INSTANTANEOUS COMMUNICATION

 

 INSTANTANEOUS COMMUNICATION

 

 

IS INSTANTANEOUS COMMUNICATION POSSIBLE?

DOES IT VIOLATE OUR UNDERSTANDING OF THE LAWS OF PHYSICS?

 

In order to visualize the communication problem let us start with one of the previously used relativistic cases (see Fermisquestion.com) shown in Fig 1.  Here we have a frame of reference moving past our frame at a velocity of 0.4C, four-tenths of the speed of light.  Our time, in Fig. 1 is exactly zero TU’s (time-units).  Note times in the other frame.  One of their clocks, for example, at X’=50 reads -20 TU’s and that clock is at approximately +46 distance units along our X axis.  This is of course a representation of a moment in time (our time) but in the moving frame this situation is spread out over an infinite amount of time.  Specifically in just the range shown in Fig.1, the scene in the moving frame is spread over 100 distance units, and 40 time units.  We learn very quickly in special relativity that what is considered simultaneous in one frame is not so in any other relatively moving frame. 

 

 

Fig. 2 shows the same relativistic situation but at a different time. Here we can see what happens as time moves ahead.   Distance here, as before, is collapsed to a singe dimension, X and the time in our frame is 40 TU’s.

 

Question:  How can we show the situation from the viewpoint of both systems on one diagram?   See Fig. 3.  Here the blue shaded lines and circles represent the moving frame.  This image, although created as a computer graphic,was not computed and plotted by software thus the plots are approximate. 

 

Consider the blue line labeled “view of x’ frame when all t’ clocks read zero”.  The blue discs along that line show a “0” next to them which represents the time at those clocks.  Thus, in the moving frame, everyone along the blue line (in our past and future) experiences the moment of simultaneous zero time.  It seems counter-intuitive that those in the moving frame (along that line), and everyone in our frame at zero time each feel certain of the validity of their own zero time particularly when the clocks in both frames seem so disconnected. 

 

In Fig. 3, the time axis is labeled “Universal Relative Time” and will be explained later.  For the present discussion the time axis represents the time in “our” frame and when we experience time=zero TU’s we consider only a horizontal line passing through zero of the time axis.  The small blue disks represent the moving frame clocks with clock readings shown next to some of them.  The red grid lines provide the time and distance coordinates in “our” frame of reference.  Note the correspondence to Fig. 1 taking note of the X distance along the bottom of the image. 

 

We are fully aware that the members of the moving frame synchronized their clocks with each other using the very same techniques we used.  They are just as confident that their time system is as valid as ours.  And yet, it seems that differences are irreconcilable. 

 

We set up Fig. 3 in the same way we set up the earlier graphics:  We assume that both frames of reference are synchronized such that when the origins of both systems cross each other, the clocks at the two origins read zero.

 

In Fig. 4 we show, in green, the time lines of signals emanating from our origin with various velocities measured in our frame of reference.  We are showing “Faster Than Light”, FTL, signals as though they are real.  We can do that as explained in Fermisquestion.com.  We wll assume that we have programmed a linear array of strobe lights to simulate those signals allowing us to observe the behavior of the signals in differing frames of reference.  Note the signal shown going in the plus X direction at 2.5 times the speed of light.  (It is seen to travel 50 X-distance units in 20 time-units).  Note that that signal lines up with the blue line which represents those clocks in the moving frame which have a zero reading.  Thus this signal, which is moving at 2.5C in our frame is also a signal which is seen in the moving frame where every clock has the same reading, zero.  That means that this signal is moving infinitely fast in the moving frame.  This is another, more convenient view, of what was discussed earlier.  The 2.5C velocity in this case is the negative reciprocal of the relative velocity of the two frames, 0.4C.  (From the viewpoint of the moving frame, our frame is moving along the negative X’ direction).

 

Next, consider Fig. 5.  Here we show the plots of signals of various speeds which were sent on their way from the moving system’s origin when the clock there read zero.  The speeds of the signals, V’, shown in Fig. 5,  are speeds measured in the moving frame.  Those speeds, except for the speed of light, C, will always be different when viewed in a different frame. 

