UPDATE September 3, 2023: Our observation today solves the mystery why the August 10, 2023 calculation had "radial corrections" but the calculation in this post does not have.
----
In this blog post we set c = 1, to simplify the notation.
M = static disk mass
K = kinetic energy of disk
I = moment of inertia of disk
J = angular momentum of disk
R₀ = disk radius
ρ = static disk mass / area
<------ ω angular velocity
dm''' dm''
• •
center of the disk
× R = distance (×, dm)
| \ α = angle (×, m) vs. (×, dm)
| \
| \ ^ v
/
• •
dm' dm
r = distance (m, dm)
| β = angle (m, ×) vs. (m, dm)
| /
|/
• m observer (or a test mass)
R_m = distance (m, ×)
^ y
|
-----> x
The diagram is much like the one on August 10, 2023, but we switched the x and y axes. We have a part of disk, dm, moving with a velocity vector v. The disk is very lightweight and rotates very slowly.
Mirror images: dm' is the mirror image of dm against the line (m, ×), dm''' is the mirror image against ×, and dm'' its image against (m, ×).
We want to determine the metric at an observer m, which is far away from the rotating disk, i.e., the distance (m, ×) is much larger than the disk radius R₀.
The Schwarzschild metric associated with dm is moving at a velocity v obliquely relative to the observer m. The calculations may be somewhat complicated.
Lorentz transformations
The boost matrix B(v) is for a frame which moves with a velocity (v_x, v_y) relative to the laboratory frame.
Let us have a position 3-vector
t
r = x
y
in the laboratory frame. The coordinates in the moving frame are
t'
r' = x' = B(v) r,
y'
where the multiplication of B(v) and r is the matrix multiplication.
We obtain the inverse Lorentz transformation by flipping the signs of v_x and v_y.
The metric perturbation caused by a part dm
Let us have a part dm moving with a velocity vector (v_x, v_y). We write the line element of the Schwarzschild metric in the moving frame, in terms of coordinates of the laboratory frame. The metric in the moving frame is
dτ² = (-1 + r_s / r') dt'² + (1 + r_s / r') dr'² + (r' dφ')²,
in polar coordinates, if we assume that r_s is small. We have
dr' = -sin(β') dx' - cos(β') dy'.
The Minkowski metric maps onto itself in a Lorentz transformation. We can concentrate on how the perturbation is mapped. In the moving frame, the perturbation to the Minkowski metric is
r_s / r' * dt'²
+ r_s / r' * (- sin(β') dx' - cos(β') dy' )².
Let us try to write the perturbation in terms of dt, dx, dy. We can drop the terms ~ v³ because we assume that v is small. We can approximate γ = 1 + v² / 2. We get:
dt' = (1 + v² / 2) dt
- v_x dx
- v_y dy,
dx' = v_x dt
+ (1 + v_x² / 2) dx
+ (v_x v_y / 2) dy,
dy' = v_y dt
+ (v_x v_y / 2) dx
+ (1 + v_y² / 2) dy.
Then
dt'² = dt² + v² dt² + v_x² dx² + v_y² dy²
- 2 v_x dt dx - 2 v_y dt dy
+ 2 v_x v_y dx dy,
and
( sin(β') dx' + cos(β') dy' )²
=
sin²(β') * (v_x² dt² + dx² + v_x² dx²
+ 2 v_x dt dx + v_x v_y dx dy)
+ cos²(β') * (v_y² dt² + dy² + v_y² dy²
+ 2 v_y dt dy + v_x v_y dx dy)
+ 2 sin(β') cos(β')
* ( v_x v_y dt² + v_x dt dy + v_y dt dx
+ v_x v_y / 2 * dx²
+ (1 + v² / 2) * dx dy
+ v_x v_y / 2 * dy²
).
We want to integrate, but what is the role of β' in the moving frame versus β in the laboratory frame? Also, r_s / r' is calculated in the moving frame, not in the laboratory frame.
