ridler_motion2

Rindler Motion in Special Relativity: Rindler Coordinates

Our destination

In our last article, Hyperbolic Trajectories, we derived some facts about the trajectory of a rocket that is undergoing constant (proper) acceleration. In this article, we will explore what these facts mean for the passengers on board a rocket undergoing Rindler motion. The goal is to come up with a coordinate system that will be convenient to use by the passengers. Because the rocket does not stay at rest (except momentarily) in any inertial reference frame, the onboard coordinate system will necessarily be noninertial. However, for the passengers themselves, it will not be too weird–they should still be able to determine locations within the rocket using the meter sticks that they have available, and they should be able to still have a time coordinate that can be used to plan meetings and so forth within the rocket.

Before getting to that coordinate system, I’m going to take a sort of long-winded path. The point of the detour is to motivate that coordinate system, to show that it really is the most convenient one to be used by our rocket travelers.

Recap of the last article

In the last article, Hyperbolic Trajectories, we derived some facts about the trajectory of a rocket that is undergoing constant (proper) acceleration. Let me repeat a few of the formulas we derived:

Equation 1: ##x = x(0) + \frac{c}{g}(V^0 – V^0(0))##

Equation 2: ##ct = ct(0) + \frac{c}{g} (V^1 – V^1(0))  ##

Equation 3: ##x = \sqrt{c^2 t^2 + \frac{c^4}{g^2}}##

Equation 4: ##\frac{d V^0}{d\tau} = \frac{g}{c} V^1##

Equation 5: ##\frac{d V^1}{d\tau} = \frac{g}{c} V^0##

In these equations, ##g## is the acceleration of the rocket, ##x## is the spatial position of the rocket at time ##t##, ##V^0## and ##V^1## are the components of the proper velocity (change in ##x## and ##ct## as a function of proper time ##\tau##, the time on the rocket’s onboard clock). ##x(0), t(0), V^0(0)## and ##V^1(0)## are the values of the position, time and proper velocity components initially (when ##\tau = 0##).

Equations 1&2 are valid in every inertial coordinate system, while Equation 3 is only assumed to be valid in the launch frame of the rocket, with the particular choices: ##x(0) = \frac{c^2}{g}, t(0) = 0, V^0(0) = c, V^1(0) = 0##. A couple of things to note: First, although the proper acceleration remains constant, the velocity stays bounded; as ##t \rightarrow \infty##, the rocket approaches the speed of light: ##x \approx ct##. Second, there is the oddness that we chose that initially ##x(0) = \frac{c^2}{g}##, rather than ##x(0) = 0##. This was simply to make the resulting formula simpler-looking–the choice of the spatial origin doesn’t have any significance–but it’s a clue as to what’s to come.

Surprising consequences of our arbitrary choice

As I said, Equation 3 was only assumed to be true in the launch frame of the rocket. But we can use the method of Lorentz invariants to show something surprising. The combination ##x^2 – c^2 t^2## is the Lorentz invariant ##X \cdot X##. As we can see from Equation 3, its value in any frame is ##\frac{c^4}{g^2}##. This combination is doubly invariant: it has the same value at all times, and in all reference frames. So this shows that as a matter of fact, Equation 3 is true in every reference frame, even though it was only derived for the launch frame. This is a miraculous consequence of the choice ##x(0) = \frac{c^2}{g}##.

Let me spell out how miraculous this choice is: Let ##F## be the initial launch frame of the rocket. That means that a time ##t=0##, the rocket is initially at rest at location ##x=\frac{c^2}{g}##. Let ##F’## be some other inertial frame—the frame of observers that are moving at some speed ##v## relative to frame ##F##. Let ##x’, t’## be the coordinates for this frame. Then the trajectory of the rocket in this frame looks exactly the same as in frame ##F##:

##x’ = \sqrt{(c t’)^2 – \frac{c^4}{g^2}}##

That means that in frame ##F’## it is also true, just as in frame ##F##, that at time ##t’=0##, the rocket is initially at rest at location ##x’ = \frac{c^2}{g}##. This might seem at first like a paradox: If the rocket is initially at rest in frame ##F##, how can it also initially be at rest in frame ##F’##?? Well, this is a miraculous consequence of the relativity of simultaneity. Clocks that are synchronized in one frame are not synchronized in another frame. Let’s identify three events:

  1. ##e_0##. This is the event with coordinates ##x = 0, t = 0## in frame ##F## and coordinates ##x’ = 0, t’ = 0## in frame ##F’##.
  2. ##e_1##. This is the event in which the rocket first launches from frame ##F##. We chose this event to have coordinates ##x= \frac{c^2}{g}, t = 0## in frame ##F##.
  3. ##e_2##. This is the event in which the rocket comes to rest (momentarily) in frame ##F’##. This event has coordinates ##x’ = \frac{c^2}{g}, t’ = 0## in frame ##F’##.

