Mixing
I need to prove a few vector identities using Cartesian Tensor Notation, and I can't figure out how!
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I have been all over the internet, but I just can't make sense of this stuff. I have done my best to learn from my textbook and different websites, but this is confusing for me. I haven't taken any calculus in years, and I'm jumping in headfirst. If anyone can help me understand how to prove these using Cartesian Tensor Notation, I would really appreciate it!
First identity: $nabla times (nabla times a) = nabla cdot (nabla cdot a) - nabla^2a$
Second identity: $nabla cdot (ab) = a cdot nabla b +b(nabla cdot a)$
Third identity: $nabla cdot (delta f) = nabla f$
Fourth identity: $delta : nabla a = nabla cdot a $
Where $delta f$ represents a small change in $f$.
Thanks everyone
calculus divergence curl
edited Jan 29 at 18:53
Victoria M
42618
asked Jan 29 at 17:13
JasenJasen
61
$endgroup$
add a comment |
$begingroup$
I have been all over the internet, but I just can't make sense of this stuff. I have done my best to learn from my textbook and different websites, but this is confusing for me. I haven't taken any calculus in years, and I'm jumping in headfirst. If anyone can help me understand how to prove these using Cartesian Tensor Notation, I would really appreciate it!
First identity: $nabla times (nabla times a) = nabla cdot (nabla cdot a) - nabla^2a$
Second identity: $nabla cdot (ab) = a cdot nabla b +b(nabla cdot a)$
Third identity: $nabla cdot (delta f) = nabla f$
Fourth identity: $delta : nabla a = nabla cdot a $
Where $delta f$ represents a small change in $f$.
Thanks everyone
calculus divergence curl
edited Jan 29 at 18:53
Victoria M
42618
asked Jan 29 at 17:13
JasenJasen
61
$endgroup$
$begingroup$
I recommend only using $cdot$ in your notation for dot products, not ordinary scalar multiplication. For example, your first identity uses them both.
$endgroup$
– J.G.
Jan 29 at 18:59
add a comment |
$begingroup$
I have been all over the internet, but I just can't make sense of this stuff. I have done my best to learn from my textbook and different websites, but this is confusing for me. I haven't taken any calculus in years, and I'm jumping in headfirst. If anyone can help me understand how to prove these using Cartesian Tensor Notation, I would really appreciate it!
First identity: $nabla times (nabla times a) = nabla cdot (nabla cdot a) - nabla^2a$
Second identity: $nabla cdot (ab) = a cdot nabla b +b(nabla cdot a)$
Third identity: $nabla cdot (delta f) = nabla f$
Fourth identity: $delta : nabla a = nabla cdot a $
Where $delta f$ represents a small change in $f$.
Thanks everyone
calculus divergence curl
edited Jan 29 at 18:53
Victoria M
42618
asked Jan 29 at 17:13
JasenJasen
61
$endgroup$
I have been all over the internet, but I just can't make sense of this stuff. I have done my best to learn from my textbook and different websites, but this is confusing for me. I haven't taken any calculus in years, and I'm jumping in headfirst. If anyone can help me understand how to prove these using Cartesian Tensor Notation, I would really appreciate it!
First identity: $nabla times (nabla times a) = nabla cdot (nabla cdot a) - nabla^2a$
Second identity: $nabla cdot (ab) = a cdot nabla b +b(nabla cdot a)$
Third identity: $nabla cdot (delta f) = nabla f$
Fourth identity: $delta : nabla a = nabla cdot a $
Where $delta f$ represents a small change in $f$.
Thanks everyone
calculus divergence curl
calculus divergence curl
edited Jan 29 at 18:53
Victoria M
42618
asked Jan 29 at 17:13
JasenJasen
61
edited Jan 29 at 18:53
Victoria M
42618
asked Jan 29 at 17:13
JasenJasen
61
edited Jan 29 at 18:53
Victoria M
42618
edited Jan 29 at 18:53
Victoria M
42618
edited Jan 29 at 18:53
Victoria M
42618
42618
asked Jan 29 at 17:13
JasenJasen
61
asked Jan 29 at 17:13
JasenJasen
61
asked Jan 29 at 17:13
JasenJasen
61
61
$begingroup$
I recommend only using $cdot$ in your notation for dot products, not ordinary scalar multiplication. For example, your first identity uses them both.
$endgroup$
– J.G.
Jan 29 at 18:59
add a comment |
$begingroup$
I recommend only using $cdot$ in your notation for dot products, not ordinary scalar multiplication. For example, your first identity uses them both.
$endgroup$
– J.G.
Jan 29 at 18:59
$begingroup$
I recommend only using $cdot$ in your notation for dot products, not ordinary scalar multiplication. For example, your first identity uses them both.
$endgroup$
– J.G.
Jan 29 at 18:59
$begingroup$
I recommend only using $cdot$ in your notation for dot products, not ordinary scalar multiplication. For example, your first identity uses them both.
$endgroup$
– J.G.
