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Momentum Heat Mass Transfer
D
     source
Dt
Rules, database,
Tensorial calculus
Rudolf Žitný, Ústav procesní a
zpracovatelské techniky ČVUT FS 2010
Introduction and rules (form of lectures,
presentations of papers by students, requirements
for exam). Preacquaintance (photography).
Working with databases and primary sources.
Tensors and tensorial calculus
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Momentum Heat Mass Transfer
3P/1C, 4 credits, exam, LS2016, room C2-83
• Lectures Prof.Ing.Rudolf Žitný, CSc. (Tuesday 8:00-10:30)
• Tutorials Ing.Karel Petera, PhD.
• Rating 30(quise in test)+30(example in test)+40(oral exam) points
A
B
C
D
E
F
90+
80+
70+
60+
50+
..49
sufficient
failed
excellent
very good
good
satisfactory
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Momentum Heat Mass Transfer
3P/2C, 4 credits, exam, LS2016, room C-434
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Momentum Heat Mass Transfer
Yeoiwon Seo
Olivia Davies
Markus Anttoni
Hansen-Haug
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Literature - sources
• Textbook:
Šesták J., Rieger F.: Přenos hybnosti tepla a hmoty, ČVUT Praha, 2004
• Monography: available at NTL
Bird, Stewart, Lightfoot: Transport Phenomena. Wiley, 2nd edition, 2006
Welty J.R.: Fundamental of momentum, heat and mass transfer,Wiley,2008
• Databases of primary sources
Direct access to databases (WoS, Elsevier, Springer,…)
knihovny.cvut.cz
Jméno DUPS
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EES Electronic sources
Information about peoples
(publications)
database of papers used
in presentations.
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Science Direct
Key words
Full text available in pdf
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Prerequisities: Tensors
Bailey
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Prerequisities: Tensors
Transfer phenomena operate with the following properties of solids and fluids
(determining state at a point in space x,y,z):
Scalars T (temperature), p (pressure),  (density), h (enthalpy), cA (concentration), k (kinetic energy)
Vectors
Tensors


f
(forces), T (gradient of scalar), and others like vorticity, displacement…
u(velocity),
  (stress),  (rate of deformation),  (gradient of vector) , deformation tensor…

u
 
Scalars are determined by 1 number.
Vectors are determined by 3 numbers
Tensors are determined by 9 numbers

u  (u x , u y , u z )  (u1 , u2 , u3 )
  xx  xy  xz    11  12  13 

 

    yx  yy  yz     21  22  23 

 



zx
zy
zz

  31  32  33 

Scalars, vectors and tensors are independent of coordinate systems (they are objective
properties). However, components of vectors and tensors depend upon the coordinate system.
Rotation of axis has no effect upon a vector (its magnitude and the arrow direction), but
coordinates of the vector are changed (coordinates ui are projections to coordinate axis). See
next slide…
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Rotation of cartesian coordinate system
Three components of a vector represent complete description (length of an arrow and its directions),
but these components depend upon the choice of coordinate system. Rotation of axis of a cartesian
coordinate system is represented by transformation of the vector coordinates by the matrix product
a1'  a1 cos(1',1)  a2 cos(1', 2)  a3 cos(1',3)
2
2’
a2'  a1 cos(2',1)  a2 cos(2', 2)  a3 cos(2',3)
a2
a3'  a1 cos(3',1)  a2 cos(3', 2)  a3 cos(3',3)
a
 a1'   cos(1',1) cos(1', 2) cos(1',3)   a1 
 ' 
  
a

cos(2
',1)
cos(2
',
2)
cos(2
',3)
 2 
   a2 
 a3'   cos(3',1) cos(3', 2) cos(3',3)   a3 
  
  
a’2
a 2 c o s (1 ', 2 ) 1’
a’1
a 1 c o s (1 ',1)
a1
1
[a ']  [[R]][a]
Rotation
matrix (Rij is
cosine of angle
between axis i’ and j’)
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Rotation of cartesian coordinate system
Example: Rotation only along the axis 3 by the angle  (positive for counter-clockwise direction)
Properties of goniometric functions
2’
2
cos(  )  cos 
cos(

