Observations versus Measurements

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Observations versus Measurements Pre-Collegiate Institutes Special and General Relativity 2015 In all discussions pertaining to relativity one needs to be careful to distinguish the reference frame from which statements are made. Another important distinction is that between an observation and a measurement. This discussion of the twin parado will be rigorous in this distinction (though earlier and later discussions may be more loose). First let s define each. Observations An observation is a more general process of looking with one s eyes and noting what is seen. It does need to be referenced to a single observer at a single event but can pertain to far away (light or time like separated) events. It is usually meant to be the observation of the light arriving from other light-like separated events. Measurements A measurement is defined as the recording of facts of a single event in a reference frames (inertial or not). It necessarily must occur at a single event. A single measurement can not relate two distinct events. A single event can be measured in an infinite number of different frames. A more loose use of the word measurement pertains to relations of two measurements performed in the same frame. A measurement is a much more strictly defined process, it is a subset of all observations. (A measurement is an observation but an observation is not necessarily a measurement) It is much easier to understand the distinction between measurements and observations by contrasting the two. The usual length contraction observation is a good eample to highlight the difference. A meter stick, at rest in frame S, is moving at speed β in another frame S. Consider an observer in the S frame, A 0, who is at the origin in frame S. At time t = t 0f = 0 the front end of the stick passes, this is event E 0f. A short time later the rear end passes: event E 0r at time t r. The recorded times by observer A 0 are two measurements and the difference in time t = t 0r t 0f is a measurement of the time it took for the meter stick to pass. If this were all that A 0 could do she could say anything about the stick s length. She needs to find the speed of the stick in order to determine (i.e. measure) its length. Thus another observer in her frame (at rest with respect to her) is required. In fact, we will set up an infinite lattice of observers in frame S, each a distance apart that gives the precision we desire (let s say each millimeter). Each observer in this frame has a clock that is synchronized with A 0 s. Now a measurement of the stick in frame S can be done. First, another observer some distance apart and along the line of travel of the stick makes two measurements of the front and rear end of the stick passing. Let s call this observer A 1 and the times noted are t 1f, t 1r. Noting the distance between the observers L 0 1 and the time it takes the front to travel from A 0 to A 1 the speed of the stick is measured: β = Having measured the speed, both A 0 and A 1 can independently measure the length of the stick. It is simply L stick = β(t 0r t 0f ) = β(t 1r t 1f ) (assuming the velocity of the stick is constant). We know from before this measured length will be shorter than 1 meter. Contrast the measurement with the observation of stick s length by observer A 0. Now at any one single time, she looks at what she sees for the length of the stick. For simplicity let s take this observation to be at time t = t 0f = 0. The front end is right under her nose (assume no time delay for light traveling from front end to her eye) while the rear end is some distance away. But at this instant in time at her eye the light she observes left the rear end of the sick some time ago. Thus where she observes the rear end to be at time t = 0 is not actually where the end is. She is observing the end at time t = /c ago, where is the position of the rear end when the light departed. The rear end is actually closer, at position + βt. Thus she might conclude that the rear end looks like it is at position at time t = 0. But an observer at that position, A, would disagree. At time t = 0, observer A would say that the L0 1 c(t 1f t 0f ). rear end passed by some time ago ( /c). On to the Twin Parado Now let s turn to the twin parado. The questions we want to answer are what each participant sees. Again, we need to make distinctions between reference frames and between observations and measurements. We will first discuss what each twin observes for the other s clock. Thus, only two observer s are involved and they will look over distant space to see each other s clock. then we will turn to measurements. We will need 3 lattice grids of observers. One for the reference frame of Alice, call it frame A, who remain at rest with respect to the Earth. For the traveler s frame, Bob, we will need two sets. One, B for the outbound journey, all at rest with respect to his ship and a second, B, at rest with respect to his ship as he returns home. We will analyze these scenarios with a series of spacetime diagrams.

First let s set the stage. Here is the view of the trip from Alice s (A) frame. The blue worldline is Bob traveling to some star and back. ct For this scenario γ = 2 The distance to the planet in this frame is 9.6 ly. From this frame, A, it takes 22 years to make the round trip. From Bob s frame the distance to the planet is 4.8 ly. Thus it takes 11/2 = 5.5 years to get to the planet. And thus it takes 11 years round trip..

