Tidal Force Tensor
Pmb
In "Gravitation and Spacetime - 2nd Ed.," Ohanian and Ruffini, the authors
explain that the Riemann tensor is also refered to as the tidal force
tensor. I guess a better name would be tidal force 4-tensor so as not to
confuse it with the Newtonian tidal force 3-tensor.
I heard that MTW also uses the term "Tidal Force Tesor" but I can't find it.
Does anyone know if this term is found in MTW?
Thanks
Pmb
WaiteDavid137
>Subject: Tidal Force Tensor
>From: "Pmb" some...@somewhere.com
>Date: 11/20/03 3:22 PM US Mountain Standard Time
>Message-id: <qWMub.49575$bQ3....@nwrdny03.gnilink.net>
It would be a bad term if they did. The Christoffel symbols are analogous to
the Newtonian "field" and it is the derivatives of those that correspond to the
Newtonian tides. Simply pick a frame according to which the derivatives locally
vanish while the Christoffel symbols don't and you've found a frame according
to which the Riemann tensor is not directly related to tides. It is only under
a particular choice of coordinates and under particular approximation that
components of the Riemann tensor are proportional to the tidal force. See the
derivation leading to equation 6.2.3
http://www.geocities.com/zcphysicsms/chap6.htm#BM73
Gauge
waited...@aol.com (WaiteDavid137) wrote
> It would be a bad term if they did. The Christoffel symbols are analogous to
> the Newtonian "field" and it is the derivatives of those that correspond to the
> Newtonian tides.
I've explained to you before why that is wrong. You've chosen a
definition of "tidal force" which is not used in modern general
relativity. You're once again confusing "tidal force" with the
non-vanishing of the derivative of the Christofel symbols. That is
*not* how tidal forces are defined in general relativity.
Gravitational tidal forces are defined in GR as "that which causes
tides" and as such are defined in terms of geodesic deviation and the
non-vanishing of the Riemann tensor.
Yes. It's true that the non-vanishing of derivatives of the
Christoffel symbols for an accelerated reference frame implies that
there will be a distortion of a solid body (at rest in accelerating
frame) unless the different parts of the solid body are "correctly"
accelerated (I seem to remember Rindler discusses this). But this is
irrelevant for tides, because tides and the tidal force are defined in
a small reference frame (or for small liquid drop) in free fall.
Obviously, if you let a small drop free fall in the accelerated
reference frame, the drop will have no acceleration with respect to an
inertial reference frame, and therefore no distortion, and it will
register no tidal force. The Riemann tensor for this case is zero,
both in the freely falling reference frame (an exactly inertial
reference frame) and in the accelerated reference frame--the spacetime
is flat!
In Newtonian mechanics the tidal force tensor is defined in this page
http://www.geocities.com/physics_world/tidal_force_tensor.htm
In GR its defined as the Riemann tensor
More generally. In the purely geometric view of general relativity (as
opposed to the 3+1 view)
Velocity v is replaced with 4-velocity U, v is part of U = (c*dt/dT,
v).
Momentum p is replaced with 4-momentum P, p is part of P = (mc,p) =
(c*gamma*m_o, p)
Force f is replaced with 4-force F, f is part if F
Notice how each 4-vector is related to a 3-vector which carries the
same name? So too with the tidal force tensor E which is replaced with
the relativistic tidal force 4-tensor R = Riemann. They too are
related. If E = Phi Newtonian gravitational potential then E_ij =
Phi,ij (comma means take partial derivatives with respect to x^i,
x^j). Then
R_j0k0 = c^2*E_jk
In Newtonian mechanics, as measured in the rest frame of the Earth,
the existance differences in gravitational acceleration is identical
to the existance of tidal forces. If one chooses a non-inertial frame
such as a rotating frame - there will still be no tidal forces present
yet there will be differences in acceleration of a freely moving
particle. Yet tidal forces cannot be induced in Newtonian gravity
simply by changing the frame of referance. In Newton's theory there
either are tidal forces present or there are no tidal forces. One
can't cause tides on a planet simply by changing the frame of
referance.
