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Is a Wick rotation a change of coordinates?

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Is a Wick rotation a change of coordinates?


Showing $I=int d^3kint dk^0frac1k^4$ to be logarithmically divergentHow to Perform Wick Rotation in the Lagrangian of a Gauge Theory (like QCD)?Is it okay to Wick rotate to give the negative of the Euclidean metric? Also, could we make the space-like coordinates imaginary instead?Curved space-time VS change of coordinates in Minkowski spaceHow does the Lorentz boost change if we introduce transformation to the minkowski metricPerforming Wick Rotation to get Euclidean action of a scalar field $Psi$Determinant of the metric tensorIntegral and Wick rotation (Srednicki ch75)How does the Equivalence Principle imply that derivatives of the metric vanish in a freely falling frame?Interpretation of Normal Coordinates






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








3












$begingroup$


My understanding is that a Wick rotation is a change of coordinates from $(t,x) rightarrow (tau , x)$ where $tau = i t$. In the $(t,x)$ coordinate system, the Minkowski metric has components $ eta_mu nu = mathrmdiag(1,-1,-1,-1)$. Using the formula for the transformation of the components under a coordinate change:



$$ eta_alpha beta = fracpartial x^mupartial x'^alphafracpartial x^nupartial x'^betaeta_mu nu $$



we find in the $(tau,x)$ coordinate system, the metric has components $eta_alpha beta = mathrmdiag(-1,-1,-1,-1)$.



In QFT for the gifted amateur by Lancaster and Blundell equation 25.4, it is stated that under a Wick rotation, the magnitude of a vector is given by



$$ x^2 = - x_E^2 $$



where $x$ is the Minkowski vector and $x_E$ is the corresponding Euclidean vector. Now I am confused by this statement, because the objects $x$ and $x_E$ are coordinate representations of a vector, say $X$, which is a geometric object independent of the coordinate system we choose, so we should expect



$$ |X|^2 = eta_mu nu x^mu x^nu = eta_alpha beta x^alpha_E x^beta_E$$



in other words, the magnitude of the vector $X$ should not depend on which coordinate system we use. So under a simple Wick rotation, how could the magnitude of a vector change?



I was thinking, maybe a Wick rotation is an active rotation into the complex plane but the book states that the metric tranforms too so we can use the Euclidean metric. If we transform both the vector and the metric then that suggests a change of coordinates, but if only the vector changes then it suggests some sort of active transformation.



My Question



Is a Wick rotation simply a change of coordinates or is it an active rotation of the vector into the complex plane?










share|cite|improve this question











$endgroup$













  • $begingroup$
    This is a horrible misunderstanding that "casual" textbooks propagate... a Wick rotation isn't a change of coordinates. Wick rotations have drastic consequences, while as you correctly point out, a change of coordinates barely does anything!
    $endgroup$
    – knzhou
    8 hours ago










  • $begingroup$
    Saying that a Wick rotation is a change of coordinates is the second most common and second worst misexplanation of it, the first being that "it's a substitution $t to it$". The vast majority of books cover this totally basic thing all wrong. It should be thought of in terms of rotating contours of integration.
    $endgroup$
    – knzhou
    8 hours ago






  • 1




    $begingroup$
    My answer here should be relevant
    $endgroup$
    – MannyC
    7 hours ago

















3












$begingroup$


My understanding is that a Wick rotation is a change of coordinates from $(t,x) rightarrow (tau , x)$ where $tau = i t$. In the $(t,x)$ coordinate system, the Minkowski metric has components $ eta_mu nu = mathrmdiag(1,-1,-1,-1)$. Using the formula for the transformation of the components under a coordinate change:



$$ eta_alpha beta = fracpartial x^mupartial x'^alphafracpartial x^nupartial x'^betaeta_mu nu $$



we find in the $(tau,x)$ coordinate system, the metric has components $eta_alpha beta = mathrmdiag(-1,-1,-1,-1)$.



