Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic Rocket Question

  1. Jun 8, 2006 #1
    If I'm on a rocket accelerating away from earth, my velocity relative to earth at any given time will be: [tex]v = \frac{at}{\sqrt{1+(\frac{at}{c})^2}}[/tex]

    This makes sense, since my velocity relative to earth will be less than the newtonian [tex]\Delta v = a \Delta t [/tex], due to relativistic effects.

    But how about later, if I decelerate relative to earth? Obviously, I can't use the same equation, since my change in velocity relative to earth will now be greater than [tex]\Delta v = a \Delta t[/tex].

    Is there a similar equation I can use to calculate my velocity relative to earth at any given time during deceleration?

    Thanks,
    Alan
     
  2. jcsd
  3. Jun 8, 2006 #2

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your velocity will be
    [tex]v = v_0-\frac{at}{\sqrt{1+(\frac{at}{c})^2}}[/tex],
    where v_0 is the velocity at the start of the deceleration, and a is the magnitude of the acceleration. There is an a^2 in the denom, so that sign doesn't matter.
     
  4. Jun 8, 2006 #3

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor


    The easiest approach, assuming a constant deacceleration, is to pick the point at which you will stop as a reference.

    Let (tstop, xstop) be the coordinates where you come to a stop.

    Then (correction) v(t) = -a(t-tstop)/sqrt(1+(a^2(t-tstop)^2)

    Thus at t=tstop, v=0
    and at t < tstop, v > 0

    You are basically time reversing the problem, and accelerating away from your stopping point.

    There are a few things about unformly accelerated notion that are extremely useful to know:

    1) all such motion is hyperbolic, i.e for the correct choice of origin x^2 - t^2 = constant
    (x^2 - c^2t^2 = constant if c is not equal to 1).

    for example: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

    (d + c^2/a) = (c^2/a) cosh(aT/c)
    t = (c/a) sinh(aT/c)

    thus (d+c^2/a)^2 - (ct)^2 = (c^2/a)^2

    as cosh^2 - sinh^2 =1

    [add]
    See also, for instance
    http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

    2) the "lines of simultaneity" for a uniformly accelerated observer all pass through one point (the origin of the coordinate system, if the choice of the origin is made as in #1)

    i.e. in the example above, all liens of simultaneity run through the point
    x = -c^2/a; t=0

    [add] Here is a proof of the later remark: assume c=1, then

    x^2 - t^2 = constant

    By taking the derivative, we get
    2*x*dx - 2*t*dt = 0

    therfore v = dx/dt = t/x. The slope of the line of simultaneity is 1/v, i.e. x/t. Thus a line from (0,0) to (x,t) is the line of simultaneity at (x,t) because it passes through (x,t) and it has the correct slope: dx/dt = x/t.
     
    Last edited: Jun 8, 2006
  5. Jun 8, 2006 #4

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I've corrected one typo and added some proofs and more references to my previous post, the added content is significant enough that I thought I'd make a note of it.
     
  6. Jun 9, 2006 #5
    What I was really looking for here was an equation for the ship's deceleration relative to earth's apparent position. This would obviously not be the same as the ship's deceleration relative to its destination.

    Thanks,
    Alan
     
  7. Jun 9, 2006 #6

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    The deacceleration relative to Earth will be the same as the deacceleration relative to its destination, as long as inertial coordinate systems are used.

    It's a simple space translation.

    Suppose the destination is a distance L away from the Earth.

    Then the coordiantes relative to the earth will be (t,x) and the coordinates relative to the destination will be (t,x-L).

    Thus when x=L, the ship is at its destination. The acceleration of the ship (d^2x/dt^2) will be the same relative to the earth as it will be to the destination, because

    d^2(x-L)/dt^2 = d^2x/dt^2

    given that L=constant.

    This simple "space translation" works the same in SR as it does in non-relativistic mechanics.

    Time translation also works in the same way.

    Note that we can also formulate the equations of motion using the principle of hyperbolic motion that I mentioned.