2
  )  sin 

cos( (   ))   sin 
2
cos(1', 2)  sin    a1 
 a1'   cos(1',1)  cos 
 ' 
 
 a2   cos(2',1)   sin  cos(2', 2)  cos    a2 
   2 ', 2
a
(
1
[[R]] [[R]]=[[I]]  [[R]]  [[ R]]
T

  )  2 ',1
2

   1' , 2
2
1
   1 ',1
therefore the rotation matrix is orthogonal and can be inverted just only
by simple transposition (overturning along the main diagonal). Proof:
1’
 cos 

 sin 
 sin   cos 

cos    sin 

cos 2   sin 2 

 sin  cos   cos  sin 
1 0


0 1
sin  

cos  
cos  sin   sin  cos  

sin 2   cos 2 

T
Stresses
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describe complete stress state at a point x,y,z
 ij
Stress tensor is a typical example of the second order tensor with a pair of
indices having the following meaning
Index of plane
(cross section)
index of force component
(force acting upon the cross section i)
y
y
xy
xz
z
y
yy
xx
x
yz
z
 zy
yx
x
zz
z
zx
x
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Tensor rotation of cartesian coordinate system
Later on we shall use another tensors of the second order describing
kinematics of deformation (deformation tensors, rate of deformation,…)
Nine components of a tensor represent complete description of state (e.g. distribution of
stresses at a point), but these components depend upon the choice of coordinate system, the
same situation like with vectors. The transformation of components corresponding to the
rotation of the cartesian coordinate system is given by the matrix product
[[ ']]  [[R]][[ ]][[R]]T
where the rotation matrix [[R]] is the same as previously
 cos(1',1) cos(1', 2) cos(1',3) 


[[ R]]   cos(2',1) cos(2', 2) cos(2',3) 
 cos(3',1) cos(3', 2) cos(3',3) 


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Tensor rotation of cartesian coordinate system
Orthogonal matrix of the rotation of coordinate system [[R]] is fully determined
by 3 parameters, by subsequent rotations around the x,y,z axis (therefore by 3
angles of rotations). The rotations can be selected in such a way that 3
components of the stress tensor in the new coordinate system disappear (are
zero). Because the stress tensor is symmetric (usually) it is possible to
anihilate all off-diagonal components
  1' 0 0 


'
T
 0  2 0   [[R]][[ ]][[R]]
 0 0  3' 


Diagonal terms are normal (principal) stresses and the axis of the rotated
coordinate systems are principal directions (there are no shear stresses in
the cross-sections oriented in the principal directions).
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Special tensors
Kronecker delta (unit tensor, components independent of rotation)
 ij  0 for i  j
 ij  1 for i  j
1 0 0


   0 1 0
0 0 1



Levi Civita tensor is antisymmetric unit tensor of the third order (with 3 indices)
In terms of the epsilon tensor the vector product will be defined
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Scalar product
Scalar product (operator ) of two vectors is a scalar
   
a  b | a || b | cos 
3
a  b  a1b1  a2b2  a3b3   aibi  aibi  ai'bi'

b

a
i 1
aibi is abbreviated Einstein notation. Repeated indices are summing (dummy) indices.
Proof that
ai bi  ai'bi'
ai'  Rim am
b 'j  R jk bk
Remark: there were used dummy indices m and k in these relations. Letters of the dummy indices can be selected arbitrary, but in this
case they must be different, so that to avoid appearance of four equal indices in a tensorial term in the following product a.b (there can be
always max. two indices with the same name indicating a summation)
T
a1' b1'  a2' b2'  a3' b3'  ai'bi'  Rim Rik ambk  Rmi
Rik ambk   mk ambk  ambm
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Scalar product


