- #1

member 728827

- TL;DR Summary
- Doing a small change in the statement of the Bell's paradox - to consider engines with constant and same F/M ratio (F is the propulsion force), instead of a constant and same proper acceleration, I think that string does not break (is my own conclusion, not being an expert in SR).

Thinking about Bell's string paradox , I understand that if both spaceships maintain the same and constant proper acceleration, the string breaks because of the non-simultaneity of the acceleration effects at both ends of the string.

But I want to introduce and consider a slight and subtle change in the statement of the original Bell’s paradox, as follows:

Where the statement of the Paradox reads that both spaceships keep the same and constant proper acceleration, I consider instead that the spaceship engines keep a constant and same F/M ratio (where F is the propulsion force). So, it's no longer the Bell's string Paradox, although at the first glance seems the same thing.

My experience is that people don't usually get the point of what this change means, and argue that it's just the Bell's scenario, so string breaks, and there's nothing to discuss about. But, it's not the Bell's scenario.

What follows is the conclusions I get, based in my limited SR background during University studies, and some online courses about the subject now that I'm retired. Don't take it too seriously.

Lets consider that both spaceships depart from their initial parking position in space, and after 1 minute proper time at leading spaceship S

At 1 hour proper time at leading spaceship S

@1 hour:

v

L

v

L

where [cm] is the middle point between spaceships, v

Being the v

##(\alpha_{cm}-\alpha_{2})*M=\frac{E \cdot A \cdot (2 \cdot \delta)}{L_{1cm}+L_{2cm}}##

where E is the Young’s module, A is the cross section area of the string, and L

Note that α

As for the theory of constant proper acceleration, I know about the Born rigidity condition. So, for two accelerating bodies S

##\alpha_2=\frac{c^2*\alpha_{cm}}{(c^2+\alpha_{cm}*D)}##

So, I predict that the [cm] observer, due to the action of the string will observe the relative speed to S

But I don’t know yet if the system can get to a static equilibrium condition.

Then, I need to consider the v

Ok, the “fictitious” relativistic force is the reason of the drifting away of the spaceships, and its as “fictitious” as it can be a “Coriolis” force due to a rotating observer’s frame. I don’t know exactly how the relativistic relative acceleration will behave during the transient period, but for [cm] observer starts as the known measured α

##\frac{1}{2} \cdot M \cdot v_{2}^2 + rw = \frac{\frac{1}{2} \cdot E \cdot A}{L_{1cm}+L_{2cm}} \cdot \delta^2##

With the values considered, I get the following results:

α

string elongation δ = 3.425e-04 m

work done by relativistic force = 2.725e-07 J

The relativistic force work done is positive, and then the static equilibrium scenario is possible. Should that value resulted negative, then static equilibrium would not be reachable, and I guess that the spaceships bounce, and the solution is then a periodic oscillation.

And, by the way, the string doesn’t break.

This first setup most resembles the Bell's scenario, in which the string is always fixed at both ends. I could also do the same, with the string fixed at both ends from the start, but I would arrive to the same conclusion: If instead of considering that the proper acceleration is always constant, I consider that is the F/M ratio of the engines what is kept constant - so proper acceleration can freely change -, then the string doesn't break.

The change in proper acceleration is really tiny, and would be very difficult even to measure it.

If instead of 1 hour, we wait 1 Year (or 6 months, or 3 months…) proper time as measured in S

@1 Year:

v

L

v

L

Being the numbers so small, the initial relative speed is only 0.2 mm/s @1 Year, the string can withstand the tension with no problem, and I guess that the solution would be probably an oscillation of both spaceships, bouncing but not breaking the string, and with a slack string most of the time.

So, if this is the case, not only the string doesn't break, but for the most part of the time, is slack, with no tension.

For the curious minds, here the Python code I made. Remark that I had to use a multi-precision numerical package, to get the required accuracy:

Some details about the scenario I considered for the Python code, and the theory behind.

The setup.-

For the journey to be comfortable for the crew, the constant propulsion force of the engine is calculated to give a 9.8 m/s

The stationary reference frame (“Earth frame” so to speak) located in outer space is where the spaceships are initially parked. For sake of simplifying the math, with reasons that I’ll explain later, I consider that the x-coordinate origin of such stationary frame is located ##\frac{c^2}{\alpha}## meters away from the parking spot, where α is the nominal proper acceleration, and c is the speed of light.

