Why is the ratio of two extensive quantities always intensive?
$begingroup$
Is this something that we observe that always happens or is there some fundamental reason for two extensive quantities to give an intensive when divided?
thermodynamics definition volume scaling
edited 2 days ago
David Z♦
63.9k23136252
asked 2 days ago
paokara moupaokara mou
1436
$endgroup$
add a comment |
$begingroup$
Is this something that we observe that always happens or is there some fundamental reason for two extensive quantities to give an intensive when divided?
thermodynamics definition volume scaling
edited 2 days ago
David Z♦
63.9k23136252
asked 2 days ago
paokara moupaokara mou
1436
$endgroup$
add a comment |
$begingroup$
Is this something that we observe that always happens or is there some fundamental reason for two extensive quantities to give an intensive when divided?
thermodynamics definition volume scaling
edited 2 days ago
David Z♦
63.9k23136252
asked 2 days ago
paokara moupaokara mou
1436
$endgroup$
Is this something that we observe that always happens or is there some fundamental reason for two extensive quantities to give an intensive when divided?
thermodynamics definition volume scaling
thermodynamics definition volume scaling
edited 2 days ago
David Z♦
63.9k23136252
asked 2 days ago
paokara moupaokara mou
1436
edited 2 days ago
David Z♦
63.9k23136252
asked 2 days ago
paokara moupaokara mou
1436
edited 2 days ago
David Z♦
63.9k23136252
edited 2 days ago
David Z♦
63.9k23136252
edited 2 days ago
David Z♦
63.9k23136252
63.9k23136252
asked 2 days ago
paokara moupaokara mou
1436
asked 2 days ago
paokara moupaokara mou
1436
asked 2 days ago
paokara moupaokara mou
1436
1436
add a comment |
add a comment |
1 Answer
1
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$begingroup$
It is mainly a mathematical reason. Extensive quantities grow with system size. If two quantities scale in the same way with a variable (in this case system size), it cancels out in the division.
Mini-example: $A$ and $B$ are extensive physical quantities both dependent on $n$. Their ratio is called $C = A / B$. If you scale the system up, $A$ and $B$ grow by a factor of $n$. What happens to $C$?
$frac{A cdot n}{B cdot n} = frac{A}{B}$
$C$ stays the same, irrespective of $n$. Hence, $C$ is intensive. The most common physical example is mass and volume, which scale with system size and still exhibit the same ratio, the density.
EDIT including the comment of probably_someone: The argumentation is particularly true since by definition an extensive quantity grows linearly with system size. This justifies the proportionality that I presented in the mini-example.
answered 2 days ago
lmrlmr
1,074520
$endgroup$
8
$begingroup$
In particular, this is true because "extensive" is specifically defined as growing linearly with system size (see e.g. en.wikipedia.org/wiki/Intensive_and_extensive_properties), which raises the question: what do we call a property that grows nonlinearly with system size (for example, as the square of the volume)?
$endgroup$
– probably_someone
2 days ago
$begingroup$
Yeah, I did not point this out explicitly. I added a few sentences to include the linearity.
$endgroup$
– lmr
2 days ago
$begingroup$
Technically couldn't the linear relations have different "slopes", so that the part dependant on the size still cancels, but there will be some extra constant factor multiplying your ratio there?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
@AaronStevens Well mathematically, it is definitely possible. I can't think of a suitable example right now though. But as you pointed out yourself, the ratio will still remain intensive.
$endgroup$
– lmr
2 days ago
4
$begingroup$
@AaronStevens Whatever the factor is is already included in $A$ and $B$ in this answer. And in particular, if the quantities have different units then not only are the slopes different, they have different units so they are clearly very different, but all that is automatically accounted for in the division. Having the same unit but different magnitude is no different.
$endgroup$
– JiK
2 days ago
|
show 2 more comments
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
It is mainly a mathematical reason. Extensive quantities grow with system size. If two quantities scale in the same way with a variable (in this case system size), it cancels out in the division.
Mini-example: $A$ and $B$ are extensive physical quantities both dependent on $n$. Their ratio is called $C = A / B$. If you scale the system up, $A$ and $B$ grow by a factor of $n$. What happens to $C$?
$frac{A cdot n}{B cdot n} = frac{A}{B}$
$C$ stays the same, irrespective of $n$. Hence, $C$ is intensive. The most common physical example is mass and volume, which scale with system size and still exhibit the same ratio, the density.
EDIT including the comment of probably_someone: The argumentation is particularly true since by definition an extensive quantity grows linearly with system size. This justifies the proportionality that I presented in the mini-example.
answered 2 days ago
lmrlmr
1,074520
$endgroup$
8
$begingroup$
In particular, this is true because "extensive" is specifically defined as growing linearly with system size (see e.g. en.wikipedia.org/wiki/Intensive_and_extensive_properties), which raises the question: what do we call a property that grows nonlinearly with system size (for example, as the square of the volume)?
