some questions

[youngone]

Hi, I'm only 15 years old and new to relativity and I'm finding it a little bit confusing. Here are some of my questions.

 

1) Consider two spaceships. Each spaceship moves away from each other at a speed of 0.9c. Now, would light from either spaceship reach the other? If possible, provide some reasons and maybe a few interesting bits along with your answer.

 

2) As we all know, if a first observer was to view another second observer travelling at 0.9c relative to him/her, he/she would view the second observer to be contracted according to the direction of travel. But how would the second observer view the first observer? Would the first observer appear contracted to the second observer? Again, if possible, provide anything interesting along with your post.

[TrueHeart]

.. away from each other at a speed of 0.9c. Now, would light from either spaceship reach the other?

Yes, light will always reach the other entity regardless of its speed. I think you might've meant to ask about the case where each spacecraft is moving away from a mutual landmark (like Earth) at .9c, in opposite directions. Because then it would seem that the two craft's velocity with respect to one another would be 1.8c... too fast for light to catch up. But relativity has an answer for that: the addition of any two speeds never results in a number greater than c; so no two entities can ever move apart with speed greater than light... it is not possible to outrun being seen.

 

I have a web site on the subject. Find it under my name/profile.

[gib65]

Well, I know enough about the speed of light to know that it does not matter how fast the object that emits the light is going, light always travels at 300,000 km/s in a vacuum relative to the observer. So even if one observer, traveling at 0.9c in one direction, is traveling away from another observer traveling at 0.9c in the opposite direction (both relative to a stationary object as lightSword pointed out), light emitted from one observer would still catch up to the other observer, and it would take (300,000 x L) seconds (where L is the distance between the two observers when the light was emitted). Now, you make think that this means the light is violating the universal speed limit of 300,000 km/s, but remember when this speed limit is applied to light, it must be applied in the context of the speed relative to the observer since, unlike material objects, the speed of light has no relative speed except to the observer.

[muad'dib]

The speed of light is not relative to an observer. If Three spaceships were traveling in different directions with different speeds and all measured the speed of the same beam of light they would all get the same result.

[swansont]

For #2: There is nothing that distinguishes the two observers or their frames of reference, so the effects they see have to be interchangeable. Thus, they both see the other one as length contracted. Each one can assume that they are at rest in their own frame, as long as those frames are inertial.

[gib65]

The speed of light is not relative to an observer. If Three spaceships were traveling in different directions with different speeds and all measured the speed of the same beam of light they would all get the same result.

 

Yes, that's what I mean to say

[youngone]

Hey, thanks to all of you for your posts!

 

Just one more thing: Same question as number 2 but involving time dilation. How would the second observer view the first observer if it is travelling at 0.9c relative to the first observer? The first observer would view time for the second observer to be moving slower. But how would the second observer view time for the first observer? Would it be faster or slower?

[cyeokpeng]

The second observer B will see the 1st observer A's time slow down also, since though the 2nd observer is moving at 0.9c wrt observer A, imagine B's frame of reference is fixed, A will appear to be moving at 0.9c in the opposite direction. Another analogy: If a train is approaching a station, observer on the platform will see passengers on the train moving say to the left. On the other hand, if you are the passenger on the train, you will see people on the platform to be moving at the same speed to the right. So B will still see A's time slow down due to this high speed 0.9c relative to B.

[youngone]

But what if they were to compare their respective times later on? Since the first observer viewed the second observer's time slow down and the second observer viewed the first observer's time slow down, how can they tell who was the slower one?

[TrueHeart]

No meaningful comparison of their respective clocks is possible until they unite in the same place. If they keep on their inertial courses then that will never happen, and the discrepancies will only grow. In order to affect a coming together, one observer will have to be accelerated, and that will have additional ramifications as to how the two reckon each other's time. Such updated reckoning will mathematically converge upon a noncontradictory truism, by the time they do meet up.

[youngone]

I'm a bit confused. Can you provide some more clarification and maybe an example?

[TrueHeart]

You've heard of the Twin Paradox perhaps?? An astronaut speeds through the cosmos and then returns to Earth to find that he has aged less than his twin brother. When the two twins first started separating from one another, each twin asserts that the other one's clock is running slowly, as compared to his own clock. That is an apparent discrepancy, but it actually has no illogical implications. When the astronaut twin makes his turnaround, the acceleration causes him to now perceive his Earthbound brother's clock to be running *fast*, as compared to his own; yet this effect is not mutual, ie. the Earthbound twin continues to reckon his astronaut brother's clock to be running slower than his own. The math works out perfectly, and when the space cadet returns to Earth, BOTH twins agree on who has aged the greater, the stay-at-home twin, and they both agree on the precise amount of that aging difference.

