# There is no gravitational dipole

#### Crosson

There is no gravitational dipole, obviously because there is no form of mass analagous to negative charge.

But think about this property of electromagnetic currents: negatives moving to the left and positives moving to the right both generate identical magnetic fields.

This means that we could make negative gravitomagnets, in addition to positive ones by using normal mass spining in the negative or positive directions, respectively.

Perhaps this is old hat for some of you, or perhaps it is false (too naive, though I know EFE take on a similarity to maxwell's equations in some particular weak field approximation).

The point is, I want to address the following claim from one of our recent sci-fi artificial gravity threads:

So you have something (your floor) that attracts objects on one side only? This is impossible.
Perhaps not so if we can generate gravitomagnetic multipoles and therefore calculate mass distrubutions that produce asymmetric fields. What do you guys think?

Related Special and General Relativity News on Phys.org

#### joecoss

I think that Gravity is Electromagnetic!

#### Crosson

I think that Gravity is Electromagnetic!
I believe in Einsteins General Theory of Relativity, which treats gravitation as a curvature in space time. According to this theory, moving masses create pseudogravitational fields. Magnetic fields are pseudoelectric fields, as seen from moving frames.

Gravitomagnetism is part of well established physics, and does not have to do with unification theories--but I understand if know one wants to discuss my point.

#### pmb_phy

Crosson said:
I believe in Einsteins General Theory of Relativity, which treats gravitation as a curvature in space time. According to this theory, moving masses create pseudogravitational fields. Magnetic fields are pseudoelectric fields, as seen from moving frames.
In relativity one has to be cautious when speaking of the electric and magnetic fields. There is such a thing called "Electric field 4-velocity " and "Magnetic field 4-velocity " each of which is "relative to observer A". One introduces a 4-vector for the observer and then the Faraday tensor is contracted with the observer's 4-velocity are contracted to find the electric field 4-velocity relative to the observer. (same with magnetism).

Pete

#### pmb_phy

Crosson said:
I believe in Einsteins General Theory of Relativity, which treats gravitation as a curvature in space time.
In GR the terms "gravitational tidal forces" and "spacetime curvature" are two different ways of speaking about the exact same thing.
According to this theory, moving masses create pseudogravitational fields. Magnetic fields are pseudoelectric fields, as seen from moving frames.
Sorry but I can't see how one could make such an assumption and how you justify the ussage of "pseudo."

Pete

#### pervect

Staff Emeritus
Crosson said:
I believe in Einsteins General Theory of Relativity, which treats gravitation as a curvature in space time. According to this theory, moving masses create pseudogravitational fields. Magnetic fields are pseudoelectric fields, as seen from moving frames.

Gravitomagnetism is part of well established physics, and does not have to do with unification theories--but I understand if know one wants to discuss my point.
There certainly is a strong anaology between gravitation and electromagnetism. The problem is that you can't exert a magnetic force on a non-moving charge with electromagnetism, so if your goal is to exert a gravitomagnetic force on a non-moving charge with gravitomagnetism, I don't quite see how you are going to do it.

When you transform to the frame of the charge (the stationary mass), there will only be electric fields acting on the charge.

There may be some ways to do this with strong fields, but I don't quite understand how they could work, by anology the method doesn't seem promising. I know Robert Forward was looking at those sorts of ideas, but his popular level writings were not explicit enough to really explain them. He probably has some more technical papers, somewhere, but I've never read them personally.

#### Antiphon

Crosson said:
There is no gravitational dipole, obviously because there is no form of mass analagous to negative charge.
Actually, the analogy between electromagnetics and gravity
is one to one IF:

1) You allow negative energy-density matter
2) You allow the time to flow backwards

Then you would have "like" gravitational "charges" repelling
(backward time) and you would have both polarities.

This only works for small masses because charge doesn't bend
space the way masses do.

#### hellfire

Antiphon said:
Actually, the analogy between electromagnetics and gravity
is one to one IF:

1) You allow negative energy-density matter
2) You allow the time to flow backwards

#### Larry717

Gravitomagnetic Poles

Magnetic Poles and Gravitomagnetic Poles

While magnetic poles are fictitious, they can be used in calculations
to determine the force between two bar magnets. The general
formula for the force between two bar magnets is:

F = (k P1 P2) / d^2

where k is the permeablity constant, P1 is the pole strength of the
first bar magnet, P2 is the pole strength of the second bar magnet,
and d is the distance between them.

