Relativity and Cosmology

We know the relativistic view: "moving clocks tick slower than stationary clocks". We also know that Einstein's Special Relativity (SR) demands: "there is no absolute movement, only relative movement". This tells us that when two clocks are moving relative to each other, we can view any one of them as the "stationary" reference and the other one as the "moving clock".

An apparent paradox arises out of this opening paragraph: if clock A is considered as the reference and clock B as moving, then B ticks slower than A. However, if B is considered as the reference and clock A as moving, then A ticks slower than B. This is a logical paradox, but it obviously arises out of the wording. It should have been: "if clock A is considered as the reference and clock B as moving, then B ticks slower than A, as observed by A. If B is considered as the reference and clock A as moving, then A ticks slower than B, as observed by B."

What does this tell us about the "real tick rate" of moving clocks? Actually nothing, except perhaps that there cannot be an absolute difference between the tick rate of clocks A and B, provided that they are moving inertially in free space. The differences are relative (or apparent) and caused by a measurement issue. What is this measurement issue?

The isotropy of the propagation of light

It originates in the second postulate of SR – Einstein's assumption that the speed of light must be constant in every direction, irrespective of the inertial movement of source or observer. From this follows the Einstein definition of simultaneity, which boils down to a convention for synchronizing static clocks inside inertial frames.

Einstein made the reasonable assumption that one can send a time-stamped light signal to any observer that is stationary relative to us. The receiver, knowing the coordinate distance from us, can then add the light travel time to that distance and set her clock accordingly. This is known as the "Einstein method" for synchronizing clocks in inertial frames. This, of course, ensures that light will propagate at c in all directions, as measured by clocks thus synchronized.

The rate of clocks

The Einstein synchronization method also ensures that we cannot measure the "real" rate of inertially moving clocks. We need two clocks (say A and C) that are separated in space to compare the rate of the moving clock (say B) to. B will however not agree that clocks A and C are synchronized and is entitled to say: "you guys can tell me nothing about the rate of my clock versus yours, because your two clocks are not correctly synchronized, IMO".

Amazingly, it seems to be impossible to measure the relative rate of two inertially moving clocks during a single pass – you need at least two passes. To achieve that, you must take one (or both) of the clocks out of being inertial – you need a gravitational field or acceleration, both of which destroy SR's applicability.

A different view

As a point for discussion I want to put forward this view: clocks in relative inertial motion all tick at some universal rate, but we have no means of detecting this. What we can detect is the relative, coordinate dependent rates of clocks, which depend on a convention for synchronization of clocks. It is a very good convention, it seems, because SR predicts exactly what we actually measure.

The situation is best described my means of a Loedel diagram^[1], as pictured here. Objects static in each frame (A or B) have wordlines progressing at the same rate up the respective time axes. The dotted red and blue lines of simultaneity do however indicate how each view the others clock as "running slow". When the blue observer's clock reads t=4.25, she observes the red clock as reading only t'=4.0 and vice-versa.

Jorrie

[1] The Loedel diagram is due to E. Loedel (Geometric Representation of the Lorentz Transformation. Am. J. Phys. 25: 327, May, 1957). It is just a special case of a Minkowski diagram, with two inertial frames in symmetrical motion in opposite directions relative to some imaginary "neutral" inertial frame. When precisely two inertial frames are required, the Loedel diagram is perhaps the easiest (rigorous) representation of SR. You can read more on Loedel diagrams on my website.

The so-called "twin paradox" remains difficult to wrap one's head around. The graphical solution to the "paradox" presented here goes a long way towards demystifying it. The secret of this solution lies in the vivid illustration of the accelerations involved, rather than just viewing inertial frames. It clearly points towards the asymmetries in the reference frames of the respective twins.

Let's just recap what the "paradox" is all about. Pam sets off on a long, fast round trip while Jim, her twin brother, remains on Earth. When Pam returns, they find Jim to have aged more than Pam due to the velocity time dilation that Pam experienced relative to Jim. To put figures to it, let's round the distance to Alpha Proxima to 4 light years (ly) and say Pam quickly accelerate to a speed of two-thirds of the speed of light. The round trip will last 12 years of Earth time. Pam experienced a time dilation factor of around 0.75, so she would have ages only 9 years during her voyage (three quarters of Jim's 12 years).