    Let the destinaton be at (tstop, xstop), and let c=1.

    then for x < (xstop/2), the hyperbolic point is at (t=0,x=-1/a) and

    (x+1/a)^2 - t^2 = 1/a^2

    and for x > xstop/2, the hyperbolic point is at (t=tstop, x=xstop+1/a)

    (x-(xstop + 1/a))^2 - (t-tstop)^2 = 1/a^2

    We can re-write the second expression to solve for x (using Maple). There are two solutions, one of which is:

    [tex]
    {\frac {{\it xstop}\,a+1-\sqrt {1+{a}^{2}{t}^{2}-2\,{a}^{2}t{\it tstop
    }+{a}^{2}{{\it tstop}}^{2}}}{a}}

    [/tex]

    the derivative of this expression is (agian using Maple)

    [tex]
    {\frac {a \left( t-{\it tstop} \right) }{\sqrt {1+{a}^{2}{t}^{2}-2\,{a
    }^{2}t{\it tstop}+{a}^{2}{{\it tstop}}^{2}}}}
    [/tex]

    the same as the expresssion I posted earlier.

    See the Wikipedia link for the hyperbolic equation (you may have to manually add in the trailing ')' to the URL, for some reason the board omits this symbol and wiki doesn't find the page.

    http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

    You can confirm that the velocity at t=tstop/2 is the same in both expressions:
     
    Last edited: Jun 9, 2006
  8. Jun 9, 2006 #7
    pervect,

    If the distance L refers to the distance between earth and the destination, according to the ship's frame, then L will not remain constant. It will get larger as the ship decelerates. Just like L got smaller as the ship accelerated away from earth. My understanding is that the relativistic rocket equation takes this into account, which is why the final velocity relative to earth (after acceleration) is less than it would be if we didn't take length contraction into account, assuming a specified acceleration and time. In other words, velocity is less than a*t.

    So, is there an equation that relates to a ship's deceleration relative to earth's apparent position, that takes this length expansion (or de-contraction) into account?

    Thanks,
    Alan
     
  9. Jun 9, 2006 #8

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    The coordinates (x,t) are the inertial coordinates of the ship in the Earth frame.

    The destination is a constant distance away in the Earth coordiantes.

    Look for instance at

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

    (I assume this is where you initally got your equations?)

    Thus d and t are in the Earth coordinate system. T is in the rocket's coordinate system (proper time). The distance to the destination according to the rocket is not given on the webpage.
     
    Last edited: Jun 9, 2006
  10. Jun 9, 2006 #9

    JesseM

    User Avatar
    Science Advisor

    The distance L is measured in an inertial frame, the frame where the rocket was at rest until it began to accelerate (the rest frame of the earth). You seem to be thinking it's the distance as seen in some non-inertial frame of the ship, or in the ship's instantaneous co-moving inertial rest frame, but that's not correct, the relativistic rocket equation describes the velocity as seen in a single inertial frame.
    No, the reason it's less as seen in an inertial frame is because "a" refers to the acceleration experienced by passengers on the ship--constant "a" means that they feel a constant G-force, which means that their acceleration in their instantaneous co-moving inertial rest frame is the same from one moment to another. Because of the way velocities transform in relativity, this would mean that the rate their coordinate velocity increases (their coordinate acceleration, which is different from 'a') as seen in a single inertial frame is constantly decreasing.
    What do you mean by "apparent"? Do you mean the distance as seen in the ship's instantaneous co-moving inertial frame?
     
    Last edited: Jun 9, 2006
  11. Jun 10, 2006 #10
    OK, then how would the ship's crew calculate their change in velocity? I assume they would have to take into account that as they accelerate, gamma would increase, and maybe we could picture the change in earth's apparent position as a kind of velocity, in the same direction of the ship's velocity. Since the distance from earth, as measured by the ship is increasing due to acceleration, but increasing less due to length contraction.

    Yes, this is what I meant. I'm looking for how to calculate everything from the ship's point of view. And I know the ship's crew cannot observe anything in real time, but they should be able to figure out their velocity, (coordinate) acceleration, and (apparent) distance from earth at any given time. Including during their deceleration. And I'm looking for how they would calculate this, not how they would observe it directly.