Then, the conceptual middle point between the spaceships, call it [cm], is at position ##x_{cm}=\frac{c^2}{\alpha}##, the trailing spaceship S

The spaceships have a mass of 10

The string is a steel cable. As to do not disturb the performance of the leading spaceship, that has to pull the string, I decided to use a very light one, whose mechanical characteristics are as follows:

Cable diameter: 1 mm, Area section: 0.6 mm

Young modulus: 130,000 N/mm

Break tension: 840 N,

Lineal density: 0,005 Kg/m

This seems odd for 10

In fact, during the journey – after settling down the spaceships at departure -, the tension of the string at the hook point in trailing spaceship S

Of course, there’s a tension in the string at S

Nonetheless, the initial distance between the spaceships when they reach the nominal acceleration is adjusted to 10,000 meters (the string is a little elongated), and with zero tension in hook point S

Theory used.-

With a constant F/M ratio from the engines, the spaceships follow a constant proper acceleration worldline in hyperbolic motion, and without any string interference they behave as in Bell’s scenario. So far, so good.

In the space-time diagram, the worldlines can be described by the following coordinate equations, considering the trailing spaceship S

##ct_1=\frac{c^2}{\alpha}\cdot\sinh(\frac{\alpha\cdot\tau_1}{c}) \quad x_1=\frac{c^2}{\alpha}\cdot\cosh(\frac{\alpha\cdot\tau_1}{c})-\frac{D_0}{2}, \quad S_1## starts at ##(0,\frac{c^2}{\alpha}-\frac{D_0}{2})##

##ct_{cm}=\frac{c^2}{\alpha}\cdot\sinh(\frac{\alpha\cdot\tau_{cm}}{c}) \quad x_{cm}=\frac{c^2}{\alpha}\cdot\cosh(\frac{\alpha\cdot\tau_{cm}}{c}), \quad S_{cm}## starts at ##(0,\frac{c^2}{\alpha})##

##ct_2=\frac{c^2}{\alpha}\cdot\sinh(\frac{\alpha\cdot\tau_2}{c}) \quad x_2=\frac{c^2}{\alpha}\cdot\cosh(\frac{\alpha\cdot\tau_2}{c})+\frac{D_0}{2}, \quad S_2## starts at ##(0,\frac{c^2}{\alpha}+\frac{D_0}{2})##

We’re free to choose the position of x=0 in space, and we can always set it where appropriate for the problem studied. In this case, if the worldline of the [cm] conceptual observer starts at (0,c2/α), then for any event-point of that worldline, the corresponding line of simultaneous events is the line that connects the (0,0) origin with the event-point, and that makes the math much easier.

To prove that, we do the derivative of (ct,x) parametric worldline with respect to proper time ##\tau##:

##[c\cdot\cosh(\frac{\alpha\cdot\tau_{cm}}{c}),c\cdot\sinh(\frac{\alpha\cdot\tau_{cm}}{c})]##

This vector is tangent to the worldline event-points, and so it represents the ct' axis for the accelerating spaceship. The x' Minkowski-orthogonal axis is the reciprocal, and then:

##[c\cdot\sinh(\frac{\alpha\cdot\tau_{cm}}{c}),c\cdot\cosh(\frac{\alpha\cdot\tau_{cm}}{c})]##

And so, a line that passes by the (0,0) origin and connects with the event-point in the worldline, is just the x' axis/line of simultaneous events for that event-point. This is a known property of the Rindler horizon, and at any proper time, the (0,0)-event is always simultaneous with the proper time in the spaceship.

But I want to introduce and consider a slight and subtle change in the statement of the original Bell’s paradox, as follows:

Where the statement of the Paradox reads that both spaceships keep the same and constant proper acceleration, I consider instead that the spaceship engines keep a constant and same F/M ratio (where F is the propulsion force). So, it's no longer the Bell's string Paradox, although at the first glance seems the same thing.

My experience is that people don't usually get the point of what this change means, and argue that it's just the Bell's scenario, so string breaks, and there's nothing to discuss about. But, it's not the Bell's scenario.