Same in GR. Tidal forces are measured in a free-fall. You're confusing
the derivative of the Christoffel symbols with tidal forces.
As I stated above => The tidal force tensor in GR is the Riemann
tensor. In general relativity tidal forces and spacetime curvature are
the exact same phenomena. Or as Kip Thorne states in his text "Black
Holes and Time warps" on page 111
"Therefore, spacetime curvature and tidal gravity must be precisely
the same thing, expressed in different languages."
WaiteDavid137
>Subject: Re: Tidal Force Tensor
>From: gau...@hotmail.com (Gauge)
>Date: 11/25/03 4:09 PM US Mountain Standard Time
>Message-id: <e7203033.03112...@posting.google.com>
>
>
>waited...@aol.com (WaiteDavid137) wrote
>
>> It would be a bad term if they did. The Christoffel symbols are analogous
>to
>> the Newtonian "field" and it is the derivatives of those that correspond to
>the
>> Newtonian tides.
>
>I've explained to you before why that is wrong.
So you claim, but in fact it was me who explained to you before why you are
wrong.
>You've chosen a
>definition of "tidal force" which is not used in modern general
>relativity.
>You're once again confusing "tidal force" with the
>non-vanishing of the derivative of the Christofel symbols.
No I haven't and I'm not confusing anything. I am telling you that this is what
it is. In relativity the Force on an object is given by
F = dP/dtau + GUP
Where I used G to represent the affine connection, the Chistoffel symbols.
Considering gravitation alone, F = 0, the ordinary derivative results in
(dP/dtau),mu = (-GUP),mu. This is the definition of the tidal force and as is
proven in the derivation leading to equation 6.2.3 at
http://www.geocities.com/zcphysicsms/chap6.htm#BM74
in only proportional to components of the Riemann tensor in particular
circumstances and for a particular choice of frame.
>R_j0k0 = c^2*E_jk
And as derived at the link is only a valid description for a particular choice
of frame.
[snipped the rest as it was all built on an unusual definition of tidal force]
Ken S. Tucker
Sorry can't help with MTW, but S. Weinberg
has a clear and crisp Eq.(6.10.1) in
Gravitation and Cosmology.
Ken S. Tucker
Alfred Einstead
> It would be a bad term if they did.
It's entirely appropriate and the understanding behind it is fairly
standard, centering on the equation of geodesic deviation:
d^2/ds^2 (J^i(q(s))) + 2/3 R^i_{jkl} q^j'(s) q^l'(s) J^k(q(s)) = 0,
that describes the deformation
J^i(q(s)) = d/da q(a,s)|a=0
that takes place along a bundle of geodesics (q(a,s): a in [-D,D];
q(0,s) = q(s)) due to the gravitational field.
This is covariant and does not disappear in any frame, and is
given in invariant form by:
(Del_q')^2 J + R(J,q') q' = 0
in terms of the curvature operator
R(u,v) = [Del_u, Del_v] - Del_{[u,v]}.
An account of this will probably be found in any standard
reference on Riemannian geometry, and is developed in Landsman [1],
section II.3.
Reference:
[1] Landsman, N.P., 1998: Mathematical Topics Between Classical
and Quantum Mechanics, Springer-Verlag.
Pmb
news:e58d56ae.03112...@posting.google.com...
> waited...@aol.com (WaiteDavid137) wrote:
> > It would be a bad term if they did.
>
> It's entirely appropriate and the understanding behind it is fairly
> standard, centering on the equation of geodesic deviation:
waite has the impression that "tidal force" is not defined as you and I have
described it. He believes that tidal forces are defined accordining to
differences in coordinate acceleration, the existance of such differences
are, in general, coordinate dependant. This definition, of course, is
untrue. The rising and falling of ocean tides can't arise from changes in
coordinate systems and that is where "tidal force" gets its name. If there
is no relative acceleration of free-test particles in a free-fall frame then
there are no tidal accelerations in any frame. Hence the reason the Riemann
tensor is also called the relativistic tidal force tensor.