    Thanks,
    Alan
     
  12. Jun 10, 2006 #11

    JesseM

    User Avatar
    Science Advisor

    "Velocity" in what coordinate system? If you want to look at a non-inertial coordinate system where they are at rest, then of course their velocity is always zero in this system, although you could calculate the way that the earth's velocity is changing in this coordinate system. But as I've said before, it's not like there's a single non-inertial coordinate system that counts as the "frame" of a non-inertial observer, you could construct such a coordinate system in a variety of ways, I don't think there's any strong physical motivation for preferring one over all the others.
    Again, I think you're confusing the issue by using commonsense words like "observe" and "point of view" and "apparent distance" for what is measured in a particular non-inertial coordinate system. Do you think there is any reason to say that this particular type of non-inertial coordinate system--one where the ship's definitions of distance and simultaneity at any given moment match those of their instantaneous co-moving inertial rest frame--represent's the ship's "point of view" in a way that some other non-inertial coordinate system with different definitions of simultaneity and distance does not?
    You only put (coordinate) before acceleration, but you really should put it before velocity and distance from earth as well; what you are really asking is for the value of these things in a particular non-inertial coordinate system (one whose definition of distance and simultaneity at each moment matches that of the ship's instantaneous co-moving inertial rest frame), but there is no reason to say that the coordinate distance of the earth in this non-inertial frame has any more right to be called the "apparent distance" than the coordinate distance in some other frame.

    Anyway, like I said earlier, in the coordinate system that I assume you're asking about, their own coordinate velocity and acceleration would be zero, because they're at rest at all moments in this coordinate system. But you can certainly ask about the coordinate velocity, acceleration and position of the earth in this coordinate system. If you can figure out the coordinate position of the earth as a function of time in this system, you could obtain the others in the normal way, by taking derivatives with respect to coordinate time. The calculation of distance as a function of time in this coordinate system seems like it'd be a bit tricky, although the formulas on the relativistic rocket page help...I'll think about it and get back to you.
     
  13. Jun 10, 2006 #12

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    To compute or measure in the rockets coordinate system the same velocity that the Earth observer measures, the rocket needs to measure its velocity relative to an object that is stationary with respect to the Earth, but at the same location as the rocket.

    "Stationary with respect to the Earth" is really defined in the Earth's frame of referece, not the rockets frame, however. There is a way to do this from the rocket's coordiante system, but it is complicated. It also requires making use of special features of the geomtery. In another post, I think I gave a longish quote about why there is no general notion of "relative velocity" of a distant object, so it should not be a surprise that finding the relative velocity of the rocket to the Earth requires us to make use of some "special features" of the geometry (namely, the fact that space-time is flat in this example).

    The advanced way of doing this from the rocket's coordinate system: the rocket "parallel transports" the velocity vector of the Earth along some path to the rocket's current location. The rocket then finds the magnitude of the resulting "parallel transported" velocity to answer the question.

    In general, this procedure of parallel transporting would give a result that depended on the exact path over which the vector was transported. But, becuase space-time in this example is flat, this process of parallel transport will give the same results, regardless of the path taken.

    To avoid the process of parallel transport, the rocket could make use of (for example) an "Earth outpost" that knows it is stationary with respect to the Earth because the outpost has been exchanging light signals with the Earth for a long time, without noticing any change in the "propagation delay".

    This "Earth outpost" appraoch is much simpler, though I don't know if it will satisfy Al68.

    If the rocket measures its velocity V with respect to the Earths' outpost, it will measure the same velocity V that the Earth (and the Earth's outpost) measures for the rocket.

    A full treatment of parallel transport requires some knowledge of tensors, which I do not believe that AL68 has.

    A somewhat popular reference:

    http://math.ucr.edu/home/baez/gr/parallel.transport.html

    Note that when you parallel transport a vector around a closed path on the Earth's surface, as in the above example, it changes direction.

    It is only when the geometry is fundamentally flat that parallel transporting a vector around a closed path does not change it's direction.

    Because parallel transporting a vector around a closed path on a flat plane does not change its direction, the result of parallel transporting a vector along a path cannot depend on the path.

    Schild's ladder would be an ideal way of explaining parallel transport, but the only reference I have which talks about it is MTW's "Gravitation", and I haven't found much on the WWW, either.

    "Gravitation" is a book that is written at the graduate level, but has largish bits that may be comprehensible to someone who is willing to ignore the rest of the book (which won't be comprehensible without graduate level physics).
     
    Last edited: Jun 10, 2006
  14. Jun 10, 2006 #13
    Jesse,

    I think your last paragraph states my question more clearly than I did. I wanted to figure out the coordinate position of earth as a function of time in the (non inertial) system in which the ship is at rest. And I agree with you that it would be tricky. That's why I was asking for help.

    pervect,

    You are correct about my knowledge of tensors. I can barely spell tensors. And I worded my question poorly, Jesse worded it better in his last paragraph.

