 12:57 GMT

>> Suppose that in K, I measure the circumference of a fast
>> merry-go-round to be 10 rod-lengths, and suppose that in K' I get a
[quoted text clipped - 7 lines]
> merry-go-round will distort or break, no matter how strong it is,
> from geometrical considerations alone.

Not just a merry-go-round. Any circle measured from a rotating frame of
reference.

You could mark it at rest and then run it for length contraction, or mark it
while moving and then slow it down. How does the circle maintain the same
number of marks?

For a line segment, it's easy: the line segment contracts along with the
intervals. Which way does a rotating circle contract?

>> How can the same not be said of any measuring-rod arbitrarily small?
>
> Not sure what you mean here.

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harry - 15 Jul 2008 15:09 GMT

>>> Suppose that in K, I measure the circumference of a fast
>>> merry-go-round to be 10 rod-lengths, and suppose that in K' I get a
[quoted text clipped - 7 lines]
>> merry-go-round will distort or break, no matter how strong it is,
>> from geometrical considerations alone.

No, neither geometry nor drawing lines can destroy a merry-go-round. In
fact, drawing lines to destroy something is witchcraft. ;-)

> Not just a merry-go-round. Any circle measured from a rotating frame of
> reference.

It doesn't really matter from where you do the measurement. The only thing
that matters is if the rod rotates with the merry-go-round or not (although
a rotating rod can be said to "belong" to the "rotating frame"). If it does
rotate with it, it is length contracted.
Harald

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Scott H - 15 Jul 2008 17:00 GMT

> It doesn't really matter from where you do the measurement. The only
> thing that matters is if the rod rotates with the merry-go-round or
> not (although a rotating rod can be said to "belong" to the "rotating
> frame"). If it does rotate with it, it is length contracted.

I still don't get it. If you line up measuring rods to rotate along with the
circle, why wouldn't the circumference itself contract along with the
measuring rods?

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Spaceman - 15 Jul 2008 17:04 GMT

>> It doesn't really matter from where you do the measurement. The only
>> thing that matters is if the rod rotates with the merry-go-round or
[quoted text clipped - 4 lines]
> with the circle, why wouldn't the circumference itself contract along
> with the measuring rods?

It would if the contraction really occured at all.
(also the diameter of the circle would change too.)
but of course.. in reality.
Length contraction is an optical illusion based non reality.
Speed (or velocity) does not change the physical length of anything.
They needed such bullshit to explain a clock malfunction mathematically
instead of finding the "physical" cause of the clock malfunction itself.
It is known as the rubber ruler kingdom of SR and the kindgom is
full of Einstein dingleberries.

Signature

James M Driscoll Jr
Spaceman

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harry - 15 Jul 2008 18:38 GMT

>> It doesn't really matter from where you do the measurement. The only
>> thing that matters is if the rod rotates with the merry-go-round or
[quoted text clipped - 4 lines]
> the circle, why wouldn't the circumference itself contract along with the
> measuring rods?

It does in principle - somewhat. Because the length contraction is
anisotropic, the disk will not freely contract and build up stress. That is,
when ignoring centrifugal force which actually expands the disk much more
than length contraction shrinks it. But even if you imagine a wheel with a
flexible middle part that freely contracts, the point is that the rod will
not length contract when measuring the radius but it will length contract
when measuring the circumference. If the circumference shrinks, so does the
radius and the true ratio is always pi. Thus, it doesn't matter what
circumference you measure, your moving radial length unit will differ from
the tangential length unit (eventhough nominally the same) and thus you
won't end up with pi as ratio.

Cheers,
Harald

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harry - 15 Jul 2008 08:33 GMT

> From The Principle of Relativity, p. 116:
>
> "We suppose that the circumference and diameter of this circle have been
> measured with a unit measure infinitely small compared with the radius,
> and that we have the quotient of the two results.

Note: traditionally this is done with a cord.

> If this experiment were performed with a measuring-rod applied at rest
> relatively to the Galilean system K, the quotient would be pi. With a
> measuring-rod at rest relatively to K' [in uniform rotation], the quotient
> would be greater than pi."
>
> I'm having a bit of trouble understanding this.

A cord would also be affected by inertial (centrifugal) forces, but an
infinitely short rod would be merely affected by length contraction. Still,
it may be easier to imagine a cord and ignore inertial forces. A shrunk cord
of originally unit length will measure more than pi for the circumference.

> Suppose that in K, I measure the circumference of a fast merry-go-round to
> be 10 rod-lengths, and suppose that in K' I get a larger number due to
> length contraction: 15 rod-lengths. What would happen if I marked off 10
> intervals with a marker?

Then you would not be full circle...

> Shouldn't the length between one mark and the next contract like a rod?

??? It's a semi-static situation. Imagine that the rod (and also the
merry-go-round somewhat) is shrunk due to a freezing cold. Then ask yourself
the same question!

> How could that happen without new marks appearing?
>
> How can the same not be said of any measuring-rod arbitrarily small?

It can't be said of any measuring-rod.

Regards,
Harald

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Natural Science Forum / Physics / General Physics / July 2008

Length contraction merry-go-round