 

Of interest in Fig. 5 is the fact that any signal emanating in the moving frame going in the increasing X’ direction will always move up, in the direction of increasing time.  That is to say those signals move into our future.  However, some FTL signals moving left are seen going into our past.  In particular, a signal going left in the moving frame with a velocity of 2.5C will appear in our frame as infinitely fast going left.  Furthermore, any signal greater than 2.5C going left in the moving frame will be seen going backward in time in our frame!  Thus we can conclude that by using two different signals traveling at two different speeds (each greater than 2.5C) transmitted from the moving frame’s origin, communication is possible between two different points in time at the same X location!  (A signal of velocity -10C is shown as an example of a signal moving backward in time). 

 

Thus, this paradox provides a strong argument against the existence of FTL signals.  Sort of.  There is however a loophole.  A very narrow one.  Of all the possible infinite number of FTL signals there is a single one for which the above presented time paradox does not exist.  If it were possible to make a case for a special frame of reference, a universal frame which encompasses the entire universe, and if it were also possible to generate an infinitely fast signal in that frame, then the general injunction against FTL signals would not apply in just that case.

 

 

 

 

 

IS A SPECIAL, UNIQUE UNIVERSAL FRAME POSSIBLE?

 

Einstein was not aware of the expansion of the universe when he formulated ideas of special relativity.  Even allowing for the expansion of the universe, special relativity is special in the sense that SR frames of reference are inertial, that is, non accelerating.  Space in SR is flat, thus SR applies to a limited region of a non-flat universe. 

 

One can, however, construct a special frame of reference that is special within a limited region.  Consider measuring an angle to every visible galaxy out to some distant limit.  Then make a spectral measurement of each radial velocity.   This might include millions of measurements.  That’s OK.  If a vector summation of every measurement yields a vector of nearly zero velocity then our velocity relative to the universe is also nearly zero and we are considered to be “co-moving”.  Considering the expanding balloon analogy … a stationary ant on the expanding balloon would be, in a sense, non-moving relative to the geometry of its balloon universe.  Wherever in the universe this vector-velocity summation is made, only a co-moving observer will obtain a zero result for that region of the universe. 

 

If the result of the measurement is a non-zero velocity vector, then that vector may be used  as a correction factor to measurements made in the measurement frame in order to find the proper co-moving values in the region. 

 

Next, What about time?  Once a co-moving frame is established, the time since the big bang may be determined by backtracking the motion of the galaxies to the beginning of time.  Clearly, however, some model of the universe and its evolution with time would be needed here and it is assumed that all sufficiently advanced societies would have worked this out and have reached the same conclusions.  Universal time standards could be established using vibrational frequencies of certain atoms which would hold true wherever in the universe the measurement is made. 

 

Thus with a universal standard time measured from the big bang, TBB, and a unique geometrical frame of reference and any location, the entire special frame of reference is defined for a local region.  What remains is to seamlessly hook up all frames everywhere in the universe and the theoreticians should be able to do that.  The seamless hookup is not, however, essential to draw conclusions from this view of geometry.

 

Fig. 6, taken from Ned Wright’s Cosmology Tutorial  http://www.astro.ucla.edu/~wright/cosmo_01.htm   is  presented here as an example of a representation of a universe having critical mass density.  Time in this diagram is shown, as in the previous diagrams,

along the vertical direction with any constant time as a horizontal line.  The present time, is the uppermost row of triangles.  In Fig. 6, all galaxies, (triangles) are considered to be stationary on this diagram, and are thus co-moving.  They are thus expanding with the universe and each has a velocity relative to every other which is exactly proportional to the distance separating them.

 

We now return to Fig.3.  At this point in the discussion, we will consider that the situation shown here represents a co-moving frame of reference.  The time axis shown here, “Universal Relative Time” is now based on absolute time, TBB, with an offset to provide a convenient zero.  Any frame moving relative to this frame is a non-co-moving frame and would be represented by tilted world lines as shown by example in Fig. 3. 

 

Now we establish some rules.  We allow for the existence of an infinitely fast signal (IFS) in the co-moving frame, and ONLY in the co-moving frame.  No other IFS signals are allowed in any other (non-co-moving) frames.  These restrictions guarantee that communication paradoxes will not exist between two points in time at the same location with any configuration of frames.  Fig. 3 shows that the IFS moves only horizontally (thus never backward in absolute time) and regardless of any tilt of a moving frame, time paradoxes will not result.

