|
|
| Line 1: |
Line 1: |
| {{More footnotes|date=May 2009}}
| | Getting a great massage is one good way to let yourself go for a short while. You'll feel energized and ready to face the day after a relaxing massage. Check out the article below to learn all you can about massages.<br><br>Scented candles should be used when you're giving a massage. It provides a warm light and relaxing fragrance. All of these factors can create a great massage that can boost a great experience.<br><br>Eat a light meal ahead of a massage. Eating a lot may make you feel uncomfortable during your massage and that can make the experience go badly. Eat as healthy as possible, mixing in fruit and vegetables.<br><br>Make an effort to wash your feet before your massage, in the event the spa does not offer a foot bath. You don't want germs that you've picked up on your feet to be spread to the rest of your body by the therapist. In the event you loved this post and you wish to receive more info about Asian massage - [http://Superbigstory.com/news/Basic_Tips_For_Giving_Someone_A_Massage superbigstory.com] - i implore you to visit our own internet site. If there isn't a foot bath option, go to a bathroom prior to a massage and clean them via the sink.<br><br>During a massage, ask your therapist to lower the lights. You want to have a relaxing and enjoyable massage, and room that's dark will provide that sort of atmosphere. It's not necessary for the room to be pitch black; however, the room shouldn't be brighter than candles would provide.<br><br>Massaging can help you with any arthritic symptoms that you may have. Medication helps, but can not always block out the aching feeling deep inside your bones. Therefore, you should consider getting yourself a massage to treat this condition. It increases circulation and awakens the muscles.<br><br>The power of a massage can not be underestimated. A massage is an effective way to rid yourself of pain, relieve stress and gain extra energy for your normal activities. No matter what health problems you have, go to a professional to learn how beneficial getting a massage is.<br><br>It is important to share any problem areas you are having with your massage therapist. The goal of your massage is muscle relaxation where you need it most. Remember that your massage therapist will not know about problem areas unless you tell them.<br><br>The relaxing benefits of neck massages should not be underestimated. It is very common for people to carry significant tension in the muscles supporting the neck. This area is an easy body part to massage and is one which can deliver great benefits. When you're massaging the neck, it's important to use lotion, and don't forget to get to the shoulder muscles too.<br><br>If you are someone who exercises, give yourself a mini-massage both before and after you exercise so you encourage blood and oxygen into those muscles to speed up recovery. Prior to exercising, pump your arms so as to increase blood flow to all your extremities. After you exercise, massage your muscles with your palm in a direction toward your heart.<br><br>You need to clean your feet before getting a good full body massage. It is important to be presentable out of respect for your massage therapist. Washing your feet with warm water will also help you relax and get ready for your massage.<br><br>The suggestions above are invaluable if you want to improve your massage skills. Ask a partner or close friend if you can practice your new skills when they have a moment to spare. Soon, you will work out all your kinks and be able to help your loved ones relieve theirs. You will be a big hit and surprise everyone with your new skills. |
| {{Technical|date=April 2012}}
| |
| In [[vector calculus]] a '''conservative vector field''' is a [[vector field]] which is the [[gradient]] of some [[function (mathematics)|function]], known in this context as a [[scalar potential]].<ref>{{cite book|title=Vector calculus|first1=Jerrold|last1=Marsden|authorlink1=Jerrold Marsden|first2=Anthony|last2=Tromba|publisher=W.H.Freedman and Company|edition=Fifth|year=2003|pp=550–561}}</ref> Conservative vector fields have the property that the [[line integral]] from one point to another is independent of the choice of path connecting the two points: it is path independent. Conversely, path independence is equivalent to the vector field being conservative. Conservative vector fields are also '''irrotational''', meaning that (in three-dimensions) they have vanishing [[curl (mathematics)|curl]]. In fact, an irrotational vector field is necessarily conservative provided that a certain condition on the geometry of the domain holds, i.e. the domain is [[simply connected]].
| |
| | |
| Conservative vector fields appear naturally in [[mechanics]]: they are vector fields representing [[force]]s of [[physical system]]s in which [[energy]] is [[conservation of energy|conserved]].<ref>George B. Arfken and Hans J. Weber, ''Mathematical Methods for Physicists'', 6th edition, Elsevier Academic Press (2005)</ref> For a conservative system, the [[work (physics)|work]] done in moving along a path in configuration space depends only on the endpoints of the path, so it is possible to define a [[potential energy]] independently of the path taken.