Pmb
WaiteDavid137
>Date: 12/2/03 5:25 AM US Mountain Standard Time
>Message-id: <e58d56ae.03112...@posting.google.com>
>
>waited...@aol.com (WaiteDavid137) wrote:
>> It would be a bad term if they did.
>
>It's entirely appropriate
As I've explained and continue to do so, no it is not.
>and the understanding behind it is fairly
>standard, centering on the equation of geodesic deviation:
> d^2/ds^2 (J^i(q(s))) + 2/3 R^i_{jkl} q^j'(s) q^l'(s) J^k(q(s)) = 0,
>that describes the deformation
> J^i(q(s)) = d/da q(a,s)|a=0
>that takes place along a bundle of geodesics (q(a,s): a in [-D,D];
>q(0,s) = q(s)) due to the gravitational field.
>
>This is covariant and does not disappear in any frame,...
>[snipped]
That is not what he is referring to, nor is it directly related to my
point(thus the snip). To exactly quote the relation that he presented and I am
discussing, pmb aka guage wrote:
>> R_j0k0 = c^2*E_jk
>> So too with the tidal force tensor E....
>> E_ij = Phi,ij
He is referring to particular elements of the Riemann tensor and calling
those elements the "tidal force tensor" and I am saying that this is a
bad thing to do as I showed in its derivation that "this" relation
depends on a particular choice of frame. Here is a simple example of
where it fails. Consider a body as observed by one using ct,x,y,z
coordinates according to which the line element is the Rindler line
element ds^2 = [(1 - alpha*z/c^2)^2]dct^2 - dx^2 - dy^2 - dz^2 The
gravitational force "felt" at some point in the body when held
stationary with respect to these coordinates varies with respect to z.
Thus this frames observer observes the body behaving as if there where a
tidal force present and indeed Phi,ij as he wrote it is not zero for
this ds^2. However the Riemann tensor is zero for this spacetime and in
fact it is merely an accelerated frame transformation away from ds^2 =
dct^2 - dx^2 - dy^2 - dz^2 Another example of a strange tidal effect in
the Rindler spacetime as I expressed it goes as follows. Consider the
pole in barn length contraction paradox for a rod traveling
perpendicular to the z direction according to an observer stationary
with respect to the Rindler space coordinates. The doors for this barn
will be suspended virtically from the ends of a support beam balanced
above the bar from the center. When the pole is in the barn the central
support will be removed so that the doors drop shut. According to the
pole frame there is a break in sumultaneity so that the far door shuts
first, but what happens relevent to this discussion is that in order for
this to happen the support beam as observed from the pole frame must
have gone from an initial horizontal orientation to a final slanted
orientation and has undergone an increase in length. So according to the
pole frame not only is there a tidal force along the z direction but
there must also be one alone the direction of motion as well. Again the
pole frame will be merely a transformation away from an inertial frame
and the Riemann tensor will still be zero but as this indicates even
with it being zero all sorts of frame dependent tides can be observed.
The Riemann tensors really only direct relevence to "tidal force" (as
I'm not talking about the geodesic deviation equation) is a relation
between a select few elements of the tensor and the tidal force "as
observed by" a local free fall frame observer.
Pmb
"Alfred Einstead" <whop...@csd.uwm.edu> wrote in message
news:e58d56ae.03112...@posting.google.com...
> waited...@aol.com (WaiteDavid137) wrote:
> > It would be a bad term if they did.
>
> It's entirely appropriate and the understanding behind it is fairly
> standard, centering on the equation of geodesic deviation:
waite has the impression that "tidal force" is not defined as you and I have
Gauge
> As I've explained and continue to do so, no it is not.