In frame ##F##, events ##e_0## and ##e_1## are simultaneous. In frame ##F’##, events ##e_0## and ##e_2## are simultaneous. Both frames think of the rocket as “initially” at rest.

An analogy

A sort-of analogy would be a train traveling west on the surface of the Earth, traveling at just the right speed so that it takes one hour to go from one time zone to the next so that every time it passes a train station, the local clocks always say 12:00.

Rewriting our equations

In the initial launch frame, we know that ##V^0(0) = c## and ##V^1(0) = 0##. Armed with these facts, and the choice ##x(0) = \frac{c^2}{g}##, we can rewrite Equation 1 and Equation 2 to make them simpler.

Equation 6: ##x = x(0) V^0/c##

Equation 7: ##ct = x(0) V^1/c##

 

Different parts of the rocket experience different accelerations

So far, we have just considered the rocket to be a single point moving through spacetime. How is the analysis affected if we take into account that the rocket is an extended object?

As we established above, if we pick an inertial coordinate system such that one point–let’s say the rear of the rocket–is initially at ##x = \frac{c^2}{g}## at time ##t=0##, according to the initial launch frame, ##F##, then that will be true in every frame ##F’## related to ##F## through the Lorentz transformations. What about the front of the rocket?

We expect that when a rocket starts accelerating, its length doesn’t change. (Okay, that’s not 100% true–the forces of acceleration will actually cause the rocket to compress a tiny bit, but I’m going to assume that that’s a small effect). That’s our experience based on nonrelativistic physics. But nonrelativistic physics works pretty well in a frame where a rocket is moving slowly. So we’re going to make the following assumption:

Assumption: If the rocket is momentarily at rest in some frame ##F’##, then all parts of the rocket are momentarily at rest in that frame at the same time, and the length of the rocket (as measured in the momentary rest frame ##F’##) is the same as it was in the “launch” frame ##F##.

Since we chose a coordinate system so that the rear of the rocket always has the same x-location in every frame in which the rocket is (momentarily) at rest, then the front of the rocket must also have this property:

Consequence: If ##x_{front}(0)## is the initial x-location of the front of the rocket in its launch frame, then its location in any other reference frame in which it is (momentarily) at rest must be the same, ##x_{front}(0)##.

If this were not the case, then the length of the rocket would not always be the same, as measured in its momentary rest frame. We have already found one trajectory that has that property, and it’s Equation 3.

##x = \sqrt{c^2 t^2 + \frac{c^4}{g^2}}##

We can show (I’ll skip the proof because it’s a little tedious) that this is the only trajectory having this property. So it must be that the location of the front of the rocket obeys this equation, as well:

##x_{front} = \sqrt{c^2 t^2 + \frac{c^4}{g^2}}##

At time ##t=0##, we find the relationship between ##x_{front}(0)## and ##g##:

##x_{front} = \frac{c^2}{g}##

Since, obviously, ##x_{front}(0) > x_{rear}(0)## this means that in order for the front of the rocket to remain at a constant distance from the rear of the rocket, the acceleration ##g## for the front of the rocket can’t be the same as that of the rear. We can solve for ##g## to see:

##g = \frac{c^2}{x_{front}}##

So ##g## must get weaker as you move forward in the rocket. So the front of the rocket is accelerating slightly less than the rear of the rocket.

How is this possible? If the rear accelerates more than the front, then doesn’t that mean that the rear will eventually catch up to the front and that the rocket will be squashed? Well, sort of.

Length-contraction and location-dependent g-forces

When you first study relativity, you learn about length contraction: If according to some frame ##F##, a rocket is traveling at speed ##v## and has length ##L## in its own rest frame, ##F’##, then it will have length ##L \sqrt{1-\frac{v^2}{c^2}}## according to frame ##F##. So, if a rocket starts at rest in frame ##F##, having length ##L##, and accelerates until it is at rest in frame ##F’##, and still has length ##L## in frame ##F’##, then from the point of view of frame ##F##, the rocket will have contracted in length. That means that the rear of the rocket will have traveled a tiny bit more than the front. This means that the rear will have had a slightly higher velocity. Which means that it will have had a slightly higher acceleration. So our conclusion, that the front of the rocket must have a smaller acceleration than the rear, is the only possibility that is consistent with length contraction.

