Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Does gravity field really exist?

  1. Nov 28, 2014 #1

    ABW

    User Avatar

    In GR gravity is associated with the curvature of space caused by the presence of massive bodies.
    Along with GR, there is an approach in which the gravitational field
    - is a real physical field, whose properties can be described by equations,
    similar to the Maxwell equations for electromagnetic field.

    Differential equations for the gravitational field
    Part 1. Maxwell's equations for the electromagnetic field can be supplemented by the differential equations for the gravitational field :
    6575583.jpg
    The second equation describes the sources of the gravitational field: It is a common substance with density "Rho" , the electromagnetic field with intensity E and H , and the gravitational field strength g , ( g is the dimension of acceleration ), p is the pressure . The fact, that the gravitational field as any kind of matter, can be a source of gravity , while nobody was paying attention . G is Newton's constant . c - is the speed of light .
    From equations we find formula for the energy density of the gravitational field :
    6414645.jpg
    Unexpected in this expression is that the gravitational constant G is in the denominator .
    Compare this expression with the energy density of electric and magnetic fields in vacuum:
    6614494.jpg
    So for the gravitational field at the surface of the Earth g = 9.81 m/sec2 , we find :
    W ~ - 10^11 Jowl/m3 . The sign “-” in the formula for the energy density of the gravitational field indicates that the energy density, and the corresponding energy of gravitational interaction is negative.

    It is easy to give a formula for refined Newton's law, taking into account, that the gravitational field is also a source of gravitation :
    6408501.jpg
    M is the mass of the body , Ro is the radius of the body.

    The gravitational field at large distances is higher compared to the classical value.
    This formula can in our opinion to explain why the space station “pioneer-10 ”, NASA , currently outside the Solar system slows down faster than it should of calculations . Here is what says about this journal Relcom.ru : “Movement "Pioneer 10" in the space of interested scientists since it was discovered that the observed deceleration cannot be explained alone by the gravitational pull of the Solar system. It can serve as evidence for the existence of an as yet unknown to science forces, or associated with any properties of the spacecraft.”
     
  2. jcsd
  3. Nov 28, 2014 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    This is not a different theory from GR; it is just a different way of expressing the theory of GR--more precisely, of GR in the weak field approximation. At least, that's what it looks like given the equations you've written down.

    If you're referring to something else, something which is actually a different theory from GR (i.e., which makes different experimental predictions), then please give a reference.
     
  4. Nov 28, 2014 #3

    ShayanJ

    User Avatar
    Gold Member

    Take a look at here! Pioneer's anomaly is solved now.
     
  5. Nov 28, 2014 #4

    Jonathan Scott

    User Avatar
    Gold Member

    I haven't checked the rest, but the Newtonian expression for the energy of the gravitational field is missing a factor of ##4 \pi##. The standard result is as follows:

    $$-\frac{g^2}{8 \pi G}$$
     
  6. Nov 28, 2014 #5

    Jonathan Scott

    User Avatar
    Gold Member

    It's unfortunately nowhere near that simple.

    The parameterised post-Newtonian (PPN) approximation used to analyse the effects of relativistic gravity in the solar system already uses a correction to the potential for non-linearity of the form ##(1 - \beta \, Gm/2rc^2)## where GR predicts ## \beta = 1 ## and that result has been confirmed to good accuracy by observations of the perihelion precession of Mercury.

    The gradient of the dimensionless scalar potential (the time dilation factor) is therefore approximately:
    $$- \left ( 1 - \frac{Gm}{rc^2} \right ) \frac{Gm}{r^2 c^2}$$
    This is approximately ##\mathbf{g}/c^2##. In coordinate acceleration units for isotropic coordinates (where the coordinate speed of light is the same in all directions), velocities scale as the square of the time dilation and the approximate result is as follows:
    $$\mathbf{g} = - \left (1 - \frac{Gm}{rc^2} \right )^5 \frac{Gm}{r^2} = - \left (1 - 5 \frac{Gm}{rc^2} \right ) \frac{Gm}{r^2}$$
    In addition, it is not thought that any modification to the law of gravity based only on radial distance could explain the Pioneer anomaly, because there is no evidence of any similar effect on planets at the same distance. Any modification would need to rely on something unusual about Pioneer, such as the fact that its motion is in a more radial direction than that of planets.

    Anyway, the Pioneer anomaly has now been satisfactorily explained as an effect of thermal recoil, due to the reflection of heat radiation from the spacecraft structure.