And you continue to be wrong.
> He is referring to particular elements of the Riemann tensor and calling
> those elements the "tidal force tensor"
Read Thorne and Blanchard -
http://www.pma.caltech.edu/Courses/ph136/yr2002/chap24/0224.1.pdf
In the *weak field limit* the Riemann tensor is *related to* (NOT
'defined by') the Newtonian tidal force tensor as
R_j0k0 = c^2*E_jk
The ony way to establish that relation is in the weak field limit.
This is NOT a definition of either tensor. It is a RELATIONSHIP
between the two. One is a 4-tensor and the other a 3-tensor. It points
out why both are called "tidal force tensor."
Again - that is the *weak field limit* as I've explained to you in the
past. In GR the *exact* expression for the tidal force tensor is the
Riemann tensor since it defines tidal accelerations, aka 'geodesic
deviation' - and tidal accelerations are defined in a locally inertial
frame.
All this is explained in "Gravitation and Spacetime - 2nd Ed.,"
Ohanian and Ruffini.
>From Thorne and Blanchard's text
http://www.pma.caltech.edu/Courses/ph136/yr2002/chap24/0224.1.pdf
"The name 'tidal gravitational field' comes from the fact that this is
the field which, generated by the moon and the sun, produces the tides
on the earth's oceans."
You don't get tides on Earth unless the Earth is in a curved
spacetime.
By the way - You've yet to prove that Kip Thorne is wrong when he
states
"spacetime curvature and tidal gravity must be precisely the same
thing, expressed in different languages"
> Consider a body as observed by one using ct,x,y,z
> coordinates according to which the line element is the Rindler line
> element ds^2 = [(1 - alpha*z/c^2)^2]dct^2 - dx^2 - dy^2 - dz^2 The
> gravitational force "felt" at some point in the body when held
> stationary with respect to these coordinates varies with respect to z.
In the *weak field limit* the relation between the Newtonian tidal
force tensor and the Einsteinian tidal force tensor (aka Riemann
tensor) holds true. In this case, i.e. weak field limit, the metric
for a uniform g-field becomes
ds^2 = c^2(1 + gz/c^2) dt^2 - dx^2 - dy^2 - dz^2
Where g is the local acceleration due to gravity. For a frame of 1g g
= -9.8m/s^2
[Moderator's note: Actually there is an analogous factor in front
of the spatial part of the metric, see for instance equation
(6.29) in Carroll's lecture notes gr-qc/9712019. The factor in front
of dt^2 comes from the Newtonian limit. The other is then implied
by the linearized Einstein equations.
-usc]
In this case the Newtonian potential is Phi = gz and therefore the
Newtonian tidal force tensor is zero.
> Thus this frames observer observes the body behaving as if there where a
> tidal force present and indeed Phi,ij as he wrote it is not zero for
> this ds^2.
That's because you weren't paying attention. I explained all of this,
i.e. the relation in the weak field limit, to you four years ago.
http://www.astronomy.net/forums/blackholes2/messages/3588.shtml?show=top&disp=0
i.e.
*****************
The tidal forces acting on a body in a gravitational field are given
in terms of this tensor. In General Relativity the existance of
spacetime curvature is determined by the non-vanishing of a tensor
caller the Riemann curvature tensor, the components of which are often
denoted as Rmnab. In the weak field limit the components Rm0a0 of the
Riemann tensor are equal to c-2 U,kl . The other components can then
be determined by measuring the components in several other frames.
Thus Rm0a0 is derived from Rk0l0. Therefore the Riemann tensor is
refered to as the tidal force tensor in general relativity since
spaceitme curvature and tidal gravity are the same thing in
differenent languages. The former in terms of relativity and the later
in terms of Newtonian gravity. In the weak field limit the tidal force
tensor reduces to the tidal force tensor in Newtonain gravity [ See OH
page 344]
*****************
I explained to you twice in that post that the relationship between
the Relativistic tidal force tensor and the Newtonian tidal force
tensor holds in the weak field limit. There is no way to establish
that relation unbless one is working in the weak field limit.
