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.