    Thanks,
    Alan
     
  15. Jun 10, 2006 #14

    JesseM

    User Avatar
    Science Advisor

    Alan, if I'm understanding you right you want to know how things would look in a particular noninertial coordinate system (call it NI) which has the following properties:

    1. If a given tick of the ship's clock is simultaneous with a distant event A in the co-moving inertial rest frame of the ship at the moment of the tick, then A should also be simultaneous with that tick in NI. Also, if D is the distance between the ship at the moment of the tick and A in the co-moving inertial rest frame of the ship at that moment, then the coordinate distance between A and the ship at the moment of the tick should also be D in NI.

    2. The time-coordinate assigned to an event on the ship's worldine by NI should be identical to the readings of clocks on the ship at the moment of the event (ie the proper time along the ship's worldine between that event and the event of leaving earth).

    I think these conditions should be enough to uniquely specify the NI coordinate system. Note, though, that this will not be a well-behaved coordinate system throughout spacetime--as the lines of simultaneity change angles at different points along the ship's worldline, then there will be points where different lines of simultaneity cross each other, and events at or beyond the crossings will be assigned two or more sets of coordinates by NI. But as long as you consider only the regions of spacetime near enough to the ship's worldline that events within those regions are assigned unique coordinates, it should be OK (although I don't know how to calculate the borders of this region).

    Nevertheless, I would say that you are incorrect to refer to this as "the (non inertial) system in which the ship is at rest"--as I keep emphasizing, over and over, and you seem to want to ignore, there is no single standard way to define the coordinate sytem of a non-inertial observer, you could certainly come up with other non-inertial coordinate systems in which the ship is at rest that do not have the properties I described for NI above, and there would be no compelling reason to treat NI as having more claim to represent the ship's "point of view" than any of these others. Please tell me, do you understand and agree with this? (and if so, will you quit using phrases like 'apparent distance' and 'the ship's point of view' to describe how things work in the NI coordinate system?)

    Anyway, to figure out the earth's distance D as a function of time T within the coordinate system NI, I think the first thing to do would be to figure out D as a function of d and v, the distance and velocity of the ship in the earth's frame. You might think that D would just be [tex]d * \sqrt{1 - v^2/c^2}[/tex] due to Lorentz contraction, but that isn't right, the distance isn't like the length of a rigid object, it's constantly changing so you have to worry about simultaneity issues. (edit: I was wrong about that, see pervect's next post and my response...the distance is like the length of a rigid object, just imagine that there was a rod attached to the earth of length d in the earth's rest frame, and when the ship reaches the end of it, both frames agree that at that moment the length of the rod in their frame is equal to the ship's distance from the earth in their frame.) So let's consider a simple inertial case where the ship is moving away from the earth at constant velocity v instead of accelerating. In the earth's frame, the ship will be at distance d at time d/v, and since in this frame the ship's clocks are running slow by a factor of [tex]\sqrt{1 - v^2/c^2}[/tex], that means the ship's clock will only read [tex](d/v)*\sqrt{1 - v^2/c^2}[/tex] at this moment. But in the ship's frame, at the moment its own clock reads this time of [tex](d/v)*\sqrt{1 - v^2/c^2}[/tex], it is the earth's clock that reads less than this because it has been slow by a factor of [tex]\sqrt{1 - v^2/c^2}[/tex] since they departed, so according to the definition of simultaneity in the ship's frame the earth's clock reads [tex](d/v)*(1 - v^2/c^2)[/tex] at this moment. And since in the earth's frame the earth is also moving away at the same speed of v (the speeds with which two inertial observers see the other moving are always reciprocal), the earth's distance at this moment can be found by multiplying velocity and time, giving [tex]v*(d/v)*(1 - v^2/c^2)[/tex], which gives [tex]d*(1 - v^2/c^2)[/tex]. So, at the point on the ship's worldline where the distance in the earth's frame is d, the distance in the ship's frame is [tex]d*(1 - v^2/c^2)[/tex].

    Now, this conclusion shouldn't be altered if we are talking about the instantaneous co-moving inertial rest frame of an accelerating observer rather than the constant inertial rest frame of an inertial ship. After all, we could imagine that we have both an inertial ship and an accelerating ship moving away from the earth, and that they cross each other's paths at the moment that they are at a distance d in the earth's frame (they must have departed earth at different times for this to work); at the point in spacetime where they cross, they both have the same instantaneous inertial rest frame, so they would both measure the same distance to the earth in this instantaneous frame.

