Higher Himalayan Shear Zone, Sutlej section: structural geology and extrusion mechanism by various combinations of simple shear, pure shear and channel flow in shifting modes

Abstract

The Higher Himalayan Shear Zone (HHSZ) in the Sutlej section reveals (1) top-to-SW ductile shearing, (2) top-to-NE ductile shearing in the upper- and the lower strands of the South Tibetan Detachment System (STDS_U, STDS_L), and (3) top-to-SW brittle shearing corroborated by trapezoid-shaped minerals in micro-scale. In the proposed extrusion model of the HHSZ, the E₁-phase during 25–19 Ma is marked by simple shearing of the upper sub-channel defined by the upper strand of the Main Central Thrust (MCT_U) and the top of STDS_U as the lower- and the upper boundaries, respectively. Subsequently, the E_2a-pulse during 15–14 Ma was characterized by simple shear, pure shear, and channel flow of the entire HHSZ. Finally, the E_2b-pulse during 14–12 Ma observed simple shearing and channel flow of the lower sub-channel defined by the lower strand of the Main Central Thrust (MCT_L) and the top of the STDS_L as the lower- and the upper boundaries, respectively. The model explains the constraints of thicknesses of the STDS_U and the STDS_L along with spatially variable extrusion rate and the inverted metamorphism of the HHSZ. The model predicts (1) shear strain after ductile extrusion to be maximum at the boundaries of the HHSZ, which crudely matches with the existing data. The other speculations that cannot be checked are (2) uniform shear strain from the MCT_U to the top of the HHSZ in the E₁-phase; (3) fastest rates of extrusion of the lower boundaries of the STDS_U and the STDS_L during the E_2a- and E_2b-pulses, respectively; and (4) variable thickness of the STDS_L and rare absence of the STDS_U. Non-parabolic shear fabrics of the HHSZ possibly indicate heterogeneous strain. The top-to-SW brittle shearing around 12 Ma augmented the ductile extruded rocks to arrive a shallower depth. The brittle–ductile extension leading to boudinage possibly did not enhance the extrusion.

Appendices

Appendix 1: Simple shear flow (line 1 in Fig. 16a)

Steady plane laminar flow of an incompressible Newtonian fluid within a very long NE–SW oriented channel with parallel- and horizontal walls is given by

$$ {\text{d}}P/{\text{d}}x = \mu \left( {{\text{d}}^{2} x/{\text{d}}y^{2} } \right) $$

(1)

(Schlichting and Gersten 1999) Here, dP/dx pressure gradient along X-direction, μ viscosity. Putting dP/dx = 0, for simple shear,

$$ \left( {{\text{d}}^{2} x/{\text{d}}y^{2} } \right) = 0 $$

(2)

Integrating twice, taking the channel thickness ‘2y _o’ and velocity of walls x = −U ₂, U ₁ at y = −y ₀, y ₀, respectively, the flow profile is:

$$ x = 0.5\left( {U_{1} - U_{2} } \right) + 0.5yy_{0}^{ - 1} \left( {U_{1} + U_{2} } \right) $$

(3)

Putting x = 0 in Eq. 3, the coordinate of the pivot, defined by intersection between the y-axis and the velocity profile, is P ₁ ≡ [0,y ₀(U ₁ − U ₂)(U ₁ + U ₂)⁻¹]

Appendix 2: Channel flow/Poiseuille flow (curve 2 in Fig. 16a)

A channel with geometry, orientation and rheology same as that of the previous case is considered. Here U ₁,U ₂ = 0 at y = + y ₀, −y ₀, dP/dx is a constant ≠0. Integrating Eq. 2 twice and using these boundary conditions, the profile as in many fluid mechanics text e.g. Pai (1956) is

$$ x = - 0.5\mu^{ - 1} \left( {dP/dx} \right)\left( {y_{0}^{2} -y^{2} } \right) $$

(4)

The vertex of this parabolic profile is [0.5μ⁻¹ y ²₀ (dP/dx), 0]

Appendix 3: Velocity profile of general shearing (Fig. 15b)

