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If force is a vector, then why is pressure a scalar?
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By definition pressure is the perpendicular force applied to a unit area. So it has a direction which is perpendicular to the area. So it should be a vector. But I did sone googling and found out that it is a scalar quantity. So it would really be helpful if someone could show me how pressure is scalar with some mathematical derivation.
pressure vectors
asked 3 hours ago
Asif Iqubal
274112
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up vote
2
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favorite
By definition pressure is the perpendicular force applied to a unit area. So it has a direction which is perpendicular to the area. So it should be a vector. But I did sone googling and found out that it is a scalar quantity. So it would really be helpful if someone could show me how pressure is scalar with some mathematical derivation.
pressure vectors
asked 3 hours ago
Asif Iqubal
274112
Possible duplicate of Define Pressure at A point. Why is it a Scalar?
– Kyle Kanos
33 mins ago
And also all linked therein
– Kyle Kanos
33 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
By definition pressure is the perpendicular force applied to a unit area. So it has a direction which is perpendicular to the area. So it should be a vector. But I did sone googling and found out that it is a scalar quantity. So it would really be helpful if someone could show me how pressure is scalar with some mathematical derivation.
pressure vectors
asked 3 hours ago
Asif Iqubal
274112
By definition pressure is the perpendicular force applied to a unit area. So it has a direction which is perpendicular to the area. So it should be a vector. But I did sone googling and found out that it is a scalar quantity. So it would really be helpful if someone could show me how pressure is scalar with some mathematical derivation.
pressure vectors
pressure vectors
asked 3 hours ago
Asif Iqubal
274112
asked 3 hours ago
Asif Iqubal
274112
asked 3 hours ago
Asif Iqubal
274112
asked 3 hours ago
Asif Iqubal
274112
asked 3 hours ago
Asif Iqubal
274112
274112
Possible duplicate of Define Pressure at A point. Why is it a Scalar?
– Kyle Kanos
33 mins ago
And also all linked therein
– Kyle Kanos
33 mins ago
add a comment |Â
Possible duplicate of Define Pressure at A point. Why is it a Scalar?
– Kyle Kanos
33 mins ago
And also all linked therein
– Kyle Kanos
33 mins ago
Possible duplicate of Define Pressure at A point. Why is it a Scalar?
– Kyle Kanos
33 mins ago
Possible duplicate of Define Pressure at A point. Why is it a Scalar?
– Kyle Kanos
33 mins ago
And also all linked therein
– Kyle Kanos
33 mins ago
And also all linked therein
– Kyle Kanos
33 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
Pressure is proportionality factor. Area is the one that gives you direction. You have to recall that pressure is defined everywhere in the bulk volume, not just at the surface. A volume of gas has pressure defined everywhere. And the force direction is determined by you - by the way you orient your surface that you put into the gas.
$$vecF=pvecA$$
Here you see that the area is the vector.
Quoting wikipedia:
It is incorrect (although rather usual) to say "the pressure is
directed in such or such direction". The pressure, as a scalar, has no
direction. The force given by the previous relationship to the
quantity has a direction, but the pressure does not. If we change the
orientation of the surface element, the direction of the normal force
changes accordingly, but the pressure remains the same.
I should clarify, that this calculates the force caused by the pressure, so it GIVES you the force perpendicular to the area, given the area vector. It's the defining equation and the only one that captures what pressure actually does, so it's always true, but needs to be understood as a formula for calculating the force from pressure.
If $vecF$ is just caused by the pressure, it can't be anything else but perpendicular to the area, otherwise you have other forces present in the system, or the liquid is not isotropic. That being said, assuming that $vecF$ is only caused by the pressure, you could calculate $p$ by taking absolute values:
$$p=fracF$$
Mathematically, you transformed a vector equation into a scalar equation assuming parallel vectors, so now you're not allowed to put in anything, but only lengths (or projections - similar argument) of F and A that are guaranteed to have been parallel, otherwise you get nonsense. You also lose the sign of pressure (for gasses, it can't happen, but for elastic solids or liquids, it can "pull" due to intermolecular forces).