What follows is the conclusions I get, based in my limited SR background during University studies, and some online courses about the subject now that I'm retired. Don't take it too seriously.

Lets consider that both spaceships depart from their initial parking position in space, and after 1 minute proper time at leading spaceship S

_{2}, both spaceships have reached the nominal 1g=9.8 m/s^{2}acceleration, and the engines are set to their nominal F/M ratio, that is kept constant and will not change thereafter.At 1 hour proper time at leading spaceship S

_{2}, the string, that until now was allowed to extend - but not to runaway - from S_{2}to S_{1}keeping a zero tension at hook point in S_{1}, is progressively stopped from further deployment. Considering an initial proper cable length of 10 Km at that proper time (according to S_{2}), I get the following results using the constant proper acceleration coordinate equations, and considering the line of simultaneous events for the observers:@1 hour:

v

_{1cm}=1.9235e-08 m/s,L

_{1cm}=5,000.000035 m,v

_{2cm}=1.9235e-08 m/s,L

_{2cm}=5,000.000035 mwhere [cm] is the middle point between spaceships, v

_{1cm}and v_{2cm}are the respective relative speeds to the [cm] and L_{1cm}+L_{2cm}is the total length of the deployed string, from spaceship to spaceship.Being the v

_{2cm}so small @1 hour when we fix the string to S_{2}, let's see if it's possible to end with a static equilibrium condition. From now on, I'll be always referring to the [cm] accelerated frame observer. I’ll only analyze the leading S_{2}spaceship, and consider that the trailing S_{1}spaceship acts symmetrically, counterbalancing the string tension. For the static equilibrium, we need Newton's 2nd law:##(\alpha_{cm}-\alpha_{2})*M=\frac{E \cdot A \cdot (2 \cdot \delta)}{L_{1cm}+L_{2cm}}##

where E is the Young’s module, A is the cross section area of the string, and L

_{1cm}and L_{2cm}are the proper distances from spaceships S_{1}and S_{2}to [cm]. All measures are as observed by [cm] in its line of simultaneous events at given proper time of 1 hour.Note that α

_{cm}is the proper acceleration of the reference frame [cm] we’re considering – which is always undisturbed by any action of the string –, and α_{2}is the final proper acceleration of S_{2}. Perhaps α_{2}should be divided by a ##\gamma^3(v_{2cm})## Lorentz factor?, but would be pretty close to 1 for any practical purpose.As for the theory of constant proper acceleration, I know about the Born rigidity condition. So, for two accelerating bodies S

_{cm}with α_{cm}and S_{2}with α_{2}to keep a constant proper distance D, α_{2}needs to have the following value:##\alpha_2=\frac{c^2*\alpha_{cm}}{(c^2+\alpha_{cm}*D)}##

So, I predict that the [cm] observer, due to the action of the string will observe the relative speed to S

_{2}decrease, until this speed vanishes, moment in which S_{2}has to be in the Born rigid motion condition respect to [cm], and so thereafter will keep a constant proper distance from [cm], and therefore also from S_{1}because S_{1}acts symmetrically.But I don’t know yet if the system can get to a static equilibrium condition.

Then, I need to consider the v

_{2cm}relative speed initial condition. From the energy-wise perspective, the string has to “take up” as elastic potential energy, the initial relative kinetic energy of the S_{2}spaceship with respect to [cm], plus the work done by the “fictitious” relativistic force acting over the spaceship’s mass along the length of the elongation δ of the string.Ok, the “fictitious” relativistic force is the reason of the drifting away of the spaceships, and its as “fictitious” as it can be a “Coriolis” force due to a rotating observer’s frame. I don’t know exactly how the relativistic relative acceleration will behave during the transient period, but for [cm] observer starts as the known measured α

_{2cm}, and ends being zero. Anyway, what I want is to know is which should be the total work done by the fictitious force to balance the energy equation.##\frac{1}{2} \cdot M \cdot v_{2}^2 + rw = \frac{\frac{1}{2} \cdot E \cdot A}{L_{1cm}+L_{2cm}} \cdot \delta^2##

With the values considered, I get the following results:

α

_{2cm}= 9.7999999999947 m/s^{2}string elongation δ = 3.425e-04 m

work done by relativistic force = 2.725e-07 J

The relativistic force work done is positive, and then the static equilibrium scenario is possible. Should that value resulted negative, then static equilibrium would not be reachable, and I guess that the spaceships bounce, and the solution is then a periodic oscillation.

