Deriving the contraction of a moving rod
Vonny N.
I'm wondering whether it is possible in the Special Theory of
Relativity to show within the confines of a single inertial frame that
a rod which is accelerating in the direction of its length will
contract. This is different to the problem of showing that a rod will
appear contracted to one observer relative to another. That is, do I
really have to switch continuously to new inertial frames to solve this
problem?
Vonny
Dirk Van de moortel
"Vonny N." <von...@hotmail.com> wrote in message news:1158977009.1...@d34g2000cwd.googlegroups.com...
See http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html
For a sufficiently short rod, use equation at the end:
gamma(t) = sqrt( 1 + (a t/c)^2 )
where
t is coordinate time
a is proper acceleration of the rod
c is light speed
Dirk Vdm
Greg Egan
"Vonny N." <von...@hotmail.com> wrote:
> Hi,
>
> I'm wondering whether it is possible in the Special Theory of
> Relativity to show within the confines of a single inertial frame that
> a rod which is accelerating in the direction of its length will
> contract. This is different to the problem of showing that a rod will
> appear contracted to one observer relative to another. That is, do I
> really have to switch continuously to new inertial frames to solve this
> problem?
>
> Vonny
There are two questions to think about with an accelerating rod. The
first is, how *exactly* is it accelerating (there are some choices), and
the second is who *exactly* is measuring its length?
If you have a rod at rest in some inertial frame, with x the direction of
its length and time coordinate t, you have to decide whether you want to
give it what's known as "Born rigid" acceleration (after Max Born), or
some other kind. In Born rigid acceleration, you actually need to apply a
different force to each point along the rod, according to the following
system: if you want the front of the rod to experience proper
acceleration a (that is, an ant riding at the front of the rod would feel
a g-force of a), you pick the origin of your coordinates so that x=1/a
for the front of the rod. You then *must* apply forces that give proper
acceleration of 1/x to each point of the rod.
What's special about doing this? Well, as the word "rigid" suggests, it
means the ant riding the rod, and all his companions, reckon that the
distances between them remain completely unchanged.
The world line of each point on the rod with initial coordinates
(t,x)=(0,x_0) can then be described by:
(t,x) = x_0 (sinh(tau/x_0), cosh(tau/x_0))
where tau is the proper time along the world line. (If you want x in
terms of t, you can eliminate tau from this relationship to get a simple
quadratic description of the world line, which is a hyperbola.) The
derivatives with respect to tau give you the 4-velocity:
u = (cosh(tau/x_0), sinh(tau/x_0))
which has (as it must) a Lorentzian squared-length of -1, and the
4-acceleration:
A = (1/x_0) (sinh(tau/x_0), cosh(tau/x_0))
which has a squared-length of (1/x_0)^2, which shows that the proper
acceleration of this point is 1/x_0. Of course we're free to shift the
origin of our coordinates to make 1/x_0 whatever we like, but once we've
done that the varying proper acceleration for the rest of the bar is
fixed. Note also that the bar can't be longer than 1/a, where a is the
proper acceleration of the front.
Now as for any length contraction, you just have to decide who is going
to measure the length of the bar according to their own notion of
simultaneity. Any ant riding at a fixed point on the bar will think the
bar's length remains constant. Why? Their unit spatial vector,
Lorentz-orthogonal to the u derived above, is:
w = (sinh(tau/x_0), cosh(tau/x_0))
Suppose the ant who started at x_0 contemplates the world line of a point
that is displaced by some fixed distance L in the w direction; this world
line will be:
(t,x) = x_0 (sinh(tau/x_0), cosh(tau/x_0)) [ant's origin]
+ L (sinh(tau/x_0), cosh(tau/x_0))
= (x_0+L) (sinh(tau/x_0), cosh(tau/x_0))
As a curve in spacetime, this is just the world line for the ant that
started out at x=x_0+L. The only difference is the way it's described:
that ant who started at x_0+L would put an argument of tau/(x_0+L) into
the hyperbolic functions. What this means is that one ant's proper time
doesn't agree with another's, even though they measure themselves as
being at rest with respect to each other. This is related to the fact
that they experience different accelerations.
