# Speed of information(2) - simple example with a mass hanging in a cable

1. Aug 10, 2007

### louk

Assume a weight hanging vertically in a strong, thin and inelastic long cable which is fixed in the top end. The top end cable is then released. When will the mass at the bottom part of the cable start moving downwards??

a) Immediately. In this case information is transferred faster than light.

b) The time corresponding to speed of light over the length of the cable – call that time t_light. In that case one have to ask what is the mechanism for transferring the information with the speed of light – and what force is keeping the mass in position during the time t_light. See also case c) below.

c) The time corresponding to speed of sound in the cable – call that time t_sound. The only force available to keep the mass up is the force created by the spring force in the cable needed to accelerate the top part of the cable mass. This force has to balance the downward force of the mass. Those forces are not balancing – see also the rough calculation below.

Or has relativistic effect to be included in this case when we have no moving objects and just one observer – say placed half way down of the cable length.

Numerical example for case c) – just for illustration.

Cable length: 1000 meters.
Cable weight: 1/1000 kg per meter.
Mass: 1000 kg.
Speed of sound in the cable: 5000 m/second.
Offset of the 1000 meter cable when loaded with the 1000 kg mass: 1/10 m.

When the top part of the cable is released then the information will be transferred downwards with the speed of sound – 5000 meters/second. There will thus be en “event-horizon” moving downward with this speed. Below that event-horizon nothing will happen and the downward force of 9810 Newtons (of the 1000 kg mass) has to be balanced by an equal upward force. The only force available is the spring force in the cable accelerating the mass above the event-horizon. But it hard make this force being equal all time.

After one millisecond the event horizon will be 5 meters down at the cable. The spring constant will be k= m*g/d = 1000*9.81/0.1 = 98100 Newton/meter. The displacement in the top 5 meter part of the cable will be 0.1*5/1000 meters 0.5*10E-3 meters. The spring force of the top part will be 0.5*10E-3 * 98100 Newton = 49 Newton. This spring force could accelerate the mass of the top part which is 25/1000 kilograms. The accelearation will be a = F/m = 49/(25/1000) = 2000 meters/second2. In one millisecond the speed will be 2 meters/second and the top part has then moved 2/1000 meter = 2 millimeters in one millisecond. The full spring displacement for the top 5 meters is however only 5/1000*0.1 = 0.5 millimeter which will probably make the effect that the major part of the spring force will add to the downward force – instead of counteract it.

Even if the “event-horizon” will move downwards with the speed of light, the spring force accelerating the part of the cable above the event-horizon will not be enough to balance the gravitational force of the mass.

2. Aug 10, 2007

### HalfThere

The thing is, there's no such thing as a truly inelastic material.

The example you gave can be used to figure out the minimum elasticity for any substance in the universe (which is still extraordinarily low, far beyond any actual known material).

Case C is the actual answer.

3. Aug 11, 2007

### louk

First of all I would like to exclude the effect of the spring force accelerating the top part of the cable downwards from the analysis to be continued. That upward force is very small compared to the gravitational downward force as the numerical example showed – the reason is of course that the mass of the cable is very small compared to the mass of the weight.

Just above the event-horizon, which is travelling downward with the speed of sound of the cable, we then have downward force caused by the weight of the mass (9810 Newton in the example) and no other force in the simplified case. This means that the cable must start accelerating with g at the top end when it is released – but just below the event-horizon the cable has to stand still. When the event-horizon reaches the mass at the bottom end, then the top end part of the cable has been accelerating for Length_of_cable/V_sound seconds – that is for 0.2 seconds in the example. The speed of the top end part of the cable will g*t^2/2 = 9.81*0.2^2/2 = 0.196 m/s. The bottom end is then still at rest.

This should also implicate that the top part of an object will always fall faster than the bottom part (when the object was hanging in the top part and then released) – calculated as:

(Length_object/V_sound_object)^2*(g/2)

I really have problems with such a conclusion. Just above the event horizon the cable should be moving and accelerating but just below the event-horizon it should stay in rest. However, the downward force below the event horizon is exactly the same as the downward force above the event-horizon. And from where comes the force that keeps the cable and mass in steady state below the event horizon?? During the time the event-horizon passes downwards the cable – I think there will be a force discrepancy and also missing energy??.

What about a stiff long rod in the same situation?? The internal forces due the speed differences in the rod could then be quite substantial??

Also, such a velocity difference effect should be quite easy to demonstrate in an experiment. Has that been done??

4. Aug 12, 2007

### PhanthomJay

In a thin rigid rod, the speed of information would approach the speed of light IF the rod was made up of a completely rigid material of infinite modulus of elasticity (unstretchable) where the speed of sound, which in solid materials is proportional to the E modulus, would approach the speed of light, as a limit, in that material. Since such a rod does not physically exist, this is a hypothetical case only. The fastest way to transmit the force (information) in your example comes from using a stiff rod like steel or beryllium, where the speed of sound is quite high, but way way short of lightspeed, and way way way short of "instantaneous".

5. Aug 13, 2007

### louk

The speed of sound in steel is about 5000 m/s.

But still - does the top part start falling before the bottom part? The spring expansion is depending on the weight of the mass in the bottom part of the table so that the contraction of the cable is limited to the string expansion. The contraction can be made small by using a less heavy weight.

But in order to pass no information before the speed of sound, the movement of the top part has to fall the distance (Length_of_cable/V_sound_cable)^2 * (g/2) meters while the bottom part is still at rest. That length can be much larger than the contraction length due the spring effect.

Also, still there is a discrepance of forces below and above the "event-horzon"

The puzzle still remains - at lest for me!