However, strictly speaking, pressure is a tensor, but for gasses, it's isotropic, so it acts as a scalar. Without going into details what a tensor is, imagine above that in $pvecA$, $p$ can also transform the direction, not just the magnitude of $vecA$, so the force does not have to point perpendicularly. This is true in elastic solids, where you can transmit sideways forces to the surface, and in viscous flowing liquids, where the viscous force is also just a stress (generalized pressure) transmitted to the surface. In this situation, $p$ has $6$ independent components, so you cannot measure it just by measuring a force on a single surface. You'd need to measure all components of force on 3 surfaces placed in different orientations. Only in gasses, you can rely on the force having the same magnitude no matter the orientation.
Further reading:
https://en.wikipedia.org/wiki/Cauchy_stress_tensor
answered 3 hours ago
orion
5,3661324
Equation I stated is the fundamental definition of force defined by pressure and is always correct. It can't fail, but if you want to invert it to calculate pressure, then it's your mistake if you put the wrong forces or areas into it (remember, you can't divide by a vector, so inversion loses information and induces assumptions). It's sort of like $y=x^2$ always gives you $y$ if you put in a value of $x$, but inversion $x=sqrty$ allows you to put in nonsense values (such as negative, if we are limited to real values), and loses the sign of the result in the process.
– orion
3 hours ago
You can't define it that way, it's not general and it's wrong if your conditions are not met. Stress tensor and pressure are always defined as the ones causing forces, not the other way around. That's just the oversimplified primary school explanation of pressure, but physically, it can't be generalized well.
– orion
2 hours ago
Especially as pressure exists without forces. Forces appear when a surface is present, they are just boundary effects. Pressure is more than that. It is more fundamental quantity, and when you describe complete motion of materials (including deformation, sound, flow, thermodynamics), the "force" definition is unhelpful and "backwards" defined.
– orion
2 hours ago
add a comment |Â
up vote
1
down vote
Pressure is a scalar because it does not behave as a vector -- specifically, you can't take the "components" of pressure and take their Pythagorean sum to obtain its magnitude. Instead, pressure is actually proportional to the sum of the components, $(P_x+P_y+P_z)/3$.
The way to understand pressure is in terms of the stress tensor, and pressure is equal to the trace of the stress tensor. Once you understand this, the question becomes equivalent to questions like "why is the dot product a scalar?" (trace of the tensor product), "why is the divergence of a vector field a scalar?" (trace of the tensor derivative), etc.
There is no physical significance to taking the diagonal components of a tensor and putting them in a vector -- there is a physical significance to adding them up, and the invariance properties of the result tells you that it is a scalar.
answered 2 hours ago
Abhimanyu Pallavi Sudhir
4,29842243
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The formula is $vecF = p vecn A$ where $vecn$ is the unit vector perpendicular to the surface.
answered 3 hours ago
md2perpe
38425
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0
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Pressure is defined by
$$
P = |F| / A
$$
where $|F|$ is the magnitude of the normal force, thus $P$ is a scalar.
answered 3 hours ago
innisfree
10.7k32655
This is incorrect. Consider the example of force being applied not perpendicular to the surface.
– Jitendra
3 hours ago
$|F|$ is the magnitude of the normal force
– innisfree
3 hours ago
by normal force do you mean taking the component along the direction perpendicular to the surface?
– Jitendra
3 hours ago
1
Yes I do mean that
– innisfree
3 hours ago
This calculates pressure based on some measurements, but doesn't define pressure. Doesn't tell you what pressure is. Pressure is there even if no surfaces or forces are present. In the middle of a gas bottle, gas has pressure. You can imagine putting a virtual surface there, and no matter how you orient it, you'd get the force of same magnitude, but that's just a side-effect.
– orion
3 hours ago
 |Â
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Pressure is proportionality factor. Area is the one that gives you direction. You have to recall that pressure is defined everywhere in the bulk volume, not just at the surface. A volume of gas has pressure defined everywhere. And the force direction is determined by you - by the way you orient your surface that you put into the gas.
$$vecF=pvecA$$
Here you see that the area is the vector.
Quoting wikipedia:
It is incorrect (although rather usual) to say "the pressure is
directed in such or such direction". The pressure, as a scalar, has no
direction. The force given by the previous relationship to the
quantity has a direction, but the pressure does not. If we change the
orientation of the surface element, the direction of the normal force
changes accordingly, but the pressure remains the same.
