Recall from my post about the quaternions that I had mentioned the split-complex (hyperbolic) numbers. I now wish to return to these and uncover the nature of calculus on the split-complex plane and determine what it says about the nature of physics. The split-complex numbers are inherently linked to special relativity and hence are linked to our understanding of the world. Therefore a foray into the world of split-complex numbers will lead to fundamental insight into how nature works. For the purposes of this post, I will consider $1+1$ dimensional spacetime (one spatial variable and one temporal variable).

By analogy, we want some sort of product that takes $\Bbb H\times \Bbb H$ into $\Bbb R$, and it turns out that the conjugation prescription is the way to do this. If we fix $z=x+jy$ its hyperbolic conjugate is $z^* = x-jy$ and their product is given by $zz^* = (x+jy)(x-jy) = x^2-y^2$. So we see that multiplying a split-complex number by its conjugate gives a real number like desired with one caveat: this number is no longer strictly positive. The form of the modulus-square is very reminiscent of the equation of a hyperbola, $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} = 1$, and this is why the split-complex numbers are often referred to as hyperbolic numbers. (Some refer to $zz^*$ as the modulus but I think it causes confusion when compared with the complex case and elect to call it the modulus-square.) One peculiar aspect of the hyperbolic numbers is that if one considers a hyperbolic number of the form $x+jx$, it has a modulus of $0$ which, as we will see, is a bit troublesome but very important.

Now that we have come up with the modulus, it is natural to ask if we can divide by hyperbolic numbers since this was the progression we took when considering the complex numbers. If we want to find an inverse for a hyperbolic number $z$, we want to determine what $z'$ in $\Bbb H$ satisfies $zz' = 1$. Since multiplication of hyperbolic numbers is commutative (meaning the order in which we multiply doesn't matter), if we find out what $z'$ satisfies $zz' = 1$, then we know that it also satisfies $z'z = 1$ and so it is the unique inverse.

So let us multiply both sides of the equality $zz' = 1$ by $z^*$ to get that $z^*zz' = z^*$, therefore $z' = \dfrac{z^*}{z^*z}$. If $z=x+jy$, then we have that $z' = \dfrac{x-jy}{x^2-y^2}$ and we have an explicit expression for the inverse of a hyperbolic number. As noted above if $z=x+jx$, then its modulus is $0$ and so the inverse of $z$ cannot hope to make sense with the above expression since we would end up dividing by $0$. One can actually show that no $z'\in\Bbb H$ exists so that $zz' = 1$ if $z=x+jx$. Geometrically one can say that the hyperbolic numbers along the lines $y=\pm\, x$ in the hyperbolic plane do not have inverses. These two lines are very important in special relativity and are known as the light cone. I will explore this further later in the post.

We have now built up hyperbolic numbers, their modulus(-square) and their inverses. It is then reasonable to ask if they have a polar-like representation like complex numbers (each complex number $z$ except $0$ can be written in the form $r\exp(i\theta)$ where $r$ is the modulus and $\theta\in [0,2\pi)$). We already have the modulus for hyperbolic numbers so we need to explore what, if anything, $\exp(j\theta)$ is. I will make the assumption that hyperbolic series converge nicely and the following proof is not meant to be rigorous but merely instructive. In the "proof" we will need the fact that $j^n = 1$ if $n$ is even and $j^n = j$ if $n$ is odd.

$$\exp(j\theta) = \sum\limits_{n=0}^{\infty} \frac{(j\theta)^n}{n!} = \sum\limits_{n=0}^{\infty} \frac{j^n\theta^n}{n!}.$$

At this point it would be prudent to separate our above series into two separate series corresponding to the case of $j^{\text{even}}$ and $j^{\text{odd}}$ giving

$$\exp(j\theta) = \sum\limits_{n=0}^{\infty} \frac{j^{2n}\theta^{2n}}{(2n)!} + \sum\limits_{n=0}^{\infty}\frac{j^{2n+1}\theta^{2n+1}}{(2n+1)!} = \sum\limits_{n=0}^{\infty} \frac{\theta^{2n}}{(2n)!}+j\sum\limits_{n=0}^{\infty} \frac{\theta^{2n+1}}{(2n+1)!}. $$

