What is the relationship between dark energy and the inertia of baryonic matter?

[kurious]

Because dark energy and baryonic matter have mass densities of the
same order -
about 10^ - 27 kg / m^3, their total mass in the universe is similar.
This would be the case if dark energy caused the inertia of baryonic
matter,wouldn't it? (analagous to Mach saying that the distant stars
contribute to the inertia of local masses).
The total amount of dark energy in the universe increases with time,
but baryonic mass would not necessarily increase with time:
protons have masses that depend mainly on their gluons and if
these became less energetic as the total amount of dark energy
increased,
then the protons would have the same inertia overall.Can anyone think
of some evidence to support these ideas?

 

[Antonio Lao]

kurious,

You might already know it but allow me to say something about the concept of inertia.

Inertia is defined as the resistance sustained by an object to a velocity changing force. The effect is seen by the manifestation of acceleration.

Every object in the universe has linear momentum (Newton's 1st law of motion). There is a mass factor within the linear momentum. It is the tendency of an object to keep the mass constant when subjected to a velocity changing force that gives its definition of inertia.

F = \frac{dp}{dt}

where F is the force, p is the linear momentum.

 

[Antonio Lao]

F = \frac{d(mv)}{dt}

where m is the mass of the object and v is the velocity of the object.

F = v \frac{dm}{dt} + m \frac{dv}{dt}

but \frac{dm}{dt} is assumed zero and a = \frac{dv}{dt} where a is the acceleration.

F = ma becomes Newton's 2nd law of motion. This law is true in classical physics, in quantum theory, in special relativity, in general relativity, in all of physics as the law of inertial force.

 

[Antonio Lao]

So for the existence of inertial force, we made the assumption that

v \frac{dm}{dt} = 0

From special theory of relativity, we can substitute m = \frac{E}{c^2} giving the following

\frac{v}{c^2} \frac{dE}{dt} = 0

 

[Antonio Lao]

For stable particles such as proton and electron the change in rest energy is zero.

\frac{dE}{dt} = 0

but the total energy of a particle includes its relativitic energy cp and rest energy E = m_0 c^2 where p=mc and m is the relativistic mass, m_0 is the rest mass. And the relativistic relation is given by

E^2 = \xi^2 - c^2 p^2

where \xi is the total energy.

 

[Antonio Lao]

The existence of inertia implies that (see note)

\frac{d}{dt} \sqrt{\xi^2 - c^2p^2} = 0

furthermore,

\xi \frac{d \xi}{dt} = c^2 p \frac{dp}{dt}

this implies an equivalence with the following

E^2 = \psi_i \times \phi_i \cdot \psi_j \times \phi_j

Note:

This is an extremum condition for finding maximum and minimum values of a function.

 

[bogdan]

Sorry for the interruption...but F=ma is valid only in classical physics...F=dp/dt is correct in general...I think...

 

[Antonio Lao]

bogdan,

F=ma defines inertial mass. In quantum mechanics, no new mass is defined and theories use the experimental values for inertial mass. Special and general relativity defined a new mass called relativistic mass and it is related to the inertial mass by \gamma where

\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

And the principle of equivalence says that inertial mass is equal to the gravitational mass.

 

[Antonio Lao]

bogdan,

And F=\frac{dp}{dt} gives linear momentum with an independent existence leading to work and kinetic energy. This is possible only if the velocity is not zero.

When the velocity is zero or only locally possible as in rotational motion, the mass becomes potential instead of kinetic. Inertial mass is another name for potential mass. We can even try to define a new mass called kinetic mass associated with each linear momentum.