But in GR its the Riemann tensor which is the tidal force tensor, not
the Newtonian tidal force tensor.
> However the Riemann tensor is zero for this spacetime and in
> fact it is merely an accelerated frame transformation away from ds^2 =
> dct^2 - dx^2 - dy^2 - dz^2 Another example of a strange tidal effect in
> the Rindler spacetime as I expressed it goes as follows.
There are no tidal effects in flat spacetime.
> Consider the
> pole in barn length contraction paradox for a rod traveling
> perpendicular to the z direction according to an observer stationary
> with respect to the Rindler space coordinates. The doors for this barn
> will be suspended virtically from the ends of a support beam balanced
> above the bar from the center. When the pole is in the barn the central
> support will be removed so that the doors drop shut. According to the
> pole frame there is a break in sumultaneity so that the far door shuts
> first, but what happens relevent to this discussion is that in order for
> this to happen the support beam as observed from the pole frame must
> have gone from an initial horizontal orientation to a final slanted
> orientation and has undergone an increase in length. So according to the
> pole frame not only is there a tidal force along the z direction but
> there must also be one alone the direction of motion as well.
There is no tidal force present since the spacetime is flat. You're
confusing tidal forces here with something else.
> Again the
> pole frame will be merely a transformation away from an inertial frame
> and the Riemann tensor will still be zero but as this indicates even
> with it being zero all sorts of frame dependent tides can be observed.
You're confusing tidal force with something else
Pmb
Gauge
Instead of repeating your claim why don't you prove it. I.e. show me
one GR text that gives that definition that you CLAIM is the correct
definition of tidal force.
You keep posting an invalid definition, invalid in the sense that
you're the only one that uses it. Nobody in the GR community holds
what you say to be true - Nobody! Therefore your claim that the
definition I've given you is wrong is a bogus claim.
Proof of definition: From "Black Holes and Time Warps," Kip S. Thorne
(also quoted in 'Exploring Black Holes' by Taylor and Wheeler)
",spacetime curvature and tidal accelerations must be precisely the
same thing, expressed in different languages."
Thorne and Blanchard
http://www.pma.caltech.edu/Courses/ph136/yr2002/chap24/0224.1.pdf
"The name 'tidal gravitational field" comes from the fact that this is
the field which, generated by the moon and the sun, produces the tides
on the earth's oceans."
Are you claiming their definition is wrong?
Prove that your definition does what a tidal force does - i.e. raises
tides. I'd like to see that. I.e. Prove that if the Earth was in flat
spacetime then you can raise ocean tides simply by chaning your frame
of referance to an accelerating frame of referance.
Once again - You're confusing "difference in force" with "tidal force"
Pmb
Do you u
Alfred Einstead
> >It's entirely appropriate
> As I've explained and continue to do so, no it is not.
It's not your call to make. To reiterate:
> >the understanding behind it is fairly standard, centering on the
> > equation of geodesic deviation:
> > d^2/ds^2 (J^i(q(s))) + 2/3 R^i_{jkl} q^j'(s) q^l'(s) J^k(q(s)) = 0,
> >that describes the deformation
> > J^i(q(s)) = d/da q(a,s)|a=0
> >that takes place along a bundle of geodesics (q(a,s): a in [-D,D];
> >q(0,s) = q(s)) due to the gravitational field.
The components of the Riemann tensor, as seen above, do indeed
describe the components of the tidal force. More precisely,
it's -2/3 R^i_{jkl}, which is the ^i_{jkl} component of the
tensor:
-2/3 (Del_J Del_{q'} - Del_{q'} Del_J - Del_{[J,q]}) q'.
This is invariant. The Rindler example has no bearing on this.