For those familiar with their hyperbolic trigonometric functions, the above can easily be recognized as

$$ \exp(j\theta) = \cosh(\theta) + j\sinh(\theta). $$

We would then like to represent any $z\in \Bbb H$ (with exception maybe to the hyperbolic numbers on the light cone since they have modulus $0$) in the form $r\exp(j\theta)$. Note: $\theta$ no longer represents an angle but is more like a parameter along the hyperbola and now varies from $-\infty$ to $\infty$, and since $r$ is a radius, it must be strictly positive. In the complex number case, $r$ was related to the geometry of the complex numbers, i.e. the level curves of $zz^*$ (circles).

We wish the same to hold for the hyperbolic case, giving that $r$

Case 1: $|x| > |y|$ and $x > y$. This implies that $x$ is positive and $y$ is between the lines $y = \pm\,x$ and so $z$ lies on the right-opening hyperbola seen in the image above

We now have a complete hyperbolic-polar description of any $z\in\Bbb H$. Since 1+1 dimensional spacetime can be viewed with the hyperbolic numbers, it is natural to ask what calculus looks like on such a space to see if we can uncover any truths about nature.

Since we can divide by hyperbolic numbers, we can potentially come up with a notion of differentiability similar to that with complex-differentiable functions (as long as we don't divide by $0$ of course). Let us try to apply the standard definition of a derivative from elementary calculus to $f$ and see what kind of coupling there is between our $u$ and $v$. From the definition of a derivative we know that if

$$ \lim_{\Delta z\rightarrow 0}\frac{f(z+\Delta z)-f(z)}{\Delta z} $$

exists, then $f$ is differentiable at $z$. Let us again look at the limit as we approach $z$ along horizontal and vertical lines. If we approach along a horizontal line, $\Delta z = \Delta x$, and if we approach along a vertical line, $\Delta z = j\Delta y$

$$ \lim_{\Delta z\rightarrow 0}\frac{f(z+\Delta z) - f(z)}{\Delta z} = \lim_{\Delta x\rightarrow 0}\frac{u(x+\Delta x,y)-u(x,y)+jv(x+\Delta x,y)-jv(x,y)}{\Delta x} $$

along horizontal lines and

$$ \lim_{\Delta z\rightarrow 0} \frac{f(z+\Delta z)-f(z)}{\Delta z} = \lim_{\Delta y\rightarrow 0}\frac{u(x,y+\Delta y)-u(x,y)+jv(x,y+\Delta y)-jv(x,y)}{j\Delta y} $$

along vertical lines. Like in the $\Bbb R^2$ case, we require that if a derivative exists at a point, the value must be the same along every path to the point and so the two expressions above must be equal if $f$ is to be differentiable at $z$. We can recognize the above expressions as the partial derivatives of $u$ and $v$ along the $x$ and $y$ directions. Since the expressions must be equal we have that

$$ \frac{\partial u}{\partial x} + j\frac{\partial v}{\partial x} = j\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} $$

since $j^{-1} = j$. If we equate the "real" and "imaginary" parts we have that

$$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} $$

and

$$ \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}. $$

These two equations are similar to the Cauchy-Riemann equations for complex-differentiable functions. If we took the partial derivative of the first equation with respect to $y$ and the second with respect to $x$ and subtracted them, we would see that $v$ satisfies the following partial differential equation

$$ \frac{\partial^2 v}{\partial x^2} - \frac{\partial^2 v}{\partial y^2} = 0 $$

assuming that

$$ \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = 0 . $$

Since both $u$ and $v$ solve this equation, so does $f$ since differentiation is a linear operator. This partial differential equation is very important though you may not recognize it at first glance. Implicitly we have been treating $x$ and $y$ on the same footing and if we wish to relate them back to physical quantities, we must recognize that if $x$ is a coordinate variable (have units of length), then $y$ must also be a coordinate variable (have units of length). If we make the change of variables $x = ct$, then we have that $f$

$$ \frac{1}{c^2}\frac{\partial^2 f}{\partial t^2} - \frac{\partial^2 f}{\partial y^2} = 0 $$

which is the wave equation in 1+1 dimensions. So we see that differentiable functions on $\Bbb H$ must solve the wave equation. The wave equation is fundamental to special relativity because it governs the behavior of light as was recognized by James Clerk Maxwell.

