Mixing
“Dividing through by Dp” in Joule Thomson effect
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In the wikipedia page for the Joule Thompson Effect, it says: 
I understand that the first formula is a 1-form on a 2-manifold. I don't understand why "Dividing by dP" is legal in this case. Is there a reference or theorem that would justify this rigorously?
differential-geometry differential-forms
asked Jan 19 at 17:32
user521247user521247
333
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In the wikipedia page for the Joule Thompson Effect, it says:
I understand that the first formula is a 1-form on a 2-manifold. I don't understand why "Dividing by dP" is legal in this case. Is there a reference or theorem that would justify this rigorously?
differential-geometry differential-forms
asked Jan 19 at 17:32
user521247user521247
333
$endgroup$
add a comment |
$begingroup$
In the wikipedia page for the Joule Thompson Effect, it says:
I understand that the first formula is a 1-form on a 2-manifold. I don't understand why "Dividing by dP" is legal in this case. Is there a reference or theorem that would justify this rigorously?
differential-geometry differential-forms
asked Jan 19 at 17:32
user521247user521247
333
$endgroup$
In the wikipedia page for the Joule Thompson Effect, it says:
I understand that the first formula is a 1-form on a 2-manifold. I don't understand why "Dividing by dP" is legal in this case. Is there a reference or theorem that would justify this rigorously?
differential-geometry differential-forms
differential-geometry differential-forms
asked Jan 19 at 17:32
user521247user521247
333
asked Jan 19 at 17:32
user521247user521247
333
asked Jan 19 at 17:32
user521247user521247
333
asked Jan 19 at 17:32
user521247user521247
333
asked Jan 19 at 17:32
user521247user521247
333
333
add a comment |
add a comment |
1 Answer
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Dividing by $dP$ is not legal - it's a case of Physicists being "sloppy". What is really happening here is that $S$ and $P$ provide a coordinate system of this manifold, and $T$ and $P$ also provide a coordinate system. Their respective dual bases of the tangent space are
$$left(frac{partial}{partial S}right)_P, left(frac{partial}{partial P}right)_S quadtext{and}quad left(frac{partial}{partial T}right)_P, left(frac{partial}{partial P}right)_T$$
Now $d H = left(frac{partial H}{partial T}right)_P d T + left(frac{partial H}{partial P}right)_T d P = T ,dS + V ,dP =
T left( left(frac{partial S}{partial T}right)_P d T + left(frac{partial S}{partial P}right)_T d Pright) + V , d P$
Since $d T$ and $d P$ are linearly independent your equation follows.
answered Jan 19 at 18:28
0x5390x539
1,460518
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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votes
$begingroup$
Dividing by $dP$ is not legal - it's a case of Physicists being "sloppy". What is really happening here is that $S$ and $P$ provide a coordinate system of this manifold, and $T$ and $P$ also provide a coordinate system. Their respective dual bases of the tangent space are
$$left(frac{partial}{partial S}right)_P, left(frac{partial}{partial P}right)_S quadtext{and}quad left(frac{partial}{partial T}right)_P, left(frac{partial}{partial P}right)_T$$
Now $d H = left(frac{partial H}{partial T}right)_P d T + left(frac{partial H}{partial P}right)_T d P = T ,dS + V ,dP =
T left( left(frac{partial S}{partial T}right)_P d T + left(frac{partial S}{partial P}right)_T d Pright) + V , d P$
Since $d T$ and $d P$ are linearly independent your equation follows.
answered Jan 19 at 18:28
0x5390x539
1,460518
$endgroup$
add a comment |
$begingroup$
Dividing by $dP$ is not legal - it's a case of Physicists being "sloppy". What is really happening here is that $S$ and $P$ provide a coordinate system of this manifold, and $T$ and $P$ also provide a coordinate system. Their respective dual bases of the tangent space are
$$left(frac{partial}{partial S}right)_P, left(frac{partial}{partial P}right)_S quadtext{and}quad left(frac{partial}{partial T}right)_P, left(frac{partial}{partial P}right)_T$$
Now $d H = left(frac{partial H}{partial T}right)_P d T + left(frac{partial H}{partial P}right)_T d P = T ,dS + V ,dP =
T left( left(frac{partial S}{partial T}right)_P d T + left(frac{partial S}{partial P}right)_T d Pright) + V , d P$
Since $d T$ and $d P$ are linearly independent your equation follows.
answered Jan 19 at 18:28
0x5390x539
1,460518
$endgroup$
add a comment |
$begingroup$
Dividing by $dP$ is not legal - it's a case of Physicists being "sloppy". What is really happening here is that $S$ and $P$ provide a coordinate system of this manifold, and $T$ and $P$ also provide a coordinate system. Their respective dual bases of the tangent space are
$$left(frac{partial}{partial S}right)_P, left(frac{partial}{partial P}right)_S quadtext{and}quad left(frac{partial}{partial T}right)_P, left(frac{partial}{partial P}right)_T$$
Now $d H = left(frac{partial H}{partial T}right)_P d T + left(frac{partial H}{partial P}right)_T d P = T ,dS + V ,dP =
T left( left(frac{partial S}{partial T}right)_P d T + left(frac{partial S}{partial P}right)_T d Pright) + V , d P$
Since $d T$ and $d P$ are linearly independent your equation follows.
answered Jan 19 at 18:28
0x5390x539
1,460518
$endgroup$
Dividing by $dP$ is not legal - it's a case of Physicists being "sloppy". What is really happening here is that $S$ and $P$ provide a coordinate system of this manifold, and $T$ and $P$ also provide a coordinate system. Their respective dual bases of the tangent space are
$$left(frac{partial}{partial S}right)_P, left(frac{partial}{partial P}right)_S quadtext{and}quad left(frac{partial}{partial T}right)_P, left(frac{partial}{partial P}right)_T$$
Now $d H = left(frac{partial H}{partial T}right)_P d T + left(frac{partial H}{partial P}right)_T d P = T ,dS + V ,dP =
T left( left(frac{partial S}{partial T}right)_P d T + left(frac{partial S}{partial P}right)_T d Pright) + V , d P$
Since $d T$ and $d P$ are linearly independent your equation follows.
answered Jan 19 at 18:28
0x5390x539
1,460518
answered Jan 19 at 18:28
0x5390x539
1,460518
answered Jan 19 at 18:28
0x5390x539
1,460518
answered Jan 19 at 18:28
0x5390x539
1,460518
1,460518
add a comment |
add a comment |
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