Relating x and t

For any part of the rocket, the relationship between its acceleration ##g## and its initially location ##x(0)## will be given by:

##x(0) = \frac{c^2}{g}##

Using this fact, we can rewrite Equation 3:

Equation 8: ##x = \sqrt{c^2 t^2 + x(0)^2}##

Clocks at different locations within the rocket run at different rates

Besides length contraction, another feature of relativity is time dilation: Roughly speaking, clocks that are moving relative to a frame ##F## will be seen to run more slowly, according to that frame. Since different parts of the rocket are traveling at slightly different speeds, it follows that clocks at different locations within a rocket will be affected differently by time dilation. We can compute how big this effect is.

The relationship between the clock time, ##tau## and coordinate time, ##t## is by definition given by:

##c \frac{dt}{d\tau} = V^0##

From Equation 6, we have the relationship between ##x## and ##V^0##:

Equation 9: ##V^0 = \frac{cx}{x(0)}##

Without solving any differential equations to find ##\tau## in terms of ##t##, we can immediately figure out an approximate fact about ##\tau##

As the rocket accelerates, it approaches the speed of light, so ##x \rightarrow ct##. So we have, approximately, for large values of ##t##:

##V^0 \approx \frac{c^2 t}{x(0)}##

So we have:

##\frac{dt}{d\tau} \approx \frac{c t}{x(0)}## (valid for large ##t##).

Or, flipping the equation:

##\frac{d\tau}{dt} \approx \frac{x(0)}{c t}##

We can see that the proper time ##\tau## depends on the location within the rocket, with higher locations (toward the front of the rocket) experiencing less time dilation. So clocks closer to the front will be seen to run faster than clocks closer to the rear, with a linear relationship between the time shown on a clock and its position, ##x(0)## within the rocket.

A Coordinate System for Rocket Travelers

Since the rocket is constantly changing inertial reference frames, the people traveling on the rocket can’t very well use any fixed inertial coordinate system to locate events that take place inside the rocket. As we have established earlier, if a part of the rocket is initially at location ##x(0)## in the “launch” frame ##F##, then it will be at the same location ##x’ = x(0)## in any reference frame ##F’## where the rocket is momentarily at rest. This means that within the rocket, passengers can use the coordinate ##R = x(0)## to locate things. Something that is “stationary” within the rocket will always maintain the same value of ##R##.

Now, what about measuring time within the rocket? Time is simple for an inertial reference frame—for any clock at rest in the inertial frame, ##d\tau = dt##, so the time can be read off from the elapsed time on the clock (after synchronizing clocks at different locations). Onboard the rocket, however, we have discovered that clocks at different locations appear to run at different rates when measured from the rocket’s momentary rest frame. So a more useful time coordinate for passengers inside the rocket is not ##\tau##, but ##\tau## scaled by the location-dependent clock rate. This motivates introducing a time coordinate, which I’ll call ##\theta##, which is defined by:

##\theta \equiv \frac{c \tau}{R}##

Since clocks that are in the front of the rocket have larger values of ##R## and larger values of ##\tau##, the ratio is approximately independent of location within the rocket, and so makes a good choice for a time coordinate.

So tentatively, we plan to use the coordinates ##R, \theta## within the rocket. Now we need to figure out mathematically how these coordinates relate to the inertial coordinates ##x,t## of the launch frame.

Going back to Equations 4&5, we can rewrite them in terms of ##R## and ##\theta## (using ##R = \frac{c^2}{g}## and ##\theta = c\tau/R##)

Equation 10: ##\frac{d V^0}{d\theta} = V^1##

Equation 11: ##\frac{d V^1}{d\theta} = V^0##

In the last article, we already established the invariant: ##(V^0)^2 – (V^1)^2 = c^2##. And we know the initial condition, that in the launch frame, when ##t = \tau = \theta = 0##, ##V^1 = 0##. These equations and the initial conditions has a unique solution:

Equation 12: ##V^0 = c cosh(\theta)##

Equation 13: ##V^1 = c sinh(\theta)##

where ##cosh## and ##sinh## are hyperbolic cosine and sine, respectively. Then we can rewrite ##x## and ##t## from Equations 6&7:

Equation 14: ##x = R cosh(\theta)##

Equation 15: ##ct = R sinh(\theta)##

These have the inverse relations:

Equation 16: ##R = \sqrt{x^2 – c^2 t^2}##

Equation 17: ##\theta = tanh^{-1}(\frac{ct}{x})##

This is almost exactly analogous, in Euclidean geometry, to the relationship between Cartesian coordinates and polar coordinates: ##x = r cos(\theta), y= r sin(\theta)##.