And, by the way, the string doesn’t break.

This first setup most resembles the Bell's scenario, in which the string is always fixed at both ends. I could also do the same, with the string fixed at both ends from the start, but I would arrive to the same conclusion: If instead of considering that the proper acceleration is always constant, I consider that is the F/M ratio of the engines what is kept constant - so proper acceleration can freely change -, then the string doesn't break.

The change in proper acceleration is really tiny, and would be very difficult even to measure it.

If instead of 1 hour, we wait 1 Year (or 6 months, or 3 months…) proper time as measured in S

_{2}, then static equilibrium state is not mechanically possible.@1 Year:

v

_{1cm}=1.9996e-04 m/s,L

_{1cm}=7,900.618106 m,v

_{2cm}=1.9996e-04 m/s,L

_{2cm}=7,900.618106 mBeing the numbers so small, the initial relative speed is only 0.2 mm/s @1 Year, the string can withstand the tension with no problem, and I guess that the solution would be probably an oscillation of both spaceships, bouncing but not breaking the string, and with a slack string most of the time.

So, if this is the case, not only the string doesn't break, but for the most part of the time, is slack, with no tension.

For the curious minds, here the Python code I made. Remark that I had to use a multi-precision numerical package, to get the required accuracy:

Python:

```
from mpmath import *
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
C =299792458 # speed of light
C2=power(C,2) # square of speed of light
M =1.0e+9 # 1000000 tons each spaceship
A0=9.8 # desired (comfortable) proper acceleration
L0=10000.0 # initial length of string/distance
"""
mechanical thread characteristics...
https://www.cablesestructurales.com/wp-content/uploads/2015/01/Catalogo-%20Cables%20-inoxidables.pdf
Diameter: 1 mm Area:0.6 mm2 Break tension: 84 Kg
E =1.3e+11 # 130000 N/mm2 Young modulus
AR=6.0e-7 # 1 mm Diameter / 0.6 mm2 section Area
"""
DAYs=24*3600
HOURs=3600
"""
Earth frame observer ct,x coordinates for constant proper accelerating bodies
tau: proper time acc: proper acceleration
Parametric (tau) world line coordinates
"""
def ct_x(tau,acc,l0):
return (C2/acc)*sinh(acc*tau/C),(C2/acc)*cosh(acc*tau/C)+l0
mp.dps = 32
def VelocityCM(tc):
# Used to find intersection of simultaneity line from tc in CM to corresponding t1 in S1
tc11 = lambda t1,k: (C2/A0)*(k*sinh(A0*tc/C)-sinh(A0*t1/C))
tc12 = lambda t1,k: (C2/A0)*(k*cosh(A0*tc/C)-cosh(A0*t1/C))+(0.5*L0)
# Used to find intersection of simultaneity line from tc in CM to corresponding t2 in S2
tc21 = lambda t2,k: (C2/A0)*(k*sinh(A0*tc/C)-sinh(A0*t2/C))
tc22 = lambda t2,k: (C2/A0)*(k*cosh(A0*tc/C)-cosh(A0*t2/C))-(0.5*L0)
t1,k=findroot([tc11,tc12],(tc,1.0))
t2,k=findroot([tc21,tc22],(tc,1.0))
# Coordinates (ct,x) for Earth frame
ct1,x1=ct_x(t1,A0,-0.5*L0)
ctc,xc=ct_x(tc,A0,0.0)
ct2,x2=ct_x(t2,A0,+0.5*L0)
