I Length contraction viewed with Heaviside equations

[learn.steadfast]

TL;DR Summary

Lorentz transformations were discovered from studying Maxwell's equations. Oliver Heavyside, and Richard Feynman, formulated an alternate version of Maxwell's Equations using retarded time. ( See Feynman Lectures on Physics, chapter 28, 28.3, & 28.4 ) Numerical study of Feynman's equations yields a simplified algebraic expression for the E field and B field of a uniformly moving charge, but expression is not elliptical as expected. Why?

Richard Feynmann, in His lectures on Physics, Vol I, chapter 28. Gives an expression for calculating the E and B field of a moving point charge based on retarded radius; eg: Knowledge of where the charge was at the time when the field was being formed. The Equation was originally derived by Oliver Heavyside and re-discovered by Feynman.

\vec E={-q \over {4 \pi \epsilon_0}} (<br /> { \vec u_{r_r} \over r_r^2}<br /> +{r_r \over c}( { d \over dt } {{ \vec u_{r_r} } \over {r_r^2} })<br /> +{1 \over c^2 } {d^2 \over {dt^2}} \vec u_{r_r}<br /> ) eq. 28.3

<br /> \vec B=\vec u_{r_r} \times { \vec E \over c }<br /> eq. 28.4

Where r_r is a scalar value, which is the distance (or time/c) back to when the particle emitted the field. \vec u_{r_r} is a unit length vector pointing toward the location where the planet was when the field began propagating to where it is in the graph.

In order to understand these two equations, I set up a numerical simulation using the Python programming language's decimal class (With 200 digits of accuracy.) I assumed a charged point particle moving in the +x direction. ( Later on, I will call this x-axis tracing particle a "point sized planet" )

For all graphs, I will plot when the particle is coincident with origin of the graph. I also simplified the Heavyside equation by setting 1 = { - q \over { 4 \pi \epsilon_0}} which allows me to focus on algebraic formulas instead of numerical values of the units of measurement.

Contours of equal field propagation time (constant r_r) from the charged particle/planet are easily graphed. They make a doppler shift circle pattern receeding at the speed of light from the uniformly moving point charge. In the following graph, the planet is at the origin and has been moving to the right from negative x values in previous times. The contour rings clearly show a doppler shift kind of diagram. Each time contour represents a constant difference in time compared to the adjacent contour. They are compressed to the right of origin, and expanded to the left.

From this diagram, I was able to figure out what the unit retarded radius vector and retarded radius value would be:

r_r= {{x^2+y^2} \over {(x^2+({1- {v^2 \over c^2} })y^2 )^{1/2} - ({v \over c}) x } }
\vec u_r = { [ { vm^2+-\sqrt{1+m^2(1-{v^2 \over c^2})} \over {1+m^2}} , {m(-v+-\sqrt{1+m^2(1-{v^2 \over c^2}})\over {1+m^2} } , 0 ]} : where m=y/x , eg: the slope.

The unit vector formula was defined for the top half of the graph, (y>0). Choose the positive sign in the equation and you will get the vector for points x>0, otherwise you get the vector for x<0. You can mirror the vector for all other quadrants.
Note carefully: The unit vector points in the direction from point x,y to where the particle was in the past and NOT to the origin of the graph.

I was able to use these algebraic expressions to write a numerical analysis routine to calculate E and B fields according to 28.3 and 28.4 using numerical differentiation.

Then I found a surprise ...
Richard Feynman pointed out that the third term of eq 28.3, is the only one responsible for radiation due to "acceleration" of particles (second derivative of position with respect to time.). Radiation magnitudes of E and B fields fall off as 1 \over r in some directions, but Lorentz contracted E and B fields of a uniformly moving charged particle is expected to fall off as 1 \over r^2 in all directions. Please review Feynman's work if you have questions.
http://www.feynmanlectures.caltech.edu/I_28.htmlWhen Feynman draws the E field of a uniformly moving point charge in another part of his lectures, he says that the drawing contracts linearly only in one axis (the axis of motion.) In essence, I thought that a linear contraction in one dimenstion should give me an elliptical outline if I were to plot a contour of constant force. Both Feynman and others tend to draw images to imply this same outcome. Lorentz contraction of a circle, looks like an ellipse.

However, careful analysis of the third term (radiation) of the Heavyside equation shows that only one component of the acceleration vector falls off with 1 \over r The other component falls off as radius squared. Even though the "planet" is moving uniformly along the x-axis with a fixed velocity, v, none the less ... the unit vector in equation 28.3 accelerates as the particle passes by and distorts the Lorentz contracted field.

When I try plotting a contour map of where the numerical solution has an arbitrary fixed force magnitude, I don't get an ellipse shape for all v. I get something that looks like a circle at very low velocities, an ellipse at medium v < 10% c velocities, but a peanut at high V>0.6C.

Constant force magnitude contour graph, for v=0.85C:

The B field at any location on the peanut graph's contour line points directly into or out of the page. (out of page for +y locations, and into page for -y locations if we assume a positively charged planet.) There is no magnetic field on the x axis. By cylindrical symmetry of the problem, the y and z axis are equivalent directions; Therefore, the magnetic field lines are making perfect circles in the y-z plane. eg: They loop around the [x,0,0] axis line.

I was able to numerically compute the magnitude of the E and B field vectors at every point at an instant of time. (eg: when the planet is at the origin in the plots). I was also able to fit the numerical results to a pair of algebraic equations. (cool!)

The following algebraic approximations are accurate to over 100 digits. I tested from v=0.000001c all the way to 0.999999c. Compared to the significant digits possible in a plank length, I'm sure the following two equations are indistinguishable from exact for practical physics problems.

| \vec{E} | = {{ {(1-{v^2 \over c^2})} \sqrt{ x^2 + y^2}} \over { ({ x^2 + {(1-{v^2 \over c^2})}y^2})^{3/2} } }
| \vec{B} | = | \vec{E} | {1 \over c } {y \over { \sqrt {x^2+y^2} } }

Note: the direction of the E vector is always colinear with a ray to/from the origin in the above plot. The E field vectors all point to or directly away from the graph's origin, where the planet is (depending on type of charge + or -).

Since the Lorentz transformation was originally discovered from Maxwell's equations, I expected to find the same thing from the Heavyside equation and in my approximations. Checking the axii for Lorentz factors in the approximations, I found these values:

| \vec{E} | = { 1 \over { \sqrt {1-{v^2 \over c^2}} }} { 1 \over y^2 } : when x=0
| \vec{E} | = {({ {1-{v^2 \over c^2}} })} { 1 \over x^2 } : when y=0

So, I do see the Lorentz (gamma) factor show up on the y axis. \gamma = { 1 \over {(1-{v^2 \over c^2})^{1/2} }} The force along the Y axis is clearly magnified by the gamma factor.

This simple discovery leads me to do a Gedanken experiment.

Suppose the charge is a point sized planet described by the previous equations. Now add a point sized satellite of opposite charge in orbit. The orbit is constrained to a y-z plane which is centered on the planet. eg: If the satellite and planet were not translating (set v=0 along the x-axis ), the satellite would orbit the planet at a fixed value of x, but a constant radius from the x axis, in y,z.

However, if the planet and satellite were both translating along the x-axis with a fixed velocity, the shape of that same satellite orbit would be a helix.

The helical motion of the sattelite orbit would follow the magnetic field lines of the charged planet. Therefore, I think that there is no force caused by the magnetic field's interaction with the sattelite. The force on the satellite comes purely from the E field of the planet.

If the system was translating along at constant velocity x=vt, then the translation would increase both the mass of the planet and satellite (each) by a factor of approxiately ~gamma. Since the force field (AKA E field) strength also increases by a factor of gamma, the increased mass and increased force ought to cancel out leaving the orbital period unchanged. F = m A becomes γF = γmA

However, I know that's not quite correct. The satellite orbital period must increase as planet and satellite translational velocity increases, and by the same factor of gamma. Otherwise, the satellite could break the speed of light becuase the orbital motion would be independent of the translational velocity.

I could just apply the formula for perpendicular addition of relativistic velocities, given a satellite velocity 'u', that is translated along the x-axis at velocity 'v'; the resulting satellite velocity ought to be time dialated:
u&#039; = {u \over { \gamma (1 - {{ u \cdot v } \over c^2} ) }}
(The dot product is zero in this example. )

But, I would prefer to see an example of how to work the problem based on the mass increase of the satellite, and then the radius of orbit re-calculated based on that mass using F=mA . The idea is to show that the orbit is indeed slower (in the rest frame, where the peanut shape is "obvserved") but the physical radius of the orbit is unchanged. (Eg: as if the satellite is orbiting on the very +-y points of the peanut shape, only. And it keeps the same radius in the y,z directions as it orbits.)

Because of Einstein's twin paradox gedanken in special relativity, I'm aware that we need to be careful of whether time dilation is an actual effect of the satellite actually orbiting slower; or an appearance effect (doppler shift) due to light propagation delay of events in the past. Simultineatiy may be also different in different reference frames.

But, in either case (let's do another gedanken) ... if I built a non-moving wall of neutrons/neutral matter along the Y axis in the rest (labarotory) frame at the origin, then I could test the actual size of the satellite orbit by making a hole of varous sizes in the non-moving neutron wall, and seeing whether or not the planet and satellite fit through the hole or "hit" the wall.

If I now place the planet with orbiting satellite at some location on the x-axis x<<0, and set it's translational velocity v=85% c (x=-10000m+vt), then the system should either go through the hole or not. That allows us to determine the "real" radius of the sattelite orbit.

