How can this work in relativity?

**russell@mdli.com** · #1 August 4th 05 posted to sci.physics.relativity

This entire thread seems to be based on an error in
calculation by dseppala. It is corrected below.

wrote:
On Fri, 22 Jul 2005 04:00:03 GMT, The Ghost In The Machine
wrote:

In sci.physics.relativity,

wrote
on Thu, 21 Jul 2005 16:30:18 GMT
:
I don't see how Einstein's length contraction notion works with this
simple gedanken experiment.

Very simply, I have two identical rods oriented along the x-axis. One
rod I'll call rod A and the other rod I'll call rod B. The two rods
have a relative velocity V along the x-axis.

Well, now they're not identical, as they're in different states.
However, that's an extremely minor problem.

As measured in their respective frames, each rod has a length L.
In Frame A (rod A's rest frame), observers according to Einstein
measure rod B to be shorter than rod A.

Yes, with varying amounts of shortness depending on precisely
how the measurement is done.

If one plays "wait for the endpoint" one gets a length-equivalent
of L/g, where g = 1/sqrt(1-v^2/c^2).

If one plays "pingy ping ping" one gets something else; I'd have
to recalculate it.

In Frame B, observers measure rod A to be shorter than rod B.

Symmetry more or less requires this.

Now here's the problem.
Let the relative velocity of the two rods be 3 meters / second.

So v = 1e-8 * c, and so gamma = 1 + 0.5e-16 to very good
approximation.

Let the length of the rods be very long say about 10**8 light-seconds.

Let's say it is exactly 1e8 light-sec.

Erm...you do realize that's a rod stretching about 74% of the way
from Earth to Alpha Centauri, I hope. 10^8 seconds = 3.169 years.
Yes I realize its a long length. I use the long length because its
much easier for me to relate to very low velocities like 3 meters /
second, and a time interval of one second because we can easily
measure those values in our everyday experience. It much harder for
me to relate to velocities approaching c and times that happen in less
than a blink of the eye when I'm trying to understand the real
physical properties of rods.

Now accelerate all points of rod A at the same time (as measured in
Frame A) to 3 meters / second such that after the acceleration rod A
has zero relative velocity wrt rod B.

And you were going to do this precisely how? The rod, in this
case, has the usual torsional, flexional, and compressional
problems. Also, the endpoints of the rod span a distance
that would take 3.169 years for a signal to cross; even if one
locates the control in the midpoint the ends won't even receive
the signal until 1.584 years have passed.
In this gedanken experiment, I would have a conveyer belt moving at 3
meters / second relative to the very long rod. I would have a long
line of people (yes, I know its a large number) holding that rod.
Then at time t0 according to their synchronized clocks, I would have
them each step on to the conveyer belt while holding the rod. These
people and the rod will accelerate to 3 meters per second over a very
short interval (I've done this experiment with a much shorter rod - a
couple of meters in length). Now prior to stepping onto the conveyer
belt, I note that there is a second rod that is on the conveyer belt
that has zero relative velocity wrt the conveyer belt. I note that it
is composed of the same number of atoms, and is the same diameter as
the rod my long line of people are holding. At time t0, when each
person in the long line of people step on to the conveyer belt, the
person at one end point of the rod steps onto the conveyer belt at the
same position as one end point of the rod that was already on the
conveyer belt happens to be. The person on the other end point must
therefore step at a position that is 3 meters beyond the other end
point of the rod that is already on the conveyer belt when he steps on
to the conveyer belt at time t0 (Einstein's equations).

No, the extra distance is about (gamma - 1)*L = .5e-16 * 1e8
= .5e-8
in light-sec, or 1.5m.

Now this
accelerated rod, does not remain stretched. Over time, the
accelerated rod and the non-accelerated rod will have identical
lengths since they have zero relative velocity and have the same
diameter, and have the same number of atoms, and no permanent
deformation of the rod occurs during acceleration (because the
acceleration rate is so low). So the rod will eventually contract
by three meters. The problem is that I cannot figure out how this
process is viewed by observers who are in the conveyer belt frame.
They measure the rod held by the long line of people to initally be 3
meters shorter than the rod in their frame.

1.5m

When the people step on
to the conveyer belt at time t0, the observers in the conveyer frame
observe that the people at one end of the rod start stepping on to the
conveyer belt one second before the people at the other end.

Yes, 1 sec is correct here. But you see, everything works
out now if you correct your earlier error.

But this
would make the end points of each rod coincident when each person
steps on to the conveyer belt (a stretching of 3 meters).

No, the physical stretch is 1.5m. And since it starts
out 1.5m *shorter* (in the belt frame) than its proper
length, the end must move 3m to give the required stretch.
In other words, exactly 1 sec as we calculated.

I think I can snip the rest.

**dseppala@austin.rr.com** · #2 August 4th 05 posted to sci.physics.relativity

On 3 Aug 2005 17:01:50 -0700, wrote:

This entire thread seems to be based on an error in
calculation by dseppala. It is corrected below.

wrote:
On Fri, 22 Jul 2005 04:00:03 GMT, The Ghost In The Machine
wrote:

In sci.physics.relativity,

wrote
on Thu, 21 Jul 2005 16:30:18 GMT
:
I don't see how Einstein's length contraction notion works with this
simple gedanken experiment.

Very simply, I have two identical rods oriented along the x-axis. One
rod I'll call rod A and the other rod I'll call rod B. The two rods
have a relative velocity V along the x-axis.

Well, now they're not identical, as they're in different states.
However, that's an extremely minor problem.