Magnetic pole strength is defined as the magnetic dipole moment of a bar
magnet divided by its length.

Sample calculations for the force between bar magnets can be
found in any good text on electricity and magnetism.

According to Papapetrou, a mass is a pole-dipole: a pole due to gravity
and a dipole due to rotation. The gravitomagnetic dipole moment is
equivalent to angular momentum. The gravitomagnetic pole is the angular
momentum of a body divided by its axial length.

The force between gravitomagnetic poles can be attractive or repulsive
just as with magnetic poles. The spin direction of one gravitomagnetic pole
interacts with the spin direction of another gravitomagnetic pole.

The force equation above can be used to find the force between
the earth and a rotating body in terms of their gravitomagnetic
pole strengths. The quantities below are helpful in this regard.

k = G / c^2 = (7.41 x 10 ^ -28)
P1 = Je / 2Re = (5.55 x 10 ^ +26)
P2 = J / s
d = distance between earth and rotating body

where k is the gravitomagnetic permeability constant, G is the
gravitational constant, c is the speed of light, P1 is the angular
momentum of the earth divided by its diameter and P2 is the
angular momentum of a rotating body divided by its axial length.

The gravitomagnetic pole strength of the earth will vary by lattitude.
Therefore, P1 should become P1 sin (latitude in degrees).

L.A.

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#### pervect

Staff Emeritus
Larry717 said:
Magnetic Poles and Gravitomagnetic Poles

While magnetic poles are fictitious, they can be used in calculations
to determine the force between two bar magnets. The general
formula for the force between two bar magnets is:

F = (k P1 P2) / d^2

where k is the permeablity constant, P1 is the pole strength of the
first bar magnet, P2 is the pole strength of the second bar magnet,
and d is the distance between them.

Magnetic pole strength is defined as the magnetic dipole moment of a bar
magnet divided by its length.

Sample calculations for the force between bar magnets can be
found in any good text on electricity and magnetism.
The force between charges is an inverse square law force, but the force between dipoles goes as the fourth power of the distance. It's the same for electric and magnetic dipoles, but it's easier to work out with electric dipoles.

Conisder a pair of dipoles oriented as follows. Let the length of the dipoles be L (the vertical distance), all charges be equal to +/- q, and the separation (horizontal distance) be x

+ -
- +

What is the attractive force betweent them? We can write the total potential easily enough

V = k*q^2*(-2/x + 2/sqrt(x^2+L^2)

Taking the derivative of this with respect to X, we get the force
$$2\,k{\it q^2}\, \left( \frac{1}{x^2}-{\frac {x}{ \left( {x}^{2} +{L}^{2} \right) ^{3/2}}} \right)$$

which can be series expanded as

F ~ 3*k*q^2*L^2 / x^4

q*L is the electric dipole moment, so the force is proortional to the product of the dipole moments as stated, but the distance dependence of the force is proportional to 1/x^4, not 1/x^2.

According to Papapetrou, a mass is a pole-dipole: a pole due to gravity
and a dipole due to rotation. The gravitomagnetic dipole moment is
equivalent to angular momentum. The gravitomagnetic pole is the angular
momentum of a body divided by its axial length.
I'm afraid I would need a much more detailed explanation to follow this. The mass has a 4-velocity, which is a rank-1 tensor. It also has an angular momentum bi-vector, which is a rank-2 tensor. (We can represent anuglar momentum as a vector in classical mechanics, but it's more easily represented by a bi-vector, a rank 2 tensor, in GR). After this point, I totally lose what you're trying to say. (I'm not terribly familair with the Papapetrou equations, so feel free to over-explain :-)).

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#### Larry717

Plenty of Resources Available

The equation,

F = k P1 P2 / d^2,

contains fictitious sources that lead to a real force. There is no mention of a field, dipole
or otherwise. As you can see the poles of two bar magnets, P1 and P2, are treated like
charges in Coulomb's law. Thus, the force goes as the inverse square of distance. Only
one dimension is involved: the line from one pole to the other.

