NEGATIVE MASS IN EINSTEIN'S ELEVATOR
by
Edmond S. Miksch
ed_miksch@yahoo.com
Copyright: May 11, 2010
In the negative mass home page, www.Negative-Mass.com , we considered a gravitational field that is produced, according to the law of gravity, by massy particles. We showed that the field so produced has a negative mass density. By the principle of equivalence, we should then expect an acceleration field seen by an observer in an accelerating frame of reference to also have a negative mass density.
In one of his thought experiments, Albert Einstein considered an observer located in a spacious chest (or elevator) in empty space, far from appreciable masses. A rope is attached to the lid of the chest, and an external being pulls on the rope. An observer inside the chest has no way of knowing whether he is in such an accelerating chest in empty space, or is in a gravitational field, as on the Earth. According to the theory of general relativity, there is no difference between a gravitational field due to massive bodies such as Earth, and such an acceleration field.
For our present work we will be considering very small differences in acceleration, and need to modify this statement slightly to say that there is no difference between a gravitational field wherein the field vectors are all parallel, and an acceleration field seen by a linearly accelerating observer. A gravitational field on the Earth having parallel field vectors could be provided in a small experimental volume by massy rings placed below and around the experimental volume.
The "principle of equivalence" states that the two fields are equivalent. We should, therefore, expect that an observer who accelerates relative to field-free space so as to experience an acceleration field, would see otherwise empty space to have a negative mass density.
We require a solution of the problem of a reference frame that is uniformly accelerted in time. We assume that the distances between points on the frame remain constant in time, but we do not assume that all points on the frame have the same acceleration. That we wish to determine to calculate the divergence of the acceleration field vector, as seen by observers on the frame.
A solution is available in James B. Hartle's book GRAVITY An introduction to Einstein's General Relativity published by Addison Wesley, 2003, page 83. That solution is modified here to include the speed of light, since we prefer to use SI units, rather than units in which the speed of light is unity.
The solution in Minkowsky space of a point that is uniformly accelerated in the x directiion is as follows:
ct = x0 sinh((c/x0) τ ) Equation EE2
The solution is plotted in Fig. EE1, which follows.
The horizontal axis is the position of the point, x. The vertical axis is the time multiplied by c. At time t = 0, the point is at positiion x0. As the point's proper time, τ runs, the point accelerates with a uniform proper acceleration, and the hyperbola shown is traced out. The line denoted L is the world thread of a flash of light emitted from the origin O at time t = 0. The world thread for the point approaches line L asymptotically because the velocity of the point approaches but never reaches the speed of light.
We are interested in the proper acceleration of the point at time t = 0 and the variation of that acceleration with x0. Differentiating the right side of Equation EE1, we obtain:
d2x/dτ2 = (x0* (c/x0)2) cosh((c/x0) τ )
d2x/dτ2 = ( c2/x0) cosh((c/x0) τ ) Equation EE3
We now note that if an observer associted with the point accelerated in accordance with Equation EE3, that observer feels it as a gravitational or acceleration field directed in the opposite direction. We denote that field as a. It has the value:
We are interested in the variation of a with x0. Accordingly, we introduce Figure EE2, which represents an accelerating train. As before, the horizontal axis is position, x. The vertical axis is cT. The line segment AB represents the position of the train at time T = 0. The hyperbola passing through A and C represents the world thread of the front end of the train. The hyperbola passing through B and D represents the world thread of the rear end of the train. The line segment CD represents the position of the train at a later time. It is noted that segment CD always points toward the origin O, and the length of the segment CD does not change as seen by observers on the train. This is because the metric in Minnkowsky space is dx2 - c2dt2. The distance from any point on the train to the singularity at O, as seen by observers on the train, does not change with time.
We imagine the train to be accelerating relative to a platform which is an inertial frame. The platform is visualized as floating in space, far from ponderable masses. The train is accelerating toward the right relative to the platform. In the bottom part of the figure, the train is moving toward the left while accelerating toward the right. The train comes to a stop at time T = 0, and then begins moving toward the right. It is noted that this is a very long train. If the acceleration of the front end of the train is 1 meter per second squared, then the train has a length of about 4.755 light years. We have ahown a train with a length equal to one half of the distance from the front end of the train to the point x0, denoted O. To obtain the variation of a with x0, we fix τ at zero, so the cosh((c/x0) τ) is unity. Hence:
We are interested in the divergence of the acceleration field vector, a. Hence:
From Equation EE5, we obtain:
Substituting Equation EE7 into EE6, we get:
To put Equation EE8 in more familiar form, we note that all derivatives in the y and z directions are zero. Hence, we get:
Thus, our uniformly accelerated observer sees space to be filled with an acceleration field having a positive divergence, and thus, space, as seen by that ovserver, has a negative mass density. To obtain an expression for the mass density, we employ the law of gravity in vector form, which is:
MV is the mass density of space, g is the gravitational field vector and G is the gravitational constant.
By the principle of equivalence, the gravitational field, g is equivalent to our acceleration field, a. Hence, the mass density of space due to the acceleration field, a is:
We can now compare these results with the calculation of the mass density of the gravitational field generated by a spherical shell of massy particles. We obtained for the mass density of the gravitational field the following:
For the divergence of the gravitational field vctor, we obtained:
Comparison of Equation EE11 with Equation G3, or Equation EE9 with Equation G5, we find that the acceleration field of the present case has twice the negative mass density as the gravitational field of the case discussed in the negtive-mass.com web page. Since we trust the principle of equivalence, we expect an error in one of the two calculatiions.
Our best guess is that the present calculation is more reliable because of its simplicity. The calculation in the negative-mass.com web page involves the acssumption that the mass of the particles is not affected by their gravitational potential relative to the observer. We look to the web page: TimeRateGradient.com to explain this.
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