About Relativity and Moving Bodies' Gravity Waves

[Lisias]

That's the problem: finally some fellow orbiters got through my thick skull what a Gravity Wave is, and how it propagates on the space-time tissue.

And then I came to this: Imagine two bodies, perhaps Quasars. QA and QB. They're not moving in their frame of reference (the distance between them never changes).

From QA, a star SA is somehow ejected towards QB near the speed of light.

From QB, another star SB is ejected towards QA near the speed of light.

SA is getting away from QA and reaching QB at the same speed - almost C.

SB is getting away from QB and reaching QA at the same speed - almost C.

SA and SB are approaching themselves at almost C since nothing can travel faster than light? But so, SA would meet SB near QB - but at the same time SB would meet SA near QA, since all speeds should be near C...

How the SA and SB gravity "tides" would collide? Being the speed of the gravity wave slightly faster than the body itself (that is near, but not at, speed of light), the wave goes a bit further ahead with time, related to his body?

But how this would affect the velocity in which this wave is approaching the other body's wave, since nothing can travel faster than light and, at least intuitively (and by Newton's laws), they would be approaching each other at 2*C?

This is a paradox or just dumbness?

 

[Keithth G]

If we replace 'gravity waves' with 'photons', this is a problem in Special Relativity. The first thing to do when considering problems in Special Relativity is think carefully about the reference frames and the simultaneity of events. This means systematically applying the Lorentz transformation, which describes how events appear in different reference frames moving at different speeds. In particular, it will tell you when and where various space-time events take place in different reference frames.

In your problem, you have one rest frame in which the QA and QB are at rest separated by some distance. You have another rest frame attached to SA and a third attached to SB.

So, assuming that you can define the 'when' and 'where' of various events in, say, the reference frame in which QA and QB are stationary, then by applying the Lorentz transformation, you can work out the 'when' and where' of the same events in the two reference frames attached to SA and SB.

The 1-D Lorentz transformation is:

[MATH]x' = \frac{x-t \, v}{\sqrt{1-\frac{v^2}{c^2}}}[/MATH]
and

[MATH]t' = \frac{t-\frac{v \,x}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}[/MATH]
where [MATH]x[/MATH] and [MATH]t[/MATH] are the position and time of a space-time event in the first reference frame; and [MATH]x'[/MATH] and [MATH]t'[/MATH] are the position and time of the space-time event in the second reference frame.

In applying the Lorentz transformation, it is best to put 'intuition' aside. Just systematically work through the arithmetic to answer the question that you want to have answered.

 

[Lisias]

I was thinking on what you said.

From the Lorentz transformation (I think the solvable parts of my problem are solved by a Lorentz Boost, since the speeds involved don't change in time), I got into Minkowski spacetime. That triggered a memory from the time I had read Hawking's "Brief Story of time" (and then I understood what Hawking means on that part of the book).

So I think I understood why intuition foul me on my proposition: i can't solve the problem in Euclidean space, I need to use Minkowski's.

From that, I get that there're events from one reference frame that simply can't be observed by another one: SB reaching QA is out of the observable "cone" for SA, this event is like division by zero on SA's reference frame, it doesn't exists there. So it will "never happens" for SA.

This is correct? I'm making any sense of this?

 

[Keithth G]

Anything to do with relativity confounds normal intuition in my view. But rather than talk about gravity waves, we can talk about photons emitted from SA towards SB; and photons emitted from SB towards SA and ask "what speed do those photons approach each other.

To begin with, let's suppose that clocks at QA and QB have been synchronised (using the Einstein clock synchronisation procedure). Because QA and QB are 'at rest' relative to each other, it makes sense to talk about synchronous events. Let's suppose that SA leaves QA and the same time as SB leaves QB (in the sense that if an observer at QA sends a light signal to an observer at QB reporting the departure time of SA according to a synchronised clock at QA then the observer at QB concurs that SB left QB at the same time according to the synchronised clock at SB).

To make things concrete, let's suppose that an observer at QA sees SA leave QA towards QB at a speed of 0.95 c; and let's suppose that an observer at QB sees SB leave QB towards QA at a speed of 0.95 c. It is pretty clear that, by symmetry (if nothing else), both an observer at QA and an observer at QB will (after taking into account the finite speed of light) agree that SA and SB encountered each other at the mid-way point between QA and QB. If the distance between the two quasars is [MATH]d[/MATH], observers at QA and QB will concur that the encounter between SA and SB took place at time [MATH]\frac{d}{1.9\,c}[/MATH] after the each star's departure from its respective quasar.

