|
|
| Line 1: |
Line 1: |
| {{Calculus |Multivariable}}
| | When you were younger it didnt appear to matter what you ate you always had a lot of vitality and for most folks which meant our fat will be kept down moreover. But because we receive older our bodies changes and the processes that you took for granted whenever we were younger dont work because well. Such is the case for metabolism. Unfortunately metabolism has become an easy target over the years as something to blame weight gain on. But that really isnt the case; you want metabolism incredibly whenever you are older. One technique to increase it happens to be from strength training.<br><br>It is wise [http://safedietplansforwomen.com/bmr-calculator bmr calculator] to have a superior scale to weigh on. I weigh everyday, however we will not wish To. I simply can't appear to help me, and since I track it on the Calorie-Count website, it is very something that I am excited to do daily.<br><br>Your basal metabolic rate (BMR)Your basal metabolic rate is significant whenever planning a fat reduction system. It shows the rate a body burns calories simply for simple metabolic functions, i.e. how countless calories would you burn for a day when you merely lie in bed. "Lying inside bed" isn't pretty exact of course, because in the event you are thinking or having conversation while in bed, the body will nonetheless burn more calories than your BMR. The BMR shows simply the minimum calories required to remain alive.<br><br>So how much water is enough? Probably not the eight glasses we've always been told. Eight 8-ounce glasses is fine should you only weigh 130 pounds. To calculate how much water you require, divide your weight inside half. You must drink that countless ounces of water each day. If you weigh 180 pounds, you need to drink 90 ounces of water.<br><br>Body Surface Area: The height plus fat lead a lot in determining bmr. The greater is the body surface region, the high is the BMR. Thin, tall persons have a higher BMR.<br><br>To calculate the activity level, see this free online calculator: http://exercise.about.com/cs/fitnesstools/l/blcalorieburn.htm which usually use your current weight, the sort of exercise, plus the amount of time you can afford to invest doing that exercise to tell you how various calories you're going to burn. Take note of the results - should you use this calculator realistically, you are able to plug it right into the BMR results.<br><br>There are 3 ways to create this deficit: diet, exercise, or perhaps a combination. If you combine techniques we have a better chance of following it. Just think, instead of completely cutting out that afternoon snack, or endangering injury by functioning out too difficult, we may really have a lighter snack and take the stairs. Its easier to create small changes which can add about 500 calories a day. |
| | |
| On a [[differentiable manifold]], the '''exterior derivative''' extends the concept of the [[pushforward (differential)|differential]] of a function to [[differential form]]s of higher degree. The exterior derivative was first described in its current form by [[Élie Cartan]]; it allows for a natural, metric-independent generalization of [[Stokes' theorem]], [[Gauss's theorem]], and [[Green's theorem]] from vector calculus.
| |
| | |
| If a ''k''-form is thought of as measuring the flux through an infinitesimal ''k''-parallelepiped, then its exterior derivative can be thought of as measuring the net flux through the boundary of a (''k''+1)-parallelepiped.
| |
| | |
| == Definition ==
| |
| | |
| The exterior derivative of a differential form of degree ''k'' is a differential form of degree {{nowrap|1=''k'' + 1.}}
| |
| | |
| If ''f'' is a smooth function (a 0-form), then the exterior derivative of ''f'' is the [[Pushforward (differential)|differential]] of ''f''. That is, d''f'' is the unique [[1-form]] such that for every smooth [[vector field]] ''X'', {{nowrap|d''f''(''X'') {{=}} d<sub>''X''</sub>''f''}}, where d<sub>''X''</sub>''f'' is the directional derivative of ''f'' in the direction of ''X''.
| |
| | |
| There are a variety of equivalent definitions of the exterior derivative of a general ''k''-form.
| |
| | |
| ===Axioms for the exterior derivative===
| |
| The exterior derivative is defined to be the unique '''R'''-linear mapping from ''k''-forms to (''k''+1)-forms satisfying the following properties:
| |
| | |
| # d''f'' is the differential of ''f'' for smooth functions ''f''.
