On Aug 28, 12:03 pm, Tom Roberts wrote:
> Sanny wrote:
>> What is the proof the Mass is always conserved?
>
> There can be no such proof, because mass is quite clearly not conserved
> for many known situations.
It depends on how you interpret the mass to be.
> Under certain general circumstances, energy is conserved, and momentum
> is conserved, and angular momentum is conserved, but not mass.
The conservation of energy is very much in the bag according to the
Noether’s theorem. However, the conservation of the angular momentum
is not a sure thing especially in a binary system.
>> Its just happens that mass do not get destroyed or created.
>
> This, too, is simply not true.
>
> The most obvious counterexample is the reaction:
>
> \gamma + A => A + e- + e+
>
> The \gamma on the left has zero mass, but the e+ and the e- on the right
> each have a clearly non-zero mass of 511 keV/c^2.
Since gamma rays have momentums, the most appropriate interpretation
is to model them with mass where this mass is a function of speed. Of
course, anyone can choose to define the mass only at rest. In doing
so, you are just making a deeper fool of yourself in the long run.
> There are literally thousands of other examples -- just look up ANY
> chemical or nuclear reaction, and with sufficient accuracy you'll find
> the masses on the two sides are not equal. Nuclear reactions typically
> have a mass difference on the order of a few MeV/c^2; chemical reactions
> typically have a mass difference on the order of a fraction of an
> eV/c^2, and it is exceedingly difficult to measure such small
> differences. The mass deficits in nuclear reactions are well established.
All these thousands of examples can easily be demystified if the mass
is defined as energy divided by the speed of light squared.
Why have you guys never thought about that? Yes, mysticism is the key
to your survival.
>> In Nuclear Fusion Mass is converted into energy.
>> E= m* c2;
>> I think this do not violate that principle of Conservation of Mass?
>
> There is no such principle. This example CLEARLY shows that mass is not
> conserved (it's converted into energy, which is not mass).
>
> Energy is conserved in nuclear fusion, but that's a different topic.
No, it is not a different topic. Your interpretation allows you to
flip-flop to the winning side whenever the situation comes into your
favor, and this is not science. The only interpretation to mass is,
once again, energy divided by the speed of light of the observer.
>> People say [...]
>
> People say lots of things. Many of them are not true.
Please stay with the topic. People (most physicists) say the only
interpretation to mass is the rest mass. That is perfectly OK.
However, in doing so, you are just adding more problems for
yourselves. Once again, the best way to interpret what mass is is
proportional to energy. Msss is energy, and energy is mass. Believe
me. In that case, mathematics get a lot simpler.
For example, the Schwarzschild metric demands the conservation of
energy. You (plural) just would not go there because interpreting
mass solely as the rest mass does not allow you to do so. In doing
so, you have no explanation why energy is conserved after declaring
that no such entity as the potential energy. From the geodesic
equations, the energy of an object with rest mass of m under the
curvature of spacetime manifested with mass M (add all other
requirements such as M >> m, etc) can be expressed as follows.
E = m c^2 sqrt(1 – 2 U) / sqrt(1 – B^2)
Where
** U = G M / c^2 / r
** B^2 c^2 = (dr/dt)^2 / (1 – 2 U)^2 + r^2 (dO/dt)^2 / (1 – 2 U)
** dO^2 = dLongitude^2 cos^2(Latitude) + dLatitude^2
The conservation of the overall mass, m sqrt(1 – 2 U) / sqrt(1 – B^2),
results in the conservation of energy as well. Under very special
circumstance such as weak gravitation (1 >> 2 U) and low speed (1 >>
B^2), the above equation reduces into the Newtonian equation
describing the conservation of energy in orbital mechanics below.
E – m^2 c^2 = m B^2 c^2 / 2 – m U c^2
Where
** m B^2 c^2 / 2 = Kinetic energy
** m U c^2 = Potential energy
> Timberwoof said:
>> There is no "proof", but beginning with Lavoisier, a French tax
>> collector who applied strict principles of accounting to his
>> observations of chemical reactions, the observationthat under ordinary
>> conditions, mass is neither created nor destroyed, has always been
>> confirmed.
>
> His measurements were insufficiently accurate to detect the mass
> differences. Indeed even today it is quite difficult to measure the mass
> deficits of chemical reactions; general theoretical arguments show they
> must be present (molecules are bound). Particle physicists measure mass
> deficits in nuclear (and sub-nuclear) reactions all the time.
Give me a break. Justifying your interpretation of mass based on a
crude 19th century (or earlier) is totally groundless.
> The Timelord said:
>> [proof mass is conserved] comes from the continuity equation [...]
>
> That equation is valid only in classical mechanics. It is not valid in
> relativity, nor in the world we inhabit.
This is nonsense. The Euler-Lagrange equation (geodesic equation)
associated with spacetime shows so time after time that the energy
must be conserved according to Noether’s theorem. This, of
course, includes your binary systems which you (plural) have claimed
gravitational radiation without trying to justify, qualify, and
understand the mathematics of geodesic variations.
> Mass conservation remains APPROXIMATELY valid in many circumstances,
> including our everyday lives. In certain circumstances it is possible to
> have mass conservation: if one had a truly closed system then the mass
> of the system as a whole would be constant. This is not possible in the
> world we inhabit (radiation always leaks in or out). In many cases,
> however, the mass transfer due to the radiation is below one's
> measurement resolution, and for most practical purposes the mass of the
> system can be considered to be conserved.
Any Lagrangian that you can derive and qualify does not indicate a non-
conservation of energy. Just because you (plural) can squeeze out a
wave equation from the crack in the geodesic variation equations, you
still have to show a violation in Noether’s theorem to justify your
gravitational radiation.
