# Dark Matter and Dark Energy

Dark Matter, Dark Energy.

Is it possible that dark matter and dark energy are illusions caused by the fact that the equations of physics are only approximations? For example, in the falling body equation, F = mg, F (the initial weight of an object) is assumed to be constant, but in fact the closer to the accelerating mass (e.g., the Earth) the greater F is. The mass, m, in the equation is also not a constant, because the farther an object is from the accelerating body (e.g., Earth) the less mass an object has. And finally, the acceleration constant, g, is not constant because it changes with time, and therefore it is actually an accelerating acceleration, which is abhorrent to physicists because it implies the concept of force per time, which implies that as time approaches zero, force approaches infinity, which means that there is no such thing as zero time, for any action-reaction, including the Big Bang. The "zero now" does not exist. Furthermore, even the universal gravitational constant, G, is really not a constant because the masses of attracting objects change with the distance between them. Of course the errors due to the fact that F, m, g, and G are not constant are normally insignificant, but are they insignificant on the cosmic scale, like when we calculate the mass of a galaxy? That is my question.

Dale
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Is it possible that dark matter and dark energy are illusions caused by the fact that the equations of physics are only approximations?
It is possible, but so far we don't have any better approximations.

The mass, m, in the equation is also not a constant, because the farther an object is from the accelerating body (e.g., Earth) the less mass an object has.
That is not true.

And finally, the acceleration constant, g, is not constant because it changes with time, and therefore it is actually an accelerating acceleration, which is abhorrent to physicists
There is nothing even mildly "abhorrent" about an accelerating acceleration. It is easily handled, although you may wind up with numerical solutions instead of closed-form solutions.

because it implies the concept of force per time, which implies that as time approaches zero, force approaches infinity, which means that there is no such thing as zero time, for any action-reaction, including the Big Bang. The "zero now" does not exist.
This seems to be completely disconnected with anything I recognize from actual physics. Please avoid personal speculation.

mfb
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Furthermore, even the universal gravitational constant, G, is really not a constant because the masses of attracting objects change with the distance between them.
It does not. Or at least no experiment ever found a deviation. What you suggest is inconsistent. A mass is a property of an object. There is no "mass of earth for the sun", "mass of earth for Jupiter", ...

Thanks for the comments. Regarding mass variance in gravitational fields: Clocks on Earth run slower than clocks in outer space, and meter sticks are not the same, so I assumed that due to general relativity, there would be a slight mass difference that would depend on elevation. Am I wrong in that respect? I realize that the mass variance is not measurable between an object on the surface of Earth and the same object in outer space. I'm afraid I lost you with the implications of accelerating acceleration, so forget that.

Orodruin
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Regarding mass variance in gravitational fields: Clocks on Earth run slower than clocks in outer space, and meter sticks are not the same, so I assumed that due to general relativity, there would be a slight mass difference that would depend on elevation. Am I wrong in that respect?

Yes. Also, what a "meter stick" is is just that, a meter stick. A clock is also always a clock which runs at the appropriate speed. That it runs faster away from a gravitational source depends on the properties of space time, not on the clock's internal properties.

PeterDonis
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Clocks on Earth run slower than clocks in outer space

Yes; more precisely, if you carefully specify what "run slower" means, yes, there is an invariant sense in which this is true.

meter sticks are not the same

No; more precisely, there is no invariant sense in which this is true. You can adopt particular coordinates and put a particular interpretation on those coordinates that makes this appear to be true, but you don't need to adopt those coordinates or that interpretation to account for any actual observables. Local measurements will always show (ideal) meter sticks to be one meter long.

The invariant fact about space around a massive body is that spacelike slices of constant time are not Euclidean. (Even that is somewhat coordinate-dependent, since "constant time" is, but this particular family of spacelike slices can be picked out by an invariant property.) However, that is a global property, not a local one; it affects how many meter sticks it takes, for example, to stretch between two 2-spheres surrounding the central body, as a function of their difference in area. But none of that affects the length of individual meter sticks.

ChrisVer
Gold Member
Is it possible that dark matter and dark energy are illusions caused by the fact that the equations of physics are only approximations? For example, in the falling body equation, F = mg, F (the initial weight of an object) is assumed to be constant, but in fact the closer to the accelerating mass (e.g., the Earth) the greater F is. The mass, m, in the equation is also not a constant, because the farther an object is from the accelerating body (e.g., Earth) the less mass an object has. And finally, the acceleration constant, g, is not constant because it changes with time, and therefore it is actually an accelerating acceleration, which is abhorrent to physicists because it implies the concept of force per time, which implies that as time approaches zero, force approaches infinity, which means that there is no such thing as zero time, for any action-reaction, including the Big Bang. The "zero now" does not exist. Furthermore, even the universal gravitational constant, G, is really not a constant because the masses of attracting objects change with the distance between them. Of course the errors due to the fact that F, m, g, and G are not constant are normally insignificant, but are they insignificant on the cosmic scale, like when we calculate the mass of a galaxy? That is my question.

There is a slight possibility that the equations of physics we have are not working fine at some regime. Such a solution is proposed by the MOND theory, which can take for example $F=ma$ not working fine at that large scale. An easy example is to take instead the force as $F= f_0 a^2$ and show that this can give the appropriate rotational curves at the galaxy's halos. *In my opinion Comment* I don't really like that idea, for other reasons however (because I want for DM to exist due to particles, because I believe in the existence of more particles than already discovered).
F in $F=mg$ is not assumed to be constant in general. $g$ itself is changing, however the change can be really small for distances much smaller than the radius of the earth. So it's assumed to be constant for displacements that cannot be compared with the earth's radius, and not in general.
The m doesn't change. The weight does [because of the change of g].
What is the problem of a time dependent g? Force per time can exist in any kind of force. For example you can have time-dependent forces without a big deal. The infinitesimal time changes will just ask for infinitesimal forces, that's what you mean by $dF/dt \equiv lim_{\epsilon \rightarrow 0} \frac{F(t+ \epsilon) - F(t)}{\epsilon}$. Derivatives are working fine in the physical world...
The masses don't change. The forces do. And that is taken into account in writting $F \sim r^{-2}$. However in that is not even true in general... In GR objects don't feel "forces", the spacetime is changing because of the existence of mass and energy, and any test-mass will travel in a "free"-path that is determined by the geodesics (geometry of spacetime).