Thus far we have seen some connections to special relativity: hyperbolic numbers of the form $x\pm jx$ denote the light cone of special relativity and differentiable functions on $\Bbb H$ solve the wave equation (and are thusly related to light in some fashion). Can we uncover any more physics by working with the hyperbolic numbers? It turns out we can!

Let us define a "rotation" in $\Bbb H$ of a hyperbolic number $z$ by "angle" $\varphi = \tanh^{-1}\left(\dfrac{v}{c}\right)$ to be given by $z\exp(j\theta)$ (we have implicitly assumed that $\left|\dfrac{v}{c}\right| < 1$ since that is the domain of $\tanh^{-1}$ - this assumption will be important physically)

If we write $\exp(j\varphi)$ as $\cosh(\varphi)+j\sinh(\varphi)$, we have that $\cosh(\varphi) = \dfrac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}\,\,$ and $\,\,\sinh(\varphi) = \dfrac{\frac{v}{c}}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$. For simplicity, I will define $\gamma$ to be $\dfrac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$. Therefore

$$z' = ct' + jx' = (ct+jx)\left(\gamma + j\gamma\dfrac{v}{c}\right) = \gamma\left(ct+x\dfrac{v}{c}\right)+\gamma\left(x+\frac{t}{c}\right)$$

and so $t' = \gamma\left(t+\frac{xv}{c^2}\right)$ and $x' = \gamma\left(t+\frac{xv}{c^2}\right)$. These are just the standard Lorentz transformations between two inertial reference frames and so it shows that Lorentz transformations are just rotations in the hyperbolic plane.

The interpretation of this is that if we compare our frame of reference to a frame of reference that is moving with uniform velocity relative to us (with $v < c$ since $\tanh^{-1}$ cannot be defined in the usual sense if $v$ is greater than or equal to $c$), then the way to relate our coordinates is just the Lorentz transformations

There is more that could be said about the connections between the hyperbolic numbers and special relativity but I will stop here. The above notion can be generalized to $3+1$ dimensions ($3$ spatial dimensions, $1$ time dimension) by including two more "imaginary" units like $j$ and these would be the place-keepers for the spatial components. In $3+1$ dimensions, the light cone makes up a cone embedded in $4$ dimensional space and the level curves of $zz^*$ are now hyperboloids but the ideas carry over somewhat naturally.

**The Hyperbolic Numbers****I will denote the set of split-complex numbers by $\Bbb H$ (for hyperbolic numbers). Recall that the split-complex numbers arise when one defines a quantity $j$ such that $j^2 = 1$ but $j$ is neither $1$ or $-1$ and taking numbers of the form $a+bj$ gives a split-complex number (in this post I assume $a$ and $b$ are real). I also called such numbers hyperbolic numbers but did not give any explanation for why they have such a moniker. To understand this description one must consider the modulus of a split-complex number $z = a+bj$. In the case of the complex numbers, the modulus-square gives the equation of a circle and so the modulus-square suggests some sort of structure on the complex numbers. The modulus-square also takes two complex numbers and outputs a real number, i.e. $|\cdot|:\Bbb C\to\Bbb R$ (recall that $|z|^2 = zz^*$).**

By analogy, we want some sort of product that takes $\Bbb H\times \Bbb H$ into $\Bbb R$, and it turns out that the conjugation prescription is the way to do this. If we fix $z=x+jy$ its hyperbolic conjugate is $z^* = x-jy$ and their product is given by $zz^* = (x+jy)(x-jy) = x^2-y^2$. So we see that multiplying a split-complex number by its conjugate gives a real number like desired with one caveat: this number is no longer strictly positive. The form of the modulus-square is very reminiscent of the equation of a hyperbola, $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} = 1$, and this is why the split-complex numbers are often referred to as hyperbolic numbers. (Some refer to $zz^*$ as the modulus but I think it causes confusion when compared with the complex case and elect to call it the modulus-square.) One peculiar aspect of the hyperbolic numbers is that if one considers a hyperbolic number of the form $x+jx$, it has a modulus of $0$ which, as we will see, is a bit troublesome but very important.

