How does relativity affect speed perception?

[Amateur001]

If I have an series of discs within each other (imagine a bunch of o's within each other, the outer one the largest, the inner-most one the smallest, and I am not limited in quantity), each spinning in the same direction and each spinning from a motion source in the disc just outside it so that it is moving relative to the disc it is within, how fast can the inner-most disc spin? It may be moving at say 500 mph relative to the disc that it is within, but to an outside observer (not in any disc) what is the maximum cumulative speed (when added to the speeds of all of the spinning discs) it could achieve, if there is one?

 

[mrspeedybob]

The maximum velocity for any object of non-zero mass in any frame of reference is the speed of light.

 

[Amateur001]

Can you elaborate? From the inner-most disc I would only be moving 500 mph. But outside the discs, the inner-most disc could appear as though it is moving at the speed of light. Are you saying simply that from outside the discs the inner-most disc cannot appear to be going faster than light? What happens if the cumulative speed is in fact faster than the speed of light (is it possible that the inner-most disc does move faster than light from the outside but that it simply cannot be perceived from the outside as doing so)?

 

[mrspeedybob]

Velocities do not actually add as you might think they do. If you have one train traveling 60 miles per hour in one direction and a second train traveling at 60 mph in the opposite direction you might think they are traveling at 120 mph relative to each other. In fact they are traveling at 119.9999999999990394 mph relative to each other. The actual formula for adding velocities can be found here. http://en.wikipedia.org/wiki/Velocity-addition_formula. The effect is more pronounced as speed increases, In fact no 2 velocities can add to more then the speed of light, hence no object can exceed the speed of light relative to any other object.

 

[Amateur001]

What if I am walking forward at 2 mph on the train that is moving forward at 60 mph? Would my velocity relative to a stationary outside observer be 62 mph or would it be less? Please explain.

 

[Janus]

What if I am walking forward at 2 mph on the train that is moving forward at 60 mph? Would my velocity relative to a stationary outside observer be 62 mph or would it be less? Please explain.

Less, about 61.99998888889 mph. Check the link given in the post before yours.

 

[mrspeedybob]

What if I am walking forward at 2 mph on the train that is moving forward at 60 mph? Would my velocity relative to a stationary outside observer be 62 mph or would it be less? Please explain.

Your velocity would be (60+2)/(1+((60x2)/670615200^2)), or about 61.99999999999998345655 mph

This is counter intuitive but here is a thought experiment that might help...

We know from experiment that the speed of light is always the same in any frame of reference.
Suppose a man is at the center of a train traveling down a track, You are standing on the ground. At the instant that the man on the train passes you light strikes both ends of the train. The train is 2 light seconds long or 372,564 miles, since you are 186,282 miles from each lightning strike you see each one 1 second after it happens and you see them simultaneously.

Remember that light must travel at the same speed for the man on the train. During the second that you waited for the light the man on the train moved toward the light from the front of the train and away from the light at the rear. He sees the bolt from the front of the train first, and then the bolt from the rear. He is also 186,282 miles from each end of the train and so exactly 1 light second from each bolt. Knowing this he says the bolt at the front of the train happened first.

The difference in perception between your simultaneous bolts and the man on the trains non simultaneous bolts is not an illusion. His measurement is just as real and valid as yours. You have to throw the concept of simultaneous events out the window. Basically you have to through your whole idea of time out the window and replace it with a much more complicated version.

Now back to the 60 mph + 2 mph problem. mph means miles per hour. We just learned that time is much more complex then we thought so that hour part of miles per hour just got a lot more complicated.