# Schwarz Inequality is your friend

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1. Dec 29, 2015

### samalkhaiat

I would like to show you how to use Schwarz inequality to prove some important general theorems and solve problems about vectors in Minkowski spacetime.
Okay, Schwarz inequality states that
$$\left| U^{k}V^{k}\right| \leq \sqrt{(U^{i})^{2}(V^{j})^{2}}. \ \ i,j,k =1,2,3 \ \ \ (1)$$
And, the equality holds if and only if $U^{i} \propto V^{i}$.
Let us see how we apply this to 4-vectors in Minkowski spacetime $M^{4}(+1, -3)$. With this signature:
i) a vector $\mathbf{U} \in M^{4}$ for which
$$\mathbf{U} \cdot \mathbf{U} \equiv \eta_{\mu\nu}U^{\mu}U^{\nu} > 0 ,$$
is called a timelike vector;
ii) a vector $\mathbf{U} \in M^{4}$ for which
$$\mathbf{U} \cdot \mathbf{U} \equiv \eta_{\mu\nu}U^{\mu}U^{\nu} < 0 ,$$
is called a spacelike vector; and
iii) a vector $\mathbf{U} \in M^{4}$ that satisfies
$$\mathbf{U} \cdot \mathbf{U} \equiv \eta_{\mu\nu}U^{\mu}U^{\nu} = 0 ,$$
is called a null (or lightlike) vector.
Now, suppose that $\mathbf{U}$ and $\mathbf{V}$ are two timelike vectors. Then, from the definition (i), we have
\begin{align*} \left(U^{i}\right)^{2} &< \left(U^{0}\right)^{2} \\ \left(V^{j}\right)^{2} &< \left(V^{0}\right)^{2} . \end{align*}
These inequalities can be combined to give
$$\sqrt{\left(U^{i}\right)^{2}\left(V^{j}\right)^{2}} < \left| U^{0} V^{0} \right| . \ \ \ \ \ \ \ (2)$$
Now, from the Schwarz inequality (1), we conclude that
$$\left| U^{k}V^{k} \right| < \left| U^{0}V^{0}\right| , \ \ \ \ \ \ \ \ \ \ \ (3)$$
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,
$$\mathbf{U}\cdot \mathbf{V} = 0 \ \ \Rightarrow \ \ \left| U^{k}V^{k}\right| = \left| U^{0}V^{0}\right| ,$$ which contradicts (3). qed.

Theorem (B): If $\mathbf{U},\mathbf{V}$ are two timelike vectors such that $U^{0} > 0, \ \ V^{0} > 0$, then $\mathbf{U}\cdot \mathbf{V} > 0$.
Proof:
$$U^{k}V^{k} \leq \left|U^{k}V^{k}\right| \leq \sqrt{(U^{i})^{2}(V^{j})^{2}} .$$
Using (2) or (3), and the fact that $U^{0}>0$ and $V^{0}>0$, we obtain
$$U^{k}V^{k} < \left|U^{0}V^{0}\right| = U^{0}V^{0} .$$
Thus
$$U^{0}V^{0} - U^{k}V^{k} = \mathbf{U}\cdot \mathbf{V} > 0 .$$ qed

Theorem (C): A timelike vector cannot be orthogonal to a non-zero null vector.
Proof:
Let $\mathbf{T}$ be a timelike vector, and $\mathbf{N}$ is a non-zero null vector. This means that
\begin{align*} \left(T^{i}\right)^{2} &< \left(T^{0}\right)^{2} \\ \left(N^{j}\right)^{2} &= \left(N^{0}\right)^{2} . \end{align*}
Thus
$$\left(T^{i}\right)^{2} \left(N^{j}\right)^{2} < \left( T^{0}N^{0} \right)^{2} .$$
Using the Schwarz identity (1), we get
$$\left( T^{k}N^{k} \right)^{2} < \left( T^{0}N^{0} \right)^{2} . \ \ \ \ \ \ (4)$$
Now, suppose $\mathbf{T}\cdot \mathbf{N} = 0$. This implies $\left( T^{k}N^{k} \right)^{2} = \left( T^{0}N^{0} \right)^{2}$ which contradicts (4). qed.