 

It’s not clear that there is any injunction against the existence of FTL signals were it not for unacceptable paradoxes.  But as explained above, there is none for the one special case of an IFS in and only in co-moving frames. 

 

How can we generate an IFS signal?  Maybe we have a starting point in quantum mechanics with entangled particles.  It is well understood that there is no known way to use the entanglement phenomenon to send manually generated information at infinite speed.  But, on the other hand, the study of entanglement might provide some clues that might lead to a different understanding of the way the elements of our universe are interconnected. 

 

An important question arises at this point.  Consider the following:

 

 A technological society, such as ourselves perhaps, discovers a technique for generating an IFS  but hasn’t worked out all the details.  All clocks are properly synchronized and the experiment is tried and indeed communication between two points is established and the delay between the time the signal was sent and a reply received is virtually zero.  However, the signal does not appear at the same clock reading wherever it is detected within the frame.   

 

After some study, it is finally realized that the reason for the different clock readings is that the frame of reference is not co-moving and the difference in the readings reveals that fact and also provides a way of actually measuring the co-moving error, and thus the local absolute velocity.

 

The question is:  What is there about the signal, or perhaps about the non-co-moving frame of reference that causes the clock readings to be different from each other?  Is there something physically different about the non-co-moving space that can provide a tangible explanation for this result?  One possible answer might be, and this might be a very undesirable pill to swallow, that there is an aether of sorts.  We would not wish to resurrect the aether that was already shown to be non-existent in the Michelson-Morley experiments.  That was a luminiferous aether … light bearing aether.  The aether we need has nothing to do with light but is needed to support, in some manner, an IFS.

 

Thus the aether we need permeates all of space throughout the universe and is itself precisely co-moving with the expansion of the universe.  Therefore all co-moving frames of reference are “stationary” with respect to this aether that they are immersed in.

 

It should be mentioned again that although IFS communication in non-co-moving frames reach back in time as measured by local clocks, paradoxes will not result because the IFS never goes backward with respect to absolute cosmic time, TBB, at any location.   

 

Now ...  What is “nowness”?

 

We don’t think too much about a feeling we might call for want of a better word ... “nowness”.  We sit and talk across the room to each other and we are quite sure we are experiencing a common feeling of the present.  When someone will be on Mars and we experience many minutes of delay we will still not have any difficulty with time awareness.  What happens, however, if we are part of a spaceship convoy, say a light-month in length and we fly past another such convoy going in the opposit direction with relative velocities between us of say half the speed of light?   Here we are then, in pretty much the same local region of the universe and we look across at each other.  Clocks are, of course, nowhere near in agreement.  We look along our convoy and we have a definite sense of nowness knowing how we synchronized our clocks.  So do the flyers in the other convoy.  Here a sense of nowness becomes a little more difficult to understand.  Suppose telepathy is real.  Telepathy, we presume works at infinite speed.  Since constant velocity frames are all equivalent can both frames have an infinitely fast telepathy?  What about nowness when we think in terms of intergalactic distance?  Can we share nowness with with someone in Andomeda?  Can we even possibly define how we would establish such a thing? 

 

Special relativity gives us the transformation equations to transform space and time values between different frames of reference.  But our inner sense of time is unprepared to operate on such a scale of space and velocity.  Can we consider that we and the Andromedans are sharing an actual nowness even tho there is no known way to confirm it. 

 

An infinitely fast signal, IFS, appears at every point in the universe at the same cosmic time ... time since the big bang.      Look at fig 4 ..  every point in any frame of reference, regardless of speed and clock readings which differ from our clocks,  every point which can be considere NOW exists only along the horizontal line at time=0.  An IFS reaches exactly those points and none other above or below.   At any later time, the IFS is still horizontal in this picture and can still only reach points along a horizontal line.but later than before.  Thus IFS signals can produce no paradoxes. 

 

Heinrich Hertz created the best electrical spark he could within the confines of a small laboratory.  He was able to observe a tiny spark in an adjacent room across a spark gap in a loop of wire.  Did he somehow imagine that a shock to an electromagnetic medium was what was needed to produce a wireless signal?   Amazingly, Hertz felt the experiment was not important and that there would be little use for the “waves” he detected.   Could we somehow produce a similar powerful shock to a spacetime medium and discover a yet unknown mechanism for communication?

 

Are the Andromedans waiting to hear from us?

