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Okay, Schwarz inequality states that

[tex]\left| U^{k}V^{k}\right| \leq \sqrt{(U^{i})^{2}(V^{j})^{2}}. \ \ i,j,k =1,2,3 \ \ \ (1)[/tex]

And, the equality holds

*if and only if*[itex]U^{i} \propto V^{i}[/itex].

Let us see how we apply this to 4-vectors in Minkowski spacetime [itex]M^{4}(+1, -3)[/itex]. With this signature:

i) a vector [itex]\mathbf{U} \in M^{4}[/itex] for which

[tex]\mathbf{U} \cdot \mathbf{U} \equiv \eta_{\mu\nu}U^{\mu}U^{\nu} > 0 ,[/tex]

is called a

*timelike*vector;

ii) a vector [itex]\mathbf{U} \in M^{4}[/itex] for which

[tex]\mathbf{U} \cdot \mathbf{U} \equiv \eta_{\mu\nu}U^{\mu}U^{\nu} < 0 ,[/tex]

is called a

*spacelike*vector; and

iii) a vector [itex]\mathbf{U} \in M^{4}[/itex] that satisfies

[tex]\mathbf{U} \cdot \mathbf{U} \equiv \eta_{\mu\nu}U^{\mu}U^{\nu} = 0 ,[/tex]

is called a

*null*(or

*lightlike*) vector.

Now, suppose that [itex]\mathbf{U}[/itex] and [itex]\mathbf{V}[/itex] are two timelike vectors. Then, from the definition (i), we have

[tex]

\begin{align*}

\left(U^{i}\right)^{2} &< \left(U^{0}\right)^{2} \\

\left(V^{j}\right)^{2} &< \left(V^{0}\right)^{2} .

\end{align*}

[/tex]

These inequalities can be combined to give

[tex]\sqrt{\left(U^{i}\right)^{2}\left(V^{j}\right)^{2}} < \left| U^{0} V^{0} \right| . \ \ \ \ \ \ \ (2) [/tex]

Now, from the Schwarz inequality (1), we conclude that

[tex]

\left| U^{k}V^{k} \right| < \left| U^{0}V^{0}\right| , \ \ \ \ \ \ \ \ \ \ \ (3)

[/tex]

holds for any two timelike vectors. Thus, we have proved the following theorem

Theorem (

**A**):

*No two timelike vectors can be orthogonal*(in the

*Minkowski sense*).

For,

[tex]\mathbf{U}\cdot \mathbf{V} = 0 \ \ \Rightarrow \ \ \left| U^{k}V^{k}\right| = \left| U^{0}V^{0}\right| ,[/tex] which contradicts (3).

__.__

**qed**Theorem (

**B**):

*If*[itex]\mathbf{U},\mathbf{V}[/itex]

*are two timelike vectors such that*[itex]U^{0} > 0, \ \ V^{0} > 0[/itex],

*then*[itex] \mathbf{U}\cdot \mathbf{V} > 0[/itex].

Proof:

[tex]U^{k}V^{k} \leq \left|U^{k}V^{k}\right| \leq \sqrt{(U^{i})^{2}(V^{j})^{2}} .[/tex]

Using (2) or (3), and the fact that [itex]U^{0}>0[/itex] and [itex]V^{0}>0[/itex], we obtain

[tex]U^{k}V^{k} < \left|U^{0}V^{0}\right| = U^{0}V^{0} .[/tex]

Thus

[tex]U^{0}V^{0} - U^{k}V^{k} = \mathbf{U}\cdot \mathbf{V} > 0 .[/tex]

**qed**Theorem (

**C**):

*A timelike vector cannot be orthogonal to a non-zero null vector*.

Proof:

Let [itex]\mathbf{T}[/itex] be a timelike vector, and [itex]\mathbf{N}[/itex] is a non-zero null vector. This means that

[tex]\begin{align*}

\left(T^{i}\right)^{2} &< \left(T^{0}\right)^{2} \\

\left(N^{j}\right)^{2} &= \left(N^{0}\right)^{2} .

\end{align*}

[/tex]

Thus

[tex]

\left(T^{i}\right)^{2} \left(N^{j}\right)^{2} < \left( T^{0}N^{0} \right)^{2} .

[/tex]

Using the Schwarz identity (1), we get

[tex]\left( T^{k}N^{k} \right)^{2} < \left( T^{0}N^{0} \right)^{2} . \ \ \ \ \ \ (4)[/tex]

Now, suppose [itex]\mathbf{T}\cdot \mathbf{N} = 0[/itex]. This implies [itex]\left( T^{k}N^{k} \right)^{2} = \left( T^{0}N^{0} \right)^{2}[/itex] which contradicts (4).

__.__

**qed**Theorem (

**D**):

*If*[itex]\mathbf{T}[/itex]

*is a timelike vector and*[itex]\mathbf{T} \cdot \mathbf{S} = 0[/itex],

*then*[itex]\mathbf{S}[/itex]

*is spacelike vector*.

Proof:

From theorem (A), [itex]\mathbf{S}[/itex] cannot be a timelike vector. And from theorem (C), [itex]\mathbf{S}[/itex] cannot be a non-zero null vector. Thus, [itex]\mathbf{S}[/itex] must be a spacelike vector.

__.__

**qed**Theorem (

**E**):

*Two non-zero null vectors are orthogonal, if and only if they are proportional*.

This theorem shows the bizarre and non-intuitive nature of

*null*vectors and “

*orthogonality*” in Minkowski spacetime.

Proof:

The “

*if*” part is trivially obvious: Assume that the null vector [itex]\mathbf{P}[/itex] is proportional to the null vector [itex]\mathbf{N}[/itex]. Then, [itex]\mathbf{P} = \lambda \mathbf{N}[/itex] for some [itex]\lambda \in \mathbb{R}[/itex]. Thus [itex]\mathbf{P} \cdot \mathbf{N} = \lambda \left(\mathbf{N} \cdot \mathbf{N} \right) = 0 .[/itex]

The “

*only if*” part:

[tex]\mathbf{P} \cdot \mathbf{P} = 0 \ \Rightarrow \left(P^{0}\right)^{2} = \left(P^{i}\right)^{2},[/tex]

[tex]\mathbf{N} \cdot \mathbf{N} = 0 \ \Rightarrow \left(N^{0}\right)^{2} = \left(N^{j}\right)^{2},[/tex]

and

[tex]\mathbf{P} \cdot \mathbf{N} = 0 \ \Rightarrow \ P^{0}N^{0} = P^{k} N^{k}.[/tex]

From these relations, we obtain

[tex]\left(P^{k}N^{k}\right)^{2} = \left(P^{0}N^{0}\right)^{2} = \left(P^{i}\right)^{2}\left(N^{j}\right)^{2} .[/tex]

Thus

[tex]\left| P^{k}N^{k} \right| = \sqrt{\left(P^{i}\right)^{2}\left(N^{j}\right)^{2}} .[/tex]

Since this equation is the equality case of Schwarz inequality, we conclude that [itex]P^{k} = \lambda N^{k}[/itex] for some scalar [itex]\lambda \neq 0[/itex]. Since [itex]N^{0} \neq 0[/itex], we have

[tex]P^{0} = \frac{P^{k}N^{k}}{N^{0}} = \lambda \frac{N^{k}N^{k}}{N^{0}} = \lambda N^{0} .[/tex]

Thus [itex]\mathbf{P} = \left(\lambda N^{0} , \lambda N^{k}\right) = \lambda \mathbf{N}[/itex].

__.__

**qed**Okay, almost all other results follow from the above theorems. So, good luck to you all.