Choice of signature important for superluminal 4-velocity? (Minkowski)

[PhDeezNutz]

TL;DR Summary

This is an extension of a homework problem. (That I already solved)

https://www.physicsforums.com/threads/superluminal-moving-points-4-velocity.988718/

See post 2

I solved the problem (got the right answer) but I would like some insight. My answer seems contingent on choice of signature. I had to specifically choose (-,+,+,+) in order for the quantity under the square root to be positive.

I guess my question boils down to "Is choice of signature important when dealing with superluminal 4-velocities"? I wanted to show for superluminal velocities that

$\tilde{U} = \left( \frac{c}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{x}}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{y}}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{z}}{\sqrt{\frac{u^2}{c^2} - 1}}\right)$

(This is Rindler Problem 5.4 btw)

And I was successful using (-,+,+,+) and unsuccessful using (+,-,-,-). I was always under the impression that choice of signature did not matter as long as you were consistent but now I have my doubts.

Thank you in advanced for your insights.

I hope I am not breaking any forum rules by asking for insight on a homework problem in a non-homework section. In my defense I already solved it and now I'm asking for general insight that goes beyond the scope of the homework problem. I only used my homework problem to motivate it.

 

[robphy]

And I was successful using (-,+,+,+) and unsuccessful using (+,-,-,-). I was always under the impression that choice of signature did not matter as long as you were consistent but now I have my doubts.

As part of being consistent with choices of convention,
there is also a need to be consistent with definitions, which also should be spelled out.

 

[PhDeezNutz]

As part of being consistent with choices of convention,
there is also a need to be consistent with definitions, which also should be spelled out.

Which definitions should I elaborate on/specify?

The only one I can think of is $ds$

$ds^2 = - c^2 dt^2 + \left(dx^2 + dy^2 + dz^2 \right)$ and therefore

$ds = \sqrt{u^2 - c^2} dt$

Am I using definitions from one signature and applying it to another?

 

[PeterDonis]

"Is choice of signature important when dealing with superluminal 4-velocities"?

First, the term "superluminal 4-velocities" is a misnomer that will only cause confusion. What you are doing is simply computing the tangent vector to a spacelike curve. All curves have tangent vectors, whether they are timelike, spacelike, or null. There is no need to bring in any talk about "superluminal 4-velocities" for the tangent vector to a spacelike curve in order to make the math work. (If that term originally came from Rindler's book--I don't have it so I can't check--then this is an example of when not to take everything said in a textbook as being valid or useful.)

That said, what you are looking for, I take it, are the components of the tangent vector $U$ to a spacelike curve, expressed in a standard Minkowski coordinate chart. What does it mean to say that $U$ is spacelike? It means the sign of its squared norm is the same as the sign of the spacelike terms in the metric signature. So if you are using signature $(- + + +)$, the squared norm of a spacelike vector is positive. If you are using signature $(+ - - -)$, the squared norm of a spacelike vector is negative. That is where the signature comes into play.

 

[PhDeezNutz]

First, the term "superluminal 4-velocities" is a misnomer that will only cause confusion. What you are doing is simply computing the tangent vector to a spacelike curve. All curves have tangent vectors, whether they are timelike, spacelike, or null. There is no need to bring in any talk about "superluminal 4-velocities" for the tangent vector to a spacelike curve in order to make the math work. (If that term originally came from Rindler's book--I don't have it so I can't check--then this is an example of when not to take everything said in a textbook as being valid or useful.)

That said, what you are looking for, I take it, are the components of the tangent vector $U$ to a spacelike curve, expressed in a standard Minkowski coordinate chart. What does it mean to say that $U$ is spacelike? It means the sign of its squared norm is the same as the sign of the spacelike terms in the metric signature. So if you are using signature $(- + + +)$, the squared norm of a spacelike vector is positive. If you are using signature $(+ - - -)$, the squared norm of a spacelike vector is negative. That is where the signature comes into play.

This gives me a lot to think about. Thank you for this post. I will report back soon.

And yes the term superluminal 4-velocity was used in Rindler's book but your explanation of merely finding the tangent of a spacelike curve makes much more sense.