Jan 29 at 18:59
add a comment |
1 Answer
1
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oldest
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Recall the following: $$ text{grad($it f$)} = nabla f = frac{partial f}{partial x}hat i + frac{partial f}{partial y}hat j + frac{partial f}{partial z}hat k \ nabla cdot f = frac{partial f}{partial x} + frac{partial f}{partial y} + frac{partial f}{partial z} \ nabla times f = begin{vmatrix} hat i & hat j & hat k \ frac{partial }{partial x} & frac{partial }{partial y} & frac{partial }{partial z} \ frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z}end{vmatrix}$$
I will prove the second identity as an example, and the others will follow via the same methods.
$text{Using the first two formulae: }nabla cdot (ab) = frac{partial (ab)}{partial x} + frac{partial (ab)}{partial y} + frac{partial (ab)}{partial z} = text{(by the product rule) } (frac{partial a}{partial x} b + afrac{partial b}{partial x}) + (frac{partial a}{partial y} b + afrac{partial b}{partial y}) + (frac{partial a}{partial z} b + afrac{partial b}{partial z}) = a(frac{partial (b)}{partial x} + frac{partial (b)}{partial y} + frac{partial (b)}{partial z}) + b(frac{partial (a)}{partial x} + frac{partial (a)}{partial y} + frac{partial (a)}{partial z}) = a cdot (nabla b) + b (nabla cdot a) space blacksquare$
The others would be good exercise, if you are stuck leave a comment and I will edit my answer to include hints.
answered Jan 29 at 19:00
Victoria MVictoria M
42618
$endgroup$
$begingroup$
Hello Victoria, thanks for your help! The only problem is that I don't think your answer proves it using Cartesian Tensor Notation. That is just proving utilizing normal Cartesian notation. Hopefully you have an idea of where to go from here!
$endgroup$
– Jasen
Jan 29 at 20:20
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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votes
$begingroup$
Recall the following: $$ text{grad($it f$)} = nabla f = frac{partial f}{partial x}hat i + frac{partial f}{partial y}hat j + frac{partial f}{partial z}hat k \ nabla cdot f = frac{partial f}{partial x} + frac{partial f}{partial y} + frac{partial f}{partial z} \ nabla times f = begin{vmatrix} hat i & hat j & hat k \ frac{partial }{partial x} & frac{partial }{partial y} & frac{partial }{partial z} \ frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z}end{vmatrix}$$
I will prove the second identity as an example, and the others will follow via the same methods.
$text{Using the first two formulae: }nabla cdot (ab) = frac{partial (ab)}{partial x} + frac{partial (ab)}{partial y} + frac{partial (ab)}{partial z} = text{(by the product rule) } (frac{partial a}{partial x} b + afrac{partial b}{partial x}) + (frac{partial a}{partial y} b + afrac{partial b}{partial y}) + (frac{partial a}{partial z} b + afrac{partial b}{partial z}) = a(frac{partial (b)}{partial x} + frac{partial (b)}{partial y} + frac{partial (b)}{partial z}) + b(frac{partial (a)}{partial x} + frac{partial (a)}{partial y} + frac{partial (a)}{partial z}) = a cdot (nabla b) + b (nabla cdot a) space blacksquare$
The others would be good exercise, if you are stuck leave a comment and I will edit my answer to include hints.
answered Jan 29 at 19:00
Victoria MVictoria M
42618
$endgroup$
$begingroup$
Hello Victoria, thanks for your help! The only problem is that I don't think your answer proves it using Cartesian Tensor Notation. That is just proving utilizing normal Cartesian notation. Hopefully you have an idea of where to go from here!
$endgroup$
– Jasen
Jan 29 at 20:20
add a comment |
$begingroup$
Recall the following: $$ text{grad($it f$)} = nabla f = frac{partial f}{partial x}hat i + frac{partial f}{partial y}hat j + frac{partial f}{partial z}hat k \ nabla cdot f = frac{partial f}{partial x} + frac{partial f}{partial y} + frac{partial f}{partial z} \ nabla times f = begin{vmatrix} hat i & hat j & hat k \ frac{partial }{partial x} & frac{partial }{partial y} & frac{partial }{partial z} \ frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z}end{vmatrix}$$
I will prove the second identity as an example, and the others will follow via the same methods.
$text{Using the first two formulae: }nabla cdot (ab) = frac{partial (ab)}{partial x} + frac{partial (ab)}{partial y} + frac{partial (ab)}{partial z} = text{(by the product rule) } (frac{partial a}{partial x} b + afrac{partial b}{partial x}) + (frac{partial a}{partial y} b + afrac{partial b}{partial y}) + (frac{partial a}{partial z} b + afrac{partial b}{partial z}) = a(frac{partial (b)}{partial x} + frac{partial (b)}{partial y} + frac{partial (b)}{partial z}) + b(frac{partial (a)}{partial x} + frac{partial (a)}{partial y} + frac{partial (a)}{partial z}) = a cdot (nabla b) + b (nabla cdot a) space blacksquare$
The others would be good exercise, if you are stuck leave a comment and I will edit my answer to include hints.
answered Jan 29 at 19:00
Victoria MVictoria M
42618
$endgroup$
$begingroup$
Hello Victoria, thanks for your help! The only problem is that I don't think your answer proves it using Cartesian Tensor Notation. That is just proving utilizing normal Cartesian notation. Hopefully you have an idea of where to go from here!