Example: scalar product of velocity u [m/s] and force F
[N] acting at a point is
power P [W] (scalar)
P  u  F  ui Fi  u x Fx  u y Fy  u z Fz

Example: Scalar product
vector of an
 2 of velocity u [m/s] and the normal
oriented surface ds [m ] is the volumetric flowrate Q [m3/s] through the
surface
 
Q  ds  u

ds

u
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Scalar product
Scalar product can be applied also between tensors or
between vector and tensor
  
n   f
3
n
i 1
i
ij
y
 ni ij  f j

n
x
i-is summation (dummy) index, while j-is
free index
This case explains how it is possible to
calculate internal stresses acting at an
arbitrary cross section (determined by outer
normal vector n) knowing the stress tensor.

f
z
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Scalar product - examples
Dot product
  of delta tensors
   
 im mj   ij
1 0 0 1 0 0 1 0 0

 
 

0 1 00 1 0  0 1 0
0 0 1 0 0 1 0 0 1

 
 

Scalar product of tensors is a tensor
   
 im mj   ij
Double dot product of tensors is a scalar
 :  
 km mk  
Trace of a tensor (tensor contraction)
tr ( )   mm
example tr ( )  3
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Vector product
Vector product (operator x) of two vectors is a vector
 


 

c  a  b  (  b )  a
3

c
  
| c || a || b | sin 
3
ci    ijk a j bk   ijk a j bk


a
j 1 k 1
for example

b
c1  123a2b3  132a3b2  a2b3  a3b2
Important relationship between the Levi Civita and the Kronecker delta
special tensors
 ijk kmn   im jn   in jm
check for i=1,j=2,(k=3),m=1,n=2
123 312 = 11 22 - 12 21 = 1
check for i=1,j=2,(k=3),m=2,n=1
123 321 = 12 21 - 11 22 = -1
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Vector product
Examples of applications
  
Moment of force (torque) M  r  F

r

F
Coriolis force

 
F  2mu  


application: Coriolis flowmeter

F

u
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Diadic product
Diadic product (no operator) of two vectors is a second order tensor
ab  
ai b j   ij
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Differential operator  (Nabla)
Bailey
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Differential operator  (Nabla)
GRADIENT – measure of spatial changes
Symbolic operator  represents a vector of first derivatives with respect x,y,z.
  
( , , )
x y x

i 
xi
 applied to scalar is a vector (gradient of scalar)
T  (
T T T
,
, )
x y z
 iT 
T
xi
 applied to vector is a tensor (for example gradient of velocity is a tensor)
 u x

 x
  u
u   x
y

 u x

 z
u y
x
u y
y
u y
z
u z
x
u z
y
u z
z









u (x  dx, y)
 iu j 
u j
xi

u (x, y)
du x  u x 
ux
dx
ux
u x
dx
x
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Differential operator 
DIVERGENCY – magnitude of sources/sinks
Scalar product  represents intensity of source/sink of a vector quantity at a point
3
 u x u y u z
u u
u 


 i  i
x
y
z
xi
i 1 xi

  u  0 source

  u  0 sink
i-dummy index, result is a scalar

  u  0 conservati on
Scalar product  can be applied also to a tensor giving a vector (e.g. source/sink
of momentum in the direction x,y,z)
  