We first consider pure shear on a rectangle ABCD where a pair of opposite walls, AD and BC, move with unequal velocities U ₃ and U ₄, respectively (U ₃ > U ₄) (Fig. 15a). Let AD and AB are of lengths ‘l’ and ‘2y ₀’, respectively. Points A (0, y ₀) and B (0, −y ₀) on the two walls, after instant ‘t’ come to A′(0, y ₀ − U₃ t) and B′(0, U ₄ t − y ₀). The line AB is deformed into a parabolic velocity profile P1 (Spurk 1993, pp 58–60), which is the velocity profile at A and B. The vertex of the parabolic profile is equidistant from these two points. Therefore, the Y-ordinate of vertex is −0.5 t (U ₃ − U ₄). Let the X-ordinate of the vertex is ‘p’. Thus the vertex has the coordinate [p, −0.5 t (U ₃ − U ₄)]. Let the equation of this parabola is

$$ y^{2} + Dx + Ey + F = 0 $$

(5)

Putting the co-ordinate of A′and B′ in Eq. 5, E and F are solved, put back in Eq. 5, and is rewritten,

$$ x = - D^{ - 1} \left[ {\left( {U_{3} - U_{4} } \right)ty + \left( {y_{0} - U_{3} t} \right)\left( {U_{4} t - y_{0} } \right) + y^{2} } \right] $$

(6)

Area bounded by this parabola with the y-axis between coordinates A′[0, (y ₀ − U₃ t)] and B′[0, (U ₄ t − y ₀)] is

$$ A_{1} = - D^{ - 1} \int\limits_{{\left( {U_{4} t - y_{0} } \right)}}^{{\left( {y_{0} - U_{3} t} \right)}} {\left[ {y^{2} + \left( {U_{3} - U_{4} } \right)ty + \left( {y_{0} - U_{3} t} \right).\left( {U_{4} t - y_{0} } \right)} \right]} {\text{dy}} $$

(7)

or,

$$ A_{1} = - 0.17D_{.}^{ - 1} \left( {2y_{0} -U_{3} t - U_{4} t} \right)\left\{ {4\left( {y_{0} - U_{3} t} \right)\left( {U_{4} t - y_{0} } \right) - t^{2} \left( {U_{3} - U_{4} } \right)^{2} } \right\} $$

(8)

Assuming a constant area deformation (Spurk 1993), it can be stated that the area bounded by the parabola P1 with the y-axis is half to that of area lost from the original rectangle due to compression, i.e.

$$ A_{1} = 0.5\left( {U_{3} + U_{4} } \right)tL $$

(9)

Eliminating ‘A ₁’ from Eqs. 8 and 9

$$ D = - 0.33t^{ - 1} L^{ - 1} \left( {U_{3} + U_{4} } \right)^{ - 1} \left( {2y_{0} - U_{3} t-U_{4} t} \right)\left[ {t\left( {U_{3} + U_{4} } \right)\left\{ {4y_{0} - \left( {U_{3} + U_{4} } \right)t} \right\} - {4y_{0} }^{2} } \right] $$

(10)

Putting this in Eq. 5 along with the values of E and F, the parabolic velocity profile for pure shear at point A is

$$ \begin{aligned} x = & \left( {2y_{0} - U_{3} t - U_{4} t} \right)^{ - 1} \left[ {\left( {U_{3} + U_{4} } \right)t\left\{ {4y_{0} - \left( {U_{3} + U_{4} } \right)t} \right\} - {4y_{0} } ^{2} } \right]^{ - 1} \\ & \quad \times 3Lt\left( {U_{3} + U_{4} } \right)\left[ {y^{2} + \left( {U_{3} - U_{4} } \right)ty + \left( {y_{0} - U_{3} t} \right)\left( {U_{4} t - y_{0} } \right)} \right] \\ \end{aligned} $$

(11)

Now a NE–SW oriented rectangle with very long length sides and full of incompressible Newtonian rheology and viscosity μ undergoing general shear by a pure shear and another component of top-to-SW sense of simple shearing (Fig. 15b) is represented by

$$ x_{\text{general\;shear}} = x_{\text{pure\;shear}} + x_{\text{simple\;shear}} $$

(12)

Putting the expressions for x _{pure shear} and x _{simple shear} from Eqs. 3 and 11 in 12.