Now that we have come up with the modulus, it is natural to ask if we can divide by hyperbolic numbers since this was the progression we took when considering the complex numbers. If we want to find an inverse for a hyperbolic number $z$, we want to determine what $z'$ in $\Bbb H$ satisfies $zz' = 1$. Since multiplication of hyperbolic numbers is commutative (meaning the order in which we multiply doesn't matter), if we find out what $z'$ satisfies $zz' = 1$, then we know that it also satisfies $z'z = 1$ and so it is the unique inverse.

So let us multiply both sides of the equality $zz' = 1$ by $z^*$ to get that $z^*zz' = z^*$, therefore $z' = \dfrac{z^*}{z^*z}$. If $z=x+jy$, then we have that $z' = \dfrac{x-jy}{x^2-y^2}$ and we have an explicit expression for the inverse of a hyperbolic number. As noted above if $z=x+jx$, then its modulus is $0$ and so the inverse of $z$ cannot hope to make sense with the above expression since we would end up dividing by $0$. One can actually show that no $z'\in\Bbb H$ exists so that $zz' = 1$ if $z=x+jx$. Geometrically one can say that the hyperbolic numbers along the lines $y=\pm\, x$ in the hyperbolic plane do not have inverses. These two lines are very important in special relativity and are known as the light cone. I will explore this further later in the post.

We have now built up hyperbolic numbers, their modulus(-square) and their inverses. It is then reasonable to ask if they have a polar-like representation like complex numbers (each complex number $z$ except $0$ can be written in the form $r\exp(i\theta)$ where $r$ is the modulus and $\theta\in [0,2\pi)$). We already have the modulus for hyperbolic numbers so we need to explore what, if anything, $\exp(j\theta)$ is. I will make the assumption that hyperbolic series converge nicely and the following proof is not meant to be rigorous but merely instructive. In the "proof" we will need the fact that $j^n = 1$ if $n$ is even and $j^n = j$ if $n$ is odd.

$$\exp(j\theta) = \sum\limits_{n=0}^{\infty} \frac{(j\theta)^n}{n!} = \sum\limits_{n=0}^{\infty} \frac{j^n\theta^n}{n!}.$$

At this point it would be prudent to separate our above series into two separate series corresponding to the case of $j^{\text{even}}$ and $j^{\text{odd}}$ giving

$$\exp(j\theta) = \sum\limits_{n=0}^{\infty} \frac{j^{2n}\theta^{2n}}{(2n)!} + \sum\limits_{n=0}^{\infty}\frac{j^{2n+1}\theta^{2n+1}}{(2n+1)!} = \sum\limits_{n=0}^{\infty} \frac{\theta^{2n}}{(2n)!}+j\sum\limits_{n=0}^{\infty} \frac{\theta^{2n+1}}{(2n+1)!}. $$

For those familiar with their hyperbolic trigonometric functions, the above can easily be recognized as

$$ \exp(j\theta) = \cosh(\theta) + j\sinh(\theta). $$

We would then like to represent any $z\in \Bbb H$ (with exception maybe to the hyperbolic numbers on the light cone since they have modulus $0$) in the form $r\exp(j\theta)$. Note: $\theta$ no longer represents an angle but is more like a parameter along the hyperbola and now varies from $-\infty$ to $\infty$, and since $r$ is a radius, it must be strictly positive. In the complex number case, $r$ was related to the geometry of the complex numbers, i.e. the level curves of $zz^*$ (circles).

We wish the same to hold for the hyperbolic case, giving that $r$

*is given by the level curves of $zz^*$ (hyperbolas)**.*If $z=x+jy$, then define $r$ to be $\sqrt{|x^2-y^2|}$ so that it is always positive. Knowing $r$ is not enough to tell us where our point $z$ lies in the hyperbolic plane since it could like on any one of four hyperbolas and could be anywhere on each of them so we must figure out which it lies on and where it is on the one it lies on. We have two cases (with two subcases each) to consider: $|x| > |y|$ (with subcases $x>y$ and $x<y$) and $|y| > |x|$ (with subcases $y>x$ and $y < x$). In each case, $\theta$ can be determined from the following: $x+jy = r(\cosh(\theta)+j\sinh(\theta))$, or $x=r\cosh(\theta)$ and $y = r\sinh(\theta)$. Therefore $\tanh(\theta) = \dfrac{y}{x}$ or $\theta = \tanh^{-1}\left(\frac{y}{x}\right)$.Here is a plot for hyperbolas (level curves of zz* in the hyperbolic plane.Notice that the light cone is shown in red. Courtesy of Wikipedia. |