The length contraction in the y direction is a factor
1 - v_y² / 2,
and in the x direction
1 - v_x² / 2.
Approximately,
sin(β') = sin(β) (1 - v_x² + v_y²),
cos(β')
= sqrt(1 - sin²(β) * (1 - v_x² + v_y²)²)
= sqrt(1 - sin²(β) * (1 - 2 v_x² + 2 v_y²))
= cos(β) + sin²(β) * (-v_x² + v_y²).
We can ignore the correction sin²(β) (-v_x² + v_y²) because
r_s / r * sin²(β) ~ 1 / R_m³,
and we are only interested in metric perturbations which are ~ 1 / R_m or ~ 1 / R_m².
\ | /
\|/
------- • dm ---
/|\
/ | \ squeezed field lines
• m test mass
---> v
How should we map r' to r? If dm is moving straight away from the observer m at a velocity v, then the moving frame, and the field of dm, appears squeezed in the direction of v.
Simultaneousness is problematic. We are interested in the metric at the observer m at the laboratory time t. The laboratory observer thinks that dm is at a specific location X at that time t. But these events are not simultaneous in the moving frame.
Let us assume that dm carries a 1 meter long ruler with it, whose far end at the laboratory time t is at the observer m. The field at the end of the ruler is, of course, the Schwarzschild metric at the distance 1 meter, as seen by dm. In this case, we have
r = (1 - v² / 2) r',
or
r' = (1 + v² / 2) r.
If the ruler is oblique, then
r'² = (1 + v²) r₁² + r₂²
= r² + v² r₁²,
where r₁ and r₂ are the components of r along v and normal to v.
We obtain:
r' = r * sqrt(1 + v² r₁² / r²)
= r * ( 1 + 1/2 v² sin²(α + β) ).
Integrals of the type 1 / r over a circle: the generalized shell theorem
Below we need to integrate several terms of the type
1 / r * dm
over a full circle, where r is the (large) distance to the observer m, and dm = ρ R₀ dα. The mass of the circle per a unit length of the circumference is ρ and its radius is R₀. We will study the sum of 1 / r for dm and dm'', as in the diagram at the start of this blog post.
Let us scale the diagram above to make calculations simpler. We keep ρ constant. If we scale by some factor X, then the length of the circle is scaled by X and 1 / r is scaled by 1 / X. The value of the integral does not change.
Let the coordinates of the observer be (0, 0), The radius of the circle is R₁ = R₀ / R_m, and its center is at (0, 1).
The coordinates of dm are (x, 1 - y), and the coordinates of dm'' are (x, 1 + y).
We will study the value of
1 / r + 1 / r'' = 1 / sqrt(x² + 1 - 2y + y²)
+ 1 / sqrt(x² + 1 + 2y + y²).
The Taylor series of 1 / sqrt(1 + a) is
1 - 1/2 a + 1 / 2! * 3/4 a² + ...
where a = x² +- 2y + y². We obtain
1 / r + 1 / r'' = 2 - 1/2 (2 x² + 2 y²)
+ 3/8 (8 y²) + ...
where we dropped all terms with a power 3 or more. Let us integrate over the semicircle for which y < 0. The integral for the term 2 is
2 π R₁ = 2 π R₀ * 1 / R_m,
while the integrals for other terms are
~ 1 / R_m³.
We are only interested in terms ~ 1 / R_m and ~ 1 / R_m², and can ignore ~ 1 / R_m³ terms.
The integral can be interpreted this way: if we are far from the circle, we can pretend that all the parts of the circle are at the distance of the center of the circle. It is like the shell theorem of Newton.
What about an integral
sin²(α) / r * dm ?
We have to integrate
sin²(α) * 2
over the lower semicircle whose radius is R₀ / R_m. The integral is
1/2 * 2 π R₀ * 1 / R_m.
Again, the integral is as if all the parts of the circle were at the distance of the center of the circle.
