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Understanding Special Relativity


Questions and Answers




 A. asks:


In the explanation about the twins paradox;  During the U-turn the difference between Traal's and Earth's clocks is changing from -7.5 to 7.5 in a short time relative to Bob. Does that mean that activity on Earth goes at tremendous speed during this period? And if it does, isn't one of the consequences of it that bodies on Earth are moving faster than light relative to Bob?




Relativity states that the relative velocity between two bodies cannot exceed the light speed. It doesn't mean that an observer on a third frame cannot see the relative speed between them beeing grater than light speed. If, for example, two spaceships are going away from an observer in opposite directions at a speed of 0.8c each, then the observer measures a relative speed of 1.6c between them, though, due to the law of velocities composition, each of the spaceships sees the other going away from it slower than the light speed. Thus, there is no problem in saying that bodies on Earth are moving at a speed greater than the light speed relative to each other in Bob's frame. The question is at what speed each body is moving relative to Bob. Special Relativity deals with inertial frames, and its rules applies to such frames. During an acceleration things are more complicated. Though at any point on the acceleration line in time-space all the rules are preserved, when we calculate the integral along a range we get strange things. One can also ask the following question: When Bob leaves Earth Traal is 8.66 light years away from him. A short time later, when he completes his acceleration Traal is a few light years closer. Does Traal approach Bob faster then light? The question is if we can call this approach a motion. If you check the relative speed between Bob and the frame of Earth and Traal at any point during the acceleration it will be lower then c (the slope of the tangent of the curve in the time-space diagram is greater then 45 degrees). Bob is continuously changing reference frames during the acceleration. Traal is getting closer to Bob due to change in the length contraction and not because it moves in space relative to any of these frames.



C. asks:


There is one thing, however, that confuses me, not in your paper but in general. Upon the explosion of a supernova, parts of it travel in all directions at high relativistic speeds. The light emitted from these parts is red shifted or blue shifted depending on the part velocity. It is measured that the red shifted light arrives to the earth just a few days after the blue shifted light. The Supernova to Earth distance is measured in millions of light years. It would take centuries to get blue shifted light after the red shifted light because according to SRT (and your explanation) the departure times (as seen by the observer on Earth) are different, they are proportional to gamma * vx/c^2 (where v is varying say from -0.9c to +0.9c). As the arrival times are virtually the same (days instead of centuries), why does Einstein state that the observer on the moving train receives the 2 flashes at different times? I see no substantial difference between the supernova and the relativistic train experiment and I suppose a real experiment should be more relevant than a thought experiment.





As you know a supernova is a huge phenomenon. Matter moves towards Earth at the part of the supernova that is closer to Earth. Matter moves away from Earth at the part that is further from Earth. The diameter of the supernova is negligible relative to the distance between Earth and the supernova, but it is still a significant distance.
The time difference between the light coming from matter that moves towards Earth (blue-shifted) and matter that moves away from Earth (red-shifted) is affected by two things: The longer way that the red shifted light pass and (as you correctly pointed) the time difference between the emission of the light relative to Earth due to the relativity of simultaneity.
Your confusion comes from choosing the wrong distance and velocity. As written in the article, simultaneity difference is proportional to the distance between the events and the relative velocity between the frames of reference. The simultaneity difference is not affected by the distance between the observer and the events. The woman in the train would measure the same time difference between the lightning bolts if the train was hundred miles further away on the track. Now, what is the reference frame on which the two events are simultaneous? This is not the reference frame of the moving matter, but the one of the center of the supernova, relative to which the pieces of matter are moving at the same (high) velocity in opposite directions. This reference frame is moving away from Earth due to the universe expansion.
So the distance you should use is the diameter of the supernova and the velocity is the velocity of the motion of the whole supernova away from Earth as it was when the supernova happened. I guess when you substitute these values in the equations you get time difference of days.

C. writes:


For the sake of simplicity, I will not take the expansion of the universe into account and I will assume that the relative speed between the Earth and the supernova is 0. Also I will assume the explosion of the supernova creates two parts moving at velocities +v and -v relative to the earth (and the center of the supernova). The first part (approaching earth with +v speed) creates blue shift and the second part (moving away from the earth with -v speed) creates red shift.

I know the events are (nearly) simultaneous relative to the supernova. It appears to me they are not simultaneous for the observer on earth, even if the earth is at rest relative to supernova. This is also true in the case of the twin paradox: When a twin passes near a distant star, he will see different times on the star and on the earth, even if the the star is at rest relative to the earth.




The velocity you put in Lorenz transformation is a relative velocity between two frames of reference, not a velocity of something that is moving relative to the reference frame. You have to decide what your reference frame is. If two events happen simultaneously for one point on a reference frame they are simultaneous everywhere on this frame. It doesn't matter if they happen on stationary or moving parts. They can only have time difference when measured from a different frame. If two pulses of light are emitted simultaneously relative to the supernova center and Earth is not moving relative to this center, then they are emitted simultaneously relative to Earth too. If you like, the relative velocity between the frames is 0, beta = 0 and gamma = 1.