    [Edited to correct r to r^2 in gradient of potential]
     
    Last edited: Nov 28, 2014
  7. Nov 28, 2014 #6

    ABW

    User Avatar

    In the system of Gauss W is:
    $$-\frac{g^2}{8 \pi G}$$
    In the SI system W is:
    $$-\frac{g^2}{2 G}$$
     
  8. Nov 28, 2014 #7
    Just a little more about the question you asked in the title. According to Einstein's GR, the gravitation field of heavy bodies really exists: it's a real physical field, or in other words, a zone of influence around such bodies. The way the gravitational field affects their motion is described by so-called field equations.
    Compare the introduction here and you will notice some similarities with your first post: https://en.wikipedia.org/wiki/Einstein_field_equations
     
  9. Nov 28, 2014 #8

    Jonathan Scott

    User Avatar
    Gold Member

    No, I'm sure that's wrong. Have you tried the integration yourself?

    (And I'm not aware of any gravitational equivalent of Gaussian units).
     
  10. Nov 29, 2014 #9

    Jonathan Scott

    User Avatar
    Gold Member

    The electrostatic force constant (using SI conventions) is ##1/4\pi\epsilon_0## and the gravitational force constant is ##G##. From straight substitution this makes ##4\pi\epsilon_0## equivalent to ##1/G## so ##\epsilon_0## is equivalent to ##1/4\pi G##.

    If you integrate the interaction term in ##\mathbf{g}^2## between the fields due to two masses, that is ##2 \mathbf{g}_1.\mathbf{g}_2##, over all space, the result is ## - 8 \pi G ## times the (negative) potential energy between those two masses, so this is what you have to divide by in order for the integral to be equal to the potential energy.

    A better semi-Newtonian model for gravitational energy distribution is to assume that the energy of each source mass is reduced by the time dilation effect of the potential of the other source, which reduces its energy by the total potential energy (so the total energy for the two masses together is reduced by twice the potential energy), and then the energy of the field in space is added back in with a positive sign. This then gives a continuous positive energy distribution which is overall equal to the energy of the original source masses minus the potential energy of the system.
     
  11. Nov 29, 2014 #10

    ABW

    User Avatar

    Well, may be I was wrong with a factor of ##4 \pi##, it is not difficult to check and correct.
    More interesting to discuss the above equations in the absence of electric and magnetic fields.
    $$\mathbf{rotg} = 0$$ $$\mathbf{divg} = - 4πGρ + \frac{g^2}{2c^2}$$
    Or in other words, whether the assumption that the gravitational field itself can be a source of gravity, is correct .
     
    Last edited: Nov 29, 2014
  12. Nov 29, 2014 #11

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    The answer to this is "it depends"--it depends on how you interpret the term "source of gravity".

    In the Einstein Field Equation, which is the central field equation of GR, the source is the stress-energy tensor, which does not include any contribution from the "gravitational field". (The equations you are writing down are just restatements of the EFE in a particular coordinate system for a particular kind of stress-energy tensor, in the weak-field approximation.) So on this interpretation of "source", the answer is no, the gravitational field is not a source of gravity.

    However, the EFE is nonlinear, which means you can't superpose two solutions to get a third solution. (Maxwell's Equations for electromagnetism, by contrast, are linear.) So if we think of the gravitational field when two gravitating bodies are present, it isn't just the sum of the fields from each body alone; there is an additional contribution which can be interpreted as due to the "gravitational field" acting as a source. (Another manifestation of this nonlinearity is that gravitational waves in GR carry energy.) So on this interpretation of "source", the answer is yes, the gravitational field is a source of gravity.

    The real answer is that whether or not the gravitational field is a source of gravity is not an assumption of GR at all. GR is based on different assumptions; the term "gravitational field" is not among them, and you don't need to find an interpretation for the term "gravitational field", or decide whether or not it's a source of gravity, to solve problems in GR.
     
  13. Nov 29, 2014 #12

    Jonathan Scott

    User Avatar
    Gold Member

    In Newtonian terms, potential energy doesn't just appear or disappear when a system of masses is changed to move masses closer together or further apart. It has to be converted to or from some other form internally, such as kinetic energy, or added or removed externally. This suggests that "gravitational energy" isn't really a separate stuff, but rather an accounting term for energy transferred by the mechanism of gravity.

    In a semi-Newtonian approach, where energy is conserved in the conventional way, it is clear that the effective source mass of a system must effectively take into account potential energy, as an adjustment relative to the energy that the same source masses would have separately, and the obvious adjustment is that the time dilation factor due to the potential of other masses. However, as I've previously mentioned, if the potential energy correction is applied to each mass separately, this reduces the energy of the system by twice the potential energy, but the energy of the field can be added back in to get the correct total energy. This is very similar to the model of electromagnetism where the Poynting Theorem shows how the flow of energy and momentum is conserved.

