Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What Lies on the Other Side of Pressure ?

  1. Aug 6, 2004 #1
    What Lies on the Other Side of Pressure ???

    Pressure is commonly defined as force per unit area.

    [tex] P = \frac{F}{A} [/tex]

    Pressure is a scalar quantity but force is a vector quantity. To make sense the equation must the scalar product of vector force and a vector which is equivalent to inverse of a vector product, [itex] r \times r [/itex].

    But the concept of pressure as a scalar is clear when it is defined as the volume rate of change of energy.

    [tex] P = \frac{\partial E}{\partial V} [/tex]

    But if pressure is properly defined within an enclosed volume, what is the definition of pressure outside this volume?

    If this volume is the universe itself, what is the pressure outside the universal volume?
  2. jcsd
  3. Aug 6, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Eeh, actually, we define the stress vector as a limiting quantity
    when A goes to zero.
    A fundamental assumption when introducing the pressure stress vector,
    is that we have isotropic conditions.
    This entails that on a given surface, the pressure stress vector can be written as:
    where p is the scalar known as pressure, and [tex]\vec{n}[/tex] is the outward normal on the surface.
  4. Aug 6, 2004 #3
    For nonrelativistic ideal gas the pressure is

    [tex] P = \frac{2}{3} u [/tex]

    and for relativistic ideal gas such as electromagnetic radiation the pressure is

    [tex] P = \frac {1}{3} u [/tex]

    where [itex] u[/itex] is the total energy density of radiation.
  5. Aug 6, 2004 #4
    Why there is twice as much pressure in nonrelativistic gas than it is in relativistic gas?
  6. Aug 6, 2004 #5
    One thing is clear that the pressure outside of an enclosed volume is always less than the pressure inside of the volume. So what is the pressure outside of the universe? The answer is that the pressure is zero. This zero pressure can also be the cause of the universal expansion.
  7. Aug 6, 2004 #6
    The boundary of the universe is then between two density values one value is always greater than zero and the other is always exactly zero. And the name for this boundary can be named as the true vacuum (a state of complete nothingness).
  8. Aug 7, 2004 #7
    From quantum field theory, we have established the fact that the vacuum we are in is not really empty. It's a false vacuum with its ceaseless fluctuations and virtual particles which could hold the key to understanding the true meaning of mass in physics.
  9. Aug 19, 2004 #8
    My belated thanks to your description of pressure. I have a question concerning how would constant density affect the equation of continuity?

    [tex] \frac{\partial \rho}{\partial t} + div \left( \rho \vec{v}\right) = 0[/tex]
  10. Aug 19, 2004 #9


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Hi, A.L:
    I think we have talked slightly beside each other in this thread:
    As far as I can see, you have been concerned with typical state equations (i.e, thermodynamical) in which the isotropic pressure is related to other important variables.
    This is obviously an extremely important issue, but I'm not sufficiently into thermodynamics&relativity to offer valuable comments.

    I gave the definition of pressure as an isotropic form of (normal) stress.
    (Clearly, when we consider the random momentum transfer molecules impart on a surface in the normal direction (i.e, as a collision), the randomness should guarantee that there is no "preferred direction", i.e, isotropic conditions).

    In the case of a constant density field, the equation of continuity implies a divergence-free (solenoidal) velocity field
  11. Aug 19, 2004 #10
    Thanks for your clarification.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook