| Topic: |
Science > Physics |
| User: |
"LC" |
| Date: |
26 Sep 2007 12:50:43 PM |
| Object: |
Definition of stress tensor in continuum mech |
I recently began studying continuum mechanics, and I'm having a bit of
trouble trying to find a mathematically precise definition of the
stress tensor. Most sources define it as some sort of nonsensical
limit like
lim_(A -> 0) \delta(F) / \delta(A)
I'm not sure how to interpret this. By definition, force seems to be a
vector valued three form:
F = \rho(x,t) a_i(x,t) dx^dy^dz
where \rho is the density at (x,t) and the vector a is the
acceleration of the particle located at (x,t). One way to interpret
the limit, I suppose, would be to require that F be expressed as the
exterior derivative of some vector valued 2-form
T = T_i1 dy^dz + T_i2 dx^dz + T_i3 dx^dy
Indeed, the equations of motion tell us precisely that the force is
equal to the divergence of the stress tensor T_ij, which is a vector-
valued 2-form.
However, this definition also seems problematic to me, because the
tensor T_ij that results is not unique. In particular, we could add
to it any divergenceless tensor and still obtain the same equations of
motion (i.e., the same force law). In other words, the definition of
T_ij involves a choice of gauge. This seems irreconcilable with the
notion of defining it once and for all through the sort of limiting
process described above.
Can someone either give me a precise mathematical account of this
discrepancy, point out where my analysis went wrong, or explain the
physical significance of the gauge? Are there some implicit
assumptions that I'm unaware of that allow us to single out a
particular choice of gauge?
Thanks
LC
.
|
|
| User: "Andy Resnick" |
|
| Title: Re: Definition of stress tensor in continuum mech |
26 Sep 2007 04:46:18 PM |
|
|
LC wrote:
I recently began studying continuum mechanics, and I'm having a bit of
trouble trying to find a mathematically precise definition of the
stress tensor.
<snip>
Can someone either give me a precise mathematical account of this
discrepancy, point out where my analysis went wrong, or explain the
physical significance of the gauge? Are there some implicit
assumptions that I'm unaware of that allow us to single out a
particular choice of gauge?
That's a different approach than what I learned; maybe I can help.
The stress tensor is usually not presented as a 'generating concept',
it's usually derived, either from the stress-strain relationship:
stress = Elastic modulus * strain
Or from a more general consideration of force and torque balance. The
strain tensor is a well-defined quantity, and the elastic modulus is
(most generally) a 4th-rank tensor.
Stress is used to describe "contact loads". These are forces acting
upon bounding surfaces, lines, or points. Put this way, the fundamental
concept is the stress vector, which is defined as the contact force per
unit area exerted upon a body at its bounding surface. This vector is
not an ordinary vector field, because stress vectors across two
different surfaces through the same point are in general different.
The stress tensor is then constructed (following Cauchy) by considering
the stress vectors acting on a tetrahedron, which leads to Cauchy's
fundamental theorem: the stress vectors are linear functions of the
stress tensor.
There are several decent derivations of all this. Landau and Lifshitz's
"Theory of Elasticity" is rather brief, Truesdell's "Classical Field
Theories" is comprehensive and dense, Marsden and Hughes' "mathematical
foundations of elasticity" is full of insight. There are others, no doubt.
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
.
|
|
|
| User: "LC" |
|
| Title: Re: Definition of stress tensor in continuum mech |
26 Sep 2007 04:26:23 PM |
|
|
Thanks for the response. I'm still not quite satisfied, and I'll
explain why below.
First of all, some notation. I'd rather refer to the stress vector as
the stress tensor, because it has 9 components, not 3, and as you say,
it depends on the normal vector to the surface element in question.
I'm going to write it as T_ij, where for example T_xy is the y-
component of the stress vector acting on the yz plane. I'll also
write div_j T_ij for the tensor d/dx_j T_ij (summation convention
used).
Stress is used to describe "contact loads". These are forces acting
upon bounding surfaces, lines, or points. Put this way, the fundamental
concept is the stress vector, which is defined as the contact force per
unit area exerted upon a body at its bounding surface. This vector is
not an ordinary vector field, because stress vectors across two
different surfaces through the same point are in general different.
Here lies the problem. It comes from asking the question:
What is meant by the statement "the stress vector is defined as
contact force per unit area"?
The only interpretation that I can see is that T_ij is a tensor with
the property that when you integrate it over the surface bounding a
body of material, you get the total force acting on that body. At
least, this is the interpretation most textbooks (including Truesdell,
for example) seem to give. By the divergence theorem, this
interpretation is equivalent to saying that the vector div_j T_ij is
equal to the force per volume acting on the material.
My concern is that T_ij is not uniquely determined by this property,
because we can always replace it with a tensor of the form T_ij +
S_ij, where div S_ij = 0.
As an analogy, consider a particle acted upon by a conservative force
F. The fact that F is conservative means that there exists a scalar U
such that F = \grad U. We might be tempted to refer to this as "the"
potential for the force F. However, we should really refer to it as
"a" potential, because it is not unique - if we set U' = U + c for
some constant c, when we still have $F = \grad U'$. So my question
is, in some sense, why do people talk about "the" stress tensor,
rather than "a" stress tensor?
-LC
.
|
|
|
| User: "Andy Resnick" |
|
| Title: Re: Definition of stress tensor in continuum mech |
27 Sep 2007 09:10:49 AM |
|
|
LC wrote:
Thanks for the response. I'm still not quite satisfied, and I'll
explain why below.
First of all, some notation. I'd rather refer to the stress vector as
the stress tensor, because it has 9 components, not 3, and as you say,
it depends on the normal vector to the surface element in question.
<snip>
Hang on- I was being quite specific when I used "vector". I postulate
the existence of a *vector* field s which depends on the time 't', the
spatial point 'x', and a unit vector 'n'. Physically, s(t,x,n)
represents the force per unit area exerted on a surface element oriented
with normal n. s is called the Cauchy stress vector, not the stress
tensor.
It is important to note that the concept of the stress vector, which is
some kind of force density, is fundamentally what sets continuum
mechanics apart from particle mechanics.
Given that there is a 'balance of momentum' (which involves integrating
s over a surface), and that s is a continuous function, there is a
unique vector field T which does not depend on a unit vector 'n':
s(t,x,n) = <T(t,x),n>.
T is the Cauchy stress tensor.