In QFT for the gifted amateur by Lancaster and Blundell equation 25.4, it is stated that under a Wick rotation, the magnitude of a vector is given by



$$ x^2 = - x_E^2 $$



where $x$ is the Minkowski vector and $x_E$ is the corresponding Euclidean vector. Now I am confused by this statement, because the objects $x$ and $x_E$ are coordinate representations of a vector, say $X$, which is a geometric object independent of the coordinate system we choose, so we should expect



$$ |X|^2 = eta_mu nu x^mu x^nu = eta_alpha beta x^alpha_E x^beta_E$$



in other words, the magnitude of the vector $X$ should not depend on which coordinate system we use. So under a simple Wick rotation, how could the magnitude of a vector change?



I was thinking, maybe a Wick rotation is an active rotation into the complex plane but the book states that the metric tranforms too so we can use the Euclidean metric. If we transform both the vector and the metric then that suggests a change of coordinates, but if only the vector changes then it suggests some sort of active transformation.



My Question



Is a Wick rotation simply a change of coordinates or is it an active rotation of the vector into the complex plane?










share|cite|improve this question











$endgroup$













  • $begingroup$
    This is a horrible misunderstanding that "casual" textbooks propagate... a Wick rotation isn't a change of coordinates. Wick rotations have drastic consequences, while as you correctly point out, a change of coordinates barely does anything!
    $endgroup$
    – knzhou
    8 hours ago










  • $begingroup$
    Saying that a Wick rotation is a change of coordinates is the second most common and second worst misexplanation of it, the first being that "it's a substitution $t to it$". The vast majority of books cover this totally basic thing all wrong. It should be thought of in terms of rotating contours of integration.
    $endgroup$
    – knzhou
    8 hours ago






  • 1




    $begingroup$
    My answer here should be relevant
    $endgroup$
    – MannyC
    7 hours ago













3












3








3


3



$begingroup$


My understanding is that a Wick rotation is a change of coordinates from $(t,x) rightarrow (tau , x)$ where $tau = i t$. In the $(t,x)$ coordinate system, the Minkowski metric has components $ eta_mu nu = mathrmdiag(1,-1,-1,-1)$. Using the formula for the transformation of the components under a coordinate change:



$$ eta_alpha beta = fracpartial x^mupartial x'^alphafracpartial x^nupartial x'^betaeta_mu nu $$



we find in the $(tau,x)$ coordinate system, the metric has components $eta_alpha beta = mathrmdiag(-1,-1,-1,-1)$.



In QFT for the gifted amateur by Lancaster and Blundell equation 25.4, it is stated that under a Wick rotation, the magnitude of a vector is given by



$$ x^2 = - x_E^2 $$



where $x$ is the Minkowski vector and $x_E$ is the corresponding Euclidean vector. Now I am confused by this statement, because the objects $x$ and $x_E$ are coordinate representations of a vector, say $X$, which is a geometric object independent of the coordinate system we choose, so we should expect



$$ |X|^2 = eta_mu nu x^mu x^nu = eta_alpha beta x^alpha_E x^beta_E$$



in other words, the magnitude of the vector $X$ should not depend on which coordinate system we use. So under a simple Wick rotation, how could the magnitude of a vector change?



I was thinking, maybe a Wick rotation is an active rotation into the complex plane but the book states that the metric tranforms too so we can use the Euclidean metric. If we transform both the vector and the metric then that suggests a change of coordinates, but if only the vector changes then it suggests some sort of active transformation.



My Question



Is a Wick rotation simply a change of coordinates or is it an active rotation of the vector into the complex plane?










share|cite|improve this question











$endgroup$




My understanding is that a Wick rotation is a change of coordinates from $(t,x) rightarrow (tau , x)$ where $tau = i t$. In the $(t,x)$ coordinate system, the Minkowski metric has components $ eta_mu nu = mathrmdiag(1,-1,-1,-1)$. Using the formula for the transformation of the components under a coordinate change:



$$ eta_alpha beta = fracpartial x^mupartial x'^alphafracpartial x^nupartial x'^betaeta_mu nu $$



we find in the $(tau,x)$ coordinate system, the metric has components $eta_alpha beta = mathrmdiag(-1,-1,-1,-1)$.



