Orthogonality of timelike and null vector

[nomather1471]

Can we show orthogonality of timelike and null vector?

 

[George Jones]

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

 

[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)&lt; (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.

 

[George Jones]

Finally, let $\{e_i \}$ be an orthonormal basis

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

 

[Meir Achuz]

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.

 

[nomather1471]

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)&lt; (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.

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)

 

[WannabeNewton]

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)

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.

 

[nomather1471]

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