$endgroup$
– probably_someone
2 days ago
$begingroup$
Yeah, I did not point this out explicitly. I added a few sentences to include the linearity.
$endgroup$
– lmr
2 days ago
$begingroup$
Technically couldn't the linear relations have different "slopes", so that the part dependant on the size still cancels, but there will be some extra constant factor multiplying your ratio there?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
@AaronStevens Well mathematically, it is definitely possible. I can't think of a suitable example right now though. But as you pointed out yourself, the ratio will still remain intensive.
$endgroup$
– lmr
2 days ago
4
$begingroup$
@AaronStevens Whatever the factor is is already included in $A$ and $B$ in this answer. And in particular, if the quantities have different units then not only are the slopes different, they have different units so they are clearly very different, but all that is automatically accounted for in the division. Having the same unit but different magnitude is no different.
$endgroup$
– JiK
2 days ago
|
show 2 more comments
$begingroup$
It is mainly a mathematical reason. Extensive quantities grow with system size. If two quantities scale in the same way with a variable (in this case system size), it cancels out in the division.
Mini-example: $A$ and $B$ are extensive physical quantities both dependent on $n$. Their ratio is called $C = A / B$. If you scale the system up, $A$ and $B$ grow by a factor of $n$. What happens to $C$?
$frac{A cdot n}{B cdot n} = frac{A}{B}$
$C$ stays the same, irrespective of $n$. Hence, $C$ is intensive. The most common physical example is mass and volume, which scale with system size and still exhibit the same ratio, the density.
EDIT including the comment of probably_someone: The argumentation is particularly true since by definition an extensive quantity grows linearly with system size. This justifies the proportionality that I presented in the mini-example.
answered 2 days ago
lmrlmr
1,074520
$endgroup$
8
$begingroup$
In particular, this is true because "extensive" is specifically defined as growing linearly with system size (see e.g. en.wikipedia.org/wiki/Intensive_and_extensive_properties), which raises the question: what do we call a property that grows nonlinearly with system size (for example, as the square of the volume)?
$endgroup$
– probably_someone
2 days ago
$begingroup$
Yeah, I did not point this out explicitly. I added a few sentences to include the linearity.
$endgroup$
– lmr
2 days ago
$begingroup$
Technically couldn't the linear relations have different "slopes", so that the part dependant on the size still cancels, but there will be some extra constant factor multiplying your ratio there?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
@AaronStevens Well mathematically, it is definitely possible. I can't think of a suitable example right now though. But as you pointed out yourself, the ratio will still remain intensive.
$endgroup$
– lmr
2 days ago
4
$begingroup$
@AaronStevens Whatever the factor is is already included in $A$ and $B$ in this answer. And in particular, if the quantities have different units then not only are the slopes different, they have different units so they are clearly very different, but all that is automatically accounted for in the division. Having the same unit but different magnitude is no different.
$endgroup$
– JiK
2 days ago
|
show 2 more comments
$begingroup$
It is mainly a mathematical reason. Extensive quantities grow with system size. If two quantities scale in the same way with a variable (in this case system size), it cancels out in the division.
Mini-example: $A$ and $B$ are extensive physical quantities both dependent on $n$. Their ratio is called $C = A / B$. If you scale the system up, $A$ and $B$ grow by a factor of $n$. What happens to $C$?
$frac{A cdot n}{B cdot n} = frac{A}{B}$
$C$ stays the same, irrespective of $n$. Hence, $C$ is intensive. The most common physical example is mass and volume, which scale with system size and still exhibit the same ratio, the density.
EDIT including the comment of probably_someone: The argumentation is particularly true since by definition an extensive quantity grows linearly with system size. This justifies the proportionality that I presented in the mini-example.
answered 2 days ago
lmrlmr
1,074520
$endgroup$
It is mainly a mathematical reason. Extensive quantities grow with system size. If two quantities scale in the same way with a variable (in this case system size), it cancels out in the division.
Mini-example: $A$ and $B$ are extensive physical quantities both dependent on $n$. Their ratio is called $C = A / B$. If you scale the system up, $A$ and $B$ grow by a factor of $n$. What happens to $C$?
$frac{A cdot n}{B cdot n} = frac{A}{B}$
$C$ stays the same, irrespective of $n$. Hence, $C$ is intensive. The most common physical example is mass and volume, which scale with system size and still exhibit the same ratio, the density.