 yx  zx  xy  yy  zy  xz  yz  zz 

f       xx 

,


,


f i   j ji
y
z
x
y
z
x
y
z 
 x
The vector fi represents resulting force of stresses acting at the surface of an
infinitely small volume (the force is related to this volume therefore unit is N/m3)
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Laplace operator 2
Divergency of gradient – measure of nonuniformity
Scalar product =2 is the operator of second derivatives (when applied to scalar
it gives a scalar, applied to a vector gives a vector,…). Laplace operator is
divergence of a gradient (gradient of temperature, gradient of velocity…)
 2T  2T  2T
 2T
  T   T  2  2  2 
x
y
z
xi xi
2
i-dummy index
2
2
2
3  2u
 2u j
  2u x  2u x  2u x  u y  u y  u y  2u z  2u z  2u z
j
  u  ( 2  2  2 , 2  2  2 , 2  2  2 )   2 
x
y
z
x
y
z
x
y
z
xi xi
i 1 xi
Physical interpretation: 2 describes diffusion processes (random molecular motion).
The term 2 appears in the transport equations for temperatures, momentum,
concentrations and its role is to smooth out all spatial nonuniformities of the
transported properties.
Laplace operator 2
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Positive value of 2T
tries to enhance the
decreasing part
1.5
T ( x)  exp(  x 2 )
1
 2T
0.5
0
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-0.5
-1
-1.5
-2
-2.5
Negative value of 2T
tries to suppress the peak
of the temperature profile
2.5
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Integrals – Gauss theorem
Volume integrals with nabla operator can be converted
to surface integrals (just only

by replacing nabla  with unit normal vector n )
Physical interpretation: accumulation in volume V is overall flux through boundary

n

Pdv

n
Pds


V Divergence of P
S
projection of P to outer normal
V
Variable P can be
Scalar (typically pressure p)  pdv   npds
V
S
Vector (vector of velocity, momentum,
heat flux). Surface integral represents flux of
vector in the direction of outer normal.
   udv   n  uds
V
S
Tensor (tensor of stresses). In this
case the Gauss theorem represents the balance
between inner stresses and outer forces acting upon the surface,
   dv   n  ds
V
S
ds
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Transcriptions
Symbolic notation, for example  2 T , is compact, unique and
suitable for definition of problems in terms of tensorial equations.
However, if you need to solve these equation (for example  2 T  f )
you have to rewrite symbolic form into index notation, giving equations
(usually differential equations) for components of vectors or tensors,
which are expressed by numbers (may be complex numbers).
Bailey
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Symbolic  indicial notation
General procedure how to rewrite symbolic formula to index
notation
Replace each arrow  by an empty place for index 
Replace each vector operator by (-1)     
Replace each dot  by a pair of dummy indices in the first free position left and right
Write free indices into remaining positions
Practice examples!!
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Coordinate systems
Previous conversion procedure can be applied only in
a cartesian coordinate systems
Formulation in the x,y,z cartesian system is not always convenient,
especially if the geometry of region is cylindrical or spherical. For
example the boundary condition of a constant temperature is difficult to
prescribe on the curved surface of sphere in the rectangular cartesian
system. In the case that the problem formulation suppose a rotational or
spherical symmetry, the number of spatial coordinates can be reduced
and the problem is simplified to 1D or 2D problem, but in a new
coordinate system.
In the following we shall demonstrate how to convert tensor terms from
symbolic notation (which is independent to a specific coordinate system)
into the cylindrical coordinate system.
Coordinate systems (cylindrical)
MHMT1
Cylindrical (and spherical) systems are defined by transformations
dxi 
x2
(x1 x2 x3)
r
(r,,z)

where
x1
s  sin 
xi
x
x
dr  i d  i dz
r

z
x1  rc  dx1   c rs 0  dr 

 
 
x2  rs  dx2    s rc 0  d 

 
 
x3  z  dx3   0 0 1  dz 
c  cos 
 c
 dr  
   s
 d     r
 dz 
   0

s 0
  dx1 
c


0   dx2 
r

dx3 


0 1
r
r

s  c
 c,
 s,
 ,

x1
x2
x1
r x2 r
Using this it is possible to express partial derivatives with respect x1,x2,x3 in
terms of derivatives with respect the coordinates of cylindrical system
T T r T  T z T
T s




c
x1 r x1  x1 z x1 r
 r
T T r T  T z T
T c




s
x2 r x2  x2 z x2 r
 r
T T r T  T z T




x3 r x3  x3 z x3 z
MHMT1
Coordinate systems (cylindrical)
In the same way also the second derivatives can be expressed
 2T  2T 2 2cs  T T T s 2  2T s 2