Now, if you want to know the contraction of the bar in the original
reference frame where it started out at rest, that's easy: just solve
the world lines I've given in terms of x_0 and tau to get a function
x(x_0, t), and compute the difference between the functions for two
different values of x_0.
If you don't want to do "Born rigid" acceleration, but some other kind,
you'll have to decide exactly what the accelerations along the bar are
going to be.
Hope this helps.
Tom Roberts
> [discussion of Born rigid motion during acceleration]
The thing missing from your discussion is why Born rigid motion is
useful -- it sounds highly artificial. But for an ordinary material, and
for a "small" acceleration coupled to a single point of a "small"
system, Born rigid motion is the natural result. Basically the
inter-molecular bonds attempt to keep the inter-molecule spacings
constant in their local rest frame, and will do so as long as the
inertial forces resulting from the acceleration are small compared to
the inter-molecular forces themselves. For practical rocket engines and
materials, Born rigid motion is a good approximation. For gedankens in
relativity in which speeds approaching c are achieved quickly, one must
either assume an unrealistically inelastic material, or acceleration
coupled all along its length as Greg discussed.
Tom Roberts
John (Liberty) Bell
Hmm.
I think the beauty of Born rigid motion is that the length of the rod
does remain constant in the proper reference frame of the rod. It is
thus a good approximation when the acceleration is induced
gravitationally, and the rod's length is too short for tidal effects to
be significant.
For a real rigid rod, when the acceleration is induced mechanically at
a point, I reckon the effects of acceleration should propagate through
that rod at the speed of sound in that material.
Whether this results in expansion or contraction of the rod then
depends on whether the induced acceleration is towards or away from the
rest of the rod.
At least at low velocities, the magnitude of this effect should be much
greater than special relativistic effects because the speed of sound is
much less than the speed of light.
In contrast, whether a Born rigid rod contracts or expands relative to
a given inertial observer, will depend on whether that acceleration
reduces or increases that rod's speed relative to that observer.
John Bell
Tom Roberts
> Tom Roberts wrote:
>> for an ordinary material, and
>> for a "small" acceleration coupled to a single point of a "small"
>> materials, Born rigid motion is a good approximation.
>
> a point, I reckon the effects of acceleration should propagate through
> that rod at the speed of sound in that material.
Sure. You're just exploring the accuracy of the approximation I discussed.
Estimate how much the space shuttle shrinks during launch due to the
acceleration-induced stress on its body. Born rigid motion is a pretty
good approximation for this, but it's obviously not exact.
Tom Roberts
tes...@um.bot
> There are two questions to think about with an accelerating rod. The
> first is, how *exactly* is it accelerating (there are some choices), and
> the second is who *exactly* is measuring its length?
[...]
> In Born rigid acceleration, you actually need to apply a different force
> to each point along the rod, according to the following system: if you
> want the front of the rod to experience proper acceleration a (that is,
> an ant riding at the front of the rod would feel a g-force of a), you
> pick the origin of your coordinates so that x=1/a for the front of the
> rod. You then *must* apply forces that give proper acceleration of 1/x
> to each point of the rod.
[...]
> If you don't want to do "Born rigid" acceleration, but some other kind,
> you'll have to decide exactly what the accelerations along the bar are
> going to be.
For the benefit of students, just thought I'd mention that we are
essentially talking about what is sometimes called "Bell's string
paradox", in which one envisions two accelerating spaceships with a taut
string stretched between them.
It is useful to consider several versions, to bring out more clearly
several insights which Greg alluded to above, but in one of the simplest
versions, we can consider two initially comoving and inertial spaceships
which both start to -accelerate- at time t=0 with constant acceleration k
(as measured in their own frames). To model the string, we must
interpolate the world lines of the bits of matter in the string between
these two hyperbolic arcs to form a "timelike congruence". To model
physical measurements along the string, we must augment this to a frame
field e_0, e_1, e_2, e_3. As Greg said, depending upon what assumptions
you make about the material laws obeyed by the string, this will result in
different choices of frame field.
