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Attempt on Twin Paradox via Minkowski diagrams

  1. Oct 2, 2012 #1

    This is an attempt to solve the twin paradox via two Minkowski diagrams with a few questions attached. Note that there might be mistakes in this drawing i will fix in the course of this thread in case someone notices any.

    First, let me explain the two Minkowski diagrams in the drawing first.

    Blue Lines: a huge blue ruler measuring 2 light-seconds in length which is at rest within system A (inertial reference system in which Bob is at rest in at all times)

    Purple Lines : a huge purple ruler measuring 2 light-seconds in system A which travels at vrel = 0.5c relative to system A. The purple ruler measures ~2.3 light-seconds in system B (inertial reference system the purple ruler is at rest in at all times )

    Green/Light Blue Lines: back/front sides of a rocket with Alice residing at the back of the rocket. The rocket is at rest primarily in system A, and accelerates instantaneously to vrel = 0.5c, system B becoming it's new rest frame.

    Yellow Lines: Those are light-beams traveling at C always, seen from within any arbitrary inertial reference frame. The x/t diagram is to be imagined in a way that ensures those light-beams having a 45° gradient.
    Last edited: Oct 3, 2012
  2. jcsd
  3. Oct 2, 2012 #2
    Explaining what Bob and Alice will experience (assuming they are both omniscient)

    System A (inertial reference frame A) is where Bob resides. He will be the twin who stays within his inertial reference frame at all times.
    Bob put a huge ruler in System A which will also remain at rest just like Bob.
    The blue lines symbolize the ruler, which is 2 Ls (lightseconds) long.

    The blue ruler will contract in System B which travels at v=0.5c by a factor of ~0.866 according to L' = L*(1-(v^2/c^2)) ~1.732 Ls.

    In System B there is also a huge purple ruler. This ruler is traveling at v=0.5c relative to Bob/Bob's System A. The purple ruler measures 2 Ls within System A as you can see from the drawing in System A. Therefore, in System B it is at rest in, it will measure ~2.3 Ls

    Alice has a rocket which is symbolized by the green and light blue colors. Green is the back and light blue is the front of the rocket.
    The rocket is also 2 Ls long within System A (Bob's rest frame). The rocket is at rest at the beginning of this experiment.
    Alice, which resides at the back of the rocket, accelerates at e1 (event). She accelerates the rocket to v=0.5c. We idealize the acceleration to be instantaneous.
    The acceleration gets Alice and the rocket into the purple's ruler rest frame. Now Alice, the rocket and the purple ruler are traveling at v=0.5c relative to Bob/his rest frame.

    Let's see what allice observes (given she was omniscient).

    Before accelerating, Alice places a long line of clocks from e1 to e2 within Bob's rest frame onto the blue ruler.
    Alice also places a long line of clocks inside the rocket parallel to the clocks she placed in Bob's rest frame (also Alice's rest frame until she accelerates).
    Now when Alice accelerates, the clock at the back of the rocket, where Alice resides, won't be affected by the acceleration/switching into System B at all.
    However, since Alice now is in System B, all events not being local are time/position shifted. The clocks in front of Alice, towards e2, display more than 0s. The further away the clocks, the higher the number of the counter will be.
    If Alice had placed clocks behind her as well, those clocks would remain at zero for quite some time, before starting to count, with the furthest away to the back showing the lowest count .
    The clocks in Bob's system A are not parallel to the clocks in the rocket anymore, because everything along the line the clocks have been placed in system A is length contracted seen from alice's new rest frame system B.
    (The clocks behind alice could also vanish for a small period of time, had she placed them right before she accelerates, with the clock furthest behind appearing last)

    After the acceleration, Alice's rocket now in System B will measure 2 Ls still. Alice's clocks seen from Alice's perspective run just as they always did.
    However Alice notices all the clocks she placed into System A are running slower by a factor of ~0.866.
    When for example alice measures 2 seconds with a clock in system B/on her rocket, on a clock in system A only ~1.732 seconds pass.

    Bob experiences the same. When a clock within Bob's system measures 2 seconds, on a clock on alice's system B only ~1.732 seconds passed.

    In short. Both see each other's clocks run slower.

    That seemingly seems to raise a paradox.
    This is not the case however if examined closer. While it is true that both see each other's clocks run slower. - When alice accelerated, all clocks she placed within Bob's system A in front of her are now displaying „the future of Bob“. Alice is basically flying towards the future of Bob, faster than the clocks placed in Bob's frame are running slower now by a factor of ~0.866, could make up for it.
    So the acceleration along with the relative velocity is that which made the difference here.