So far, so good. But what do observers at SA and SB see? An observer at SA sees QA recede at 0.95 c; and QB approach at 0.95 c. Similarly, an observer at SB sees QB recede at 0.95 c; and QA approach at 0.95 c. Nothing particularly problematic there. But what speed does SA see SB approach at? Well, if you work through the various Lorentz transformation algebra, you will eventually conclude that the general form for the speed of SB as seen by SA is given by the general relativistic expression:

[MATH]v' =\frac{u-v}{1-\frac{u\,v}{c^2}}[/MATH]
where [MATH]u = 0.95 c[/MATH] is the speed of SA in the rest frame of QA and QB; and [MATH]v = -0.95 c[/MATH] is the speed of SB in the rest frame of QA and QB. In other words,

[MATH]v' = 0.99868 c [/MATH]
So, SA sees SB approach it at 0.99868 c - i.e., very close to the speed of light. Similarly, SB sees SA approach it at 0.99868 c.

Before the encounter between SA and SB, let's suppose that SA emits a stream of photons towards SB; and, likewise, SB emits a stream of photons towards SB. Using the same equation for adding speeds, it is easy to see that SA sees the stream of photons approach from SB at the speed of light; and likewise for SB. What happens of course, is that SA sees the frequency of light from SB highly 'blue shifted' and equally SB sees the light stream from SA to be highly blue-shifted.

As with photons, so with gravity waves.

 

[Lisias]

That was interesting...

If SA and SB are approaching each other at 0.99868c, after the middle of the path they will be distancing each other at that very same speed. The u and v values don't change over space or time, v' is constant.

So after the crosspoint, and in the QA/QB rest frame, SB is distancing from SA faster than SB is distancing from QB. But just a little bit.

From that point, that photon beans SA and SB are exchanging shifts to red (and something similar happens to the gravity waves). No big news.

What's interesting is that SA and SB are now distancing from each other only slightly faster than they are distancing their origin quasars. But the distance they will go is still the same, and QA will see SB arriving at the same time QB will see SA doing it.

I'm having a hard time on shutting up the intuition (who's yelling "WTF? WTF? WTF?" on my head).

I'll try to redo the math using SA as a rest frame this week to see what's happens (or where I will err), I'll go back to you on weekend.

Thanks.

 

[Keithth G]

What's interesting is that SA and SB are now distancing from each other only slightly faster than they are distancing their origin quasars. But the distance they will go is still the same, and QA will see SB arriving at the same time QB will see SA doing it.

In Special Relativity, you have to be a little bit careful in making statements like this. Underpinning that thought process are some assumptions about simultaneity which break down badly when velocities approach the speed of light.

This isn't necessarily a complete list, but Special Relativity rests upon two fundamental pillars. The first is that light travels at the speed of light in all inertial reference frames; and the second is there exists a procedure for synchronising clocks within the same inertial reference frame, the Lorentz transformation then specifies how clocks in another inertial reference frame moving relative to the first will record the time of events given a set of synchronised clocks in the firs. Synchronous events in one reference frame will not be recorded as synchronous events in another reference frame moving relative to the first.

So, although we go through a process of synchronising clocks in the rest frame of QA and QB, and consequently an observer at QA and an observer at QB will agree on the synchronicity of space-time events (e.g., the arrival of SA at QB; and the arrival of SB at QA), once we move to a reference frame co-moving with SA or SB, those events will no longer be recorded by observers as being synchronous.

All very problematic and puzzling, I know. But Nature seems to agree that this is how things operate - so I guess we are stuck with grappling with the consequences.

The best way that I know of of dealing with the failure of intuition to comprehend this weirdness is just to set up the Lorentz transformations between all of the rest frames of interest, and the grind through the tedious calculations to work out what any given observer actually would 'see'.

 

[Lisias]

In Special Relativity, you have to be a little bit careful in making statements like this. Underpinning that thought process are some assumptions about simultaneity which break down badly when velocities approach the speed of light.

I only dare to make such statement because QA and QB are on the same rest frame, and (implicitly) there's nothing more on that "Universe". And you hinted that, too.

Would the Quasars be approximating or distancing each other, we would have one additional rest frame to deal with., and it appears that different rest frames have different "realities".

An additional interesting (and this one is not hypothetical) problem is to consider two photons (PA and PB) being emitted at the same time by one star S, one going to the exact opposite direction from the other.

These two photons *will be* travelling from S at c speed in S rest frame. What I would get by using PA's rest frame, I will see later while trying to apply the equations.