Given that the experiment uses translational motion only in the x direction, I would not expect the satellite orbit's radius to change much (if at all). Lorentz contraction is not supposed to happen perpendicular to the direction of motion, and both the planet and the satellite are point sized objects.

So, if the planet and satellite would just barely fit inside the hole without hitting when v of the system is 0, then the planet and satellite system should still just barely fit inside the same hole when passing through it along the x-axis an arbitrary velocity v.

The same experiment should even work if the satellite is orbiting in the x-y plane instead of the x-z plane. The analysis will become far more complicated, but Lorentz length contraction should still happen only along the x direction, and not the y direction.

Any suggestions on how to attack the shape of the acutal sattelite orbit (x-y plane) with respect to the peanut shape equations for |E| and the magnitude of |B| is appreciated. I think the satellite orbit ought to be an ellipse relative to the planet, but I'm unsure how to derive that result given the algebraic equations I have already come up with.

 

[learn.steadfast]

I made two mistakes; First, the reference frame I am computing the Heavyside equations in is at rest with respect to "me" the observer, but both the planet and satellite are moving; Therefore the satellite is moving forward through a magnetic field, (x+ direction), and the magnetic field DOES affect the force of attraction between the - charged satellite and the +charged planet.

Secondly, I described the original satellite orbit parallel to the y-z plane. But, when I changed the orbit's orientation to test Lorentz contraction, I made a type-o and called the original orbit the x-z plane. Although the expression will be complicated to compute, if the orbit is cahnged to the x-y plane, or the x-z plane (instead of the y-z plane), Lorentz contraction should exist for any orbits extended in the x direction.

 

[Nugatory]

Note: “Heaviside” not “Heavyside”

 

[vanhees71]

Well, I don't know, who was first, and it may well be that Heaviside got the formula independently earlier, but it goes back to the physicists Lienard (1898) and Wiechert (1901).

The first use of a retarded potential in electromagnetic theory goes back to Riemann (1858) and Lorenz (1861).

 

[learn.steadfast]

Note: “Heaviside” not “Heavyside”

Yes, and the magnetic field equation is missing a constant of v/c^2 which is worse than a spelling error of a proper name. Spell check on my computer wanted Heavyside... I don't know why. I changed the equation four times during the editing period, and it still showed up as 1/c. If there was a way to edit the original post, I'd fix all the typos... but there isn't. If anyone actually tries to use my equations, they'll soon figure out that it's off by a scale factor, but otherwise it's correct.

 

[learn.steadfast]

Well, I don't know, who was first, and it may well be that Heaviside got the formula independently earlier, but it goes back to the physicists Lienard (1898) and Wiechert (1901).

The first use of a retarded potential in electromagnetic theory goes back to Riemann (1858) and Lorenz (1861).

Feynman credits Heviside's equation as 1902 in his lectures on Physics, see footnotes at bottom of page:

http://www.google.com/url?sa=t&rct=...u/II_21.html&usg=AOvVaw3pQVbKX9-sSnXuAQtyRrbB
Feynman also notes the names Lienard and Wiechert, but not as the inventors of the particular formula that correctly encompasses all knowledge about EM fields in one equation.

 

[learn.steadfast]

I'm trying to figure out if the orbital of the satellite in the yz plane has the same radius as if it were not moving in the x direction; eg: Does Lorentz contraction affect satellite orbit radius or not.

I want to try approaching this by balancing masses and and forces; if the ratio of force to mass is the same in the relativistically transformed case as when the satellite was orbiting a stationary planet, then the physical orbit size ought to be the same.

So, I wan't a general way to compute the mass of the moving satellite and planet as a pair.

The Lorentz factor, gamma, is computed as:
\gamma = { 1 \over \sqrt { 1 - {{ v_x^2 + v_y^2 + v_z^2} \over c^2 }} }

And effective mass goes as rest mass times gamma: m=m_0 \gamma

But, when I did additions of orthogonal velocity (relativistic equation), time dilation was explicitly taken into account to reduce one of the velocities being added:

{v_y}&#039; = { v_y \over { {\gamma}_x ( 1 - 0 ) } }
Where gamma_x is computed component wise for the x direction only (eg: setting z velocity to zero.)

So, the mass scaling factor for two added velocities (relativistic) is:
\gamma_{xy} = { 1 \over \sqrt { 1 - {{ v_x^2 + {v_y^2 \over \gamma_x^2 }} \over c^2 }} }
\gamma_{xy} = { 1 \over \sqrt { 1 - {{ v_x^2 + {v_y^2(1- {v_x^2 \over c^2}) }} \over c^2 }} }
\gamma_{xy} = { 1 \over \sqrt { 1 - {{ v_x^2 + v_y^2 - {v_y^2v_x^2 \over c^2 }} \over c^2 }}}

However, isn't that the same as just multiplying the gamma factors without time dilation correction?
\gamma_{xy} = { 1 \over \sqrt { 1 - {{ v_y^2} \over c^2 }} } { 1 \over \sqrt { 1 - {{ v_x^2} \over c^2 }} }
\gamma_{xy} = { 1 \over \sqrt{ ({ 1 - {{ v_y^2} \over c^2 }})({ 1 - {{ v_x^2} \over c^2 }}) }}
\gamma_{xy} = { 1 \over \sqrt{ 1 - { v_x^2 + v_y^2 - {v_x^2v_y^2\over c^2} \over c^2 } }}

So, I wonder if the proper (time dialated) gamma factor for three dimensions can also be computed by a simple product. eg:
\gamma_{xyz} = { 1 \over \sqrt { 1 - {{ v_z^2} \over c^2 }} } { 1 \over \sqrt { 1 - {{ v_y^2} \over c^2 }} } { 1 \over \sqrt { 1 - {{ v_x^2} \over c^2 }} }

That would mean that if I know the total mass of an object (rest + motion), and I acccelerate it in an orthogonal direction to it's present motion; that the final mass will be the product of the original mass I measured, times the gamma factor of the change in velocity that I measure.

Have I overlooked anything?

 

[Ibix]

I'm not sure I completely followed this on a small phone screen. With that in mind, however...

You seem to be saying that $\vec F=\gamma m\vec a$, which is incorrect. $\vec F=d\vec p/dt=d(\gamma(t)mv(t))/dt$.

I don't understand why you are inserting gamma factors into gamma factors. $\gamma=(1-\vec v\cdot\vec v/c^2)^{-1/2}$, where the velocity is the velocity measured in the frame you are working in. The devices doing the measurement are not subject to time dilation.

 

[vanhees71]

I'm not sure, what $\vec{F}$ you are using. There are unfortunately old-fashioned expressions still used in modern textbooks, where the clear Lorentz-covariant notation already found by Minkowski in 1908 isn't used. Even Feynman uses "relativistic mass" and sins like that.

The most easy way to write down the equations of a point particle in an external magnetic field, neglecing radiation reactions are
$$m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2} = \frac{q}{c} F^{\mu \nu} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau},$$
where $m$ is the invariant (!) scalar mass, $x^{\mu}(\tau)$ the world line of the particle, $\tau$ its proper time, $q$ its charge, and $F^{\mu \nu}$ the components of the Faraday tensor.

Writing this in terms of four-momentum, you get
$$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=K^{\mu},$$
where $K^{\mu}=q F^{\mu \nu} p^{\nu}/(m c)$ is the Minkowski four-force.

Sometimes you find mixed notation, i.e., people use the four-momentum spatial components $\vec{p}$ and write the EoM in terms of derivatives with respect to coordinate time (which is of course formally correct, but utmost confusing for the student). To get this equation just use the spatial part of the covariant equation and write (in Heaviside-Lorentz or Gaussian units alike)
$$\frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} = \gamma \frac{\mathrm{d} \vec{p}}{\mathrm{d} t} =\vec{K}=\frac{q}{c} \left (\frac{\mathrm{d} x^0}{\mathrm{d} \tau} \vec{E} +\frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} \times \vec{B} \right) = q \gamma (\vec{E}+\vec{\beta} \times \vec{B}),$$
where $\vec{\beta}=\mathrm{d} \vec{x}/\mathrm{d} t/c$. Now $\gamma$ cancels on both sides, and you get with the non-covariant (!) force components $\vec{F}=\vec{K}/\gamma$
$$\frac{\mathrm{d} \vec{p}}{\mathrm{d} t} = \vec{F}=q (\vec{E}+\vec{\beta} \times \vec{B}).$$
Now you can also write the momentum in terms of coordinate-time derivatives,
$$\vec{p} =m \gamma \frac{\mathrm{d} \vec{x}}{\mathrm{d} t},$$
and you get the (1+3)D non-covariant formulation of the EoM. Sometimes, but rarely, it can be of advantange when solving the equations of motion in comparison to the covariant formulation.

Note that, of course, also in the covariant formalism, there are only 3 independent EoM since
$$p_{\mu} p^{\mu}=m^2 c^2=\text{const}$$
and thus
$$p_{\mu} \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=0,$$
which is compatible with the EoM since the Minkowski force fulfills this constraint,
$$p_{\mu} K^{\mu}=0.$$

 

[jartsa]

That would mean that if I know the total mass of an object (rest + motion), and I acccelerate it in an orthogonal direction to it's present motion; that the final mass will be the product of the original mass I measured, times the gamma factor of the change in velocity that I measure.

No, it's larger, if the acceleration really is exactly orthogonal.