I should clarify, that this calculates the force caused by the pressure, so it GIVES you the force perpendicular to the area, given the area vector. It's the defining equation and the only one that captures what pressure actually does, so it's always true, but needs to be understood as a formula for calculating the force from pressure.
If $vecF$ is just caused by the pressure, it can't be anything else but perpendicular to the area, otherwise you have other forces present in the system, or the liquid is not isotropic. That being said, assuming that $vecF$ is only caused by the pressure, you could calculate $p$ by taking absolute values:
$$p=fracF$$
Mathematically, you transformed a vector equation into a scalar equation assuming parallel vectors, so now you're not allowed to put in anything, but only lengths (or projections - similar argument) of F and A that are guaranteed to have been parallel, otherwise you get nonsense. You also lose the sign of pressure (for gasses, it can't happen, but for elastic solids or liquids, it can "pull" due to intermolecular forces).
However, strictly speaking, pressure is a tensor, but for gasses, it's isotropic, so it acts as a scalar. Without going into details what a tensor is, imagine above that in $pvecA$, $p$ can also transform the direction, not just the magnitude of $vecA$, so the force does not have to point perpendicularly. This is true in elastic solids, where you can transmit sideways forces to the surface, and in viscous flowing liquids, where the viscous force is also just a stress (generalized pressure) transmitted to the surface. In this situation, $p$ has $6$ independent components, so you cannot measure it just by measuring a force on a single surface. You'd need to measure all components of force on 3 surfaces placed in different orientations. Only in gasses, you can rely on the force having the same magnitude no matter the orientation.
Further reading:
https://en.wikipedia.org/wiki/Cauchy_stress_tensor
answered 3 hours ago
orion
5,3661324
Equation I stated is the fundamental definition of force defined by pressure and is always correct. It can't fail, but if you want to invert it to calculate pressure, then it's your mistake if you put the wrong forces or areas into it (remember, you can't divide by a vector, so inversion loses information and induces assumptions). It's sort of like $y=x^2$ always gives you $y$ if you put in a value of $x$, but inversion $x=sqrty$ allows you to put in nonsense values (such as negative, if we are limited to real values), and loses the sign of the result in the process.
– orion
3 hours ago
You can't define it that way, it's not general and it's wrong if your conditions are not met. Stress tensor and pressure are always defined as the ones causing forces, not the other way around. That's just the oversimplified primary school explanation of pressure, but physically, it can't be generalized well.
– orion
2 hours ago
Especially as pressure exists without forces. Forces appear when a surface is present, they are just boundary effects. Pressure is more than that. It is more fundamental quantity, and when you describe complete motion of materials (including deformation, sound, flow, thermodynamics), the "force" definition is unhelpful and "backwards" defined.
– orion
2 hours ago
add a comment |Â
up vote
3
down vote
Pressure is proportionality factor. Area is the one that gives you direction. You have to recall that pressure is defined everywhere in the bulk volume, not just at the surface. A volume of gas has pressure defined everywhere. And the force direction is determined by you - by the way you orient your surface that you put into the gas.
$$vecF=pvecA$$
Here you see that the area is the vector.
Quoting wikipedia:
It is incorrect (although rather usual) to say "the pressure is
directed in such or such direction". The pressure, as a scalar, has no
direction. The force given by the previous relationship to the
quantity has a direction, but the pressure does not. If we change the
orientation of the surface element, the direction of the normal force
changes accordingly, but the pressure remains the same.
I should clarify, that this calculates the force caused by the pressure, so it GIVES you the force perpendicular to the area, given the area vector. It's the defining equation and the only one that captures what pressure actually does, so it's always true, but needs to be understood as a formula for calculating the force from pressure.
If $vecF$ is just caused by the pressure, it can't be anything else but perpendicular to the area, otherwise you have other forces present in the system, or the liquid is not isotropic. That being said, assuming that $vecF$ is only caused by the pressure, you could calculate $p$ by taking absolute values:
$$p=fracF$$
Mathematically, you transformed a vector equation into a scalar equation assuming parallel vectors, so now you're not allowed to put in anything, but only lengths (or projections - similar argument) of F and A that are guaranteed to have been parallel, otherwise you get nonsense. You also lose the sign of pressure (for gasses, it can't happen, but for elastic solids or liquids, it can "pull" due to intermolecular forces).