    So, we know that at the point on the ship's worldline that corresponds to time t in the earth's frame, when the distance in the earth's frame is d, the NI system should say the distance of the earth at that same point on the ship's worldline is [tex]d*(1 - v^2/c^2)[/tex]. And from the relativistic rocket page we have an expression for d as a function of a and t: [tex]d = (c^2/a)*(\sqrt{1 + (at/c)^2} - 1)[/tex]. So this means that the distance D in NI as a function of time t in the earth's frame is [tex]D = (1 - v^2/c^2)*(c^2/a)*(\sqrt{1 + (at/c)^2} - 1)[/tex]. And the page also gives us a function for the time t as measured in the earth's frame as a function of the time T as measured by clocks on the ship: [tex]t = (c/a)*sh(aT/c)[/tex], where the function sh x is the hyperbolic sine function (also called sinh -- you can find it on most calculators). So, substituting this into the formula for D as a function of t gives D as a function of T:

    [tex]D = (1 - v^2/c^2)*(c^2/a)*(\sqrt{1 + (a/c)^2 * ((c/a)*sh(aT/c))^2} - 1)[/tex]

    simplifying a bit,

    [tex]D = ((c^2 - v^2)/a)*(\sqrt{1 + (sh(aT/c))^2} - 1)[/tex]

    and this page mentions there's a hyperbolic function identity that says (ch(x))^2 = 1 + (sh(x))^2, so this simplifies to:

    [tex]D = ((c^2 - v^2)/a)*(ch(aT/c) - 1)[/tex]

    There's a good chance I made an error somewhere along the line, but if not, this should be the formula for the coordinate distance of the earth D as a function of coordinate time T in the NI coordinate system. To find the coordinate velocity of earth you'd take the first derivative of D with respect to T, while to find the coordinate acceleration of earth you'd take the second derivative. The page with the hyperbolic function identities above says that d sh(x) / dx = ch(x) and d ch(x) / dx = sh(x). So, taking these derivatives is just a matter of using the chain rule of calculus:

    [tex]dD/dT = ((c^2 - v^2)/a)*(sh(aT/c) * (a/c))[/tex]

    or

    [tex]dD/dT = ((c^2 - v^2)/c)*sh(aT/c)[/tex]

    Taking another derivative gives:

    [tex]d^2 D /dT^2 = ((c^2 - v^2)/c)*(ch(aT/c) * (a/c))[/tex]

    or

    [tex]d^2 D /dT^2 = (a*(c^2 - v^2)/c^2)*ch(aT/c)[/tex]

    Again, buyer beware, but if I haven't made any errors these should be the formulas for the coordinate velocity and coordinate acceleration of the earth in the NI coordinate system.
     
    Last edited: Jun 11, 2006
  16. Jun 10, 2006 #15

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm think that the distance in the ship's coordinate system should be just sqrt(1-v^2/c^2) times the distance in the Earth's coordinate system.

    Basically, the angle betweent the line of simultaneity and the horizontal 'x' axis is just the rapidity, theta, therfore the length in the Earth frame divided by the length in the instantaneous ship frame is just cosh(theta).

    Another way of putting this: letting c=1

    The slope of the "line of simultaneity" is 1/v , thus delta-t = v*delta-x

    If the Earth is at the origin of the coordinate system and the spaceship is at (t,d) in the Earth coordinate system

    The point on the Earth's trajectory (t=*, x=0) simultaneous with (t,x) in the ship frame is just

    (t-v*d,0)

    The lorentz interval betweent these points is d^2 - (v*d)^2

    The distance is the square root of the Lorentz interval, i.e d*sqrt(1-v^2)

    The following diagram might have helped, if it came out :-(. (I was trying gnuplot this time around). I'll keep it, in case I can get it to work.

