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B Some basic questions about length contraction in SR

  1. Jun 9, 2016 #1

    rede96

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    Imagine there are two space ships, Ship A and Ship B. Ship A sees Ship B travelling at 0.865c towards him. (And of course vice versa) From Ship A's FOR there is a marker 9,000,000 km away which is at rest wrt Ship A. The marker, Ship A and the path of Ship B therefore form a straight line.

    When Ship B reaches the marker, it will detach a shuttle that will accelerate to become at rest wrt Ship A, which should coincide with it reaching Ship A exactly, so it can dock safely. Once Ship B dispatches the shuttle, it will accelerate away in a different direction.

    So someone has to work out the acceleration the shuttle must go though so it can be at rest wrt Ship A after the it's travelled the distance from the marker to ship A.

    Without going into the actual calculations, my question are:

    1) What distance is used to work out the rate of acceleration for the shuttle? (As from B's FOR, it will see the distance between the marker and Ship A length contracted by about half. So only 4,500,000 km. But from Ship A's FOR the marker is still 9,000,000 km away.)

    2) If Ship B was travelling at the same speed relative to A but from left to right compared to line of the marker / Ship A, and still dispatched the shuttle when it reached the marker, what distance between the marker and Ship A would used to make the calculation? (I take it as the straight line the marker and Ship A make is no longer in the direction of travel of Ship B, then the distance between the marker and Ship A would not be length contracted?)

    Thanks in advance for any help on this.
     
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  3. Jun 9, 2016 #2

    Ibix

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    You can use either frame as long as you do so consistently - distances, velocities, accelerations and times all measured in that frame. Note that the shuttle's acceleration will not be constant in any inertial frame because the shuttle is changing velocity.
     
  4. Jun 9, 2016 #3

    rede96

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    Ok, thanks. But does that mean Ship A would measure a different velocity for Ship B and vice versa? The reason I am asking is from Ship A's FOR the shuttle has 9,000,000 km to slow down from 0.865 c to 0 but in Ship B's FOR the shuttle has only 4,500,000 km to speed up from 0 to 0.865 c so it can match the speed of ship A.

    That's the part that is confusing me.

    EDIT: Is it because the amount of Fuel / Thrust required to make the manoeuvre is the same measured for both FOR's? E.g. it is the same amount of fuel / thrust to slow down from 0.865c to 0 in 9,000,000 km in A's FOR as it is to speed up from 0 to 0.865c in 4,500,000 km in B's FOR? It's just that different frames will measure different times, velocities etc for the shuttle?
     
    Last edited: Jun 9, 2016
  5. Jun 9, 2016 #4

    Orodruin

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    Well, this depends on if the proper acceleration is assumed constant or not. Obviously it is possible to have a constant coordinate acceleration as well.
     
  6. Jun 9, 2016 #5

    Ibix

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    In A's frame, the shuttle launches at at distance D and decelerates to zero in that distance. In B's frame the shuttle launches at D/2 and accelerates away from ship A, so the rendevous point is not where ship A is at launch time - A has overflown B and meets the shuttle somewhere downrange. So your D/2 is something of a red herring.

    Look up the maths for vessels under constant acceleration and work it out if you want.

    @Orodruin - true, I was assuming constant acceleration there.
     
  7. Jun 9, 2016 #6

    PeterDonis

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    No, it doesn't; it has more distance than that. In Ship B's FOR, ship A and the marker are moving in some direction, call it the positive x direction. The shuttle accelerates in this same direction, increasing speed from zero to the speed of Ship A as Ship A catches up with it. When Ship A finally catches up with the shuttle, the point in space, in Ship B's FOR, at which that happens is not 4,500,000 km from where Ship A started out. That point would be the point where the shuttle originally launched from, but the shuttle itself is moving in the positive x direction, so it will be further along in that direction when Ship A catches up to it.

    In fact, the scenario as you have set it up does not have a unique solution; there are an infinite number of possible solutions, characterized by different possible ways that ship B could accelerate. To get a unique solution, you have to impose an additional constraint. The simplest one would be to specify that the shuttle's proper acceleration is constant--i.e., the crew of the shuttle feels a constant acceleration.
     
  8. Jun 9, 2016 #7

    rede96

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    Ok, got that now thanks.


    If I specify the shuttles proper acceleration is constant, can I assume that the two ships would observe the shuttle travelling the same distance?
     
  9. Jun 9, 2016 #8

    Ibix

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    Maybe this will help. It's a slightly simplified version of your problem - in ship A's frame, the shuttle simply drops instantly to 0.5c and stays at constant velocity until intercept. Here is the Minkowski diagram (simply a displacement-time graph) for the scenario as observed by ship A:
    minkowski1.png
    I've drawn ship A's worldline in red. It stays stationary at the origin. I've shown ship B's worldline in blue - it's moving at ##\sqrt 3c/2##. I've drawn the shuttle in blue, launched at -3 light seconds and moving at a constant 0.5c. The horizontal scale is 1 light second per grid square; the vertical is 1 second per grid square.