However, strictly speaking, pressure is a tensor, but for gasses, it's isotropic, so it acts as a scalar. Without going into details what a tensor is, imagine above that in $pvecA$, $p$ can also transform the direction, not just the magnitude of $vecA$, so the force does not have to point perpendicularly. This is true in elastic solids, where you can transmit sideways forces to the surface, and in viscous flowing liquids, where the viscous force is also just a stress (generalized pressure) transmitted to the surface. In this situation, $p$ has $6$ independent components, so you cannot measure it just by measuring a force on a single surface. You'd need to measure all components of force on 3 surfaces placed in different orientations. Only in gasses, you can rely on the force having the same magnitude no matter the orientation.
Further reading:
https://en.wikipedia.org/wiki/Cauchy_stress_tensor
answered 3 hours ago
orion
5,3661324
Equation I stated is the fundamental definition of force defined by pressure and is always correct. It can't fail, but if you want to invert it to calculate pressure, then it's your mistake if you put the wrong forces or areas into it (remember, you can't divide by a vector, so inversion loses information and induces assumptions). It's sort of like $y=x^2$ always gives you $y$ if you put in a value of $x$, but inversion $x=sqrty$ allows you to put in nonsense values (such as negative, if we are limited to real values), and loses the sign of the result in the process.
– orion
3 hours ago
You can't define it that way, it's not general and it's wrong if your conditions are not met. Stress tensor and pressure are always defined as the ones causing forces, not the other way around. That's just the oversimplified primary school explanation of pressure, but physically, it can't be generalized well.
– orion
2 hours ago
Especially as pressure exists without forces. Forces appear when a surface is present, they are just boundary effects. Pressure is more than that. It is more fundamental quantity, and when you describe complete motion of materials (including deformation, sound, flow, thermodynamics), the "force" definition is unhelpful and "backwards" defined.
– orion
2 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Pressure is proportionality factor. Area is the one that gives you direction. You have to recall that pressure is defined everywhere in the bulk volume, not just at the surface. A volume of gas has pressure defined everywhere. And the force direction is determined by you - by the way you orient your surface that you put into the gas.
$$vecF=pvecA$$
Here you see that the area is the vector.
Quoting wikipedia:
It is incorrect (although rather usual) to say "the pressure is
directed in such or such direction". The pressure, as a scalar, has no
direction. The force given by the previous relationship to the
quantity has a direction, but the pressure does not. If we change the
orientation of the surface element, the direction of the normal force
changes accordingly, but the pressure remains the same.
I should clarify, that this calculates the force caused by the pressure, so it GIVES you the force perpendicular to the area, given the area vector. It's the defining equation and the only one that captures what pressure actually does, so it's always true, but needs to be understood as a formula for calculating the force from pressure.
If $vecF$ is just caused by the pressure, it can't be anything else but perpendicular to the area, otherwise you have other forces present in the system, or the liquid is not isotropic. That being said, assuming that $vecF$ is only caused by the pressure, you could calculate $p$ by taking absolute values:
$$p=fracF$$
Mathematically, you transformed a vector equation into a scalar equation assuming parallel vectors, so now you're not allowed to put in anything, but only lengths (or projections - similar argument) of F and A that are guaranteed to have been parallel, otherwise you get nonsense. You also lose the sign of pressure (for gasses, it can't happen, but for elastic solids or liquids, it can "pull" due to intermolecular forces).
However, strictly speaking, pressure is a tensor, but for gasses, it's isotropic, so it acts as a scalar. Without going into details what a tensor is, imagine above that in $pvecA$, $p$ can also transform the direction, not just the magnitude of $vecA$, so the force does not have to point perpendicularly. This is true in elastic solids, where you can transmit sideways forces to the surface, and in viscous flowing liquids, where the viscous force is also just a stress (generalized pressure) transmitted to the surface. In this situation, $p$ has $6$ independent components, so you cannot measure it just by measuring a force on a single surface. You'd need to measure all components of force on 3 surfaces placed in different orientations. Only in gasses, you can rely on the force having the same magnitude no matter the orientation.