In the next article, we will ignore, for the moment, where this noninertial coordinate system came from and instead will explore what the universe looks like when described in these coordinates. This coordinate system will turn out to have analogies to features of General Relativity, including gravitational time dilation and event horizons.

Coming up: The Universe According to Rindler

27 replies
  1. craigthone says:
    stevendaryl

    Yes, you're right.

    [Edit]

    I guess there are three different types of acceleration, with components given by:

    1. ##frac{d^2 x^mu}{dt^2}##
    2. ##frac{d^2 x^mu}{dtau^2}##
    3. ##frac{d^2 x^mu}{dtau^2} + Gamma^mu_{nu lambda} frac{dx^nu}{dtau} frac{dx^lambda}{dtau}##

    I guess #3 is "proper acceleration", but I'm not sure whether "coordinate acceleration" means 1 or 2.For the 2nd one,
    $$frac{d^2x^mu}{dtau^2}=frac{dU^mu}{dtau}$$
    Generally ##dU^mu## does not give the relative velocity, so we need something can be directly subtracted, i.e. ##theta##
    For the uniformly accelerating object, the proper acceleration is defined as
    $$alpha=frac{dtheta}{dtau}$$

  2. Eli Pankiewicz says:

    Numerical Simulation with Riedler Coordinates:

    The Equations of Motions (geodetic Equations) are:
    $$frac{d^2 x^alpha}{d t^2}=-Gamma^alpha_{betagamma}frac{d x^beta}{d t}frac{d x^gamma}{d t}quad quad (1)$$
    With c=1 we get for the Riedler Coordinates :
    $$begin{bmatrix}
    t_I\
    x_I \
    y_I \
    z_I \
    end{bmatrix}=
    begin{bmatrix}
    left(x+frac{1}{a}right)sinh(a,tau)\
    left(x+frac{1}{a}right)cosh(a,tau)\
    y\
    z
    end{bmatrix} $$
    $$Rightarrow $$

    $$frac{d^2 s^alpha}{d tau^2}=- left( {frac {d}{dt}}tau left( t right) right) ^{2}{x(t)}^{2}{a}^
    {2}-2, left( {frac {d}{dt}}tau left( t right) right) ^{2}x(t)a-
    left( {frac {d}{dt}}tau left( t right) right) ^{2}+ left( {
    frac {d}{dt}}x left( t right) right) ^{2}quad quad (2)
    $$
    and the Metric g is:
    $$g=begin{bmatrix}
    -left(x,a+1right)^2 & 0 & 0 & 0 \
    0 & 1 & 0 & 0 \
    0 & 0 & 1 & 0 \
    0 & 0 & 0 & 1 \
    end{bmatrix}
    $$
    So the EQM with equation (1) and the metric g are :

    Equation (4)
    $${frac {d^{2}}{d{t}^{2}}}tau left( t right) +2,{frac {a left( {
    frac {d}{dt}}x left( t right) right) {frac {d}{dt}}tau left( t
    right) }{x left( t right) a+1}}=0\
    {frac {d^{2}}{d{t}^{2}}}x left( t right) + left( {frac {d}{dt}}
    tau left( t right) right) ^{2}a left( x left( t right) a+1
    right) =0\
    frac{d^2}{d t^2}y(t)=0\
    frac{d^2}{d t^2}z(t)=0\
    $$
    Numerical Simulation

    To do the simulation we need the initial conditions for the differential equations (4). These must satisfied also the equation (3).
    Data :

    $$a=10,, x(0)=2,,frac{d x}{d t}big|_{t=0}=0.1 ,, frac{d y}{d t}big|_{t=0}=0.0
    ,, frac{d z}{d t}big|_{t=0}=0.0 $$
    we can calculate with equation (3) and
    $$frac{d^2 s}{d tau^2}=0quad Rightarrowquad frac{d tau}{d t}big|_{t=0}=pm 0.004761904762 $$

    Result:

    can i put hier JPEG Format?

  3. stevendaryl says:
    Mentz114

    ==

    The definition of proper acceleration is ##dot{u}_a =nabla_b u_a u^b## and for the metric ##diag(-g_{00}, 1,1,1)## the acceleration of ##u^a=1/sqrt{g_{00}} partial_t## is just what you got in your post, ##-partial_x g_{00}/(2,g_{00})## ( in the x direction ).