    Now here's the same scenario as drawn by ship B:
    minkowski2.png
    The scale is the same (apologies - the axes should be labelled x' and t', but I've only just seen that the primes are missing). In this case, ship A is moving with velocity ##-\sqrt 3c/2##, ship B is stationary at x'=-6. As you can see, at the time of launch, the separation between ships A and B is indeed 1.5 ls, half the value in ship A's frame. However, this is largely irrelevant as the shuttle never crosses this gap - it's moving to the left by about 4.5 ls.

    I realise that the shuttle has moved different distances in the two frames. This is essentially a function of the velocity of the shuttle. I suspect, although I've not done the maths, that it would be equal if the shuttle had been under constant proper acceleration for the whole journey. (As I see you've just asked).
     
  10. Jun 9, 2016 #9

    PeterDonis

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    No. The two ships are still in relative motion, so they still have different notions of distance. Distance is coordinate-dependent; it's not absolute. We've been over this in previous threads.
     
  11. Jun 9, 2016 #10

    PeterDonis

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    I don't think so. Regardless of how the shuttle moves, it has a start event and an end event, and since we're in flat spacetime, there will be a unique invariant spacetime interval between those two events. That interval will split up into a "spatial separation" and a "time separation" in different ways for different inertial frames, and since the "spatial separation" is the distance the OP is looking for, that distance can't be the same in different frames.
     
  12. Jun 9, 2016 #11

    Ibix

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    @PeterDonis - I like your argument in general. But isn't there a special case where ##\Delta x'=-\Delta x##? I have the intuition that the shuttle decelerating from rest in one frame to rest in the other at constant proper acceleration leads to this case, basically because the bottom half of a hyperbolic worldline is the mirror image of the top half.

    I'm going to have to do the maths, aren't I?
     
  13. Jun 9, 2016 #12

    Ibix

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    According to Susskind, the x and t coordinates of an object that passes through the origin undergoing constant proper acceleration ##\alpha##, in the frame where it is instantaneously at rest at the origin, are $$\begin{eqnarray*}
    t(v)&=&\frac 1\alpha\frac v{\sqrt{1-v^2}}\\
    x(t)&=&\frac 1\alpha\left(\sqrt{1+\alpha^2t^2}-1\right)
    \end{eqnarray*}$$In this case we are interested in the coordinates when it is moving with velocity v, so we can simplify the expression for t and substitute it into the expression for x to get:$$\begin{eqnarray*}
    t(v)&=&\frac {\gamma v}\alpha\\
    x(v)&=&\frac {\gamma-1}\alpha
    \end{eqnarray*}$$We can then run that through the LT to get that the final position in the frame moving at speed v, where the object started in motion and decelerated to rest:$$\begin{eqnarray*}
    x'&=&\gamma\left(\frac{\gamma-1}\alpha-\frac{\gamma v^2}\alpha\right)\\
    &=&\frac{1-\gamma}\alpha\\
    &=&-x\end{eqnarray*}$$
    So rede96 is correct in this case because it's a highly symmetrical special case.
     
  14. Jun 9, 2016 #13

    PeterDonis

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    Yes, but there are still complications. You have two worldlines that need to be transformed, not just one.

    Also, we have been using the term "distance traveled by the shuttle" inconsistently in this discussion. You are using it to mean the difference in ##x## coordinate between where the shuttle starts and where it ends. But the OP was using it to mean the difference in ##x## coordinate between where the shuttle is when it starts, and where ship A is when the shuttle starts. The OP was thinking of that as "the distance the shuttle has to cover". See further comments below.

    In the unprimed frame, ship A is moving at speed ##v## in the positive ##x## direction. Since it is at ##x(v)## at time ##t(v)##, then at time ##t = 0##, when the shuttle starts out, ship A will be at (using capital letters for ship A's coordinates)

    $$
    X(0) = x(v) - v t(v) = \frac{\gamma - 1 - \gamma v^2}{\alpha} = \frac{1}{\alpha} \left( \frac{1}{\gamma} - 1 \right)
    $$

    This is the distance that the shuttle is from ship A at time ##t = 0##, when it starts out from rest in this frame. So we have two possible meanings of "the distance the shuttle has to cover". One is the distance ##x(v)## that you calculated. The second is the distance ##X(0)## that I just calculated. And the way the OP was thinking about the scenario, the second distance is the one he was implicitly assuming.

