I Length contraction viewed with Heaviside equations

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.

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

\vec B=\vec u_{r_r} \times { \vec E \over c }
[/itex] eq. 28.4

Where [itex]r_r[/itex] is a scalar value, which is the distance (or time/c) back to when the particle emitted the field. [itex]\vec u_{r_r}[/itex] 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 [itex] 1 = { - q \over { 4 \pi \epsilon_0}} [/itex] which allows me to focus on algebraic formulas instead of numerical values of the units of measurement.

Contours of equal field propagation time (constant [itex]r_r[/itex]) 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:

[itex]r_r= {{x^2+y^2} \over {(x^2+({1- {v^2 \over c^2} })y^2 )^{1/2} - ({v \over c}) x } } [/itex]
[itex]\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 ]}[/itex] : 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 [itex] 1 \over r [/itex] in some directions, but Lorentz contracted E and B fields of a uniformly moving charged particle is expected to fall off as [itex] 1 \over r^2 [/itex] in all directions. Please review Feynman's work if you have questions.

When 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 [itex]1 \over r[/itex] 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.

[itex] | \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} } } [/itex]
[itex] | \vec{B} | = | \vec{E} | {1 \over c } {y \over { \sqrt {x^2+y^2} } } [/itex]

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:

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

So, I do see the Lorentz (gamma) factor show up on the y axis. [itex] \gamma = { 1 \over {(1-{v^2 \over c^2})^{1/2} }} [/itex] 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:
[itex]u' = {u \over { \gamma (1 - {{ u \cdot v } \over c^2} ) }} [/itex]
(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.
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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.
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Note: “Heaviside” not “Heavyside”


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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).
Note: “Heaviside” not “Heavyside”
Yes, and the magnetic field equation is missing a constant of [itex]v/c^2[/itex] 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 [itex]1/c[/itex]. 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.
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:

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.
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:
[itex]\gamma = { 1 \over \sqrt { 1 - {{ v_x^2 + v_y^2 + v_z^2} \over c^2 }} }[/itex]

And effective mass goes as rest mass times gamma: [itex]m=m_0 \gamma[/itex]

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:

[itex] {v_y}' = { v_y \over { {\gamma}_x ( 1 - 0 ) } } [/itex]
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:
[itex]\gamma_{xy} = { 1 \over \sqrt { 1 - {{ v_x^2 + {v_y^2 \over \gamma_x^2 }} \over c^2 }} }[/itex]
[itex]\gamma_{xy} = { 1 \over \sqrt { 1 - {{ v_x^2 + {v_y^2(1- {v_x^2 \over c^2}) }} \over c^2 }} }[/itex]
[itex]\gamma_{xy} = { 1 \over \sqrt { 1 - {{ v_x^2 + v_y^2 - {v_y^2v_x^2 \over c^2 }} \over c^2 }}} [/itex]

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

So, I wonder if the proper (time dialated) gamma factor for three dimensions can also be computed by a simple product. eg:
[itex]\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 }} } [/itex]

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?


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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.


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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,
$$p_{\mu} K^{\mu}=0.$$

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