Further reading:
https://en.wikipedia.org/wiki/Cauchy_stress_tensor
answered 3 hours ago
orion
5,3661324
Pressure is proportionality factor. Area is the one that gives you direction. You have to recall that pressure is defined everywhere in the bulk volume, not just at the surface. A volume of gas has pressure defined everywhere. And the force direction is determined by you - by the way you orient your surface that you put into the gas.
$$vecF=pvecA$$
Here you see that the area is the vector.
Quoting wikipedia:
It is incorrect (although rather usual) to say "the pressure is
directed in such or such direction". The pressure, as a scalar, has no
direction. The force given by the previous relationship to the
quantity has a direction, but the pressure does not. If we change the
orientation of the surface element, the direction of the normal force
changes accordingly, but the pressure remains the same.
I should clarify, that this calculates the force caused by the pressure, so it GIVES you the force perpendicular to the area, given the area vector. It's the defining equation and the only one that captures what pressure actually does, so it's always true, but needs to be understood as a formula for calculating the force from pressure.
If $vecF$ is just caused by the pressure, it can't be anything else but perpendicular to the area, otherwise you have other forces present in the system, or the liquid is not isotropic. That being said, assuming that $vecF$ is only caused by the pressure, you could calculate $p$ by taking absolute values:
$$p=fracF$$
Mathematically, you transformed a vector equation into a scalar equation assuming parallel vectors, so now you're not allowed to put in anything, but only lengths (or projections - similar argument) of F and A that are guaranteed to have been parallel, otherwise you get nonsense. You also lose the sign of pressure (for gasses, it can't happen, but for elastic solids or liquids, it can "pull" due to intermolecular forces).
However, strictly speaking, pressure is a tensor, but for gasses, it's isotropic, so it acts as a scalar. Without going into details what a tensor is, imagine above that in $pvecA$, $p$ can also transform the direction, not just the magnitude of $vecA$, so the force does not have to point perpendicularly. This is true in elastic solids, where you can transmit sideways forces to the surface, and in viscous flowing liquids, where the viscous force is also just a stress (generalized pressure) transmitted to the surface. In this situation, $p$ has $6$ independent components, so you cannot measure it just by measuring a force on a single surface. You'd need to measure all components of force on 3 surfaces placed in different orientations. Only in gasses, you can rely on the force having the same magnitude no matter the orientation.
Further reading:
https://en.wikipedia.org/wiki/Cauchy_stress_tensor
answered 3 hours ago
orion
5,3661324
edited 2 hours ago
answered 3 hours ago
orion
5,3661324
answered 3 hours ago
orion
5,3661324
answered 3 hours ago
orion
5,3661324
5,3661324
Equation I stated is the fundamental definition of force defined by pressure and is always correct. It can't fail, but if you want to invert it to calculate pressure, then it's your mistake if you put the wrong forces or areas into it (remember, you can't divide by a vector, so inversion loses information and induces assumptions). It's sort of like $y=x^2$ always gives you $y$ if you put in a value of $x$, but inversion $x=sqrty$ allows you to put in nonsense values (such as negative, if we are limited to real values), and loses the sign of the result in the process.
– orion
3 hours ago
You can't define it that way, it's not general and it's wrong if your conditions are not met. Stress tensor and pressure are always defined as the ones causing forces, not the other way around. That's just the oversimplified primary school explanation of pressure, but physically, it can't be generalized well.
– orion
2 hours ago
Especially as pressure exists without forces. Forces appear when a surface is present, they are just boundary effects. Pressure is more than that. It is more fundamental quantity, and when you describe complete motion of materials (including deformation, sound, flow, thermodynamics), the "force" definition is unhelpful and "backwards" defined.
– orion
2 hours ago
add a comment |Â
Equation I stated is the fundamental definition of force defined by pressure and is always correct. It can't fail, but if you want to invert it to calculate pressure, then it's your mistake if you put the wrong forces or areas into it (remember, you can't divide by a vector, so inversion loses information and induces assumptions). It's sort of like $y=x^2$ always gives you $y$ if you put in a value of $x$, but inversion $x=sqrty$ allows you to put in nonsense values (such as negative, if we are limited to real values), and loses the sign of the result in the process.