$endgroup$
– Jasen
Jan 29 at 20:20
add a comment |
$begingroup$
Recall the following: $$ text{grad($it f$)} = nabla f = frac{partial f}{partial x}hat i + frac{partial f}{partial y}hat j + frac{partial f}{partial z}hat k \ nabla cdot f = frac{partial f}{partial x} + frac{partial f}{partial y} + frac{partial f}{partial z} \ nabla times f = begin{vmatrix} hat i & hat j & hat k \ frac{partial }{partial x} & frac{partial }{partial y} & frac{partial }{partial z} \ frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z}end{vmatrix}$$
I will prove the second identity as an example, and the others will follow via the same methods.
$text{Using the first two formulae: }nabla cdot (ab) = frac{partial (ab)}{partial x} + frac{partial (ab)}{partial y} + frac{partial (ab)}{partial z} = text{(by the product rule) } (frac{partial a}{partial x} b + afrac{partial b}{partial x}) + (frac{partial a}{partial y} b + afrac{partial b}{partial y}) + (frac{partial a}{partial z} b + afrac{partial b}{partial z}) = a(frac{partial (b)}{partial x} + frac{partial (b)}{partial y} + frac{partial (b)}{partial z}) + b(frac{partial (a)}{partial x} + frac{partial (a)}{partial y} + frac{partial (a)}{partial z}) = a cdot (nabla b) + b (nabla cdot a) space blacksquare$
The others would be good exercise, if you are stuck leave a comment and I will edit my answer to include hints.
answered Jan 29 at 19:00
Victoria MVictoria M
42618
$endgroup$
Recall the following: $$ text{grad($it f$)} = nabla f = frac{partial f}{partial x}hat i + frac{partial f}{partial y}hat j + frac{partial f}{partial z}hat k \ nabla cdot f = frac{partial f}{partial x} + frac{partial f}{partial y} + frac{partial f}{partial z} \ nabla times f = begin{vmatrix} hat i & hat j & hat k \ frac{partial }{partial x} & frac{partial }{partial y} & frac{partial }{partial z} \ frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z}end{vmatrix}$$
I will prove the second identity as an example, and the others will follow via the same methods.
$text{Using the first two formulae: }nabla cdot (ab) = frac{partial (ab)}{partial x} + frac{partial (ab)}{partial y} + frac{partial (ab)}{partial z} = text{(by the product rule) } (frac{partial a}{partial x} b + afrac{partial b}{partial x}) + (frac{partial a}{partial y} b + afrac{partial b}{partial y}) + (frac{partial a}{partial z} b + afrac{partial b}{partial z}) = a(frac{partial (b)}{partial x} + frac{partial (b)}{partial y} + frac{partial (b)}{partial z}) + b(frac{partial (a)}{partial x} + frac{partial (a)}{partial y} + frac{partial (a)}{partial z}) = a cdot (nabla b) + b (nabla cdot a) space blacksquare$
The others would be good exercise, if you are stuck leave a comment and I will edit my answer to include hints.
answered Jan 29 at 19:00
Victoria MVictoria M
42618
answered Jan 29 at 19:00
Victoria MVictoria M
42618
answered Jan 29 at 19:00
Victoria MVictoria M
42618
answered Jan 29 at 19:00
Victoria MVictoria M
42618
42618
$begingroup$
Hello Victoria, thanks for your help! The only problem is that I don't think your answer proves it using Cartesian Tensor Notation. That is just proving utilizing normal Cartesian notation. Hopefully you have an idea of where to go from here!
$endgroup$
– Jasen
Jan 29 at 20:20
add a comment |
$begingroup$
Hello Victoria, thanks for your help! The only problem is that I don't think your answer proves it using Cartesian Tensor Notation. That is just proving utilizing normal Cartesian notation. Hopefully you have an idea of where to go from here!
$endgroup$
– Jasen
Jan 29 at 20:20
$begingroup$
Hello Victoria, thanks for your help! The only problem is that I don't think your answer proves it using Cartesian Tensor Notation. That is just proving utilizing normal Cartesian notation. Hopefully you have an idea of where to go from here!
$endgroup$
– Jasen
Jan 29 at 20:20
$begingroup$
Hello Victoria, thanks for your help! The only problem is that I don't think your answer proves it using Cartesian Tensor Notation. That is just proving utilizing normal Cartesian notation. Hopefully you have an idea of where to go from here!
$endgroup$
– Jasen
Jan 29 at 20:20
add a comment |
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$begingroup$
I recommend only using $cdot$ in your notation for dot products, not ordinary scalar multiplication. For example, your first identity uses them both.
$endgroup$
– J.G.
Jan 29 at 18:59