    However, when one starts looking the sort of tiny corrections to gravity which this scheme might produce, it is immediately clear that they are so small that one would also need to take into account other GR aspects such as the curvature of space, and we already know that GR predictions match experiment very precisely at least in the solar system. This means that there isn't any obvious reason to try to stretch Newtonian theory into this area.

    Personally, I like to look into the relationship between Newtonian and GR viewpoints of gravity, as I feel it helps me to gain a more intuitive understanding.
     
  14. Nov 29, 2014 #13

    Dale

    Staff: Mentor

    Well said.
     
  15. Nov 30, 2014 #14

    Jonathan Scott

    User Avatar
    Gold Member

    I've just spotted a MNRAS paper "Gravitational field energy density for spheres and black holes" from 1985 by D Lynden-Bell and J Katz which takes the GR Schwarzschild solution (for the static spherical case) and applies the model where the effective energy density of the field is ##\mathbf{g}^2/8\pi G## (positive) and the effective energy of the central mass is reduced by the time dilation (which as they point out reduces it by twice the potential energy). In this model the entire energy of a black hole resides in the field outside the event horizon!

    This is exactly the same model that up to now I've only ever seen discussed in Newtonian terms. They even use isotropic coordinates (which I've always preferred to Schwarzschild).

    It is however rather weird that in this model the location of the "effective energy" is clearly not the same thing as the source of the field!
     
  16. Nov 30, 2014 #15

    bhobba

    Staff: Mentor

    Interesting question.

    Without delving into it you will find the following book by Ohanian explores it very thoroughly:
    https://www.amazon.com/Gravitation-Spacetime-Hans-C-Ohanian/dp/1107012945

    I do not 100% endorse his views - I do not agree that tidal forces are always the give away of gravitational forces vs acceleration. It was the first serious book I learned GR from before moving onto my bible - Wald.

    But on these type of issues it really has no equal.

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
  17. Nov 30, 2014 #16

    bhobba

    Staff: Mentor

    Not so fast Kemosabe :p:p:p:p:p:p:p:p

    I didn't want yo get into it because the book I pointed to by Ohanian examines the issue in detail - but it's not quite that simple.

    You can model gravity in exactly the same way as EM and you get linearised gravity which actually is perfectly OK in explaining many phenomena.

    The issue though is it contains its own destruction and inevitably leads to GR. First, it can be shown, particles do not move in flat space-time - but rather as if space-time had an infinitesimal curvature. Most importantly however it ignores the issue of gravity gravitating so it must be non-linear.

    The interesting thing however is GR is rather strange, because, as Ohanion shows, the linear equations imply the non linear ones and you get full GR.

    This is also related to the issue - is space-time curved or does the gravitational field simply make clocks and rulers behave as if it was curved. Experimentally there is no way to tell the difference.

    BTW my bible on GR is Wald - do you can guess which side on that debate I come down on - but it's debatable 100% for sure.

    Thanks
    Bill
     
    Last edited by a moderator: Dec 1, 2014
  18. Nov 30, 2014 #17

    bhobba

    Staff: Mentor

    Including gravity itself - which is why the full equations of GR are non-linear.

    Thanks
    Bill
     
  19. Nov 30, 2014 #18

    ShayanJ

    User Avatar
    Gold Member

    Sorry, but that doesn't make sense to me!
    Considering that possibility means accepting that gravitational fields know how to make our instruments to go wrong. But because we have different methods for time and distance measurements, it means gravity knows how to deal with different types of measurement devices (which means "he" knows the mechanisms we use!!!) and, because we can still make new devices, it means gravity can learn how to fool our new instruments too. But...come on...who are we talking about here? Frank Abagnale?
     
  20. Nov 30, 2014 #19

    bhobba

    Staff: Mentor

    Weird hey.

    It's more reasonable to assume its curved.

    The guy who first clued me into this stuff was Steve Carlip when he posted a lot on sci.physics. relativity. He is no crank by a long shot.

    Thanks
    Bill
     
  21. Nov 30, 2014 #20

    ShayanJ

    User Avatar
    Gold Member

    Yeah...that's also much more beautiful!
     
    Last edited: Nov 30, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Does gravity field really exist?
  1. Gravity does not exist! (Replies: 14)

Loading...