Looking for help understanding scalars vs. vectors

[Physics_is_beautiful]

Why are pressure and energy not considered as vectors?

 

[Physics_is_beautiful]

http://webphysics.iupui.edu/JITTworkshop/152Basics/vectors/vectors.html
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/2-1-scalars-and-vectors/

content here is roughly my understanding level as well

anything upto tenth grade level,

 

[Physics_is_beautiful]

Why are pressure and energy not considered as vectors?

Even so, momentum is considered a vector, yet kinetic energy, and other energies are considered scalar. Why so?

Moreover, why is pressure not a vector? wouldn't a direction be required to show where pressure acts upon?

 

[PeroK]

Even so, momentum is considered a vector, yet kinetic energy, and other energies are considered scalar. Why so?

Kinetic energy is conserved in an elastic collision, but not "directional" kinetic energy.

More fundamentally, conservation of energy is related to time translation invariance, hence has only one component.

Moreover, why is pressure not a vector? wouldn't a direction be required to show where pressure acts upon?

Pressure acts in all directions. Not just one.

 

[vanhees71]

Why are pressure and energy not considered as vectors?

Obviously you are talking about Newtonian physics, where you deal with vectors and tensors in 3D Euclidean space. Energy is a scalar, since there is no direction involved in its meaning. It first occurs in point-particle mechanics as the kinetic energy of a particle, which is given by $E_{\text{kin}}=m \vec{v}^2/2$, and that's by definition of the scalar product of vectors a scalar, having no direction in any way.

Pressure is a bit more subtle. It's indeed a scalar, but that's not so intuitively clear as with energy, since there is indeed some "direction" hidden in a pressure, because it's meaning is that if you have a fluid (gas or liquid, where pressure is a useful quantity) it describes a "force per unit area", i.e., it's a force imposed on, e.g., the container walls this fluid is contained in.

Physically such "surface forces" are described by what's called a stress tensor. It's most easily explained when introducing a Cartesian basis and talk in terms of components of vectors and tensors. The stress tensor is a socalled symmetric 2nd-rank tensor with components $\sigma_{jk}=\sigma_{kj}$. The meaning of this tensor is that it is supposed to describe the force on some surface imposed through interactions of this surface with the fluid. Consider a very small ("infinitesimally small") part of the surface, and define a vector with components $\mathrm{d}^2 f_j$, whose magnitude is the ("infinitesimally small") area of this "surface element" and which is directed perpendicular (in either of the two possible directions) to the surface element. Then the fluid imposes a force on this surface element given by
$$\mathrm{d} F_k=\sum_{j=1}^3 \sigma_{kj} \mathrm{d}^2 f_j.$$
A usual fluid is (at least in "infinitesimally small fluid elements") isotropic, i.e., there's no preferred direction, and the only tensor, which does not define any preferred direction is proportional to the only invariant tensor components, where invariant refers to the behavior of the tensor components under rotations, and thus you get
$$\sigma_{jk}=-P \delta_{jk},$$
where
$$\delta_{jk}=\begin{cases} 1 & \text{for} \quad j = k, \\ 0 & \text{for} \quad j \neq k, \end{cases}$$
where the pressure, $P$, must also be invariant under rotations, i.e., it's a scalar and thus a quantity which in no way defines any direction. Still the physical meaning is given in terms of the stress tensor it defines, i.e., if you have a surface element, described by a surface-normal vector $\mathrm{d}^2 \vec{f}$, then the force on this surface element is given by
$$\mathrm{d} F_{k} = -\sum_{j=1}^3 P \delta_{jk} \mathrm{d}^2 f_k = -P \mathrm{d}^2 f_j,$$
i.e., due to the fluid's pressure the forces imposed on the surface element is always in direction of the surface normal, and indeed in this description of surface forces the only direction available is the surface normal vector, because there's no other direction somehow defined by the fluid. That's why pressure is a scalar, although some "vector-like" meaning in the sense that it describes forces on surfaces due to the presence of the fluid.

 

[Mister T]

Moreover, why is pressure not a vector? wouldn't a direction be required to show where pressure acts upon?

It's the force associated with the pressure that has a direction.

 

[DaveE]

Obviously you are talking about Newtonian physics, where you deal with vectors and tensors in 3D Euclidean space. Energy is a scalar, since there is no direction involved in its meaning. It first occurs in point-particle mechanics as the kinetic energy of a particle, which is given by $E_{\text{kin}}=m \vec{v}^2/2$, and that's by definition of the scalar product of vectors a scalar, having no direction in any way.

