# Clarifications on proving lightlike vectors must be orthogonal with themselves

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?

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).

• could you explain this line please? $(−𝑑𝑡^2+𝑑𝑥^2+𝑑𝑦^2+𝑑𝑧^2)(𝑣)=−[𝑑𝑡(𝑣)]^2+[𝑑𝑥(𝑣)]^2+[𝑑𝑦(𝑣)]^2+[𝑑𝑧(𝑣)]^2$ Jun 14 at 17:14
• @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. Jun 14 at 20:04
• But, having said this, note that writing $(-dt^2+dz^2+dy^2+dz^2)(v)$ is just a different notation for writing $g(v,v)$. Jun 14 at 20:18