The purpose of my presentation was to explain in a simple and compact way the
real force between fictitious gravitomagnetic poles. It was not intended to treat
gravitomagnetism as a whole. For the interested reader there are resources
that provide a background for the topics mentioned here.

Magnetic Poles Force Law:

Papapetrou's Equations and Discussion of Rotating Masses in the Gravitational Field:
Proceedings of the Royal Society A (64) 57-75; 302-310 (1951); A (209) 248-258 (1951)

The following contains an introduction to gravitomagnetism in flat spacetime. The
statement that "there is no clear picture of the gravitomagnetic moment" is incorrect as
a Google search for "gravitomagnetic dipole moment" and "angular momentum" definitely
equates the two.

arxiv.org/abs/gr-qc/0304084
(click on .pdf to see the entire article)

The following contains an introduction to gravitomagnetism in curved spacetime:
arxiv.org/pdf/quant-ph/0301095

L.A.

#### pervect

Staff Emeritus
My favorite article on gravitomagnetism which unfortunately is not AFAIK available on the internet for free is:

"Anology between general relativity and electromagnetism for slowly moving particles in weak gravitational fields" by Edward G Harris.

It turns out you can make a very strong anaolgy between gravity and E&M for weak fields and low velocities.

mass density rho acts almost exactly like charge density rho
muss current rho*u (u=v/sqrt(1-v^2)) acts almost exactly like current density J

The result derived in this paper is a set of 3-vector quantites g and H, g is analogous to E, and H is analogous to B in Maxwell's equations. The formal results (eq 16a-16d + 5c) wind up being:

$$\nabla \cdot g = -4 \pi G \rho$$
$$\nabla \cdot H = 0$$
$$ \nabla \times g = -(\frac{1}{c}) \frac{\partial H}{\partial t}$$
$$\nabla \times H - (\frac{}{c}) \, \frac{\partial g}{\partial t} = (\frac{4 \pi}{c}) (- G \rho u})$$

and the equations of motion are

$$m\frac{d^2 x}{dt^2} = m(g+\frac{v \times H}{c})$$

These are *extremely* close to the cgs versions of Maxwell's equations,
compare to

http://scienceworld.wolfram.com/physics/MaxwellEquations.html

except for an odd factor of 2 and 4 here and there, which I've set off with brackets, and the minus signs. The minus signs account for the fact that the force between like charges is attractive, not repulsive - since the sign of the g-field is reversed from that of the E-field, the sign of the force is reversed as the force law is the same.

Of course terms of order (v/c)^2 or higher were dropped, this is linearized gravity.

It's also worth noting that g and H do not transform in the same manner as E and B under Lorentz boosts.

#### pervect

Staff Emeritus
Larry717 said:
The equation,

F = k P1 P2 / d^2,

contains fictitious sources that lead to a real force.
Apparently, from your quoted paper, P1 and P2 are NOT dipole moments, as I had assumed, but something roughly equivalent to charge.

#### Larry717

Inverse Square Law

Previously, the force between the poles of two different bar magnets was given as:

F = k P1 P2 / d^2 (magnitude only).

The poles are fictitious because they are not independent entities. It was shown
that like magnetic poles, there are also (fictitious) gravitomagnetic poles. This is
drawn from the close analogy between electromagnetism and gravitomagnetism.

Coulomb defined a unit magnetic pole as one which repels an identical pole at a
distance of 1 cm in a vacuum with a force of 1 dyne.

Newton's gravitational law and Coulomb's law for charges are both inverse square
laws. Inverse square laws usually mean that the line connecting two bodies is from
the center of one to the center of the other.

While the force equation above is inverse square, the line connecting the poles is
from edge to edge. This is the case because each pole is precisely at the end of
a bar magnet. The pole is not the bar magnet; the line connecting poles does not
connect the centers of two bar magnets:
epsc.wustl.edu/classwork/ 454/pdfs/lecture_white_magnetic.pdf
(Refer to pp.2,3 especially fig.2)

Given the above, what does it mean when two opposite poles become coupled?
The distance between them seems to be zero; the force appears to be infinite.
But this can't be. What is the remedy?

L.A.

#### Creator

Larry717 said:
.

