# Orthogonality of timelike and null vector

1. Oct 28, 2013

### nomather1471

Can we show orthogonality of timelike and null vector?

Last edited: Oct 28, 2013
2. Oct 28, 2013

### George Jones

Staff Emeritus
If X is timelike, and Y is orthogonal to X, then Y is spacelike.

3. Oct 28, 2013

### WannabeNewton

I just want to expand upon what George said. Let $(M,g)$ be a space-time and $p\in M$. Furthermore, let $X$ be a time-like vector in $T_p M$ and $Y$ a non-zero vector in $T_p M$ such that $g_p(X,Y) = 0$. Finally, let $\{e_i \}$ be an orthonormal basis for $T_p M$ so that $X = X^i e_i$ and $Y = Y^i e_i$.

Then $(X^2)^2 + (X^3)^2 + (X^4)^2 < (X^1)^2$ and $X^1 Y^1 = X^2 Y^2 + X^3Y^3 + X^4Y^4$.
Note this immediately implies that $(X^1)^2 > 0$ and that $(Y^2)^2 + (Y^3)^2 + (Y^4)^2 > 0$.

Therefore $$(X^1 Y^1)^2 \leq ((X^2)^2 + (X^3)^2 + (X^4)^2)((Y^2)^2 + (Y^3)^2 + (Y^4)^2)< (X^1)^2((Y^2)^2 + (Y^3)^2 + (Y^4)^2)$$ thus $(Y^1)^2 < (Y^2)^2 + (Y^3)^2 + (Y^4)^2$ i.e. $Y$ is space-like.

4. Oct 28, 2013

### George Jones

Staff Emeritus
WLOG, choose $X = \{e_0 \}$.

5. Oct 28, 2013

### clem

A time-like vector can be Lorentz transformed to be purely time-like:$x^\mu=(t,0)$.
A null vector always has x=t, so they can't be orthogonal.

Last edited: Oct 28, 2013
6. Oct 28, 2013

### nomather1471

I think you have to show that:
$$(X^1 Y^1)^2 \leq ((X^2)^2 + (X^3)^2 + (X^4)^2)((Y^2)^2 + (Y^3)^2 + (Y^4)^2)$$

7. Oct 28, 2013

### WannabeNewton

I had forgotten completely that one has the freedom to Lorentz boost to a frame wherein $X = e_0$ as stated by George and Clem above, which makes the calculation much simpler. Nevertheless, the above inequality follows directly from the Cauchy-Schwarz inequality.

8. Oct 28, 2013

### nomather1471

I am student at geometry so your notation is quite different for me but thanks for everyone for all replies...