Also, if you say that the two events happen at approximately the same point, they will be simultaneous relative to any reference frame, no matter how fast it moves. So the only thing that makes difference between the times they arrive to Earth is the diameter of the supernova.



S. asks:


During the U-turn in the twins paradox, 15 years pass at Earth from Bob's point of view. What happens if Bob changes his mind, continues his turn and return to his original course? Is the time on Earth goes backwards for him? Does his brother Al earn 15 years back, or perhaps this time cannot be returned and time continues from where it reached?




When Bob gets to Traal as described in the twins paradox, there are two reference frames: This of the spaceship and this of the plants Traal and Earth. In the frame of the planets the distance to Earth is 8.66 light years and the time on Earth is 10 years after the launch. In the frame of the spaceship the distance to Earth is 4.33 light years and the time on it is 2.5 light years from launch. If Bob suddenly stops (makes a half of the U-turn) he moves to the reference frame of the planets, and thus time on Ears turns to be 10 years from launch for him, that is 7.5 years later. If he then immediately accelerate and return to his original speed, he is back in his old frame again (actually he is a little behind in the time axis because it took him time to stop and accelerate, but this difference is negligible), and thus time on Earth is again only 2.5 years after launch. Apparently his brother gained 7.5 years back. Actually, these 7.5 years do not really pass, not forward nor backward. What changes is only what Bob calls "now". It is a different thing in the two frames.

Imagine a man standing in a wood. He defines a coordinate system where the line left-right is the x axis and the line backward-forward is the y axis. All the trees that are exactly to his left or exactly to his right has a value y=0. Trees that are behind this line relative to his position have negative value of y and trees that are in front of it have positive y value. Now the man turns a little to his left and set a new coordinate system relative to his new position. All the trees to his left have now grater y values and the trees to his right have smaller y values. Clearly neither the trees nor the man have moved in any direction. The new y values are only due to the change in the coordinate system.

In relativity time is also an axis in time-space. When Bob moves from one frame to another the line he calls "now" changes, just as line y=0 have changed when the man changed his coordinate system. The line "now" in the new frame goes through different set of events in time-space. Al don't gain years just as the trees in the wood will not move back if the man will return to his original position.

The laws of nature also make sure that Bob will not be able to see this change. Signals that are sent from Earth in light speed will always get to Bob in the order they left. If he has a strong telescope, he will see life on Earth moving always forward. He might see them in slow-motion or in fast forward, but never in rewind.




M Asks:


I am trying to understand what happens when a very long body accelerates. In the explanation about the twin paradox we assume that after the acceleration the spaceship remains approximately at the same point and the same time as it was before, relative to Earth. This was good enough for a point mass. But in the case of a very long solid spaceship that accelerates, we cannot assume the same about both ends of it. After the end of the acceleration, if we look from the original frame, there is a time difference between the ends of the spaceship and its length is smaller than it was before the acceleration. If we place clocks at fixed distances along the spaceship and in front of every clock we place another clock on the inertial space station on which it anchors, after the acceleration the distance between the clocks on the spaceship as observed from the space station will be shorter than the distance between the clocks on the station. If the two bodies are long enough, one of the clocks must be approximately in front of the same clock as it was before the acceleration. How can we determine which clock it will be? What is the difference between the points? Also, the time difference between the clocks changes from point to point. Again, how can we determine what is the time difference at each point?

It seems to me that there are two degrees of freedom here while in reality the state of the spaceship after the acceleration must be well defined.




According to relativity strange things happen during acceleration. During the acceleration, time doesn’t pass at the same rate at all the points of the spaceship. Clocks that are closer to the front of the spaceship will move faster than clocks closer to the back of it. When the spaceship stops accelerating and becomes inertial again, the clocks are synchronized in the new reference frame. Clocks at the rear will show earlier time than clocks at the front. We can't speak about the time that passed in the spaceship during the acceleration. Different time passed at every point of the spaceship. If we want to resynchronize the clocks we can freely choose which clock shows the correct time. Apparently there is a degree of freedom here, but it is all about the origin of one reference frame relative to the other.

And what about distances? As observed from the space station, all the points along the spaceship start accelerating synchronously. But in this reference frame the acceleration is not constant. Due to the non linear composition of velocities the acceleration is reduced with time. In addition, in the reference frame of the spaceship all the points of it stop accelerating simultaneously. But this frame is moving relative to the frame of the station so in this frame it will not happen simultaneously. Rear points will stop accelerating before frontal points. In the frame of the station the acceleration ends when the front end of the spaceship stops accelerating. Since all the points end up at the same speed, the rear points had grater acceleration (they reached the final speed faster) and thus moved longer distance then the front points. At the end of the acceleration all the points on the spaceship will be shifted from the points on the station in the direction of the motion of the spaceship. The front end of it will be closest to its original position. As we get closer to the back of the spaceship, there will be greater shift of the point. The distance between two points on the spaceship will be smaller then the distance of the matching points on the spaceship (as observed from the station). This is the expected length contraction.



P Asks:


I am trying to find a formula for relativistic perfect elastic collision. I know that the total energy and the total momentum are preserved in elastic collision, but I can't solve the equations for the velocities after the collision as we do in classic mechanics.



The answer is too long for this page. I have put it here.