Homework Help: Future-pointing and past-pointing time-like vectors

1. Dec 30, 2015

spaghetti3451

1. The problem statement, all variables and given/known data

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 $V^{a}W_{a}>0$.

2. Relevant 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.

3. 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?

2. Dec 30, 2015

Staff: Mentor

Where is the definition for that?
Not by definition, but via (a).

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

Schwarz Inequality is your friend (from today!)

3. Dec 30, 2015

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?

Last edited: Dec 30, 2015
4. Dec 30, 2015

Staff: Mentor

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

5. Jan 1, 2016

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?

6. Jan 1, 2016

Staff: Mentor

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

7. Jan 1, 2016

spaghetti3451

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

8. Jan 1, 2016

Staff: Mentor

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

9. Jan 1, 2016

spaghetti3451

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

10. Jan 1, 2016

spaghetti3451

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.

Last edited: Jan 1, 2016
11. Jan 1, 2016

Staff: Mentor

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.

12. Jan 1, 2016

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?

13. Jan 1, 2016

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.

14. Jan 1, 2016

Staff: Mentor

Correct.

15. Jan 1, 2016

TSny

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.

16. Jan 1, 2016

spaghetti3451

Yes.

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 realise 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.

17. Jan 1, 2016

TSny

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.

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 $x$.

18. Jan 1, 2016

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 $P^{0}Q^{0}>0$.

19. Jan 1, 2016

spaghetti3451

Is it correct?

20. Jan 1, 2016

TSny

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

Yes. Nice.

21. Jan 1, 2016

Staff: Mentor

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