<snip>
The only interpretation that I can see is that T_ij is a tensor with
the property that when you integrate it over the surface bounding a
body of material, you get the total force acting on that body. At
least, this is the interpretation most textbooks (including Truesdell,
for example) seem to give. By the divergence theorem, this
interpretation is equivalent to saying that the vector div_j T_ij is
equal to the force per volume acting on the material.
My concern is that T_ij is not uniquely determined by this property,
because we can always replace it with a tensor of the form T_ij +
S_ij, where div S_ij = 0.
Ah. I see what you are asking. I'm not sure how best to answer that
question, but I think it's related to the other stress tensors (see below)
<snip>
So my question
is, in some sense, why do people talk about "the" stress tensor,
rather than "a" stress tensor?
There are other stress tensors: the first and second Piola-Kirchoff
stress tensors, the convected and co-rotational stress tensors, and
another one used for Cosserat materials.
My (very limited) understanding is that the Cauchy tensor (or rather
<T,n>) is the force per unit *deformed* area. The Piola tensors are
used for *undeformed* areas, but beyond that I'm not clear on the details.
Check out Marsden and Hughes. It's a Dover book, so it's a no-brainer
to have around.
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
.
|
|
|
| User: "LC" |
|
| Title: Re: Definition of stress tensor in continuum mech |
27 Sep 2007 03:10:45 PM |
|
|
On Sep 27, 10:10 am, Andy Resnick <andy.resn...@op.case.edu> wrote:
Hang on- I was being quite specific when I used "vector". I postulate
the existence of a *vector* field s which depends on the time 't', the
spatial point 'x', and a unit vector 'n'. Physically, s(t,x,n)
represents the force per unit area exerted on a surface element oriented
with normal n. s is called the Cauchy stress vector, not the stress
tensor.
My fault. I apologize for the misunderstanding.
There are other stress tensors: the first and second Piola-Kirchoff
stress tensors, the convected and co-rotational stress tensors, and
another one used for Cosserat materials.
My (very limited) understanding is that the Cauchy tensor (or rather
<T,n>) is the force per unit *deformed* area. The Piola tensors are
used for *undeformed* areas, but beyond that I'm not clear on the details.
Check out Marsden and Hughes. It's a Dover book, so it's a no-brainer
to have around.
Hm. I'm not sure these are related to the problem I'm pointing out.
It seems that they all have to do with looking at the cauchy stress
tensor from different frames of reference. As you say, the Cauchy
tensor is what you get when you look at it from the moving frame, and
the Piola tensor is what you get when you look at it from the
reference frame, etc. I'd have to learn more about them, but it seems
that they would all suffer from the same ambiguity, although in a
different frame the ambiguity might look different, e.g. you wouldn't
be adding a divergenceless tensor, you'd be adding a tensor with some
other property.
Incidentally, when I was googling this yesterday I came across
Feynman's bachelor thesis (!):
http://dspace.mit.edu/handle/1721.1/10786
He's talking about a stress tensor in a very different context (forces
in an atom, rather than a continuous material), but he notes the same
ambiguity (on p. 19). His conclusion regarding these multiple stress
tensors is that "there is no reason to choose one rather than the
other" (p 22).
By the way, another cause for concern here is that there are
asymmetric divergenceless tensors, and therefore the property of a
material having a symmetric stress tensor is not well defined! In
fact, any arbitrary tensor can be made into a symmetric (even
diagonal) tensor by adding a divergenceless tensor (!!). To see why,
consider an arbitrary tensor T_ij. Define a vector V_i by
V_i = div T_ij
Integrating with respect to x_i, we can then find functions W_i such
that:
d/d_{x_i} ( W_i ) = V_i
for all i = 1,2,3. Now consider the diagonal tensor:
S_ij = 0 for i not equal to j
S_ii = W_i for all i
By definition, div(S_ij) = V_i = div(T_ij). Thus div(S_ij - T_ij) =
0. But this shows that we can turn T_ij into a **diagonal** tensor by
adding the divergenceless tensor R_ij = S_ij - T_ij.
So at this point I'm totally and thoroughly confused. Shear stress
doesn't exist? Something must have gone wrong, right?
-LC
.
|
|
|
| User: "Andy Resnick" |
|
| Title: Re: Definition of stress tensor in continuum mech |
01 Oct 2007 03:06:20 PM |
|
|
LC wrote:
Sorry for the delay, our news server went down last week... Anyhow, I
wrote this just as it barfed:
<snip>
He's talking about a stress tensor in a very different context (forces
in an atom, rather than a continuous material), but he notes the same
ambiguity (on p. 19). His conclusion regarding these multiple stress
tensors is that "there is no reason to choose one rather than the
other" (p 22).
By the way, another cause for concern here is that there are
asymmetric divergenceless tensors, and therefore the property of a
material having a symmetric stress tensor is not well defined! In
fact, any arbitrary tensor can be made into a symmetric (even
diagonal) tensor by adding a divergenceless tensor (!!). To see why,
consider an arbitrary tensor T_ij. Define a vector V_i by
<snip>
Several of us had a discussion about asymmetric stress tensors a few
months ago- I'm not sure we got every issue resolved, but we did make
some progress.
If the (Cauchy) stress tensor is symmetric, that means that both
momentum and moment of momentum (torque) are balanced. There's also a
requirement for no "assigned couples and couple-stresses", which Edward
Green and Mati Meron understand better than I.
In any case, the stress tensor is not a material property any more than
pressure is a property of water.
But I do now understand your question about "gauge invariance" of the
stress tensor. A divergence-free stress tensor corresponds to zero
force, so maybe it doesn't matter. I think your question involves
possible coordinate rotations to the principal axes of stress. The
scalar invariants of stress are symmetric functions of the principal
stresses.
Section 46 in the Appendix of "Classical Field Theories" ("Tensor
fields") has something to say about this point, but I haven't had the
time to really go through it (yet...).. Subsection 'c' is entitled
'Decompositions', so maybe I'll learn something there.
(n.b.- I had a chance to skim it, and yes, it seems that decomposing a
stress tensor as you do above simply expresses the total field as a sum
of normal and shear components.)
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
.
|
|
|
|
|
|
| User: "Edward Green" |
|
| Title: Re: Definition of stress tensor in continuum mech |
01 Oct 2007 08:10:18 PM |
|
|
On Sep 26, 5:26 pm, LC <luc...@gmail.com> wrote:
<...>
By the divergence theorem, this
interpretation is equivalent to saying that the vector div_j T_ij is
equal to the force per volume acting on the material.
BTW, to preserve this property, we have to include all momentum fluxes
in the stress tensor: e.g., if x-rays are absorbed inside a body then
we must include the flux of x-rays across any bounding surface in the
stress tensor if we want to preserve your claimed relation. This leads
to the "stress-energy tensor".
