General Caption: In this figure, we analysis the tides on Earth in the ideal limit: the ideal tide. The ideal tide gives the main behavior---all the other behaviors are perturbations---but very noticeable ones on the human scale when you are NOT in the open ocean.
The analysis is well beyond the scope of an introductory astronomy course and yours truly wishes someone had explained it to yours truly in 1977. Read it as poetry.
Features:
Image 1 vastly exaggerates the size of the tidal bulges. As we show below, the ideal tidal range is 1.07145 m and the observed open ocean tidal range is typically ∼ 0.6 m.
The classical limit sounds very restrictive, but pretty much everything from cosmic dust to the large scale structure can be analyzed as in the classical limit in good to excellent approximation. Black holes CANNOT of course.
Proof:
m_i*a_i = F_i = F_i' + m_i*g_i where F_i is the net force on particle i, F_i' is the net internal force on particle i (including internal gravitational fields), and g_i is the external gravitational field at particle i due to the whole rest of the universe. In fact, in most cases, g_i can be calculated by integrating up the gravitational fields due to relatively local sources of gravity using just Newtonian physics. More remote sources can be assumed to cancel due to the uniform density of the universe on average: a source far off in one direction will be canceled by one far off in the opposite direction. This procedure is invalid for the observable universe as whole for which general relativistic cosmology is needed. General relativistic cosmology gives us the comoving frames that expand with the observable universe and it confirms the astrophysical 2nd law formula (A2LF) we derive here given the assumptions we made above for large astro-bodies (e.g., galaxies and larger) that are just moving with a comoving frame (i.e., participating in the mean expansion of the universe). This is the extra confirmation alluded to above. Smaller astro-bodies are part of larger ones, and so the confirmation propagates down to all smaller scales from the comoving frame scale. F_i' = sum_j(F_ji) where F_ji is the force of particle j on particle i. We sum on i to get sum_i(m_i*a_i) = (1/2)sum_(ij)F_ij + sum_i(m_i*g_i) = sum_i(m_i*g_i) where the (1/2) is to correct for double counting and (1/2)sum_(ij)F_ij = 0 by pairwise cancellation of forces by Newton's 3rd law. Now the CM is defined by r = sum_(m_i*r_i)/m, where m = sum_i(m_i), and thus we get v = sum_(m_i*v_i)/m and a = sum_(m_i*a_i)/m. Note that origin and absolute size scale for r_i, v_i, and a_i are left unspecified since we do not need those specifications, but they could be specified relative to a local comoving frame if needed. Now, we define g_ave = sum_i(m_i*g_i)/m = a to be mass-weighted average external gravitational field. We can now write Newton's 2nd law for particle i as m_i*a_i = sum_j(F_ji) + m_i*g_i m_i*(a_i-a) = sum_j(F_ji) + m_i*(g_i - a) m_i*(a_i-a) = sum_j(F_ji) + m_i*(g_i - g_ave) m_i*a_i' = sum_j(F_ji) + m_i*[ (g_i-g) - sum_j[(m_j)*(g_j-g)/m] ] , where m_i*a_i' = sum_j(F_ji) + m_i*gt(i) is our final astrophysical 2nd law formula (A2LF), where a_i' is acceleration relative to the CM, g is the external gravitational field at the CM, gt(i) = [ (g_i-g) - sum_j[(m_j)*(g_j-g)/m] ] is the effective tidal field (i.e., force per unit mass) on particle i, and the A2LF makes no use of absolute position, velocity, and acceleration. So those can all be calculated relative to the CM. A tidal force is just the difference in gravity between two points.The existence of the astrophysical 2nd law formula (A2LF) completes the proof. If a form of the Newton's 2nd law of motion (AKA F=ma) holds for the CM defined reference frame, then it is an inertial frame in which Newtonian physics holds.
Note the 2nd term in the effective tidal field is annoying. However, it is probably often small or negligible due to cancellation by significant symmetry of the astro-body and of the tidal field. In fact, for the ideal tide, it is zero by symmetry as we show below.