 

[vanhees71]

No matter in which context you treat superluminal particles (tachyons), the choice of the sign convention of the Minkowski pseudometric (it's NOT a metric and $x \cdot x$ is NOT a norm for four-vectors $x$) doesn't matter at all. Neither the physics nor the mathematics depend on this choice of convention. All you have to do is to use the correct signs everywhere.

 

[PhDeezNutz]

Suppose I use signature $(+,-,-,-)$ and I am dealing with a spacelike curve $ds^2 \lt 0$.

$ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 \lt 0$

Does that imply $ds$ is imaginary/complex?

If so that has me worried because the answer I'm supposed to get does not have any imaginary units in it.

 

[PeterDonis]

Does that imply $ds$ is imaginary/complex?

No. It just means you picked a signature that is more suited to handling timelike curves than spacelike ones. In many books the signature $(+, -, -, -)$ is called the "timelike convention" and the signature $(-, +, +, +)$ is called the "spacelike convention" for that very reason. But you can still do computations either way.

Note, btw, that the $ds$ that appears in $ds^2$ in the line element is not the same as the $ds$ that appears when you take derivatives with respect to the curve parameter $s$ in the tangent vector $U$. The double use of the same notation is unfortunate but unavoidable at this point.

 

[Nugatory]

If so that has me worried because the answer I'm supposed to get does not have any imaginary units in it.

Whether $ds$ is real along a timelike vector and complex along a spacelike one or the other way around around depends on the sign convention you choose for the metric.

When you've done everything right in your calculation, the only difference will be the sign of the squared interval (or equivalently the appearance of a factor of $i$ in the unsquared interval). When that's the only difference between your answer and the one you're "supposed to get" your answer isn't wrong, just written using a different convention.

 

[PeterDonis]

Whether $ds$ is real along a timelike vector and complex along a spacelike one or the other way around around depends on the sign convention you choose for the metric.

This is true for line elements, but the OP doesn't need to calculate any line elements. He just needs to take derivatives of coordinates with respect to a curve parameter. Since the coordinates and the curve parameter are all real, there is no need for imaginary numbers anywhere, regardless of which metric signature is being used.

 

[PhDeezNutz]

@PeterDonis , forgive me if I am wrong, you mentioned that the $ds$ in $\tilde{U} = c \frac{d\tilde{R}}{ds}$ and the general line element in $ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$ are different. I thought the typical definition of 4-velocity

$\frac{d\tilde{R}}{d \tau}$ requires that the two aforementioned $ds$'s are the same.

$d \tau^2 = \frac{ds^2}{c^2} \Rightarrow ds = c d\tau$

plugging $ds$ back into $\tilde{U} = c \frac{d\tilde{R}}{ds} = \frac{d \tilde{R}}{d \tau}$

 

[PeterDonis]

I thought the typical definition of 4-velocity

$\frac{d\tilde{R}}{d \tau}$ requires that the two aforementioned ds's are the same.

No. Again, the tangent vector components are the derivatives of the coordinates with respect to the curve parameter. The $ds$ in the denominator of that derivative is not the same as the $ds$ in the line element. They're two different things.

Thinking of derivatives as fractions is a common sloppy shortcut that works in some contexts, but is best avoided.

 

[robphy]

Which definitions should I elaborate on/specify?

The only one I can think of is $ds$

$ds^2 = - c^2 dt^2 + \left(dx^2 + dy^2 + dz^2 \right)$ and therefore

$ds = \sqrt{u^2 - c^2} dt$

Am I using definitions from one signature and applying it to another?

Since it seems (in your example) the goal is to determine
the unit vector associated with a general non-null tangent vector, $\tilde W =(c,\dot x, \dot y, \dot z)$,
it seems to me that, for spacelike vectors,
$\tilde{U} = \left( \frac{c}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{x}}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{y}}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{z}}{\sqrt{\frac{u^2}{c^2} - 1}}\right)$
and
for timelike vectors,
$\tilde{U} = \left( \frac{c}{\sqrt{1-\frac{u^2}{c^2} }}, \frac{\dot{x}}{\sqrt{1 -\frac{u^2}{c^2} }}, \frac{\dot{y}}{\sqrt{1-\frac{u^2}{c^2} }}, \frac{\dot{z}}{\sqrt{1 - \frac{u^2}{c^2}}}\right)$

needs to be generalized to
$\tilde{U} = \frac{1}{\sqrt{\left| \tilde W\cdot \tilde W \right|/c^2}} \left( c, \dot x, \dot y, \dot z \right) \frac{1}{\sqrt{\left| 1-\frac{u^2}{c^2} \right|}} \left( c, \dot x, \dot y, \dot z \right)$
to handle both cases.