    As one can see from the Minkowski diagram above, if Alice for example, accelerated (instantaneously) back to Bob's rest frame, System A, when the clock at the back of her rocket displayed ~3.464s, the clock she placed in System A before accelerating parallel to the clock at the back of the rocket, would now show 4 seconds.

    It follows that Alice aged less than Bob.

    An example:

    Let's check the clock Alice placed on the ground at the far end of the blue ruler, into Bob's frame system A, next to her rocket's nose, while the rocket was in Bob's frame still (before accelerating).
    After the acceleration, entering System B, that specific clock displays 1 second. The blue ruler, Alice placed the clocks on along it's line, is length contracted down to 1.732 Ls, seen from Alice’s rest frame System B.
    Alice sees that specific clock coming towards her at a relative speed of v=0.5c. It will take about 1.732Ls/0.5c = 3.464s for the clock to reach Alice at the back of the rocket.
    That clock being in Bob's rest frame however, will be going slower than clocks placed within Alice’s rocket by a factor of about 0.866. So while a clock inside the rocket would count 3.464s, a clock within Bob's frame would count about 3.464s * 0.866 ~ 3s. So the specified clock reaching Alice, would display 4 seconds total.
    If Alice accelerates back to Bob's frame instantaneously at this event, when the specified clock reaches her, then locally the specified clock retains it's current count.
    Alice now in the rest frame of Bob System A, would compare the clocks she initially synced and see that hers, inside the rocket, shows 3.464s while the one which was always in Bob's system would show 4s.

    Alice is now younger than Bob.

    As you can see, it is indeed possible for Alice to see all clocks within Bob's system to go slower and yet Alice not being able to get the clocks within her rocket to show higher digits than the clocks in Bob's frame flying by her at a relative speed of v=0.5c.
    That is, when comparing the counts to clocks in the rocket right at the moment when they are next to each other.

    The two causes for this is the initial acceleration/switch of the reference frame, resulting in all clocks placed in Bob's system, in front of Alice to display higher counts, more the further away in front they are AND the relative speed which has clocks in Bob's frame moving towards Alice at vrel = 0.5c.
    Last edited: Oct 3, 2012
  4. Oct 2, 2012 #3
    How to draw this:

    Draw system A:
    First of course you compose System A. This shouldn't be an issue when given vrel of the rocket and the sizes of the rulers/rocket as done above.

    Draw system B:
    Once system A is done, we have to try and map the events of system A into system B in accord with SR.
    So first we pick an arbitrary point of reference, in this case e1'.
    e1' is the point where Alice accelerates to vrel= 0.5c which puts her and her rocket into system B, her new rest frame.
    This means that ABOVE e1' we should see a green straight line, meaning that the rocket's back is at rest relative to system B.
    Below e1', the green line should have a gradient according to it's speed, being vrel=0,5c relative to B. (Easiest done when drawn on grid-paper. Lightbeams have a 45° gradient, meaning 1 square right/left per square up/down. Vrel=0.5c would be 1 square left/right per 2 squares up/down. Picking vrel = 0.5c makes this easier to draw even without grid paper)

    We also know that the purple ruler's left side will at some point with the rocket's back at e1'. We also said that the purple ruler is always at rest within system B. So we know it will be a straight line which goes through e1'.

    Now we draw the left side and middle of the blue ruler. We know of the left side that it has the same speed with the rocket's back BEFORE the rocket started to accelerate at e1'. We also know that it goes through e1' at some point. So we draw a blue line going through e1' at the same gradient the green line (back of rocket) has before e1'.
    The middle worldline of the ruler is just parallel to the front.
    It is not really important how thick the blue ruler appears when you draw it. What matters are the relations and not the „zoom level“. In other words, you do not have to draw this at the same zoom level as system A but only make sure causality is preserved along with the two postulates of SR.

    Once we have the left side and middle of the ruler, we draw a lightbeam going from the middle side (e4') towards the left, crossing e1' at some point. From the middle of the ruler, this lightbeam started from (e4'), we draw another beam moving top right, crossing the purple ruler's right side.
    We also draw a lightbeam from e1' that moves to the right and crosses the middle (e5'). From this point we draw a lightbeam crossing the purple ruler's right side, coming from bottom and moving top left to cross this point (e5').
    Where the beam moving top right, and the beam coming from bottom right cross, is e2', similar to where e2 is in system A.