    Now let's look at this in the primed frame. In this frame, the shuttle starts out moving at speed ##-v## at time ##t' = 0##, and ends up at rest. You have shown that the shuttle's ##x'## coordinate at this event is ##- x(v)##, and it's easy to show that the shuttle's ##t'## coordinate at this event is equal to ##t(v)##. But now, where is ship A in this frame? In this frame, ship A is at rest at ##X' = x'(v) = -x##. So in this frame, there is only one possible meaning to "the distance the shuttle has to cover", namely, the distance ##x'##. That is the distance the shuttle travels in this frame, and it is also the distance between the shuttle and ship A, in this frame, when the shuttle starts out.

    You are correct that, for this special case, we have ##x' = - x## (and ##t' = t##) for the shuttle. But we do not have that the distance from ship A to the shuttle, when the shuttle starts out, is the same in both frames, i.e., we do not have ##X'(0) = X(0)##.
     
  15. Jun 10, 2016 #14

    Ibix

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    @PeterDonis - I agree with your maths, but had read the OP's #7 as conceding that the initial question was flawed and asking a revised question, which I think I answered. However, I've now connected this thread to the proper distance thread from the other day, so a clarification might be in order.

    @rede96 - In ship A's frame the shuttle launches as ship B passes the marker, 9,000,000km from ship A, and crosses the 9,000,000km decelerating steadily to rest.

    In ship B's frame the shuttle launches as the marker passes ship B, 4,500,000km ahead of ship A. It accelerates in the direction away from ship A, eventually matching position and velocity with it 9,000,000km from ship B.

    This is basically drawn in my diagrams, the only corrections being that the purple line should be a hyperbola and that the launch-to-ship-A distance in the first diagram should equal the ship-B-to-intercept distance in the second.

    The important thing to note is that these 9,000,000km distances are not measuring the same thing. In ship A's frame it is the distance from the marker to ship A. In ship B's frame it is the distance from ship B to the intercept point. Neither of these distances is 9,000,000km in the other frame - in fact, the second one isn't unambiguously defined in ship A's frame.

    Finally getting back to mission planning, one could work in either frame. In either frame the coordinate distance between marker and ship A is a relevant datum. In the rest frame of ship A that happens to be the proper distance between the two, and that is the only frame where the coordinate distance is the same as the distance travelled by the shuttle.

    I'm inclined to think that this whole "proper distance" concept is unhelpful because, at least for this kind of scenario, it's irrelevant to practical calculations. Proper time is useful because it's a "distance travelled" measure along time-like world lines. I'm not sure how often one needs the equivalent concept of "distance travelled" along space-like world lines.
     
  16. Jun 10, 2016 #15

    rede96

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    Thanks, yes got this now.

    Again, understood now thanks. Although not sure how B would make the calculation, as the distance covered by the shuttle still has to be 9,000,000 km to get to the intercept point but there is natural reference as there is in A's frame. So just wasn't sure on the Math.

    Yes, as regards my other posts I can see now that what all I was doing was just stating if you measure an object (which I was using to illustrate a distance between two bodies moving wrt each other) in a rest frame then the length of that object will always be measured to be the same. Which is correct, however I realise now that the distance I was representing by the object doesn't correspond to a proper distance, as there is no absolute rest frame, which was the unwitting assumption I was making.

    Thanks to all for your help. :-)
     
  17. Jun 11, 2016 #16

    Ibix

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    It's just an intercept calculation. In Newtonian physics, B would declare that ship A is at ##x_A=x_0-v_At## moving at speed ##v_A## and the shuttle is at ##x_s=-\alpha t^2/2## moving at speed ##v_s=-\alpha t## and accelerating at constant acceleration ##\alpha##. At intercept you require ##x_A=x_s## and ##v_A=v_s## and solve for ##\alpha## (modulo possible sign errors).

    In our relativistic form the only difference is that you have a slightly more complex expression for the position of the shuttle. See the third and fourth equations in #12 for the relationship between ##x##, ##t##, and ##v##. Solve as in the last paragraph.

    Actually this is a special case of a slightly more complex equation. You can set up the maths in a similar way for an arbitrary frame where both A and B are moving. Since ##\alpha## is a proper acceleration the solution will always be the same. You will find it simplifies to the calculation above in the case where B's velocity is zero, and to the distance-travelled-is-distance-between-ships-at-launch form in the case where A's velocity is zero.

    You're welcome. The most useful tool for SR for me has been the spacetime diagram, aka Minkowski diagram, as drawn in #8. They gave me an intuition for what's going on - and all they are really is displacement-time graphs.
     
    Last edited: Jun 11, 2016
  18. Jun 11, 2016 #17

    Dale

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    @rede96 I encourage you to pay close attention to this comment by Ibix

    A point object at rest is represented as a line parallel to the time axis. A point in motion is represented by a diagonal line. Two points at rest relative to each other are represented by parallel lines. There is a unique notion of the distance between two parallel lines, the length of a perpendicular segment from one line to the other, this corresponds to the proper distance. The length of a non perpendicular segment is different from a perpendicular segment, this corresponds to length contraction. There is no unique notion of distance between non parallel lines. Etc.
     
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