| |
| # {{nowrap|1=d(d''f'') = 0}} for any smooth function ''f''.
| |
| # {{nowrap|1=d(''α''∧''β'') = d''α''∧''β'' + (−1)<sup>''p''</sup> (''α''∧d''β'')}} where α is a ''p''-form. That is to say, d is an [[derivation (algebra)|antiderivation]] of degree 1 on the [[exterior algebra]] of differential forms.
| |
| | |
| The second defining property holds in more generality: in fact, {{nowrap|1=d(d''α'') = 0}} for any ''k''-form ''α''; more succinctly, {{nowrap|1=d<sup>2</sup> = 0}}. The third defining property implies as a special case that if ''f'' is a function and α a ''k''-form, then {{nowrap|1=d(''fα'') = d(''f''∧''α'') = d''f''∧''α'' + ''f''∧d''α''}} because functions are 0-forms, and scalar multiplication and the exterior product are equivalent when one of the arguments is a scalar.
| |
| | |
| ===Exterior derivative in local coordinates===
| |
| Alternatively, one can work entirely in a [[local coordinate system]] (''x''<sup>1</sup>, ..., ''x''<sup>''n''</sup>). First, the coordinate differentials d''x''<sup>1</sup>, ..., d''x''<sup>''n''</sup> form a basic set of one-forms within the [[coordinate chart]]. The formulas in this section rely on the [[Einstein summation convention]]. Given a [[multi-index]] {{nowrap|''I'' {{=}} (''i''<sub>1</sub>, ..., ''i''<sub>''k''</sub>)}} with {{nowrap|1 ≤ ''i<sub>p</sub>'' ≤ ''n''}} for {{nowrap|1 ≤ ''p'' ≤ ''k''}}, the exterior derivative of a ''k''-form
| |
| :<math>\omega = f_I\mathrm{d}x^I=f_{i_1,i_2\cdots i_k}\mathrm{d}x^{i_1}\wedge \mathrm{d}x^{i_2}\wedge\cdots\wedge \mathrm{d}x^{i_k}</math>
| |
| over '''R'''<sup>''n''</sup> is defined as
| |
| :<math>\mathrm{d}{\omega} = \sum_{i=1}^n \frac{\partial f_I}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}x^I.</math>
| |
| | |
| For general ''k''-forms <math>\omega = \sum_I f_I\; \mathrm{d}x^I</math>, where the components of the multi-index ''I'' run over all the values in {1, ..., ''n''}, the definition of the exterior derivative is extended [[linear]]ly. Note that whenever ''i'' is one of the components of the multi-index ''I'' then {{nowrap|d''x<sup>i</sup>''∧d''x<sup>I</sup>'' {{=}} 0}} (see [[wedge product]]).
| |
| | |
| The definition of the exterior derivative in local coordinates follows from the preceding definition. Indeed, if {{nowrap|''ω'' {{=}} ''f''<sub>''I''</sub> d''x''<sup>''i''<sub>1</sub></sup>∧···∧d''x''<sup>''i''<sub>''k''</sub></sup>}}, then
| |
| :<math>\begin{align}
| |
| \mathrm{d}{\omega} &= \mathrm{d} (f_I \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k} ) \\
| |
| &= \mathrm{d}f_I \wedge (\mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k}) +
| |
| f_I \mathrm{d}(\mathrm{d}x^{i_1}\wedge \cdots \wedge \mathrm{d}x^{i_k}) \\
| |
| &= \mathrm{d}f_I \wedge \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k} +
| |
| \sum_{p=1}^k (-1)^{(p-1)}f_I \mathrm{d}x^{i_1} \wedge \cdots
| |
| \wedge \mathrm{d}x^{i_{p-1}}\wedge \mathrm{d}^2x^{i_p} \wedge \mathrm{d}x^{i_{p+1}}\wedge \cdots
| |
| \wedge \mathrm{d}x^{i_k} \\
| |
| &= \mathrm{d}f_I \wedge \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k} \\
| |
| &= \sum_{i=1}^n \frac{\partial f_I}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k} \\
| |
| \end{align}</math>
| |
| | |
| Here, we have here interpreted ''f''<sub>''I''</sub> as a 0-form, and then applied the properties of the exterior derivative.