The purpose of my presentation was to explain in a simple and compact way the
real force between fictitious gravitomagnetic poles.
I think you did a good job answering Pervect's objection that a stationary test mass would have no gravitomagnetic force acting upon it in a gravitomagnetic field, (only gravitoelectric).
His initial objection is correct since the gravitomagnetic equations show terms analogous to the EM Lorentz force (involving qv X B).

However having a spinning test mass circumvents this objection by providing a gravitomagnetic dipole-dipole interaction, which should provide force even if the test mass is originally not moving through the gravitomagnetic field.

I'm not sure exactly what you are getting at in your last post. Maybe you can clarify.
P.S.Typical derivation shows that the gravitomagnetic field falls of as the inverse cube of distance; 1/r^3.

Creator

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#### Larry717

Looking for a Finite Force

For those who are new to my posts on magnetic poles and gravitomagnetic poles
please review them carefully along with the comments by other authors. I am not
going to summarize them here.

As I have shown before, the line of reasoning concerning magnetic poles can be
easily extended to gravitomagnetic poles. Right now, let's stay with the example of
bar magnets and their poles.

Coulomb's law for charges is inverse square and assumes a center to center line
between the charges.

The force law for magnetic poles (not dipoles) is also inverse square. But the line
connecting the poles is not from the center of one bar magnet to the center of
another. A pole only exists at the end of a bar magnet. So the line connecting two
poles is from the edge of one bar magnet to the edge of another.

Place two identical bar magnets colinear to each other. Have the north pole of one
joined to the south pole of the other.

For a separation of zero distance units between the poles, what force results? If
it is infinite, something is wrong. Is there a way to obtain a finite force when the
poles are joined together?

L.A.

#### Antiphon

Larry717,

I checked for your posts and journal but couldn't find your posts on gravitomagnetism.

In any case, bar magnets are dipoles, which is fine. But in any case the more direct
analogy to writing the equations of gravitiation as an electrodynamics-like formulation
would be to use draw the analogy between electric charge and gravitational mass.

If you do this then there are gravitational dipoles just like a bar magnet. They are
like "bar-gravnets". They would occur around any spinning mass in direct analogy to
the magnetic eqivalent current that circulates around the periphery of a bar magnet.

#### Larry717

Antiphon said:
Larry717,

I checked for your posts and journal but couldn't find your posts on gravitomagnetism.
My first post is titled, "Gravitomagnetic Poles" in the Gravitomagnetism thread. The second
is titled, "Plenty of Resources Available." If you continue through the thread you will see
some comments to my posts. I don't put the word Gravitomagnetism in the subject line.

You will often find the following equation in my posts:

F = k P1 P2 / d^2

This is the force between one pole on one bar magnet and another pole on another bar magnet.
Presently, I'm asking if there is a way to find a finite force between a coupled north pole and
south pole (i.e., d = 0).

L.A.

#### Larry717

Solution?

Finite force between poles

The force at zero distance between point poles would be infinite,
although all available bar magnets have some size and so you
can never get to zero distance between the poles of coupled bar
magnets. The force gradient between the poles flattens out as
the distance between the poles gets smaller instead of continuing
to obey the inverse square law.

L.A.

#### Meir Achuz

Homework Helper
Gold Member
"Presently, I'm asking if there is a way to find a finite force between a coupled north pole and south pole (i.e., d = 0)."

The force between two touching (d=0) bar magnets is (in Gaussian units).
F=2pi MM'A, where M and M' are the magnetizations, and A is the common touching area. If M and M' are in gauss, and A in cm^2, F will be in dynes.

#### Larry717

What about 0 < d < 1 ? (inverse square law)

The formula in Gaussian units describing the force between two
coupled bar magnets (d=0 between the opposite poles) is:

F = 2pi M M' A

The same formula can be converted to MKSA units, and after
the smoke clears, is:

F = k P1 P2 / A (d=0)

where A is the area of one face of one bar magnet. This is very
similar to the formula for d>0:

F = k P1 P2 / d^2

I have stated the definitions of the above quantities previously.
----------------------------------------------------------------------------------------

The maximum force is at (d=0). This force is constant until (d>1).
Finally, at large enough (d) the force reaches a minimum. This is
similar to the modified inverse square law for the intensity of light.