.
|
|
|
|
| User: "Edward Green" |
|
| Title: Re: Definition of stress tensor in continuum mech |
01 Oct 2007 08:02:19 PM |
|
|
On Sep 26, 5:26 pm, LC <luc...@gmail.com> wrote:
<...>
The only interpretation that I can see is that T_ij is a tensor with
the property that when you integrate it over the surface bounding a
body of material, you get the total force acting on that body. At
least, this is the interpretation most textbooks (including Truesdell,
for example) seem to give. By the divergence theorem, this
interpretation is equivalent to saying that the vector div_j T_ij is
equal to the force per volume acting on the material.
My concern is that T_ij is not uniquely determined by this property,
because we can always replace it with a tensor of the form T_ij +
S_ij, where div S_ij = 0.
The stress tensor may not be uniquely determined by an interpretation
of its divergence, but that is not its only property. It describes
forces acting across any surface, which in principle can be measured.
We are not free to add aribtrary increments, divergence free or not.
The divergence free part of the stress tensor is what we normally
think of as static "stress". If all static stresses were equivalent,
there would be no difference between siting under 1 atm of pressure,
and sitting under 10,000 atms. Let's try that out. :-)
As an analogy, consider a particle acted upon by a conservative force
F. The fact that F is conservative means that there exists a scalar U
such that F = \grad U. We might be tempted to refer to this as "the"
potential for the force F. However, we should really refer to it as
"a" potential, because it is not unique - if we set U' = U + c for
some constant c, when we still have $F = \grad U'$. So my question
is, in some sense, why do people talk about "the" stress tensor,
rather than "a" stress tensor?
To pursue the analogy, we need a situation where the "arbitrary" part
of the potential has some physical interpretation: say, in GR, where
it determines (what the cognoscenti would shudder to call) the local
rate of time.
Sometimes it helps to be an idiot.
.
|
|
|
|
| User: "GSS" |
|
| Title: Re: Definition of stress tensor in continuum mech |
29 Sep 2007 11:03:03 PM |
|
|
On Sep 27, 2:26 am, LC <luc...@gmail.com> wrote:
.........
Stress is used to describe "contact loads". These are forces acting
upon bounding surfaces, lines, or points. Put this way, the fundamental
concept is the stress vector, which is defined as the contact force per
unit area exerted upon a body at its bounding surface. This vector is
not an ordinary vector field, because stress vectors across two
different surfaces through the same point are in general different.
Here lies the problem. It comes from asking the question:
What is meant by the statement "the stress vector is defined as
contact force per unit area"?
Stress by definition is a force per unit area.
A stress tensor with nine components is defined at every point P of
the continuum and will be a function of space coordinates of P and
time.
At point P consider a unit area with specified normal n. On this unit
area there there will be two shear stress components and one normal
stress component acting. Vector sum of these two shear forces and one
normal force acting on this unit area will define the stress vector
which is thus dependent on the stress tensor components at P as well
as the orientation of the unit normal n.
The only interpretation that I can see is that T_ij is a tensor with
the property that when you integrate it over the surface bounding a
body of material, you get the total force acting on that body. At
least, this is the interpretation most textbooks (including Truesdell,
for example) seem to give. By the divergence theorem, this
interpretation is equivalent to saying that the vector div_j T_ij is
equal to the force per volume acting on the material.
My concern is that T_ij is not uniquely determined by this property,
because we can always replace it with a tensor of the form T_ij +
S_ij, where div S_ij = 0.
Agreed, the stress tensor at P is not uniquely determined from the
divergence theorem alone. Incidentally your divergence theorem
corresponds to the equilibrium equations of elasticity in the
continuum. The stress tensor components are uniquely determined only
when either the distribution of forces or the displacements are
specified at the bounding surface of the region under consideration as
per the standard boundary value problem of elasticity.
For a little better illustration of the representation of the stress
tensor in an elastic continuum, kindly refer to,
http://www.geocities.com/gurcharn_sandhu/pdf_art/stress_tensor.pdf
Here I am using the term 'elastic continuum' for the continuum merely
to distinguish it from a rigid, non-deformable continuum. That is to
highlight the deformability characteristics of the continuum.
In this regard, kindly let me know whether the spacetime continuum as
modeled in GR is assumed to be a rigid or a deformable continuum?
GSS
.
|
|
|
| User: "GSS" |
|
| Title: Re: Definition of stress tensor in continuum mech |
02 Oct 2007 06:21:45 AM |
|
|
On Sep 30, 9:03 am, GSS <gurcharn_san...@yahoo.com> wrote:
On Sep 27, 2:26 am, LC <luc...@gmail.com> wrote:
........> > Stress is used to describe "contact loads". These are forces acting
upon bounding surfaces, lines, or points. Put this way, the fundamental
concept is the stress vector, which is defined as the contact force per
unit area exerted upon a body at its bounding surface. This vector is
not an ordinary vector field, because stress vectors across two
different surfaces through the same point are in general different.
Here lies the problem. It comes from asking the question:
What is meant by the statement "the stress vector is defined as
contact force per unit area"?
Stress by definition is a force per unit area.
A stress tensor with nine components is defined at every point P of
the continuum and will be a function of space coordinates of P and
time.
At point P consider a unit area with specified normal n. On this unit
area there there will be two shear stress components and one normal
stress component acting. Vector sum of these two shear forces and one
normal force acting on this unit area will define the stress vector
which is thus dependent on the stress tensor components at P as well
as the orientation of the unit normal n.
The only interpretation that I can see is that T_ij is a tensor with
the property that when you integrate it over the surface bounding a
body of material, you get the total force acting on that body. At
least, this is the interpretation most textbooks (including Truesdell,
for example) seem to give. By the divergence theorem, this
interpretation is equivalent to saying that the vector div_j T_ij is
equal to the force per volume acting on the material.
My concern is that T_ij is not uniquely determined by this property,
because we can always replace it with a tensor of the form T_ij +
S_ij, where div S_ij = 0.
Agreed, the stress tensor at P is not uniquely determined from the
divergence theorem alone. Incidentally your divergence theorem
corresponds to the equilibrium equations of elasticity in the
continuum. The stress tensor components are uniquely determined only
when either the distribution of forces or the displacements are
specified at the bounding surface of the region under consideration as
per the standard boundary value problem of elasticity.
