Can Two Future-Pointing Null Vectors Sum to a Time-Like Vector?

[spaghetti3451]

Homework Statement

Show that the sum of two future-pointing null vectors is a future-pointing time-like vector, except when the two null vectors have the same direction. Conversely, show that any time-like vector can be expressed as a sum of two null vectors. For a given time-like vector the two null vectors are not uniquely determined: what is the nature of the freedom in their choice?

Homework Equations

Under the 'mostly-minus' convention, a null vector $N^{a}$ satisfies $N^{a}N_{a}=0$, a time-like vector $T^{a}$ satisfies $T^{a}T_{a}>0$, and a space-like vector $S^{a}$ satisfies $S^{a}S_{a}<0$.

Given a future-pointing time-like vector $P^{a}$ (with $P^{0}>0$), any other future-pointing time-like or null vector $Q^{a}$ satisfies $P^{a}Q_{a}>0$.

The Attempt at a Solution

Consider two future-pointing null vectors $P^{a}$ and $Q^{a}$. These vectors satisfy the following relations:

$P^{a}P_{a}=0,\ Q^{a}Q_{a}=0,\ P^{a}Q_{a}>0$.

Therefore, $(P^{a}+Q^{a})(P_{a}+Q_{a}) = P^{a}P_{a}+ 2\ P^{a}Q_{a} + Q^{a}Q_{a} = 2\ P^{a}Q_{a}$.

Now, if $P^{a} \not\propto Q^{a}$, then $(P^{a}+Q^{a})(P_{a}+Q_{a}) = 2\ P^{a}Q_{a} > 0$ and

if $P^{a} \propto Q^{a}$, then $(P^{a}+Q^{a})(P_{a}+Q_{a}) \propto 2\ P^{a}P_{a} = 0$.

Therefore, the sum of two future-pointing null vectors is a time-like vector, except when the two null vectors have the same direction, in which case the sum is again a null vector (regardless of the time-orientation of the original two vectors).Consider a time-like vector $S^{a}+T^{a}$, where the nature of $S^{a}$ and $T^{a}$ are not yet known. The vector $S^{a}+T^{a}$ satisfies the following relation:

$(S^{a}+T^{a})(S_{a}+T_{a})>0$, which means that

$S^{a}S_{a}+T^{a}T_{a}+2S^{a}T_{a}>0$

Where do I go from here?

 

[Fightfish]

Consider a time-like vector $S^{a}+T^{a}$, where the nature of $S^{a}$ and $T^{a}$ are not yet known. The vector $S^{a}+T^{a}$ satisfies the following relation:

$(S^{a}+T^{a})(S_{a}+T_{a})>0$, which means that

$S^{a}S_{a}+T^{a}T_{a}+2S^{a}T_{a}>0$

Where do I go from here?

What happens if you now choose S and T to be null-like?

 

[spaghetti3451]

If $S^{a}$ and $T^{a}$ are null-like, then $S^{a}T_{a}>0$

which means that $S^{a}T_{a}$ has to be future-pointing as well.But then, the relation $S^{a}S_{a}+T^{a}T_{a}+2S^{a}T_{a}>0$ is also satisfied for $S^{a}$ and $T^{a}$ future-pointing time-like?

 

[Fightfish]

Yes, the question asks you to show that a time-like vector can be expressed as a sum of two null vectors; it did not say that that is the only way to decompose it. I don't see why you can't express a time-like vector as a sum of two time-like vectors if you wanted to.

 

[spaghetti3451]

Thanks! Got it!

Let me now answer the third part of the question:

For a given time-like vector, the two null vectors are not uniquely determined: what is the nature of the freedom in their choice?

The condition for the vectors $T^{a}$ and $S^{a}$ to be null is that $T^{a}S_{a}>0$,

so either both vectors are future-pointing, or both are past-pointing.

Is this correct?

 

[Fightfish]

Yup, that seems right to me.