$$ \begin{aligned} x_{\text{general\;shear}} = & \left( {2y_{0} - U_{3} t - U_{4} t} \right)^{ - 1} \left[ {\left( {U_{3} + U_{4} } \right)t\left\{ {4y_{0} - t\left( {U_{3} + U_{4} } \right)} \right\} - {4y_{0} } ^{2} } \right]^{ - 1} 3Lt\left( {U_{3} + U_{4} } \right) \\ & \quad \left[ {y^{2} + \left( {U_{3} - U_{4} } \right)yt + \left( {y_{0} - U_{3} t} \right)\left( {U_{4} t - y_{0} } \right)} \right] + 0.5\left( {{\text{U}}_{1} -{\text{U}}_{2} } \right) + {0.5{\text{yy}}_{0}^{ - 1} \left( {{\text{U}}_{1} + {\text{U}}_{2} } \right)} \\ \end{aligned} $$

(13)

As this is a quadratic expression of y, it represents a parabola. The Y-ordinate of the vertex is

$$ \begin{gathered} Y \equiv 0.5[\left( {U_{4} - U_{3} } \right)t - \left( {U_{1} + U_{2} } \right)\{ \left( {2y_{0} - U_{3} t - U_{4} t} \right) \hfill \\ \times\left\{ {\left( {U_{3} + U_{4} } \right)t\left( {4y_{0} - U_{3} t - U_{4} t} \right) - 4y_{0}^{2} } \right\}\left\{ {1.5Lt\left( {U_{3} + U_{4} } \right)} \right\}^{ - 1} {\text{y}}_{0}^{ - 1} ] \hfill \\ \end{gathered} $$

(14)

Therefore, the thickness of the STDS_U characterized by a top-to-NE sense of shearing within the top of the deformed rectangle is given by

$$ T^{\prime} = \left( {y_{0} - U_{3} t} \right) - Y $$

(15)

The zone of top-to-NE shearing does not form when

$$ T^\prime \le 0 $$

(16)

Appendix 4: A combination of simple shear and channel flow in a shifting mode (curve 3 in Fig. 17)

A channel with geometry, orientation and rheology same as that of "Appendix 1" is considered. The velocities are x = U ₁₁, –x = U ₂₁ at y = y ₀, −y ₀, respectively, at the boundaries; also a pressure gradient (dP ₁/dx) acts. Using these in Eq. 2, the velocity profile, defined within y = y ₀,−y ₀, is

$$ x = 0.5\left( {U_{11} -U_{21} } \right) + 0.5yy_{0}^{ - 1} \left( {U_{11} + U_{21} } \right)-0.5\mu^{ - 1} \left( {{\text{d}}P_{1} /{\text{d}}x} \right)\left( {y_{0}^{2} -y^{2} } \right) $$

(17)

From Eqs. 3, 4 and 17, we note that x _{combined flow} = x _{simple shear} + x _{channel flow}.

The parabolic profile given by Eq. 17 has its vertex at:

• X-ordinate: 0.5 (U₁₁ − U₂₁) – 0.125 y ⁻²₀ (U₁₁ + U₂₁)² μ (dx/dP₁) – 0.5 μ⁻¹y ²₀ (dP₁/dx)

• Y-ordinate: −0.5y ⁻¹₀ μ (U₁₁ + U₂₁)(dx/dP₁)

The maximum value of the X-ordinate on profile 17 is obtained at the vertex. Therefore, the fluid attains highest velocity U ₃ = X-ordinate at the vertex. For another combined simple shear and channel flow in the sub-channel defined by the walls y = −t, −y ₀ due to their shear velocities x = U ₁₂, −U ₂₂, respectively, and a pressure gradient (dP ₂/dx < 0), the velocity profile is to be deduced. Transforming O′[0,−0.5 (y ₀ + t)] as the new origin [0,0], the old coordinates [0,−t] and [0,0] become [0,0.5 (y ₀ − t)], and [0,0.5 (y ₀ + t)], respectively. Substituting y ₀ = 0.5 (y ₀ − t), U ₁₁ = U ₁₂, U ₂₁ = U ₂₂ and (dP ₁/dx) = (dP ₂/dx) in Eq. 17, the velocity profile, defined within y = 0.5 (y ₀ − t), 0.5 (y ₀ + t), with reference to the new coordinate axes, is

$$ x = 0.5\left( {U_{12} - U_{22} } \right) + 0.5\left( {U_{12} + U_{22} } \right)\left( {y_{0} - t} \right)^{ - 1} y-0.5\mu^{ - 1} \left( {{\text{d}}P_{2} /{\text{d}}x} \right)\left[ {\left\{ {0.5\left( {{\text{y}}_{0} - {\text{t}}} \right)} \right\}^{2} - {\text{y}}^{2} } \right] $$