c 
(  )

x12 r 2
r  r r
r r  2 r 2
 2T  2T 2 2cs  T T T c 2  2T c 2

s 
(  )

x22 r 2
r  r r
r r  2 r 2
 2T  2T

x32 z 2
giving expression for the Laplace operator in the cylindrical coordinate system
(use goniometric identity s2+c2=1)
 2T  2T  2T  2T T 1  2T 1  2T
 2  2  2 
 2 2 2
2
x1 x2 x3 r
r r  r
z
MHMT1
Coordinate systems (cylindrical)
Previous example demonstrated how to solve the problem of transformations to
cylindrical coordinate system with scalars. However, how to calculate gradients
or divergence of a vector and a tensor field? Vectors and tensors are described
by 3 or 3x3 values of components now expressed in terms of new unit vectors
u  umim  ur e ( r )  u e ( )  uz e ( z )
Einstein summation applied in
the cartesian coordinate system
unit vectors in the cylindrical
coordinate system
   mn im in 
  rr e ( r ) e ( r )   r e ( r ) e ( )   rz e ( r ) e ( z )  ...   zz e ( z ) e ( z )
Unit vectors in orthogonal coordinate system are orthogonal, which means that
e ( r ) e ( r )  1 e ( r ) e ( )  0
... e ( z ) e ( z )  1
and therefore the components in the new coordinate system are for example
u  u e ( )  umem( )
ui  ur ei ( r )  u ei ( )  uz ei ( z )
 rz  e
(r)
 e


ei( z )  e ( z )  ii
( z)
 em( r ) mnen( z )
this is i-th cartesian component
of the unit vector
Coordinate systems (cylindrical)
MHMT1
Transformation of unit vectors
x2
(x1 x2 x3)
i2
e ( )
r
e (r )
(r,,z)

i1
x1
 e ( r )   c s 0   i1 
 ( )  
 
e


s
c
0

 
  i2 
 e ( z )   0 0 1   i3 
 

 
 i1   c  s 0   e ( r ) 
  
  ( ) 
i

s
c
0
 2 
e 
 i   0 0 1   e (z) 

 3 

Example: gradient of temperature can be written in the following way
(alternatively in
the cartesian and the cylindrical coordinate system)
substitute by using
previously derived
T 
T T
T s

c
x1 r
 r
T
T
T
T
T ( r )
T
T ( ) T ( z )
i1 
i2 
i3  (c
s
)e  ( s
c
)e 
e 
x1
x2
x3
x1
x2
x1
x2
x3
T ( r ) 1 T ( ) T ( z )

e 
e 
e
r
r 
z
Nabla operator in the
cylindrical coordinate system
e
(r )

( ) 1 
(z) 
e
e
r
r 
z
MHMT1
Coordinate systems (cylindrical)
Example: Divergence of a vector u  umim  ur e ( r )  u e ( )  uz e ( z )
(r )
( )
( z)
um  (ur em  u em  u z em )
u


xm
xm
em( r ) u ( )
em( ) u z ( z )
em( z )
ur ( r )

em  ur

em  u

em  u z
xm
xm xm
xm xm
xm
components of unit vectors follow from the previously derived
e1( r )  c
e1( )   s
e1( z )  0
e2( r )  s
e2( )  c
e2( z )  0
e3( r )  0
e3( )  0
e3( z )  1
 e ( r )   c s 0   i1 
 ( )  
 
 e     s c 0   i2 
 e ( z )   0 0 1   i3 
 

 
Note the fact, that the partial derivatives of these components with respect to xm
are not zero and can be calculated using the previously derived relationships