No problem, except that one may ask: Pam could have considered herself as stationary and that it was Jim that flew away from her. In such a case Pam should have ended up older than Jim. This is the essence of the "paradox". The solution lies in the asymmetry due to the fact that Pam did the physical acceleration at the start and end and also at the turnaround point, while Jim remained at rest (or inertial) for the whole time. This is a correct interpretation, yet it can also be confusing, because it has been proven that acceleration per se does not influence the rate of clocks. The bulk of the voyage is in any case inertial (cruising at constant speed), so where does the difference in the ages of the twins come from?

In the variant of the twin paradox presented here, Pam accelerates or decelerates all the way to the turnaround point and back. She maintains a positive acceleration for the first quarter (halfway to the turnaround) and then decelerates until the turaround point. The return trip is essentially just the reverse of the outbound leg.

The round trip to Alpha Proxima and back (8 ly distance) is chosen to again last about 12 Earth years, which requires an acceleration/deceleration of 0.8g (80% of normal Earth gravity) as measured by the spaceship accelerometers. This is called "proper acceleration" and is not the same as the coordinate acceleration as observed by Jim. Relativistic effects will cause the coordinate acceleration to be less than the proper acceleration that Pam feels.

Pam will reach a maximum relative speed of 0.924c at the halfway point (on both the outbound and the inbound legs), resulting in Pam aging only 8 years during the 12 year voyage (Earth time). The math for calculating this is slightly complex, but do not worry about it. The graphical representation below is much more user-friendly!^[1]

The left graph shows how Pam gradually accelerates and decelerates at 0.8g, as viewed from Jim's inertial frame near Earth. She reaches about 60% of light speed after one year of Jim's time and 92% of light speed after three of Jim's years, which is the halfway point.

The blue bullets on Pam's worldline show how she ages relative to Jim. At the halfway point, her calendar reads only 2 years, meaning a total of 8 years during will elapse for the total voyage. In this type of diagram, the more the worldline deviates from the vertical, the more the spacing between the "year-ring" markers is stretched. If the slope becomes 45 degrees (the speed of light), the spacing is stretched to infinity, i.e., no aging takes place. At the 92% of light speed that Pam reaches, the "time stretch factor", also called gamma", is about 2.55.

The green worldline belongs to a hypothetical star near Alpha Proxima that is at rest relative to Earth, meaning there is no stretching of the time scale. The star time runs in step with Earth time, or more technically, they have a common inertial frame of reference.

The right-hand graph shows the same scenario from Pam's accelerating frame of reference. Her acceleration (and the reversal thereof) has interesting effects on the worldlines of Jim and the star, as viewed by Pam. We see sudden changes in the slope of the worldlines, as well as loops in spacetime! A sudden change in the slope of a worldline indicates a sudden jump in apparent velocity of the object. Loops indicate apparent spacetime movement backwards in time. How can this be?

Let's take the two issues one by one. The sudden changes in the slope of the worldines of the right hand graphs happen where the acceleration reverses (e.g. at the red 3-year marker). A positive acceleration means that the relative velocity increases and so does the Lorentz contraction. As Pam flies away from Earth at increasing speed, the distance between Pam and the Earth appears Lorentz contracted more and more. The result is that the apparent speed of recession of Earth is less than what it would have been without Lorentz contraction. Therefore the slope if the red worldline from 0 to 3 years is less than what it would have been.

When Pam's spaceship starts to decelerate, the relative speed between her ship and Earth starts to decrease. The Lorentz contraction then becomes less and it appears as if Earth is moving away form her at a greater speed. In fact the apparent speed eventually becomes superluminal, with a slope "flatter" than that of light (in this case, with a slope of less than -1). This is not a "real speed", but simply as things appear from an accelerating frame of reference. One must remember that even the speed of light is not the same in all directions when measured in an accelerating frame of reference.

The second interesting effect is the apparent time reversal of the right-hand curves. This comes from the differences in the definition of simultaneity between Pam's accelerating reference frame and the Jim/star inertial frame. As Pam picks up speed relative to Jim and as the distance between them increases, their clocks get more and more out of synchronization. The dotted lines connecting Jim's frame to the star frame for corresponding times illustrate that - they are lines of synchronization for Jim's frame, while Pam's lines of synchronization are all horizontal. So, no clocks go backwards in time; it is just when Pam projects Jim's clock along his line of simultaneity that the "loops in time" appear.

