Future-pointing and past-pointing time-like vectors

[spaghetti3451]

Homework Statement

I need to prove the following:

a) If $P^{a}$ and $Q^{a}$ are time-like and $P^{a}Q_{a}>0$, then either both are future-pointing or both are past-pointing.

b) If $U^{a}$, $V^{a}$ and $W^{a}$ are time-like with $U^{a}V_{a}>0$ and $U^{a}W_{a}>0$, then $V^{a}W_{a}>0$.

Homework Equations

Using the 'mostly minus' convention, $A^a$ is time-like, null and space-like if $A^{a}A_{a}$ is $>0,=0,<0$ respectively.

A time-orientation is chosen by taking at will some time-like vector, say $U^{a} = (1,0,0,0)$, and designating it to be future-pointing. Any other time-like or null vector $V^{a}$ such that $g_{ab}U^{a}V^{b}>0$ is also future-pointing, whereas if $g_{ab}U^{a}V^{b}<0$, then $V^{a}$ is past-pointing.

The Attempt at a Solution

a) If $P^{a}$ is time-like future-pointing and $P^{a}Q_{a}>0$, then (by definition) $Q^{a}$ is also future-pointing.

Simialarly, for $P^{a}$ past-pointing.

b) If $U^{a}$ is time-like future-pointing and $U^{a}V_{a}>0$, then (by definition) $V^{a}$ is also future-pointing.

If $U^{a}$ is time-like future-pointing and $U^{a}W_{a}>0$, then (by definition) $W^{a}$ is also future-pointing.

Since both $V^{a}$ and $W^{a}$ are future-pointing, $V^{a}W_{a}>0$.

Similar argument for $U^{a}$ past-pointing.

Are my proofs sound?

 

[mfb]

a) If $P^{a}$ is time-like future-pointing and $P^{a}Q_{a}>0$, then (by definition) $Q^{a}$ is also future-pointing.

Where is the definition for that?

b) If $U^{a}$ is time-like future-pointing and $U^{a}V_{a}>0$, then (by definition) $V^{a}$ is also future-pointing.

Not by definition, but via (a).

(b) is fine once you fix (a).

Schwarz Inequality is your friend (from today!)

 

[spaghetti3451]

I've read the post. It is very helpful.

I can see that the following theorem is useful:

For $U^{a}$ and $V^{a}$ timelike and $U^{0}, V^{0} >0$, we have that $U^{a}V_{a}>0$.

How do I relate the fact that $U^{0}, V^{0} >0$ with future-pointing and past-pointing vectors?

 

[mfb]

How do I relate the fact that $U^{0}, V^{0} >0$ with future-pointing and past-pointing vectors?

That you can get from the definition now, if you evaluate the sums there explicitely.

 

[spaghetti3451]

Alright, here's my proof.

We are given that $P^{a}$ and $Q^{a}$ are timelike and that $P^{a}Q_{a}>0$.

If $P^{a}$ is timelike, then $P^{a}P_{a}>0 \implies (P^{0})^{2}>(\vec{P})^{2}$. Similarly, for $Q^{a}$.
Therefore, $(P^{0})^{2}(Q^{0})^{2}>(\vec{P})^{2}(\vec{Q})^{2}$.
So, by the Cauchy-Schwarz inequality, $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$, which implies that either $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$, where the RHS is positive or $P^{0}Q^{0}<\vec{P}\cdot{\vec{Q}}$, where the RHS is negative.

On the other hand, if $P^{a}Q_{a}>0$, then $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$. Therefore, $P^{0}Q^{0}>0$.
So, either both $P^{0}$ and $Q^{0}$ are positive and hence future-pointing, or both $P^{0}$ and $Q^{0}$ are negative and hence past-pointing.Is my proof correct?

 

[mfb]

Correct.
You can add a 1-line proof that future-pointing <=> P⁰>0.

 

[spaghetti3451]

Isn't it a convention that a four-vector $P^{0}$ is future pointing iff $P^{0}>0$?