Generalized Shell Theorem. If we have to integrate
f(α) / r,
or
f(α) / r²
over a full circle, where the distance r is measured from a point P far away from the circle relative to the radius, and P is at the angle α = 0 relative to the center of the circle, and
f(π - α) = f(α) (symmetry),
f(π + α) = f(-α) (symmetry),
for any 0 ≤ α ≤ π / 2, and for any α, holds
f(2 π + α) = f(α) (periodicity),
then we can pretend that r = the distance to the center of the circle.
The error in the approximation is at most ~ 1 / r³. The conditions above say that the function f is symmetric relative to a diameter of the circle where the diameter is normal to the angle 0, and that f is periodic.
/ α = angle vs. (×, P)
___ /_
/ \ r
| × | ----------------------------- P
\_______/
<------->
symmetry
Proof. We can calculate as above in this section, that the integral of f(α) / r over the whole circle is
π / 2
∫ f(α) * 2 * R₀ / R_m * dα.
-π / 2
By the symmetry of f, this is equal to
3/2 π
∫ f(α) R₀ / R_m * dα.
-π / 2
The periodicity of f implies that the above is equal to
2 π
∫ f(α) / R_m * R₀ dα,
0
which is the integral of f(α) / R_m over the whole circle.
The proof for f(α) / r² is similar to f(α) / r. Q.E.D.
The integral of the dt'² term
r_s / r * ( 1 + 1/2 v² sin²(α + β) )
*
( dt² + v² dt² + v_x² dx² + v_y² dy²
- 2 v_x dt dx - 2 v_y dt dy
+ 2 v_x v_y dx dy),
v = ω R,
v_x = ω R cos(α),
v_y = ω R sin(α),
r = sqrt( R² sin²(α) + (R_m - R cos(α))² )
= R sin(α) / sin(β).
We must integrate the formula above for all dm. There should be a lot of canceling of different terms for dm and its mirror images dm' and dm'', just as we calculated on August 10, 2023.
The term for dt² is the attractive gravity force of the static mass of the disk. The mixed terms dt dx and dt dy are "magnetic" forces.
1. r_s / r * dt² : the integral is essentially 2 G M / R_m * dt², or the Schwarzschild / newtonian gravity attraction of the mass M of the static disk. We used the generalized shell theorem.
2. r_s / r * v² dt² : the integral is 2X the gravity of the kinetic energy K of the disk. We can use the generalized shell theorem.
3. r_s / r * 1/2 v² sin²(α + β) dt² : if R_m / R₀ is large, then β is small and the integral is 0.5X the gravity of the kinetic energy K of the disk. We can use the generalized shell theorem.
4. r_s / r * v_x² dx² : 2 G K / R_m * dx².
5. r_s / r * v_y² dy² : 2 G K / R_m * dy².
6. r_s / r * -2 v_x dt dx : the symmetry between dm and dm'' almost cancels this term, but not entirely, because dm is closer. This is the "dm moves to the right" term in our August 10, 2023 calculation. The sum is
dt dx * -2 G dm / (R_m / cos(β))²
* 2 R * cos(α) cos(β) * 2 v cos(α).
The value of cos³(β) differs from 1 by ~ 1 / R_m². That makes ~ 1 / R_m⁴ and we can ignore cos³(β). The mass element is
dm = ρ dR R dα.
The integral of cos²(α) over 0 < α < π / 2 is π / 4. We have to multiply by 2 to take into account the left side of the disk. We have to integrate
dt dx *
R₀
∫ -2 G / R_m² * π / 2 * 2 R * 2 ω R ρ R dR
0
= dt dx * -G / R_m² * ρ π R² * R² ω
= dt dx * -G / R_m² * M * R² ω
= dt dx * -G / R_m² * 2 I ω
= dt dx * G / R_m² * -2 J.
7. r_s / r * 2 v_y dt dy : the symmetry between dm' and dm makes this zero.
8. r_s / r * 2 v_x v_y dx dy : the symmetry between dm' and dm makes this zero.
Summary: We have to add 2.5 K to M to get the gravity attraction. There is some stretching in the x and y metrics. There is a magnetic term which "corresponds to" -2 J.