| |
| | |
| ==Intuitive explanation==
| |
| | |
| [[File:Ascending and Descending.jpg|right|thumb|500px|M. C. Escher's painting ''Ascending and Descending'']]
| |
| | |
| [[M. C. Escher|M. C. Escher's]] painting ''[[Ascending and Descending]]'' illustrates a non-conservative vector field, impossibly made to appear to be the gradient of the varying height above ground as one moves along the staircase. It is "rotational" in that one can keep getting higher or keep getting lower while going around in circles. It is non-conservative in that one can return to one's starting point while ascending more than one descends or vice-versa. In reality, the height above the ground is a scalar potential field: if one returns to the same place, one goes upward exactly as much as one goes downward. Its gradient would be a conservative vector field, and is irrotational. The situation depicted in the painting is impossible.
| |
| | |
| ==Definition==
| |
| | |
| A [[vector field]] [[vector field|<math> \mathbf{v} </math>]] is said to be ''conservative'' if there exists a [[scalar field]] [[phi|<math> \varphi </math>]] such that
| |
| :<math> \mathbf{v}=\nabla\varphi.</math>
| |
| Here [[Dipole|<math>\nabla\varphi</math>]] denotes the [[gradient]] of <math>\varphi</math>. When the above equation holds, <math>\varphi</math> is called a [[scalar potential]] for <math>\mathbf{v}</math>.
| |
| | |
| The [[Helmholtz decomposition|fundamental theorem of vector calculus]] states that any vector field can be expressed as the sum of a conservative vector field and a [[solenoidal field]].
| |
| | |
| ==Path independence==
| |
| | |
| {{main|Gradient theorem}}
| |
| | |
| A key property of a conservative vector field is that its integral along a path depends only on the endpoints of that path, not the particular route taken. Suppose that <math> S\subseteq\mathbb{R}^3
| |
| </math> is a region of three-dimensional space, and that <math> P </math> is a rectifiable path in <math> S </math> with start point <math> A </math> and end point <math> B </math>. If <math> | |
| \mathbf{v}=\nabla\varphi </math> is a conservative vector field then the [[gradient theorem]] states that
| |
| | |
| :<math> \int_P \mathbf{v}\cdot d\mathbf{r}=\varphi(B)-\varphi(A). </math>
| |
| | |
| This holds as a consequence of the [[Chain Rule]] and the [[Fundamental theorem of calculus|Fundamental Theorem of Calculus]].
| |
| | |
| An equivalent formulation of this is to say that
| |
| | |
| :<math> \oint \mathbf{v}\cdot d\mathbf{r}=0 </math>
| |
| | |
| for every closed loop in ''S''. The [[Gradient theorem#Converse of the gradient theorem|converse of this statement]] is also true: if the [[circulation (fluid dynamics)|circulation]] of '''v''' around every closed loop in an [[open set]] ''S'' is zero, then '''v''' is a conservative vector field.
| |
| | |
| ==Irrotational vector fields==
| |
| [[File:Irrotationalfield.svg|thumb|300px|The above field {{nowrap|1='''v''' = (−''y''/(''x''<sup>2</sup> + ''y''<sup>2</sup>), +''x''/(''x''<sup>2</sup> + ''y''<sup>2</sup>), 0)}} includes a vortex at its center, so it is ''non''-irrotational; it is neither conservative, nor does it have path independence. However, any simply connected subset that excludes the vortex line (0,0,''z'') will have zero curl, {{nowrap|1=∇ × '''v''' = 0}}. Such vortex-free regions are examples of irrotational vector fields.]]
| |
| A vector field <math> \mathbf{v} </math> is said to be ''irrotational'' if its [[Curl (mathematics)|curl]] is zero. That is, if
| |
| | |
| :<math> \nabla\times\mathbf{v} = \mathbf{0}. </math>
| |
| | |
| For this reason, such vector fields are sometimes referred to as '''curl free field''' ('''curl-free vector field''') or ''curl-less'' vector fields.
| |
| | |
| It is an [[Vector calculus identities#Curl of the gradient|identity of vector calculus]] that for any scalar field <math>\varphi</math>:
| |
| | |
| :<math> \nabla \times \nabla \varphi=\mathbf{0}. </math>
| |
| | |
| Therefore every conservative vector field is also an irrotational vector field.
| |
| | |
| Provided that <math> S </math> is a [[simply-connected]] region, the converse of this is true: every
| |
| irrotational vector field is also a conservative vector field.
