290
26 Eigen Stresses and Dislocations
This supplies us with equations familiar to us from the relativistic field theories
with one single signal velocity, here the transversal speed of sound c T . This has farreaching consequences that we will now illustrate using our above, uniform moving
straight screw dislocation, by explicitly calculating the field of this dislocation. In
order to do this,we insert (392) and (394) into (397) and (398). After elementary
calculations,we find the following field equation, the
always signifies the derivative
of the immediately following variables in the brackets,
∂
2
∂x 2 +
∂
2
∂ y 2 −
1
c
2
T
∂
2
∂t 2
ε ik =
=−
b
2
⎛
⎜
⎜
⎜
⎜
⎝
0
0
δ(x −V t) δ(y)
0
0
−
1−
V
2
c
2
T
δ
(x −V t) δ(y)
δ(x −V t) δ
(y) −
1−
V
2
c
2
T
δ
(x −V t) δ(y)
0
⎞
⎟
⎟
⎟
⎟
⎠
(399)
and
∂
2
∂x 2 +
∂
2
∂ y 2 −
1
c
2
T
∂
2
∂t 2
v i = V b
⎛
⎝
0
0
δ(x − V t) δ
(y)
⎞
⎠ .
(400)
This is mathematically seen, due to the fact that all derivatives with respect to z
disappear, a plane problem. For the mathematical discussion of these strict relativistic
field equations,we refer the reader to Günther [34].
The only solution of equations (399) and (400) using natural boundary conditions
is,
ε 13 = ε 31 = −
b
4π
1
1 − V 2 /c
2
T
∂
∂ y
ln
(x − V t) 2 +
1 −
V 2
c
2
T
y 2 ,
ε 23 = ε 32 = −
b
4π
1
1 − V 2 /c
2
T
∂
∂x
ln
(x − V t) 2 +
1 −
V 2
c
2
T
y 2 ,
v 3
= −
V b
2π
1
1 − V 2 /c
2
T
∂
∂ y
ln
(x − V t) 2 +
1 −
V 2
c
2
T
y 2 .
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(401)
All other components of ε and v are zero. Equation (401) is therefore the elastic
deformation field, which accompanies the plastic deformation, which in turn creates
a straight screw dislocation moving with a constant velocity through the crystal.
One can calculate the elastic energy E that such a dislocation moving through the
medium with the velocity V causes.
2 If E o is the energy of the stationary dislocation,
then one discovers the relationship E = E o /
1 − v 2 /c
2
T . Such a field possesses,
2 The question concerning the ’cutoff radius’ will not answered here.
26 Eigen Stresses and Dislocations
This supplies us with equations familiar to us from the relativistic field theories
with one single signal velocity, here the transversal speed of sound c T . This has farreaching consequences that we will now illustrate using our above, uniform moving
straight screw dislocation, by explicitly calculating the field of this dislocation. In
order to do this,we insert (392) and (394) into (397) and (398). After elementary
calculations,we find the following field equation, the
always signifies the derivative
of the immediately following variables in the brackets,
∂
2
∂x 2 +
∂
2
∂ y 2 −
1
c
2
T
∂
2
∂t 2
ε ik =
=−
b
2
⎛
⎜
⎜
⎜
⎜
⎝
0
0
δ(x −V t) δ(y)
0
0
−
1−
V
2
c
2
T
δ
(x −V t) δ(y)
δ(x −V t) δ
(y) −
1−
V
2
c
2
T
δ
(x −V t) δ(y)
0
⎞
⎟
⎟
⎟
⎟
⎠
(399)
and
∂
2
∂x 2 +
∂
2
∂ y 2 −
1
c
2
T
∂
2
∂t 2
v i = V b
⎛
⎝
0
0
δ(x − V t) δ
(y)
⎞
⎠ .
(400)
This is mathematically seen, due to the fact that all derivatives with respect to z
disappear, a plane problem. For the mathematical discussion of these strict relativistic
field equations,we refer the reader to Günther [34].
The only solution of equations (399) and (400) using natural boundary conditions
is,
ε 13 = ε 31 = −
b
4π
1
1 − V 2 /c
2
T
∂
∂ y
ln
(x − V t) 2 +
1 −
V 2
c
2
T
y 2 ,
ε 23 = ε 32 = −
b
4π
1
1 − V 2 /c
2
T
∂
∂x
ln
(x − V t) 2 +
1 −
V 2
c
2
T
y 2 ,
v 3
= −
V b
2π
1
1 − V 2 /c
2
T
∂
∂ y
ln
(x − V t) 2 +
1 −
V 2
c
2
T
y 2 .
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(401)
All other components of ε and v are zero. Equation (401) is therefore the elastic
deformation field, which accompanies the plastic deformation, which in turn creates
a straight screw dislocation moving with a constant velocity through the crystal.
One can calculate the elastic energy E that such a dislocation moving through the
medium with the velocity V causes.
2 If E o is the energy of the stationary dislocation,
then one discovers the relationship E = E o /
1 − v 2 /c
2
T . Such a field possesses,
2 The question concerning the ’cutoff radius’ will not answered here.