On a previous web page, I used the Minkowski Interval to give the correct experimental result to the Michelson-Morley experiment. The Interval can be considered as having the form of an average called the geometric mean. This average was used instead of the more obvious arithmetic mean, which Michelson and Morley used in their calculation, which I believe to be wrong. Indeed their experiment gave the famous null result that light does not vary its time of travel according to being in line (longitudinally with) or across (transverse to) the Earth's motion.
I expressed the M-M expt. in terms of the Interval, as follows, equation 1:
I² = t²(c² - u²) = t²(c² + {iu}²).
The letter, I, stands for the Interval, a constant distance that both the longitudinal and transverse parts of a split light beam have to cover. The time, t, for both journeys is the same in accord with the experimental result. The (c² - u²) part of the equation refers to the light beam going with and against the Earth velocity, u. That is: (c² - u²) = (c - u)(c + u). The light beam is reflected back and forth in line with Earth orbit. And the average time is found, according to the geometric mean formula by multiplying distance over velocity back by distance over velocity forth, and taking the square root. Or eqn. 2:
t = I/(c² - u²)^1/2.
Because the Interval is in geometric mean form, the geometric mean also has to be used to average the earth-transversely reflected light beam. I argued how this was justifiable but show the working here, eqn. 3:
t² = I²(c² + {iu}²) = (I/{(-c)² + (iu)²}^1/2)(I/{c² + (iu)²}^1/2).
This is actually working backwards to show where the transverse geometric mean comes from. It is equal to eqn. 2, but for the formal differences.
Moving to the right side of eqn. 3, these two factors can be considered as two Pythagorean hypotenuses multiplied together, before the square root of that multiple is taken to give the geometric mean time, t.
Those familiar with the geometry of the Michelson-Morley experiment may recognise these two hypotenuses as the time taken for the Earth-transverse light beam to go from source to reflector and the same time taken to return.
If a graph were drawn of a half-circle, the value, -c, would be on the negative x-axis, with c on the positive x-axis. The two hypotenuses are otherwise undistinguished as they both relate to the positive y-axis, here assigned the imaginary variable, iu. No especial significance is attached to light having an alternative negative value or Earth velocity having a so-called imaginary value. These are just ways to label a half-circle's co-ordinates to track the experiment.
But what if we modify the Michelson-Morley experiment so that it has to be
expressed beyond the half-circle, to bring-in the full circle of
co-ordinates?
We could do this very easily by a mere change of sign in eqn. 1, making it
eqn. 4:
I² = t²(c² - u²) = t²(c² + {-iu}²).
Eqn. 4 makes no practical difference to the experiment. It is just a conventional difference, between iu and -iu, to signify that a transverse light beam can be transverse to the Earth's motion in either of two ways, at 180 degrees to each other.
But then there might be the following equation 5:
I² = t²(c² - u²) = t'²(c² - {iu}²).
This right side factorises in terms of: t'²(c² + {iu}²) = t'²(c + iu)(c -
iu). This brings in both the positive and the negative y-axis in terms of +iu
and -iu, as well as the positive and negative x-axes in terms of +u and
-u.
The time, t', has been indexed not to assume it is the same as t.
Notice of eqn. 5 that the two sides of the Interval are essentially the same in form. Geometricly interpreted, they would be doing the same thing except that they are doing it perpendicularly to each other, which is what the imaginary symbol, i, signifies.
It is already known what the left side of eqn. 5 means. It is the light beam at speed, c, going back and forth in line with the Earth's motion. Hence the right side must mean the light beam going back and forth transversely to the Earth's motion.
The question is: how does this new arrangement differ from the
conventional Michelson-Morley experiment? Well, it is apparent that the two
sides are on the same footing with regard to going back and forth. Therefore
a change in the M-M expt. must represent this.
Suppose that the source of the light beam is put at the center or origin of
the co-ordinate circle. The x-axis is in line with Earth motion. The y-axis
is transverse to Earth motion. The split light beams move perpendicularly
from the origin. Each is reflected once from both their positive and negative
circumference positions back to the center.