That's because that orthogonal acceleration will cause cause the momentum in the "original direction" to increase.(The final total energy will be larger than the product of the original total energy you measured, times the gamma factor of the change in velocity that you measure)

(Orthogonal change of momentum would be a different story)

 

[learn.steadfast]

I'm not sure I completely followed this on a small phone screen. With that in mind, however...

You seem to be saying that $\vec F=\gamma m\vec a$, which is incorrect. $\vec F=d\vec p/dt=d(\gamma(t)mv(t))/dt$.

I don't understand why you are inserting gamma factors into gamma factors. $\gamma=(1-\vec v\cdot\vec v/c^2)^{-1/2}$, where the velocity is the velocity measured in the frame you are working in. The devices doing the measurement are not subject to time dilation.

I thought force and acceleration in my rest frame was scaled by the gamma factor, exactly as you show. In several texts, exactly that argument is made. It doesn't change the equations that I derived, but your comment may affect my analysis of orbital size.

Good question. Let me elucidate about why I am putting in time dilation:

Einstein's Gedanken of the twin paradox reveals that there are two distinct possibilites when it comes to time dialation. Given two spaceships A and B, each containing an identical clock A and B. Should the ships separate by being one ship being instantly accelerated (which ship accelerated, we don't know for sure), then we still know something about what would be observed: The people in each spaceship could both measure the relative speed of the other ship by reflecting light pulses from one ship off the other and timing the round trip time. Both observers in A and B frames of reference (FOR) would agree that the velocity of separation between spaceships was the same. If they were to watch each other's clocks using telescopes, both observers would conclude the other's clock is running slower by the exact same time dilation factor. However, if the ship which was acutally accelerated away is later returned by "deceleration"; the result is that clock which was both accelerated and decelerated will register less time elapse than the other clock when both clocks are brough back together.

So, let's say 'B' was the accelerated and decelerated clock. In that case, during the whole experiment in my lab frame 'A' I KNOW that B's clock hands, light beam, etc. whatever was used to measure time in frame B made less revoltuions per second than my 'A' clock. B made less revolutions per second in my measurement frame of reference during the experiment.

If the B clock was oriented perpendicular to the direction of travel, we also know that there is no length change in either the y or the z size of the clock.

I believe I am justified in attributing a real time dilation to 'B', because they were the object which were actually accelerated. Is this not correct?

In the experiment I originally set up, we have a satellite orbiting a planet in a yz plane. That system is a "clock." We are going to really accelerate both satellite and planet equally in the x direction to some fixed speed. Therefore, the satellite going around that planet must in fact be orbiting at a slower rate. For we can use the number of times the satellite makes a revolution around the planet as a "clock". The appearance of the satellite, if observed through a telescope, must also be slower than it runs in reality. So, even though my lab frame is NOT accelerated (by definition) the ACTUAL time dilation effect (not the appearance, but the actual effect) must be happening with the accelerated planet and satellite.

I believe I am justified in attributing a real time dilation to the sattelite and planet, because they were the object which were actually accelerated. Is this not correct?

I see you give a definition for momentum, and that's good. I'll spend a day thinking about it before continuing. (I also see the next post is about four vectors and does an excellenct job.)

I'm trying to work the problem from the lab (rest) frame with the knowledge that the planet and satellite were accelerated. So, I wan't to compute their actual position, and not the "apparent" position as viewed through a telescope. Eg: my fomulas will be entirely based on x,y,z and t coordinates in the lab 'A' frame.

Let me point out that Maxwell's equations, and Heviside's, were developed before Minkowski space ever was conceived. The older idea is that the fields surrounding a charged particle are "force" fields.

If I know the direction a force is pointing and I do a line integral from infinity (parallel to the direction of the force); what I get is the total potential energy that a particle would have if moved against that force.

Work = force x distance.

The definition of work is the same as energy, and energy and mass are equivalent.
I would like to see if integrating the force of maxwell's equations, heaviside's equations, etc. yields the same result as your analysis does. I'm not saying you're wrong, I'm saying we ought to be able to get the same result either way.

There are two sources of forces, in my equations, the coloumb field (delayed and distorted), and the magnetic field.

Magnetic fields, only generate force based on the velocity component perpendicular to the magnetic filed lines of force. In the case of a charged satellite and planet, only the componenent of velocity in the X direction matters. So, the question of time dialation/velocity vector component slow down in Y and Z should have no affect on the forces computed.

 

[learn.steadfast]

I'm not sure, what →F\vec{F} you are using. There are unfortunately old-fashioned expressions still used in modern textbooks, where the clear Lorentz-covariant notation already found by Minkowski in 1908 isn't used. Even Feynman uses "relativistic mass" and sins like that.

I'm following along with Feynman. The Lorentz covarient notation you use wasn't taught at my college for undergraduate work. I can pull out a book on tensors, but it will take me a long time to decipher your post.

With vectors, I'm simply using a three vector \vec F = [ F_x, F_y, F_z ] ; to denote a direction in cartesian coordinates. It's a pair of three real numbers representing a magnitude and direction in a square bracket list. In the same manner, \vec v=[V_x, V_y, V_z ] is simply the velocity in the x,y, and z directions in a 3 vector.
So, F with an arrow on top in my first post \vec F, is the traditional force vector of Maxwell's, Feynmans/Heviside's equation(s) when done in cartesian coordinates. Neither Maxwell, Feynman, nor Heaviside used Minkowski's notation nor tensors for the equation I'm studying.

The magnitude of a force vector is simply it's absolute length:
| \vec F | = \sqrt {( F_x^2 + F_y^2 + F_z^2 )} = \sqrt { \vec F \cdot \vec F }

There is a problem with your approach; I think it misses the core question I'm attempting to probe.
I'm not trying to solve the problem in the easiest way possible, I'm trying to explore historical equations and alternate representations.

The core issue is that the unit vector shown by Feynman/Heavyside, accelerates with respect to time even when a particle is moving at a constant speed. That is to say, when a particle passes an arbitrary point, and we define a unit length vector following the direction of the particle from that arbitrary point, the unit vector *often* has non-zero acceleration even though the particle has no acceleration whatsoever:

Eg: to demonstrate! Define a particle's motion as: x=vt, y=0, z=0. AKA P=[ vt,0,0 ]
I can make a vector that points at that particle from an arbitrary location of x=0,y=1,z=0 AKA R=[0,1,0]
It's easily done by normalizing the difference between the two points:

\vec u_{PR} = { [vt,-1,0] \over \sqrt{ v^2t^2 + 1^2 + 0^2 }}

So, now I have a unit vector pointing in the direction of the particle, P, from arbitrary point R.
If you take the second derivative of that unit vector's components with respect to time, it's non zero.

Because the second derivative of \vec u_{PR} with respect to time is non-zero, there are acceleration effects in any equation based on the acceleration of a unit vector even if the particle is NOT accelerating.

These effects may not all be radiation effects, because not all components of the vector show the same fall off with respect to distance. Some components fall off as 1/r (the radius to the particle), some fall off as 1/r^2 the square of the distance.

Feynman calls components that fall off as 1/r, radiation. Those that fall off as 1/r^2 are part of the Coloumb field, and the Magnetic field.

So, Feynman's analysis of the third term of Heaviside's equation is incomplete.

That's why I'm exploring Heaviside carefully, rather than solving a problem using special relativity or using Minkowski Tensors alone.

I should be able to get the same answer using any of the above methods as to what the radius of an orbiting charge is in the yz plane. eg: when given that the whole system is uniformly accelerated perpendicular to the direction of satellite yz motion in the x direction.

That's the goal, to explore the interchangibility of the equations; and if they are not interchangable, to isolate what is different and why it exists.

 

[PeterDonis]

now I have a unit vector pointing in the direction of the particle, P, from arbitrary point R

First, Feynman's unit vector $\mathbf{e}_{r'}$ points from the point where the field is being measured, in the direction of the charge that is producing the field; it doesn't point towards some arbitrarily chosen point. So your unit vector is pointing backwards from the way it should be pointing, and it should be pointing towards a specific point, the point where the charge is, not an arbitrarily chosen point.

Second, the unit vector does not point in the direction of the charge "now". It points in the direction where the charge was one light-travel time ago. And the light travel time changes as the particle moves. So your calculation of a unit vector is incorrect; you are calculating a unit vector that points (once the first issue above is corrected) in the direction of the charge "now". If you re-do the calculation so that the unit vector points towards where the charge was one light-travel time ago, and take into account that the light travel time is changing as the particle moves relative to the charge, I think you will find that there is zero second derivative of the unit vector with respect to time if the particle is moving in a straight line relative to the charge.

 

[learn.steadfast]

No, it's larger, if the acceleration really is exactly orthogonal.

That's because that orthogonal acceleration will cause cause the momentum in the "original direction" to increase.

OK. I think Orthogonal means at 90 degrees or pi \over 2 radians to the direction of motion.
Parallel means in the same or opposite direction of motion.

But the velocity of something, and it's momentum are a little different...
I was only talking about the velocity, so it's possible the velcocity components shrink but the momentum componets increase in the same directions. The reason is that the mass has changed (increased).

My primary assumption was that an orbiting satellite in a yz plane is every bit as much a "clock" as light bouncing between two mirrors. There is a period or cycle time of orbit. If the whole clock is accelerated as a unit (say by gravity, so both satellite and planet accelerate equally in the X direction) then the time of the clock needs to be dialated by the lorentz factor, because it's a clock...

I'm not sure about the orbit size.

