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mysearch
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Hi,
In the process of reading up about dark energy in the field of cosmology, I felt I was beginning to lose sight of what basic physics might be telling me about energy density and pressure assumed to be associated with this form of energy. This is why I am raising these questions in this forum and not the cosmology forum.
From a classical physics perspective, energy seems to reduce to just 3 forms, i.e. rest, kinetic and potential. However, potential (gravitational) energy seems to imply a negative scalar quantity, while kinetic energy seems to imply a positive scalar quantity in order to maintain the conservation of energy under the free fall example to be outlined. However, on reflection, it was not clear to me where rest energy fits on this scale, although I do attempt to answer this question below?
In part, the following example linked to the attached diagram simply tries to highlight some of these issues along with the relationship between energy density and pressure. I would appreciate any insights/clarifications or corrections.
Thanks
Example:
If we initially consider [m] as a stationary unit mass infinitely far from [M] it would have zero kinetic energy and zero potential energy. However, as [m] free-falls towards [M] it gains positive kinetic energy, i.e. velocity, while still maintaining the conservation of energy. It does this because potential energy is defined as a form of negative energy. However, as we approach [M], we realize that [M] is a composite mass made up of [n] unit particles contained within a spherical shell. If we assume that each unit particle is stationary within this spherical container, might we define its energy density in terms of its rest energy divided by its volume, i.e. [\rho=nm_pc^2/(4/3) \pi r^3]?
On the basis that each unit particle contained with the shell is stationary, we might intuitive understand that they would exert no pressure [P] on the shell. This tells us that the presence of an energy density does not automatically imply an associated pressure. This case is reflected in the equation P=\omega \rho c^2 when \omega=0. Of course, if we said that each unit particle within the shell had a velocity [v], we might guess that the collective process of the unit particles colliding with the shell over time might be inferred as a pressure. In fact if we dismantle the units of pressure into force per unit area, we can actually visualise the process. The unit of force reduces to [kg.m/s2], but can equally be described as the momentum [mv] per unit time. So, over time, the unit particles collide with the shell, change direction and hence transfer momentum per unit time to the shell over a given area, i.e. pressure. However, the inference is that [\omega] is reflective of only the kinetic energy and not the rest energy, although the two are linked via the mass of the particles.
If we were to define the energy density of the shell when the unit particles have velocity, accepting energy to be a scalar quantity and not a vector quantity, it is assumed that the new energy density is now the sum of the kinetic energy and rest energy of the particles?
However, does the kinetic energy within the shell change the effective mass of [M] with respect to [m], i.e. does [m] experience a larger gravitational force?
Initially, I assume it would not, but was then not so sure. Rest energy is a function of m_oc^2 while non-relativistic kinetic energy is approximated by 1/2mv^2. Clearly, rest energy of the unit particles would be much greater than the kinetic energy for non-relativity speeds. Of course, if we introduce relativistic speeds, the introduction of [\gamma] increases the effective mass.
So are rest energy and kinetic energy both positive scalar quantities that combined to create an effective gravitational mass, i.e. m = (E_R+E_K)/c^2?
Returning briefly to dark energy. With [\omega=-1], dark energy is said to have a negative pressure. Normally, in relative terms, negative pressure might be associated with a region of lower pressure, i.e. a vacuum, which sucks things towards it. However, in this case, negative pressure is said to act like anti-gravity.
- However, by virtue of its energy, does dark energy also have an effective mass and therefore a gravitational effect?
- If [\omega=-1] then P=-\rho c^2 it seems to imply that the gravitational effect would be comparable to the pressure effect, which would then cancel the expansion associated darke energy?
- Finally, dark energy is said not to act as a force on an object but rather expands the space between two objects, analogous to a force but different. Of course, you can’t help wondering as to the underlying physics of this process – any thoughts?
In the process of reading up about dark energy in the field of cosmology, I felt I was beginning to lose sight of what basic physics might be telling me about energy density and pressure assumed to be associated with this form of energy. This is why I am raising these questions in this forum and not the cosmology forum.
From a classical physics perspective, energy seems to reduce to just 3 forms, i.e. rest, kinetic and potential. However, potential (gravitational) energy seems to imply a negative scalar quantity, while kinetic energy seems to imply a positive scalar quantity in order to maintain the conservation of energy under the free fall example to be outlined. However, on reflection, it was not clear to me where rest energy fits on this scale, although I do attempt to answer this question below?
In part, the following example linked to the attached diagram simply tries to highlight some of these issues along with the relationship between energy density and pressure. I would appreciate any insights/clarifications or corrections.
Thanks
Example:
If we initially consider [m] as a stationary unit mass infinitely far from [M] it would have zero kinetic energy and zero potential energy. However, as [m] free-falls towards [M] it gains positive kinetic energy, i.e. velocity, while still maintaining the conservation of energy. It does this because potential energy is defined as a form of negative energy. However, as we approach [M], we realize that [M] is a composite mass made up of [n] unit particles contained within a spherical shell. If we assume that each unit particle is stationary within this spherical container, might we define its energy density in terms of its rest energy divided by its volume, i.e. [\rho=nm_pc^2/(4/3) \pi r^3]?
On the basis that each unit particle contained with the shell is stationary, we might intuitive understand that they would exert no pressure [P] on the shell. This tells us that the presence of an energy density does not automatically imply an associated pressure. This case is reflected in the equation P=\omega \rho c^2 when \omega=0. Of course, if we said that each unit particle within the shell had a velocity [v], we might guess that the collective process of the unit particles colliding with the shell over time might be inferred as a pressure. In fact if we dismantle the units of pressure into force per unit area, we can actually visualise the process. The unit of force reduces to [kg.m/s2], but can equally be described as the momentum [mv] per unit time. So, over time, the unit particles collide with the shell, change direction and hence transfer momentum per unit time to the shell over a given area, i.e. pressure. However, the inference is that [\omega] is reflective of only the kinetic energy and not the rest energy, although the two are linked via the mass of the particles.
If we were to define the energy density of the shell when the unit particles have velocity, accepting energy to be a scalar quantity and not a vector quantity, it is assumed that the new energy density is now the sum of the kinetic energy and rest energy of the particles?
However, does the kinetic energy within the shell change the effective mass of [M] with respect to [m], i.e. does [m] experience a larger gravitational force?
Initially, I assume it would not, but was then not so sure. Rest energy is a function of m_oc^2 while non-relativistic kinetic energy is approximated by 1/2mv^2. Clearly, rest energy of the unit particles would be much greater than the kinetic energy for non-relativity speeds. Of course, if we introduce relativistic speeds, the introduction of [\gamma] increases the effective mass.
So are rest energy and kinetic energy both positive scalar quantities that combined to create an effective gravitational mass, i.e. m = (E_R+E_K)/c^2?
Returning briefly to dark energy. With [\omega=-1], dark energy is said to have a negative pressure. Normally, in relative terms, negative pressure might be associated with a region of lower pressure, i.e. a vacuum, which sucks things towards it. However, in this case, negative pressure is said to act like anti-gravity.
- However, by virtue of its energy, does dark energy also have an effective mass and therefore a gravitational effect?
- If [\omega=-1] then P=-\rho c^2 it seems to imply that the gravitational effect would be comparable to the pressure effect, which would then cancel the expansion associated darke energy?
- Finally, dark energy is said not to act as a force on an object but rather expands the space between two objects, analogous to a force but different. Of course, you can’t help wondering as to the underlying physics of this process – any thoughts?