| |
| | |
| The above statement is ''not'' true if <math> S </math> is not [[simply-connected]]. Let <math> S </math> be the usual 3-dimensional space, except with the <math>z</math>-axis removed; that is <math> S=\mathbb{R}^3\setminus\{(0,0,z)~|~z\in\mathbb{R}\} </math>. Now define a vector field by
| |
| | |
| :<math> \mathbf{v}= \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}, 0 \right). </math>
| |
| | |
| Then <math> \mathbf{v} </math> exists and has zero curl at every point in <math> S </math>; that is
| |
| <math>\mathbf{v}</math> is irrotational. However the circulation of <math>\mathbf{v}</math> around the unit circle in the <math> x,y</math>-plane is equal to <math> 2\pi</math>. Indeed we note that in [[polar coordinates]] <math> \mathbf{v}= \mathbf{e_{\phi}}/r </math>, so the integral over the unit circle is equal <math> \int \mathbf{v}\mathbf{e_{\phi}}\mathrm{d}\phi=2\pi </math>. Therefore <math> \mathbf{v}</math> does not have the path independence property discussed above, and is not conservative. (However, in any connected subregion of S, it is still true that it is conservative. In fact, the field above is the gradient of <math>\arg (x + iy)</math>. As we know from complex analysis, this is a multi-valued function which requires a branch cut from the origin to infinity to be defined in a continuous way; hence, in a region that does not go around the ''z''-axis, its gradient is conservative.)
| |
| | |
| In a [[simply-connected]] region an irrotational vector field has the path independence property. This can be seen by noting that in such a region an irrotational vector field is conservative, and conservative vector fields have the path independence property. The result can also be proved directly by using [[Stokes' theorem]]. In a connected region any vector field which has the path independence property must also be irrotational.
| |
| | |
| More abstractly, a conservative vector field is an [[exact form|exact 1-form]]. That is, it is a 1-form equal to the [[exterior derivative]] of some 0-form (scalar field) <math>\phi</math>. An irrotational vector field is a [[closed and exact differential forms|closed 1-form]]. Since ''d''<sup>2</sup> = 0, any exact form is closed, so any conservative vector field is irrotational. The domain is [[simply connected]] if and only if its first [[homology group]] is 0, which is equivalent to its first [[cohomology group]] being 0. The first [[de Rham cohomology]] group <math>H_{\mathrm{dR}}^{1}</math> is 0 if and only if all closed 1-forms are exact.
| |
| | |
| ==Irrotational flows==<!-- [[Irrotatational flow]] redirects to here -->
| |
| {{main|Vorticity}}
| |
| The [[flow velocity]] <math> \mathbf{v} </math> of a fluid is a vector field, and the [[vorticity]] <math>\boldsymbol{\omega}</math> of the flow can be defined by
| |
| | |
| :<math>\boldsymbol{\omega}=\nabla\times\mathbf{v}. </math>
| |
| | |
| A common alternative notation for ''vorticity'' is <math>\zeta\;</math>.<ref>Clancy, L.J., ''Aerodynamics'', Section 7.11, Pitman Publishing Limited, London. ISBN 0-273-01120-0</ref> | |
| | |
| If <math> \mathbf{v} </math> is irrotational, with <math>\nabla\times\mathbf{v} =\mathbf{0}</math>, then the flow is said to be an '''irrotational flow'''. The vorticity of an irrotational flow is zero.<ref>{{citation | title=Elements of Gas Dynamics | first1=H.W. | last1=Liepmann | authorlink1=Hans W. Liepmann | first2=A. | last2=Roshko | authorlink2=Anatol Roshko | publisher=Courier Dover Publications | year=1993 | origyear=1957 | isbn=0-486-41963-0 }}, pp. 194–196.</ref>
| |
| | |
| [[Kelvin's circulation theorem]] states that a fluid that is irrotational in an [[inviscid flow]] will remain irrotational. This result can be derived from the [[vorticity transport equation]], obtained by taking the curl of the Navier-stokes equations.
| |
| | |
| For a two-dimensional flow the vorticity acts as a measure of the ''local'' rotation of fluid elements. Note that the vorticity does ''not'' imply anything about the global behaviour of a fluid. It is possible for a fluid traveling in a straight line to have vorticity, and it is possible for a fluid which moves in a circle to be irrotational.
| |
| | |
| ==Conservative forces==
| |
| [[File:Conservative fields.png|thumb|upright=1.5|Examples of potential and gradient fields in physics<br/>''Scalar fields (scalar potentials) (<span style="color:yellow">yellow</span>)'': '''V<sub>G</sub>''' - gravitational potential; '''W<sub>pot</sub>''' - potential energy; '''V<sub>C</sub>''' - Coulomb potential; ''Vector fields (gradient fields) (<span style="color:cyan">cyan</span>)'': '''a<sub>G</sub>''' - gravitational acceleration; '''F''' - force; '''E''' - electric field strength]]
| |
| | |
| If the vector field associated to a force <math> \mathbf{F} </math> is conservative then the force is said to be a [[conservative force]].