For a little better illustration of the representation of the stress
tensor in an elastic continuum, kindly refer to,http://www.geocities.com/gurcharn_sandhu/pdf_art/stress_tensor.pdf
Here I am using the term 'elastic continuum' for the continuum merely
to distinguish it from a rigid, non-deformable continuum. That is to
highlight the deformability characteristics of the continuum.
In this regard, kindly let me know whether the spacetime continuum as
modeled in GR is assumed to be a rigid or a deformable continuum?
GSS
I invite all readers to give their opinion, if any, whether the
spacetime continuum as modeled in GR is assumed to be a rigid or a
deformable continuum?
If it is assumed to be a deformable continuum, then how come that the
strained state of this continuum has never been examined in terms of
associated strain tensors?
GSS
.
|
|
|
| User: "Edward Green" |
|
| Title: Re: Definition of stress tensor in continuum mech |
02 Oct 2007 04:06:18 PM |
|
|
On Oct 2, 7:21 am, GSS <gurcharn_san...@yahoo.com> wrote:
I invite all readers to give their opinion, if any, whether the
spacetime continuum as modeled in GR is assumed to be a rigid or a
deformable continuum?
If it is assumed to be a deformable continuum, then how come that the
strained state of this continuum has never been examined in terms of
associated strain tensors?
No answers, but a few platitudes:
Clearly "opinion" is not worth to much here: what is needed is some
definition of what it means for a continuum to be deformable or rigid.
Opinion would enter more into a choice of what definition might be a
reasonable one to adopt: there may be more than one.
Given such a definition, then yours is a technical question, not a
matter of opinion.
In terms of what might be a reasonable definition, we might wonder
whether we want to ask the question of a four dimensional something,
or a three dimensional something. There may be no natural definition
of universal time, but I'm not convinced there might not be a natural
idea of a three dimensional continuum which does its thing locally,
and dynamically, and doesn't give a figo whether or not we want to
search for a universal t coordinate.
Secondly, even given a sense of "deformation", by asking for an
accompanying "stress" you are taking things a notch further. Stress
and strain go together in ordinary material: it's plausible that if
there is something like strain in space then there is also something
like stress also, and something like a constitutive relation, but
merely plausible (whether the whole analogical package can be put
through might be a useful test of "reasonableness" in the
identification).
All these something-likes are things to be searched for in the math of
General Relativity, or in the class of maths which are sufficiently
something-like General Relativity to reproduce its success. I guess we
may offer our "opinion" about whether this might be a worthwhile
search. I think it would.
.
|
|
|
| User: "GSS" |
|
| Title: Re: Definition of stress tensor in continuum mech |
03 Oct 2007 06:23:37 AM |
|
|
On Oct 3, 2:06 am, Edward Green <spamspamsp...@netzero.com> wrote:
On Oct 2, 7:21 am, GSS <gurcharn_san...@yahoo.com> wrote:
I invite all readers to give their opinion, if any, whether the
spacetime continuum as modeled in GR is assumed to be a rigid or a
deformable continuum?
If it is assumed to be a deformable continuum, then how come that the
strained state of this continuum has never been examined in terms of
associated strain tensors?
No answers, but a few platitudes:
Clearly "opinion" is not worth to much here: what is needed is some
definition of what it means for a continuum to be deformable or rigid.
Opinion would enter more into a choice of what definition might be a
reasonable one to adopt: there may be more than one.
Let me attempt to distinguish between a rigid and a deformable
continuum of points. Let P be *any* point in this continuum. Let P1,
P2, P3, .... Pn be n points in the neighborhood of P. Let ds_1 be the
separation distance between points P and P1, ds_2 be the separation
distance between points P and P2, ..., ds_n be the separation distance
between points P and Pn. If these separation distances ds_1, ds_2,
ds_3,...., ds_n from point P to all of its neighborhood points, remain
constant and invariant with time under all circumstances, then the
continuum under consideration can be regarded as rigid. If under
certain circumstances, these separation distances change to say ds'_1,
ds'_2, ds'_3,...., ds'_n then the continuum under consideration can
be regarded as deformable.
Since the separation distance dS between two neighborhood points of
the spacetime continuum does change under the influence of
gravitational field (as per GR), obviously the spacetime continuum is
assumed to be deformable in GR.
Given such a definition, then yours is a technical question, not a
matter of opinion.
In terms of what might be a reasonable definition, we might wonder
whether we want to ask the question of a four dimensional something,
or a three dimensional something. There may be no natural definition
of universal time, but I'm not convinced there might not be a natural
idea of a three dimensional continuum which does its thing locally,
and dynamically, and doesn't give a figo whether or not we want to
search for a universal t coordinate.
Yes, instead of starting with a 4-D spacetime continuum, it is quite
appropriate to consider its subset, the 3-D space continuum, as
modeled in GR, to examine its deformation characteristics.
Secondly, even given a sense of "deformation", by asking for an
accompanying "stress" you are taking things a notch further. Stress
and strain go together in ordinary material: it's plausible that if
there is something like strain in space then there is also something
like stress also, and something like a constitutive relation, but
merely plausible (whether the whole analogical package can be put
through might be a useful test of "reasonableness" in the
identification).
No, I didn't ask for an accompanying "stress". I only mentioned strain
tensors associated with deformation of the continuum. We could take
the things a notch further at a later stage.
All these something-likes are things to be searched for in the math of
General Relativity, or in the class of maths which are sufficiently
something-like General Relativity to reproduce its success. I guess we
may offer our "opinion" about whether this might be a worthwhile
search. I think it would.
In this regard, you may kindly examine the article on the strained
state of a continuum at,
http://www.geocities.com/gurcharn_sandhu/pdf_art/continuum_strain.pdf
Last paragraph of this article is reproduced below for information.
"As shown above, the notion of deformed or strained state of the
continuum under study is derived from the variability or invariance of
arc element ds. Whenever the arc element ds changes over to ds' under
certain situations, the changed state of the continuum will be termed
the deformed or strained state. The strained state can be considered
fully defined or fully determined once we know or uniquely determine
the displacement vector field at all points of the continuum. The
strained state can also be defined through specification of strain
tensor components provided these components satisfy Saint Venant's
compatibility equations. Finally the strained state can also be
defined through specification of modified metric coefficients from
which the required strain tensor components can be computed subject to
the compatibility conditions. However, the compatibility conditions
require that the modified metric must be Euclidean to ensure that the
resulting strained state of the continuum corresponds to smooth,
finite and continuous displacement components and to avoid
discontinuities within the continuum. This fact is of crucial
importance for examining the validity of the current mathematical
model of General Relativity."