 

[PhDeezNutz]

Since it seems (in your example) the goal is to determine
the unit vector associated with a general non-null tangent vector, $\tilde W =(c,\dot x, \dot y, \dot z)$,
it seems to me that, for spacelike vectors,
$\tilde{U} = \left( \frac{c}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{x}}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{y}}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{z}}{\sqrt{\frac{u^2}{c^2} - 1}}\right)$
and
for timelike vectors,
$\tilde{U} = \left( \frac{c}{\sqrt{1-\frac{u^2}{c^2} }}, \frac{\dot{x}}{\sqrt{1 -\frac{u^2}{c^2} }}, \frac{\dot{y}}{\sqrt{1-\frac{u^2}{c^2} }}, \frac{\dot{z}}{\sqrt{1 - \frac{u^2}{c^2}}}\right)$

needs to be generalized to
$\tilde{U} = \frac{1}{\sqrt{\left| \tilde W\cdot \tilde W \right|}} \left( c, \dot x, \dot y, \dot z \right) \frac{1}{\sqrt{\left| 1-\frac{u^2}{c^2} \right|}} \left( c, \dot x, \dot y, \dot z \right)$
to handle both cases.

So it seems the magnitude bars make all the difference. It makes sense in hindsight; we want to maintain direction of the tangent vector after normalizing it so the magnitude bars are a necessity.

 

[robphy]

So it seems the magnitude bars make all the difference. It makes sense in hindsight; we want to maintain direction of the tangent vector after normalizing it so the magnitude bars are a necessity.

Either that, or else use separate definitions for each case...

 

[PhDeezNutz]

No. Again, the tangent vector components are the derivatives of the coordinates with respect to the curve parameter. The $ds$ in the denominator of that derivative is not the same as the $ds$ in the line element. They're two different things.

Thinking of derivatives as fractions is a common sloppy shortcut that works in some contexts, but is best avoided.

I agree, it was one of the oppositions I had to my own solution (treating derivatives as fractions). I wish I had a working knowledge of more formal mathematics/analysis.

 

[PhDeezNutz]

Either that, or else use separate definitions for each case...

There's just something that doesn't sit right with me about using different definitions in the same space. I much prefer your way of generalizing.

 

[robphy]

There's just something that doesn't sit right with me about using different definitions in the same space. I much prefer your way of generalizing.

Yeah... it's as if there is a bigger single idea out there,
generalizing from a familiar but only-now-seen-to-be-just-a-specialized case.

 

[PeterDonis]

There's just something that doesn't sit right with me about using different definitions in the same space.

In the interest of pointing out inconvenient facts , I will note that none of the definitions given in this thread will work for tangent vectors to null curves.

 

[PhDeezNutz]

In the interest of pointing out inconvenient facts , I will note that none of the definitions given in this thread will work for tangent vectors to null curves.

you are of course correct. I clearly don’t know enough differential geometry to truly understand the topic at hand. Do you have suggestions for books or list of topics I need to look into?

 

[robphy]

In the interest of pointing out inconvenient facts , I will note that none of the definitions given in this thread will work for tangent vectors to null curves.

Or even curves that smoothly change from a
segment that is timelike to one that is spacelike, or the reverse.

 

[PeterDonis]

Do you have suggestions for books or list of topics I need to look into?

Sean Carroll's online lecture notes on GR give a decent introduction to differential geometry (Chapters 2 and 3):

https://arxiv.org/abs/gr-qc/9712019

 

[vanhees71]

There are no complex numbers in Minkowski space, because it's a real affine pseudo-Euclidean space. I don't know what you want to calculate with superluminal particles (tachyons).

