Genady

\(\cos(\frac {\pi} {4})=\frac {1} {\sqrt{2}}\) \(\cos(\frac {\pi} {5})=\frac {1+\sqrt{5}} {4}\) \(\cos(\frac {\pi} {6})=\frac {\sqrt{3}} {2}\) Is cosine of any fraction of π an algebraic number?

-4 - 4 = -8 -4 - 3 = -7 -4 - 2 = -6 -4 - 1 = -5 -4 - 0 = -4 -4 - -1 = -3 -4 - -2 = -2 -4 - -3 = -1 -4 - -4 = ?

I think, it is. Two observers measure distance between two events. Per the model, they count number of nodes in the graph on the path which connects two events. Do they count different number of nodes?

Does the graph distance depend on observer like the spatial distance does?

a=1.8954942670...

What is a justification of identifying the graph distance with a spatial distance? Is it a postulate of the model?

This is the same as the hyperlink in the OP: Anyway, is this model falsifiable?

Here is some: Claims in 'Duty to Warn' Letter to Harris Alleging Compromised Election Are Misleading | Snopes.com

Misconceptions.

They are entangled.

@dqqd, this analysis in Wikipedia is quite clear: Variable-mass system - Wikipedia

I disagree: the force has not been redefined. In the formula \(F=\frac {dP}{dt}\), \(P\) is the momentum of the entire system. If the system is composed of parts, \(P=P_1+P_2+...\), then \(F=\frac {d(P_1+P_2+...)}{dt}=m_1 \frac {dv_1}{dt}+v_1 \frac {dm_1}{dt}+m_2 \frac {dv_2}{dt}+v_2 \frac {dm_2}{dt}+...\), where no mass crosses the boundary of the composed system.

Force, however, does not depend on the inertial frame you choose.

Morning with the family

But in the OP, v is not "the velocity of the ejected mass relative to the bulk mass," but rather (my emphasis), It depends on reference frame.

There is no rocket in my scenario. My scenario demonstrates that by applying the quoted formula wrongly, we get a ridiculous result that a force on the bucket is proportional to velocity with which an observer runs by it.

Remove the bucket and consider a ball of water in free fall. The molecules leave in all directions. The total force on the ball is zero.

Consider a bucket of water with mass \(m\) and the water evaporating at rate \(\frac {dm}{dt}\). No force is applied to the bucket. Consider now the same bucket in a reference frame where it moves with velocity \(v\). If we apply the formula "F=dp/dt=mdv/dt+vdm/dt" in this reference frame, we get a ridiculous result that there is a force \(F=v \frac {dm}{dt}\) on the bucket. Conclusion: this is a wrong way to apply the above formula.

Local inhomogeneities and anisotropies do not matter on cosmological scales. The cosmological principle's assumption is that on some large scale, currently about 100+ Mpc AFAIK, the universe is homogeneous and isotropic. All manifestations of the DE so far are on such scales.

If it does, the cosmological principle needs to be reconsidered.

That is perhaps what he did and now tries to convince himself that he should stop feeling guilty. In his own words,

The Lizard.

Breakfast

They cannot land as anything else just by the fact that they are primes.

That will be good. I don't want to die young, healthy, and/or dirty.