Case 1: $|x| > |y|$ and $x > y$. This implies that $x$ is positive and $y$ is between the lines $y = \pm\,x$ and so $z$ lies on the right-opening hyperbola seen in the image above

*.**Case 2: $|x| > |y|$ and $x < y$*

*.*This implies that $x$ is negative and $y$ is between the lines $y = \pm\,x$ and so $z$ lies on the left-opening hyperbola. One can easily see how the other two cases pan out.We now have a complete hyperbolic-polar description of any $z\in\Bbb H$. Since 1+1 dimensional spacetime can be viewed with the hyperbolic numbers, it is natural to ask what calculus looks like on such a space to see if we can uncover any truths about nature.

**Calculus on $\Bbb H$****The study of calculus on $\Bbb H$ will be very similar in methodology to that of the study of calculus on $\Bbb C$. Suppose we have a hyperbolic function $f:\Bbb H\to\Bbb H$ (again, I should restrict myself to sets in $\Bbb H$, but this will do for simplicity's sake), then we write $f(z) = f(x+jy) = f(x,y)$**

*.*$f$ associates each $(x,y)\in \Bbb H$ with a pair $(x',y')\in \Bbb H$ (here I'm using the notion that each $x+jy$ can be viewed as a coordinate pair $(x,y)$ in the plane). We can then write $f(x,y) = u(x,y)+jv(x,y)$, where $u$ associates $(x,y)$ with $x'$ and $v$ associates $(x,y)$ with $y'$.Since we can divide by hyperbolic numbers, we can potentially come up with a notion of differentiability similar to that with complex-differentiable functions (as long as we don't divide by $0$ of course). Let us try to apply the standard definition of a derivative from elementary calculus to $f$ and see what kind of coupling there is between our $u$ and $v$. From the definition of a derivative we know that if

$$ \lim_{\Delta z\rightarrow 0}\frac{f(z+\Delta z)-f(z)}{\Delta z} $$

exists, then $f$ is differentiable at $z$. Let us again look at the limit as we approach $z$ along horizontal and vertical lines. If we approach along a horizontal line, $\Delta z = \Delta x$, and if we approach along a vertical line, $\Delta z = j\Delta y$

*.*We then have two expressions for the derivative:$$ \lim_{\Delta z\rightarrow 0}\frac{f(z+\Delta z) - f(z)}{\Delta z} = \lim_{\Delta x\rightarrow 0}\frac{u(x+\Delta x,y)-u(x,y)+jv(x+\Delta x,y)-jv(x,y)}{\Delta x} $$

along horizontal lines and

$$ \lim_{\Delta z\rightarrow 0} \frac{f(z+\Delta z)-f(z)}{\Delta z} = \lim_{\Delta y\rightarrow 0}\frac{u(x,y+\Delta y)-u(x,y)+jv(x,y+\Delta y)-jv(x,y)}{j\Delta y} $$

along vertical lines. Like in the $\Bbb R^2$ case, we require that if a derivative exists at a point, the value must be the same along every path to the point and so the two expressions above must be equal if $f$ is to be differentiable at $z$. We can recognize the above expressions as the partial derivatives of $u$ and $v$ along the $x$ and $y$ directions. Since the expressions must be equal we have that

$$ \frac{\partial u}{\partial x} + j\frac{\partial v}{\partial x} = j\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} $$

since $j^{-1} = j$. If we equate the "real" and "imaginary" parts we have that

$$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} $$

and

$$ \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}. $$

These two equations are similar to the Cauchy-Riemann equations for complex-differentiable functions. If we took the partial derivative of the first equation with respect to $y$ and the second with respect to $x$ and subtracted them, we would see that $v$ satisfies the following partial differential equation

$$ \frac{\partial^2 v}{\partial x^2} - \frac{\partial^2 v}{\partial y^2} = 0 $$

assuming that

*u*is continuously twice differentiable. Similarly,*u*solves the following equation$$ \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = 0 . $$