No. Again, the tangent vector components are the derivatives of the coordinates with respect to the curve parameter. The $ds$ in the denominator of that derivative is not the same as the $ds$ in the line element. They're two different things.

Thinking of derivatives as fractions is a common sloppy shortcut that works in some contexts, but is best avoided.

Well, for normal massive particles you usually indeed choose the proper time of the particle as the affine parameter of the curve, because it's convenient and defines energy and momentum in a physical way.

For me the most convenient way is to use Hamilton's principle (in Lagrangian form). There are two strategies.

(a) parameter independent form

You start with an arbitrary parameter $\lambda$ for the world line of the particle (which can also be the coordinate time of an arbitrary inertial frame ("Galilean coordinates"); I restrict myself to special relativity here). To write the Lagrangian in manifestly covariant form you make it homogeneous in $\dot{x}^{\mu}$, where the dot means derivatives wrt. $\lambda$. Then you get a Lorentz-invariant action
$$A[x]=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda L(x,\dot{x}).$$
Noether's theorem for the free point particle tells you that the "kinetic term" is (in the signature (+---))
$$L_0=-m c^2\sqrt{\dot{x}^{\mu} \dot{x}_{\mu}}.$$
The most simple example for an interaction part is the motion of a charged particle in an em. field,
$$L_{\text{em}}=-\frac{q}{c} A_{\mu}(x) \dot{x}^{\mu}.$$
Now since $L$ doesn't depend explicitly on $\lambda$ and due to the homogeneity of rank 1 wrt. $\dot{x}^{\mu}$ there's a constraint
$$\dot{x}^{\mu} \frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial L}{\partial \dot{x}^{\mu}}= \dot{x}^{\mu} \frac{\partial L}{x^{\mu}},$$
that is fulfilled for all (!) trajectories not only the solutions of the Euler-Lagrange equations, which implies that the particle has only three not four independent degrees of freedom as it should be.

This enables you to choose $\lambda$ as an affine parameter, i.e., imposing the constraint
$$\dot{x}_{\mu} \dot{x}^{\mu}=\text{const}.$$
For massive particles (time-like curves) the constant is $>0$ and can be chosen to be $c^2$. Then $\lambda=\tau$ with $\tau$ the proper time of the particle. This simplifies the equations of motion significantly.

The advantage of this approach is that it is using a general worldline parameter without changing the physics content and having a manifest covariant formalism. It's only a bit inconvenient since you have to impose the affinity of the parameter to simplify the equations as an additional constraint.

That's why I like the alternative formulation most

(b) Affine-parameter form of the Lagrangian

The trick here is to formulate the Lagrangian from the very beginning under the constraint that $\lambda$ is an affine parameter. You can do this with help of a Lagrange parameter, but the upshot is to use
$$L_0=-\frac{m}{2} \dot{x}_{\mu} \dot{x}^{\mu}$$
but keep the $L_{\text{int}}$ from the previous formalism, i.e., a rank-1 homogeneous expression wrt. $\dot{x}^{\mu}$.

Then the fact that $L$ doesn't depent explicitly on $\lambda$ automatically leads to the constraint $L=\text{const}$ along the solutions of the Euler-Lagrange equations, and you can make $L=-m/2 c^2$ such that $\lambda=\tau$ automatically without imposing it as additional constraint.

For more details, see

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[PhDeezNutz]

I was merely solving a homework problem. I have no familiarity with Tachyons.

In my class we didn’t get into the Lagrangian/Hamiltonian formulation of SR but I’m sure it is something I will need to eventually learn so thank you for the link.

I think for now I have to move on to the next question and newer material. I have to know my limits and for now the question in the OP is beyond me but I’d like to thank everyone for contributing. It will take time for me to digest and understand all of this.

In the meantime I will have to just remember when to use either signature. Or to use @robphy ’s general formulation that works for pure time-like and pure space-like trajectories.When I’m ready to come back to this problem I will bump this thread.

 

[vanhees71]

You should use one and only one signature. Otherwise you confuse yourself unnecessarily. There's no principle difference in this choice concerning the physics anyway. I'm used to the (+---) convention, because that's what the majority in my field (relativistic heavy-ion/particle physics) uses. It's just by chance, which one you get used to first. In your case, I'd use the one used by the lecturer/textbook.