I'm curious that you would say the orbit increases in size. A rigid rod, or an approximation of it, is made up of atoms. Atoms, too, are made up of + nuclei (massive) and - charges (light), that can have angular momentum like a planet and satellite. So, if a stick of atoms is a "rigid rod", and rigid rods don't shrink or expand in any direction orthogonal to the acceleration, why would a satellite orbit radius and planet do so?

What did I miss?

 

[learn.steadfast]

First, Feynman's unit vector $\mathbf{e}_{r'}$ points from the point where the field is being measured, in the direction of the charge that is producing the field;

As usual, you are wrong. Feynman's unit vector points from an arbitrary point of measurement to where the charge WAS in the past, when it created the field.

it doesn't point towards some arbitrarily chosen point.

Neither does my vector. It points FROM an arbitrarily chosen point called R, AT a moving charge called P.
Subtract the tail of the vector from the tip, to get the direction of the vector.
I subtracted R from P.
It's simple linear algebra. The vector points TOWARDS P.

So your unit vector is pointing backwards from the way it should be pointing, and it should be pointing towards a specific point, the point where the charge is, not an arbitrarily chosen point.

Duh. Look again. And even if it was, that would merely be a sign reversal ... it wouldn't change the form of the equation.

Second, the unit vector does not point in the direction of the charge "now". It points in the direction where the charge was one light-travel time ago.

Feynman's unit vector does, my example does not. My example points at the charge. There is a reason for this, if you had read the first post carefully ... and you didn't. There are two equations, Feynman's equation has three terms and the unit vector points at the retarded position of the charge. The algebraic simplification of Feynman's equation, for a uniformly moving charge, has a net force field that points in the direction of the charge where it is "now."

In either case, I was only trying to show that a vector tracking a a linearly moving object yields acceleration terms when you take a second derivative. It doesn't even matter if the it's the retarded position or the current position. Try working it out for Feynman's unit vector, yourself.

I'm still having trouble believing you think EM waves can't reflect off a mirror for QED. Feynman himself, in his book on QED for laymen says that the field result is the same "as if" it had reflected off the front and back surface. You bring up objections to split hairs that are irrelevant. If real mirrors aren't made out of atoms ... I don't know what they are made out of. If my vector doesn't accelerate whether retarded or NOW, then I don't know what acceleration is.

Peter, really, try asking questions rather than putting other people down all the time.

 

[PeterDonis]

As usual, you are wrong.

This attitude towards someone who is trying to help you will get you a warning and a closed thread.

I'm still having trouble believing you think EM waves can't reflect off a mirror for QED.

I have no idea what you're talking about. But clearly you don't want my help so I'll confine myself to moderating the thread and not respond further.

 

[jartsa]

I'm curious that you would say the orbit increases in size.

Well I didn't actually say anything about orbit size.

I'll say something now: Orbit size contracts in the direction of the motion of the center of the orbit. There is no change in the direction perpendicular to the motion of the center of the orbit. I'm confident that's correct too.Imagine two masses oscillating, one horizontally other one vertically, this system's center of mass orbits around in a circle, if the phases of the oscillations are correct.

Now imagine accelerating those two oscillators together to high speed. The orbit should be a contracted circle now, I guess, I haven't thought about this very carefully.

 

[Ibix]

I thought force and acceleration in my rest frame was scaled by the gamma factor, exactly as you show.

That's not what I show. You need to differentiate the gamma as well - the result id that force and acceleration are not typically parallel.

I believe I am justified in attributing a real time dilation to 'B', because they were the object which were actually accelerated. Is this not correct?

Yes and no. Your analysis of the twin paradox leading to "real" and "not real" time dilation is wrong, or at least confusing. Time dilation is a coordinate effect, boiling down to clock synchronisation issues, while the differential aging seen in the twin paradox is invariant and boils down to "the distance between two points depends on the route taken". These are rather different phenomena. However, this isn't really relevant here. You will, indeed, measure moving clocks such as your orbiting charge to be ticking slowly by a factor of $\gamma$, and this is true whether you accelerated or your charges accelerated.

I'm trying to work the problem from the lab (rest) frame with the knowledge that the planet and satellite were accelerated. So, I wan't to compute their actual position, and not the "apparent" position as viewed through a telescope. Eg: my fomulas will be entirely based on x,y,z and t coordinates in the lab 'A' frame.

I didn't think you were suggesting otherwise.

I expect the orbital period to increase by a factor of $\gamma$ compared to the orbital period when the central charge is at rest. This will be the gamma factor related to the motion of your central charge, which is why I have no idea why you were introducing gammas inside gammas - you simply measure the motion of the central charge with clocks and rulers at rest in your frame.

You will need to use the correct formula for the association between force if you wish to balance centripetal forces with the Lorentz force. The general formula may reduce to your simpler form in this case, but I wouldn't bet on it. Off the top of my head, I think it does so only if $\vec a.\vec v=0$ and I don't think that is the case here.

I think jartsa's analysis of the orbital diameters in the case of motion in the plane of the orbit is correct. It may be difficult to see, though, since the path of the orbiting charge will be a cycloid of some kind. The easiest way to check your answer is to write down the coordinates of the orbiting charge in the rest frame $(t,R\cos(\omega t),R\sin(\omega t),0)$, and Lorentz transform this. You can also write down the electromagnetic field strength tensor in the rest frame and transform this, which will tell you the correct electric and magnetic fields fairly simply.

 

[learn.steadfast]

I'm not sure I completely followed this on a small phone screen. With that in mind, however...

You seem to be saying that $\vec F=\gamma m\vec a$, which is incorrect. $\vec F=d\vec p/dt=d(\gamma(t)mv(t))/dt$.

I thought about it, over night. You're correct, I was misapplying that equation.
I see where that mistake is; There's an extra contraction factor associated with the force. When taking the chain rule for the derivative on Gamma, I see that Gamma isn't independent of time.

Defining V={ v \over c}, to simplify the algebra:
<br /> F= γma ({V^2 \over { 1- V^2 }} + 1 )=\gamma^3 ma<br />

The change I need to make doesn't affect the equation I curve fit in the original post ; but it does affect the balance of forces in the orbit for when the "clock" of satellite and planet is in motion.

 

[vanhees71]

As I said, it's more difficult to use non-covariant quantities than to use covariant ones. I'm not particularly happy with Feynman's treatment of relativistic mechanics in his famous books, because to my surprise he uses very old-fashioned non-covariant objects. This in my experience rather confuses the student than helping him or her. Nevertheless I still don't have a clue what you are aiming at. I've given the correct equations of motion of a point particle in an external electromagnetic field, neglecting radiation reaction, in both the covariant and the (1+3) form above. I don't understand, what is unclear with this pretty straight-forward calculation.

Last but not least, if you want to deal with gravity in relativity theory the right thing is Einstein's General theory of relativity, but that seems to me far off your mathematical expertise. To understand physics you have to learn the adequate mathematics, and for SRT that's 4D tensor calculus in Minkowski space and for GRT it's 4D tensor calculus in pseudo-Riemannian (Lorentz) space.

 

[learn.steadfast]

Your analysis of the twin paradox leading to "real" and "not real" time dilation is wrong, or at least confusing. Time dilation is a coordinate effect, boiling down to clock synchronisation issues, while the differential aging seen in the twin paradox is invariant and boils down to "the distance between two points depends on the route taken". These are rather different phenomena. However, this isn't really relevant here. You will, indeed, measure moving clocks such as your orbiting charge to be ticking slowly by a factor of γ\gamma, and this is true whether you accelerated or your charges accelerated.

Fair enough. It is confusing, and I think it is worth looking at a tiny bit more.
I'm curious what you mean by the idea of "route taken"

Every clock measurement seems to require three reference frames to complete. The observer of the clock's reference frame, the outgoing (leaving) reference frame, and the returning reference frame.

Clocks are usually defined by a change in position that is cyclic; Therefore one or more moving objects must go away from the observer and come back in order to be measured. (Or the observer must go to it, which is the same thing.) A poor example is the motion of a pendulum which leaves and then returns to a starting position. Light or electrons bouncing between two stationary mirrors is a better example.

The gamma that I "measure" is always going to be some kind of average of distance traveled by one or more objects in two opposite directions. For example, a ship takes off toward Bernard's star 6 Light years away, at 0.6 C. If I watch the clock on the ship through a telescope, I'm going to see the clock on the ship runs at half the speed of the clock on my desk, but only during the time that I see the ship leaving me. When I see the ship come back toward me, I'm going to see that the clock on the ship runs twice as fast as mine.

(incidentally, the person on the ship will see the exact same thing.)

These measurements are both higher and lower than the theoretical Lorentz factor of 1.25, but they are round trip averages and light is one of the objects that is moving. However, round trip measurements are the only thing scientists can actually "measure" with complete certainty.

I've seen authors try to break the twin paradox explanation into three inertial reference frames, so that no acceleration is involved. Usually, it goes along the lines of A,B,C are intertial objects moving relative to each other. A is earth, B is somthing going away from earth, C is something going toward earth.
Twin paradox: the real explanation (no math) - YouTube

The beautiful result is that the time dilation (which is integration of the lambda factor) for B and C is achieved without any actual acceleration in the thought experiment. The actual "accumuation of time" on clocks A,B,C can be proven to happen during constant relative motion.

The problem with that type of approach is that we could just as well artificially break A's motion into two different inertial frames, and pretend to synchronize A's clock during the reversal of direction. If accleration has nothing to do with it, who cares if the person on Earth never feels acceleration?

If A and B have relative velocity 0.6C , and A and C have relative velocity to each other of 0.6C, then B and C either have to have a relative velocity of 0 or ~0.8863C. That's why time accumulation is slowest in B and C frames of reference. Their relative velocity of B to C is much higher when B and C must leave and return to Earth (~0.8663C is the result.)