# Proper (=real) distance from CM to...
lc1=sqrt(power(x1-xc,2)-power(ct1-ctc,2))
lc2=sqrt(power(x2-xc,2)-power(ct2-ctc,2))
# With the proper times, calculate the speeds relative to Earth frame
vc=C*tanh(A0*tc/C)
v1=C*tanh(A0*t1/C)
v2=C*tanh(A0*t2/C)
# Using relativistic speed addition, calculate de relative speeds viewed by CM frame
v1c=(vc-v1)/(1-(vc*v1/C2))
v2c=(v2-vc)/(1-(vc*v2/C2))
gmc=C/sqrt(C2-power(vc,2))
# Elapsed time in Earth frame
tearth=(C/A0)*sinh(A0*tc/C)
# Distance traveled according to Earth frame
dearth=(C2/A0)*cosh(A0*tc/C)
return v1c,v2c,t1,t2,lc1,lc2,gmc
def Graphic(tau_ini,tau_end,stp):
tCM=np.arange(tau_ini,tau_end,stp,dtype=float)
v1c=np.full_like(tCM,0.0)
v2c=np.full_like(tCM,0.0)
a2c=np.full_like(tCM,0.0)
l2c=np.full_like(tCM,0.0)
gmc=np.full_like(tCM,0.0)
i=0
for tc in tCM:
r_v1c,r_v2c,r_t1,r_t2,r_l1c,r_l2c,r_gmc = VelocityCM(tc)
l2c[i]=r_l2c-(0.5*L0)
v1c[i]=r_v1c
v2c[i]=r_v2c
gmc[i]=r_gmc
# Calculate acceleration from speed change
# don't know how to obtain this value otherway!
if (i==0):
v_2=r_v2c
elif (i==1):
v_1=r_v2c
dt=2.0*(tc-tCM[0])
else:
acc=(r_v2c-v_2)/dt
a2c[i-1]=acc
v_2=v_1
v_1=r_v2c
i+=1
fig, ax1 = plt.subplots()
# ax1 ...
ax1.set_xlabel('γ (Lorentz)')
color = 'tab:blue'
ax1.set_ylabel('a2c m/s²',color=color)
ax1.tick_params(axis='y',labelcolor=color)
ax1.plot(gmc[1:-1],a2c[1:-1],color=color,linestyle='dotted')
"""
# ax2 ...
ax2 = ax1.twinx()
color = 'tab:green'
ax2.set_ylabel('ΔL m',color=color)
ax2.tick_params(axis='y',labelcolor=color)
ax2.plot(gmc[1:-1],l2c[1:-1],color=color)
"""
plt.title('[cm] Start=%4.1f End=%4.1f (days)'%(tau_ini/DAYs,tau_end/DAYs))
fig.tight_layout()
plt.grid()
plt.show()
def EqSystem(tbrk):
eq1 = lambda a,d,rw: a-(A0*C2/(C2+A0*(lc2+d)))
eq2 = lambda a,d,rw: (A0-a)*M-(E*AR*(2*d)/(lc1+lc2))
eq3 = lambda a,d,rw: 0.5*M*power(v2c,2)+rw-(0.5*E*AR/(lc1+lc2)*power(d,2))
v1c,v2c,t1,t2,lc1,lc2,gmc = VelocityCM(tbrk)
print('v1c=%.9e'%(v1c))
print('v2c=%.9e'%(v2c))
print('lc1=%.9e'%(lc1))
print('lc2=%.9e'%(lc2))
a2born,dl,relw=findroot([eq1,eq2,eq3],(A0,1,1),Verify=True)
print('SOLUTION IS:')
print('α2b=%.13e m/s² dl=%.3e m relw=%.3e J'%(a2born,dl,relw))
# A negative relw means "no static equilibrium" but bouncing?
# EqSystem(365*DAYs)
Graphic(365*DAYs,395*DAYs,8*HOURs)
```

Some details about the scenario I considered for the Python code, and the theory behind.

The setup.-

For the journey to be comfortable for the crew, the constant propulsion force of the engine is calculated to give a 9.8 m/s

^{2}(1g) proper acceleration to the mass of the spaceship.The stationary reference frame (“Earth frame” so to speak) located in outer space is where the spaceships are initially parked. For sake of simplifying the math, with reasons that I’ll explain later, I consider that the x-coordinate origin of such stationary frame is located ##\frac{c^2}{\alpha}## meters away from the parking spot, where α is the nominal proper acceleration, and c is the speed of light.