(18)

Now going back from new origin O′[0,0] to the old origin O [0,0.5 (y ₀ + t)], Eq. 18 is rewritten

$$ \begin{aligned} x = & 0.5\left( {U_{12} - U_{22} } \right) + \left( {U_{12} + U_{22} } \right)\left( {y_{0} - t} \right)^{ - 1} \left[ {y - 0.5\left( {y_{0} + t} \right)} \right] \\ & \quad -0.5\mu^{ - 1} \left( {{\text{d}}P_{2} /{\text{d}}x } \right)\left[ {\left( {y_{0} + t} \right) - \left\{ {0.5\left( {y_{0}^{2} + t^{2} } \right)} \right\} - y^{2} } \right] \\ \end{aligned} $$

(19)

The profiles given by Eqs. 17 and 19 have zones of top-to-NE shearing, the STDS_U and the STDS_L, respectively, with thicknesses

$$ T_{1} = y_{0} -0.5y_{0}^{ - 1} \left( {U_{11} + U_{21} } \right)\mu_{1} \left( {{\text{d}}P_{1} /{\text{d}}x} \right)^{ - 1} $$

(20)

$$ T_{2} = 0.5\left( {y_{0} - t} \right)-\left( {y_{0} - t} \right)^{ - 1} \left( {U_{12} + U_{22} } \right)\mu_{2} \left( {{\text{d}}P_{2} /{\text{d}}x} \right)^{ - 1} $$

(21)

The thickness of the HHSZ outside the respective detachments are

$$ T_{1} {}^{\prime} = \left[ {y_{0} + 0.5y_{0}^{ - 1} \left( {U_{11} + U_{21} } \right)\mu_{1} \left( {{\text{d}}P_{1} /{\text{d}}x} \right)^{ - 1} } \right] $$

(22)

$$ {\text{T}}_{2}^{\prime} = \left[ {\left( {{\text{y}}_{0} - {\text{t}}} \right) + 0.5\left( {{\text{y}}_{0} - {\text{t}}} \right)^{ - 1} \left( {{\text{U}}_{12} + {\text{U}}_{22} } \right)\mu_{2} \left( {{\text{dP}}_{2} /{\text{dx}}} \right)^{ - 1} } \right] $$

(23)

From Eqs. 20–23, we note T ₁ < T ₁′and T ₂ < T ₂′. From Eq. 19, the STDS_U does not form when

$$ \left( {U_{11} + U_{21} } \right)\left( {{\text{d}}P_{1} /{\text{d}}x} \right)^{ - 1} = {\text{or}} > \left( {2y_{0}^{2} \mu_{1}^{ - 1} } \right);\quad T_{1} = {\text{or}} < 0 $$

(24)

From Eq. 21, the STDS_L does not form when

$$ \left( {U_{12} + U_{22} } \right)\left( {{\text{d}}P_{2} /{\text{d}}x} \right)^{ - 1} = {\text{or}} > \left[ {2\left( {y_{0} - {\text{t}}} \right)^{2} \mu_{2}^{ - 1} } \right];\quad T_{2} = {\text{or}} < 0 $$

(25)

Note that for certain combination of flow parameters (y ₀, U ₁₁, U ₂₁, μ₁, dP ₁/dx, t, U ₁₂, U ₂₂, μ₂ and dP ₂/dx), T ₂ is <T ₁.