    I don't know if that helps or hinders.Yes, so in the particular case in which only ##g_{00}## is different from ##pm 1##, it works like a potential. Okay. Good to know when it works.

  4. sweet springs says:

    Thanks to post #17, an addition to my post #19. Say at the top end of rocket proper acceleration is g, the length of rocket cannot exceed g/c^2. Infinite tidal force will tear top end part as it reaches event horizon.
    Acceleration has no upper limit as speed has c. But if we consider finite size of body L and keeping it sound, there seems to be limit of c^2/L.

  5. PeterDonis says:
    stevendaryl

    I guess #3 is "proper acceleration",Yes.

    stevendaryl

    I'm not sure whether "coordinate acceleration" means 1 or 2.It means #1. #2 is a sort of "in between" acceleration that in general doesn't get used; the only time I've seen it is for the case of inertial coordinates in flat spacetime (or at least a locally flat patch), where the connection coefficients all vanish so #2 is equivalent to #3.

  6. Mentz114 says:

    ==

    stevendaryl

    Yes, you're right.

    [Edit]

    I guess there are three different types of acceleration, with components given by:

    1. ##frac{d^2 x^mu}{dt^2}##
    2. ##frac{d^2 x^mu}{dtau^2}##
    3. ##frac{d^2 x^mu}{dtau^2} + Gamma^mu_{nu lambda} frac{dx^nu}{dtau} frac{dx^lambda}{dtau}##

    I guess #3 is "proper acceleration", but I'm not sure whether "coordinate acceleration" means 1 or 2.The definition of proper acceleration is ##dot{u}_a =nabla_b u_a u^b## and for the metric ##diag(-g_{00}, 1,1,1)## the acceleration of ##u^a=1/sqrt{g_{00}} partial_t## is just what you got in your post, ##-partial_x g_{00}/(2,g_{00})## ( in the x direction ).

    I don't know if that helps or hinders.

  7. stevendaryl says:
    PeterDonis

    I think you mean coordinate acceleration. The geodesic equation describes a worldline with zero proper acceleration, by definition.Yes, you're right.

    [Edit]

    I guess there are three different types of acceleration, with components given by:

    1. ##frac{d^2 x^mu}{dt^2}##
    2. ##frac{d^2 x^mu}{dtau^2}##
    3. ##frac{d^2 x^mu}{dtau^2} + Gamma^mu_{nu lambda} frac{dx^nu}{dtau} frac{dx^lambda}{dtau}##

    I guess #3 is "proper acceleration", but I'm not sure whether "coordinate acceleration" means 1 or 2.

  8. PeterDonis says:
    stevendaryl

    the geodesic equation gives us the proper accelerationI think you mean coordinate acceleration. The geodesic equation describes a worldline with zero proper acceleration, by definition.

  9. stevendaryl says:

    Okay, I see where a logarithmic "potential" can come into play, in general.

    The equations of motion for a particle freefalling in curved spacetime are given by the geodesic equations:

    ##ddot{x^mu} = -Gamma^mu_{nu lambda} dot{x^nu} dot{x^lambda}##

    where a dot above a variable means a derivative with respect to proper time. I'm going to assume that there is one time coordinate, ##x^0##, and three spatial coordinates, ##x^j## (##j## ranges from 1 to 3). I'm going to assume that the components of the metric tensor are independent of time. I'm also going to assume that the metric tensor is diagonal (so ##g_{mu nu} = 0## unless ##mu = nu##).

    Let's figure out the spatial acceleration for a test particle that is initially at rest: ##dot{x^j} approx 0##. We can compute ##dot{x^0}## in terms of the metric tensor:

    ##dtau^2 = g_{mu nu} dx^mu dx^nu Rightarrow 1 = g_{mu nu} dot{x^mu} dot{x^nu}## Since ##dot{x^0}## is the only nonzero component of the 4-velocity, we have:

    ##1 = g_{00} (dot{x^0})^2##

    So ##dot{x^0} = frac{1}{sqrt{g_{00}}}##.

