ANSWERING SOME TOUGH QUESTIONS

by Miles Mathis

Abstract: I will answer some penetrating questions concerning relativity from a reader, clarifying variable assignments.

I haven't written about Special Relativity in a while, but I got some recent emails that led me to publish this extension of my papers. It may be useful to some. Answering direct questions is almost always very useful. This is his first email:

Miles,

I'm reading your paper "The Discovery of First-Degree Relativity and the Refutation of Gamma."  I am confused by one of your definitions and would like a clarification.  In your gedanken experiment using the blinker in the tunnel you make the following definitions:

t' = the period of each clock, from its own vicinity.
x' = distance blinker has gone relative to tunnel marks, according to its (i.e. the blinker’s) own visual measurements.
v '= velocity blinker is going, by its own calculation.
You then give the following example of how the observer on the blinker measures his velocity:

If you are with the blinker, then you will measure your own velocity like this

v' = x'/t'

Let us say that your first blink is at the 1km mark.  Your second at the 2km mark, and so on. Obviously, your v' = 1km/s

I am confused by the use of the tunnel marks as the measure of x’ but the blinker as the measure of time.  For example, if I am stationary in the tunnel, I would measure the velocity of the blinker relative to me by measuring how long it took to for a single point on the blinker to go from one tunnel mark to the next using my local clock.  On the other hand, if I am in the blinker and make a similar measurement, I would measure the time it takes a single point on the tunnel wall to move between two markers in the blinker.  In other words I use my own x and my own t in both cases.

By using primes on all the variables in v’=x’/t’, your implication is that they are all from the same inertial reference frame—which they are not.

Was it your intention to measure local velocity by mixing x and t from two different frames?  If so, I would appreciate it if you could explain to me why?

And in a follow-up email, he said this,

I am not saying that you are not measuring a valid velocity.  In fact, in relativity jargon you are measuring the "proper" velocity.  This is defined as the ratio of distance in the other frame divided by time in the observer’s frame.  What I described to you using the train as the measure of distance is the "coordinate" velocity where time and distance are both measured in the observer’s frame.  The unique thing about proper time is that it has no upper limit.  As your train’s speed approaches the speed of light, the blinker rate will approach zero while the rate that marks are passing on the tunnel wall will continue to increase.  Clearly this means that in the limit the measured velocity approaches infinity as the speed of your train approaches c.

What is confusing is that you called the measured distance x’, which is the observer’s distance.  But if it were really x’ then the measured velocity could not approach infinity as the train speed approaches c.

Here was my answer:

OK, I see what your problem is now.  You think real relativity is x/t.  I have been told before that I am not doing relativity, since I am finding x'/t.  But, as I say in my papers, you can't define the observed velocity as x/t, since you can't be given x, and you can't get it from the given v, either.  Since all you start with in Einstein's own proof is the given v and the constant c, you can't claim that v = x/t.  Here's why, once again.  I will show it in three ways.  First, we use my thought problem, which we have already been discussing.  You are trying to use the tunnel marks as x, while I am pointing out that they are really x'.  You claim the tunnel is x because it is not moving with regard to the eye or the zero point.  Therefore the tunnel is part of S, not S'.  But remember that it is the blinker that is seeing the marks on the tunnel from no distance.  The marks on the tunnel are some very large distance from the eye.  Since distance alone is enough to cause the need for transforms (time separation), it is the blinker that needs no transform on the tunnel marks.  The blinker can read the marks right off the wall, as you have admitted.

But of course this means that the eye CANNOT read the marks right off the wall and call that x; if the eye reads the marks right off the wall, the eye is reading x', not x.  I have defined x' as the way the blinker sees or measures itself, and x as the way the eye sees or measures the blinker.  If the eye reads the marks right off the wall, that is x', and that is because the tunnel marks are at the same distance as the blinker.   Look at it this way: let us say the blinker is passing the mark “1 million km.”  The marks are labeled, right on the wall in big painted letters. The eye, looking at the blinker and the tunnel with a telescope, sees the writing that says "1 million km mark".  So he writes that down as his x for his v equation.  But is that the way he sees or measures x, or isn't he just writing down someone else's previous measurement?  The eye is not measuring that distance with a ruler, or seeing that distance, he is just writing down what someone else measured before him.  It is a pre-existing number.  That is why it isn't x.  It is x'.  For the eye to discover x, he would have to do a transform.  Since the eye is ALREADY trying to calculate a transform, he cannot use x to do it.