[gib65]

OK, so let me get this straight: time slows down for both observers when travelling uniformly, but then all of a sudden time speeds up for the stay-at-home when the traveller suddenly turns around (effectively accelerating). For the whole while that the traveller is accelerating, the stay-at-home ages rapidly enough to surpass the amount of time that the traveller has experienced, plus enough extra to resume aging slowly upon the travellers ride home at uniform velocity to still be older by the time they both reunite. Is this correct?

 

In order for this to work out, I presume that the farther away the traveller has gone before accelerating to return home, the faster the stay-at-home will have to age. This follows because the act of accelerating in order to turn around and head home takes the same amount of time to complete regardless of how far the traveller has gone before doing this. But if the gap between their ages keeps getting wider as the traveller moves at uniform velocity (in either direction), then the amount of aging the stay-at-home has to "catch-up" on depends on how long/far the traveller has been travelling. Therefore, if the act of accelerating to return home takes constant time, then the RATE at which the stay-at-home will have to age during this act cannot be constant. Have I got this right?

 

Essentially, what this boils down to is that in a gravitational field, the rate at which time slows down is directly proportional to the depth with which you are in the gravitational field. Is this right?

 

This also means that when the stay-at-home begins to increase his rate of aging (when the traveller turns around) the rate change is a discontinuous jump. That is, relative to the traveller before he turns around, the stay-at-home's rate of aging is slow, but since the act of accelerating begins instantaneously, the gravitational field that suddenly appears (in the context of GR) is also instantaneous, and the difference between the traveller's and the stay-at-home's depth within this field likewise appears instantaneously. Therefore, the rate of aging for the stay-at-home must make an instantaneous leap from slow to fast (relative to the traveller). Is this what happens?

[swansont]

Time doesn't slow down for the observer - the observer sees time slow down for the other person.

[TrueHeart]

gib65:

You are 100% correct about your second paragraph! Bravo, very astute! The time hastening factor is proportional to the distance away. But you've got something else in your post horribly wrong: You must watch how you speak and phrase things when talking about Relativity! Nothing is absolutely true, nothing of the supposed "big picture" should be cited.. tsk tsk. You can only deal with observer and observed and with what is true of each of their viewpoints. No distortion inarguably affects any entity, it's all about each observer's reckoning of alien clockworks and alien measures. Please, carefully reread my post before yours and you'll understand what pains one must go to to make oneself clear, in describing relativistic situations.

 

But even so, YES! Congratulations !! Your description is perfectly right about how the clock rate speeding-up works out mathematically, it needs to be more than enough to compensate, like you cited, because the rudimentary time dilation attributable to speeding entities must also be continually applied to the brother, per either twin's reckoning.

 

I just read your 3rd paragraph and it is correct.

 

I just read your 4th paragraph and it is correct.

[youngone]

Does this mean that when the second observer who is travelling changes his/her velocity and makes the return trip to the first observer who is at rest, the second observer will see the first observer's age increase dramatically when he/she starts to return back to the first observer?

 

What are the reasons for this? Can you please explain it to me?

 

PS Please be patient with me! I know I am very slow on this topic!

[swansont]

Go here. Pay special attention to the "Doppler shift explanation" section.

[TrueHeart]

Does this mean that when the second observer who is travelling changes his/her velocity and makes the return trip to the first observer who is at rest, the second observer will see the first observer's age increase dramatically when he/she starts to return back to the first observer? ..

To best understand the subject, perhaps you might have to temporarily forsake trying to figure what either observer actually sees with their eyes. The dynamics at high speeds are so complex, compounded by other-than relativistic effects -- like the Doppler effect and like simple transmission delay times -- that it may not be truly helpful (to you) to imagine what is seen. Instead, imagine what is reckoned by each party, assuming that both observers do understand Relativity.

[youngone]

Sorry for not posting in a while. I was travelling with my family recently.

 

OK, I think I am starting to get it now. Apparently the math predicts that the twin who makes the return journey to the other twin ages more slowly. This is because the twin has to accelerate in the other direction resulting in a change of his/her frame of reference, am I right? This means that if the twin on earth was to jump into a spaceship and catch up with his/her twin, he/she would have aged less when compared to the twin, right?

 

But what I still don't get is what each twin computes (not sees) throughout the whole event. OK, so at first, the one twin computes that time for the other twin runs slowly (let's say at half the speed) and vice versa. Now the earth twin will calculate that the first part of the journey takes 12 years in his/her own time. This means that the astronaut twin will take 6 years according to his/her own time and he/she will reckon that during the first part of the journey the earth twin will age only 3 years, right? Now what happens after this, when the astronaut twin makes the turnaround?

[youngone]

Sorry for not posting in a while. I was travelling with my family recently.