The following par is hogwash---correct it.
This is good because the g_i CANNOT be calculated correctly by
Newtonian physics
as noted above.
However, the tidal forces can be.
Asymptotically
with increasing distance r (i.e., distance from the CM recall)
from an external gravity source,
the tidal force declines in size as 1/r**3 (which we actually show
below for the special case of the
ideal tide).
Now given the 1/r**3 decline,
one can integrate up the tidal forces
extending to infinity
in a universe
of uniform density on average
and get a finite value.
(We do NOT know if the
universe
is infinite, but
it does seem to have uniform
density on average.
See Wikipedia:
Cosmological principle.)
In fact, usually cosmologically remote contributions are negligible (which is good since we CANNOT calculate them correctly by Newtonian physics) and for many (but NOT all) systems, all one needs is calculate the tidal forces from a few nearby sources of gravity. In fact, for many important systems, the tidal forces are completely negligible. If the astro-body is small enough, the external gravitational field is approximately uniform and the tidal forces approximately zero. This is true of spacecraft in orbit.
Actually, if you want very accuracy/precision, you need general relativity even for systems where the A2LF gives very good results. The tiny perihelion shift of Mercury required general relativistic treatment to be accounted for. The that accounting was the first evidence for general relativity (see Wikipedia: Tests of general relativity: Perihelion shift of Mercury).
EXCEPT that very strong gravitational fields (like those very near black holes) may cause the rotation of free-fall frames relative to the observable universe, but this is a tricky point for which yours truly cannot find a clear explication. The best so far (and it does NOT say much) is Wikipedia: Inertial frame of reference: General relativity.
The upshot is that there is an absolute rotation (i.e., one with respect to the observable universe) even though there is NO absolute space in the sense used by Isaac Newton (1643--1727).
Nowadays, we do know inertial frames (i.e., free-fall frames) attached to centers of mass (CMs) do accelerate with respect to each other. This follows from the derivation and form of the A2LF above.
But local inertial frames (i.e., those at the same place) do NOT accelerate with respect to each other: they only differ in velocity. Reference frames that accelerate relative to local inertial frames are our modern non-inertial frames.
However, despite original Newtonian physics view of inertial frames and non-inertial frames that was apparently held up until the advent of general relativity in 1915 people were able before 1915 to use Newtonian physics to calculate to high accuracy/precision the behavior Solar System, Earth-Moon system, moon systems and the tides (see Wikipedia: Tide: History of tidal theory; Theory of tides: History).
Yours truly has to think those old old-time celestial mechanics practitioners were somehow using our derived A2LF implicitly assuming it was right or that non-inertial-frame effects were negligible. But this is a fine point of the history of celestial mechanics.
Well, if the only external force on the astro-body is gravity (i.e., the only significant one), then Newtonian physics (which is good enough for our understanding here) dictates that its center of mass is in free fall. The internal forces on the astro-body (e.g., self-gravity, pressure force, etc.) cancel out pairwise by Newton's 3rd law of motion as far as the motion of the center of mass is concerned. Internal forces AND external forces too can cause motions relative to the center of mass, of course.
To do so one must regard deviations in the external gravitational field (see definition below) from the external gravitational field at the CM_⊕ as forces in their own right. They are called tidal forces. Clearly tidal forces are combinations of gravitational forces and inertial forces, but as follows from the discussion above that in itself causes no errors in analysis.