THE RELATIVITY TWIN PARADOX

THE TWIN PARADOX

 

 

 

 

The Twin Paradox, sometimes referred to as the Clock Paradox, is frequently used as an attempt to illustrate an error with special relativity.  In SR, two frames in motion relative to each other are in a sense equivalent.  Viewing the other from either one requires the same transformation of time and length.  Thus a dimension on B viewed from A should be exactly the same as the corresponding dimension on A viewed from B.  Furthermore,  B’s velocity measured in A using A’s instruments is exactly the same as A’s velocity measured in B.

 

Therefore, because of this symmetry, if one member of a set of twins takes a ride on a spaceship then he/she should not gain or lose any time resulting from the trip.

 

Not so.  The very short reason is that the space-time paths taken by the twins are not symmetrical with respect to each other as would be required for the elapsed time to remain the same for both frames.  Twin A, the traveler, starts on Earth, boards a moving frame for the trip out, then stops and boards another frame in the opposite direction, then finally stops moving (relative to the Earth) and debarks.  Twin B, on the other hand, does none of these movements.

 

In the following, a twin will travel out for 150 days at 2/10 of the speed of light and return at the same speed.   

 

DETAILS OF THE JOURNEY

 

Twin B jumps on a space ship travelling at 0.2c, two-tenths of the speed of light.  We can use a version of the “Relativity Window” to visualize the transformations involved.  Fig. 1 shows the scene at time zero just after twin B boarded the ship and is travelling to the right at 0.2c.  The clocks in our, A’s, frame of reference are seen all synchronized at t = 0.  The clocks In B’s frame are seen from our frame with the readings shown.  Remember that in B’s frame, B will see synchronized clocks.  If a photograph is made in B’s frame at a great distance along an axis perpendicular to the two axes shown with a camera stationary in B’s frame, the clocks in B’s frame will all have the same readings and our clocks will differ. 

 

Next, allow twin B to travel for 150 time units.   See Fig. 2.   As described in the Fermi’s Question section, these views are normalized so that time can be any chosen units.  Let’s say one TU is a day so the twin travels for 150 days at 0.2c.  At that time we observe that twin B is located at a distance of 30 light-days away and his clock is seen to read 146.97.  (146.969)

 

Red notes below are from twin appendix

 

     From (1)  t’ = beta (150 - .2 * 30) = 146.969

             Where beta = 1.0206

 

Twin B however, is not concerned with the clock opposite him.  He is concerned with the clock of his twin, back at x = 0.  He will conduct an observation which is exactly symmetrical to the one described in the previous paragraph.  See Fig. 3 in which we adjusted time in the stationary frame to the value 146.97.  In Fig. 3 one of B’s clocks reads 150 (days) and is directly opposite x = 0 which is, of course, the location of twin A.  This occurs when the clock at x = 0 reads 146.97.  From twin B’s viewpoint, all clocks in his frame read the same and so he concludes that his twin’s clock is reading lower than his clock.  Note carefully the symmetry here:  twin A observes twin B’s clock to be lower than his by exactly the same amount twin B notes A’s clock to be less than his!

 

But the symmetry stops here.

 

Let us return to the conditions of Fig. 2 and direct the twin to return to Earth.  He does this by jumping from his frame onto another frame moving left at the same speed, 0.2 of the speed of light.  We will assume the transition takes place in a very brief period of time so that we may neglect any small transitional time accumulation (or loss) and we will also assume he can miraculously survive the acceleration.  Fig. 4 shows the scene just after the “jump”.  In Fig 4 we use the same transformations we’ve used all along but we assume, for simplicity of calculation we establish a set of auxiliary frames of reference:  One frame is stationary and has its origin at twice the distance of the “jump” point and another frame, a moving frame which becomes the return frame for the traveler.  Two conditions are required for this frame;  (1) its velocity with respect to the home frame be the same as twin B’s velocity was going out and (2) the clock at the jump point in the new frame read exactly what twin B’s clock read just before the jump.  (See Twin Appendix for details)

 

We can now advance the time to t=300, the time we expect the traveler to arrive home.  See Fig. 5. 

 

Using the last equation in the twin appendix;  t’’’ = beta * 300 – 2 * 150 * 0.2 * beta = 293.94

 

Thus, if  the Lorentz transformation equations are believed,  it’s hard to escape the conclusion that the twin returns younger.  It might still be argued by some that despite the mathematics, the twins remain the same age regardless of the behavior of the clocks.  This would be a dangerous proposal because:

 

(1)  The observant traveler will then notice that his clock seems to be running slowly with respect to his metabolism.