Then, the conceptual middle point between the spaceships, call it [cm], is at position ##x_{cm}=\frac{c^2}{\alpha}##, the trailing spaceship S

_{1}is at ##x_1=\frac{c^2}{\alpha}-\frac{D_0}{2}##, and the leading spaceship S_{2}is at ##x_2=\frac{c^2}{\alpha}+\frac{D_0}{2}##.The spaceships have a mass of 10

^{9}Kg each. It’s a lot, but the effects that we’ll observe are really tiny, so we need some big numbers. For the same reason, the distance between spaceships at the start position is 10 Km.The string is a steel cable. As to do not disturb the performance of the leading spaceship, that has to pull the string, I decided to use a very light one, whose mechanical characteristics are as follows:

Cable diameter: 1 mm, Area section: 0.6 mm

^{2},Young modulus: 130,000 N/mm

^{2},Break tension: 840 N,

Lineal density: 0,005 Kg/m

This seems odd for 10

^{9}Kg spaceships, but keep in mind that we’re not towing the trailing spaceship. Initially, the cable has only to withstand the inertial forces due to the acceleration and its mass, which is 50 Kg for 10,000 meters of cable. From now on, I call it “string” again.In fact, during the journey – after settling down the spaceships at departure -, the tension of the string at the hook point in trailing spaceship S

_{1}is considered always zero, if not stated contrary for some reason.Of course, there’s a tension in the string at S

_{2}leading spaceship, because it’s pulling the string with an acceleration. So, there’s a certain amount of “pretension” in the string that is about half the load it can withstand without breaking.Nonetheless, the initial distance between the spaceships when they reach the nominal acceleration is adjusted to 10,000 meters (the string is a little elongated), and with zero tension in hook point S

_{1}.Theory used.-

With a constant F/M ratio from the engines, the spaceships follow a constant proper acceleration worldline in hyperbolic motion, and without any string interference they behave as in Bell’s scenario. So far, so good.

In the space-time diagram, the worldlines can be described by the following coordinate equations, considering the trailing spaceship S

_{1}, a middle point [cm] and the leading spaceship S_{2}:##ct_1=\frac{c^2}{\alpha}\cdot\sinh(\frac{\alpha\cdot\tau_1}{c}) \quad x_1=\frac{c^2}{\alpha}\cdot\cosh(\frac{\alpha\cdot\tau_1}{c})-\frac{D_0}{2}, \quad S_1## starts at ##(0,\frac{c^2}{\alpha}-\frac{D_0}{2})##

##ct_{cm}=\frac{c^2}{\alpha}\cdot\sinh(\frac{\alpha\cdot\tau_{cm}}{c}) \quad x_{cm}=\frac{c^2}{\alpha}\cdot\cosh(\frac{\alpha\cdot\tau_{cm}}{c}), \quad S_{cm}## starts at ##(0,\frac{c^2}{\alpha})##

##ct_2=\frac{c^2}{\alpha}\cdot\sinh(\frac{\alpha\cdot\tau_2}{c}) \quad x_2=\frac{c^2}{\alpha}\cdot\cosh(\frac{\alpha\cdot\tau_2}{c})+\frac{D_0}{2}, \quad S_2## starts at ##(0,\frac{c^2}{\alpha}+\frac{D_0}{2})##

We’re free to choose the position of x=0 in space, and we can always set it where appropriate for the problem studied. In this case, if the worldline of the [cm] conceptual observer starts at (0,c2/α), then for any event-point of that worldline, the corresponding line of simultaneous events is the line that connects the (0,0) origin with the event-point, and that makes the math much easier.

To prove that, we do the derivative of (ct,x) parametric worldline with respect to proper time ##\tau##:

##[c\cdot\cosh(\frac{\alpha\cdot\tau_{cm}}{c}),c\cdot\sinh(\frac{\alpha\cdot\tau_{cm}}{c})]##

This vector is tangent to the worldline event-points, and so it represents the ct' axis for the accelerating spaceship. The x' Minkowski-orthogonal axis is the reciprocal, and then:

##[c\cdot\sinh(\frac{\alpha\cdot\tau_{cm}}{c}),c\cdot\cosh(\frac{\alpha\cdot\tau_{cm}}{c})]##

And so, a line that passes by the (0,0) origin and connects with the event-point in the worldline, is just the x' axis/line of simultaneous events for that event-point. This is a known property of the Rindler horizon, and at any proper time, the (0,0)-event is always simultaneous with the proper time in the spaceship.