The null result of the Michelson-Morley experiment suggested light moved at the same speed regardless of earth-orientation, or any bearing this orientation might have on the light beam's orientation to some prefered orientation of the the larger world or whole universe: the problem of a supposed light wave medium, the "universal ether" and its possible prevailing wind.
If light moves at constant speed, then two simultaneous light emissions, over the same distance, should take the same time. A modified M-M experiment, that still observes these conditions, one would expect to take the same time. But, if my interpretation is correct, the Interval shows that my suggested axial version of the experiment would give different times.
From eqn. 5, the longitudinal journey time, t = I/(c² - u²)^1/2.
The transverse journey time, t' = I/(c² - {iu}²)^1/2 = I/(c² + u²)^1/2.
However, from a practical experimental point of view, this poses all sorts of problems. The algebra essentially represents two back and forth journeys crossing each other, in the middle of each leg, in effect, at the origin of rectangular co-ordinates. The beam could not be split from the same source, as in the conventional Michelson-Morley experimental set-up. Two beams would have to start simultaneously at the ends of the x and y axes respectively. And the experimenters would have to know whether or not they did, before they could know whether or not they arrived back at the same time, to their respective starting points.
I dont know whether it would be possible to have an entangled system of two quantum particles at a distance to each other, such that acting on one particle in a given direction, sent the other off, automaticly, at right angles to it.
There are progressively less simple options, which might be more practical, tho they have their own problems. Another possible scenario for a non-null M-M type expt. is shown in figure 1.
Figure 1: Extended M-M expt. with non-null result.
Figure 1 is a simple extension of the Michelson-Morley experiment. The perpendicularly split beams are reflected back to their origin but instead of being stopped there, are allowed to continue for the same length they have already covered in that direction. This means that the split beams no longer have the same destination, but, in figure 1's case, the left and bottom poles, opposite the two mirrors.
The original Michelson-Morley experiment split a beam perpendicularly with one beam aligned to the Earth's motion and reflected them back to source to time any difference in their arrival. The apparatus used is called a Michelson-Morley interferometer, because the measure or "meter" depended on the interference properties of light waves. If the light waves travel over unequal distances, the waves will be out of phase and cause interference bands of bright crests and dark trofs, where the light waves have respectively reinforced and neutralised each other.
The classical experiment produced a null result. This was unexpected because the calculation of times used the arithmetic mean to average the two-way journey in line with Earth motion. The geometric mean, as I keep pointing out, gives the correct prediction: same time taken.
You cannot use the Michelson-Morley interferometer when one of the beams
does not return to meet the other. But it might be possible to use the
simultaneity property of quantum entangled particles. These are two
sub-atomic particles that form a conserved system, so that when they are
moved apart, an action on one will produce an automatic reaction on the
other, tho it may have been removed far away. This automatic reaction is a
way of measuring time simultaneity.
The idea is that when the split beams arrive at their destinations, they each
strike one of the two entangled particles, imparting a momentum. If one knows
the joint quantum effect of a simultaneous strike, then one knows when the
light beams have taken the same time.
I dont know, not knowing more than popular physics book explanations. Ive
supposed a classical effect of resultant momentums, which may be wrong for
quantum mechanics. Never mind, the basic idea may still be usable in
conformity with actual quantum principles.
Anyway, to measure whether the beams arrive at the same time, it might be possible to use the properties of a quantum-entangled two-particle system. When each split beam reaches its destination, say it is a high energy foton, it hits one of the two entangled particles, passing on its momentum in the same direction.
Quantum-entangled particles are non-locally conserved systems. So, a
momentum applied to one particle in one direction will prompt an equal and
opposite momentum to the entangled particle. Suppose the two beams arrive at
the same time, to exert two equal perpendicular forces on their respective
entangled particles, which would - if only in classical mechanics - be sent
in equi-angular resultant directions. (See figure 1.)
