# Four vectors

1. Mar 26, 2017

### koustav

are spacelike and timelike orthogonal?what is the mathematical proof

2. Mar 26, 2017

### Orodruin

Staff Emeritus
Not in general, no.

3. Mar 26, 2017

### SiennaTheGr8

Not exactly, no, but any four-vector that's orthogonal to a spacelike vector must be timelike, and vice versa.

Spacelike just means that the magnitude of the three-vector spatial component is greater than the time component. Timelike means the opposite. And two vectors are orthogonal if their dot product is zero. So let's say we have the following two four-vectors:

$\mathbf{A} = (A_t, \mathbf{a}) \qquad \mathbf{B} = (B_t, \mathbf{b})$.

If they are orthogonal, then their dot product is zero:

$\mathbf{A} \cdot \mathbf{B} = (A_t)(B_t) - (a)(b) = 0$,

which means that:

$\dfrac{A_t}{a} = \dfrac{b}{B_t}$.

Assuming that $\mathbf{A}$ and $\mathbf{B}$ aren't lightlike (magnitude of zero), then this result can only occur if one of them is spacelike and the other is timelike. In other words, it must be the case that if $A_t > a$, then $b > B_t$ (and vice versa).

4. Mar 26, 2017

### Orodruin

Staff Emeritus
This is incorrect. Let $V = (0,1,0,0)$ and $W = (0,0,1,0)$. Then $V\cdot W = 0$ and both $V$ and $W$ are space-like.

5. Mar 26, 2017

### SiennaTheGr8

Retracted!

I need to rethink the conditions/exceptions for that "rule."

6. Mar 26, 2017

### Orodruin

Staff Emeritus
Any vector orthogonal to a time-like vector is space-like. Any vector orthogonal to a light-like vector is either proportional to the light-like vector itself or space-like. A vector orthogonal to a space-like vector may be time-like, space-like, or light-like.

7. Mar 26, 2017

### SiennaTheGr8

Thanks for setting me straight on that. I hadn't thought through it rigorously enough.

8. Mar 26, 2017

### koustav

can you help me with the mathematical proof

9. Mar 26, 2017

### SiennaTheGr8

Aside from the error I made that @Orodruin pointed out, I'd add that I assumed here that $a \neq 0$ and $B_t \neq 0$, which certainly may not be the case.

10. Mar 26, 2017

### Orodruin

Staff Emeritus
In the case of a time-like vector, go to a reference frame where it is proportional to (1,0,0,0). Do the corresponding thing for space-like and null vectors.

11. Mar 26, 2017

### DrGreg

This should be
$$\mathbf{A} \cdot \mathbf{B} = A_tB_t - \mathbf{a} \cdot \mathbf{b} = A_tB_t - a_x b_x - a_y b_y- a_z b_z=0$$

12. Mar 26, 2017

### SiennaTheGr8

Yes, dot product.

I got a lot wrong there!

13. Mar 26, 2017

### vanhees71

I'm using the mainly-minus convention for the Minkowski product ("west-coast convention" a la Bjorken+Drell). Then you have for the components $x^{\mu}$ and $y^{\mu}$ with respect to a Minkowski-orthonormal basis,
$$x \cdot y=\eta_{\mu \nu} x^{\mu} y^{\nu}=x^0 y^0-\vec{x} \cdot \vec{y},$$
where $\vec{x} \cdot \vec{y}$ is the usual Euclidean product built with the three spatial components.

A four-vector $x$ is now called timelike if $x \cdot x>0$, lightlike if $x \cdot x=0$, and spaclike if $x \cdot x<0$.

You can prove that any non-zero vector which is Minkowski-orthogonal to a timelike vector is spacelike (the opposite is of course wrong as shows above). So let $x$ be space like and $y$ and arbitrary vector such that $x \cdot y=0$. We have to show that from $x \cdot x>0$ one necessarily then has $y \cdot y<0$. The trick is to write the product out and use that the usual scalar product of the spaclike components is positive definite. From that you get
$$|\vec{x} \cdot \vec{y}| \leq |\vec{x}| |\vec{y}| \; \Rightarrow \; (\vec{x} \cdot \vec{y})^2 \geq \vec{x}^2 \vec{y}^2,$$
which is known as the Cauchy-Schwarz inequality. Now we have
$$x \cdot y=x^0 y^0 -\vec{x} \cdot \vec{y}=0 \; \Rightarrow \; x^0 y^0=\vec{x} \cdot \vec{y} \; \Rightarrow y^0=\frac{\vec{x} \cdot \vec{y}}{x^0} \qquad (*)$$
and
$$x \cdot x=(x^0)^2-\vec{x}^2>0 \; \Rightarrow \; |x^0|>|\vec{x}|.$$
This implies that $|x^0|>0$ and thus we can indeed use (*):
$$y \cdot y=(y^0)^2-\vec{y}^2=\frac{(\vec{x} \cdot \vec{y})^2}{(x^0)^2}-\vec{y}^2=\frac{(\vec{x} \cdot \vec{y})^2-(x^0)^2 \vec{y}^2}{(x^0)^2}.$$
Now use the Cauchy-Schwarz inequality above to get
$$y \cdot y \leq \frac{(|\vec{x}|^2 -(x^0)^2) \vec{y}^2}{(x^0)^2}.$$
If $\vec{y} \neq 0$, then since $x$ is assumed to be timelike we get $y \cdot y<0$, i.e., $y$ is indeed spacelike.

Now assume $\vec{y}=0$, but then because of $x \cdot y=0$ we have $x^0 y^0=\vec{x} \cdot \vec{y}=0$. Since $x$ was assumed to be timelike, this implies that then also $y^0=0$, i.e., then $y=0$, but we assumed $y$ not to be the null vector. So $y$ must be timelike. QED.