– orion
3 hours ago
You can't define it that way, it's not general and it's wrong if your conditions are not met. Stress tensor and pressure are always defined as the ones causing forces, not the other way around. That's just the oversimplified primary school explanation of pressure, but physically, it can't be generalized well.
– orion
2 hours ago
Especially as pressure exists without forces. Forces appear when a surface is present, they are just boundary effects. Pressure is more than that. It is more fundamental quantity, and when you describe complete motion of materials (including deformation, sound, flow, thermodynamics), the "force" definition is unhelpful and "backwards" defined.
– orion
2 hours ago
Equation I stated is the fundamental definition of force defined by pressure and is always correct. It can't fail, but if you want to invert it to calculate pressure, then it's your mistake if you put the wrong forces or areas into it (remember, you can't divide by a vector, so inversion loses information and induces assumptions). It's sort of like $y=x^2$ always gives you $y$ if you put in a value of $x$, but inversion $x=sqrty$ allows you to put in nonsense values (such as negative, if we are limited to real values), and loses the sign of the result in the process.
– orion
3 hours ago
Equation I stated is the fundamental definition of force defined by pressure and is always correct. It can't fail, but if you want to invert it to calculate pressure, then it's your mistake if you put the wrong forces or areas into it (remember, you can't divide by a vector, so inversion loses information and induces assumptions). It's sort of like $y=x^2$ always gives you $y$ if you put in a value of $x$, but inversion $x=sqrty$ allows you to put in nonsense values (such as negative, if we are limited to real values), and loses the sign of the result in the process.
– orion
3 hours ago
You can't define it that way, it's not general and it's wrong if your conditions are not met. Stress tensor and pressure are always defined as the ones causing forces, not the other way around. That's just the oversimplified primary school explanation of pressure, but physically, it can't be generalized well.
– orion
2 hours ago
You can't define it that way, it's not general and it's wrong if your conditions are not met. Stress tensor and pressure are always defined as the ones causing forces, not the other way around. That's just the oversimplified primary school explanation of pressure, but physically, it can't be generalized well.
– orion
2 hours ago
Especially as pressure exists without forces. Forces appear when a surface is present, they are just boundary effects. Pressure is more than that. It is more fundamental quantity, and when you describe complete motion of materials (including deformation, sound, flow, thermodynamics), the "force" definition is unhelpful and "backwards" defined.
– orion
2 hours ago
Especially as pressure exists without forces. Forces appear when a surface is present, they are just boundary effects. Pressure is more than that. It is more fundamental quantity, and when you describe complete motion of materials (including deformation, sound, flow, thermodynamics), the "force" definition is unhelpful and "backwards" defined.
– orion
2 hours ago
add a comment |Â
up vote
1
down vote
Pressure is a scalar because it does not behave as a vector -- specifically, you can't take the "components" of pressure and take their Pythagorean sum to obtain its magnitude. Instead, pressure is actually proportional to the sum of the components, $(P_x+P_y+P_z)/3$.
The way to understand pressure is in terms of the stress tensor, and pressure is equal to the trace of the stress tensor. Once you understand this, the question becomes equivalent to questions like "why is the dot product a scalar?" (trace of the tensor product), "why is the divergence of a vector field a scalar?" (trace of the tensor derivative), etc.
There is no physical significance to taking the diagonal components of a tensor and putting them in a vector -- there is a physical significance to adding them up, and the invariance properties of the result tells you that it is a scalar.
answered 2 hours ago
Abhimanyu Pallavi Sudhir
4,29842243
add a comment |Â
up vote
1
down vote
Pressure is a scalar because it does not behave as a vector -- specifically, you can't take the "components" of pressure and take their Pythagorean sum to obtain its magnitude. Instead, pressure is actually proportional to the sum of the components, $(P_x+P_y+P_z)/3$.
The way to understand pressure is in terms of the stress tensor, and pressure is equal to the trace of the stress tensor. Once you understand this, the question becomes equivalent to questions like "why is the dot product a scalar?" (trace of the tensor product), "why is the divergence of a vector field a scalar?" (trace of the tensor derivative), etc.