| |
| | |
| ===Invariant formula===
| |
| Alternatively, an explicit formula can be given for the exterior derivative of a ''k''-form ''ω'', when paired with {{nowrap|''k'' + 1}} arbitrary smooth [[vector field]]s ''V''<sub>0</sub>,''V''<sub>1</sub>, ..., ''V''<sub>k</sub>'':
| |
| | |
| :<math>\mathrm{d}\omega(V_0,...,V_k) = \sum_i(-1)^{i} V_i\left(\omega(V_0, \ldots, \hat V_i, \ldots,V_k)\right)</math>
| |
| | |
| ::<math>+\sum_{i<j}(-1)^{i+j}\omega([V_i, V_j], V_0, \ldots, \hat V_i, \ldots, \hat V_j, \ldots, V_k)</math>
| |
| | |
| where <math>[V_i,V_j]</math> denotes [[Lie bracket of vector fields|Lie bracket]] and the hat denotes the omission of that element:
| |
| :<math>\omega(V_0, \ldots, \hat V_i, \ldots,V_k) = \omega(V_0, \ldots, V_{i-1}, V_{i+1}, \ldots, V_k).</math>
| |
| | |
| In particular, for 1-forms we have: {{nowrap|1=d''ω''(''X'',''Y'') = ''Xω''(''Y'') − ''Yω''(''X'') − ''ω''([''X'',''Y''])}}, where ''X'' and ''Y'' are vector fields.
| |
| | |
| == Stokes' Theorem on manifolds ==
| |
| | |
| If ''M'' is a compact smooth orientable ''n''-dimensional manifold with boundary, and ''ω'' is an (''n''−1)-form on ''M'', then the generalized form of [[Stokes' Theorem]] states that:
| |
| | |
| :<math>\int_M{\mathrm{d}\omega} = \int_{\partial{M}}\omega</math>
| |
| | |
| Intuitively, if one thinks of ''M'' as being divided into infinitesimal regions, and one adds the flux through the boundaries of all the regions, the interior boundaries all cancel out, leaving the total flux through the boundary of ''M''.
| |
| | |
| == Examples ==
| |
| | |
| ===1===
| |
| Consider {{nowrap|1=''σ'' = ''u'' d''x''<sup>1</sup>∧d''x''<sup>2</sup>}} over a 1-form basis {{nowrap|1=d''x''<sup>1</sup>,...,d''x''<sup>''n''</sup>}}.
| |
| The exterior derivative is:
| |
| | |
| :<math>\begin{align}
| |
| \mathrm{d} \sigma &= \mathrm{d}(u) \wedge \mathrm{d}x^1 \wedge \mathrm{d}x^2 \\
| |
| &= \left(\sum_{i=1}^n \frac{\partial u}{\partial x^i} \mathrm{d}x^i\right) \wedge \mathrm{d}x^1 \wedge \mathrm{d}x^2 \\
| |
| &= \sum_{i=3}^n \left(\frac{\partial u}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}x^1 \wedge \mathrm{d}x^2\right)
| |
| \end{align}</math>
| |
| | |
| The last formula follows easily from the properties of the [[wedge product]]. Namely, {{nowrap|1=d''x''<sup>''i''</sup> ∧ d''x''<sup>''i''</sup> = 0}}.