The simplest if not the most realistic choice is probably to assume that
the string is accelerated with constant acceleration k all along its
length, so that the world lines of the bits of matter in the string all
look like hyperbolic arcs with vertical tangents at t=0 and with path
curvature k. (It might be better to speak of a "rod" rather than a
"string" here, and to imagine rocket engines attached all along the rod.)
Now that you have a frame field (exercise for the reader), you can use a
nifty tool introduced in many good gtr textbooks (the very best account is
probably in Eric Poisson, A Relativist's Toolkit, Cambridge University
Press, 2004). Namely, the expansion tensor defined by the timelike unit
vector e_0, which we compute in terms of the frame, following e.g. the
discussion of computing the Riemann tensor in coframe components in
Misner, Thorne, and Wheeler, Gravitation, Freeman, 1973. These are the
"physical components" corresponding to what observers riding on the string
(or rod) will measure. The result is that the string is expanding. So in
this "Bell acceleration scenario", eventually, the string must break!
Alternatively, we could assume that the two spaceships and all the bits of
matter on the string act like Rindler observers. Then, as you easily see
from the hyperbolic trigonmetry implicit in the pictures at Greg's site or
in MTW, the frame field has vanishing expansion, so the string has--
infinitesimally speaking-- constant length. So in this "Born rigid
acceleration scenario", the string will -not- break.
Note that in the Born rigid acceleration scenario, the -trailing- observer
is accelerating -faster-. If this isn't clear, note that the world lines
of the Rindler observers in Greg's plots form a family of nested
"concentric" hyperbolas, which is analogous to a family of nested
concentric circles. In the latter case, of course, you know that inner
circles bend faster (have larger path curvature) than outer ones. The
"paradox", I suppose, is that this is consistent (in relativistic
dynamics) with the claim that the string has vanishing expansion tensor,
whereas in Newtonian dynamics, if we push harder on the trailing end of an
elastic rod, it must be compressed.
We could also attempt to consider a scenario in which the leading ship
accelerates with constant acceleration k after time t=0 and simply tows
the second ship at the end of the string (rod). Then of course a wave of
tension must travel down the rod, and we might appeal to the relativistic
theory of elasticity, which is rather tricky (serious students can look
for a recent Ph.D. dissertation which was posted to the arXiv).
One way to understand why elastic bodies are tricky in relativity is to
notice that strictly speaking Hooke's law cannot be compatible with
Lorentz symmetry, since in Hooke's theory, all the bits of a spring
respond instantaneously when one end is pulled! Nonetheless it is quite
illuminating to consider -Newtonian- models of an accelerated elastic rod
in which we use a linear chain of springs to model the rod. This is a
fine exercise in using the Laplace transform method! It also turns out to
be closely related to the fascinating "Fermi-Ulam-Pasta phenomenon".
See the FAQ
http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html
for another discussion of Bell's "spaceships on a string paradox".
"T. Essel"
Tom Roberts
> For the benefit of students, just thought I'd mention that we are
> essentially talking about what is sometimes called "Bell's string
> paradox", in which one envisions two accelerating spaceships with a taut
> string stretched between them.
There is a much simpler way to resolve this paradox, using just
elementary SR. The level of discussion should match the level the
student is studying.
As you say, start the two rockets simultaneously from rest in an
inertial frame, and give them identical proper accelerations. Then after
some elapsed proper time have them stop accelerating so they are again
both at rest in an inertial frame. Now compute their spatial separation
in the new inertial frame, and compare it to their separation in the
original frame. You'll find they are further apart, so the string must
break if they accelerate long enough.
I'll leave the actual computation to the reader, but point out that
because of the way it is set up, the distance between the rockets
remains constant in their initial inertial frame, and they stop
accelerating simultaneously in that frame. With this observation plus a
Lorentz transform the computation can easily be done.
Tom Roberts
cma...@yahoo.com
That's only because the back of the shuttle mechanically pushes its
front. That's why you have to take account of the speed of sound
propagating through the shuttle's material structure.
