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There are 2 parts to my question:


1) Say we choose the metric signature to be (-+++), as in the Wikipedia page. Then the invariant interval in Minkowski space is written:

$ds^{2} = -(dt^{2}) + dx^{2} + dy^{2} + dz^2$

Taking $t=-i\tau$ gives:

$ds^{2} = d\tau^{2} + dx^{2} + dy^{2} + dz^2$

which is just the Euclidean invariant interval. This works out very nicely.

However, consider the conventions used by Peskin and Schroeder in their book ("An Introduction to Quantum Field Theory"). They use the metric signature (+---) (page xix, in "Notations and Conventions"). On page 193 they perform a Wick rotation on a linear combination of momenta (defined on page 191) called "$l$". In their transformation they take the zeroth component $l^{0} = il^{0}_{E}$, where on the LHS it's Minkowski formalism and on the RHS it's Euclidean. Using this prescription we'd have for the invariant interval, $t=i\tau$, so the Minkowski interval:

$ds^{2} = dt^{2} - (dx^{2}) - (dy^{2}) - (dz^2)$

would become:

$ds^{2} = -(d\tau^{2}) - (dx^{2}) - (dy^{2}) - (dz^2)$

So the first question is:

Is it a problem that the result here is a negative of what we'd usually use for the Euclidean invariant interval?

My guess is that it doesn't matter, but it would be nice to get confirmation of this.


2) If we're willing to except that negative versions of the Euclidean interval are okay, could we not define our Wick rotation in a different manner also?

Take, for example, the first Minkowski metric signature (-+++), such that:

$ds^{2} = -(dt^{2}) + dx^{2} + dy^{2} + dz^2$

Is there anything wrong with defining our Wick rotation as $t=\tau$, $x^{1} = ix^{1}$, $x^{2} = ix^{2}$, and $x^{3} = ix^{3}$, such that we get for the Euclidean invariant interval:

$ds^{2} = -(d\tau^{2}) - (dx^{2}) - (dy^{2}) - (dz^2)$

Of course, the same Wick rotation could be applied to the Minkowski invariant interval which uses the metric signature (+---):

$ds^{2} = dt^{2} - (dx^{2}) - (dy^{2}) - (dz^2)$

to give the original (non-negative) Euclidean invariant interval:

$ds^{2} = d\tau^{2} + dx^{2} + dy^{2} + dz^2$

Both methods of Wick rotation (making the time-like coordinate imaginary, or making the space-like components imaginary) seem viable to me. It's perhaps more convenient to alter the time-like components as it involves less algebra. For example, if we make the space-like components imaginary, an integral over Minkowski space $\int d^{d}x = \int dx^{0}, dx^{1},..., dx^{d}$ would acquire either an i, -1, -i, or 1 in front when we Wick rotate, depending on how many spatial coordinates we work with.

My thanks for any feedback.

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  • $\begingroup$ Could someone clarify why the result 6.50 in page 193 of Peskin is not valid for m=3 and only for m>3? Wick rotation uses only:1.integrand should rapidly go to zero.2.enclosing any one pole.I don't see how this has anything to do with overall power of the integrand. It also doesn't make sense if I use counting of powers of momenta l. $\endgroup$ – quarkonium Apr 9 '17 at 7:32
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If I remember well, this section of Peskin & Schroeder uses the Wick rotation to solve integrals of type $$\int \frac{1}{k^n + ...} \cdot ... $$ Where $k^n = k \cdot k \cdot k ... (\rm n \, times)$. By performing the Wick rotation we suddenly get a spherically symmetric problem which enables us to use the well known tricks for such a case. But notice that $(p \cdot k)_{+---}=-(p \cdot k)_{-+++}$ - by performing the Wick rotation in the inverse signature we will only get an overall minus sign and that does not change the applicability of our trick. So it poses no problem.

As you propose, as long as you take care of not crossing any singularities with your integration contour, you can perform a Wick rotation in other coordinates. If you do this properly without forgetting the factors from differentials or changed orientation of the contour, the result should be the same.

Do note that a number of other formulas also change sign when switching between signatures. E.g. $$\{\gamma^\mu,\gamma^\nu\} = 2 g^{\mu \nu},\, (+---) \to \{\gamma^\mu,\gamma^\nu\} = -2 g^{\mu \nu},\,(-+++)$$ and so on. So to compare the results of different signatures, you would have to trace back all these signs and correct them. In the end, it really does not change any of the tricks of the calculations, let even the experimental predictions.

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  • $\begingroup$ Thanks for the response, I found it very helpful. My original problem was that I'm working in the opposite signature to Peskin and Schroeder. In their equation above (6.49) they have $l_{E}^{2} + \Delta$ in the denominator (both with the same signs), where $l_{E}$ is the Wick rotated quantity. This allows the use of the beta function in evaluating the integral. However with opposite metric signature the $l_{E}^{2}$ becomes negative making it difficult to use the beta function. Wick rotating the spatial coordinates instead solves this problem as $l_{E}^{2}$ and $\Delta$ obtain the same sign. $\endgroup$ – Siraj R Khan Sep 17 '14 at 9:32
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    $\begingroup$ I never went through the opposite signature calculations explicitly, but in the -+++ you have $p^2=-m^2$, so the propagator will have $1/(p^2+m^2)$ instead of the usual $1/(p^2-m^2)$. This will change your $\Delta$ and other things, so I believe you will use the same trick in the end. But as I said, I have not tried it. $\endgroup$ – Void Sep 17 '14 at 10:09
  • $\begingroup$ Hi, just to let you know. I was being a little silly. It's perfectly possible to use the beta function even if $l_{E}^{2}$ has a different sign. $\endgroup$ – Siraj R Khan Sep 17 '14 at 12:20

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