    [tex]
    \]
    \begin{picture}(1500,900)(0,0)
    \setlength{\unitlength}{0.240900pt}
    \sbox{\plotpoint}{\rule[-0.200pt]{0.400pt}{0.400pt}}%
    \put(100.0,82.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(80,82){\makebox(0,0)[r]{ 0}}
    \put(1419.0,82.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(100.0,238.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(80,238){\makebox(0,0)[r]{ 1}}
    \put(1419.0,238.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(100.0,393.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(80,393){\makebox(0,0)[r]{ 2}}
    \put(1419.0,393.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(100.0,549.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(80,549){\makebox(0,0)[r]{ 3}}
    \put(1419.0,549.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(100.0,704.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(80,704){\makebox(0,0)[r]{ 4}}
    \put(1419.0,704.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(100.0,860.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(80,860){\makebox(0,0)[r]{ 5}}
    \put(1419.0,860.0){\rule[-0.200pt]{4.818pt}{0.400pt}}
    \put(100.0,82.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(100,41){\makebox(0,0){ 0}}
    \put(100.0,840.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(368.0,82.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(368,41){\makebox(0,0){ 1}}
    \put(368.0,840.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(636.0,82.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(636,41){\makebox(0,0){ 2}}
    \put(636.0,840.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(903.0,82.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(903,41){\makebox(0,0){ 3}}
    \put(903.0,840.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(1171.0,82.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(1171,41){\makebox(0,0){ 4}}
    \put(1171.0,840.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(1439.0,82.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(1439,41){\makebox(0,0){ 5}}
    \put(1439.0,840.0){\rule[-0.200pt]{0.400pt}{4.818pt}}
    \put(100.0,82.0){\rule[-0.200pt]{322.565pt}{0.400pt}}
    \put(1439.0,82.0){\rule[-0.200pt]{0.400pt}{187.420pt}}
    \put(100.0,860.0){\rule[-0.200pt]{322.565pt}{0.400pt}}
    \put(100.0,82.0){\rule[-0.200pt]{0.400pt}{187.420pt}}
    \put(1279,820){\makebox(0,0)[r]{sqrt(x*x-1)}}
    \put(1299.0,820.0){\rule[-0.200pt]{24.090pt}{0.400pt}}
    \put(379,127){\usebox{\plotpoint}}
    \multiput(379.58,127.00)(0.499,0.643){109}{\rule{0.120pt}{0.614pt}}
    \multiput(378.17,127.00)(56.000,70.725){2}{\rule{0.400pt}{0.307pt}}
    \multiput(435.00,199.58)(0.583,0.498){93}{\rule{0.567pt}{0.120pt}}
    \multiput(435.00,198.17)(54.824,48.000){2}{\rule{0.283pt}{0.400pt}}
    \multiput(491.00,247.58)(0.655,0.498){81}{\rule{0.624pt}{0.120pt}}
    \multiput(491.00,246.17)(53.705,42.000){2}{\rule{0.312pt}{0.400pt}}
    \multiput(546.00,289.58)(0.701,0.498){77}{\rule{0.660pt}{0.120pt}}
    \multiput(546.00,288.17)(54.630,40.000){2}{\rule{0.330pt}{0.400pt}}
    \multiput(602.00,329.58)(0.758,0.498){71}{\rule{0.705pt}{0.120pt}}
    \multiput(602.00,328.17)(54.536,37.000){2}{\rule{0.353pt}{0.400pt}}
    \multiput(658.00,366.58)(0.758,0.498){71}{\rule{0.705pt}{0.120pt}}
    \multiput(658.00,365.17)(54.536,37.000){2}{\rule{0.353pt}{0.400pt}}
    \multiput(714.00,403.58)(0.779,0.498){69}{\rule{0.722pt}{0.120pt}}
    \multiput(714.00,402.17)(54.501,36.000){2}{\rule{0.361pt}{0.400pt}}
    \multiput(770.00,439.58)(0.788,0.498){67}{\rule{0.729pt}{0.120pt}}
    \multiput(770.00,438.17)(53.488,35.000){2}{\rule{0.364pt}{0.400pt}}
    \multiput(825.00,474.58)(0.826,0.498){65}{\rule{0.759pt}{0.120pt}}
    \multiput(825.00,473.17)(54.425,34.000){2}{\rule{0.379pt}{0.400pt}}