Interestingly, the green star worldline does not have a loop, but still displays the "backwards time" characteristic (see the main graph above). In effect, the bottom-right green curve (from 0 to 6 years) is a left-right mirror image of the red curve from 6 to 12 years. This is simply because Pam's turnaround happens at the location of the star, but it happens 4 ly away from Jim. If Pam would have turned around at Jim's location and headed out to the star again, the green worldline would have displayed an identical loop to the red one.

How does this solve the "twin paradox"? Recall that the "paradox" is about viewing the situation from the two different inertial frames and apparently getting different (paradoxical) results. The above analysis shows that the results are the same when viewed from either Jim's or Pam's frame of reference - Jim ages 12 years while Pam ages 8 years. The fact that Pam accelerates and Jim stays inertial makes the situation asymmetrical and there is no valid argument for a paradox. The inclusion of the acceleration phases in the graphs makes this a lot more visible.

Do you agree? Just click "Reply" below and let us know...

Jorrie

[1] The math can be found in the downloads from this web page. The main graph is also available on this page, where it is viewable on a larger scale than what the CR4 editor allows.

Having just read Janna Levin's excellent book "How the Universe got its spots", I'm still pondering her statement that the universe probably appears to be infinite, while it is most likely to be quite compact. "Quite compact" she admits, may mean that it is possibly larger than what we can observe, but on the other side of that horizon, it may just repeat itself endlessly, without actually being infinite. Huh?

OK, I don't quite understand all that she wrote about the topology that makes such a scenario possible. What I do understand is that there should be a limit to the size of the universe - infinite things are not really describable by math, science or in words. If the universe is now infinite, it must have started out infinitely large at the Big Bang. Huh? Again!

In principle, something that is finite can never become infinite in size through growing bigger, because then it must have a size and a thing with size cannot be infinite in size, not so? A finite universe is the only one that makes sense.

Fortunately, the best data that we have suggest that the cosmos may be just on the closed side of flat. Closed means it is compact, however big that "compact" may be. It also means that it can be finite without an edge, because it can fold around on itself, much like Earth's surface has no edge because it is folded into a sphere. So how big must such a 'spherical cosmos' be? Here's how big:^[1]

This picture suppresses one spatial dimension, so that the universe is portrayed as a two-dimensional surface that is curved into some fictitious hyperspace, hence forming a hypersphere. We have no access to the other dimension (the inside) of the hypersphere. Our observable cosmos is pictured as the small yellow circle on the surface of the hypersphere, around 28 Giga lightyears (Gly) in diameter, with us in the center of course. The observable 'horizon' appears to be 14 Gy from us in light travel time. Actually, it was about a 300 times smaller (~0.045 Gly radius) when the light that we now observe left the horizon. Due to the rapid early expansion, the light took 14 Gy to reach us.

During the 14 Gy that the light was in transit towards us, that region expanded ~1000-fold, from 0.045Gly to around 45 Gly radius today (the white circle, diameter 90 Gly). We will never see the horizon as it "looks" today, because light will take 45 Gy to reach us! This "observable universe" can however not be the total cosmos - observations of the visible part point to more of the same on the other side of that horizon.

Measurements of the overall curvature of space (terribly difficult due to the slight curvature) indicate that a closed universe must be at least 800 GLy in circumference, almost 10 times the present diameter of the observable universe.^[2] We use circumference here and not diameter, because we are talking of something similar to the circumference of the Earth. The previous two regions (circles) are characterized by their diameters, because they are situated on the surface of the incredibly large expanding "ball" of hyperspace. So large that on the scale of our human experience, the hypersphere is pretty much "infinite" in circumference.

In the end, it appears as if the cosmos is so big that we don't need to care about its size. Also, the cosmos probably doesn't care much about us either – we are just too insignificant in comparison. We should rather care more about the three-dimensional oblate spheroid that we call home, Earth, Gaia, or whatever. At least we know what size it is and it is surely not big enough to ignore our wrongdoings…

Jorrie

[1] Background artwork is the Cosmic Microwave Background as measured by WMAP. My circles are not quite drawn to scale.

[2] From latest WMAP data by NASA. I have used rounded values, just to illustrate the points. The 800 Gly minimum circumference comes from the maximum value of the curvature parameter today, Ω_c ≤ 0.246+0.1 ≤ 0.346. Then circumference today ≥ Observable universe diameter today x Π / 0.346 ≥ 800 Gly.

This is an 'engineering-view' of relativistic orbits, written in the terms and mathematics that most engineers are comfortable with, yet it is scientifically rigorous. It is an attempt to demystify the rather complex behavior of relativistic orbits.

