# Is their a mass to gravity proportion?

Discussion in 'Physics & Math' started by too-open-minded, Jun 10, 2012.

1. ### hansdaValued Senior Member

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In Newtonian Model, Gravitational Force is a function of mass and distance. So F=f(m,r).

3. ### Confused2Registered Senior Member

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The thread seems to lack any attempt to explain the concept of a field ... in this case a gravitational field caused by a mass exists whether or not another mass is present.

Last edited: Nov 16, 2017

5. ### DaveC426913Valued Senior Member

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Gary, your two posts did not parse correctly; they are unreadable. Did you have something to say, and did you want some guidance about the quote feature?

7. ### QuarkHeadRemedial Math StudentValued Senior Member

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This is perhaps the because concept of a field is context-dependent.. For example, in abstract algebra, a field is defined as a set equipped with two closed operations (usually denoted $+$ and $\times$) each of which has an inverse, respectively $-x$ and $\frac{1}{x}$ and an identity such that $x+(-x)=0$ and $x \times \frac{1}{x}=1$.

The definition of a field in linear algebra is the assignment to every point in some chosen space (usually a manifold) a scalar, a vector or a tensor.

Of course, to a mathematician, this is somewhat problematic - how exactly is this assignment made? Specifically, it invokes the infamous Axiom of Choice, which mathematicians try to avoid where possible. There is a solution, but that is not relevant here

8. ### Confused2Registered Senior Member

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Great post. A shame you chose not to continue. I think what you are trying to say is that when a tree falls in the forest and there is no one there to hear it fall - it still falls. Likewise the gravitational field created by a mass exists whether or not there is another mass in the vicinity.

9. ### DaveC426913Valued Senior Member

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This pretty much goes without saying.
Unless you somehow have a zero-dimensional point source of mass, that gravitational field is exerted by a mass that is of non-zero extent in 3 spatial dimensions. It is attracted to itself, even if nothing else. (Every atom of Earth attracts every other atom).

10. ### QuarkHeadRemedial Math StudentValued Senior Member

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Er no. I was waiting for a response, and I thank you for yours
Well it is far "worse" than that - to a mathematician, the gravitational field exists even in the absence of any mass. I argue as follows.......

Physics assumes the presence of a metric field over spacetime. Experiment (gravitational redshift, for example) shows that this field is not constant in the presence of a source (mass-energy) i.e. the metric tensor is different at every spacetime point.

Now the curvature field is defined as the 1st derivative (with respect to the coordinates) of the connection, which is itself the 1st derivative of the metric (also with respect to the coordinates), hence the curvature field is 2nd order derivative of the metric field.

Now it is an elementary fact from the differential calculus that if the 2nd derivative of a function is constant, the function is described by a straight line. In the present context, this means that if and only if the metric field is constant, then spacetime is in some sense globally "flat" or, if you prefer, strictly Minkowskian.

Finally, stare awhile at the gravitational field equations as written down by Einstein in 1915, namely
$T_{jk}=R_{jk}-g_{jk}R$ where the LHS is the source, the 1st term of the RHS is the curvature, the metric is $g_{jk}$ and $R$ is the trace of the curvature (in its matrix representation). Since the curvature field depends on the metric field, and the curvature scalar (the trace) depends on the curvature itself, you should be able to convince yourself that, from my argument above,

a) an everywhere zero curvature does NOT imply an everywhere zero metric

b) a constant metric implies the absence of a source

c) the absence of a source does NOT imply the complete absence of a metric field.

d) since everything in the theory depends on the metric, you may as well call it the gravitational field (but it's probably better if you didn't).

This as nice as it could be, since the French mathematician Pierre de Laplace showed in the 18th century that the divergence of the gradient of a scalar field being zero implies the absence of a source, and conversely, but NOT the absence of a scalar field. He writes, for the scalar field $\phi$, that $\del^2\phi =0$ implies ONLY the absence of a field source

11. ### Confused2Registered Senior Member

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Thank you Quarkhead - better than I expected (or deserved).
Is it possible we are using 'field' in different ways? By analagy with electrostatics the field (my field anyway) is dV/dx where V is the potential (a scalar) which is (by definition) the work done moving a unit charge from out point of interest to infinity - for convenience (and only convenience) we assume the potential at infinity is zero and the work done in the absence of any sources will be zero. Your field looks very much like my potential where dQuarkfield/dx = Confusedfield where your field is a scalar and mine is a vector.

The enquirer is most likely to encounter 'gravity' or the Earth's gravitational field or 'g' expressed in units m/s/s as a vector rather than a scalar (J/kg). In the case of the Higgs 'field' I would certainly accept your definition of field and I'm sure your useage is generally correct but in the local context of this thread I stand by my use of 'field' as appropriate for the context.

12. ### Confused2Registered Senior Member

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499
Oops - from memory field is probably -dV/dx ...