 

[Keithth G]

I only dare to make such statement because QA and QB are on the same rest frame, and (implicitly) there's nothing more on that "Universe". And you hinted that, too.

OK, I agree: observers at QA and QB will agree on which events occur simultaneously. (The safest thing, though, is always to preface comments about which observer is observing what in which reference frame).

These two photons *will be* travelling from S at c speed in S rest frame. What I would get by using PA's rest frame

.

Strictly speaking, there is no rest frame for a photon: no matter which inertial reference frame you choose, it is always travelling at the speed of light. What you can do, though, is choose a reference frame that is very, very close to the speed of light and then take the limit as v -> c and see what happens.

 

[Linguofreak]

A useful resource is https://en.wikipedia.org/wiki/Relativity_of_simultaneity

 

[Keithth G]

As an example of how to apply the Lorentz transformations, consider a rod of length [MATH]d[/MATH] that is stationary in an inertial reference frame. One end of the rod is located at the origin of that reference frame; and the other end of the rods is located at a distance [MATH]d>0[/MATH] to the right of the origin.

Now, consider another inertial reference frame that is moving from the left to the right with a speed [MATH]v[/MATH] relative to the first inertial reference frame. Space-time events in the first reference frame with coordinates [MATH]{x,t}[/MATH] map to spacetime events in the second reference frame [MATH]{x',t'}[/MATH] by the Lorentz transformation:

[MATH]x' = \frac{x-v\,t}{\sqrt{1-v^2/c^2}}[/MATH]
[MATH]t' = \frac{t-v\,x/c^2}{\sqrt{1-v^2/c^2}}[/MATH]
So, the series of space-time events [MATH]\left\{0,t\right\}[/MATH] that mark one end of the rod located at the origin in the first reference frame is given by the series of space-time events in the second reference frame:

[MATH]\left\{-\frac{t v}{\sqrt{1-\frac{v^2}{c^2}}},\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}\right\}[/MATH]
But, if we re-label [MATH]\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}[/MATH] with [MATH]\tau[/MATH] we see that the series of space-time points in the second reference frame that marks that end of the rod is just [MATH]\left\{-v\,\tau,\tau\right\}[/MATH] - i.e., a series of points moving from right to left with velocity, [MATH]v[/MATH]. Not a surprising result.

But wait, let's now consider the other end of the rod. In the first reference frame, this is marked by a series of space-time events [MATH]\left\{d,t\right\}[/MATH]. using the Lorentz transformation, in the second (moving) reference frame, this maps to a series of points:

[MATH]\left\{\frac{d-t v}{\sqrt{1-\frac{v^2}{c^2}}},\frac{t-\frac{d v}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}\right\}[/MATH]
Again, if we re-label [MATH]\frac{t-\frac{d v}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}[/MATH] as [MATH]\tau[/MATH], we find that the series of space-time points corresponding to the other end of the rod is now [MATH]\left\{d \sqrt{1-\frac{v^2}{c^2}}+\tau \, v,\tau\right\}[/MATH]. This again is a series of points moving from right to left with speed [MATH]v[/MATH] but now offset by a constant distance [MATH]d \sqrt{1-\frac{v^2}{c^2}}[/MATH] from the other end of the rod.

The length of the rod
In the first reference frame, by construction, it is easy to see that the length of the rod is just [MATH]d[/MATH]. But what is the 'length' of the rod in the second reference frame? Well, our normal definition of length is usually to calculate the distance separating two points at the same time - [MATH]\tau[/MATH], say. By this measure of distance, an observer in the second reference frame would say that the rod has length [MATH]d \sqrt{1-\frac{v^2}{c^2}}[/MATH].

This is of course the well-known relativistic length contraction. But has the rod really changed length? Well, yes and no. Certainly there is no new force that is pushing atoms in the rod closer together. What has happened is that there has been a sleight-of-hand here so that the space-time events used to measure the length of the rod in the first reference frame is not the same as the space-time events used to measure the length of the rod in the second reference frame. So, it is hardly surprising then that the two observers will not agree on the usual measures of 'length'.

Proper distance
To calculate a concept of length that is invariant between the two reference frames, one needs a new measure of the 'distance' between two distance between events. That measure is known as the 'proper distance' and is calculated as

[MATH]d_p^2=(x_2-x_1)^2-c^2\,(t_2-t_1)^2=(x'_2-x'_1)^2-c^2\,(t'_2-t'_1)^2[/MATH].

And one can indeed confirm that the proper distance of the rod in both reference frames is just [MATH]d[/MATH] - i.e., the normal measure of the length of the rod.