I'm trying to prove that lightlike vectors in Minkowski space must be orthogonal to themselves, and I have two questions about this.

I tried two different approaches:

  1. If lightlike, $ds^2=0$
    By definition of $ds$ in Minkowski space, we have $ds^2 = dx^2 + dy^2 + dz^2 - dt^2$ where I let c=1.
    So, $ds^2 = dx^2 + dy^2 + dz^2 = dt^2$

but this means that $dt$ has to be zero, right? so the vectors would be something like
$\begin{pmatrix} 0 \\ dx\\ dy\\ dz \end{pmatrix}$
but this doesn't seem right to me.

  1. say that vectors are orthogonal if their scalar product is zero, meaning $g_{\mu \nu}A^{\mu}A^{\nu} = 0$
    for lightlike in Minkowski,
    $N_{\mu \nu}A^{\mu}A^{\nu} = 0$
    so, $A_{\nu}A^{\nu} = 0$
    but doesn't this just imply $A_{\nu}$ & $A^{\nu}$ must be orthogonal. These don't seem to be the same vector, so I'm not sure it's fair to conclude from this that a lightlike vector is orthogonal to itself.

So my questions are what is wrong with approach 1, and what conclusions can we make from approach 2?


1 Answer 1


We say two vectors $v,w$ are orthogonal with respect to $g$ if $g(v,w)=0$. That's just by definition. In components, this is written as $g_{ab}v^aw^b=0$. If you're working in Minkowski with standard basis vectors, this is equivalent to $-(v^0w^0)+v^1w^1+v^2w^2+v^3w^3=0$.

Now, by definition a (non-zero) vector $v$ is said to be lightlike if $g(v,v)=0$, i.e as you wrote, $g_{ab}v^av^b=0$. Clearly, a lightlike vector $v$ is orthogonal (with respect to $g$) to itself according to these two definitions; there's nothing to be proven.

Now, recall the definition of the symbol $ds^2$, it means for any vector $v$, we define $ds^2(v)=g(v,v)$. So, saying a vector $v$ is lightlike is equivalent to saying $ds^2(v)=0$. Again, in Minkowski, if we write this out, this says \begin{align} 0=ds^2(v)&=(-dt^2+dx^2+dy^2+dz^2)(v)\\ &=-[dt(v)]^2+[dx(v)]^2 + [dy(v)]^2+ [dz(v)]^2\\ &=-(v^0)^2+ (v^1)^2+(v^2)^2+(v^3)^2. \end{align} Which means the vector $v$ has components $(v^0,v^1,v^2,v^3)\in\Bbb{R}^4$ which lie on a certain cone (in the above, writing $dt(v)$ for instance means the $t$-component of the vector $v$, and $dx(v)$ means the $x$-component of the vector $v$ etc).

Where you went wrong in the first step is the final step

... So $ds^2= dx^2+dy^2+dz^2=dt^2$

No. As I've written above, it is $ds^2$ (when applied to $v$) which is $0$, and thus $dx^2+dy^2+dz^2=dt^2$ (everything applied on a given lightlike vector $v$). I'm not sure how you concluded $dt=0$ from here.

For example, in Minkowski, the vector $v=e_0+e_1= (1,1,0,0)$ is lightlike (among infinitely many other possible examples).

  • $\begingroup$ could you explain this line please? $(βˆ’π‘‘π‘‘^2+𝑑π‘₯^2+𝑑𝑦^2+𝑑𝑧^2)(𝑣)=βˆ’[𝑑𝑑(𝑣)]^2+[𝑑π‘₯(𝑣)]^2+[𝑑𝑦(𝑣)]^2+[𝑑𝑧(𝑣)]^2$ $\endgroup$ Jun 14 at 17:14
  • $\begingroup$ @Relativisticcucumber $-dt^2+dx^2+dy^2+dz^2$ is a quadratic form. It eats a vector $v$ and outputs a number. $(-dt^2+dx^2+dy^2+dz^2)(v)$ means the value of the quadratic form on the vector. This equals $-(dt^2)(v)+ (dx^2)(v)+(dy^2)(v)+(dz^2)(v)$. And the meaning of these symbols is for example $(dx)^2(v):= (dx\otimes dx)(v,v)=(dx(v))\cdot (dx(v))= [dx(v)]^2$. So really it's a matter of unwinding definitions. $\endgroup$
    – peek-a-boo
    Jun 14 at 20:04
  • $\begingroup$ But, having said this, note that writing $(-dt^2+dz^2+dy^2+dz^2)(v)$ is just a different notation for writing $g(v,v)$. $\endgroup$
    – peek-a-boo
    Jun 14 at 20:18

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