Note the tidal force at the CM_⊕ is by definition zero
Note the extension of the CM_⊕ inertial frame to the whole Earth for the analysis is valid, but extending it further it becomes increasingly less convenient for analysis to the point that it becomes horrifically inconvenient and as relativistic effects become important increasingly wrong. The perspective of the expanding universe with its continuum of comoving frames (i.e., the basic of inertial frames) is the best perspective for the analysis of the observable universe so far as we know.
g_x = (-GM_ext/r**2)*cos(γ) = [-GM_ext/((r_cm+x)**2+y**2)]*[(r_cm+x)/sqrt((r_cm+x)**2+y**2)] g_y = (-GM_ext/r**2)*sin(γ) = [-GM_ext/((r_cm+x)**2+y**2)]*[y/sqrt((r_cm+x)**2+y**2)]
The relevant expansions and resulting relevant formulae are:
r**2 = (r_cm+x)**2+y**2) = r_cm**2*[1+2*(x/r_cm)+(x/r_cm)**2+(y/r_cm)**2] ≅ r_cm**2*[1+2*(x/r_cm)] 1/r**2 ≅ (1/r_cm)**2*[1-2*(x/r_cm)] r ≅ r_cm*[1+(x/r_cm)] = r_cm + x cos(γ) ≅ 1 sin(γ) ≅ y/r_cm g_x_1st = (-GM_ext/r_cm**2)*[1-2*(x/r_cm)] g_y_1st = (-GM_ext/r_cm**2)*(y/r_cm) gt_x_1st = (2GM_ext/r_cm**3)*R_eq_⊕*(x/R_eq_⊕) 1st order x-coordinate tidal force gt_y_1st = (2GM_ext/r_cm**3)*R_eq_⊕*(-1/2)*(y/R_eq_⊕) 1st order y-coordinate tidal force gt_strength = (2GM_ext/r_cm**3)*R_eq_⊕ fiducial tidal force strength gt_strength_Moon = 1.09753*10**(-6) N/kg = 1.11992*10**(-7)*g fiducial tidal force strength for the Moon. gt_strength_☉ = 0.505695*10**(-6) N/kg = 0.516015*10**(-7)*g fiducial tidal force strength for the Sun.where g is Earth's surface gravitational field g (fiducial value 9.8 N/kg). We can see that gt_strength_Moon and gt_strength_☉ are very small compared to g which means local variations on the surface of the Earth of the external gravitational field have very small and usually negligible effect. However, over the extend of the Earth, they give the tides.
We need a differential equation for pressure along the solid Earth from the point (x=R_e_⊕,y=0) to the point (x=0,y=R_e_⊕,y=0). The appropriate differential formula for volume element of volume dV and cross-sectional area dA is
(p_b - p_a)*dA = ρ*g_θ*dV where ρ is water density assumed constant, θ is the angle subtended at the CM_⊕ by the arc length along the solid Earth measured from the positive x axis counterclockwise toward the positive y axis, b and a are points on the solid Earth, θ_b > θ_a. g_θ = -ρ*gt_x_1st*sin(θ) + ρ*g_y_1st*cos(θ) is the tidal field along the solid Earth increasing θ as the positive direction. Thus g_θ points in the negative direction.The differential equation follows and solution follows:
(dp/dθ) = -R_eq_⊕*ρ*gt_strength*[cos(θ)*sin(θ) + (1/2)*sin(θ)*cos(θ)] p(θ) = p_x_axis - (3/2)*R_eq_⊕*ρ*gt_strength*sin(θ)**2To convert to sea level h, we assume the only significant gravitational field which points radially downward and that air pressure p_air is constant. The formula for pressure with depth is
p = p_air + ρ*g*h(see Wikipedia: Hydrostatics: Pressure in fluids at rest). Since pressure in hydrostatic equilibrium is isotropic (i.e., the same in all directions), combining the last two equations we obtain the following expressions
h(θ) = h_x_axis - (3/2)*R_eq_⊕(gt_strength/g)*sin(θ)**2 h_x_axis - h_y_axis = (3/2)*R_eq_⊕*(gt_strength/g) tidal range in general = 1.07145 m for the Moon alone = 0.493682 m for the Sun alone.The values for the ideal tidal range we have just calculated agree to better than 2 significant figures with those cited by Wikipedia: Tide: Amplitude and cycle time which were doubtless calculated in a similar way, but probably with slightly different input numbers.
The observed open ocean tidal range is typically ∼ 0.6 m. Given the vast simplifying assumptions made for our ideal tidal ranges