As measured in their respective frames, each rod has a length L.
In Frame A (rod A's rest frame), observers according to Einstein
measure rod B to be shorter than rod A.

Yes, with varying amounts of shortness depending on precisely
how the measurement is done.

If one plays "wait for the endpoint" one gets a length-equivalent
of L/g, where g = 1/sqrt(1-v^2/c^2).

If one plays "pingy ping ping" one gets something else; I'd have
to recalculate it.

In Frame B, observers measure rod A to be shorter than rod B.

Symmetry more or less requires this.

Now here's the problem.
Let the relative velocity of the two rods be 3 meters / second.

So v = 1e-8 * c, and so gamma = 1 + 0.5e-16 to very good
approximation.

Let the length of the rods be very long say about 10**8 light-seconds.

Let's say it is exactly 1e8 light-sec.

Erm...you do realize that's a rod stretching about 74% of the way
from Earth to Alpha Centauri, I hope. 10^8 seconds = 3.169 years.
Yes I realize its a long length. I use the long length because its
much easier for me to relate to very low velocities like 3 meters /
second, and a time interval of one second because we can easily
measure those values in our everyday experience. It much harder for
me to relate to velocities approaching c and times that happen in less
than a blink of the eye when I'm trying to understand the real
physical properties of rods.

Now accelerate all points of rod A at the same time (as measured in
Frame A) to 3 meters / second such that after the acceleration rod A
has zero relative velocity wrt rod B.

And you were going to do this precisely how? The rod, in this
case, has the usual torsional, flexional, and compressional
problems. Also, the endpoints of the rod span a distance
that would take 3.169 years for a signal to cross; even if one
locates the control in the midpoint the ends won't even receive
the signal until 1.584 years have passed.
In this gedanken experiment, I would have a conveyer belt moving at 3
meters / second relative to the very long rod. I would have a long
line of people (yes, I know its a large number) holding that rod.
Then at time t0 according to their synchronized clocks, I would have
them each step on to the conveyer belt while holding the rod. These
people and the rod will accelerate to 3 meters per second over a very
short interval (I've done this experiment with a much shorter rod - a
couple of meters in length). Now prior to stepping onto the conveyer
belt, I note that there is a second rod that is on the conveyer belt
that has zero relative velocity wrt the conveyer belt. I note that it
is composed of the same number of atoms, and is the same diameter as
the rod my long line of people are holding. At time t0, when each
person in the long line of people step on to the conveyer belt, the
person at one end point of the rod steps onto the conveyer belt at the
same position as one end point of the rod that was already on the
conveyer belt happens to be. The person on the other end point must
therefore step at a position that is 3 meters beyond the other end
point of the rod that is already on the conveyer belt when he steps on
to the conveyer belt at time t0 (Einstein's equations).

No, the extra distance is about (gamma - 1)*L = .5e-16 * 1e8
= .5e-8
in light-sec, or 1.5m.

Now this
accelerated rod, does not remain stretched. Over time, the
accelerated rod and the non-accelerated rod will have identical
lengths since they have zero relative velocity and have the same
diameter, and have the same number of atoms, and no permanent
deformation of the rod occurs during acceleration (because the
acceleration rate is so low). So the rod will eventually contract
by three meters. The problem is that I cannot figure out how this
process is viewed by observers who are in the conveyer belt frame.
They measure the rod held by the long line of people to initally be 3
meters shorter than the rod in their frame.

1.5m

When the people step on
to the conveyer belt at time t0, the observers in the conveyer frame
observe that the people at one end of the rod start stepping on to the
conveyer belt one second before the people at the other end.

Yes, 1 sec is correct here. But you see, everything works
out now if you correct your earlier error.

But this
would make the end points of each rod coincident when each person
steps on to the conveyer belt (a stretching of 3 meters).

No, the physical stretch is 1.5m. And since it starts
out 1.5m *shorter* (in the belt frame) than its proper
length, the end must move 3m to give the required stretch.
In other words, exactly 1 sec as we calculated.

I think I can snip the rest.
Thanks for the reply. I found a high precision calculator which also
gives the result 0.5 * 10**-8 light-sec. So the problem I had was
due to the calculator I used. Thank you for correcting the
arithmetic.
David

**russell@mdli.com** · #3 August 4th 05 posted to sci.physics.relativity

wrote:

[snip]

Thanks for the reply. I found a high precision calculator which also
gives the result 0.5 * 10**-8 light-sec. So the problem I had was
due to the calculator I used. Thank you for correcting the
arithmetic.

Or you could have used a simple taylor expansion.
If we define
u = v^2/c^2
We have by the Chain Rule:
dg/du = (1/2)(1-u)^(-3/2)
which evaluates to 1/2 at u=0. So the first-order
expansion in u is
g = 1 + (1/2)u
i.e.
g = 1 + (1/2)v^2/c^2

This approximation is excellent for speeds up to,
oh, a million m/s or so. If you insist on working
examples at familiar velocities, you might as well
commit this formula to memory as your "practical"
definition of gamma.

Work smarter, not harder. Please.

Thread Tools
Show Printable Version Email this Page
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
How Does This Work?	Roy L. Fuchs	Physics - General Discussion	6	February 21st 06 03:30 PM
How can this work in relativity?	dseppala@austin.rr.com	The Theory of Relativity	59	August 14th 05 02:08 PM
How does this work, then?	Androcles	The Theory of Relativity	3	June 5th 04 08:27 PM
Why won't this work??	Timtro	Physics - General Discussion	0	November 9th 03 10:52 AM