Further the validity of the current mathematical model of GR has also
been examined at,
http://www.geocities.com/gurcharn_sandhu/pdf_art/invalidity_gr.pdf
GSS
.
|
|
|
| User: "Edward Green" |
|
| Title: Re: Definition of stress tensor in continuum mech |
04 Oct 2007 09:13:04 PM |
|
|
On Oct 3, 7:23 am, GSS <gurcharn_san...@yahoo.com> wrote:
On Oct 3, 2:06 am, Edward Green <spamspamsp...@netzero.com> wrote:
On Oct 2, 7:21 am, GSS <gurcharn_san...@yahoo.com> wrote:
I invite all readers to give their opinion, if any, whether the
spacetime continuum as modeled in GR is assumed to be a rigid or a
deformable continuum?
If it is assumed to be a deformable continuum, then how come that the
strained state of this continuum has never been examined in terms of
associated strain tensors?
No answers, but a few platitudes:
Clearly "opinion" is not worth to much here: what is needed is some
definition of what it means for a continuum to be deformable or rigid.
Opinion would enter more into a choice of what definition might be a
reasonable one to adopt: there may be more than one.
Let me attempt to distinguish between a rigid and a deformable
continuum of points. Let P be *any* point in this continuum. Let P1,
P2, P3, .... Pn be n points in the neighborhood of P. Let ds_1 be the
separation distance between points P and P1, ds_2 be the separation
distance between points P and P2, ..., ds_n be the separation distance
between points P and Pn. If these separation distances ds_1, ds_2,
ds_3,...., ds_n from point P to all of its neighborhood points, remain
constant and invariant with time under all circumstances, then the
continuum under consideration can be regarded as rigid. If under
certain circumstances, these separation distances change to say ds'_1,
ds'_2, ds'_3,...., ds'_n then the continuum under consideration can
be regarded as deformable.
Yes, that sounds reasonable. Applying this to GR requires some more
work. For one thing, if "spacetime" is your continuum, the thing is
invariant for all time. We would have to resort to some other
consideration to say that one part of spacetime is distended wrt
another, than "changes in time".
If we restrict ourselves to "space", then again we are faced with some
difficulties regarding "changing in time". Which time? How do we track
a "point in space" as it evolves? Do we paint it? And of course we
must give meaning to "distance".
Since the separation distance dS between two neighborhood points of
the spacetime continuum does change under the influence of
gravitational field (as per GR), obviously the spacetime continuum is
assumed to be deformable in GR.
I don't know what that means. Assuming by "separation" you mean
"under the spacetime metric", nothing "changes" under the influence of
the gravitational field, since (as someone once put it) "nothing moves
in spacetime".
I'm not ridiculing your idea, just pointing out we have to be very
methodical about giving it meaning. My opinion is, yes, it's probably
possible to make a well-motivated case that something is behaving
"elastically" in GR; there may be more than one mapping.
Given such a definition, then yours is a technical question, not a
matter of opinion.
In terms of what might be a reasonable definition, we might wonder
whether we want to ask the question of a four dimensional something,
or a three dimensional something. There may be no natural definition
of universal time, but I'm not convinced there might not be a natural
idea of a three dimensional continuum which does its thing locally,
and dynamically, and doesn't give a figo whether or not we want to
search for a universal t coordinate.
Yes, instead of starting with a 4-D spacetime continuum, it is quite
appropriate to consider its subset, the 3-D space continuum, as
modeled in GR, to examine its deformation characteristics.
Secondly, even given a sense of "deformation", by asking for an
accompanying "stress" you are taking things a notch further. Stress
and strain go together in ordinary material: it's plausible that if
there is something like strain in space then there is also something
like stress also, and something like a constitutive relation, but
merely plausible (whether the whole analogical package can be put
through might be a useful test of "reasonableness" in the
identification).
No, I didn't ask for an accompanying "stress". I only mentioned strain
tensors associated with deformation of the continuum. We could take
the things a notch further at a later stage.
All these something-likes are things to be searched for in the math of
General Relativity, or in the class of maths which are sufficiently
something-like General Relativity to reproduce its success. I guess we
may offer our "opinion" about whether this might be a worthwhile
search. I think it would.
In this regard, you may kindly examine the article on the strained
state of a continuum at,
http://www.geocities.com/gurcharn_sandhu/pdf_art/continuum_strain.pdf
Last paragraph of this article is reproduced below for information.
"As shown above, the notion of deformed or strained state of the
continuum under study is derived from the variability or invariance of
arc element ds. Whenever the arc element ds changes over to ds' under
certain situations, the changed state of the continuum will be termed
the deformed or strained state. The strained state can be considered
fully defined or fully determined once we know or uniquely determine
the displacement vector field at all points of the continuum. The
strained state can also be defined through specification of strain
tensor components provided these components satisfy Saint Venant's
compatibility equations. Finally the strained state can also be
defined through specification of modified metric coefficients from
which the required strain tensor components can be computed subject to
the compatibility conditions. However, the compatibility conditions
require that the modified metric must be Euclidean to ensure that the
resulting strained state of the continuum corresponds to smooth,
finite and continuous displacement components and to avoid
discontinuities within the continuum. This fact is of crucial
importance for examining the validity of the current mathematical
model of General Relativity."
Further the validity of the current mathematical model of GR has also
been examined at,
http://www.geocities.com/gurcharn_sandhu/pdf_art/invalidity_gr.pdf
GSS- Hide quoted text -
- Show quoted text -
.
|
|
|
| User: "GSS" |
|
| Title: Re: Definition of stress tensor in continuum mech |
05 Oct 2007 09:54:08 AM |
|
|
On Oct 5, 7:13 am, Edward Green <spamspamsp...@netzero.com> wrote:
On Oct 3, 7:23 am, GSS <gurcharn_san...@yahoo.com> wrote:
On Oct 3, 2:06 am, Edward Green <spamspamsp...@netzero.com> wrote:
On Oct 2, 7:21 am, GSS <gurcharn_san...@yahoo.com> wrote:
I invite all readers to give their opinion, if any, whether the
spacetime continuum as modeled in GR is assumed to be a rigid or a
deformable continuum?
If it is assumed to be a deformable continuum, then how come that the
strained state of this continuum has never been examined in terms of
associated strain tensors?
No answers, but a few platitudes:
Clearly "opinion" is not worth to much here: what is needed is some
definition of what it means for a continuum to be deformable or rigid.
Opinion would enter more into a choice of what definition might be a
reasonable one to adopt: there may be more than one.