In QFT for the gifted amateur by Lancaster and Blundell equation 25.4, it is stated that under a Wick rotation, the magnitude of a vector is given by



$$ x^2 = - x_E^2 $$



where $x$ is the Minkowski vector and $x_E$ is the corresponding Euclidean vector. Now I am confused by this statement, because the objects $x$ and $x_E$ are coordinate representations of a vector, say $X$, which is a geometric object independent of the coordinate system we choose, so we should expect



$$ |X|^2 = eta_mu nu x^mu x^nu = eta_alpha beta x^alpha_E x^beta_E$$



in other words, the magnitude of the vector $X$ should not depend on which coordinate system we use. So under a simple Wick rotation, how could the magnitude of a vector change?



I was thinking, maybe a Wick rotation is an active rotation into the complex plane but the book states that the metric tranforms too so we can use the Euclidean metric. If we transform both the vector and the metric then that suggests a change of coordinates, but if only the vector changes then it suggests some sort of active transformation.



My Question



Is a Wick rotation simply a change of coordinates or is it an active rotation of the vector into the complex plane?







special-relativity metric-tensor coordinate-systems complex-numbers wick-rotation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









Qmechanic

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asked 8 hours ago









Matt0410Matt0410

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  • $begingroup$
    This is a horrible misunderstanding that "casual" textbooks propagate... a Wick rotation isn't a change of coordinates. Wick rotations have drastic consequences, while as you correctly point out, a change of coordinates barely does anything!
    $endgroup$
    – knzhou
    8 hours ago










  • $begingroup$
    Saying that a Wick rotation is a change of coordinates is the second most common and second worst misexplanation of it, the first being that "it's a substitution $t to it$". The vast majority of books cover this totally basic thing all wrong. It should be thought of in terms of rotating contours of integration.
    $endgroup$
    – knzhou
    8 hours ago






  • 1




    $begingroup$
    My answer here should be relevant
    $endgroup$
    – MannyC
    7 hours ago
















  • $begingroup$
    This is a horrible misunderstanding that "casual" textbooks propagate... a Wick rotation isn't a change of coordinates. Wick rotations have drastic consequences, while as you correctly point out, a change of coordinates barely does anything!
    $endgroup$
    – knzhou
    8 hours ago










  • $begingroup$
    Saying that a Wick rotation is a change of coordinates is the second most common and second worst misexplanation of it, the first being that "it's a substitution $t to it$". The vast majority of books cover this totally basic thing all wrong. It should be thought of in terms of rotating contours of integration.
    $endgroup$
    – knzhou
    8 hours ago






  • 1




    $begingroup$
    My answer here should be relevant
    $endgroup$
    – MannyC
    7 hours ago















$begingroup$
This is a horrible misunderstanding that "casual" textbooks propagate... a Wick rotation isn't a change of coordinates. Wick rotations have drastic consequences, while as you correctly point out, a change of coordinates barely does anything!
$endgroup$
– knzhou
8 hours ago




$begingroup$
This is a horrible misunderstanding that "casual" textbooks propagate... a Wick rotation isn't a change of coordinates. Wick rotations have drastic consequences, while as you correctly point out, a change of coordinates barely does anything!
$endgroup$
– knzhou
8 hours ago












$begingroup$
Saying that a Wick rotation is a change of coordinates is the second most common and second worst misexplanation of it, the first being that "it's a substitution $t to it$". The vast majority of books cover this totally basic thing all wrong. It should be thought of in terms of rotating contours of integration.
$endgroup$
– knzhou
8 hours ago




$begingroup$
Saying that a Wick rotation is a change of coordinates is the second most common and second worst misexplanation of it, the first being that "it's a substitution $t to it$". The vast majority of books cover this totally basic thing all wrong. It should be thought of in terms of rotating contours of integration.
$endgroup$
– knzhou
8 hours ago




1




1




$begingroup$
My answer here should be relevant
$endgroup$
– MannyC
7 hours ago




$begingroup$
My answer here should be relevant
$endgroup$
– MannyC
7 hours ago










1 Answer
1






active

oldest

votes


















5














$begingroup$

[The following is a half-remembered comment that my doctoral advisor told me some years ago, so I may have garbled it. I welcome corrections in the comments; feel free to tell me I'm full of it as well.]