    So the geodesic equation gives us the proper acceleration for a spatial coordinate:
    ##ddot{x^j} = -Gamma^j_{0} dot{x^0} dot{x^0} = -Gamma^j_{0} frac{1}{g_{00}}##

    Now, we can use a definition of ##Gamma## in terms of the metric tensor:

    ##Gamma^j_{00} = frac{1}{2} g^{jk} (partial_0 g_{k0} + partial_0 g_{0k} – partial_k g_{00})##

    Since the metric components are independent of ##x^0##, this simplifies to:
    ##Gamma^j_{00} = -frac{1}{2} g^{jk} partial_k g_{00}##

    Plugging this into the acceleration equation:
    ##ddot{x^j} = frac{1}{2} g^{jk} partial_k g_{00} frac{1}{g_{00}} = g^{jk} partial_k ln sqrt{g_{00}} = partial^j ln sqrt{g_{00}} ##

    So the initial proper acceleration is analogous to Newtonian physics with a potential of ##phi = ln sqrt{g_{00}}##. However, that's only true initially, when you can ignore terms involving ##dot{x^j}##.

  10. sweet springs says:

    In my post #15 I said about rear engine case. For front engine, no built on but hanged down take place. No compression but torsion works.
    This reminds me that A. Gaudi used a model of church in Spain hanged upside down to investigate structure. No compression but torsion of the same magnitude work on every parts, I think.
    If all parts has its own engine producing g(x) and tied with threads, no push and pull work on threads as I noted in my post #11.

  11. PeterDonis says:
    stevendaryl

    In nonrelativistic physics, the significance of ##phi## is that there is a total energy that is conserved: ##frac{1}{2} mdot{x}^2 + phi(x)##. But there doesn't seem to be anything similar to that in the Rindler problem.Curves of constant ##x, y, z## in Rindler coordinates are orbits of a timelike Killing vector field, so there is a conserved energy associated with that KVF that is a constant of geodesic motion. The problem is that, unlike non-relativistic Newtonian mechanics or Schwarzschild spacetime in GR, the potential associated with this conserved energy does not approach a finite limit at infinity; it increases without bound. That is what prevents it from having a useful physical interpretation globally.

    But if you restrict to a finite region, things work just the way you are describing: the potential is constant along each worldline in the Rindler congruence, and the kinetic energy of a freely falling object, relative to observers on each Rindler worldline, varies with the potential on each worldline in just the right way. For example, suppose we drop an object from the top of a Rindler rocket of length ##L##. We work in an inertial frame in which the dropped object is at rest; so at the instant the object is dropped, the bottom of the rocket is at ##x = x_0## and the top of the rocket is at ##x = x_0 + L##. Then the kinetic energy of the bottom of the rocket, relative to the dropped object, is just its kinetic energy in the frame we have chosen, which is simply ##left( gamma – 1 right) m## (I'm using units where ##c = 1##), where ##gamma## is what you call ##V^0## in your article. This will also turn out to be the same as the change in potential from the top to the bottom of the rocket.

  12. PeterDonis says:
    sweet springs

    I am sure that experts have well established skill to estimate and calculate on structural mechanics so that it is practically rigid and does not hardly collapse.The conditions any structure is subjected to on Earth are extremely mild compared to what we consider in relativity problems. 1 g acceleration is a very small acceleration in relativistic terms; calculate ##c^2 / g## for a 1 g acceleration and see what length ##L## it corresponds to.

  13. jbriggs444 says:
    sweet springs

    Born rigid body is theoretical one that cannot be made up of real matters.Born rigidity characterizes the relationship between the positions in parts of an extended object over time. Indeed, there is no physical object that can maintain exact Born rigidity in the face of a momentary stress applied at a single point. However, there are many physical objects that can maintain Born rigidity in the face of a continuous stress applied on a single surface. e.g. the Eiffel tower. It is supported by its base and maintains Born rigidity in the face of that continuously applied force.

    You are correct that the Eiffel tower will only maintain approximate rigidity in the face of a minor earthquake or strong and gusting wind.

  14. sweet springs says:
    PeterDonis

    There is no such thing as "infinite elastic coefficient". The finite speed of light places finite limits on material properties (basically that the speed of sound in the material cannot exceed the speed of light).Going back to reality, I have scarce idea of architechtur or structural technology of towers in various shape and size, I am sure that experts have well established skill to estimate and calculate on structural mechanics so that it is practically rigid and does not hardly collapse. If tower is extremely high change of gravity according to height or distance from the Earth with acceleration g(x) might be considered together with time dilation with it. Same thing we will do in construction of rockets with Rindler's g(x) L<x<L+H above mentioned where L corresponds to rear end and H corresponds to top end.