There is no physical significance to taking the diagonal components of a tensor and putting them in a vector -- there is a physical significance to adding them up, and the invariance properties of the result tells you that it is a scalar.
answered 2 hours ago
Abhimanyu Pallavi Sudhir
4,29842243
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Pressure is a scalar because it does not behave as a vector -- specifically, you can't take the "components" of pressure and take their Pythagorean sum to obtain its magnitude. Instead, pressure is actually proportional to the sum of the components, $(P_x+P_y+P_z)/3$.
The way to understand pressure is in terms of the stress tensor, and pressure is equal to the trace of the stress tensor. Once you understand this, the question becomes equivalent to questions like "why is the dot product a scalar?" (trace of the tensor product), "why is the divergence of a vector field a scalar?" (trace of the tensor derivative), etc.
There is no physical significance to taking the diagonal components of a tensor and putting them in a vector -- there is a physical significance to adding them up, and the invariance properties of the result tells you that it is a scalar.
answered 2 hours ago
Abhimanyu Pallavi Sudhir
4,29842243
Pressure is a scalar because it does not behave as a vector -- specifically, you can't take the "components" of pressure and take their Pythagorean sum to obtain its magnitude. Instead, pressure is actually proportional to the sum of the components, $(P_x+P_y+P_z)/3$.
The way to understand pressure is in terms of the stress tensor, and pressure is equal to the trace of the stress tensor. Once you understand this, the question becomes equivalent to questions like "why is the dot product a scalar?" (trace of the tensor product), "why is the divergence of a vector field a scalar?" (trace of the tensor derivative), etc.
There is no physical significance to taking the diagonal components of a tensor and putting them in a vector -- there is a physical significance to adding them up, and the invariance properties of the result tells you that it is a scalar.
answered 2 hours ago
Abhimanyu Pallavi Sudhir
4,29842243
answered 2 hours ago
Abhimanyu Pallavi Sudhir
4,29842243
answered 2 hours ago
Abhimanyu Pallavi Sudhir
4,29842243
answered 2 hours ago
Abhimanyu Pallavi Sudhir
4,29842243
4,29842243
add a comment |Â
add a comment |Â
up vote
0
down vote
The formula is $vecF = p vecn A$ where $vecn$ is the unit vector perpendicular to the surface.
answered 3 hours ago
md2perpe
38425
add a comment |Â
up vote
0
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The formula is $vecF = p vecn A$ where $vecn$ is the unit vector perpendicular to the surface.
answered 3 hours ago
md2perpe
38425
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The formula is $vecF = p vecn A$ where $vecn$ is the unit vector perpendicular to the surface.
answered 3 hours ago
md2perpe
38425
The formula is $vecF = p vecn A$ where $vecn$ is the unit vector perpendicular to the surface.
answered 3 hours ago
md2perpe
38425
answered 3 hours ago
md2perpe
38425
answered 3 hours ago
md2perpe
38425
answered 3 hours ago
md2perpe
38425
38425
add a comment |Â
add a comment |Â
up vote
0
down vote
Pressure is defined by
$$
P = |F| / A
$$
where $|F|$ is the magnitude of the normal force, thus $P$ is a scalar.
answered 3 hours ago
innisfree
10.7k32655
This is incorrect. Consider the example of force being applied not perpendicular to the surface.
– Jitendra
3 hours ago
$|F|$ is the magnitude of the normal force
– innisfree
3 hours ago
by normal force do you mean taking the component along the direction perpendicular to the surface?
– Jitendra
3 hours ago
1
Yes I do mean that
– innisfree
3 hours ago
This calculates pressure based on some measurements, but doesn't define pressure. Doesn't tell you what pressure is. Pressure is there even if no surfaces or forces are present. In the middle of a gas bottle, gas has pressure. You can imagine putting a virtual surface there, and no matter how you orient it, you'd get the force of same magnitude, but that's just a side-effect.
– orion
3 hours ago
 |Â
show 1 more comment
up vote
0
down vote
Pressure is defined by
$$
P = |F| / A
$$
where $|F|$ is the magnitude of the normal force, thus $P$ is a scalar.
answered 3 hours ago
innisfree
10.7k32655
This is incorrect. Consider the example of force being applied not perpendicular to the surface.