I must admit I dont know whether this would work even in theory. The older
Wolfgang Pauli said he knew too much. But a little knowledge is a dangerous
thing. Quantum phenomena tend not to change too gradually but in quantum
jumps, so the vector "clocks" with resultant pointers, I imagine in the
diagram, might be wrong in quantum, as distinct from classical, mechanics.
However, the basic concept of harnassing the simultaneity property of
entangled particles might still be valid. Please consider such diagrams not
as definitive but as no more than suggestive of the sort of things that might
be done for the purpose in hand.
And we cannot be sure what will happen to anything before it has been tried.
That is the point of trying.
A less important caution is that I have used linear momentum, which is simple to show, whereas experimenters might prefer to use, say, angular momentum or spin.
The most obvious prediction of the result would be that since the light beams cover the same distance and light always moves at a constant speed, irrespective of the frames of reference, then the split beams should take the same times to arrive at their destinations. But a geometric mean calculation, that correctly predicts the null result of the classic Michelson-Morley experiment, says otherwise of this extended version in figure 1.
The classic experiment is called a null result because it contradicted the expectation of different times, from Michelson and Morley's calculation by averaging the reflection times with an arithmetic mean.
The prediction, for figure 1, is that the split beams would take different times. Using the geometric mean, consider the split beams from the source at the center of the cross-ways beams. The horizontal beam starts in the same direction as the Earth's motion, u. The time it takes to reach the first reflecting mirror is given by the distance, I, from source to mirror, divided by the light speed, c, plus Earth motion, u, it is being carried along with.
Similar considerations apply for the times given by the other three ratios in figure 1.
The geometric mean is used for the total time each split beam takes, because each time a beam is reflected, there is a change in motion or acceleration (or decelleration) of the beam relative to earth motion. (The arithmetic mean should only be used to average constant velocity.) The geometric mean time for each beam is found by multiplying their respective two times and taking the square root.
So, the total horizontal time, T = ({I/(c+u)}{2I/(c-u)})^1/2
= {2I²/(c²-u²)}^1/2.
Or so I suppose, that being the simplest continuation from using the geometric mean on the classic M-M expt.
Likewise, the total vertical time, T' = ({I/(c+iu)}{2I/(c-iu)})^1/2
= {2I²/(c²+u²)}^1/2.
It can be seen that the vertical time is less than the horizontal time. In other words, the beam split-off at right angles to the earth's motion will take less time than the beam in line. This is a non-null result for a simple extension of journeys by the split light beam in the Michelson-Morley experiment. It might seem contrary to intuition unless explained in geometric mean terms.
It may be asked why not make this experiment much simpler by having double
reflections of the split light beams, so that they meet again at the center,
without having to use quantum entanglement at all?
Unfortunately, the simplest way I could think-of, to do this, is with a
5-times reflection for each split light beam. Figure 2 only shows this
schematicly for the vertical beam to keep the diagram less cluttered.
Figure 2: 5-reflection extension of the M-M expt.
But the figure 2 set-up has its own problems. The "beam," shown in figure
2, even if no larger than a split photon, might be supposed to collide with
itself on its first reflections back to the origin - this half-way house
being essentially the original Michelson-Morley experiment. That is before
the split photons get to their next mirror reflections.
Or, would the beams "know" not to collide, on crossing thru the origin, over
a longer multiply reflected journey, which gives them different times, as
averaged by the geometric mean? Given the eccentric behavior of light quanta
(or a confused idea of them), that seems a distinct possibility, that only
experiment can reveal.
Or, is the whole relative motion of light to the earth simply irrelevant
and will the split light beam take the same time, anyway?
It cannot be entirely irrelevant if the reflection of light motion with
respect to earth motion creates an acceleration effect, which Einstein's
equivalence principle says is equivalent to a gravitational effect on the
light beam. If large enough to be measurable, there should be a distinct
difference in the slowing down effects on the earth-transverse and
earth-longitudinal light beams.