Let me attempt to distinguish between a rigid and a deformable
continuum of points. Let P be *any* point in this continuum. Let P1,
P2, P3, .... Pn be n points in the neighborhood of P. Let ds_1 be the
separation distance between points P and P1, ds_2 be the separation
distance between points P and P2, ..., ds_n be the separation distance
between points P and Pn. If these separation distances ds_1, ds_2,
ds_3,...., ds_n from point P to all of its neighborhood points, remain
constant and invariant with time under all circumstances, then the
continuum under consideration can be regarded as rigid. If under
certain circumstances, these separation distances change to say ds'_1,
ds'_2, ds'_3,...., ds'_n then the continuum under consideration can
be regarded as deformable.
Yes, that sounds reasonable. Applying this to GR requires some more
work. For one thing, if "spacetime" is your continuum, the thing is
invariant for all time. We would have to resort to some other
consideration to say that one part of spacetime is distended wrt
another, than "changes in time".
Yes, for the spacetime continuum your point is valid. The problem
however is that when we assume the spacetime to be a physical entity,
we tend to visualize it as a sort of 4-D 'Block'. It is in this
'Block' view of spacetime that "nothing moves". Actually the notion of
spacetime is a mathematical abstract notion used to model the dynamic
phenomenon in 3-D physical space. The 3-D space with all its material
content does not physically exist for all past and future times. One
convenient view to visualize the situation could be to regard the 3-D
space, with all its interacting material content, as 'moving' along
the time coordinate just as a train moves along its tracks!!
I have referred to the spacetime continuum "as modeled in GR". As per
GR some regions of spacetime are *flat* and some are *curved*
depending on the matter-energy distribution in the vicinity. When we
shift our focus from a flat region to a curved region of spacetime we
can notice the *change* in spacetime characteristics. What I have
shown is that in the GR model of spacetime, when we shift our focus
from a *flat* region to a *curved* region of spacetime we can notice
the *change in separation distances between the neighborhood spacetime
points* thereby implying deformation of the spacetime continuum.
If we restrict ourselves to "space", then again we are faced with some
difficulties regarding "changing in time". Which time? How do we track
a "point in space" as it evolves? Do we paint it? And of course we
must give meaning to "distance".
Meaning to "distance" is given by the very definition of metric of
space. As far as time is concerned, we can always use the
internationally accepted standard notion of time as UTC or TAI. While
considering the deformation of space as per the GR model, I have only
considered 'static gravitational fields' produced by spherically
symmetric bodies of matter and for which the Schwarzschild solution is
considered valid.
Since the separation distance dS between two neighborhood points of
the spacetime continuum does change under the influence of
gravitational field (as per GR), obviously the spacetime continuum is
assumed to be deformable in GR.
I don't know what that means. Assuming by "separation" you mean
"under the spacetime metric", nothing "changes" under the influence of
the gravitational field, since (as someone once put it) "nothing moves
in spacetime".
As explained above, here "change" is from *flat* region to *curved*
region of spacetime continuum. Let me illustrate the notion of change
under the influence of gravitational field, through an example.
When a surface is represented in the parametric form by 2-d surface
coordinates, the intrinsic geometry of the surface is described by its
2-d metric tensor. The Riemann tensor composed from the 2-d metric
components is non-zero for a curved surface and zero for a plane or
flat surface.
Let us consider a large circular metal ring of radius R, filled inside
with a plane thin film membrane (rubber membrane or soap film). The
intrinsic geometry of any small region of this thin film can be
represented by a 2-d flat metric with zero Riemann tensor.
Let us now imagine that we exert a steady pressure over a small
localized region of this film (say by impinging an air jet) in such a
way that a small hemispherical bubble of radius r<<R is formed in this
local region. The 2-d surface of this hemispherical bubble can be
represented by a modified 2-d metric with non-zero Riemann tensor.
**Obviously it is not difficult to visualize that the localized
hemispherical bubble induced by a steady external pressure is actually
a deformed (elongated/stretched) membrane with a curved surface in
comparison to the undeformed plane membrane in the surrounding
region.**
By moving the impinging air jet sideways, location of the
hemispherical bubble on the large plane membrane can be easily
shifted. The state of deformation of the curved membrane in comparison
to the plane membrane can be studied in detail by comparing the
Riemannian metric of the curved surface with the Euclidean metric of
the plane surface. It can be easily shown that all displacements
produced on the curved surface of the membrane are continuous and
finite.
Let us now come back to our familiar 3-d space continuum and consider
the gravitational field of our solar system. Kindly imagine the
trajectory of the barycenter of our solar system within our galaxy.
Let S be the location of the barycenter on this trajectory at certain
time t. Let V be the volume of the solar system within which the
gravitation of the sun is significant. Let D be the diameter of this
volume. Now imagine that at certain different time t+T the barycenter
of the solar system is located at a different point S1 on the
trajectory such that the distance SS1 is greater than 2D. It is quite
obvious that at time t, the volume V of space around point S1 is free
from any gravitational influence and hence as per GR it can be
regarded as 'flat' or Euclidean space. But at time t+T the volume V of
space around S1 is occupied by the solar system and hence as per GR
the metric of this space will get modified to Riemannian metric to
produce physical deformations in the space continuum.
The essential point I am stressing here is that a plane membrane
surface with Euclidean metric does get deformed into a curved surface
with Riemannian metric under the influence of external pressure. It is
precisely in the same way it has been postulated in GR that 'flat'
space with Euclidean metric gets deformed to a 'curved' space with
Riemannian metric under the influence of a steady state gravitational
field. Accordingly the corresponding deformation characteristics have
been examined in detail.
I'm not ridiculing your idea, just pointing out we have to be very
methodical about giving it meaning. My opinion is, yes, it's probably
possible to make a well-motivated case that something is behaving
"elastically" in GR; there may be more than one mapping.
No, you have missed the point. I have not indicated anywhere that
'something' is behaving "elastically" in GR. What I have actually
shown is that the space *continuum* as modeled in GR gets *deformed*
under the influence of a gravitational field to the state of *dis-
continuum* leading to the *invalidity of GR*. You may like to have a
re-look at this inference.
http://www.geocities.com/gurcharn_sandhu/pdf_art/invalidity_gr.pdf
GSS
.
|
|
|
| User: "Edward Green" |
|
| Title: Re: Definition of stress tensor in continuum mech |
05 Oct 2007 06:39:16 PM |
|
|
On Oct 5, 10:54 am, GSS <gurcharn_san...@yahoo.com> wrote:
On Oct 5, 7:13 am, Edward Green <spamspamsp...@netzero.com> wrote:
... if "spacetime" is your continuum, the thing is
invariant for all time. We would have to resort to some other
consideration to say that one part of spacetime is distended wrt
another, than "changes in time".
