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

Rindler Horizon and Lorentz contraction

  1. Jan 2, 2015 #1
    Hello,

    Onhttp://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] website it says in the section "below the rocket, something strange is happening" that the distance of an object which passes the accelerating observer never increases -c2/α. I think this means, that the Rindler Horizon is always located at that distance in the momentarily comoving frame. So I would expect that from my perspective on earth the distance between the horizon and the accelerating observer is lorentzcontracted (although the horizon doesn't exist in my inertial frame, I should be able to calculate the distance at which a light signal will never reach the observer at a given point in time).

    I've done an exemplary calculation by computing the distance between horizon and accelerating observer and then compared it to c2/(αγ). Unfortunately I didn't get the same result. So before I write down my calculation, does it make sense at all to assume the distance is contracted in my frame or am I missing something?

    Thanks in advance for your help!
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 2, 2015 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Yes.

    Yes, but be careful about interpreting what that means. If ##L'## is the distance in the MCIF (i.e., ##L' = c^2 / a## ), then ##L## , the distance in the "earth" frame, will be larger than ##L'## by the factor ##\gamma## (which will be a function of time in the earth frame since the rocket is accelerating), i.e., ##L = \gamma c^2 / a## .
     
  4. Jan 2, 2015 #3

    WannabeNewton

    User Avatar
    Science Advisor

    In the standard background global inertial frame ##(t,x)## the Rindler horizon is just the line ##t = x## so yes the distance between the Rindler observer and ##t = x## will decrease as ##t\rightarrow \infty## when measured by the inertial observer in the ##t = \text{const.}## simultaneity lines since the Rindler observer's worldline is just a hyperbola with the ##t = x## line as the asymptote. This much is clear from a space-time diagram.
     
  5. Jan 2, 2015 #4
    Mhh, but if the horizon stays at the same distance in the MCIF, isn't he measuring the rest lenght and since both the horizon and the observer are moving in my inertial frame, shouldn't I be the one who observes the contracted distance?

    I think it's probably better to post my calculation:
    As WannabeNewton said I used ct=x for the Rindler Horizon and for the accelerated observer I used the formula x=c2/α√(1+(αt)2/c2). I computed Δx for α=1,031ly/y2(≈1g) and t=1y. This gave me 0,39ly. In order to compute γ I calculated the speed of the observer at t=1y using the formula v=αt/√(1+(αt)2/c2), which gave 0,72c and a lorentz factor of 1,44. And 1,44c2/α gives 1,4ly.
     
    Last edited: Jan 2, 2015
  6. Jan 2, 2015 #5

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    This is true, but again, we need to be careful interpreting what it means and how it relates to the distances ##L'## and ##L##.

    The distance I called ##L'##, in the MCIF, is a spacelike interval between the point ##(t, x) = (0, 0)## in the Earth frame, and any point on the hyperbola ##x^2 - c^2 t^2 = c^4 / a^2## , which is the worldline of the accelerating rocket. The fact that ##L'## is constant is just a reflection of the fact that the equation of the hyperbola is, as should be obvious, ##x^2 - c^2 t^2 = ( L' )^2##, and that the LHS of this equation is just the expression for the interval from ##(0, 0)## to ##(t, x)##. Or, to put it another way, we are just expressing the fact that the line of simultaneity through any point on the rocket's worldline, in the MCIF at that point, passes through the origin ##(0, 0)## of the Earth frame.

    The distance I called ##L## is a spacelike interval between the point ##(t, x) = (k, 0)## , which is on the time axis of the Earth frame ( ##k## is any positive number), and the point ##(k, \sqrt{c^2 k^2 + c^4 / a^2})## , i.e., the point on the hyperbola with the same ##t## value in the Earth frame. Physically, this is the distance in the Earth frame between an observer who remains at rest at the spatial origin (and who, at time ##t = 0## , emits the light ray that marks out the Rindler horizon) and the rocket, i.e., it is how far the rocket has traveled from its launch point at that instant in the Earth frame, plus the distance of the launch point from the origin (where the light ray that marks the horizon is launched).

    The distance you are describing, which I will call ##L^*## , is a spacelike interval between the point ##(t, x) = (k, k)## , which is on the ##t = x## line in the Earth frame ( ##k## is, again, any positive number), and the point ##(t, x) = (k, \sqrt{c^2 k^2 + c^4 / a^2})##, i.e., the point on the hyperbola with the same ##t## value in the Earth frame. Physically, this is the distance in the Earth frame between the light ray that marks out the Rindler horizon (which is emitted at ##(0, 0)## by the observer at the spatial origin) and the rocket, i.e., it is how far behind the rocket the light ray is at that instant in the Earth frame.

    From the above we can easily write down formulas for ##L## and ##L^*## (I will now use units where ##c = 1## to save typing):

    $$
    L = \sqrt{k^2 + ( L' )^2}
    $$

    $$
    L^* = \sqrt{k^2 + ( L' )^2} - k = L - k = k \left( \frac{L}{k} - 1 \right)
    $$

    Using the fact that the rockets ##\gamma## factor, in the Earth frame, is given by ##\gamma = x / L' = \sqrt{t^2 + ( L' )^2} / L'##, we find that ##L = \gamma L'## , as I said in my previous post, and ##L^* = \gamma L' - k## . Either of these values could plausibly be interpreted as the "distance to the horizon" in the Earth frame; but neither one is equal to ##L' / \gamma##, which is what the OP was expecting.
     
  7. Jan 2, 2015 #6

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    The full horizon, as seen in the Earth frame, is a lightlike surface, so thinking of it as "at rest" is going to cause problems, since a light ray can never be at rest.