    \multiput(881.00,508.58)(0.802,0.498){67}{\rule{0.740pt}{0.120pt}}
    \multiput(881.00,507.17)(54.464,35.000){2}{\rule{0.370pt}{0.400pt}}
    \multiput(937.00,543.58)(0.826,0.498){65}{\rule{0.759pt}{0.120pt}}
    \multiput(937.00,542.17)(54.425,34.000){2}{\rule{0.379pt}{0.400pt}}
    \multiput(993.00,577.58)(0.811,0.498){65}{\rule{0.747pt}{0.120pt}}
    \multiput(993.00,576.17)(53.449,34.000){2}{\rule{0.374pt}{0.400pt}}
    \multiput(1048.00,611.58)(0.851,0.497){63}{\rule{0.779pt}{0.120pt}}
    \multiput(1048.00,610.17)(54.384,33.000){2}{\rule{0.389pt}{0.400pt}}
    \multiput(1104.00,644.58)(0.826,0.498){65}{\rule{0.759pt}{0.120pt}}
    \multiput(1104.00,643.17)(54.425,34.000){2}{\rule{0.379pt}{0.400pt}}
    \multiput(1160.00,678.58)(0.851,0.497){63}{\rule{0.779pt}{0.120pt}}
    \multiput(1160.00,677.17)(54.384,33.000){2}{\rule{0.389pt}{0.400pt}}
    \multiput(1216.00,711.58)(0.826,0.498){65}{\rule{0.759pt}{0.120pt}}
    \multiput(1216.00,710.17)(54.425,34.000){2}{\rule{0.379pt}{0.400pt}}
    \multiput(1272.00,745.58)(0.836,0.497){63}{\rule{0.767pt}{0.120pt}}
    \multiput(1272.00,744.17)(53.409,33.000){2}{\rule{0.383pt}{0.400pt}}
    \multiput(1327.00,778.58)(0.851,0.497){63}{\rule{0.779pt}{0.120pt}}
    \multiput(1327.00,777.17)(54.384,33.000){2}{\rule{0.389pt}{0.400pt}}
    \multiput(1383.00,811.58)(0.851,0.497){63}{\rule{0.779pt}{0.120pt}}
    \multiput(1383.00,810.17)(54.384,33.000){2}{\rule{0.389pt}{0.400pt}}
    \put(1279,779){\makebox(0,0)[r]{.866*x}}
    \multiput(1299,779)(20.756,0.000){5}{\usebox{\plotpoint}}
    \put(1399,779){\usebox{\plotpoint}}
    \put(100,82){\usebox{\plotpoint}}
    \multiput(100,82)(18.564,9.282){4}{\usebox{\plotpoint}}
    \multiput(156,110)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(212,138)(18.497,9.416){3}{\usebox{\plotpoint}}
    \multiput(267,166)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(323,194)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(379,222)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(435,250)(18.431,9.545){3}{\usebox{\plotpoint}}
    \multiput(491,279)(18.497,9.416){3}{\usebox{\plotpoint}}
    \multiput(546,307)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(602,335)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(658,363)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(714,391)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(770,419)(18.497,9.416){3}{\usebox{\plotpoint}}
    \multiput(825,447)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(881,475)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(937,503)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(993,531)(18.497,9.416){3}{\usebox{\plotpoint}}
    \multiput(1048,559)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(1104,587)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(1160,615)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(1216,643)(18.431,9.545){3}{\usebox{\plotpoint}}
    \multiput(1272,672)(18.497,9.416){3}{\usebox{\plotpoint}}
    \multiput(1327,700)(18.564,9.282){3}{\usebox{\plotpoint}}
    \multiput(1383,728)(18.564,9.282){3}{\usebox{\plotpoint}}
    \put(1439,756){\usebox{\plotpoint}}
    \sbox{\plotpoint}{\rule[-0.400pt]{0.800pt}{0.800pt}}%
    \sbox{\plotpoint}{\rule[-0.200pt]{0.400pt}{0.400pt}}%
    \put(1279,738){\makebox(0,0)[r]{.866}}
    \sbox{\plotpoint}{\rule[-0.400pt]{0.800pt}{0.800pt}}%
    \put(1299.0,738.0){\rule[-0.400pt]{24.090pt}{0.800pt}}
    \put(100,217){\usebox{\plotpoint}}
    \put(100.0,217.0){\rule[-0.400pt]{322.565pt}{0.800pt}}
    \sbox{\plotpoint}{\rule[-0.200pt]{0.400pt}{0.400pt}}%
    \put(100.0,82.0){\rule[-0.200pt]{322.565pt}{0.400pt}}
    \put(1439.0,82.0){\rule[-0.200pt]{0.400pt}{187.420pt}}
    \put(100.0,860.0){\rule[-0.200pt]{322.565pt}{0.400pt}}
    \put(100.0,82.0){\rule[-0.200pt]{0.400pt}{187.420pt}}
    \end{picture}
    \[
    [\tex]
     