 

[PhDeezNutz]

I am also used to (+,-,-,-) and it is the one the book/lecturer use. I guess I'll use that with the approach outlined in post 13 in order to maintain direction/orientation of curve.

 

[robphy]

Yes, stick to one signature. I use $(+,-,-,-)$ for pedagogical reasons,
although relativity favors the opposite convention.

My approach with the absolute values applies only to normalization.
I don't use absolute values in general.
In particular, the square-interval is [always] signed.

 

[vanhees71]

One should also note that you don't get too far with tachyons. I'm not aware of a detailed worked out theory for classical (point-particle) tachyons. So I can't say much about how far you can get with that to make physical sense. Within QFT they are problematic due to problems with microcausality principle to be valid. For newer ideas, see e.g.,

https://arxiv.org/abs/0709.0414

One should note however, that this paper is not published!

A published one, with the microcausality condition intact:

https://arxiv.org/abs/hep-ph/0105153

 

[PhDeezNutz]

Yes, stick to one signature. I use $(+,-,-,-)$ for pedagogical reasons,
although relativity favors the opposite convention.

My approach with the absolute values applies only to normalization.
I don't use absolute values in general.
In particular, the square-interval is [always] signed.

Hmm I thought your approach with absolute values was also to maintain direction i.e making a tangent vector parallel instead of anti-parallel.I hope Peter Donis forgives me for what I'm about to say next but it's the only way I know how to get the correct answer. I'm going to conflate parameter variable $ds$ with "arclength" $ds$ and I'm going to treat derivatives as mere quotients of differentials. I'm also going to use signature (+,-,-,-).

In doing so I think the new definition of 4-velocity should be

$\tilde{U} = c \frac{d\tilde{R}}{\left|ds\right|}$

$ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$

$ds = \sqrt{c^2 - \dot{x}^2 - \dot{y}^2 - dot{z}^2} dt$

We know that $\dot{x}^2 + \dot{y}^2 + \dot{z}^2 > c^2$

$ds = \sqrt{c^2 - u^2} \text{ } dt = \sqrt{(-1)\left( u^2 - c^2\right)} \text{ }dt = i c \sqrt{\frac{u^2}{c^2} - 1} \text{ }dt$

$\left| ds \right| = c \sqrt{\frac{u^2}{c^2} - 1}$

$d\tilde{R} = \left( c dt, dx, dy, dz \right)$

My answer is

$\tilde{U} = c \frac{d\tilde{R}}{\left|ds\right|} = \frac{c \left( c dt, dx, dy, dz \right) }{c \sqrt{\frac{u^2}{c^2} - 1} \text{ } dt} = \left( \frac{u^2}{c^2} - 1\right)^{-\frac{1}{2}} \left( c , \dot{x}, \dot{y}, \dot{z} \right)$

 

[PeterDonis]

for normal massive particles you usually indeed choose the proper time of the particle as the affine parameter of the curve

Along timelike curves, yes. Along spacelike curves, proper distance along the curve is the natural affine parameter.

 

[PeterDonis]

I think the new definition of 4-velocity should be

All you are doing is treating derivatives as fractions (bad) and taking an absolute value, which throws away information (bad), in order to reach the same conclusion you could reach by just using the standard definition of the tangent vector components, which are, as I've already said, just the derivatives of the coordinates with respect to the curve parameter.

it's the only way I know how to get the correct answer

You don't know how to take derivatives with respect to the curve parameter?

If so, learning how to do that would seem to be a much better idea than what you are doing now.

 

[PhDeezNutz]

All you are doing is treating derivatives as fractions (bad) and throwing away information about whether a curve is timelike or spacelike (bad) in order to reach the same conclusion you reach by just using the standard definition of the tangent vector components, which are, as I've already said, just the derivatives of the coordinates with respect to the curve parameter.
You don't know how to take derivatives with respect to the curve parameter?

If so, learning how to do that would seem to be a much better idea than what you are doing now.