"If ... measurable" is the operative phrase, which renders this discussion academic. Presumably, to carry out the suggested extended M-M tests, the LISA experiment, with three laser-carrying satellites, triangulating the solar system, would need two more satellites, even if one of them was the Earth or the Moon as a laser base.
Another consideration is that the original Michelson-Morley test only used
single reflections that turned-out not to spoil the test. Tho, there was also
the semi-mirror that split off part of the original beam into a perpendicular
direction.
But what of five reflections to each beam? Surely that leaves a lot of room
for inaccurate measurement. Actually, light beams would not be used nowadays,
but lasers, which are single-wavelength beams of light, whose concentration
would reduce somewhat the measurement problem of the beam degrading from
being bounced about. The material of the reflectors also is a
consideration.
On a small scale, it looks necessary to move the beams slightly off
perpendicular to each other, so that each beam's mirror at the origin, for
the third of the five reflections, does not obstruct any of the end-to-end
mirror reflections. The margin of error, this would introduce, would then
have to be worked out.
The experiment would have to take place over a large scale to minimise any
distortion to the perpendicular beam-splitting from mid-mirror obstruction.
But an extremely large scale should be needed, anyway, to measure a
gravitational effect, so this mirror-setting issue is dwarfed by the
gravitational-scale issue.
If all such practical difficulties could be over-come or at least allowed-for, the basic calculation for 5-reflection M-M test is as follows:
Taking the horizontal beam thru six back and forth trips, the total Earth-longitudinal time, T = ({I/(c+u)}{2I/(c-u)}{I/(c+u)}{I/(c-u)}{2I/(c+u)}{I/(c-u)})^1/6
= {(2)^1/3}I/(c²-u²)^1/2.
Likewise the Earth-transverse time, T' =
({I/(c+iu)}{2I/(c-iu)}{I/(c+iu)}{I/(c-iu)}{2I/(c+iu)}{I/(c-iu)})^1/6 =
{(2)^1/3}I/(c²+u²)^1/2.
As in the case of figure 1, the figure 2 case shows that the Earth-transverse time is quicker than the Earth-horizontal time.
Variations, on the classic experiment, might give a non-null result but the slightness of the non-null effect presumably is so small as to make it disappear, if not on an extra-terrestial scale.
More tests are variations on the idea of having only one of the beams reflected, with the other beam not reflected but allowed to go the same distance in one direction, as the reflected beam.
Figure 3: Measuring non-simultaneity of reflected and non-reflected split beam over same distance.
The M-M expt. in terms of the Interval, I, is:
I² = t²(c² - u²) = t²(c² + {iu}²).
The times, t, are the same. The only difference between the two sides of the equation is one of form, with the imaginary term, iu, expressing perpendicular velocity. Suppose either of the two sides is made non-reflective.
For example, the time taken by the reflected beam in line with Earth is:
t² = {I/(c-u)}{I/(c+u)} = I²/(c² - u²).
Suppose a time, t', taken by a non-reflected beam in line with Earth, whose
relative velocity light speed is less earth's motion, (c-u'), or plus earth's
motion, (c+u').
Then t' = 2I/(c-u') or 2I/(c+u').
This scenario is a genuine variation on the M-M test, in that the split light beams have to travel the same distance, but they take different times.
From the two equations: I² = t²(c² - u²) and 2I = t'(c - u'), this equation:
I² = t²(c² - u²) = t'²(c -u')²/4.
This equation simplifies assuming the local times are the same, t = t'.
Solving for u' and taking the geometric mean (symbolised as [u']) of its
solutions in two roots, positive and negative. Then further particularising
by letting u = c, this would give a possible solution [u'] = ±u.
At any rate, there is a range of values in which the reflected and
non-reflected split beams over the same distance would take the same time,
tho in general they would not.
At this point, I broke-off from this web page, because I stumbled on an
interesting property of a statistical version of the Interval, which suggests
it has a local invariance in a radius (or amplitude) symmetry, as the
conventional Interval has a global invariance in space-time rotational (or
phase) symmetry.
It is more convenient to explain this on a further web page.
Richard Lung.
December 2009.