Theorem (D): If $\mathbf{T}$ is a timelike vector and $\mathbf{T} \cdot \mathbf{S} = 0$, then $\mathbf{S}$ is spacelike vector.
Proof:
From theorem (A), $\mathbf{S}$ cannot be a timelike vector. And from theorem (C), $\mathbf{S}$ cannot be a non-zero null vector. Thus, $\mathbf{S}$ 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 $\mathbf{P}$ is proportional to the null vector $\mathbf{N}$. Then, $\mathbf{P} = \lambda \mathbf{N}$ for some $\lambda \in \mathbb{R}$. Thus $\mathbf{P} \cdot \mathbf{N} = \lambda \left(\mathbf{N} \cdot \mathbf{N} \right) = 0 .$
The “only if” part:
$$\mathbf{P} \cdot \mathbf{P} = 0 \ \Rightarrow \left(P^{0}\right)^{2} = \left(P^{i}\right)^{2},$$
$$\mathbf{N} \cdot \mathbf{N} = 0 \ \Rightarrow \left(N^{0}\right)^{2} = \left(N^{j}\right)^{2},$$
and
$$\mathbf{P} \cdot \mathbf{N} = 0 \ \Rightarrow \ P^{0}N^{0} = P^{k} N^{k}.$$
From these relations, we obtain
$$\left(P^{k}N^{k}\right)^{2} = \left(P^{0}N^{0}\right)^{2} = \left(P^{i}\right)^{2}\left(N^{j}\right)^{2} .$$
Thus
$$\left| P^{k}N^{k} \right| = \sqrt{\left(P^{i}\right)^{2}\left(N^{j}\right)^{2}} .$$
Since this equation is the equality case of Schwarz inequality, we conclude that $P^{k} = \lambda N^{k}$ for some scalar $\lambda \neq 0$. Since $N^{0} \neq 0$, we have
$$P^{0} = \frac{P^{k}N^{k}}{N^{0}} = \lambda \frac{N^{k}N^{k}}{N^{0}} = \lambda N^{0} .$$
Thus $\mathbf{P} = \left(\lambda N^{0} , \lambda N^{k}\right) = \lambda \mathbf{N}$. qed.
Okay, almost all other results follow from the above theorems. So, good luck to you all.

2. Jan 1, 2016

### andrewkirk

What notation are you using? The C-S inequality has two sums over indices on the right-hand side and one sum on the left-hand side. But the statement above has no explicit sums, and no implicit sums under any notation system I have yet learned. If one relaxes the rule in the Einstein summation convention that identical indices are summed over only when one is up and one down, then there will be a sum on the left-hand side, but I have not yet encountered a notation convention or summation rule that will imply the necessary sums on the right-hand side. Could you please explain what convention you are using and how it implies the three sums? Thank you.

3. Jan 1, 2016

### strangerep

Heh, physicists often play loose with the Einstein summation convention in that way. I.e., repeated indices are summed, even if both are up or both are down. Afaict, Sam's formulas are correct under this convention.

BTW, it's notable that Sam's post got 8 "likes" very quickly. I wonder if (or how many) other posts in the technical forums have done so well.

4. Jan 2, 2016

### samalkhaiat

The vectors in Schwarz inequality are Euclidean 3-vectors. So, there is no difference between upper and lower indices. So, in Euclidean space, the Einstein summation convention is simply “repeated indices are summed over”:
$$U^{k}V^{k} \equiv \sum_{k =1}^{3} U^{k}V^{k} ,$$
$$(U^{i})^{2} = U^{i}U^{i} \equiv \sum_{i=1}^{3} U^{i}U^{i} ,$$
and
$$(V^{j})^{2} = V^{j}V^{j} \equiv \sum_{j =1}^{3} V^{j}V^{j} .$$
For, 4-vectors $\mathbf{V} = (V^{0}, V^{j})$ some care is needed:
\begin{align*} \mathbf{U} \cdot \mathbf{V} &= \eta_{\mu \nu} U^{\mu}V^{\nu} \\ &= U^{0}V^{0} - \sum_{j = 1}^{3} U^{j}V^{j} . \end{align*}
So, I used the above mentioned summation convention, i.e., dropped the summation sign, and wrote
$$\mathbf{U} \cdot \mathbf{V} = U^{0}V^{0} - U^{i}V^{i} ,$$
which is correct because if you use $U^{0} = U_{0} , \ U^{j} = - U_{j}$ in the above, you get
\begin{align*} \mathbf{U} \cdot \mathbf{V} &= U_{0}V^{0} + U_{i}V^{i} \\ &= U_{\mu}V^{\mu} \\ &= \eta_{\mu\nu}U^{\mu}V^{\nu} . \end{align*}
I hope it is clear enough for you now. Let me know if it is not.