One way to think about a Wick rotation is that the "Euclidean" and "Lorentzian" manifolds (both of which are four-dimensional real manifolds, with a particular metric) can be viewed as hypersurfaces lying in an underlying four-dimensional complex manifold. For example, in the complex manifold $mathbbC^4$ with the obvious metric, you can find hypersurfaces with four (real) dimensions that are diffeomorphic to Euclidean four-space, and hypersurfaces with four (real) dimensions that are diffeomorphic to Minkowski space. The reason that Wick rotations are successful in flat spacetime is because the functions we're looking at are generally holomorphic, and so they can be analytically continued from one "cross-section" to another.



In this picture, a vector lying in a Euclidean cross-section of $mathbbC^4$ must be actively "rotated" into the Lorentzian cross-section. Simply changing the coordinates on your cross-section will not magically "pull in" a vector that doesn't already lie in that cross-section.



This picture, by the way, does not necessarily carry over to the analysis in curved spacetimes. We might think that if the Lorentzian metric is of the form
$$
ds^2 = - f(r,t) dt^2 + g^ij dx_i dx_j
$$

in some set of coordinates, then we can define a Euclidean analog
$$
ds_E^2 = f(r,t) dt^2 + g^ij dx_i dx_j
$$

and do the analysis there. However, there is no guarantee that there exists a complex manifold having these two cross-sections, and so we cannot rely on the Euclidean results to tell us anything about the Lorentzian physics.






share|cite|improve this answer











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    1 Answer
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    active

    oldest

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    5














    $begingroup$

    [The following is a half-remembered comment that my doctoral advisor told me some years ago, so I may have garbled it. I welcome corrections in the comments; feel free to tell me I'm full of it as well.]



    One way to think about a Wick rotation is that the "Euclidean" and "Lorentzian" manifolds (both of which are four-dimensional real manifolds, with a particular metric) can be viewed as hypersurfaces lying in an underlying four-dimensional complex manifold. For example, in the complex manifold $mathbbC^4$ with the obvious metric, you can find hypersurfaces with four (real) dimensions that are diffeomorphic to Euclidean four-space, and hypersurfaces with four (real) dimensions that are diffeomorphic to Minkowski space. The reason that Wick rotations are successful in flat spacetime is because the functions we're looking at are generally holomorphic, and so they can be analytically continued from one "cross-section" to another.



    In this picture, a vector lying in a Euclidean cross-section of $mathbbC^4$ must be actively "rotated" into the Lorentzian cross-section. Simply changing the coordinates on your cross-section will not magically "pull in" a vector that doesn't already lie in that cross-section.



    This picture, by the way, does not necessarily carry over to the analysis in curved spacetimes. We might think that if the Lorentzian metric is of the form
    $$
    ds^2 = - f(r,t) dt^2 + g^ij dx_i dx_j
    $$

    in some set of coordinates, then we can define a Euclidean analog
    $$
    ds_E^2 = f(r,t) dt^2 + g^ij dx_i dx_j
    $$

    and do the analysis there. However, there is no guarantee that there exists a complex manifold having these two cross-sections, and so we cannot rely on the Euclidean results to tell us anything about the Lorentzian physics.






    share|cite|improve this answer











    $endgroup$



















      5














      $begingroup$

      [The following is a half-remembered comment that my doctoral advisor told me some years ago, so I may have garbled it. I welcome corrections in the comments; feel free to tell me I'm full of it as well.]