Observation of Alice s clock by Bob on his ship. On the outbound journey he sees Alice s clock to tick very slow. On the return trip he sees her clock to tick faster than his own. These are observations, not measurements. He sees her clock to tick only 4 years on his outbound journey but 18 on the return trip. ct

Alice s frame, what does Alice see for Bob s clock. ct

Observation of Bob s clock by Alice on Earth. On the outbound journey she sees Bob s clock to tick very slow. On the return trip she sees his clock to tick faster than her own. These are observations, not measurements. She sees his clock to tick 3 years and some on his outbound journey but a little under 7 years on the return trip. ct ct

Thus we see that by observing the far away twin, the rate at which the clocks tick depends on whether they are moving away or towards each other. Note this is different than the usual time dilation result which pertains to measurements. In fact, as Bob returns home, traveling at β = 0.87, γ = 2 Alice will observe him to approach at a speed faster than light. This was a problem asked earlier. Thus, these observations need not obey the principles that we have put forth because they are not measurements. You might say that they are optical illusions. However, when measurements are performed, these are not illusions. If a meter stick is measured to be less than one meter then in that frame it is less than one meter. Measurements in the twin parado Let s now turn to measurements that our twins can perform. By themselves, they can only relate times for events that occur in their immediate vicinity. In order to carry out measurements further afield a lattice work of friends will need to be created. For simplicity, we will only add observers along the direction of travel. In the first diagram on the net page, which is the view from Alice s frame, we have established three friends (set {A}: Alicia, Amanda, Andrea, say) at rest and spaced out between Earth and the distant planet (we would really need many more but this will keep the diagrams from getting too busy). We have also established several friends of Bob (set {B}) who are at rest with respect to him and are spaced out along the line of travel. They are represented by the set of parallel blue lines. We will eventually need yet another set for when Bob returns (set {B }). In the respective rest frames of each set, each member of that set s clocks are synchronized. With this construction Alice can measure Bob s clock by receiving data from each of her friends. They report what Bob s clock reads as he passes by them. Bob can likewise measure Alice s clock with his set of friends, each reporting back.

Measurements in frame A ct

Now you might suggest that Alice can measure Bob s clock by simply watching the clocks of Bob s friends as they race by. She knows how fast Bob is traveling, the distance to the planet, and thus how long it will take Bob to reach the planet in her frame (11 years). Thus after 11 years, she notes the clock of the passing friend of Bob to see what Bob s clock reads. This scheme will not work. The reason is that in Alice s frame, the clocks of set {B} are not synchronized! Thus she needs to get the information from her friend at the far away planet (or, knowing relativity, calculate back what it should read -that s not a measurement though). Notice in the net diagram we have added the two lines of simultaneity for Bob as he departs to and arrives at the planet. The blue dashed lines are points where the clocks read the same value. Thus upon departure Alice measures all of the clocks along the horizontal time slice that is the ais. Since this is not coincident with the blue dotted line, the clocks do not have the same value at Alice s time t = 0. You can see that at this time the clock that is passing by the far away planet has a reading much earlier than Bob s (t B = 0 if we set the departure event at t = t B = 0). The line of simultaneity for when Bob reaches the planet intersects Alice s own worldline at a time before he reaches the planet in her frame. Thus, at t = 11y the clock passing Alice will read a significant time more than 5.5y (the value that Bob s clock reads upon arrival).

ct This is the time t A = 11y This is the time t B = 5.5y This is the time t B = 0

The switch The key point in the twin parado is that for the return trip Bob needs to introduce a new set of observers that are at rest with respect to him, the set {B }. Notice the line of simultaneity now, it slopes upwards and the intersection with Alice is much later than with set {B}. The story now proceeds just as before, each observer in {A} sees the clock of each member of {B } to run half the rate of their own as they pass by. Likewise, each member of {B } sees the {A} clocks to run half the rate of their own as they pass. Thus at the point where the red dashed line intersects Alice s worldline, the clock would read the same value as where the blue dashed line intersects. (This assumes that the time for Bob to accelerate is negligible, otherwise it would have a slightly later time). Thus from Alice s perspective, while Bob accelerates in transforming from frame B to B his clock changes very little (not at all if change is instantaneous). This is represented by the length of time between the two intersection points. Recall, this is what Alice measures of the frame B and B by eamining the local members passing by her. Those members of {A} in the vicinity of the planet would see Bob s clock running at half-speed, start to pass at a greater rate as he decelerated on approach to the planet, be momentarily running at the same speed while he is at rest with respect to A and then slow down to half speed as he accelerated up to speed for his return journey. In order for Bob to measure Alice s clock, he needs to change his set of observers for his return trip. This is what makes everything come out correctly and is the difference between Alice and Bob. Alice can keep the same set of observers throughout the entire journey while Bob can not.

This is the time t A = 22y ct This is the time t B = 11y This is the time t A = 11y This is the time t B = 5.5y This is the time t B = 5.5y

In this last scenario we change the story slightly to have a finite acceleration and respite on the planet. ct Bob returns home at γ = 2 Bob changing frames This is the time t B = 7y t A = 12y, t B = 6.5y One year for both A and B t A = 11y, t B = 5.5y Bob accelerates for 0.5 y locally. Bob stays for 1 y. Bob decelerates for 0.5 y locally*. Bob changing frames This is the time t B = 5y *Note: We ignored any time dilation during the acceleration periods. This would only change the value of Bob s clock by a small amount. Bob travels off to planet at γ = 2