    Last edited: Jun 11, 2006
  17. Jun 10, 2006 #16

    JesseM

    User Avatar
    Science Advisor

    Yes, I think you're correct. I confused myself by thinking about what the earth's clock would read in the ship's frame at the moment that the ship's clock reads [tex](d/v)*\sqrt{1 - v^2/c^2}[/tex], but what's important is the time in the ship's own frame at this moment--since in its frame the earth has been moving at velocity v for a time [tex](d/v)*\sqrt{1 - v^2/c^2}[/tex], at this time the earth must be a distance [tex]v*(d/v)*\sqrt{1 - v^2/c^2}[/tex] in this frame, or [tex]d*\sqrt{1 - v^2/c^2}[/tex]. So a corrected version of my argument after that point would look like this:
     
    Last edited: Jun 10, 2006
  18. Jun 11, 2006 #17
    Jesse,

    Thanks for all your work. I have to say, I liked your first answer. It fit in with what I was thinking. But I'd better reread and absorb everything better before I ask any more questions.

    pervect, I appreciate your help, too. I just don't have the math skills to use some of it.

    Thanks,
    Alan
     
    Last edited: Jun 11, 2006
  19. Jun 11, 2006 #18
    This makes sense to figure out D as a function of d at any moment. But if we were to have an event, like a signal sent out by the ship's crew, this signal would not be simultaneous in both frames. Is it safe to say, that if we wanted to figure out D, d, T, and t for this event, that your earlier post would be correct? At this event, the earth's clock would read [tex](d/v)*(1 - v^2/c^2)[/tex], and the rest of your earlier post would apply?

    Thanks,
    Alan
     
  20. Jun 11, 2006 #19

    JesseM

    User Avatar
    Science Advisor

    Simultaneous with what? It only makes sense to ask if two events are simultaneous within a single frame, you can't really ask whether a single event is simultaneous in two different frames (you can ask if it occurs at the same time-coordinate in two coordinate systems, but this is different, and it depends on the arbitrary choice of where you put the origin of each coordinate system, not just their relative velocity).
    The earth's clock would read [tex](d/v)*(1 - v^2/c^2)[/tex] at the time of this event in the ship's frame (but you shouldn't say 'At this event, the earth's clock would read...' because an 'event' in relativity refers only to a single point in spacetime, so anything happening elsewhere like on earth would not be 'at this event'). But my calculation of the distance D in the ship's frame was wrong, the distance would be the time of the event according to the ship's clock (ie [tex](d/v)*\sqrt{1 - v^2/c^2}[/tex]) multiplied by v, since in the ship's frame the ship's own clock is ticking at the normal rate while the earth is moving away at velocity v. Thus the distance D would be [tex]v*(d/v)*\sqrt{1 - v^2/c^2}[/tex], or [tex]d*\sqrt{1 - v^2/c^2}[/tex].

    To make this more clear, think in terms of the example I mentioned in the edit to that earlier post--suppose there was a rod attached to the earth, of length d in the earth's frame. If we look at the event of the ship passing next to the far end of the rod, then obviously in the earth's frame, the distance between that event and an event happening "simultaneously" on earth (like the event of the earth's clock ticking d/v) would be d, the length of the rod. So if you now look at the same event of the ship passing the end of the rod in the ship's frame, and you want to calculate the distance between this event and an event on earth which is simultaneous with it in the ship's frame (in this case, the event of the earth's clock ticking [tex](d/v)*(1 - v^2/c^2)[/tex]), then the distance would just be equal to the length of the rod in the ship's frame, or [tex]d*\sqrt{1 - v^2/c^2}[/tex] due to Lorentz contraction. This would be the distance of the earth at the moment the ship passes the end of the rod in the ship's frame.
     
    Last edited: Jun 11, 2006
  21. Jun 11, 2006 #20
    OK, if the event of the ship sending out the signal would be simultaneous with the earth's clock ticking d/v, in earth's frame, wouldn't this signal also be simultaneous with the event of a rod attached to the ship (length d in earth's frame) reaching earth? And the length of this rod would be [tex]d/\sqrt{1 - v^2/c^2}[/tex] in the ship's frame? It looks to me like we can't consider distance to be like a rigid rod, because it is constantly changing, which is what you said in your earlier post, before the edit. I have to say again that your earlier post matches up with what I was thinking. Maybe I'm biased because your first conclusion matched my own thoughts, but it still looks right to me.

    Thanks,
    Alan
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relativistic Rocket Question
  1. Relativistic rockets (Replies: 4)

Loading...