  15. sweet springs says:

    In Rindler system
    [tex]proper phi(x)=c^2 ln frac{x}{L}=c^2 ln(1+frac{h}{L})=c^2frac{h}{L}+o((frac{h}{L})^2)[/tex]
    so in case |h|<<L only the first term survives and it is familiar "mgh".

  16. stevendaryl says:
    sweet springs

    According to text of Landau-Lifshitz section 88 problem 1,
    for bodies staying still,
    [tex] proper g=-c^2 nabla ln sqrt{g_{00}}[/tex]
    thus
    [tex] proper g=-frac{c^2}{x}[/tex] or
    [tex]proper g(x)=proper g(L+h)=frac{g(L)}{1+hfrac{g(L)}{c^2}}[/tex]
    holds strictly.

    From this I suspect the relation
    [tex]phi(x)=c^2 ln sqrt{g_{00}}[/tex]
    for stationary gravity field at least. But I am not sure of this.The issue is exactly what is the significance of that ##phi(x)##. In nonrelativistic physics, the significance of ##phi## is that there is a total energy that is conserved:##frac{1}{2} mdot{x}^2 + phi(x)##. But there doesn't seem to be anything similar to that in the Rindler problem. There are conserved quantities involved, but not involving that, as far as I can see.

  17. sweet springs says:

    According to text of Landau-Lifshitz section 88 problem 1,
    for bodies staying still,
    [tex] proper g=-c^2 nabla ln sqrt{g_{00}}[/tex]
    thus
    [tex] proper g=-frac{c^2}{x}[/tex] or
    [tex]proper g(x)=proper g(L+h)=frac{g(L)}{1+hfrac{g(L)}{c^2}}[/tex]
    holds strictly.

    From this I assume the relation
    [tex]phi(x)=c^2 ln sqrt{g_{00}}[/tex]
    holds for still object in stationary gravity field at least. But I am not sure of this yet.

  18. sweet springs says:
    sweet springs

    proper g(x)=-(frac{L}{x})^2nabla phi(x)=-frac{c^2}{x}Obviously when we say x=L+h h is height from the "ground" x=L this formula becomes
    [tex]proper g(x)=proper g(L+h)=frac{g(L)}{1+hfrac{g(L)}{c^2}}[/tex]
    as I wrote in other thread. There stevendaryl derived this by approximation. I would like to know how accurate this relation is. This is the background of my question in previous post#9.
    So in other words I would like to know how to get proper gravity acceleration g(x), not through phi but directly from g_00. Best.

  19. sweet springs says:
    PeterDonis

    I'm not sure what you're claiming here, but if you are claiming that you can have all parts of the rocket have exactly the same proper acceleration and not separate as in the Bell spaceship paradox, you are incorrect.No. I intend to claim the case all parts are tied with threads and will move with their proper acceleration g(x) , function of positions. I apologize my saying "Bell's thread" in post might have misled you.

  20. sweet springs says:

    I agree with your saying about ordinary real matters. Born rigid body is theoretical one that cannot be made up of real matters.

    sweet springs

    Approximately,
    ϕ(x)=(g00−1)c^2/2I said approximately but I am not sure about it. I should appreciate it if you would teach me this relation of phi and g_00 is strict one or only an approximation.

  21. PeterDonis says:
    sweet springs

    Say we divide rocket into parts according to x and each part move with proper g(x), Bell's paradox threads connecting parts wil not be torn apart.I'm not sure what you're claiming here, but if you are claiming that you can have all parts of the rocket have exactly the same proper acceleration and not separate as in the Bell spaceship paradox, you are incorrect.

  22. PeterDonis says:
    sweet springs

    Born rigidity body can do it with infinite elastic coefficientThere is no such thing as "infinite elastic coefficient". The finite speed of light places finite limits on material properties (basically that the speed of sound in the material cannot exceed the speed of light).

  23. sweet springs says:

    Let me add some more.

    [tex]proper g(x) * small proper time at x= -frac{c^2dt}{L}[/tex]

    It is relative speed to instantaneous IFR after small time. As it does not depend on x, local instantaneous IFRs are integrated to a common global instantaneous IFR. Rindler system holds global instantaneous IFRs and changes them every moment of time. The above formula gives relative speed of Old instantaneous IFR and New instantaneous IFR.

    Say we divide rocket into parts according to x and each part move with proper g(x), Bell's paradox threads connecting parts wil not be torn apart. If rocket body is made of Born rigid material, not engine but all the parts of rocket follow proper g(x).
    Stress works on body so that proper g(x) take place everywhere on the body. Born rigidity body can do it with infinite elastic coefficient so no change of proper length.