Appendix 5: Combined flow through a two-layer HHSZ (Fig. 20)

A channel of same geometry and geographic orientation as "Appendix 1" is considered. The channel is assumed to be full with two immiscible Newtonian fluids- the top and the bottom layers with viscosities μ₁ and μ₂, respectively (μ₁ < μ₂). The line y = t (0 < t < −y ₀) is taken as the fluid interface. The top- and the bottom walls are sheared with velocities U ₁ and −U ₂, respectively, and (dP/dx) < 0. The distribution of velocities in the top- and the bottom layers of the fluid are x₁ and x₂, respectively. The following deductions are based on steps similar to those in Papanastasiou et al. (2000). From Navier–Stokes equation, for the top and the bottom fluid layers

$$ \left( {{\text{d}}P/{\text{d}}x} \right) = \mu_{1} \left( {{\text{d}}^{2} x_{1} /{\text{d}}y_{2} } \right) $$

(26)

$$ \left( {{\text{d}}P/{\text{d}}x} \right) = \mu_{2} \left( {{\text{d}}^{2} x_{2} /{\text{d}}y_{2} } \right) $$

(27)

At the fluid interface, i.e. at the line y = t, the momentum flux is continuous through fluid–fluid interface. In other words, at y = t,

$$ \mu_{1} \left( {{\text{d}}x_{1} /{\text{d}}y} \right) = \mu_{2} \left( {{\text{d}}x_{2} /{\text{d}}y} \right) $$

(28)

Integrating Eqs. 26 and 27:

$$ \left( {{\text{d}}x_{1} /{\text{d}}y} \right)\mu_{1} = y\left( {{\text{d}}P/{\text{d}}x} \right) + C $$

(29)

$$ \left( {{\text{d}}x_{2} /{\text{d}}y} \right)\mu_{2} = y\left( {{\text{d}}P/{\text{d}}x} \right) + C^{\prime} $$

(30)

From Eqs. 28, 29, and 30, C = C′. Integrating Eqs. 29 and 30

$$ x_{1} = 0.5y^{2} \mu_{1}^{ - 1} \left( {{\text{d}}P/{\text{d}}x} \right) + Cy\mu_{1}^{ - 1} + C_{1} $$

(31)

$$ x_{2} = 0.5y^{2} \mu_{2}^{ - 1} \left( {{\text{d}}P/{\text{d}}x} \right) + Cy\mu_{2}^{ - 1} + C_{2} $$

(32)

Now at y = t, x ₁ = x ₂; at y =+y ₀, x ₁ = U ₁; and at y = −y ₀, x ₂ = −U ₂. Using these in Eqs. 31 and 32, C ₁ = C ₂. Using this back in Eqs. 31 and 32, C and C ₁ are solved and put in those equations:

$$ \begin{aligned} x_{1} = & 0.5\mu_{1}^{ - 1} y^{2} \left( {{\text{d}}P/{\text{d}}x} \right) \\ & + y\mu_{1}^{ - 1} \left[ {\left( {U_{1} + U_{2} } \right)y_{0}^{ - 1} \left( {\mu_{1}^{ - 1} + \mu_{2}^{ - 1} } \right)^{ - 1} - 0.5y_{0} \left( {{\text{d}}P/{\text{d}}x} \right)\left( {\mu_{1}^{ - 1} - \mu_{2}^{ - 1} } \right)\left( {\mu_{1}^{ - 1} + \mu_{2}^{ - 1} } \right)^{ - 1} } \right] \\ & + \left[ {\left\{ {\left( {U_{1} \mu_{1} - U_{2} \mu_{2} } \right)-{\text{y}}_{0}^{2} \left( {{\text{d}}P/{\text{d}}x} \right)} \right\}\left( {\mu_{1} + \mu_{2} } \right)^{ - 1} } \right] \\ \end{aligned} $$

(33)

$$ \begin{aligned} x_{2} = & 0.5y^{2} \mu_{2}^{ - 1} \left( {{\text{d}}P/{\text{d}}x} \right) \\ & + y\mu_{2}^{ - 1} \left[ {\left( {U_{1} + U_{2} } \right)y_{0}^{ - 1} \left( {\mu_{1}^{ - 1} + \mu_{2}^{ - 1} } \right)^{ - 1} - 0.5y_{0} \left( {{\text{d}}P/{\text{d}}x} \right)\left( {\mu_{1}^{ - 1} - \mu_{2}^{ - 1} } \right)\left( {\mu_{1}^{ - 1} + \mu_{2}^{ - 1} } \right)^{ - 1} } \right] \\ & + \left[ {\left\{ {\left( {U_{1} \mu_{1} - U_{2} \mu_{2} } \right) - y_{0}^{2} \left( {{\text{d}}P/{\text{d}}x} \right)} \right\}\left( {\mu_{1} + \mu_{2} } \right)^{ - 1} } \right] \\ \end{aligned} $$

