# Clear up confusion on Minkowski signature

All given metrics are for orthonormal-basis.

**2 dimensional spacetime :**

I saw that Minkowski Metric looks like this : $$\pmatrix{-1 & 0 \\ 0 & 1}$$ or $$\pmatrix{1 & 0 \\ 0 & -1}$$

I was wondering why it's not written like identity metric. To define a vector in spacetime it's written like this $dS^2 = -(ct)^2+(dx)^2=(ict)^2+(dx)^2$. I know it's same as Minwkowski metric. But in Euclidean metric I was just using identity metric. Is there really any derivation of Minkowski metric? Or he just wrote it curiously. What the negative explains? I know that $ct$ is coordinate. Particle physicist write vector in spacetime (I had read it 3-4 hours ago, I can't find the source again so can't add reference here) like this $ dS^2 = (ct)^2-(dx)^2$.

**4 dimensional spacetime :**

The minkowski signature is $\pmatrix{- & + & + & +}$ or negative for space and positive for time. I read in the question that, the tensor for spacetime can be written like this : $ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}$ while $g_{\mu \nu}$ is Minkowski signature and $dx^{\mu}$ and $dx^{\nu}$ represent space coordinate. But in PSE question he wrote that $dx^0=ict$ where $i$ is imaginary number and $dx^0$ is time coordinate. I wonder what's the main reason of taking $ict$ as coordinate. $ct$ looks good to me to be a coordinate. At first I thought, the tensor got a negative sign for time for Minkowski signature. But when I saw $ict$ as coordinate my mind changed again.

I know the question is too confusing and I am confused about time coordinate, it is $ct$ or $ict$, I don't know. Why Minkowski wrote signature like that?

$g_{xx}=\vec e_x \cdot \vec e_x=1$ That's what we exactly know. But it's different for Minkowski spacetime, when dot product of time basis vector is 1 then dot product of space basis vector is -1 and vice versa, why?

I forgot to mention which actually made me curious to write the question, we know that dot product of any basis vector to itself is $1$ (they always not 1) $\vec e \cdot \vec e = |e| \ |e| cos\theta$ $\theta=0$ hence $\vec e \cdot \vec e = |e| \ |e|$. To write $\vec e \cdot \vec e = -|e|^2$, we must make them perpendicular but which is never true. Angle between dot product of vector of itself is $0$ but in Minkowski spacetime diagram, it is something different.

## 1 answer

As you probably know, one of the postulates Einstein's Special Relativity is based on is that all observers see light in vacuum go at the same speed $c$.

Now consider a lamp at rest relative to John being switched on. The light is going through vacuum from then on, until after some time it arrives at some object, also at rest relative to John.

Now John observes this, and measures that the time between the lamp being switched on, and the light reaching the object, is $\Delta t$. Obviously the distance the light has travelled in that time is $\Delta s = c \Delta t$.

Now John has a coordinate system, and in this coordinate system, since light in vacuum goes in a straight line, the distance $\Delta s$ is given by the Euclidean formula $$\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2.$$ Inserting the previous equation we therefore get $c^2 \Delta t^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$, or if we bring everything on one side, $$c^2 \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 = 0.$$

Clara, who moves with considerable speed relative to John, will observe a time difference $\Delta t'$ and a travelled distance $\Delta s'$, which are different from John's values. But since the speed of light is the same for John and Clara, we must of course also have the relation $\Delta s' = c \Delta t'$. Thus with the same argumentation as above, we have $$c^2 \Delta t'^2 - \Delta x'^2 - \Delta y'^2 - \Delta z'^2 = 0.$$

Thus we see that the term $c^2 \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$ arises naturally from the constancy of the speed of light. Indeed, one can check that Lorentz transformations don't change its value even if it is not zero. This invariant quantity is called spacetime interval.

Now for the calculation above obviously it isn't relevant whether we multiply the full term with $-1$. However the *relative* sign of the $c^2\Delta t^2$ term is fixed by the calculation.

Now conversely one can ask what exactly are the linear transformations that leave the spacetime interval invariant, and one finds that these are exactly the Lorentz transformations. Thus the spacetime interval, with that relative minus sign between space and time directions, captures the essence of special relativity.

Now if you look at the spacetime interval, you see that it *almost* looks like an Euclidean metric, except for this relative sign. Now if we choose the global sign of the metric so that spacelike distances (those with $\Delta s > c \Delta t$) are positive, we see that we can make the metric *formally* look like an Euclidean metric by simply adding an imaginary unit to the time coordinate, as $(ic\Delta t)^2 = -(c\Delta t)^2$. Therefore some authors like to add that imaginary unit. But note that this is just a mathematical trick to make the whole thing look more familiar. I don't think you can draw any deep insights from it. Indeed, I think it actually distracts from the fundamental difference between Euclidean metric and Minkowski metric.

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