As was mentioned in an earlier post, you have to envisage an
acceleration applied to all points of the structure. It's not too hard
to imagine how it can be done: add rockets to the top of the shuttle,
and put a power switch at its center. Turn on the switch, the signal
reaches both ends of the shuttle at the same epoch, and both the bottom
and top rockets ignite simultaneously. So the role of the speed of
sound is eliminated and you get a shuttle that doesn't contract. Okay
the acceleration is not yet applied to all points of the structure,
only the two endpoints; but the rest is left to imagination.
Chris
Nicolaas Vroom
<dirkvand...@ThankS-NO-SperM.hotmail.com.telenet-ops.be> schreef in
bericht news:XXeRg.92257$OV7.1...@phobos.telenet-ops.be...
This reply raises certain comments.
1. The url shows a sketch based on the coordinates
x horinzontal and t vertical,
however the sketch shows the position x(T) of an
observer i.e. as a function of proper time T.
My first reaction is that the vertical line should show T,
but ofcourse there are also different solutions.
2. The text shows 2 equations.
a) dV(T) = a(T) dT
b) dv(T)/dT = a(T) [ 1 - v^2(T) ]
if you combine those 2 you get:
a(T)dT/dT = a(T) [ 1 - v^2(T) ] or
1 = [ 1 - v^2(T) ]
which does not seem correct.
3. The above mentioned question in short is:
Do I really have to switch continuously to new inertial frames
to show length contraction when accelerations are involved.
IMO this is the same problem as:
Do I really have to switch continuously to new inertial frames
to show length contraction when v is variable.
IMO the answer is no.
When v is constant
gamma = 1 / sqrt(1- v^2/c^2)
When v is variable
gamma(t) = 1 / sqrt(1- v(t)^2/c^2)
I have made a sketch
http://users.pandora.be/nicvroom/dirk4.jpg
to show a simple example based on t.
(T is not used)
That means you do not have to calculate a(t) exclusive.
in order to show length contraction when accelerations
are involved.
Anyway the best you can do is to state that
a(t) = v(t)/dt
or that v(t) = int a(t)*dt over 0 to t
but that is all
Nicolaas Vroom
Dirk Van de moortel
"Nicolaas Vroom" <nicolaa...@pandora.be> wrote in message news:JwmZg.135706$p85.2...@phobos.telenet-ops.be...
You can also look at the position as a set of 'points'
{ ( x(T), t(T) ) where T in some interval } .
If you like, eliminating T, you can translate this to a set
{ ( x(t), t ) where t in some interval }
This is elementary analytic geometry.
On the drawing I used the proper time T as labels for the
worldline, because in the frame of the accelerating observer,
the *proper* acceleration a(T) is given as a function of
*proper* time. The idea of the exercise is to find relations
between this proper time T and proper acceleration a as
functions of coordinate time t, coordinate position x, and
coordinate velocity v.
>
> 2. The text shows 2 equations.
> a) dV(T) = a(T) dT
meaning, I quote:
"At each time T, observer A can write the amount with which
his velocity has changed during a small (infinitesimal) proper time
interval dT during which the acceleration does not change, as
dV(T) = a(T) dT".
This is the increase of the velocity as seen in the ICRF.
I carefully used *uppercase* V for this, to avoid that the reader
would do exactly what you have done below.
I had thought about using another symbol delta instead of
the d for differential, but I assumed the reader would understand.
You can ignore the symbolic form of the LHS of the equation.
Just look at the RHS and at the physical meaning of it. *That* is
what will be used a few lines later, when the standard relativistic
velocity composition is used to express the increased velocity as
seen in the inertial frame.
> b) dv(T)/dT = a(T) [ 1 - v^2(T) ]
>
> if you combine those 2 you get:
> a(T)dT/dT = a(T) [ 1 - v^2(T) ] or
> 1 = [ 1 - v^2(T) ]
> which does not seem correct.
Of course it's not correct.
Dirk Vdm