A A very peculiar emergent definition of gravity

arivero
Gold Member
Messages
3,481
Reaction score
187
Newtonian gravity is the force law such that for any mass M, the circular orbit of any test particle m at a distance equal to the Compton length of M has the same areal speed, independent of M and m.

I wonder, have you seen this sort of definition online in the literature? I guess that independency of m is used in most arguments, but independency of M is less usual.
 
  • Like
Likes haushofer
Physics news on Phys.org
arivero said:
Newtonian gravity is the force law such that for any mass M, the circular orbit of any test particle m at a distance equal to the Compton length of M has the same areal speed, independent of M and m.
Where are you getting this from? Do you have a reference? I'm thinking not, since you say:

arivero said:
have you seen this sort of definition online in the literature?
Please note that personal speculation is off limits here.
 
arivero said:
the old exercise we did here
That was 20 years ago and our rules on speculation, even in the BTSM forum, are much stricter now than they were then.

Thread will remain closed.
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

Similar threads

Back
Top