s  c
 ,

x1
r x2 r
giving
e1( r ) c c 
s


 s
x1 x1  x1
r
e1( )
s
s 
s


c
x1
x1
 x1
r
…and so on
MHMT1
Coordinate systems (cylindrical)
Substituting these expression we obtain
u s 2
u1 ur 2 ur cs
s 2 u
cs

c 
 ur 
cs 
 u
x1 r
 r
r
r
 r
r
u c 2
u2 ur 2 ur cs
c 2 u
cs

s 
 ur 
cs 
 u
x2
r
 r
r
r
 r
r
u3 u z

x3
z
Summing together, the final form of divergence in the cylindrical coordinate
system is obtained
 ur ur 1 u u z
u 
 

r
r r  z
MHMT1
Coordinate systems (cylindrical)
Example: Gradient of a vector
u  umim  ur e ( r )  u e ( )  uz e ( z )
  u
 rr  em( r )
 (ur en( r )  u en( )  u z en( z ) )
xm
en( r )
 r  em( r )
 (ur en( r )  u en( )  u z en( z ) )
xm
en( ) ...
m and n are dummy indices (summing is required)
Substituting previous expressions for unit vector and their derivatives results
to final expression for the velocity gradient tensor in a cylindrical coordinate
system
u

ur
u z 


r
r
r 

1 u
1 u z 
  1 ur
u   (
 u )
(
 ur )
r 
r 
r  



u
ur
u z 




z
z
z 

MHMT1
Coordinate systems (general)
Procedure how to derive tensorial equations in a general
coordinate system.
1. Rewrite equation from the symbolic notation to the index notation for cartesian coordinate
u j
system, for example
u   
xi
  ij
r
i
 f ij (r1 , r2 , r3 ) therefore also the first and
2. Define transformation x1(r1,r2,r3),… r1(x1,x2,x3),… and
x j
the second derivatives of scalar values, for example
u j
x1

u j
r1
f11 
u j
r2
f 21 
u j
r3
f31 ,...
( ri )
( ri )
3. Define unit vectors of coordinate systems ri and express their cartesian coordinates em  g m (r1 , r2 , r3 )
Calculate their derivatives with respect to the cartesian coordinates
em( r ) g mr g mr r1 g mr r2 g mr r3




xi
x1
r1 xi r2 xi r3 xi
4. In case that the result is a vector, for example the gradient of scalar, calculate its components from
u  u e ( )  umem( )
In the case that result is a tensor calculate its components
 rz  e
(r)
 e
(z)
 ei( r ) ij e(j z )
Remark: This suggested procedure (transform everything, including new unit vectors, to the cartesian coordinate system) is
straightforward and seemingly easy. This is not so, it is „crude“, lenghty (the derivation of velocity gradient is on several lists of
paper) and without finesses. Better and more sophisticated procedures are described in standard books, e.g. Aris R: Vectors,
tensors…N.J.1962, or Bird,Stewart,Lightfoot:Transport phenomena.
MHMT1
EXAM
Tensors
MHMT1
What is important (at least for exam)
You should know what is it scalar, vector, tensor and transformations at
rotation of coordinate system
[a ']  [[R]][a ]
[[ ']]  [[R]][[ ]][[R]]T
(and how is defined the rotation matrix R?)
Scalar and vector products
3
a  b   ai bi  ai bi
i 1
 

  

c  a  b  (  b )  a
3
3
ci    ijk a j bk   ijk a j bk
j 1 k 1
(and what is it Kronecker delta and Levi Civita tensors?)
MHMT1
What is important (at least for exam)
Nabla operator. Gradient
(
  
, , )
x y x
i 

xi
Divergence
3
 u x u y u z
u u
u 


 i  i
x
y
z
xi
i 1 xi
Laplace operator
 2T  2T  2T
 2T
  T   T  2  2  2 
x
y
z
xi xi
2
MHMT1
What is important (at least for exam)
Gauss integral theorem
   Pdv   n  Pds
V
S
(demonstrate for the case that P is scalar, vector, tensor)