Now the second way to show this.  Let us say that you stomp your feet and say, "NO! v= x/t and that is all there is to it."  Then I ask you to define x.  You say that x is the distance traveled in the system S, which belongs to the eye or the zero-point or the stationary system.  Well, according to relativity, that should be the x at no great distance from the eye.  Since time separation causes the need for transforms (as is admitted), any point a great distance away from the eye will require transforms.  So, aren't most points on the tunnel at a distance from the eye?  Yes.  So they should require transforms.  If they require transforms, then they aren't x.  You don't require transforms on your own system variables.

As I show, you can't be given x.  You can only be given x'.  Your given would be some pre-existing measurement, like my tunnel, already marked off.  You have to do a transform to discover x.  You can't be given x, you have to calculate it.  And you don't calculate it with the equation vt, you calculate it with a transform, as I show.

One more run at it, for good measure.  Perhaps you say, "Well, we can get x right from vt, since that is just the definition of velocity.  Einstein gives us a velocity to start with, which is the velocity of the object relative to the eye."  And I say, no, if you are given a velocity, it can only be x'/t, and you seek transforms from that.  v= x/t only works for objects nearby and going slowly, which is why we are finding transforms in the first place.  If v=x/t worked for objects at a great distance, we wouldn't need relativity, would we?

You may say, "Well, you yourself find values for both x and t, so we can then combine them afterwards to find some v, can't we?"  We can, but that velocity is pretty much meaningless.  It isn't a relative velocity, since it isn't how things will look to anyone in the problem.  If you put x over t, you are putting a transformed distance over a local time, which isn't anyone's data.  You use a transform to go from one set of data to another, or to "normalize" an incoming set of data.  But you can't "observe" x/t from a great distance, so it won't ever be your data from a distance.  And you won't ever have a spacecraft sending you data in that form, since the spacecraft's data will be in the form v'=x'/t'.

Also, you are using your own methods to interpret my equations.  My velocity transforms for one degree of relativity in this particular problem show that as the blinker approaches c, the eye sees its velocity approach c/2.  The relative velocity is not higher than the local velocity, it is lower.  Not only does the observed velocity not approach infinity, it is actually less than the real velocity.  Doubtless you will think this conflicts with experimental data, but remember that the blinker is on an outbound trajectory.  It is the inbound trajectory where v approaches infinity as v' approaches c.  This is the situation that we find in accelerators, since the particles have to be approaching a detector.

And so I have shown that it was not me that was "mixing" variables, it was the mainstream. This reader, following the normal but faulty procedure, was misapplying his variables, specifically the variable x. The variable assignment in this problem has not been rigorous enough, from the time of Einstein, and the faulty variable assignments were enough to cause major problems in applying the math to satellites and so on. This is what caused the Pioneer Anomaly and other anomalies.

What this means is that in the "current jargon", coordinate velocity is a ghost. It is meaningless. We can only calculate what this reader calls proper velocity. Real transforms must be performed between proper velocity (x'/t) and local velocity (x'/t'), since those are the only velocities that can come up in data. When we normalize incoming data, we are changing it from x'/t to x'/t'. We are changing it from data that has been garbled by distance and velocity into local data. The fraction x/t never comes up in relativity, since x/t only applies to nearby data. We don't do transforms on nearby data.

Now that that is cleared up, we can look at the rest of his email:

I have no problem with the fact that the blinker is correctly seeing the marks on the tunnel because there is no distance involved.  You are absolutely correct that he sees the marks on the tunnel correctly and that each mark on the tunnel that he sees represents 1 Km of distance moved IN THE STATIONARY FRAME.  In other words, you are using a stationary measuring rod (the tunnel) to measure distance and a moving clock (blinker) for measuring time.  Stated mathematically the velocity you are measuring is dx/dt'.

To illustrate this, let's assume that the blinker has a rest length of 1 km.  Then x=x'=0 represents the back of the blinker and x=x'=1 km represents the front.  Now have the same blinker moving at 1 km per second when the back of the blinker (x'=0) passes the zero mark on the tunnel (x=0).  The front of the blinker is still at x'=1 km and is therefore a distance of 1 km from the origin as measured in the moving frame.  But because of length contraction the front has not yet reached the first mark on the tunnel so how can its distance measured when it does reach that mark also be x'=1 km?