 

OK, I think I am starting to get it now. Apparently the math predicts that the twin who makes the return journey to the other twin ages more slowly. This is because the twin has to accelerate in the other direction resulting in a change of his/her frame of reference, am I right? This means that if the twin on earth was to jump into a spaceship and catch up with his/her twin, he/she would have aged less when compared to the twin, right?

 

But what I still don't get is what each twin computes (not sees) throughout the whole event. OK, so at first, the one twin computes that time for the other twin runs slowly (let's say at half the speed) and vice versa. Now the earth twin will calculate that the first part of the journey takes 12 years in his/her own time. This means that the astronaut twin will take 6 years according to his/her own time and he/she will reckon that during the first part of the journey the earth twin will age only 3 years, right? Now what happens after this, when the astronaut twin makes the turnaround?

[TrueHeart]

.. This means that if the twin on earth was to jump into a spaceship and catch up with his/her twin, he/she would have aged less when compared to the twin, right?

Right!

But what I still don't get is what each twin computes (not sees) throughout the whole event. OK, so at first, the one twin computes that time for the other twin runs slowly (let's say at half the speed) and vice versa. Now the earth twin will calculate that the first part of the journey takes 12 years in his/her own time. This means that the astronaut twin will take 6 years according to his/her own time and he/she will reckon that during the first part of the journey the earth twin will age only 3 years, right? Now what happens after this, when the astronaut twin makes the turnaround?

The space-travelling twin computes that during his/her turnaround, Earth and its residents age 18 years. Earth ages 3 years during the first leg, then 18 at turnaround, then 3 more, totalling 24 years for the round trip. This agrees with the reckoning of the earthbound twin: 24 years will elapse, but the astronaut twin ages only 12... 6 on the outbound leg and the same again for the return.

[TrueHeart]

.. This means that if the twin on earth was to jump into a spaceship and catch up with his/her twin, he/she would have aged less when compared to the twin, right?

Right!

But what I still don't get is what each twin computes (not sees) throughout the whole event. OK, so at first, the one twin computes that time for the other twin runs slowly (let's say at half the speed) and vice versa. Now the earth twin will calculate that the first part of the journey takes 12 years in his/her own time. This means that the astronaut twin will take 6 years according to his/her own time and he/she will reckon that during the first part of the journey the earth twin will age only 3 years, right? Now what happens after this, when the astronaut twin makes the turnaround?

The space-travelling twin computes that during his/her turnaround, Earth and its residents age 18 years. Earth ages 3 years during the first leg, then 18 at turnaround, then 3 more, totalling 24 years for the round trip. This agrees with the reckoning of the earthbound twin: 24 years will elapse, but the astronaut twin ages only 12... 6 on the outbound leg and the same again for the return.

[TrueHeart]

Of course, you might wonder what assumptions had to be made in order to arrive at those simple whole numbers, the 18, the 3 the 12... I mean, acceleration is always a gradual event, and this means gradually changing speed, and consequences for the earthbound twin's reckoning of the astronaut's time dilation. What acceleration strength did I choose? or did I somehow compute an instantaneous turnaround? In other words, how does one eliminate any slop factor??

 

Answer: I allowed for instantaneous turnaround by using only the basic Lorentz transform. Read my brief explanation here and replace my numbers with yours: .866c speed, which makes the gamma=2; and make the earth-synchronous buoy be 10.4 light-years away, which makes for a 12-year outbound leg per earth reckoning.

[TrueHeart]

Of course, you might wonder what assumptions had to be made in order to arrive at those simple whole numbers, the 18, the 3 the 12... I mean, acceleration is always a gradual event, and this means gradually changing speed, and consequences for the earthbound twin's reckoning of the astronaut's time dilation. What acceleration strength did I choose? or did I somehow compute an instantaneous turnaround? In other words, how does one eliminate any slop factor??

 

Answer: I allowed for instantaneous turnaround by using only the basic Lorentz transform. Read my brief explanation here and replace my numbers with yours: .866c speed, which makes the gamma=2; and make the earth-synchronous buoy be 10.4 light-years away, which makes for a 12-year outbound leg per earth reckoning.

[youngone]

Ok, just a few more questions. I will use the same situation as previous but eliminate the turnaround and the first half of the journey.

 

So, we say there are two points, A and B, 0.86 lightyears apart. On A is the earth twin while on B is the astronaut twin. The astronaut twin would then instantly accelerate (or consider him already moving) to 0.866c where gamma=2 (the time dilation factor, right?). The whole journey will take 1 year according to the earth twin and will see the other twin's time run at half speed (6 months) while the astronaut twin will take 6 months to complete the journey and see the earth twin's time run at half the speed (3 months). Now, what is the solution to the situation?

 

Oh, and sorry for constantly pestering you like this. I would personally do the math but I don't know anything about the mathematics of relativity and if possible can you provide a website which teachs everything about the math in a simple manner?