– Jitendra
3 hours ago
$|F|$ is the magnitude of the normal force
– innisfree
3 hours ago
by normal force do you mean taking the component along the direction perpendicular to the surface?
– Jitendra
3 hours ago
1
Yes I do mean that
– innisfree
3 hours ago
This calculates pressure based on some measurements, but doesn't define pressure. Doesn't tell you what pressure is. Pressure is there even if no surfaces or forces are present. In the middle of a gas bottle, gas has pressure. You can imagine putting a virtual surface there, and no matter how you orient it, you'd get the force of same magnitude, but that's just a side-effect.
– orion
3 hours ago
 |Â
show 1 more comment
up vote
0
down vote
up vote
0
down vote
Pressure is defined by
$$
P = |F| / A
$$
where $|F|$ is the magnitude of the normal force, thus $P$ is a scalar.
answered 3 hours ago
innisfree
10.7k32655
Pressure is defined by
$$
P = |F| / A
$$
where $|F|$ is the magnitude of the normal force, thus $P$ is a scalar.
answered 3 hours ago
innisfree
10.7k32655
answered 3 hours ago
innisfree
10.7k32655
answered 3 hours ago
innisfree
10.7k32655
answered 3 hours ago
innisfree
10.7k32655
10.7k32655
This is incorrect. Consider the example of force being applied not perpendicular to the surface.
– Jitendra
3 hours ago
$|F|$ is the magnitude of the normal force
– innisfree
3 hours ago
by normal force do you mean taking the component along the direction perpendicular to the surface?
– Jitendra
3 hours ago
1
Yes I do mean that
– innisfree
3 hours ago
This calculates pressure based on some measurements, but doesn't define pressure. Doesn't tell you what pressure is. Pressure is there even if no surfaces or forces are present. In the middle of a gas bottle, gas has pressure. You can imagine putting a virtual surface there, and no matter how you orient it, you'd get the force of same magnitude, but that's just a side-effect.
– orion
3 hours ago
 |Â
show 1 more comment
This is incorrect. Consider the example of force being applied not perpendicular to the surface.
– Jitendra
3 hours ago
$|F|$ is the magnitude of the normal force
– innisfree
3 hours ago
by normal force do you mean taking the component along the direction perpendicular to the surface?
– Jitendra
3 hours ago
1
Yes I do mean that
– innisfree
3 hours ago
This calculates pressure based on some measurements, but doesn't define pressure. Doesn't tell you what pressure is. Pressure is there even if no surfaces or forces are present. In the middle of a gas bottle, gas has pressure. You can imagine putting a virtual surface there, and no matter how you orient it, you'd get the force of same magnitude, but that's just a side-effect.
– orion
3 hours ago
This is incorrect. Consider the example of force being applied not perpendicular to the surface.
– Jitendra
3 hours ago
This is incorrect. Consider the example of force being applied not perpendicular to the surface.
– Jitendra
3 hours ago
$|F|$ is the magnitude of the normal force
– innisfree
3 hours ago
$|F|$ is the magnitude of the normal force
– innisfree
3 hours ago
by normal force do you mean taking the component along the direction perpendicular to the surface?
– Jitendra
3 hours ago
by normal force do you mean taking the component along the direction perpendicular to the surface?
– Jitendra
3 hours ago
1
1
Yes I do mean that
– innisfree
3 hours ago
Yes I do mean that
– innisfree
3 hours ago
This calculates pressure based on some measurements, but doesn't define pressure. Doesn't tell you what pressure is. Pressure is there even if no surfaces or forces are present. In the middle of a gas bottle, gas has pressure. You can imagine putting a virtual surface there, and no matter how you orient it, you'd get the force of same magnitude, but that's just a side-effect.
– orion
3 hours ago
This calculates pressure based on some measurements, but doesn't define pressure. Doesn't tell you what pressure is. Pressure is there even if no surfaces or forces are present. In the middle of a gas bottle, gas has pressure. You can imagine putting a virtual surface there, and no matter how you orient it, you'd get the force of same magnitude, but that's just a side-effect.
– orion
3 hours ago
 |Â
show 1 more comment
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Possible duplicate of Define Pressure at A point. Why is it a Scalar?
– Kyle Kanos
33 mins ago
And also all linked therein
– Kyle Kanos
33 mins ago