  24. sweet springs says:

    Let me say some more.

    Metric of Rindler system is

    [tex]ds^2=(frac{x}{L})^2 c^2 dt^2 – dx^2 -dy^2 -dz^2[/tex]
    where x is distance from E.H. and d is positive value L=c^2/g where g is acceleration or gravity observed there g(L). t time there is chosen as world time.

    Approximately,
    [tex]phi(x)=(g_{00}-1)frac{c^2}{2}=[(frac{x}{L})^2-1]frac{c^2}{2}[/tex]
    [tex]g(x)=-nabla phi(x)=-xfrac{c^2}{L^2}[/tex]

    This is a acceleration in world time. In proper time there we put square of coefficient
    [tex]dtau(x)=frac{x}{L}dtau(L)=frac{x}{L}dt[/tex]

    [tex]proper g(x)=-(frac{L}{x})^2nabla phi(x)=-frac{c^2}{x}[/tex]

    Integration I did seems wrong due to the fact proper acceleration changes according to g_00(x) in different x. Integration of proper acceleration seems meaningless.

  25. stevendaryl says:
    sweet springs

    Any proper acceleration systems of different acceleration values are the same one. Their distance from E.H. ,x, make apparent difference i.e. values of proper acceleration.So even though I'm not exactly sure what the significance of your "gravitational potential" is, you're absolutely right that all accelerated rockets look the same. The only difference from one case to another is where the "height" is measured from.

  26. stevendaryl says:
    sweet springs

    I have questions.
    I suppose
    [tex]g(x)=-frac{c^2}{x}[/tex]
    where x is distance from Event Horizon which is at backward of rocket and is shared by every part of the rocket. Gravity potential is by integrating this,
    [tex]phi(x)=c^2 logfrac{x}{x_0}+phi_0[/tex]

    Any proper acceleration systems of different acceleration values are the same one. Their distance from E.H. ,x, make apparent difference i.e. values of proper acceleration.

    Do these make sense?I have to think about the significance of the potential that you are deriving. Normally, the significance of a potential is that the combination of kinetic energy plus potential is constant for an object dropped into that potential. Right off the bat, I don't see how that comes into play.

    A freely falling object in the accelerated ##R, theta## coordinate system will satisfy two conservation equations:

    1. ##frac{1}{2} m (frac{dR}{dtau})^2 – frac{1}{2} m R^2 (frac{dtheta}{dtau})^2 = -frac{1}{2} m c^2##
    2. ##R^2 frac{dtheta}{dtau} = mathcal{E}##

    where ##tau## is proper time for the object. The first equation is really just the equation ##frac{dtau}{dt} = sqrt{1 – frac{1}{c^2}(frac{dx}{dt})^2}## squared and multiplied by ##frac{1}{2} mc^2## to look more like a Newtonian conservation of energy equation. The second is a consequence of the fact that the "gravitational potential" is independent of the coordinate ##theta##. ##mathcal{E}## is an arbitrary constant. If we assume that the object is dropped from "rest" at position ##R_0##, then the relationship between ##theta## and ##tau## is initially: ##theta = frac{c tau}{R_0}##, so ##frac{dtheta}{dtau} = frac{c}{R_0}##. So the constant ##mathcal{E}## is just ##c R_0##. So the constant ##mathcal{E}## turns out to be ##c R_0##.

    We can stick the second into the first to get a new conservation equation:

    ##frac{1}{2} m (frac{dR}{dtau})^2 – frac{1}{2} m c^2 frac{(R_0)^2}{R^2} = -frac{1}{2} m c^2##

    This is an exact equation,

    So in terms of ##tau##, the path of a freefalling object seems more like an inverse-square potential, instead of a logarithmic potential.

    I don't immediately see how the potential you derive can be understood in terms of the accelerated coordinate system. Maybe I need to look at the nonrelativistic limit of the exact equation. I'll have to think about it more.

  27. sweet springs says:

    I have questions.
    I suppose
    [tex]g(x)=-frac{c^2}{x}[/tex]
    where x is distance from Event Horizon which is at backward of rocket and is shared by every part of the rocket. Gravity potential is by integrating this,
    [tex]phi(x)=c^2 logfrac{x}{x_0}+phi_0[/tex]

    Any proper acceleration systems of different acceleration values are the same one. Their distance from E.H. ,x, make apparent difference i.e. values of proper acceleration.

    Do these make sense?

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