| |
| | |
| ===2===
| |
| For a 1-form {{nowrap|1=''σ'' = ''u'' d''x'' + ''v'' d''y''}} defined over '''R'''<sup>''2''</sup>. We have, by applying the above formula to each term (consider {{nowrap|1=''x''<sup>1</sup> = ''x''}} and {{nowrap|1=''x''<sup>2</sup> = ''y''}}) the following sum,
| |
| | |
| :<math>\begin{align}
| |
| \mathrm{d} \sigma
| |
| &= \left( \sum_{i=1}^2 \frac{\partial u}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}x \right) + \left( \sum_{i=1}^2 \frac{\partial v}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}y \right) \\
| |
| &= \left(\frac{\partial{u}}{\partial{x}} \mathrm{d}x \wedge \mathrm{d}x + \frac{\partial{u}}{\partial{y}} \mathrm{d}y \wedge \mathrm{d}x\right) + \left(\frac{\partial{v}}{\partial{x}} \mathrm{d}x \wedge \mathrm{d}y + \frac{\partial{v}}{\partial{y}} \mathrm{d}y \wedge \mathrm{d}y\right) \\
| |
| &= 0 - \frac{\partial{u}}{\partial{y}} \mathrm{d}x \wedge \mathrm{d}y + \frac{\partial{v}}{\partial{x}} \mathrm{d}x \wedge \mathrm{d}y + 0 \\
| |
| &= \left(\frac{\partial{v}}{\partial{x}} - \frac{\partial{u}}{\partial{y}}\right) \mathrm{d}x \wedge \mathrm{d}y
| |
| \end{align}</math>
| |
| | |
| == Further properties ==
| |
| | |
| ===Closed and exact forms===
| |
| {{main|Closed and exact forms}}
| |
| A ''k''-form ''ω'' is called ''closed'' if d''ω'' is 0; closed forms are the [[Kernel (algebra)|kernel]] of d. ''ω'' is called ''exact'' if ''ω'' = d''α'' for some ''(k-1)''-form ''α''; exact forms are the [[Image (mathematics)|image]] of d. Because {{nowrap|1=d<sup>2</sup> = 0}}, every exact form is closed. The [[Poincaré lemma]] states that in a contractible region, the converse is true.
| |
| | |
| ===de Rham cohomology===
| |
| Because the exterior derivative d has the property that {{nowrap|1=d<sup>2</sup> = 0}}, it can be used as the [[differential]] (coboundary) to define [[de Rham cohomology]] on a manifold. The ''k''-th de Rham cohomology (group) is the vector space of closed ''k''-forms modulo the exact ''k''-forms; as noted in the previous section, the Poincaré lemma states that these vector spaces are trivial for a contractible region, for ''k'' > 0. Integration of forms gives a natural homomorphism from the de Rham cohomology to the singular cohomology over the real numbers '''R''' of a [[smooth manifold]]. The theorem of de Rham shows that this map is actually an isomorphism, a far-reaching generalization of the Poincaré lemma. As suggested by the generalized Stokes' theorem, the exterior derivative is the "dual" of the [[Chain_complex#Formal_definition|boundary map]] on singular simplices.
| |
| | |
| ===Naturality===
| |
| The exterior derivative is natural in the technical sense: if {{nowrap|1=''f'': ''M'' → ''N''}} is a smooth map and ''Ω''<sup>''k''</sup> is the contravariant smooth [[functor]] that assigns to each manifold the space of ''k''-forms on the manifold, then the following diagram commutes
| |
| [[Image:Exteriorderivnatural.png|center]]
| |
| | |
| so {{nowrap|1=d(''f''*''ω'') = ''f''*d''ω'',}} where ''f''* denotes the [[pullback (differential geometry)|pullback]] of ''f''. This follows from that ''f''*''ω''(·), by definition, is ''ω''(''f''<sub><sub>*</sub></sub>(·)), ''f''<sub><sub>*</sub></sub> being the [[Pushforward (differential)|pushforward]] of ''f''. Thus d is a [[natural transformation]] from Ω<sup>''k''</sup> to ''Ω''<sup>''k''+1</sup>.
| |
| | |
| == Exterior derivative in vector calculus ==
| |
| | |
| Most [[vector calculus]] operators are special cases of, or have close relationships to, the notion of exterior differentiation.
| |
| | |
| ===Gradient===
| |
| A [[smooth function]] ''f'': '''R'''<sup>''n''</sup> → '''R''' is a 0-form. The exterior derivative of this 0-form is the 1-form
| |
| | |
| :<math>\mathrm{d}f = \sum_{i=1}^n \frac{\partial f}{\partial x^i}\, \mathrm{d}x^i = \langle \nabla f,\cdot \rangle.</math>
| |
|
| |
| That is, the form d''f'' acts on any vector field ''V'' by outputting, at each point, the [[scalar product]] of ''V'' with the gradient ∇''f'' of ''f''.