The relativistic orbit of a particle with mass m at distance r from an uncharged, non-rotating black hole with mass M can be obtained from a set of 'flat space' equations of pseudo-forces^[1], a-la Newton's gravity:

Here g_tt = 1-2GM/(rc²) , the gravitational redshift factor squared, valid for r > 2GM/c². Force F_r is the radial gravitational pseudo-force on a particle moving with radial speed v_r and transverse speed v_t. It is easy to spot that it is the Newtonian force, modified by some relativistic terms. Force F_t is an apparent transverse gravitational pseudo-force on the particle, as is evident from its trajectory - it does not exist in Newtonian gravity. Numerical integration of equations (1) and (2) gives the relativistic orbit.

The origins^[2] of these equations will be discussed later, but in brief: the normal Newtonian gravitational force is modified by the curved spacetime through which the orbiting particle moves. It causes additional apparent pseudo-forces on the particle, as viewed in the Schwarzschild coordinate system. Below we will discuss the pseudo-forces term by term and show their effects on the orbit of a particle around a black hole.

For a pure Newton orbit, we simply set c => infinity, hence g_tt = 1, F_r = -GMm/r², F_t = 0 and we have the familiar Kepler ellipse that repeats over and over. The black circle at the focus of the ellipse represents the Schwarzschild (event horizon) radius R_H=2GM/(rc²) of the point mass, drawn to scale. The small blue circle represents the particle at time t. The periapsis of the orbit is at 5R_H and the (transverse) velocity at that point is v_t = 0.3744c, chosen for a convenient orbit size and some other esthetical considerations. All the orbits shown here have these initial conditions and are drawn to the same scale.

If we set c to its normal value, but ignore the influence of velocity, i.e., we only consider the effect of gravitational time dilation (or redshift) g_tt = 1-2GM/(rc²), the radial force F_r becomes:

The Newtonian force has been multiplied through for clarity. That positive 2(GMm)²/(r³c²) term reduces the magnitude of the negative Newtonian force and modifies the orbit as shown below.

Apart from a larger ellipticity due to the lesser gravity, there is also a retrograde precession of the orbital ellipse. The closer to the event horizon, the more the Newtonian force is reduced, meaning that the orbit curves less when close to the black hole. This causes the orbit to precess in a retrograde direction.

If we now also factor in the effect or radial velocity v_r, while ignoring transverse velocity, we have the radial force as:

The radial velocity term reduces the magnitude of the negative Newtonian force further and gives the orbit an increased ellipticity. The radial velocity does not change the orbital precession appreciably, so the retrograde periapsis shift remains. At the periapsis, the radial velocity is zero, so there the third term vanishes. Near the periapsis it is small, explaining the small effects on ellipticity and precession.

That third term can be viewed as a sort of 'anti-gravity' effect that is proportional to the inverse square of the distance from the hole and to the radial speed of the particle squared. It is however a pseudo force that is due to the curvature of space and the relativistic effects of velocity and sadly, it cannot be used to propel a spaceship.

Next we also factor in the effect or transverse velocity v_t, but we keep the transverse force F_t = 0 for now. The radial force equation is now complete:

Note the differences between the radial and transverse velocity terms - apart from the factors +3 and -2 respectively, there is a g_tt below the line in the radial term, caused by the fact that radial movement encounters changing spatial curvature, while pure transverse movement stays at a constant curvature.

The fact that the last (negative) term works with the normal negative Newtonian force, drastically reduces the ellipticity of the orbit and surprisingly, it causes the orbital precession to become prograde. This is due to the large value of v_t near periapsis, causing a stronger inward force and hence the orbit curves more than in the previous cases. The prograde precession for the particular initial conditions is about 25 degrees per orbit, periapsis to periapsis.

One may think that this is it - there can after all only be radial forces in the symmetrical gravitational field of a 'point source'. Not so in the curved spacetime around a black hole.

Only when we also consider the transverse force, eq. (2) above, do we have the full relativistic orbit. It has an increased ellipticity and an increased pro-grade precession, caused mainly by the transverse pseudo-force. The particular orbit chosen precesses by exactly 90 degrees per orbit, so that after four revolutions it repeats itself, creating a simple four-leaf pattern.