But, if I Define A as the ship, and B and C as the Earth. Then the situation totally reverses, and the paradox comes back to light. The time on Earth must be slower becasue it is the one travelling.

In all cases, there has to be something to justify a frame of reference change.
Special relativity doesn't give that "something", but in general relativity there is gravity/acceleration.

So, when you say "route taken", what do you mean?

 

[Dale]

we could just as well artificially break A's motion into two different inertial frames

Not really. In that case v=0, so the two different inertial frames are the same and the Lorentz transform is the identity transformation.

 

[Ibix]

Every clock measurement seems to require three reference frames to complete. The observer of the clock's reference frame, the outgoing (leaving) reference frame, and the returning reference frame.

No. Something moving does not require you either to define or use its rest frame.

You gave the example of a pendulum. Yes, it swings backwards and forwards. In the frame where it has no net motion, time at its position is defined in terms of multiples of the period it takes to return to its centrepoint. Full stop. No need for other frames. If I want measurements at different spatial locations, I simply replicate my clock at those locations and synchronise these clocks with the origin clock. Still no other frame required.

The gamma that I "measure" is always going to be some kind of average of distance traveled by one or more objects in two opposite directions. For example, a ship takes off toward Bernard's star 6 Light years away, at 0.6 C. If I watch the clock on the ship through a telescope, I'm going to see the clock on the ship runs at half the speed of the clock on my desk, but only during the time that I see the ship leaving me. When I see the ship come back toward me, I'm going to see that the clock on the ship runs twice as fast as mine.

Here, you are involving the Doppler effect. Time dilation is what's left over once you subtract out the effects of changing distance from the Doppler effect. Or better, just use a network of synchronised clocks. Note the time on your (stationary according to you) clock that is right beside the ship when the ship's clock reads zero. Note the time on your other clock that is right beside the ship when the ship's clock reads one. The difference in your readings is $1/\gamma$ - and all done in one frame.

However, round trip measurements are the only thing scientists can actually "measure" with complete certainty.

No they aren't - I've just shown you a way you can measure time dilation. It's completely true that there are assumptions built into my measurements - they are in how I determine that a remote clock is at rest with respect to me and synchronised. But that is the nature of the beast. You cannot avoid it if you want to consider measurements at different locations, as you are doing with your moving charges. How you combine your Doppler measurements to construct "what is really happening" contains those same assumptions.

The problem with that type of approach is that we could just as well artificially break A's motion into two different inertial frames, and pretend to synchronize A's clock during the reversal of direction.

As Dale says, no. The problem with naively stitching frames together to predict (incorrectly) that the stay-at-home will be younger emerges from failing to realize that you need to account for clock synchronisation changes. It's rather like crossing from one time zone to another, not adjusting your watch, and trying to explain the mismatch between your wristwatch and the wall clock by claiming you've gone back in time an hour.

Sure you can split up the stay-at-home's experience into two identical inertial frames. But because those two frames are identical they have identical clock synchronisation procedures, so no problem arises. Acceleration has nothing to do with it. The only reason you need to accelerate at all is because there's no other way for you to meet your twin again in flat spacetime.

The explanation for the twin paradox is that the interval, the sum of $\Delta s$, where $\Delta s^2=c^2\Delta t^2-(\Delta x^2+\Delta y^2+\Delta z^2)$, over each path is invariant and different for the two paths. If I pick a coordinate system where one of the twins is at rest then it is clear that $\Delta s=c\Delta t$, the time elapsed for that twin for the segment of the experiment where they are at rest in that frame. Since the value is invariant this is true in any coordinate system, and it's obvious working in the rest frame of the stay-at-home that the total is different for the two twins, and lower for the traveller.

Note that I have not invoked time dilation anywhere. And the different paths through spacetime (different (x,t) coordinates) occupied by the traveller and stay-at-home between departure and return is what I mean by "different routes".

 

[Dale]

No. Something moving does not require you either to define or use its rest frame.

@learn.steadfast you should pause a minute to let this sink in. This is the essence of the principle of relativity: any frame is equally valid so there is no need to use any particular one.

 

[learn.steadfast]

@learn.steadfast you should pause a minute to let this sink in. This is the essence of the principle of relativity: any frame is equally valid so there is no need to use any particular one.

Yes, it is; and you've given good advice.
But that's also precisely the issue I'm trying to point out with respect to the twin paradox.
If acceleration is ignored, then either twin can claim the other is "moving" and they are not.
It is totally up to me, as a person defining the problem to choose which frames of reference I wish to use.

When I talk about clock's needing three reference frames, I over stated the case; I meant, there is always the need to consider that reversals or exchanges of frame are always "legal" and we "need" to avoid choosing a preferred frame of reference based on our own prejudices. There has to be somthing distinct to break the symmetry of the problem with respect to "who" is moving more than the other.

At each "event", where a change in direction, speed, or synchronization of clocks happens; it' is always possible to claim that the speed change did not occur at the location where I intuitively believe it happened, but rather that the speed change happened at another location.

In the clock problems that I am talking about, I'm trying to keep the idea simplified to special relativity.
I don't have the certainty of general relativity mathematics needed to handle the problem, yet. I'm not proficient at Gaussian coordinates; I'm limited to understanding the issue in terms of world line diagrams, and rotations from the Lorentz transform in order to work out problems.

I do accept the measurement of Ct as equivalent to a distance, and computing invariant lengths between two events. eg: I allow also for differences in simultanuousness between two different observers/frames of reference. Feel free to use these ideas in explaining things to me.

 

[Ibix]

It is totally up to me, as a person defining the problem to choose which frames of reference I wish to use

True. But if you define the traveling twin as "not moving" then you need to use a non-inertial frame. And time dilation (and all the neat symmetries) only applies between inertial frames, so naively saying "I'm stationary therefore he must be younger because time dilation" will not work.

 

[Dale]

we "need" to avoid choosing a preferred frame of reference based on our own prejudices

No, we don't. We can totally choose a reference frame based on our own prejudices. If it is an inertial frame that we have chosen then it is the same as any other inertial frame. If we have chosen a non-inertial frame then we have to use the appropriate math for that frame, but we are free to do that also. Whether it is motivated by prejudices or expediency or convenience or whatever is irrelevant. As long as we do the math correctly then we can use any frame chosen regardless of the reason for choosing it.

The problem with the twin paradox is not that anyone frame is better than any other. It is that people do the math wrong. Either they try to apply formulas for an inertial frame to a scenario described in a non-inertial frame or they incorrectly transform between inertial frames.

I do accept the measurement of Ct as equivalent to a distance, and computing invariant lengths between two events. eg: I allow also for differences in simultanuousness between two different observers/frames of reference. Feel free to use these ideas in explaining things to me.

Personally, my explanation would be essentially identical to the one by @vanhees71 in post 9. It will be easier to learn the notation and then do the problem with the easy notation than to do it with poorly-adapted notation to begin with.

If you insist on doing it that way then my recommendation would be as follows:
1) work the problem in the rest frame of the planet
2) then boost to the frame where the planet is moving
3) check to see that the various equations are correct in the moving frame also

 

[Ibix]

Either they try to apply formulas for an inertial frame to a scenario described in an inertial frame

I think you mean "non-inertial" at the end there.

 

[Dale]

I think you mean "non-inertial" at the end there.

Oops, yes. I have fixed it.

 

[learn.steadfast]

True. But if you define the traveling twin as "not moving" then you need to use a non-inertial frame. And time dilation (and all the neat symmetries) only applies between inertial frames, so naively saying "I'm stationary therefore he must be younger because time dilation" will not work.

OK, I'll leave this point after this post. I need to plug my naiieve gamma cubed mass acceleration formula in and see if it's consistent. The force equation is really only valid in one dimension, because I took the derivative that way. But, the formula is a start. I can at least Guess and Check my results.

I'll now respond to you: True, but How do you know the frame is non-inertial by experiment and not by definition of A,B,C?
The key point is that DOC is denying that Gravity or acceleration had anything to do with the result.

Doc is a bit like a person doing quantum mechanics trying to label a photon going through slit B or C.
The problem is that watching the slit the or naming the photon (like an angel), will change the outcome of the experiment. It's as if photons are intelligent and watching you, just to mess with you. (Or if you prefer, some angels ARE the Beast. )

I think that computing an invariant is the same as saying that once the experiment is over and we do find evidence of which "slit" A/B the angel really chose to go through (or which frame of reference "B/C" was really inertial), THEN, we can post-dict, what the age difference has to be. We know the relative difference in time, or the relative difference in Length, by calculation ; but we don't know which object "A" or "B" to apply the age decrease to until after the experiment.

( After all an electron is just an electron, supposedly not even God knows one from another ... why should we allow a scientist to have a brain if angels of light and electrons don't? )

I agree with you on how to calculate Lorentz contraction from Doppler effect. But, my point was that Doppler effect is something that is totally watched. There is (usually) light from a clock going to an observer, even here on Earth -- so there is a Doppler shift (tiny) when doing experiments on earth.
That is the nature of the Beast. :)

But the experiment can be done just as well by simply using the doppler shifted measurements without computing the Lorentz factor, and still get understandable results. People may even be more inclined to believe the results because the "clocks" were watched all the time. Human psychology plays a part in physics interpretation.

\gamma_{doppler} = \sqrt{ 1-v/c \over 1+v/c }
For 0.6C, the gamma doppler is 0.5 or 2.0, depending only on whether the objects are traveling toward each other or apart from each other.