 

[mfb]

The definition you have in post 1 expresses it in a slightly different way, although you get this result by evaluating the sums there.

 

[spaghetti3451]

So, you mean that a four-vector $P^{0}$ being future pointing iff $P^{0}>0$ is not a convention?

 

[spaghetti3451]

The definition you have in post 1 expresses it in a slightly different way, although you get this result by evaluating the sums there.

The definition says that we choose an arbitrary time-like vector $P^{a}=(1,0,0,0)$ with $P^{0}>0$ and designate it to be future-pointing. Then, if $g_{ab}P^{a}Q_{a}>0$ for some time-like vector $Q^{a}$, then $Q^{a}$ is also future-pointing.

What strikes me is that the definition arbitrarily assigns a time-like vector $P^{a}$ with $P^{0}>0$ as being future-pointing. That's why I ask if it's a convention that a time-like vector $P^{a}$ with $P^{0}>0$ is called future-pointing.

And if so, I don't see why we need to use the $g_{ab}P^{a}Q_{a}>0$ to evaluate sums.

 

[mfb]

It is an arbitrary convention, right - you could also assign positive time-components to past-pointing and negative ones to future-pointing. Would be odd, but it would lead to consistent physics as well.

You need to use it because you are given this definition of future-pointing.

 

[spaghetti3451]

Ah! I see.

So, it is indeed an arbitrary definition. But what I do need to show is that the class of future-pointing vectors $P^{a}$ all have $P^{0}>0$ if at least one of them have $P^{0}>0$.

That is the reason why we choose $P^{a}=(1,0,0,0)$ because then, when we evaluate $P^{a}Q_{a}>0$, all the space-components of $Q_{a}$ are eliminated and we are left with the simple relation that $Q^{a}>0$.

Thus, we have shown basically that the class of future-pointing vectors and the class of past-pointing vectors do not overlap, and that each class can be identified by the sign of the temporal component of the four-vectors.

Am I correct?

 

[spaghetti3451]

Alright, so now let me show the equivalence between statements 1 and 2.

1. A time-orientation is chosen by taking at will some time-like vector, say $U^{a}=(1,0,0,0)$ and designating it to be future-pointing. Any other time-like or null vector $V^{a}$ such that $U^{a}V_{a}>0$ is also future-pointing, whereas if $U^{a}V_{a}<0$, then $V^{a}$ is past-pointing.

2. A four-vector $P^{a}$ is future-pointing if and only if $P^{0}>0$.If some time-like vector $U^{a}=(1,0,0,0)$ is chosen to be future-pointing, then for any other time-like vector $V^{a}$ with $U^{a}V_{a}>0$, we have

$U^{a}V_{a}>0 \implies U^{0}V_{0}+U^{i}V_{i}>0 \implies U^{0}V^{0}>0 \implies V^{0}>0$.

Therefore, all time-like vectors $V^{a}$ with $V^{0}>0$ are future-pointing.

Similarly, for any other time-like vector $W^{a}$ with $U^{a}W_{a}<0$, we have

$U^{a}W_{a}<0 \implies U^{0}W_{0}+U^{i}W_{i}<0 \implies U^{0}W^{0}<0 \implies W^{0}<0$.

Therefore, all time-like vectors $W^{a}$ with $W^{0}<0$ are future-pointing.

 

[mfb]

Correct.

 

[TSny]

We are given that $P^{a}$ and $Q^{a}$ are timelike and that $P^{a}Q_{a}>0$.

If $P^{a}$ is timelike, then $P^{a}P_{a}>0 \implies (P^{0})^{2}>(\vec{P})^{2}$. Similarly, for $Q^{a}$.
Therefore, $(P^{0})^{2}(Q^{0})^{2}>(\vec{P})^{2}(\vec{Q})^{2}$.
So, by the Cauchy-Schwarz inequality, $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$, which implies that either $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$, where the RHS is positive or $P^{0}Q^{0}<\vec{P}\cdot{\vec{Q}}$, where the RHS is negative.