| |
| | |
| The most prominent examples of conservative forces are the force of gravity and the electric field associated to a static charge. According to [[Newton's law of universal gravitation|Newton's law of gravitation]], the [[gravitational force]], <math>\mathbf{F}_G</math>, acting on a mass <math>m</math>, due to a mass <math>M</math> which is a distance <math>r</math> away, obeys the equation
| |
| | |
| :<math> \mathbf{F}_G=-\frac{GmM\hat{\mathbf{r}}}{r^2}, </math>
| |
| | |
| where <math>G</math> is the [[Gravitational Constant]] and <math>\hat{\mathbf{r}}</math> is a unit vector pointing from <math>M</math> towards <math>m</math>. The force of gravity is conservative because <math>\mathbf{F}_G=-\nabla\Phi_G</math>, where
| |
| | |
| :<math> \Phi_G=-\frac{GmM}{r} </math>
| |
| | |
| is the [[Gravitational potential energy]].
| |
| | |
| For [[conservative force]]s, ''path independence'' can be interpreted to mean that the [[work done]] in going from a point <math>A</math> to a point <math>B</math>
| |
| is independent of the path chosen, and that the work ''W'' done in going around a closed loop is zero: | |
| | |
| :<math> W=\oint \mathbf{F}\cdot d\mathbf{r}=0. </math>
| |
| | |
| The total [[conservation of energy| energy]] of a particle moving under the influence of conservative forces is conserved, in the sense that a loss of potential energy is converted to an equal quantity of kinetic energy or vice versa. | |
| | |
| == See also ==
| |
| * [[Beltrami vector field]]
| |
| * [[Complex lamellar vector field]]
| |
| * [[Helmholtz decomposition]]
| |
| * [[Laplacian vector field]]
| |
| * [[Longitudinal and transverse vector fields]]
| |
| * [[Potential field]]
| |
| * [[Solenoidal vector field]]
| |
| | |
| ==References and notes==
| |
| ;General
| |
| * D. J. Acheson, ''Elementary Fluid Dynamics'', Oxford University Press (2005)
| |
| ;Citations
| |
| {{reflist}}
| |
| ; Notes
| |
| {{reflist|group=note}}
| |
| | |
| [[Category:Vector calculus]]
| |
| [[Category:Force]]
| |
| | |
| [[de:Potentialfeld]]
| |
Getting a great massage is one good way to let yourself go for a short while. You'll feel energized and ready to face the day after a relaxing massage. Check out the article below to learn all you can about massages.
Scented candles should be used when you're giving a massage. It provides a warm light and relaxing fragrance. All of these factors can create a great massage that can boost a great experience.
Eat a light meal ahead of a massage. Eating a lot may make you feel uncomfortable during your massage and that can make the experience go badly. Eat as healthy as possible, mixing in fruit and vegetables.
Make an effort to wash your feet before your massage, in the event the spa does not offer a foot bath. You don't want germs that you've picked up on your feet to be spread to the rest of your body by the therapist. In the event you loved this post and you wish to receive more info about Asian massage - superbigstory.com - i implore you to visit our own internet site. If there isn't a foot bath option, go to a bathroom prior to a massage and clean them via the sink.
During a massage, ask your therapist to lower the lights. You want to have a relaxing and enjoyable massage, and room that's dark will provide that sort of atmosphere. It's not necessary for the room to be pitch black; however, the room shouldn't be brighter than candles would provide.
Massaging can help you with any arthritic symptoms that you may have. Medication helps, but can not always block out the aching feeling deep inside your bones. Therefore, you should consider getting yourself a massage to treat this condition. It increases circulation and awakens the muscles.
The power of a massage can not be underestimated. A massage is an effective way to rid yourself of pain, relieve stress and gain extra energy for your normal activities. No matter what health problems you have, go to a professional to learn how beneficial getting a massage is.
It is important to share any problem areas you are having with your massage therapist. The goal of your massage is muscle relaxation where you need it most. Remember that your massage therapist will not know about problem areas unless you tell them.
The relaxing benefits of neck massages should not be underestimated. It is very common for people to carry significant tension in the muscles supporting the neck. This area is an easy body part to massage and is one which can deliver great benefits. When you're massaging the neck, it's important to use lotion, and don't forget to get to the shoulder muscles too.
If you are someone who exercises, give yourself a mini-massage both before and after you exercise so you encourage blood and oxygen into those muscles to speed up recovery. Prior to exercising, pump your arms so as to increase blood flow to all your extremities. After you exercise, massage your muscles with your palm in a direction toward your heart.
You need to clean your feet before getting a good full body massage. It is important to be presentable out of respect for your massage therapist. Washing your feet with warm water will also help you relax and get ready for your massage.
The suggestions above are invaluable if you want to improve your massage skills. Ask a partner or close friend if you can practice your new skills when they have a moment to spare. Soon, you will work out all your kinks and be able to help your loved ones relieve theirs. You will be a big hit and surprise everyone with your new skills.