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.