Yes, for the spacetime continuum your point is valid. The problem
however is that when we assume the spacetime to be a physical entity,
we tend to visualize it as a sort of 4-D 'Block'. It is in this
'Block' view of spacetime that "nothing moves".
Of course.
Actually the notion of
spacetime is a mathematical abstract notion used to model the dynamic
phenomenon in 3-D physical space. The 3-D space with all its material
content does not physically exist for all past and future times. One
convenient view to visualize the situation could be to regard the 3-D
space, with all its interacting material content, as 'moving' along
the time coordinate just as a train moves along its tracks!!
Certainly. I agree with you in spirit. However, the problem comes up
that there is seldom a natural choice for a universal time coordinate,
along whose tracks things are moving. For this reason many start
thinking of the four dimensional continuum as the real stuff. The
three dimensional stuff may indeed be wiggling around like a jelly,
but our local "times" are conditioned on the state of this jelly,
which makes things complicated.
I have referred to the spacetime continuum "as modeled in GR". As per
GR some regions of spacetime are *flat* and some are *curved*
depending on the matter-energy distribution in the vicinity. When we
shift our focus from a flat region to a curved region of spacetime we
can notice the *change* in spacetime characteristics. What I have
shown is that in the GR model of spacetime, when we shift our focus
from a *flat* region to a *curved* region of spacetime we can notice
the *change in separation distances between the neighborhood spacetime
points* thereby implying deformation of the spacetime continuum.
You have of course carefully defined all your terms.
Meaning to "distance" is given by the very definition of metric of
space. As far as time is concerned, we can always use the
internationally accepted standard notion of time as UTC or TAI. While
considering the deformation of space as per the GR model, I have only
considered 'static gravitational fields' produced by spherically
symmetric bodies of matter and for which the Schwarzschild solution is
considered valid.
Well, that restriction simplifies things somewhat.
<...>
I'm not ridiculing your idea, just pointing out we have to be very
methodical about giving it meaning. My opinion is, yes, it's probably
possible to make a well-motivated case that something is behaving
"elastically" in GR; there may be more than one mapping.
No, you have missed the point. I have not indicated anywhere that
'something' is behaving "elastically" in GR. What I have actually
shown is that the space *continuum* as modeled in GR gets *deformed*
under the influence of a gravitational field to the state of *dis-
continuum* leading to the *invalidity of GR*.
Oops. By elastic behavior I intended more or less a synonym to
deformation, though I agree this is sloppy, since even with continuum
solids we must distinguish between elastic and plastic deformations,
which also requires us to consider whether along with our mapping of
"strain" we are going to have something mapping to "stress".
But more oopsy is your use of "discontinuum", and you claim to
invalidate GR. You've lost your audience here. Sorry. :-/
.
|
|
|
| User: "GSS" |
|
| Title: Re: Definition of stress tensor in continuum mech |
06 Oct 2007 11:44:28 AM |
|
|
On Oct 6, 4:39 am, Edward Green <spamspamsp...@netzero.com> wrote:
On Oct 5, 10:54 am, GSS <gurcharn_san...@yahoo.com> wrote:
On Oct 5, 7:13 am, Edward Green <spamspamsp...@netzero.com> wrote:
... if "spacetime" is your continuum, the thing is
invariant for all time. We would have to resort to some other
consideration to say that one part of spacetime is distended wrt
another, than "changes in time".
Yes, for the spacetime continuum your point is valid. The problem
however is that when we assume the spacetime to be a physical entity,
we tend to visualize it as a sort of 4-D 'Block'. It is in this
'Block' view of spacetime that "nothing moves".
Of course.
Actually the notion of
spacetime is a mathematical abstract notion used to model the dynamic
phenomenon in 3-D physical space. The 3-D space with all its material
content does not physically exist for all past and future times. One
convenient view to visualize the situation could be to regard the 3-D
space, with all its interacting material content, as 'moving' along
the time coordinate just as a train moves along its tracks!!
Certainly. I agree with you in spirit. However, the problem comes up
that there is seldom a natural choice for a universal time coordinate,
along whose tracks things are moving. For this reason many start
thinking of the four dimensional continuum as the real stuff. The
three dimensional stuff may indeed be wiggling around like a jelly,
but our local "times" are conditioned on the state of this jelly,
which makes things complicated.
Yes, 'things' always appear complicated as long as we don't possess
relevant information to make 'things' clear.
Well, instead of viewing the 3-D space continuum as "wiggling around
like a jelly", we might view it as an elastic continuum capable of
supporting infinitely many modes of vibrations. In 4-D spacetime
continuum it might correspond to infinitely many modes of 'spacetime
distortions'. Pushing this viewpoint a 'notch further' we might even
suggest that all material phenomenon, all elementary particles might
be 'existing' as specific modes of spacetime distortions. This line of
thinking has been explored and presented in 'outline form only' at,
http://www.geocities.com/gurcharn_sandhu/pdf_art/spacetime_deformation.pdf
I have referred to the spacetime continuum "as modeled in GR". As per
GR some regions of spacetime are *flat* and some are *curved*
depending on the matter-energy distribution in the vicinity. When we
shift our focus from a flat region to a curved region of spacetime we
can notice the *change* in spacetime characteristics. What I have
shown is that in the GR model of spacetime, when we shift our focus
from a *flat* region to a *curved* region of spacetime we can notice
the *change in separation distances between the neighborhood spacetime
points* thereby implying deformation of the spacetime continuum.
You have of course carefully defined all your terms.
Meaning to "distance" is given by the very definition of metric of
space. As far as time is concerned, we can always use the
internationally accepted standard notion of time as UTC or TAI. While
considering the deformation of space as per the GR model, I have only
considered 'static gravitational fields' produced by spherically
symmetric bodies of matter and for which the Schwarzschild solution is
considered valid.
Well, that restriction simplifies things somewhat.
<...>
I'm not ridiculing your idea, just pointing out we have to be very
methodical about giving it meaning. My opinion is, yes, it's probably
possible to make a well-motivated case that something is behaving
"elastically" in GR; there may be more than one mapping.
No, you have missed the point. I have not indicated anywhere that
'something' is behaving "elastically" in GR. What I have actually
shown is that the space *continuum* as modeled in GR gets *deformed*
under the influence of a gravitational field to the state of *dis-
continuum* leading to the *invalidity of GR*.
