this is a very straightforward thread say you have a space ship moving at .9 c relative to earth. inside this space ship, in its inertial frame, the laws of physics are the same as on earth, right? on this space ship, say you have a train moving at .9 c. To an outside observer this train on the ship is moving in the same direction as the ship. What appears to be the velocity of the train? The train is going to be both time dilated and length dilated, meaning it will be lorentz factor squared.. but this still comes out to be faster than c. it would be 2.294^2 = 5.263 .9/5.263 = .171 .9+.171 = 1.071 , faster than light. What am I doing wrong? Please Register or Log in to view the hidden image!
The relativistic equation for the addition of velocities is V = v1 + v2 / 1 + (v1v2/c^2) So you've got .9c + .9c / 1 + ((.9c*.9c)/c^2) 1.8c/1.81c = 0.99447513812154696132596685082873c
To add what has already been said, the reason that the addition of velocities formula gives a different answer than the one that the OP got is that it accounts for the relativity of simultaneity and the OP doesn't. To illustrate, let assume that there are two points on the Ship, A and B, and that the train travels from A to B. In the Ship frame, we measure the time that it takes for the train to travel from A to B and get an answer of T`. The OP assumes that we can find the time it takes for the train to go from A to B in the Earth frame by simply applying the time dilation formula to T`. This is not correct. If we place clocks at A and B that are synchronized in the Ship frame, we find that the train leaves Clock A when it reads 0 and arrives at Clock B when it reads T`. In the Ship frame, where the clocks are synchronized, both clocks tick off a time of T`during the trip. This means that the time to go from A to B is T`in the ship frame However, in the Earth frame Clock A and B are not synchronized and Clock B will not read the same as Clock A. It will, in fact, lag behind clock A. The train still leaves clock A when it reads 0 and arrives at Clock B when it reads T`. If we call the time difference between Clocks A and B t`, then the time that ticks off on clock A and B between the moment the train leaves A and arrives at B is T`+t`. This is the time that you must apply the time dilation formula to in order to get the time it takes in the Earth frame. Once you take this into account, you will find that the Train always travels at less than c relative to the Earth frame as measured from the Earth frame.
Thank you Alex for this very straightforward answer. I will have to digest this before I'm sure what it means - it sounds right. I remember reading this somewhere before but I didn't know where to look to refresh so I just decided to post here. Please Register or Log in to view the hidden image!
Let's say that A and B are a distance of X apart in the ship frame. We want to synchronize the clocks in the ship frame, so we emit a flash of light from a point halfway between A and B which starts each clock when it reaches it. Since light travels at the same speed in both directions and A and B are stationary with respect to the ship, it reaches both A and B at the same time, they start at the same time and are synchronized in the ship frame. In the Earth frame, the light also travels at the same speed in both directions, however, the points A and B are moving with respect to the frame. Thus the front of the ship is running away from the light and the rear of the ship is rushing towards the light. The light heading towards A meets up with A before the light headed towards B can catch up with B. Thus, from the Earth frame, the clock at A starts before the Clock at B, and once both clocks are running, Clock B lags behind Clock A. If the distance between A and B is x in the ship frame, and v is the speed of the ship relative to the Earth then it is \(x \sqrt{1-\frac{v^2}{c^2}}\) in the Earth frame. If the light was emitted at a point halfway between A and B, then the distance from midpoint to A or B is \(\frac{x}{2} \sqrt{1-\frac{v^2}{c^2}}\) The light heading for B will then take a time of \(\frac{x \sqrt{1-\frac{v^2}{c^2}}}{2(c-v)}\) to reach B and the light heading for A will take a time of \(\frac{x \sqrt{1-\frac{v^2}{c^2}}}{2(c+v)}\) to reach A The difference in time between the light reaching A and B is \(\frac{x \sqrt{1-\frac{v^2}{c^2}}}{2(c-v)}-\frac{x \sqrt{1-\frac{v^2}{c^2}}}{2(c+v)}\) \(\frac{x\sqrt{1-\frac{v^2}{c^2}}}{2}\left ( \frac{1}{c-v}- \frac{1}{c+v}\right )\) \(\frac{x\sqrt{1-\frac{v^2}{c^2}}}{2}\left (\frac{(c+v)-(c-v)}{c^2-v^2} \right )\) \(\frac{x\sqrt{1-\frac{v^2}{c^2}}}{2}\left (\frac{2v}{c^2-v^2} \right )\) \(\frac{xv \sqrt{1-\frac{v^2}{c^2}}}{{c^2-v^2}}\) This is the time difference in the Earth frame. You have to apply the time dilation factor to get the difference in the ship frame: \(\frac{xv \sqrt{1-\frac{v^2}{c^2}}}{{c^2-v^2}} \sqrt{1-\frac{v^2}{c^2}}\) \(\frac{xv \left (1-\frac{v^2}{c^2} \right)}{c^2-v^2} \) \(\frac{xv \left (1-\frac{v^2}{c^2} \right)}{c^2 \left (1-\frac{v^2}{c^2} \right )} \) Giving the difference in the readings between the clocks at A and B as \(\frac{xv}{c^2}\) If the velocity of the train is u, then the reading on Clock B when the train arrives will be \(\frac{x}{u}\) However, in the Earth frame, Clock B got a late start, meaning that if clock A read 0 when the train left, it will read \(\frac{x}{u}+\frac{xv}{c^2}\) \(x\left( \frac{c^2+uv}{uc^2} \right )\) When the train arrives at B This equals a time in the Earth frame of \(\frac{x\left( \frac{c^2+uv}{uc^2} \right )}{\sqrt{1-\frac{v^2}{c^2}}}\) The distance the ship travels in this time is \(v\frac{x\left( \frac{c^2+uv}{uc^2} \right )}{\sqrt{1-\frac{v^2}{c^2}}}\) The train in addition travels from A to B so it travels a distance of \(v\frac{x\left( \frac{c^2+uv}{uc^2} \right )}{\sqrt{1-\frac{v^2}{c^2}}}+ x\sqrt{1-\frac{v^2}{c^2}}\) To get the velocity of the train in the Earth frame we divide this by the time: \(\frac{v\frac{x\left( \frac{c^2+uv}{uc^2} \right )}{\sqrt{1-\frac{v^2}{c^2}}}+ x\sqrt{1-\frac{v^2}{c^2}}}{v\frac{x\left( \frac{c^2+uv}{uc^2} \right )}{\sqrt{1-\frac{v^2}{c^2}}}}\) The 'x's cancel out: \(\frac{v\frac{\left( \frac{c^2+uv}{uc^2} \right )}{\sqrt{1-\frac{v^2}{c^2}}}+ \sqrt{1-\frac{v^2}{c^2}}}{v\frac{\left( \frac{c^2+uv}{uc^2} \right )}{\sqrt{1-\frac{v^2}{c^2}}}}\) \(\frac{v\frac{\left( \frac{c^2+uv}{c^2} \right )}{u\sqrt{1-\frac{v^2}{c^2}}}+ \sqrt{1-\frac{v^2}{c^2}}}{v\frac{\left( \frac{c^2+uv}{c^2} \right )}{u\sqrt{1-\frac{v^2}{c^2}}}}\) \(\frac{v\frac{\left( 1+\frac{uv}{c^2} \right )}{u\sqrt{1-\frac{v^2}{c^2}}}+ \sqrt{1-\frac{v^2}{c^2}}}{v\frac{\left( 1+\frac{uv}{c^2}\right )}{u\sqrt{1-\frac{v^2}{c^2}}}}\) \(v + \frac{ \sqrt{1-\frac{v^2}{c^2}}}{\left ( \frac{1+\frac{uv}{c^2}}{u\sqrt{1-\frac{v^2}{c^2}}} \right )}\) \(v + \frac{u \left( 1-\frac{v^2}{c^2} \right )}{1+\frac{uv}{c^2}}\) \(\frac{v\left(1+\frac{uv}{c^2} \right)}{1+\frac{uv}{c^2}}+\frac{u \left( 1-\frac{v^2}{c^2} \right )}{1+\frac{uv}{c^2}}\) \(\frac {v+\frac{uv^2}{c^2}+ u - \frac{uv^2}{c^2}}{1+\frac{uv}{c^2}}\) \(\frac{u+v}{1+\frac{uv}{c^2}}\) Which leaves us with the velocity addition equation given by AlexG
Janus, I appreciate this monument to physics that you have erected in this thread. I can almost follow it. Please Register or Log in to view the hidden image! But I have another question, given the answer that I got. It is related to this thread, in which two cars approach each other at 50 mph and one car which slams into a wall at 100 mph. Say we have a mass 2kg approaching a stationary space station at the speed of .9c. It is composed of two 1kg masses bound by a spring that holds an enormous amount of energy - 2.33e17 joules. This is the same amount of energy it takes to accelerate a 2kg object to .9c. When this spring is released, the 1kg masses are thrown in opposite directions on their line of motion towards the space station. One now appears to travel at .9945 c towards the space station and the other is stationary relative to the space station. The problem is - energy seems to have been created. The spring + initial kinetic energy of the 2kg mass is only 4.66e17 j while the .9945 1 kg mass carries 7.69e17 j towards the space station. This is really a matter of reference frames, and as I know four-momentum is conserved, but is there such a thing as four-kinetic energy?? Edit- now that I think about it, the spring carrying 2.33e17 joules will certainly have a mass to start with due to e=mc^2. So then, would this spring then carry extra kinetic energy when travelling at .9c? Edit 2- so I went back and checked how much mass 2.33e17 joules would make and it works out quite perfectly. Carrying a spring with 2.33e17 joules the object would weigh an additional 2.588 kg . Awesome! Please Register or Log in to view the hidden image!