Well, I have already shown that the velocity of the blinker is not dx/dt' from any vantage. From the blinker, the velocity is dx'/dt'. From the eye, the velocity is dx'/dt. But his second paragraph brings in a further complication, for it is here that he brings in length contraction. Remember that my thought problem with the blinker has nothing to do with the length of the blinker. I am calculating how the total distance travelled is different from two vantages, not how the length of the blinker is different. Of course, they are related and proportional, but they aren't equivalent. My reader also wants to look at the first interval, which is a bit confusing, as we see. It is much cleaner to look at more distant intervals, but we will let him set our table here. His mistake is all in his last sentence. He is assuming length contraction before he has calculated it, and this assumption messes up his logic. He assumes that because the blinker is length contracted, the front of it is not at 1km. But it is. The front of the blinker is at 1km in both systems. You will say, “How can the observer see contraction then? If the front of the blinker is at 1km in both systems, no one should be able to measure contraction.” Unfortunately, that question betrays a complete misunderstanding of relativity. Relativity is not caused by the fact that the front of the blinker is at different places in the two systems, it is caused by the fact that light has to travel back to the eye from the front of the blinker. That light arrives late, so if the eye assumes all the light is arriving together, he comes to the false conclusion that the blinker is shorter than it is. We then need a transform between that false conclusion and the actual length of the blinker.

We see from this that if the blinker is very long, say a million miles long, we will need a transform even with a zero velocity. The light from the front of the blinker will arrive late, therefore we will mismeasure its length. The velocity only adds to this mismeasurement. This is also true of a very long blinker or train parked a long way away from us. With no velocity, it will already appear length contracted, since the light from the front end will arrive late. It is also true of the marks on the tunnel, which are stationary. They are parked, with no velocity. But we would still see large gaps at a large distance as contracted, due to the way light arrives. A million miles at that distance wouldn't look to us like a million miles.

You will say, if the light from the more distant points arrives late, shouldn't we see the object as longer, not shorter? No. I have explained this in detail elsewhere, but I will gloss it again. To see an object, we must create an image on our retina or in our instruments. This image will be made of light arriving at our eye all at the same time. That is what an image is. So we won't be able to build an image from that late arriving light, because it is late. We must join our light from the near part of the object to light from the far part of the object, so we are using older light from the far part. To arrive at our eye at the same time, it must have been emitted earlier. It has farther to go, you see, so it has to start out earlier. Since it is earlier, the far end of the object wasn't as far away as it is now. We are seeing the near end now and the far end some time in the past, you see. So the object is squashed.

And that is the answer to my reader's other question. Although the object is at the same place in both systems, it appears to be nearer in the S system, since the S system is seeing the front end with old light. My reader says, "because of length contraction the front has not yet reached the first mark on the tunnel." But the front has reached the first mark, in both systems. The observer cannot see that it has, but it has. The problem is that the observer is not seeing the front end now. He is seeing a slightly older front end.

My reader will now say, "But x is what the eye sees, right? If the eye sees old light, then so be it. The variable x must be that contracted length!" Yes, I never disagreed with that. I do find an x, and that is precisely what it is. My difference with standard theory is that I don't put that x into any velocity equation. I don't because that is not the way we measure or calculate velocity. To develop the right transforms, we must manipulate our givens in the right way, and x is never given to us. What we are given is x', so we have to develop our equations from that. The variable x can be a piece of data, but x/t cannot be a piece of data in a relativity problem, by definition. x/t is not relative, and it cannot be relativized.

Finally, this reader points out that my equations show no limit at c for the variable v. You can see that I agree with that, though we have to let the object approach us, not flee us as in the blinker example. The limit at c is for v', not v. There is no limit for v. Although I admit that, I warn that that does not imply that information can travel faster than c. Even when I let the object approach, achieving an apparent velocity over c, no real object is going over c. Again, the calculation of v from the eye or zero-point is a false calculation. It is a miscalculation. It is the calculation of an optical or data illusion, that is all. No part of the object is outrunning light. No part of the object is beating the light back to the eye. This is very clear in my thought problem, since we find values for v over c without ever letting the object move faster than light. If the object never moves faster than light, then any message contained in the data carried by the object cannot outrun light. The object only appears to move over c, and that only because the data gets compressed. We misread the compression as a higher velocity. I encourage you to study the problem more closely, if you don't comprehend this point. It is very clear once you understand the actual motions.

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