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Why pressure is a scalar?

  1. Dec 29, 2014 #1
    Would you please explain in details , why pressure is a scalar, though, [itex]pressure = \frac {force}{area}[/itex] and force is a vector ?
  2. jcsd
  3. Dec 29, 2014 #2
    When calculating the pressure on a surface you assume that the force is applied perpendicularly to the surface itself. That is, you consider the force which has the same direction of the vector normal to the surface.
  4. Dec 29, 2014 #3


    Staff: Mentor

    Rewrite the expression as ##f=p~a##. In that expression force is a vector and pressure is a scalar, so what kind of quantity is area?
  5. Dec 29, 2014 #4

    Stephen Tashi

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    Looking at the "talk" page associated with the Wikipedia article on pressure makes one gun shy of trying to define pressure. http://en.wikipedia.org/wiki/Talk:Pressure (see 3. Pressure definition).

    How about the idea that "pressure is a scalar part of a tensor"?
  6. Dec 29, 2014 #5


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    The simple explanation is in post #2, the complicated one (in terms of the Cauchy stress-energy tensor) is hinted to in post#4. As soon as one goes to study continuum (classical) dynamics, then the true nature of 'p' results. For high-school physics, post#2 will suffice.
    Last edited: Dec 31, 2014
  7. Dec 31, 2014 #6


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    The formal explanation of the answers above is as follows. Take an arbitrary volume ##V## with boundary ##\partial V## out of a fluid. Then there's a force corresponding to the interaction of the fluid outside of the volume on the surface of the volume. The total force (neglecting bulk forces like gravity) on the fluid volume then is given by
    $$F_j=\int_{\partial V} \mathrm{d}^2 f_k \sigma_{kj},$$
    where ##\sigma_{kj}## is the stress tensor of the fluid and ##\mathrm{d}^2 \vec{f}## the surface-normal vectors along the boundary of the volume pointing by convention outward of the volume.

    Pressure is one part of the stress tensor of a fluid
    $$\sigma_{jk}=s_{jk}-P \delta_{jk}.$$
    By definition ##s_{jk}##, the "stress deviator", is traceless and thus
    $$\sigma_{jj}=-3 P$$
    Since ##\sigma_{jk}## is a 2nd rank tensor, the pressure is a scalar.

    The geometrical meaning is that the pressure tries to change the magnitude of the volume, while the traceless stress deviator tends to deform it.

    The Wikipedia article on this topic is very nice with very good figures, making the thing pretty intuitive:

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