Last edited: Jan 2, 2016
5. Jan 2, 2016

### samalkhaiat

Is that means "wow"?

6. Jan 2, 2016

### strangerep

But you're using the "west coast" (hep) metric convention (+,-,-,-), hence $U^j = -U_j$, as you noted later in your post #4. (I didn't mention this before because it doesn't affect your original post -- for which I see the "like" count is now up to 10.)

This is one of the reasons why, some years ago, I became a convert to the "east coast" (relativist) convention (-,+,+,+).
For other readers: these conventions are somewhat tangential to the main topic, but you can read more about them in Peter Woit's blog article.

7. Jan 3, 2016

8. Jan 3, 2016

### samalkhaiat

Well, the inequality I wrote deals with Euclidean vectors, so it is meaningless to bother about the Minkowski metric signature. However, one needs to be careful when one applies it to the spatial components of 4-vectors, as I mentioned.
As for the mostly-plus or mostly-minus signatures, personally I don’t mind working with any of them. I believe people prefer one on the other because they are lazy. Put it this way, had Weinberg written his GR text in the mostly-minus metric, you would see all relativists using it now. And the same (I believe) applies to the Bjorken & Drell texts: had they used the mostly-plus signature, we would see particle physicists using it now.

9. Jan 4, 2016

### pyroartist

May the Schwarz be with you.

(Sorry, couldn't resist.)

10. Jan 4, 2016

### George Jones

Staff Emeritus
How do you prove an axiom/definition?

11. Jan 4, 2016

### samalkhaiat

Well, if you call that an axiom, then I obviously did. My only axiom was the Schwarz inequality and the only definition used in the proof is that of timelike vectors.

12. Jan 5, 2016

### George Jones

Staff Emeritus
Minkowski spacetime:

Minkowski spacetime $\left( V,\mathbf{g}\right)$ is a 4-dimensional vector space $V$ together with a symmetric, non-degenerate, bilinear mapping $g:V\times V\rightarrow\mathbb{R}$. A vector in $V$ is called a 4-vector, and a 4-vector $v$ is called timelike if $g\left(v,v\right) >0$, lightlike if $g\left(v,v\right) =0$, and spacelike if $g\left(v,v\right) <0$. $\left( V,g\right)$ is such that: 1) timelike vectors exist; 2) $v$ is spacelike whenever $u$ is timelike and $g\left( u,v\right)=0$.

13. Jan 5, 2016

### samalkhaiat

Aren’t these the definitions (i)-(iii) in #1?
The “existence” of these three types of vectors follows from the fact that the quadratic form $Q: M^{4} \to \mathbb{R}$, associated with the inner product $g$, $\left( Q(v) = g(v,v) = v^{2} \right)$, on the Minkowski vector space $M^{4}$, is indefinite.
This statement is provable. It is the content of Theorem D.

*****
My purpose from this thread was to show people the usefulness of the Schwarz inequality (on the spatial subspace $V^{3} \equiv \big \{ \mathbf{v} \in M^{4}, \ v^{0} = 0 \big \}$) for proving statements about Lorentz vectors. Clearly, people liked what I did.
This is an [I-type] thread. So, I had no intention to lay down, in here, the axiomatic structure of the Minkowski vector space $M^{4}$ or that of the Minkowski spacetime (as 4-dimentional differentiable manifold). And, if I wanted to, I could have done a pretty good job in that as well.