      One way to think about a Wick rotation is that the "Euclidean" and "Lorentzian" manifolds (both of which are four-dimensional real manifolds, with a particular metric) can be viewed as hypersurfaces lying in an underlying four-dimensional complex manifold. For example, in the complex manifold $mathbbC^4$ with the obvious metric, you can find hypersurfaces with four (real) dimensions that are diffeomorphic to Euclidean four-space, and hypersurfaces with four (real) dimensions that are diffeomorphic to Minkowski space. The reason that Wick rotations are successful in flat spacetime is because the functions we're looking at are generally holomorphic, and so they can be analytically continued from one "cross-section" to another.



      In this picture, a vector lying in a Euclidean cross-section of $mathbbC^4$ must be actively "rotated" into the Lorentzian cross-section. Simply changing the coordinates on your cross-section will not magically "pull in" a vector that doesn't already lie in that cross-section.



      This picture, by the way, does not necessarily carry over to the analysis in curved spacetimes. We might think that if the Lorentzian metric is of the form
      $$
      ds^2 = - f(r,t) dt^2 + g^ij dx_i dx_j
      $$

      in some set of coordinates, then we can define a Euclidean analog
      $$
      ds_E^2 = f(r,t) dt^2 + g^ij dx_i dx_j
      $$

      and do the analysis there. However, there is no guarantee that there exists a complex manifold having these two cross-sections, and so we cannot rely on the Euclidean results to tell us anything about the Lorentzian physics.






      share|cite|improve this answer











      $endgroup$

















        5














        5










        5







        $begingroup$

        [The following is a half-remembered comment that my doctoral advisor told me some years ago, so I may have garbled it. I welcome corrections in the comments; feel free to tell me I'm full of it as well.]



        One way to think about a Wick rotation is that the "Euclidean" and "Lorentzian" manifolds (both of which are four-dimensional real manifolds, with a particular metric) can be viewed as hypersurfaces lying in an underlying four-dimensional complex manifold. For example, in the complex manifold $mathbbC^4$ with the obvious metric, you can find hypersurfaces with four (real) dimensions that are diffeomorphic to Euclidean four-space, and hypersurfaces with four (real) dimensions that are diffeomorphic to Minkowski space. The reason that Wick rotations are successful in flat spacetime is because the functions we're looking at are generally holomorphic, and so they can be analytically continued from one "cross-section" to another.



        In this picture, a vector lying in a Euclidean cross-section of $mathbbC^4$ must be actively "rotated" into the Lorentzian cross-section. Simply changing the coordinates on your cross-section will not magically "pull in" a vector that doesn't already lie in that cross-section.



        This picture, by the way, does not necessarily carry over to the analysis in curved spacetimes. We might think that if the Lorentzian metric is of the form
        $$
        ds^2 = - f(r,t) dt^2 + g^ij dx_i dx_j
        $$

        in some set of coordinates, then we can define a Euclidean analog
        $$
        ds_E^2 = f(r,t) dt^2 + g^ij dx_i dx_j
        $$

        and do the analysis there. However, there is no guarantee that there exists a complex manifold having these two cross-sections, and so we cannot rely on the Euclidean results to tell us anything about the Lorentzian physics.






        share|cite|improve this answer











        $endgroup$



        [The following is a half-remembered comment that my doctoral advisor told me some years ago, so I may have garbled it. I welcome corrections in the comments; feel free to tell me I'm full of it as well.]



        One way to think about a Wick rotation is that the "Euclidean" and "Lorentzian" manifolds (both of which are four-dimensional real manifolds, with a particular metric) can be viewed as hypersurfaces lying in an underlying four-dimensional complex manifold. For example, in the complex manifold $mathbbC^4$ with the obvious metric, you can find hypersurfaces with four (real) dimensions that are diffeomorphic to Euclidean four-space, and hypersurfaces with four (real) dimensions that are diffeomorphic to Minkowski space. The reason that Wick rotations are successful in flat spacetime is because the functions we're looking at are generally holomorphic, and so they can be analytically continued from one "cross-section" to another.



        In this picture, a vector lying in a Euclidean cross-section of $mathbbC^4$ must be actively "rotated" into the Lorentzian cross-section. Simply changing the coordinates on your cross-section will not magically "pull in" a vector that doesn't already lie in that cross-section.