Since both $u$ and $v$ solve this equation, so does $f$ since differentiation is a linear operator. This partial differential equation is very important though you may not recognize it at first glance. Implicitly we have been treating $x$ and $y$ on the same footing and if we wish to relate them back to physical quantities, we must recognize that if $x$ is a coordinate variable (have units of length), then $y$ must also be a coordinate variable (have units of length). If we make the change of variables $x = ct$, then we have that $f$

*solves the following partial differential equation*$$ \frac{1}{c^2}\frac{\partial^2 f}{\partial t^2} - \frac{\partial^2 f}{\partial y^2} = 0 $$

which is the wave equation in 1+1 dimensions. So we see that differentiable functions on $\Bbb H$ must solve the wave equation. The wave equation is fundamental to special relativity because it governs the behavior of light as was recognized by James Clerk Maxwell.

**Connections to Einstein's Theory of Special Relativity**Thus far we have seen some connections to special relativity: hyperbolic numbers of the form $x\pm jx$ denote the light cone of special relativity and differentiable functions on $\Bbb H$ solve the wave equation (and are thusly related to light in some fashion). Can we uncover any more physics by working with the hyperbolic numbers? It turns out we can!

Let us define a "rotation" in $\Bbb H$ of a hyperbolic number $z$ by "angle" $\varphi = \tanh^{-1}\left(\dfrac{v}{c}\right)$ to be given by $z\exp(j\theta)$ (we have implicitly assumed that $\left|\dfrac{v}{c}\right| < 1$ since that is the domain of $\tanh^{-1}$ - this assumption will be important physically)

*.*What this does physically to $z$ is moves $z$ along the hyperbola it lies on by an "angle" $\varphi$. If we write $z = ct+jx$, can we represent $z' = z\exp(j\varphi)$ in terms of $v$, $t$, and $x$?If we write $\exp(j\varphi)$ as $\cosh(\varphi)+j\sinh(\varphi)$, we have that $\cosh(\varphi) = \dfrac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}\,\,$ and $\,\,\sinh(\varphi) = \dfrac{\frac{v}{c}}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$. For simplicity, I will define $\gamma$ to be $\dfrac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$. Therefore

$$z' = ct' + jx' = (ct+jx)\left(\gamma + j\gamma\dfrac{v}{c}\right) = \gamma\left(ct+x\dfrac{v}{c}\right)+\gamma\left(x+\frac{t}{c}\right)$$

and so $t' = \gamma\left(t+\frac{xv}{c^2}\right)$ and $x' = \gamma\left(t+\frac{xv}{c^2}\right)$. These are just the standard Lorentz transformations between two inertial reference frames and so it shows that Lorentz transformations are just rotations in the hyperbolic plane.

The interpretation of this is that if we compare our frame of reference to a frame of reference that is moving with uniform velocity relative to us (with $v < c$ since $\tanh^{-1}$ cannot be defined in the usual sense if $v$ is greater than or equal to $c$), then the way to relate our coordinates is just the Lorentz transformations

*and*that we lie on the same hyperbola in the hyperbolic plane (since one is just a rotation of the other).There is more that could be said about the connections between the hyperbolic numbers and special relativity but I will stop here. The above notion can be generalized to $3+1$ dimensions ($3$ spatial dimensions, $1$ time dimension) by including two more "imaginary" units like $j$ and these would be the place-keepers for the spatial components. In $3+1$ dimensions, the light cone makes up a cone embedded in $4$ dimensional space and the level curves of $zz^*$ are now hyperboloids but the ideas carry over somewhat naturally.

thanks cameron :)

ReplyDeleteNice article. I have a small question: how can we divide by hyperbolic numbers? I mean, yes, if we are not along y=x or y=-x, but surely the limit defining the derivative includes the troublesome zero divisors... so.. what is meant by "limit" here?

ReplyDeleteThanks for the comment and praise! This is a valid concern however, remember that your numerator is also going to zero by continuity. As long as you assure that you never approach your point along shifted light cones emanating from that point, the quotient is well defined since the denominator will never be zero.

Delete