One way the twin experiment might proceed is:

"A" observed "B"'s clock turn reddish, and run at half the speed of A's clock for 16 years. A writes down 8 years for "B"'s clock. A then sees "B"'s clock turn blue OR maybe "C" passed "B", (they aren't sure') A sees the blue colored system run at twice the speed of A's clock for 4 more years of A's time.

At the experiment conclusion: 'A' computes: 16*0.5 + 4*2 = 16 years observed for B/C (whoever it was.)

B,or C, writes down:

8*2.0 = 16 years + a note, from someone else that could be pranking you, that 8 years passed before that.
Everybody is happy. A KNOWS they are 20 years older, but B/C isn't sure when the experiment actually began as they struggle with Schizophrenia. But B did write down that 8 years passed since their last episode of hearing voices from angels, and the whole time they saw an extra blue shifted "A" through the telescope.

Still, 'B' writes down their results, because there is enough information that they are sure of what happened for the last 8 years, so the previous 8 years can be figured out by subtraction logic.

8*0.5 + 8*2.0 = 20 years have passed on "A" through the telescope.

A and B (or C?) are happily in agreement.

-----------

Now A, B /C decide to repeat the experiment, because of the schizophrenia issue. They want to see if the messages from angels are repeatable.

A sees "B" go by, and 8 years elapse for A while A watches half that time elapse in the telescope.
A notices that the LIGO detector went blip during a test run, but the blip didn't repeat once the computers were online properly. So A assumes two black holes must have collided. A insists their own frame is still inertial. But then A has a shock, A looks in the telescope and sees "C" (or "B?") blue shifted for the next 8 years. All the sudden, "B" or "C" pass Earth when only 16 years have passed on "A"'s clock.

A computes that the time on B's clock is 0.5*8 years + 2.0*8 years = 20 years, total.
So, A concludes someone pulled a prank on her and that she was the one who "moved" more than the other "B" or "C" (whoever it really was that came back to earth.)

The only thing that observers A,B,C can actually measure is red shift or blue shift, which verifies that they are approaching or receeding from each other at 0.6C, and therefore they can check the rates of their own clock realtive to another. But if we don't allow B or C to know for sure whether they are the same person or not there's a problem:

If we claim that "B" and "C" must have "synchronized" their clocks, we could just as well say that it just miraculously happened that their clocks were identical when they passed each other and nobody synchronized anything, and nobody passed any messages. (we can't distinguish "B" and "C" by experiement, anymore. )

The only way to know before the 8 year mark, which frame of reference will register less years at the end of the experiment, is to see if blue light comes toward the "A" Earth from "C". Angels from "C" had to arrive at Earth before the 8 year mark, to show that objects "B" and "C" are distinct.

But scientist A doesn't like to put down "and then a miracle of blue lights happened..." in his journal.
For that is the same as saying that some "DOC" rigged the experiment before it ever began. DOC put a shizophrenic angel "C" somewhere far out in the universe so they could just "happen" to be passing by at the appropriate time.

Doc has to write the results up in a peer reviewed journal. But, if the experiment is not rigged, then the experimenter is not allowed to see extra blue light's in the sky before 8 years have passed. They must wait until 8 years pass on their clock, to KNOW, who will accumulate less total time at the end of the experiment.

 

[Dale]

How do you know the frame is non-inertial by experiment and not by definition of A,B,C?

With accelerometers.

 

[vanhees71]

The lesson learned from this overly complicated thread is that the best you can do in relativity is to work with manifestly covariant expressions.

E.g., the socalled "twin paradox" becomes a triviality if one just keeps in mind that for any twin (idealized to a point particle for the matter of this discussion) his aging is giving by his proper time, and that's a Minkowski invariant,
$$\Delta \tau = \int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} \lambda}}.$$
Here $\lambda$ is an arbitrary world-line parameter. It can be chosen as a Lorentz scalar or also as the coordinate time of an arbitrary inertial reference frame ("lab frame"). Then
$$\left (\frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \right)=\begin{pmatrix}c \\ \vec{v} \end{pmatrix}$$
and
$$\Delta \tau = \int_{t_1}^{t_2} \mathrm {d} t \sqrt{c^2-\vec{v}^2}.$$

 

[SiennaTheGr8]

Recommendation to the OP: check out the explanation of the twin paradox in Taylor and Wheeler's Spacetime Physics. Read it over and over again until you completely understand it. You're overlooking exactly the aspects of the relativity of simultaneity that EVERYONE misses at first. David Morin calls it the "rear clock ahead" principle.

If you take this advice, you're in store for a profound and empowering "aha!" moment, and you'll fully appreciate that the situation with the twins isn't symmetrical. Until you really comprehend the relativity of simultaneity, you'll struggle with rather simple SR problems, let alone the more complicated scenarios you've put forth in this thread.

Cheers!

 

[vanhees71]

For a very nice graphical explanation, using Minkowski diagram for the most simple case of one twin traveling back and forth along one direction, see

https://aapt.scitation.org/doi/abs/10.1119/1.4947152

 

[PAllen]

For a very nice graphical explanation, using Minkowski diagram for the most simple case of one twin traveling back and forth along one direction, see

https://aapt.scitation.org/doi/abs/10.1119/1.4947152

That is paywalled. I cannot view without paying $30.

 

[learn.steadfast]

I have other posts to catch up on, but I'd like to comment on this point for I think it clarifies the special relativity issue / assumption that is causing me trouble.

The only reason you need to accelerate at all is because there's no other way for you to meet your twin again in flat spacetime.

For future reference, if I make statements that something is "like" something else, I do not MEAN they are identical. Carefully note that they are likely Not the same in sense that identical "twins" are meant to be the same. For example, if I say "A duck is like an airplane" please do not jump to the conclusion that I think a duck IS an airplane or that an airpane must have feathers. etc.

In Special Relativity, the idea that light is accelerated by gravity is not yet accounted for. The purpose of using "twins" in special relativity experiments is to simplify the mathematics, for since they are "identical" in how they act, or measure time, etc. we are invoking the scientific method of investigation and attemption to limit what might interfere with a thought experiment; eg: I am trying to isolate what assumptions (false/true) we may be making that we don't realize we are making.

In the twin paradox I linked to by Dr. Don Lincoln, Femilab:


The experiment is broken into a "triplet's" paradox, in order to avoid acceleration. Please watch the video carefully. I think Doc does an admirable job.

The implicit assumption of the show, however, is that calling something an "inertial reference frame" is sufficient to solve the paradox. This is why I tried to call your attention to the doppler effect and measurements of acceleration; I wanted to try and isolate what "acceleration" is.

There are at least two ways I know to measure acceleration. I am not certain they are the same under the assumptions of Special Relativity.

One method is that if two "identical" lasers (twins) are in motion toward (or away) from each other, that they can measure a color change (doppler shift) of each other. Therefore, they can measure acceleration between the two lasers by measuing the time derivative of the laser's color change.

The second method of measuring acceleration is to use two masses (inertias) that are in torsion with each other. We attach one mass rigidly to a "reference" frame, and measure the shear force required to keep the other mass moving rigidly next to the first mass. When I accelerate the reference frame (eg: by shoving a cart with my hand that the accelerometer is attached to) the other mass is "dragged" along with the reference mass by a spring or other energy transfer device. We can therefore measure the amount of force the transfer device uses in order to measure/sense the "acceleration" of the cart.

An inertial accelerometer is the kind of accelerometer that a computer hard disk drive uses to detect when it is dropped or shoved suddenly, to protect the heads from "crashing." A doppler accelerometer is the kind of accelerometer used to measure whether a distant sun of the same spectra as our own, is moving (OR) accelerating toward or away from us.

However, I do not thnk an intertial accelerometer necessarily gives the same measurements in Special Relativity that a doppler shift accelerometer will give. The reason is that "free fall" can not be detected by an inertial accelerometer, but it can be detected by a doppler accelerometer.

When we did the "twin" paradox, using minkowski space, I suspect we are assuming that somehow we can magically tell the difference between inertial acceleration and gravitational acceleration in Special Relativity by the mere labeling of "frames of reference."

But it seems to me that we can always use gravitational acceleration to "hide" acceleration from an inertial accelerometer.

A special television show was aired last week about General Relativy's 100th birthday. At the end of the show they posed a Gedanken "Suppose the Sun disappeared from our solar system, never mind "HOW" it happened, what would be the effect on the earth?"

The television show's point was that it would take 8 minutes for the gravitational wave from the diappearing sun to reach the Earth, and for Earth to stop "curving" around the region where the sun used to be. The Earth would then move in a straight line flying off at a tangent...

The issue I see is that the Earth is normally in "Free" fall around the sun (in Special Relativity) and an inertial accelerometer can not detect the acceleration of the Earth by the sun (accurately). Whether the sun is there (or not), an inertial accelerometer will measure approximately 1g because of the accelerometer's close proximity to Earth's surface. I am unsure what the accelerometer would meausre if it was some-how moved magically to the center of the Earth ... but I suspect it would measure zero.

Both before and after the "sun" went out, the Earth has the same mass , and therefore an inertial accelerometer will still measure 1g if oriented parallel to the Earth's gravitational field at the surface of the Earth, or probably zero if at the center. The inertial accelerometer is not going to measure the same amount of acceleration as a doppler accelerometer will measure because an inertial accelerometer doesnt' "know" the total mass of the entire universe. (Especially if the sun suddenly just "disappeared.).