On the other hand, if $P^{a}Q_{a}>0$, then $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$. Therefore, $P^{0}Q^{0}>0$.
So, either both $P^{0}$ and $Q^{0}$ are positive and hence future-pointing, or both $P^{0}$ and $Q^{0}$ are negative and hence past-pointing.

I'm not following how you claim that the statement $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$ implies that either $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$, where the RHS is positive or $P^{0}Q^{0}<\vec{P}\cdot{\vec{Q}}$, where the RHS is negative.

Are you saying that the statement $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$ by itself implies that if $\vec{P}\cdot{\vec{Q}}$ is positive then $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$?

Suppose $P^{\alpha} = (2, 1, 0, 0)$ and $Q^{\alpha} = (-2, 1, 0, 0)$. Both vectors are timelike. $P^{0}Q^{0} = -4$ while $\vec{P}\cdot{\vec{Q}} = 1$. Then $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$ and $\vec{P}\cdot{\vec{Q}} > 0$, but it is not true that $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$.

Hope I'm not overlooking something or misinterpreting your statement.

 

[spaghetti3451]

I'm not following how you claim that the statement $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$ implies that either $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$, where the RHS is positive or $P^{0}Q^{0}<\vec{P}\cdot{\vec{Q}}$, where the RHS is negative.

Are you saying that the statement $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$ by itself implies that if $\vec{P}\cdot{\vec{Q}}$ is positive then $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$?

Yes.

Suppose $P^{\alpha} = (2, 1, 0, 0)$ and $Q^{\alpha} = (-2, 1, 0, 0)$. Both vectors are timelike. $P^{0}Q^{0} = -4$ while $\vec{P}\cdot{\vec{Q}} = 1$. Then $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$ and $\vec{P}\cdot{\vec{Q}} > 0$, but it is not true that $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$.

Ah! I see! Thanks for pointing out the mistake.

I thought I could use the fact that $x^{2}>y^{2} \implies x>y$ for $y$ positive or $x<y$ for $y$ negative. But, I realize now that it's not so simple, because $x$, in this case $P^{0}Q^{0}$, is itself a product of two numbers.

So, I have to go back to the drawing board and rethink a new proof, or at least, modify the existing proof.

 

[TSny]

I thought I could use the fact that $x^{2}>y^{2} \implies x>y$ for $y$ positive or $x<y$ for $y$ negative. But, I realize now that it's not so simple, because $x$, in this case $P^{0}Q^{0}$, is itself a product of two numbers.

Hmm. Even if $x$ were not the product of two numbers, it would still not be true in general that $x^{2}>y^{2} \implies x>y$ for $y$ positive.

So, I have to go back to the drawing board and rethink a new proof, or at least, modify the existing proof.

You are close. Along with $x^{2}>y^{2}$ you have $x > y$ (from $P^{\alpha}Q_{\alpha} > 0$). From both of these together you can conclude something important about $x$.

 

[spaghetti3451]

Got it!

$x^{2}>y^{2} \implies (x+y)(x-y)>0$.

Now, $x>y \implies x-y>0 \implies x+y>0 \implies x>-y$.

The only way for $x>y$ and $x>-y$ are both valid is if $x>0$, that is $P^{0}Q^{0}>0$.

 

[spaghetti3451]

Is it correct?

 

[TSny]

Got it!

$x^{2}>y^{2} \implies (x+y)(x-y)>0$.

Now, $x>y \implies x-y>0 \implies x+y>0 \implies x>-y$.

The middle $\implies$ is of course using both $x-y>0$ and $(x+y)(x-y)>0$.

The only way for $x>y$ and $x>-y$ are both valid is if $x>0$, that is $P^{0}Q^{0}>0$.

Yes. Nice.

 

[mfb]

You need the sign condition for x anyway (you compare the product of the time-component later), where the argument works.