(34)

As these are quadratic equations of ‘y’, they represent parabolas with the same ‘Y’-ordinate of their vertices

$$ Y \equiv - \left( {U_{1} + U_{2} } \right)y_{0}^{ - 1} \left( {\mu_{1}^{ - 1} + \mu_{2}^{ - 1} } \right)^{ - 1} \left( {{\text{d}}P/{\text{d}}x} \right)^{ - 1} + 0.5y_{0} \left( {\mu_{1}^{ - 1} - \mu_{2}^{ - 1} } \right)\left( {\mu_{1}^{ - 1} + \mu_{2}^{ - 1} } \right)^{ - 1} $$

(35)

Appendix 6: Flow restricted within the upper parts of the HHSZ (Fig. 21)

We choose a channel with a geometry, geographic orientation and rheology same as that of case- (I). A simple shear of top-to-SW sense with velocities of its top and bottom walls as U ₁ and −U ₂, respectively, and that only top portion of the channel (y = t to y ₀) have an additional flow component imparted by the pressure gradient (dP/dx). Replacing ‘t’ in ‘y’ in Eq. 3, velocity along the X-direction at coordinate (0,t) is

$$ x = 0.5\left( {U_{1} - U_{2} } \right) + 0.5y_{0}^{ - 1} {\text{t}}\left( {U_{1} + U_{2} } \right) $$

(36)

Thus, in the upper sub-channel, velocities of its boundaries are: x =+U ₁, at y = y ₀ and at y = t, x = 0.5 (U ₁ − U ₂) +0.5 y ⁻¹₀ t (U ₁ + U ₂).

Using these conditions and parameters in Eq. 2, the velocity profile in the upper sub-channel is

$$ \begin{aligned} {{x}} = & 0.5\mu^{ - 1} y^{2} \left( {{\text{d}}P/{\text{d}}x} \right) + \left\{ {0.5yy_{0}^{ - 1} \left( {U_{1} + U_{2} } \right)} \right\}-0.5y\mu^{ - 1} \left( {{\text{d}}P/{\text{d}}x} \right)\left( {y_{0} + t} \right) \\ & \quad + 0.5\left( {U_{1} - U_{2} } \right) + 0.5t\mu^{ 1} y_{0} \left( {{\text{dP}}/{\text{dx}}} \right) \\ \end{aligned} $$

(37)

The Y-ordinate of the vertex of the parabolic profile is

$$ {{Y}} \equiv 0.5\left[ {({\text{y}}_{0} + {\text{t}}) - \mu ({\text{d}}P/{\text{d}}x)^{ - 1} y_{0}^{ - 1} ({\text{U}}_{1} + {\text{U}}_{2} )} \right]. $$

(38)

Three profiles 3, 2, and 1 are presented in Fig. 21 for Y>, =, and <y ₀, respectively. Only in the former case, an STDS_U forms with a thickness

$$ T = (y_{0} - Y) $$

(39)

Appendix 7: Calculations for analogue models

The Reynolds Number is defined as

$$ R_{e} = \rho {\text{vy}}_{0} \mu^{ - 1} $$

(40)

where ρ density of the fluid, μ dynamic viscosity of the fluid, y ₀ half the width of the channel, and v velocity of flow. In the real situation of the HHSZ, taking crustal density ρ = 2.7 gm cm⁻³, viscosity of partially molten mid-crustal rocks μ = 10¹⁹ Pa s (as taken by Jamieson et al. 2004), v = 4.8 mm year⁻¹ (the maximum value for the Zanskar Shear Zone, a continuation of the STDS_U, as given by Dèzes et al. 1999), R_e comes out in the order of 10⁻²¹.

In the model, taking the density of the PDMS ρ = 0.95 gm cm⁻³, μ = 10⁵ Pa s (Talbot and Aftabi 2004 and references therein), the slowest velocity of extrusion attained v = 3 mm per 10 min, half the width of the model HHSZ (y ₀) = 1.25 cm in few experiments, R_e comes out in the order of 10⁻¹⁴.