The interval (0,0) to (1,0) is not used in the modified inverse square
law for the intensity of light. It represents the finite size of the source.
The intensity decreases from a maximum value to a minimum value.

-----------------------------------------------------------------------------------------

#### Creator

pervect said:
The result derived in this paper is a set of 3-vector quantites g and H, g is analogous to E, and H is analogous to B in Maxwell's equations. The formal results (eq 16a-16d + 5c) wind up being:

$$\nabla \cdot g = -4 \pi G \rho$$
$$\nabla \cdot H = 0$$
$$ \nabla \times g = -(\frac{1}{c}) \frac{\partial H}{\partial t}$$
$$\nabla \times H - (\frac{}{c}) \, \frac{\partial g}{\partial t} = (\frac{4 \pi}{c}) (- G \rho u})$$

and the equations of motion are

$$m\frac{d^2 x}{dt^2} = m(g+\frac{v \times H}{c})$$

These are *extremely* close to the cgs versions of Maxwell's equations,.....
Of course terms of order (v/c)^2 or higher were dropped, this is linearized gravity.
These are, of course, the formal 'linearized' equations analogous to Maxwell, which are very helpful in describing the gravitoelectric (g) and gravitomagnetic (H) fields in weak field approximation.

However, if you don't mind the nit pick, your third eqn. [ltex] \nabla \times g = -(\frac{1}{c}) \frac{\partial H}{\partial t}[/ltex], is actually [ltex]\nabla \times g = 0[/ltex] when using only v/c terms. The [ltex]-(\frac{1}{c}) \frac{\partial H}{\partial t}[/ltex] term only appears when you do include the (v/c)^2 terms, but doing so also introduces other non-Maxwellian terms elsewhere in field eqns., and is one reason to consider the eqns. 'almost' exactly analogous to Maxwell's; (see Braginsky, Caves, & K. Thorne, Phys. Rev.D, 15, p.2047,(1977)). Just thought you ought to know. The other reason, as you indicated, is the factor of 2 and 4 . Some years ago this factor of 4 was believed to be due to the spin 2 field associated with gravity. However, I think it is merely another factor pointing to the appoximate nature of the linearized equation. In deriving the wave equations, the speed of the gravity wave (when leaving the factor of 4 in the strength of H) turns out to be only c/2....which is further indication that we are only dealing with order of magnitude estimates, provided we take the true value to be c from full GR analysis, which I think can be done.
(see Forward) :shy:

In any case; I'm glad you presented the equations here in differential form; a form which, to me, allows for easier mental manipulation and analysis, but only when coupled with two more eqns. of similar caliber.

I recommend you add to your dozier the definition of the gravitomagnetic (H) and gravitoelectric (g) fields in terms of the gravitomagnetic vector and scalar potentials, A and [ltex]\phi[/ltex], respectively, which are given below (and which are exactly analogous to the EM vector & scalar potentials).

$$g = \nabla \phi -(\frac{1}{2c})\frac{\partial A}{\partial t}$$

$$H = \nabla \times A$$

(Here, I'm only using the same [ltex]A[/ltex] and [ltex]\phi[/ltex] as in EM, but it should be understood they are not the same as in EM).

From these: 1. the wave equation is derivable (gravity waves),
and 2. other interesting facts can be easily disciphered;

For ex., the first eqn. shows clearly that a time rate of change in the gravito vector potential (A) results in Newtonian gravity, g.
Since the gravito magnetic field H, and thus its potential A, are created by mass currents, it is obvious that an abrupt change in the rotation rate of a spinning object creates a change in g in the space around the object.

Creator Last edited:

#### robphy

Homework Helper
Gold Member
A while back (post #3) , I posted a side-comment that spawned no discussion. I'll reproduce it here...in case anyone in this thread has some insight into it.