| |
| | |
| The 1-form d''f'' is a section of the [[cotangent bundle]], that gives a local linear approximation to ''f'' in the cotangent space at each point.
| |
| | |
| ===Divergence===
| |
| A vector field ''V = (v<sub>1</sub>, v<sub>2</sub>, ... v<sub>n</sub>)'' on '''R'''<sup>n</sup> has a corresponding (''n-1'')-form
| |
| | |
| :<math>\omega _V = v_1 \; (\mathrm{d}x^2 \wedge \mathrm{d}x^3 \wedge \cdots \wedge \mathrm{d}x^n) - v_2 \; (\mathrm{d}x^1 \wedge \mathrm{d}x^3 \cdots \wedge \mathrm{d}x^n) + \cdots + (-1)^{n-1}v_n \; (\mathrm{d}x^1 \wedge \cdots \wedge \mathrm{d}x^{n-1})
| |
| </math>
| |
| :<math> =\sum_{p=1}^n{(-1)^{(p-1)}v_p(\mathrm{d}x^1 \wedge \cdots \wedge \mathrm{d}x^{p-1} \wedge \widehat{\mathrm{d}x^{p}} \wedge \mathrm{d}x^{p+1} \wedge \cdots \wedge \mathrm{d}x^n)}.</math>
| |
| where <math> \widehat{\mathrm{d}x^{p}}</math> denotes the omission of that element.
| |
| | |
| (For instance, when ''n'' = 3, in three-dimensional space, the 2-form ω<sub>''V''</sub> is locally the [[scalar triple product]] with ''V''.) The integral of ω<sub>''V''</sub> over a hypersurface is the [[flux]] of ''V'' over that hypersurface.
| |
| | |
| The exterior derivative of this (''n''−1)-form is the ''n''-form
| |
| | |
| :<math>\mathrm{d} \omega _V = \operatorname{div}(V) \; (\mathrm{d}x^1 \wedge \mathrm{d}x^2 \wedge \cdots \wedge \mathrm{d}x^n).</math>
| |
| | |
| ===Curl===
| |
| A vector field ''V'' on '''R'''<sup>''n''</sup> also has a corresponding 1-form
| |
| | |
| :<math>\eta_V = v_1 \; \mathrm{d}x^1 + v_2 \; \mathrm{d}x^2 + \cdots + v_n \; \mathrm{d}x^n.</math>,
| |
| | |
| Locally, η<sub>''V''</sub> is the dot product with ''V''. The integral of η<sub>''V''</sub> along a path is the [[Mechanical work|work]] done against −''V'' along that path.
| |
| | |
| When ''n'' = 3, in three-dimensional space, the exterior derivative of the 1-form η<sub>''V''</sub> is the 2-form
| |
| | |
| :<math>\mathrm{d} \eta_V = \omega _{\operatorname{curl}(V)}.</math>
| |
| | |
| ===Invariant formulations of grad, curl, div, and Laplacian===
| |
| The [[vector calculus]] operators above can be written in coordinate-free notation as follows:
| |
| | |
| : <math>
| |
| \begin{array}{rcccl}
| |
| \operatorname{grad}(f) &=& \nabla f &=& \left( {\mathbf d} f \right)^\sharp \\
| |
| \operatorname{div}(F) &=& \nabla \cdot F &=& \star {\mathbf d} \left( \star F^\flat \right) \\
| |
| \operatorname{curl}(F) &=& \nabla \times F &=& \left[ \star \left( {\mathbf d} F^\flat \right) \right]^\sharp, \\
| |
| \Delta f &=& \nabla^2 f &=& \star {\mathbf d} \left( \star {\mathbf d} f \right) \\
| |
| \end{array}
| |
| </math>
| |
| | |
| where <math>\star</math> is the [[Hodge dual|Hodge star operator]] and <math>\flat</math> and <math>\sharp</math> are the [[musical isomorphism]]s. These equations define the grad, div, curl, and Laplacian operators as returning vectors rather than forms, in contrast to the previous section.