That leads me back to the assumption I'm trying to explore:
Whenever we do measurments in Minkowski space, for special relativity, we always measure with time of travel via the length "ct", to make the measurement co-variant. But this automatically implies that light is some-how involved in the measurement and acceleration is inherently a doppler measurement.

The second issue is that Minkowski space uses the generalized Pythagorean theorem, which is the same as taking the square root of an absolute value. In every case where a square root is taken, there are mathematically TWO solutions. It's not clear to me that one can't use Newtonian gravity in Special Relativity, to artificially manipulate which soultion applies to path length, or time length, etc.

For example, in the Twin paradox, at the 8 year mark ... what happens if two massive (but tiny) black holes moving in opposite directions were to move approximately perpendicular to both A and B at the "midway" point between them? The inertial accelerometers on A and B are not going to measure much change (if any!) ... But the doppler shift accelerometer is going to suddenly blue shift on both A and B very distinctively. This change will happen under the assumptions of Special Relativity, as well.

Therefore, I think that "inertial reference frames" are a necessary condition to claim that something is not accelrating, but merely being in an inertial reference frame is not a *sufficient* condition to guarantee that measurable acceleration is not happening in Special Relativity.

I realize that General Relativity may come to a different conclusion, but I think General Relativity uses different assumptions.

I have completed my calculations for the orbit of the planet-satellite system based on F=ma. One very important point is that the effective mass is different in different directions for 3 vectors. It goes as gamma in directions perpendicular to the direction of motion, and gamma cubed in the direction of motion. Therefore, the direction of the force vector is only the same as the direction of the acceleration vector when going exactly perpendicular or exactly parallel to the direction of motion. In all other cases, the direction of acceleration is not exactly the same as the direction of the applied force.

 

[PeterDonis]

One method is that if two "identical" lasers (twins) are in motion toward (or away) from each other, that they can measure a color change (doppler shift) of each other. Therefore, they can measure acceleration between the two lasers by measuing the time derivative of the laser's color change.

More precisely, these will measure relative acceleration between two different objects.

The second method of measuring acceleration is to use two masses (inertias) that are in torsion with each other. We attach one mass rigidly to a "reference" frame, and measure the shear force required to keep the other mass moving rigidly next to the first mass. When I accelerate the reference frame (eg: by shoving a cart with my hand that the accelerometer is attached to) the other mass is "dragged" along with the reference mass by a spring or other energy transfer device. We can therefore measure the amount of force the transfer device uses in order to measure/sense the "acceleration" of the cart.

This is one possible implementation of an accelerometer, which measures proper acceleration of a single object. There are multiple ways of making an accelerometer:

https://en.wikipedia.org/wiki/Accelerometer
The key point is that an accelerometer is a measurement on a single object, whereas the Doppler shift measurement you describe is a measurement involving two objects, the light source and the receiver. So they're not measuring the same thing.

I do not thnk an intertial accelerometer necessarily gives the same measurements in Special Relativity that a doppler shift accelerometer will give. The reason is that "free fall" can not be detected by an inertial accelerometer, but it can be detected by a doppler accelerometer.

No, you have it backwards. The inertial accelerometer does detect free fall: you're in free fall if and only if your inertial accelerometer reads zero.

The Doppler shift accelerometer does not detect free fall, because knowing that your velocity relative to something else is changing, which is what the Doppler shift accelerometer is measuring, does not tell you that either your or the something else are or are not in free fall. In flat spacetime, if your Doppler shift accelerometer reads nonzero, then either you or the something else must not be in free fall, and it's possible that you both are not. In curved spacetime (i.e., when gravitating masses are present), your Doppler shift accelerometer can read nonzero even if both you and the something else are in free fall (this is a manifestation of tidal gravity, i.e., spacetime curvature).

When we did the "twin" paradox, using minkowski space, I suspect we are assuming that somehow we can magically tell the difference between inertial acceleration and gravitational acceleration in Special Relativity by the mere labeling of "frames of reference."

No. The twin paradox can be resolved without even using reference frames at all, and does not require any violation of the equivalence principle.

it seems to me that we can always use gravitational acceleration to "hide" acceleration from an inertial accelerometer.

No. An inertial accelerometer measures proper acceleration directly; there's no way to "hide" it.

At the end of the show they posed a Gedanken "Suppose the Sun disappeared from our solar system, never mind "HOW" it happened, what would be the effect on the earth?"

And this is an excellent example of why you should not use pop science shows to discuss science. This scenario violates stress-energy conservation so it is inconsistent with the laws of physics. You can't validly conclude anything from an inconsistent scenario.

the Earth is normally in "Free" fall around the sun (in Special Relativity) and an inertial accelerometer can not detect the acceleration of the Earth by the sun (accurately).

Of course not, since free fall means zero proper acceleration.

Whether the sun is there (or not), an inertial accelerometer will measure approximately 1g because of the accelerometer's close proximity to Earth's surface.

It will if it is sitting at rest on the Earth's surface, yes. This is due to the Earth's surface pushing up on the accelerometer and preventing it from freely falling.

But if you drop the inertial accelerometer off a cliff (or, better, from a platform at the top of the inside of a tall tower which has been evacuated to remove air resistance), it will read zero while it is falling.

Whenever we do measurments in Minkowski space, for special relativity, we always measure with time of travel via the length "ct", to make the measurement co-variant. But this automatically implies that light is some-how involved in the measurement

It implies no such thing. The speed of light is a universal constant, which really means it's just a unit conversion factor. Using units in which $c = 1$, which is what you're describing, works just fine for measurements that have nothing to do with light.

The second issue is that Minkowski space uses the generalized Pythagorean theorem, which is the same as taking the square root of an absolute value. In every case where a square root is taken, there are mathematically TWO solutions.

And since we are dealing with lengths in time and space, which are obviously positive, the positive square root is the only one that is relevant.

what happens if two massive (but tiny) black holes moving in opposite directions were to move approximately perpendicular to both A and B at the "midway" point between them?

Nothing much. See below.

The inertial accelerometers on A and B are not going to measure much change (if any!)

Yes.

But the doppler shift accelerometer is going to suddenly blue shift on both A and B very distinctively.

No, it won't. The light going between A and B might be bent by the black holes, but its frequency as seen by A or B won't change; it falls into the gravity well and then climbs back out again, and the two frequency shifts cancel.

This change will happen under the assumptions of Special Relativity, as well.

No, it won't, because you can't use SR to analyze a scenario that includes black holes, since SR only works if spacetime is flat and black holes require curved spacetime.

I think that "inertial reference frames" are a necessary condition to claim that something is not accelrating

Neither inertial acceleration nor Doppler acceleration have anything to do with reference frames. They are direct measurements and will be the same regardless of any choice of frame, or even if you make no choice of frame at all.

merely being in an inertial reference frame is not a *sufficient* condition to guarantee that measurable acceleration is not happening in Special Relativity

If by "being in an inertial reference frame", you mean "being in free fall", then this is exactly the sufficient (and necessary) condition for zero inertial acceleration, and it tells you nothing by itself about Doppler acceleration. See above.

I realize that General Relativity may come to a different conclusion, but I think General Relativity uses different assumptions.

I have no idea what you mean by this.

 

[learn.steadfast]

I thought you were going to stay out of the thread and watch IBEX and others answer my questions. Couldn't stand not to inject again?

The key point is that an accelerometer is a measurement on a single object, whereas the Doppler shift measurement you describe is a measurement involving two objects, the light source and the receiver. So they're not measuring the same thing.

The inertial accelerometer example that I gave was attached to a cart. It was accelerating with respect to my hand pushing it. The measurement was not a measurement on one object but on two masses that are connected by a spring relative to 'me' who pushed the cart.

I have never seen an example where anyone can truly make measurement (even in a gedanken) with a single object. There is always a reference of some kind used as a standard, whether it be a point in space, a clock, or something !

So, I have no idea what you are talking about.

No, you have it backwards. The inertial accelerometer does detect free fall: you're in free fall if and only if your inertial accelerometer reads zero.

Fair enough, but you've missed something as usual because there were two objects being considered...

The accelerometer reads NON zero, on Earth while both the accelerometer and Earth freely fall toward the Sun. There is no string, or material object, keeping the accelerometer positioned with respect to the sun ... The meter is still not detecting Free fall even though Free fall is going on.

If you removed the sun, there would still be the gravity of Earth being measured by the inertial accelerometer. The sensitivity of the accelerometer is ovewhelmed by the Earth's gravity regardless of whether the sun exists or not; because the accelerometer is closer to the earth.

If you put the accelerometer exactly at the center of gravity between the sun and earth, ONLY then it will read zero but it will be not be falling (in any intelligible sense of the english words) toward EITHER mass, especially when measured relative to the stars.

I'm talking only about special relativity, but since you bring up the point:

In special relativity, if there aren't "black holes" according to you, then substitude "very large mass" for what I said. The point is that because of Newtonian gravity, when a large mass is (temporarily) placed between two free floating objects (lasers) that are a fixed distance apart; the gravity will cause those objects to accelerate toward each other. Therefore, a doppler shift will occur even though the inertial accelerometers do not detect the relative acceleration.

The doppler shift has nothing to do with a gravity well itself, as you pointed out ... So I have no idea why you brought it up.measure zer

 

[PeterDonis]

I thought you were going to stay out of the thread and watch IBEX and others answer my questions. Couldn't stand not to inject again?

The thread has been dormant for a few weeks, and the items I responded to in this latest post of yours are very different from the original topic of the thread anyway.