I know in Multi-variable calculus

say we have a curve

$\vec{r} \left( t \right) = \left( x(t) , y(t), z(t) \right)$

that

$\vec{r}' \left( t \right) = \left( x'(t) , y'(t), z'(t) \right)$

should I just set $t = s$ for our problem?

 

[PeterDonis]

should I just set $t = s$ for our problem?

No, since $t$ is not the arc length along the curve.

In this case, you have a coordinate 4-vector $X(s) = (t(s), x(s), y(s), z(s))$. Just apply the same method you did for a 3-vector to that; the tangent vector $U$ is just $X'(s)$.

 

[vanhees71]

Well, take the approach with the Lagrangian as for usual subluminal massive particles for tachyons, which by definition travel along space-like world lines. Then you can define
$$L_0=-m c^2 \sqrt{-\dot{x}_{\mu} \dot{x}^{\mu}}.$$
with the same argument that this leads to a parameter-independent Lorentz-invariant action. Then the "four momentum" is
$$p_{\mu}=\frac{1}{c} \frac{\partial L_0}{\partial \dot{x}^{\mu}} = m c \dot{x}_{\mu} (-\cdot{x}_{\mu} \dot{x}^{\mu})^{-1/2}$$
and the on-shell condition is
$$p_{\mu} p^{\mu}=-m^2 c^2; \Rightarrow \; \vec{p}^2=(p^0)^2+m^2 c^2$$
This is then in accordance with the tachyonic feature that the particle moves in any inertial frame with a constant speed larger than the speed of light, i.e.,
$$\vec{v}=c \frac{\vec{p}}{p^0}=c \frac{\vec{p}}{\sqrt{\vec{p}^2-m^2 c^2}} \; \Rightarrow \; |\vec{v}| \geq c.$$
I think the real trouble with the tachyons starts when it comes to interacting ones and causality.

 

[PhDeezNutz]

No, since $t$ is not the arc length along the curve.

In this case, you have a coordinate 4-vector $X(s) = (t(s), x(s), y(s), z(s))$. Just apply the same method you did for a 3-vector to that; the tangent vector $U$ is just $X'(s)$.

Alright I've used your approach (or at least what I think is your approach) to come up with a more rigorous solution. However I am getting a factor of $\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}$ in front instead of $\frac{1}{\sqrt{\frac{u^2}{c^2} - 1}}$. I'm going to use signature (+,-,-,-) if that's okay.

The only questionable mathematical maneuver that might be troublesome is asserting $\frac{dt}{ds} = \frac{1}{\frac{ds}{dt}}$ but I think there's a version of the inverse function theorem that supports this.

My solution is as follows;

$\tilde{U} = c \frac{d \tilde{R}}{ds}$

$\frac{d \tilde{R}}{ds} = \left(c \frac{dt}{ds}, \frac{dx}{ds}, \frac{dy}{ds}, \frac{dz}{ds} \right)$

Using the chain rule we have

$\frac{d\tilde{R}}{ds} = \left( c \frac{dt}{ds}, \dot{x} \frac{dt}{ds}, \dot{y} \frac{dt}{ds}, \dot{z} \frac{dt}{ds} \right) = \frac{dt}{ds} \dot{\tilde{R}}$

So I guess we need to know $\frac{dt}{ds}$

$\frac{ds}{dt} = \sqrt{c^2 - \dot{x}^2 - \dot{y}^2 - \dot{z}^2} = \sqrt{c^2 - u^2}$

I believe from the inverse function theorem of single-variable calculus we can say

$\frac{dt}{ds} = \frac{1}{\sqrt{c^2 - u^2}}$

Giving us

$\tilde{U} = c \frac{d \tilde{R}}{ds}= c \frac{dt}{ds} \dot{\tilde{R}} = \frac{c}{\sqrt{c^2 - u^2}} \left( c, \dot{x},\dot{y},\dot{z} \right)$Your way (or at least what I perceive to be your way) is more satisfying than naively assuming derivatives are mere quotients of differentials. That said if I am wrong about the inverse function theorem then all of what I typed is pretty much bunk. I think if used (-,+,+,+) I would have gotten the correct answer, but I am still conflicted because the general impression I get from this thread is that signature should not matter.