    Also, as you can see from my post in response to WBN, the "distance" in the MCIF (what I called ##L'## ) is a spacelike interval from the rocket to the point ##(0, 0)## in the Earth frame--i.e., the "distance" to the horizon in the MCIF never changes because the "horizon" in the MCIF is really just a single point in spacetime, not a worldline. So this "horizon" can't really be thought of as either "moving" or "at rest", since both of those concepts apply to a worldline, not a point. That means ##L'## can't really be thought of as a "rest length".

    You can interpret the distance I called ##L^*## as the distance between the moving horizon and the moving observer in the Earth frame, but as you can see from my calculations, this does not equal ##L' / \gamma## . My comments above may help to explain why.
     
  8. Jan 2, 2015 #7
    Thank you very much for taking the time to type it out in general. After reading your comment I now see where my confusion came from and also noticed that adding 1ly to 0,39ly (to calculate the distance according to your definition) gives the number I calculated for γc2

    I didn't think about the fact, that the accelerating observer is always measuring the proper distance between the events (t|x)=(0|0) and (t'|x')=(k'|0). This seems very interesting to me, because wouldn't this imply, that the MCIF is always observing the clock of an observer at the origin of my inertial frame to read 0? If this is correct, is that how the website comes to the conclusion that "time comes to a complete halt there"?
     
  9. Jan 2, 2015 #8

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    "Observing" is the wrong word to use; for one thing, since light emitted at that origin marks out the Rindler horizon, that light will never reach the rocket, so the rocket never actually sees that event!

    A better way to put it is that, in the MCIF of the rocket, the current reading of the rocket's clock is always simultaneous with the reading "zero" on the clock at the origin of the Earth frame.

    I can't say for sure since I didn't write the website, but I suspect you are right.
     
  10. Jan 2, 2015 #9

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    The "distance" to an event horizon (such as the Rindler horizon) transforms differently than you might think, because it's a distance between a lightlike worldline (some point on the horizon) and a timelike worldline (some timelike observer). So it doesn't follow the rules of Lorentz contraction, which apply when you have a ruler (a spacelike interval in some inertial frame).

    If you are in flat space-time (like in the Rindler case, which also approximates a massive black hole), the distance of a lightbeam to an observer is just the amount of time it takes light to reach a momentarily co-moving inertial observer.

    I believe the distance transforms in the same way as the doppler factor does, but I'd have to review the calculation to be positive. It doesn't follow the Lorentz contraction law of changing by gamma, though.
     
  11. Jan 2, 2015 #10
    @pervect: That's very interesting, thanks for mentioning that! I've multiplied c2/α by the doppler factor and it indeed gives the distance between the front of the horizon (in my frame) and the MCIF as calculated in my second post, so I guess you're right. Now of course I am curious why that works, but I can imagine that the derivation is too complicated for me, at least with my current level of understanding.
     
  12. Jan 2, 2015 #11

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Using the formulas from post #5, this is easily confirmed: we have

    $$
    L' \sqrt{\frac{1 - v}{1 + v}} = \gamma L' \left( 1 - v \right) = L \left( 1 - v \right) = L \left( 1 - \frac{k}{L} \right) = L - k = L^*
    $$

    where we have used the fact that ##v = k / L## gives the velocity of the rocket in the Earth frame at the point ##(t, x) = (k, L)## on its worldline.
     
  13. Jan 3, 2015 #12

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    It falls out easily if you use Bondi's 's K-calculus. See https://archive.org/details/RelativityCommonSense
    doppler.png

    We've got three worldlines on the above diagram. The 45 degree slanted worldline is a lightlike worldline. The vertical wordline and the less slanted worldline are both timelike worldlines. Time runs up the page, space (distance) runs left-right.

    By definition the "distance" from the lightlike worldline to a timelike worldline from some particlular event on a timelike worldline (on the diagram, this event is labelled T=0 and is present on both timelike worldlines) is the proper time between elapsed said event and the arrival of the lightbeam.

    This may sound technical, but it means if a light beam is headed right for you, and is one light-second away from you, it will arrive in one second, because light travels at the speed of light,.

    So in the diagram above, the time of arrival of the light beam for the non-moving observer (vertical worldline on the diagram) is T = L/c, L being the "distance" for the non-moving worldline. The event occurs at T=0, reception of the light signal occurs at T=L/c on the clock carried by the vertical worldline, the proper time elapsed is L/c.

    Now, we want to find the distance for the second observer moving with respect to the first. From the diagram we see if if this non-moving observer emits a pulse at the same time he receives the light signal, this re-emitted light signal will be coincident with the lightlike worldline, and by the Bondi formula will arrive at some doppler shifted time kT, k being the doppler shift factor. (If you're familiar with the Bondi k-calculus, this will be obvious, if not - well, I suppose you'd have to read the relevant sections of the book. It's not a bad way to start learning SR. Basically, though, k is the interval of time of transmission to reception, so if we send a signal at some proper time tau from a stationary observer, the moving observer receives it at a time k*tau, k being by definition the doppler shift factor).

    But the "distance" to the lightbeam is just the proper time of arrival multipled by c. Since the arrival time is kL/c, the distance to the second observer (the slanted timelike worldline) is kL, k being the "doppler shift factor", in this diagram k is greater than one representing a redshift.
     
    Last edited: Jan 3, 2015
  14. Jan 3, 2015 #13
    @pervect That was very helpful and easy to follow, thanks.
    Or I simply imagine the proper time to be the period of an electromagnetic wave, but anyway I will take a look at the Bondi k-calculus.

    And also thanks to PeterDonis for the quick mathematical proof!
    I guess I have no further questions now :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rindler Horizon and Lorentz contraction
  1. Rindler Horizon (Replies: 0)

  2. Lorentz contraction (Replies: 2)

Loading...