    Now we can draw the purple ruler's right side which we know crosses e2' and moves straight up (vrel = 0). We also know that this ruler should be around 2.3Ls long, seen from an observer at rest in system B, as it is 2Ls long seen from an observer at rest in system A (length contraction). We can add this to the drawing.

    We also can draw the blue ruler's right side now, of which we know that it crosses e2' and moves parallel to it's right/middle side (vrel = 0.5c).

    We know that Alice, once she is done accelerating, will measure the blue ruler to be at 1.732 Ls. This is why we know that the length measured from e1' to the right side of the blue ruler should be 1.732 Ls

    The front of the rocket is drawn in as an estimate, going by knowing that Bob will see the rocket shrink down to ~1.732 Ls. The only way I saw to establish this, is by postponing the instantaneous acceleration until the distance between the back and the front is 1.732Ls in system A.
    An arbitrary omniscient observer in System B watching Alice's rocket accelerating into system B to become at rest with system B, will see the front side of the rocket be at rest first, with the back side coming to rest postponed.

    ...to be continued
    Last edited: Oct 2, 2012
  5. Oct 2, 2012 #4
    A few questions:

    1) The light blue line was NOT drawn precisely, and is just an estimate, going by the fact that the rocket will contract after the acceleration down to about 1.732 Ls. As i thought, rightly i believe, the only way for the rocket to contract, is for the front to postpone the instantaneous acceleration until the distance between the green line (back of rocket) and the light blue line (front of rocket) are at a 1.732Ls distance to each other.
    However, i am not 100% sure if i got that right. One of the reasons is that the distance between the back and front is 2 Ls. Imagining the back side starts accelerating the rocket, it would take 2Ls/0.5c = 4sec for the information about the back having accelerated to reach the front. Yet the front accelerates at an estimate of 0.5sec already.
    Does that mean that one would also have to put „accelerators“ all along the rocket, in order to ensure a smooth acceleration?

    2) Can we say that all acceleration does is cause a change of the inertial reference frame on the object it works on? If not, what else does acceleration do which cannot be explained solely with a change of the reference frame?

    3) As far as i understand acceleration, while there are devices named accelerometers, and i hear often the phrase „he feels the acceleration“, acceleration itself cannot really be felt i think. If we could accelerate every part of an object equally within free space, each part of the object would not feel the acceleration. Similar to free fall.

    However there is a twist to this. If I cut a long ruler into several arbitrary small pieces which is at rest in a given inertial reference frame A, and then accelerated each of those pieces with the same acceleration seen from observers which are at rest in A until the ruler reaches a given vrel relative to A, then logically, once those pieces stop accelerating, the total length of the ruler's pieces would be the same still. It would not have length contracted. Is that thinking correct?
    Given above is correct, and given that that each part of the ruler is not special when compared to the other parts, then the only way I see to establish all parts going through the same acceleration process to reach final velocity vrel and yet the ruler parts as a whole length contracting, is to postpone the acceleration process of parts in front, more the further the part is in front.

    All parts of the ruler would go through the same acceleration process, but because of the back part of the rocket being always a bit faster than the front parts, because of it accelerating first seen from system A, the back part would close distance to the front parts until final velocity is reached for all parts.

    4) Again if above is correct, then a force field caused by gravity of a huge mass is not really the perfect field to accelerate a rocket to high speeds, as it does exactly the opposite of what is required. A gravity field would accelerate the front of a rocket stronger than the back. This is because the back of the rocket is further away of the mass generating the gravity field.
    So the gravity field is not really an ideal free fall situation. If I think right, then a very strong gravity field would tear a long rocket apart, even in so called free fall.

    The perfect force field to accelerate the rocket in, would, if above is correct, be some kind of anti-gravity like field. It would be a field which gets weaker the further to the front of the rocket you move. At which rate it would get weaker I did not calculate (therefore anti-gravity LIKE but not the same possibly).
    In such a field, any observer inside the rocket, would not be able to detect any acceleration by “feeling” it.
    That is if we consider infinitesimal small observers which for an infinitesimal small volume of space can consider all their parts being within exactly one inertial reference system.
    Each observer keeps switching the inertial reference system constantly but there is not really any experiment he can do to figure this out WITHIN his current inertial reference system. He does NOT “feel” acceleration.
    Only by observing other objects moving relative to him, NOT within the same inertial reference frame, he would know that others consider him to be accelerating.

    Acceleration should be relative therefore??
    Last edited: Oct 3, 2012
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