Oops. By elastic behavior I intended more or less a synonym to
deformation, though I agree this is sloppy, since even with continuum
solids we must distinguish between elastic and plastic deformations,
which also requires us to consider whether along with our mapping of
"strain" we are going to have something mapping to "stress".
Pure deformation and the associated strain are *geometrical*
properties of the continuum. In the spacetime continuum as modeled in
GR, the gravitational field induced 'curvature' or deformations are
purely geometrical attributes. For studying these deformations and the
associated strains, we don't need to invoke the elasticity property of
the continuum. Of course when we further consider the situation that
on removal of the gravitational field, the state of curvature or
deformation of the continuum gets back to the original undeformed
state, the property of elasticity of the continuum could be implied.
But more oopsy is your use of "discontinuum", and you claim to
invalidate GR. You've lost your audience here. Sorry. :-/
Well, I don't mind losing some of my audience as long as I don't lose
my argument!!
Essentially what I have shown is the incompatibility of gravitation
induced strain components that transforms the space continuum to the
discontinuous state of 'discontinuum' thereby implying invalidity of
GR as a physical theory. Of course GR may still be retained as a
'useful' mathematical model, but it can no longer be treated as a
valid theory of Physics!
Let us therefore agree to disagree on the validity of GR as a theory
of Physics and close our discussions herewith.
GSS
-----------------------------------------------------------------
Finally to assert this point, I have repeated my arguments below.
Let us consider a spherical polar coordinate system with origin at
point O and the coordinate parameters r, theta and phi. Further for
the sake of convenience let us represent theta by letter q and phi by
letter p. The metric coefficients for this coordinate system in the un-
deformed or gravitation free space continuum are given as,
g_rr = 1 ; g_qq = r^2 ; g_pp= r^2.Sin^2 (q) ..... (1)
The arc element or the separation distance ds between two neighboring
space points P and Q in this region will be given by:
(ds)^2 = g_rr (dr)^2 + g_qq (dq)^2 + g_pp (dp)^2
= 1.(dr)^2 + r^2.(dq)^2 + r^2.Sin^2(q).(dp)^2 ..... (2)
Now let us assume that a spherically symmetric body of mass M and
radius r0, is located at the origin O of this coordinate system. Due
to the gravitational field in its vicinity (i.e. r > r0 > 0), the
modified metric coefficients h_ij are given by the Schwarzschild
solution as:
h_rr = 1/(1 - 2GM/c^2r) ; h_qq = r^2 ; h_pp= r^2.Sin^2 (q) .... (3)
Thus the modified radial metric coefficient h_rr at any particular
space point P(r, q, p) can be taken as a function of M and its value
in the region under consideration is always greater than unity for
M>0. The arc element or the modified separation distance ds' between
two neighboring space point positions P' and Q' in this region will be
given by:
(ds')^2 = h_rr (dr)^2 + h_qq (dq)^2 + h_pp (dp)^2
= (1/(1 - 2GM/c^2 r)).(dr)^2 + r^2.(dq)^2 + r^2.Sin^2 (q).
(dp)^2 .... (4)
Therefore, now we can compute the induced strain tensor components
e_ij from the modified metric coefficients h_ij as,
2 e_rr = h_rr - g_rr = (1/(1 - 2GM/c^2 r)) - 1 ...... (5)
with the factor 2GM/c^2 r <<1, equation (5) will get simplified to,
e_rr = GM/c^2 r .................. (6)
and e_qq = h_qq - g_qq = 0 ; e_pp = h_pp - g_pp = 0 ..... (7)
and e_qp = e_pq = e_rq = e_qr = e_rp = e_pr = 0 ...... (8)
This set of strain tensor components constitutes the strain field
induced in the region of space continuum where the gravitational field
of M has modified the metric coefficients to h_ij.
Incompatibility of the Induced Strain Components
------------------------------------------------
For complete description of the strained state of the space
continuum, we must be able to uniquely determine the displacement
vector field U from the specified strain tensor components. For this
the strain tensor components are required to satisfy Saint Venant's
integrability or compatibility conditions. The displacement vector
components obtained from the integration of the relevant partial
differential equations must be single valued, finite and continuous
functions of coordinates and must satisfy physical constraints over
the boundary of the region of space under consideration.
It can be easily seen that the radial strain components e_rr
given by equation (6), with all other components being zero, cannot
satisfy the required compatibility conditions. In order to illustrate
and highlight this problem, let us consider the relative displacement
vector U that gives rise to the strain components e_rr, e_qq and
e_pp . If u^r is the only non-zero component of the displacement
vector U, then the strain components dependent on u^r are given by,
e_rr = Du^r/Dr ; e_qq = u^r/r and e_pp = u^r/r .... (9)
Obviously, if the radial strain component e_rr is non-zero, the radial
displacement component u^r must be non-zero. But once the radial
displacement component u^r is non-zero, the tangential strain
components e_qq and e_pp cannot be zero. This precisely is the
incompatibility of the strain components e_rr, e_qq and e_pp induced
by the static gravitational field of a spherically symmetric
gravitating body of mass M. This incompatibility is not limited to
the strain components induced by the Schwarzschild metric of
spherically symmetric, static gravitational fields but is applicable
to all strain components induced by the Riemannian metric obtained
from EFE. In fact one of the essential requirements imposed by the
standard compatibility conditions on strain components e_ij is that
the Riemann tensor composed from e_ij must be a zero tensor. This can
be true only if both metrics namely g_ij and h_ij are Euclidean which
however contradicts the basic postulate of General Relativity.
Therefore, the specification of metric coefficients (3) as per the
Schwarzschild solution is physically invalid and unacceptable.
Further, even if we overlook the compatibility conditions for a
while, from equations (6) and (9) we get,
e_rr = du^r/dr = GM/c^2 r ....... (10)
which appears to be easily integrable. A simple integration of
equation (10) yields,
u^r = (GM/c^2).Ln(r/r0) .......... (11)
Apparently we seem to have obtained a nice looking solution for the
radial displacement component u^r in spite of the fact that the
compatibility conditions were not satisfied by the strain components.
Well, a closer look at equation (11) will show that this finite value
of u^r will give rise to a finite value of tangential strain
components e_qq and e_pp? (equation (9)) whereas these components are
required to be zero as per the modified metric h_ij (equation (7)) of
the space continuum. This accounts for the incompatibility. Further
the radial displacement u^r given by equation (11) tends to infinity
as r tends to infinity. This is invalid since as per physical
constraints, u^r must tend to zero when the gravitational field tends
to zero at infinitely large r.
GSS
.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Related Articles |
|
|