        This picture, by the way, does not necessarily carry over to the analysis in curved spacetimes. We might think that if the Lorentzian metric is of the form
        $$
        ds^2 = - f(r,t) dt^2 + g^ij dx_i dx_j
        $$

        in some set of coordinates, then we can define a Euclidean analog
        $$
        ds_E^2 = f(r,t) dt^2 + g^ij dx_i dx_j
        $$

        and do the analysis there. However, there is no guarantee that there exists a complex manifold having these two cross-sections, and so we cannot rely on the Euclidean results to tell us anything about the Lorentzian physics.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 hours ago

























        answered 7 hours ago









        Michael SeifertMichael Seifert

        18k2 gold badges33 silver badges60 bronze badges




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            Tom Holland Mục lục Đầu đời và giáo dục | Sự nghiệp | Cuộc sống cá nhân | Phim tham gia | Giải thưởng và đề cử | Chú thích | Liên kết ngoài | Trình đơn chuyển hướngProfile“Person Details for Thomas Stanley Holland, "England and Wales Birth Registration Index, 1837-2008" — FamilySearch.org”"Meet Tom Holland... the 16-year-old star of The Impossible""Schoolboy actor Tom Holland finds himself in Oscar contention for role in tsunami drama"“Naomi Watts on the Prince William and Harry's reaction to her film about the late Princess Diana”lưu trữ"Holland and Pflueger Are West End's Two New 'Billy Elliots'""I'm so envious of my son, the movie star! British writer Dominic Holland's spent 20 years trying to crack Hollywood - but he's been beaten to it by a very unlikely rival"“Richard and Margaret Povey of Jersey, Channel Islands, UK: Information about Thomas Stanley Holland”"Tom Holland to play Billy Elliot""New Billy Elliot leaving the garage"Billy Elliot the Musical - Tom Holland - Billy"A Tale of four Billys: Tom Holland""The Feel Good Factor""Thames Christian College schoolboys join Myleene Klass for The Feelgood Factor""Government launches £600,000 arts bursaries pilot""BILLY's Chapman, Holland, Gardner & Jackson-Keen Visit Prime Minister""Elton John 'blown away' by Billy Elliot fifth birthday" (video with John's interview and fragments of Holland's performance)"First News interviews Arrietty's Tom Holland"“33rd Critics' Circle Film Awards winners”“National Board of Review Current Awards”Bản gốc"Ron Howard Whaling Tale 'In The Heart Of The Sea' Casts Tom Holland"“'Spider-Man' Finds Tom Holland to Star as New Web-Slinger”lưu trữ“Captain America: Civil War (2016)”“Film Review: ‘Captain America: Civil War’”lưu trữ“‘Captain America: Civil War’ review: Choose your own avenger”lưu trữ“The Lost City of Z reviews”“Sony Pictures and Marvel Studios Find Their 'Spider-Man' Star and Director”“‘Mary Magdalene’, ‘Current War’ & ‘Wind River’ Get 2017 Release Dates From Weinstein”“Lionsgate Unleashing Daisy Ridley & Tom Holland Starrer ‘Chaos Walking’ In Cannes”“PTA's 'Master' Leads Chicago Film Critics Nominations, UPDATED: Houston and Indiana Critics Nominations”“Nominaciones Goya 2013 Telecinco Cinema – ENG”“Jameson Empire Film Awards: Martin Freeman wins best actor for performance in The Hobbit”“34th Annual Young Artist Awards”Bản gốc“Teen Choice Awards 2016—Captain America: Civil War Leads Second Wave of Nominations”“BAFTA Film Award Nominations: ‘La La Land’ Leads Race”“Saturn Awards Nominations 2017: 'Rogue One,' 'Walking Dead' Lead”Tom HollandTom HollandTom HollandTom Hollandmedia.gettyimages.comWorldCat Identities300279794no20130442900000 0004 0355 42791085670554170004732cb16706349t(data)XX5557367