EDIT including the comment of probably_someone: The argumentation is particularly true since by definition an extensive quantity grows linearly with system size. This justifies the proportionality that I presented in the mini-example.
answered 2 days ago
lmrlmr
1,074520
edited 2 days ago
answered 2 days ago
lmrlmr
1,074520
answered 2 days ago
lmrlmr
1,074520
answered 2 days ago
lmrlmr
1,074520
1,074520
8
$begingroup$
In particular, this is true because "extensive" is specifically defined as growing linearly with system size (see e.g. en.wikipedia.org/wiki/Intensive_and_extensive_properties), which raises the question: what do we call a property that grows nonlinearly with system size (for example, as the square of the volume)?
$endgroup$
– probably_someone
2 days ago
$begingroup$
Yeah, I did not point this out explicitly. I added a few sentences to include the linearity.
$endgroup$
– lmr
2 days ago
$begingroup$
Technically couldn't the linear relations have different "slopes", so that the part dependant on the size still cancels, but there will be some extra constant factor multiplying your ratio there?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
@AaronStevens Well mathematically, it is definitely possible. I can't think of a suitable example right now though. But as you pointed out yourself, the ratio will still remain intensive.
$endgroup$
– lmr
2 days ago
4
$begingroup$
@AaronStevens Whatever the factor is is already included in $A$ and $B$ in this answer. And in particular, if the quantities have different units then not only are the slopes different, they have different units so they are clearly very different, but all that is automatically accounted for in the division. Having the same unit but different magnitude is no different.
$endgroup$
– JiK
2 days ago
|
show 2 more comments
8
$begingroup$
In particular, this is true because "extensive" is specifically defined as growing linearly with system size (see e.g. en.wikipedia.org/wiki/Intensive_and_extensive_properties), which raises the question: what do we call a property that grows nonlinearly with system size (for example, as the square of the volume)?
$endgroup$
– probably_someone
2 days ago
$begingroup$
Yeah, I did not point this out explicitly. I added a few sentences to include the linearity.
$endgroup$
– lmr
2 days ago
$begingroup$
Technically couldn't the linear relations have different "slopes", so that the part dependant on the size still cancels, but there will be some extra constant factor multiplying your ratio there?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
@AaronStevens Well mathematically, it is definitely possible. I can't think of a suitable example right now though. But as you pointed out yourself, the ratio will still remain intensive.
$endgroup$
– lmr
2 days ago
4
$begingroup$
@AaronStevens Whatever the factor is is already included in $A$ and $B$ in this answer. And in particular, if the quantities have different units then not only are the slopes different, they have different units so they are clearly very different, but all that is automatically accounted for in the division. Having the same unit but different magnitude is no different.
$endgroup$
– JiK
2 days ago
8
8
$begingroup$
In particular, this is true because "extensive" is specifically defined as growing linearly with system size (see e.g. en.wikipedia.org/wiki/Intensive_and_extensive_properties), which raises the question: what do we call a property that grows nonlinearly with system size (for example, as the square of the volume)?
$endgroup$
– probably_someone
2 days ago
$begingroup$
In particular, this is true because "extensive" is specifically defined as growing linearly with system size (see e.g. en.wikipedia.org/wiki/Intensive_and_extensive_properties), which raises the question: what do we call a property that grows nonlinearly with system size (for example, as the square of the volume)?
$endgroup$
– probably_someone
2 days ago
$begingroup$
Yeah, I did not point this out explicitly. I added a few sentences to include the linearity.
$endgroup$
– lmr
2 days ago
$begingroup$
Yeah, I did not point this out explicitly. I added a few sentences to include the linearity.
$endgroup$
– lmr
2 days ago
$begingroup$
Technically couldn't the linear relations have different "slopes", so that the part dependant on the size still cancels, but there will be some extra constant factor multiplying your ratio there?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
Technically couldn't the linear relations have different "slopes", so that the part dependant on the size still cancels, but there will be some extra constant factor multiplying your ratio there?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
@AaronStevens Well mathematically, it is definitely possible. I can't think of a suitable example right now though. But as you pointed out yourself, the ratio will still remain intensive.
$endgroup$
– lmr
2 days ago
$begingroup$
@AaronStevens Well mathematically, it is definitely possible. I can't think of a suitable example right now though. But as you pointed out yourself, the ratio will still remain intensive.
$endgroup$
– lmr
2 days ago
4
4
$begingroup$
@AaronStevens Whatever the factor is is already included in $A$ and $B$ in this answer. And in particular, if the quantities have different units then not only are the slopes different, they have different units so they are clearly very different, but all that is automatically accounted for in the division. Having the same unit but different magnitude is no different.
$endgroup$
– JiK
2 days ago
$begingroup$
@AaronStevens Whatever the factor is is already included in $A$ and $B$ in this answer. And in particular, if the quantities have different units then not only are the slopes different, they have different units so they are clearly very different, but all that is automatically accounted for in the division. Having the same unit but different magnitude is no different.
$endgroup$
– JiK
2 days ago
|
show 2 more comments
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