Pressure is a bit more subtle. It's indeed a scalar, but that's not so intuitively clear as with energy, since there is indeed some "direction" hidden in a pressure, because it's meaning is that if you have a fluid (gas or liquid, where pressure is a useful quantity) it describes a "force per unit area", i.e., it's a force imposed on, e.g., the container walls this fluid is contained in.

Physically such "surface forces" are described by what's called a stress tensor. It's most easily explained when introducing a Cartesian basis and talk in terms of components of vectors and tensors. The stress tensor is a socalled symmetric 2nd-rank tensor with components $\sigma_{jk}=\sigma_{kj}$. The meaning of this tensor is that it is supposed to describe the force on some surface imposed through interactions of this surface with the fluid. Consider a very small ("infinitesimally small") part of the surface, and define a vector with components $\mathrm{d}^2 f_j$, whose magnitude is the ("infinitesimally small") area of this "surface element" and which is directed perpendicular (in either of the two possible directions) to the surface element. Then the fluid imposes a force on this surface element given by
$$\mathrm{d} F_k=\sum_{j=1}^3 \sigma_{kj} \mathrm{d}^2 f_j.$$
A usual fluid is (at least in "infinitesimally small fluid elements") isotropic, i.e., there's no preferred direction, and the only tensor, which does not define any preferred direction is proportional to the only invariant tensor components, where invariant refers to the behavior of the tensor components under rotations, and thus you get
$$\sigma_{jk}=-P \delta_{jk},$$
where
$$\delta_{jk}=\begin{cases} 1 & \text{for} \quad j = k, \\ 0 & \text{for} \quad j \neq k, \end{cases}$$
where the pressure, $P$, must also be invariant under rotations, i.e., it's a scalar and thus a quantity which in no way defines any direction. Still the physical meaning is given in terms of the stress tensor it defines, i.e., if you have a surface element, described by a surface-normal vector $\mathrm{d}^2 \vec{f}$, then the force on this surface element is given by
$$\mathrm{d} F_{k} = -\sum_{j=1}^3 P \delta_{jk} \mathrm{d}^2 f_k = -P \mathrm{d}^2 f_j,$$
i.e., due to the fluid's pressure the forces imposed on the surface element is always in direction of the surface normal, and indeed in this description of surface forces the only direction available is the surface normal vector, because there's no other direction somehow defined by the fluid. That's why pressure is a scalar, although some "vector-like" meaning in the sense that it describes forces on surfaces due to the presence of the fluid.

Wow, you were a lot smarter than I was in tenth grade!

 

[Haborix]

Wow, you were a lot smarter than I was in tenth grade!

If you want a simple answer from @vanhees71, you must invoke philosophy.

 

[DaveE]

Imagine a container (ballon, cylinder & piston, whatever...) with a pressurized gas or fluid. The force on any small area of the containment surface has a direction, always normal to the surface, but the magnitude depends only on the area of that piece. So the fluid itself has pressure, but no preferred direction. The direction of the force is only determined by the surface in question.

 

[DaveE]

If you want a simple answer from @vanhees71, you must invoke philosophy.

I believe the art of teaching intro physics requires a little bit of lying and a little bit of "trust me for now, you'll learn that later". I truly believe that nearly all of the physics I know, as an EE, is actually wrong if you question it deeply enough. But there is utility, in both learning and practice, of using simple models. There is also utility in describing models in simple, less rigorous, language. OK, there is a stress tensor, but do you have to say it that way to someone that has probably never studied tensors?

 

[vanhees71]

If you want a simple answer from @vanhees71, you must invoke philosophy.

My answer is the simplest one I can give. Of course you have to learn linear algebra and vector calculus to understand physics.

 

[vanhees71]

I believe the art of teaching intro physics requires a little bit of lying and a little bit of "trust me for now, you'll learn that later". I truly believe that nearly all of the physics I know, as an EE, is actually wrong if you question it deeply enough. But there is utility, in both learning and practice, of using simple models. There is also utility in describing models in simple, less rigorous, language. OK, there is a stress tensor, but do you have to say it that way to someone that has probably never studied tensors?

For sure as an electrical engineer you also had to learn the adequate math, and vector calculus is for sure a necessary prerequisite for this business too.

 

[Haborix]

The point is that the OP doesn't have that background knowledge and most of us would resist saying you can't understand physics without that background.

 

[DaveE]

My answer is the simplest one I can give. Of course you have to learn linear algebra and vector calculus to understand physics.

And yet they still manage to teach it in High Schools around the world.

 

[vanhees71]

Really? That's progress. Unfortunately, for my taste, there's a lack of continuum mechanics even in the university physics curriculum.

 

[DaveE]

oy. This is pointless, I'm done.