The integral of the dx'² term
r_s / r * sin²(β) (1 - 2 v_x² + 2 v_y²)
* (v_x² dt² + dx² + v_x² dx²
+ 2 v_x dt dx + v_x v_y dx dy).
We can ignore this because
1 / r * sin²(β) ~ 1 / R_m³.
The integral of the dy'² term
r_s / r *
cos²(β) * (v_y² dt² + dy² + v_y² dy²
+ 2 v_y dt dy + v_x v_y dx dy).
We have
cos²(β) = 1 - sin²(β),
and 1 / r * sin²(β) ~ 1 / R_m³. We can assume that cos(β) = 1. Since v_y² has the same value for dm and dm'', the symmetry condition of the generalized shell theorem is satisfied and we can pretend that r = R_m.
We have
v_y = ω R sin(α).
The integral of the v_y² term over the whole disk is
2 G / R_m * ω²
R₀ 2 π
* ∫ ∫ R² sin²(α) ρ dR R dα
0 0
= 2 G / R_m * ω² * ρ π R₀⁴ / 4
= 1/2 G / R_m * ω² * 2 I
= 4 G K / R_m.
1. r_s / r * v_y² dt² : 4 G K / R_m * dt².
2. r_s / r * dy² : 2 G M / R_m * dy².
3. r_s / r * v_y² dy² : 4 G K / R_m * dy².
4. r_s / r * 2 v_y dt dy : the values for dm' and dm cancel each other out because the sign of v_y flips. We can ignore the term.
5. r_s / r * v_x v_y dx dy : the sign of v_y flips for dm' and dm, and the term is canceled. We can ignore the term.
The integral of the dx' dy' term
r_s / r
* 2 cos(β) sin(β)
* (1 - v_x² + v_y²)
* ( v_x v_y dt² + v_x dt dy + v_y dt dx
+ v_x v_y / 2 * dx²
+ (1 + v² / 2) * dx dy
+ v_x v_y / 2 * dy²
).
We can assume that cos(β) = 1, because the difference from 1 only would bring a 1 / R_m⁴ term. Most terms above are ~ v³ or higher, and we can ignore them.
1. r_s / r * 2 sin(β) v_x v_y dt²
= 4 G dm / r * v² sin(β) cos(α) sin(α).
We have
sin(β) = R / r * sin(α).
The integrand becomes
4 G dm R / r² * v² cos(α) sin²(α).
The values for dm and dm'' cancel each other almost exactly:
1 / (R_m - R)² - 1 / (R_m + R)²
= 4 R_m R / (R_m² - R²)²
~ 1 / R_m³.
We can ignore the term.
2. r_s / r * 2 sin(β) v_x dt dy
= r_s / r * 2 v sin(β) cos(α) dt dy.
Again the values for dm and dm'' cancel each other almost exactly. We can ignore the term.
3. r_s / r * 2 sin(β) v_y dt dx
= 4 G dm / r * R / r * sin²(α) v dt dx
= 4 G / R_m² * R sin²(α) ω R dm dt dx.
The integral
4 G / R_m² * ω
R₀ 2 π
* ∫ ∫ R² sin²(α) ρ dR R dα
0 0
= G / R_m² * ω ρ π R₀⁴
= G / R_m² * 2 J.
The term is G / R_m² * 2 J dt dx.
This term probably corresponds to the "dm' approaches faster" term of the August 10, 2023 blog post.
4. r_s / r * sin(β) v_x v_y dx² : we can ignore this like we did in case 1.
5. r_s / r * 2 sin(β) (1 + v² / 2) dx dy : the values for dm' and dm cancel each other exactly. We can ignore the term.
6. r_s / r * sin(β) v_x v_y dy² : we can ignore this like we did in case 1.
7. r_s / r * 2 sin(β) * (- v_x² + v_y²) dx dy : the values for dm' and dm cancel each other exactly. We can ignore the term.
Sanity checks
There probably are errors in the long calculations above.
Let us check if we can find the newtonian attraction term of the rotating mass. It is in the dt'² section above.
The dy² metric should be stretched like in the standard Schwarzschild metric. It is in the dy'² section above.
The dx² metric should be very little perturbed. There are no large dx² terms.
The "magnetic" component dt dx is zero. The magnetic term in dt'² is canceled by a term in dx' dy'. The canceling term probably is the "dm' approaches faster" term of August 10, 2023. It did cancel the magnetic component there, too.
On August 10, 2023 we saw that in the radial motion toward the test mass m, there is some positive frame dragging which comes from radial corrections. Let us try to spot that in this blog post.
It looks like that the frame dragging caused by the radial movement of dm' and dm toward / away from the observer m is "lost" in the metric. The reason is that the metric does not understand the handedness of the gravitomagnetic field?
Or is it so that we have to calculate the x derivatives of the metric explicitly?
There should probably be a non-zero x derivative of g_ty because if the observer m moves slightly to the positive x direction, then he is closer to the right side of the disk where mass is moving away from him.
The gravitomagnetic field caused by the x movement of dm, dm'' is preserved in the metric because even though they have opposite effects, the effect of dm'' is smaller because it farther away.
A metric cannot express the "handedness" rules for a gravitomagnetic field?
On August 16, 2023 we claimed that the Einstein approximation formula loses "crucial information" about the movement of masses. We retracted the claim on August 20, 2023 when we realized that in the particular configuration, the metric g_ty (y is defined like in this blog post today), the value of g_ty varies with x.
But g_ty cannot vary with x for a rotating disk, since the metric has to have a rotational symmetry? No, since we are calculating with a fixed y axis which does not turn, there cannot be rotational symmetry. The metric for a neighboring observer should have the same y axis as the observer m, so that we can compare the values of g_ty.
If the neighbor uses his own y axis, then we must rotate his metric to use our own y axis. Let us calculate if we can get the right gravitomagnetic effect in that way.
Our present calculation method of the metric only calculates values for varying R_m. To get the values neighbors of the observer r, we have to use a rotation, or devise some other way.
A static metric around a rotating disk?
How can we model magnetism if there is no magnetic term?
This reminds us of the Ehrenfest paradox. One cannot define a fully consistent metric on a rotating disk itself?
Could it be that one cannot define a consistent metric around a rotating disk?
We have in earlier blog posts remarked that if a very heavy neutron star moves faster than what is the local speed of light inside it, then we cannot use static spatial coordinates at all. The neutron star must drag the coordinates along with it. There might be a similar thing happening around a rotating disk, even though the velocities are much less than the speed of light. Note that if we use rotating coordines, then the speed of the spatial coordinates does exceed the speed of light when we go to a great distance.
If one takes a fixed y coordinate line and adds the metric perturbations for each part of the rotating disk, then we end up in a metric which probably can reproduce the accelerations which we calculated on August 10, 2023. The left side of the disk is approaching the observer m and the right side receding from him. There is no rotational symmetry. The observer should see phenomena which are not symmetric.
But if one demands that the metric has to have rotational symmetry, then g_tx is the only component which can affect the x acceleration of a test mass m approaching along the y axis, and above we calculated that the dt dx, or g_tx is essentially zero.
It is as if we would be unable to define a metric which extends around the disk. This is like the Ehrenfest paradox.
We need to check carefully, if abandoning the rotational symmetry makes the metric sensible.
Conclusions
We will write a new blog post about this confusing situation. We may have made a calculation error on August 10, or in this blog post.
Since the left side of the disk is moving toward the observer m and the right receding,
dg_ty / dx
should be non-zero. But if the metric is rotationally symmetric, the derivative must be zero.
The metric has to be given in rotating coordinates? But there should be a mapping to static coordinates.
The Gödel rotating metric contains closed time-like curves, which can be regarded as proving that the solution is not reasonable. There may be something similar going on around any rotating mass.
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