The measurement was not a measurement on one object but on two masses that are connected by a spring relative to 'me' who pushed the cart.

Yes, but one mass is held fixed to the cart, and you say you're measuring the force necessary to keep the second mass rigidly attached to the first mass, and "rigidly attached" means "not moving relative to", so you're not measuring the relative motion of the masses (which is what a Doppler accelerometer measures), you're measuring the proper acceleration of the cart. Which in the case you describe is caused by your hand pushing the cart: but you could just as easily attach your apparatus to a rocket in empty space and measure the proper acceleration of the rocket when its engine fires.

I have never seen an example where anyone can truly make measurement (even in a gedanken) with a single object.

It depends on what you mean by "a single object". The point is that an accelerometer (an inertial accelerometer, in your terminology) measures proper acceleration locally, without regard to any external reference. For example, in the case of the rocket in empty space that I just described, it measures the proper acceleration of the rocket without reference to anything outside the rocket (and the apparatus, which is attached to the rocket). This is a common and well understood measurement, for example in inertial navigation systems.

The accelerometer reads NON zero, on Earth while both the accelerometer and Earth freely fall toward the Sun.

It reads nonzero sitting at rest on the Earth's surface because, as I said, the Earth's surface is pushing up on it. Take that away by dropping the accelerometer off a cliff (or, as I said, inside an evacuated tunnel to eliminate air resistance) and it will read zero: everything is in free fall, so there is no need to exert any force on the second mass to keep it at the same distance from the first mass. The spring will sit in its unstressed equilibrium position and the masses will simply fall towards the Earth. The fact that the Earth is free falling in orbit about the Sun doesn't change that. Or you could put the masses in a spaceship well away from the Earth in a free-fall orbit, and the whole apparatus would just float along with the ship reading zero.

If you removed the sun

This is inconsistent with the laws of physics, so there is no way to even discuss what would happen. It's like asking what would happen if two plus two were five.

In special relativity, if there aren't "black holes" according to you, then substitude "very large mass" for what I said.

Doesn't matter as far as using SR is concerned. SR is only valid if spacetime is flat. If any mass (and therefore gravity) is present, spacetime is not flat. You can certainly set up such a scenario, but you can't analyze it using SR; you need GR. However, for the particular question you're interested in, the GR answer is pretty simple and is basically what you say; see below.

when a large mass is (temporarily) placed between two free floating objects (lasers) that are a fixed distance apart; the gravity will cause those objects to accelerate toward each other.

Yes, that's true. I was confused because you talked about the twin paradox, and there's no need to bring in the twin paradox for a scenario like this. Just have two free floating lasers and have a large mass fly by between them. Yes, in that case the lasers will start falling towards each other and a Doppler accelerometer attached to each one will read nonzero.

Therefore, a doppler shift will occur even though the inertial accelerometers do not detect the relative acceleration.

Yes, in the case just described, inertial accelerometers attached to each laser would read zero, since both lasers are in free fall. The nonzero Doppler acceleration is due to the spacetime geometry.

 

[Dale]

One method is that if two "identical" lasers (twins) are in motion toward (or away) from each other, that they can measure a color change (doppler shift) of each other.

This is most definitely not a measurement of proper acceleration. This measurement could be zero when the proper acceleration is nonzero or it could be nonzero when the proper acceleration is zero.

I recommend reading about proper acceleration to get a better understanding of its meaning and importance. Accelerometers, by definition, measure proper acceleration. Your Doppler measurement does not measure proper acceleration. Therefore it is not an accelerometer.

The second method of measuring acceleration is to use two masses (inertias) that are in torsion with each other. We attach one mass rigidly to a "reference" frame, and measure the shear force required to keep the other mass moving rigidly next to the first mass.

This is essentially correct. A few details are off, but the basic concept is right. It doesn’t need to be torsion, usually it is compression or bending. In either case you usually measure the strain rather than the force, but of course they are related.

However, I do not thnk an intertial accelerometer necessarily gives the same measurements in Special Relativity that a doppler shift accelerometer will give.

That is correct because your Doppler shift device is not an accelerometer.

When we did the "twin" paradox, using minkowski space, I suspect we are assuming that somehow we can magically tell the difference between inertial acceleration and gravitational acceleration in Special Relativity by the mere labeling of "frames of reference."

There is no gravitational acceleration in special relativity. If you have gravitational acceleration then you need to use general relativity, not special relativity.

Therefore, I think that "inertial reference frames" are a necessary condition to claim that something is not accelrating, but merely being in an inertial reference frame is not a *sufficient* condition to guarantee that measurable acceleration is not happening in Special Relativity

Remaining at rest in an inertial frame in SR is both a necessary and a sufficient condition for having zero proper acceleration. In GR this still holds except that inertial frames are only local rather than global as they are in SR

I'm talking only about special relativity ... In special relativity, if there aren't "black holes" according to you, then substitude "very large mass" for what I said. The point is that because of Newtonian gravity

Newtonian gravity is inconsistent with talking only about special relativity. That is the whole reason that general relativity was developed. If you are talking about gravity (which you keep doing) then you are not taking about SR.

You are contradicting yourself every time you talk about gravity and then claim you are only talking about SR. If you want others to stop talking about GR then you need to stop bringing it up by discussing gravity. You cannot discuss gravity at all if you want to limit the conversation to SR.

 

[PeterDonis]

If you have gravitational acceleration then you need to use general relativity

And strictly speaking, "gravitational acceleration" is a poor choice of terminology since a body moving solely under the influence of gravity is in free fall, with zero proper acceleration.

 

[Ibix]

The implicit assumption of the show, however, is that calling something an "inertial reference frame" is sufficient to solve the paradox.

No frames are needed at all. The only reason you need to introduce frames explicitly is to explain how the "paradox" arises from using frames implicitly in a careless manner.

There are at least two ways I know to measure acceleration. I am not certain they are the same under the assumptions of Special Relativity.

You specify a mass-on-a-spring accelerometer and a Doppler accelerometer. These do, indeed, measure two different things, usually called proper acceleration and coordinate acceleration (assuming the coordinates in question treat the laser accelerometer as at rest). The distinction is present in Newtonian physics as well, although we don't usually make a point of it.

However, I do not thnk an intertial accelerometer necessarily gives the same measurements in Special Relativity that a doppler shift accelerometer will give. The reason is that "free fall" can not be detected by an inertial accelerometer, but it can be detected by a doppler accelerometer.

They won't give the same readings in general, no. For example, fit two carts with an inertial and a laser accelerometer each and point the laser at the other cart. Push one cart. The inertial accelerometer on that cart will kick, but the Doppler accelerometer on both carts will kick.

When we did the "twin" paradox, using minkowski space, I suspect we are assuming that somehow we can magically tell the difference between inertial acceleration and gravitational acceleration in Special Relativity by the mere labeling of "frames of reference."

If we are working in Minkowski spacetime then there is no gravity by hypothesis. If there is gravity then it's not Minkowski spacetime.

But it seems to me that we can always use gravitational acceleration to "hide" acceleration from an inertial accelerometer

This is backwards. General relativity does not treat gravity as a force so does not regard it as causing acceleration. It simply modifies the concept of "moving in a straight line".

they posed a Gedanken "Suppose the Sun disappeared from our solar system, never mind "HOW" it happened, what would be the effect on the earth?"

As Peter points out, this violates the local conservation of energy and is therefore not a situation that can be described in general relativity, since local conservation of energy follows directly from the Einstein Field Equations. Attempting to reason from this point is, therefore, nothing to do with GR.

Whenever we do measurments in Minkowski space, for special relativity, we always measure with time of travel via the length "ct", to make the measurement co-variant. But this automatically implies that light is some-how involved in the measurement and acceleration is inherently a doppler measurement

The constant c in special relativity does not have to be the speed of light. It's perfectly possible to model photons as having non-zero mass, in which case light does not travel at c (or indeed, any unique speed). There is no effect on relativity. I believe we currently place the photon mass at less than 10^-54kg, and we usually assume it to be zero, but we do not actually know that. In modern treatments of relativity you derive the existence of a constant with dimensions of velocity directly from symmetry requirements. Only by experiment do you then discover that it is the same as the speed of light.

The second issue is that Minkowski space uses the generalized Pythagorean theorem, which is the same as taking the square root of an absolute value. In every case where a square root is taken, there are mathematically TWO solutions. It's not clear to me that one can't use Newtonian gravity in Special Relativity, to artificially manipulate which soultion applies to path length, or time length, etc.

I don't understand how you think your last sentence relates to the first two, but Newtonian gravity is fundamentally incompatible with relativity. It requires a notion of global simultaneity that isn't present.

 

[Ibix]

I have never seen an example where anyone can truly make measurement (even in a gedanken) with a single object. There is always a reference of some kind used as a standard, whether it be a point in space, a clock, or something !

I agree to a point. However, the point about the inertial accelerometer is that it is a "closed room" experiment. You can build one in a sealed opaque box and it will work. Not so your laser accelerometer (unless you use it on a free-floating mass inside the box, which is simply using the device as another form of inertial accelerometer).

The accelerometer reads NON zero, on Earth while both the accelerometer and Earth freely fall toward the Sun.

The accelerometer is not in free fall. Free fall is exactly what it says - falling freely. Sitting on the floor, it is not falling. Thus it shows the acceleration due to the force exerted on it by the floor.

And can I suggest you leave off the personal attacks? It just makes you look foolish, especially when you are incorrect as in this case.

I'm talking only about special relativity,

...yet you keep bringing up gravity. Special relativity only works where gravity can be neglected.