Note from eq. (2) that the transverse force is dependent on the product of the radial and transverse velocities. If the product v_rv_t is positive, i.e., during the outwards portions of the orbit, the transverse pseudo-force works with the orbital movement. During the inwards portions of the orbit, v_rv_t is negative and the transverse pseudo-force works against the orbital movement. This balances the energy equations nicely, but it must obviously cause deviations from Newtonian orbits.

It is clear that the transverse pseudo-force causes the bulk of the periapsis precession of relativistic orbits. It is however not a real force, but again an artifact of the Schwarzschild coordinates chosen. While the particle moves in the curved spacetime, its coordinate positions over time correspond to all the pseudo forces discussed above.

It is hoped that this post and possible discussions that may follow will make relativistic orbits a little less mysterious!^[3]

Jorrie

[1] All Newtonian gravitational forces on a point particle are pseudo-forces, because a freely moving particle does not 'feel' accelerations. It is only when the particle's movement is measured in some static coordinate system that one can deduce that there is a Newtonian forces acting on it. In general relativity, such motion is called geodesic movement and is force-free.

[2] The mathematical details (and a lot of explaining) of the relativistic orbital accelerations are contained in Appendix C of Relativity-4-Engineers.

[3] Also look at the beautiful patterns of the highly relativistic orbit in the previous post in this CR4 Blog again. Note that that 'four-leaf-clover' performs a few close-by, nearly circular orbits around the black hole before zooming out to the apoapsis and back again. The same equations and principles do however apply.

-J

In my previous Blog entry, the pretty chaotic orbit around a pair of binary black holes was shown. Below is a more sedate orbit, perhaps useful in the far, far future. It may offer spacecraft an "easy escape route" from binary black holes.

Consider orbiting black holes, each of mass M, separated by distance 50M in geometric units^[1], oriented as shown at time t=0, with constant orbital speed V_M = 0.1c (orbit period T=1571M). Let a particle fall from rest at a coordinate distance x=75M. Since the gravitational field is not isotropic, one can expect the particle to fall in with a curved path, as shown in the next figure.

The particle is initially dragged towards the top hole and as the bottom hole comes closer, the path bends downward. When the particle crosses the orbital radius of the holes (25M), it reaches a speed of 0.252c. This is in fact slightly less than what it would have been in a pure Newtonian case, as will be explained later. The normalized 'geometric time' of t=494M can be converted to seconds by multiplying it by G/c³, where M is the mass of the holes in kg.

At t=575M, the particle has taken a 'slingshot' around the hole and is crossing the hole orbit radius again, with a significant gain in velocity (now 0.374c). This is the relativistic 'gravity assist' or 'flyby' maneuver, providing considerably more 'delta-V' (almost 50%) than what a 'Newtonian flyby' would have suggested. The reasons for the extra gain in the relativistic case will be discussed later.

Finally, at t=754M, the particle reaches its original distance (75M) from the center again, but with enough velocity (0.27c) to escape from the black holes (the escape velocity at distance 75M is 0.233c). As can be deduced from the diagrams, the 'test' lasted for slightly less than half a full orbit of the black holes. So, starting from rest in the coordinate system, the particle gained enough energy from the moving black holes to exceed escape energy - quite impressive!

The principle of gravity assist is simple: pass behind an orbiting body (without entering a closed orbit) and you will gain energy from the body; pass in front of it and you will lose energy to the body. If you have to pass on both sides, like in this example and want to gain energy, pass closer behind the body than what you pass in front of it. This is a 'juggle' that NASA has to do almost every time with their interplanetary probes - in order to save fuel and enable more payload to be carried.

I will post some equations in the next Blog entry and attempt to explain the differences between the Newtonian and relativistic slingshot effects. BTW, this does not cause the "flyby anomaly" experienced by many spacecraft. In the weak field, low velocity limit of solar system exploration, the relativistic corrections to Newton's equations are many orders of magnitude smaller than the measured anomaly.

Jorrie

Notes:

[1] Geometric units normalize c and G to unity, so that mass, energy, distance and time are all expressed in meters (or cm in older books and papers) and velocity is dimensionless. This simplifies equations and calculations considerably and it is very easy to convert the results back to SI units. The conversion factors are: G/c² = 7.41 x 10^-28 m/kg and G/c³ = 2.47 x 10^-36 s/kg, or the inverses if you need to convert the other way around. As an exercise, you may want to plug in one solar mass (~2 x 10³⁰ kg, or ~1480 meter) for each black hole and check the real times of the experiment.