There is a curious equation in Hawking/Ellis (p. 85) that has always intrigued me. Sometimes these are called "quasi-maxwellian equations".
Hawking/Ellis said:
...the Bianchi Identities
$$R_{ab[cd;e]}=0$$
They can be rewritten as
$$C^{abcd}{}_{;d}=J^{abc}$$ (4.28)
where
$$J^{abc}=R^{c[a;b]}+\frac{1}{6}g^{c[b}R^{;a]}.$$ (4.29)
These equations are rather similar to Maxwell's equations in electrodynamics:
$$F^{ab}{}_{;b}=J^a$$
where $$F^{ab}$$ is the electromagnetic field tensor and $$J^a$$ is the source current. Thus in a sense one could regard the Bianchi Identities (4.28) as field equations for the Weyl tensor giving that part of the curvature at a point that depends on the matter distribution at other points.
I've been toying around with that J-tensor but haven't found a satisfactory physical and geometric interpretation for it. Has anyone enountered this J-tensor or the quasi-Maxwellian equations?

Note that the equations above are not approximations. They are algebraically related to the Einstein Equations. In addition, like the electromagnetic tensor $$F$$, the Weyl tensor $$C$$ can be decomposed into "electric" and "magnetic" parts using an observer's 4-velocity.

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#### pervect

Staff Emeritus
Creator said:
However, if you don't mind the nit pick, your third eqn. [ltex] \nabla \times g = -(\frac{1}{c}) \frac{\partial H}{\partial t}[/ltex], is actually [ltex]\nabla \times g = 0[/ltex] when using only v/c terms. The [ltex]-(\frac{1}{c}) \frac{\partial H}{\partial t}[/ltex] term only appears when you do include the (v/c)^2 terms, but doing so also introduces other non-Maxwellian terms elsewhere in field eqns., and is one reason to consider the eqns. 'almost' exactly analogous to Maxwell's; (see Braginsky, Caves, & K. Thorne, Phys. Rev.D, 15, p.2047,(1977)). Just thought you ought to know. I didn't know - thanks! That explains why I had to transplant equation 5c in my reference to get the complete set of Maxwell's equations :-).

I recommend you add to your dozier the definition of the gravitomagnetic (H) and gravitoelectric (g) fields in terms of the gravitomagnetic vector and scalar potentials, A and [ltex]\phi[/ltex], respectively, which are given below (and which are exactly analogous to the EM vector & scalar potentials).

$$g = \nabla \phi -(\frac{1}{2c})\frac{\partial A}{\partial t}$$

$$H = \nabla \times A$$

(Here, I'm only using the same [ltex]A[/ltex] and [ltex]\phi[/ltex] as in EM, but it should be understood they are not the same as in EM).

From these: 1. the wave equation is derivable (gravity waves),
and 2. other interesting facts can be easily disciphered;

For ex., the first eqn. shows clearly that a time rate of change in the gravito vector potential (A) results in Newtonian gravity, g.

Since the gravito magnetic field H, and thus its potential A, are created by mass currents, it is obvious that an abrupt change in the rotation rate of a spinning object creates a change in g in the space around the object.

Creator [/quote]

Aha - I think I may be getting some insight as to how some of Forward's gravitomagnetic gravity proposals work, thanks to your comments.

#### Larry717

For Easy Reference

APPENDIX-- SI (MKS) Dimensisons of the Gravitomagnetic Field.

Mass Current = Kg/Sec = (Weber/Meter)(Coul/Meter)

Gravitomagnetic Dipole Moment = (Kg)(Meter-Squared)/Sec
= Angular Momentum
= (Coulomb)(Weber)

Gravitomagnetic Charge = (Velocity)(Meter) = Square-Meter/Sec

Gravitomagnetic Field = (Mass Current)/Meter = Kg/Sec-Meter
= ((Kg)(Meter^2)/Sec)/Meter^3
= Spin Density = Angular Momentum/Cubic-Meter
= (Coulomb)(Weber)/Cubic-Meter

Gravitomagnetic Flux Density = (Gravitomagnetic Charge)/Meter^2
= Velocity/Meter
= 1/Sec = Angular Velocity

Gravitomagnetic Vector Potential = (Gravitomagnetic Charge)/Meter
= Velocity = Meter/Sec

Gravitomagnetic Permeability = Gravitomagnetic Flux per Gravitomagnetic Field
= Meter/Kg

Assuming Transverse Gravitational Waves Propagate at Light Speed --
= 1/(c-squared)(epsilon0)
= 9.316E-27 Meter/Kg

L.A.

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