| |
| | |
| == See also ==
| |
| *[[Exterior covariant derivative]]
| |
| *[[de Rham complex]]
| |
| *[[Discrete exterior calculus]]
| |
| *[[Green's theorem]]
| |
| *[[Lie derivative]]
| |
| *[[Stokes' theorem]]
| |
| *[[Fractal derivative]]
| |
| | |
| == References ==
| |
| * {{cite book |author=Flanders, Harley |title=Differential forms with applications to the physical sciences |publisher=Dover Publications |location=New York |year=1989 |pages=20 |isbn=0-486-66169-5 |oclc= |doi=}}
| |
| * {{cite book |author=Ramanan, S. |title=Global calculus |publisher=American Mathematical Society |location=Providence, Rhode Island |year=2005 |pages=54 |isbn=0-8218-3702-8 |oclc= |doi=}}
| |
| * {{cite book |author=Conlon, Lawrence |title=Differentiable manifolds |publisher=Birkhäuser |location=Basel, Switzerland |year=2001 |pages= 239 |isbn=0-8176-4134-3 |oclc= |doi=}}
| |
| * {{cite book |author=Darling, R. W. R. |title=Differential forms and connections |publisher=Cambridge University Press |location=Cambridge, UK |year=1994 |pages=35 |isbn=0-521-46800-0 |oclc= |doi=}}
| |
| | |
| | |
| {{Tensors}}
| |
| | |
| [[Category:Differential forms]]
| |
| [[Category:Differential operators]]
| |
| [[Category:Generalizations of the derivative]]
| |
| | |
| [[pl:Forma różniczkowa#Różniczka zewnętrzna formy]]
| |
| [[ru:Дифференциальная форма#Связанные определения]]
| |
When you were younger it didnt appear to matter what you ate you always had a lot of vitality and for most folks which meant our fat will be kept down moreover. But because we receive older our bodies changes and the processes that you took for granted whenever we were younger dont work because well. Such is the case for metabolism. Unfortunately metabolism has become an easy target over the years as something to blame weight gain on. But that really isnt the case; you want metabolism incredibly whenever you are older. One technique to increase it happens to be from strength training.
It is wise bmr calculator to have a superior scale to weigh on. I weigh everyday, however we will not wish To. I simply can't appear to help me, and since I track it on the Calorie-Count website, it is very something that I am excited to do daily.
Your basal metabolic rate (BMR)Your basal metabolic rate is significant whenever planning a fat reduction system. It shows the rate a body burns calories simply for simple metabolic functions, i.e. how countless calories would you burn for a day when you merely lie in bed. "Lying inside bed" isn't pretty exact of course, because in the event you are thinking or having conversation while in bed, the body will nonetheless burn more calories than your BMR. The BMR shows simply the minimum calories required to remain alive.
So how much water is enough? Probably not the eight glasses we've always been told. Eight 8-ounce glasses is fine should you only weigh 130 pounds. To calculate how much water you require, divide your weight inside half. You must drink that countless ounces of water each day. If you weigh 180 pounds, you need to drink 90 ounces of water.
Body Surface Area: The height plus fat lead a lot in determining bmr. The greater is the body surface region, the high is the BMR. Thin, tall persons have a higher BMR.
To calculate the activity level, see this free online calculator: http://exercise.about.com/cs/fitnesstools/l/blcalorieburn.htm which usually use your current weight, the sort of exercise, plus the amount of time you can afford to invest doing that exercise to tell you how various calories you're going to burn. Take note of the results - should you use this calculator realistically, you are able to plug it right into the BMR results.
There are 3 ways to create this deficit: diet, exercise, or perhaps a combination. If you combine techniques we have a better chance of following it. Just think, instead of completely cutting out that afternoon snack, or endangering injury by functioning out too difficult, we may really have a lighter snack and take the stairs. Its easier to create small changes which can add about 500 calories a day.