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May 19, 2008, 15:46 
Divergence of Tensor (physical meaning)

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what is physical meaning of divergence of a second order tensor (is a vector)?
what about div^2 second order tensor (is scaler)? 

May 22, 2008, 15:48 
Re: Divergence of Tensor (physical meaning)

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In a simplistic manner one can say that the use of a second rank tensor is a convenient way to write down generalized equations in vector form. A 2nd rank tensor can represent for example an "energy momentum", say rho*v_i*v_j, where i,j=1,2,3 (or x,y,z if you prefer). Then the derivative of such a tensor would enter the vectorial equation
say T_ij = rho v_i v_j Then dT_ij/dx_i = v_j d/dx_i (rho v_i) + rho v_i dv_j/dx_i where the first term usually appears in the equation of continuity. Using such a "notation" the momentum equation in Fluid Dynamics can be written d/dt(rho v_i) =  dp/dx_i  dT_ij/dx_j (inviscid equations) and one can even define the momentum flux tensor as say S_ij = p delta_ij + rho v_i v_j where delta_ij = 1 if i=j, =0 othewise. THen the momentum equation becomes d/dt(rho v_i) =  dS_ij/dx_j The derivative here is really a whole multiplications and summation: dS_ij / dx_j = dS_i1/dx_1 + dS_i2/dx_2 + dS_i3/dx_3 (or if you prefer it is a divergence sign). And the index i takes the values 1 for x, 2 for y, and 3 for z. So this is really 3 equations written as one (vectorial form). THis is really convenient when you want to write lengthy equations in a compact form or when you work (say) in 3 or more dimensions (e.g. in General Relativity you add the time as the 4 dimension; in Riemanian Geometry you can even assume an infinite number of dimensions in a coordinate system that is not Cartesian, nor orthogonal, in that case it is extremely convenient to use tensor "notation"). So the Divergence of a 2nd rank tensor is a vector and its second derivative (divergence) would be a scalar. It can represent forces, flux of energy, flux of momentum, &c... Consider this intersting example. In orthogonal coordinates (Cartesian) the length of an element ds is just given using the Pythagoras formula for a triangle with a right angle: ds^2 = dx^2 + dy^2 + dz^2 or in two dimension ds^2 = dx^2 + dy^2 However, if the system of coordinates does not have right angles (say there is an angle alpha between the X and Y axis) then instead of Pythagoras on has to use ds^2 = dx^2 + dy^2 + 2*dx*dy*cos(alpha) or in TENSOR notation ds^2 = g_ij dx_i dx_j where the metric tensor g_ij is just g_11 =1 g_22 =1 g_12=g_21 = 2 cos(alpha) and you carry out a summation over the indices (the summation sign is usually ignored). (it would be a little more complex in 3D but similar). So this is just a convenient way of writing the equation, though it has tremendous implications when considering much more general noneuclidian space (Riemanian Geometry) as this includes derivative of the metric tensor itself (e.g. in polar and spherical coordinates the metrix tensor is responsible for the terms giving rise to the Coriolis and centrifugal acceleration as one has to put the metrix tensor under the derivative). So in a more general sens derivatives of tensor give us insight into the geometry of space itself (e.g. on the surface of sphere). I hope this helps. 

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