rest mass

[AntonioLao]

Physics defines rest mass of an object as that measured by a relatively motionless observer. In this sense rest mass in no way implies absoluteness. On the other hand, since absolute motion is almost impossible to measure, the best surmise is that rest mass indicates parallel displacements of equal absolute magnitudes between object and observer. However, in a universe of positive or negative curvature, parallel displacements are impossible without dimensional extension. Fortunately, these curvatures could be demonstrated effectively within localized massive objects such as stars, galaxies, and black holes. Globally, the cosmological flatness problem would not go away. This calls for a more plausible spatiotemporal theory of perfect symmetry.

[mkirkpatrick]

If I understand this correctly there cannot be any mass that is at rest within this universe
in other words motionless,that can only exist as potential in an unexpressed universe,is that not so?


regards michael.

[AntonioLao]

The entire universe is at rest therefore only the universe has a rest mass.

[mkirkpatrick]

The entire universe is at rest therefore only the universe has a rest mass.

Well then what does this tell us?



regards michael.

[AntonioLao]

Well then what does this tell us?

The total mass of the universe.

[mkirkpatrick]

The total mass of the universe.

Rest the way it is used in this context,is really though a misnomer,as all mass vibrates
at a very high frequency,it is hardly resting,in the literal sense!



regards michael.

[AntonioLao]

as all mass vibrates at a very high frequency,it is hardly resting,in the literal sense!

This is the distinction between angular frequency and linear frequency. In physics, the distinction is a factor of 2 pi. The angular frequency equals the product of linear frequency and 2pi.

[mkirkpatrick]

This is the distinction between angular frequency and linear frequency. In physics, the distinction is a factor of 2 pi. The angular frequency equals the product of linear frequency and 2pi.

I understand this to a point,it is just that it seems an inappropriate term to use,that's
all.



regards michael.

[AntonioLao]

I understand this to a point,it is just that it seems an inappropriate term to use,that's all.

In vector notation, I would express them as normal and tangential real components, a new upcoming thread title.