 

(2)  It will seem to run slowly regardless of his direction or speed.

 

(3)  Therefore, it will always appear to run fastest only when he is stationary with the Earth.

 

(4)  Therefore, there is something special about the velocity of the Earth.

 

Not a very good line of reasoning. 

 

= ==============================================

 

 

TWIN PARADOX APPENDIX

 

We can simplify the picture so as to make the solution more intuitive.

 

For the first half of the journey, the outgoing half, we will simply employ the methods discussed earlier.  The return trip is a little more difficult and will be described here. 

 

The approach will be to establish a frame of reference moving from right to left which will be the frame the traveler twin moves into when he/she reverses direction to return home.  The reversed frame, O’’’, will “start” from a point on the x axis twice as far as the reversal point.  We will also define another frame, an auxiliary frame, O’’, which is stationary and also has its origin at the same point along the x axis.  The two frames, O’’’ and O’’ are related through simple Lorentz equations.  Then converting x’’ values from O’’ back to x in the home frame, O, requires nothing more than a simple translation. 

 

 

We start with the simplified, normalized Lorentz equations from the “Fermi’s Question” section.  (Refer to that section for the special initial conditions required for the simplifications)

 

(1)  t’ = beta(t – vx)

 

(2)  x’ = beta(x – vt)

 

(3)  x = (x’/beta) + vt

 

where beta = 1/sqr(1 – vv)

 

 

To find x’, t’ opposite x = 0:

 

     x’(at x=0) = -beta v t

     t’(at x=0)  = beta t

 

What is the travelers frame immediately after reversal when his/her velocity has become –v? 

 

(Reversal time is tr, reversal x is xr, the return speed will be the same as the outgoing speed, v).

 

The reversal frame is equivalent to a frame, O’’’, which had a t’’’ = x’’’ = 0 crossover at a point x = 2(xr) and with a velocity of -v.  We view the frame O’’’ as moving relative to another, (stationary), frame O’’ whose origin is also located at 2(xr).  See Fig. App 1.

 

We again use the normalized Lorentz transforms, following equations (1), (2), and (3), after reversal, to convert from O’’’ to O’’:

 

Note that v is now reversed,

 

(4)  t’’’ = beta(t’’ + v x’’)

 

(5)  x’’’ = beta(x’’ + v t’’)

 

We now convert O’’ values to O values, x, t after reversal time, tr:

 

          Converting from O’’ to O:       x = x’’ + 2 v tr

 

                   Where tr is the reversal time in the O frame.

                    Note that the last term in the above equation is just double xr; 2xr.

 

                    x = x’’ + 2xr ,       x’’ = x – 2xr

 

                   Furthermore, since O’’ is stationary relative to O,   t’’ = t.

 

          Converting from O’’’ to O’’:    

 

                   If x’’’ = 0, and since t’’ = t, then from (5):    x’’ = -v t

                   from (4), also if x’’’=0:   t’’’ = beta (t – v*v t) = t/beta = t’’/beta

 

 t’’’ (at x’’’=0) has the same value as the traveler had before the reversal.

 

                   (    t’(x’=0) = t / beta    (obtained from equation (1) above.)    )

 

 

What is t’, x’  opposite  x=0 before reversal?   t’ (opp x=0)  = beta t    (from (1))

                 x’ (opp x=0) = -beta v t   (from (2))

 

 

Next, some general results after the velocity reversal, after tr.

 

What are x’’’ and t’’’ opposite any x at any time t  > tr ?  (note that v is –v after t=tr)

 

           Since        x’’ = x – 2xr

           and            t’’ = t

 

         using (5)     x’’’ = beta(x – 2 xr + v t)

            thus          x = (x’’’/beta)  + 2xr - vt

             and          x = 2xr - vt   opposite x’’’=0

 

Next, using (4)    t’’’ = beta(t + v(x – 2 xr))          (note,  (x-2 xr) is x’’ opposite x

            Then        t’’’ = beta(t + vx – v 2xr)

 

Next, what are the values in O’’’ opposite x=0?

            x’’’ = -2xr beta  + vt beta          and          t’’’ = t beta - 2xr v beta