Are Space-like and Time-like Vectors Related in Orthogonality?

[spaghetti3451]

Homework Statement

Prove the following:

a) If $P^a$ is time-like and $P^{a}S_{a}=0$, then $S^{a}$ is space-like.

b) If $P^a$ is null and $P^{a}S^{a}=0$, then $S_{a}$ is space-like or $S^{a} \propto P^{a}$.

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.

The Attempt at a Solution

a) $S^{a}S_{a} = S^{a}S_{a} + P^{a}S_{a} = (S^{a} + P^{a})S_{a} = (S^{a} + P^{a})(S_{a} + P_{a}) = S^{a}S_{a} + P^{a}S_{a} + P_{a}S^{a} + P^{a}P_{a} = S^{a}S_{a} + P^{a}P_{a} = ?$

 

[TSny]

a) $S^{a}S_{a} = S^{a}S_{a} + P^{a}S_{a} = (S^{a} + P^{a})S_{a} = (S^{a} + P^{a})(S_{a} + P_{a})$

How do you justify the last equality shown above?

Note that $P^{a}S_{a}$ is an invariant. It has the same value in all reference frames. Try to pick a nice reference frame using what you know about $P^{a}$.

 

[spaghetti3451]

Well, in the 'mostly minus' convention, $P^{a} = \{1,0,0,0 \}$ is one possibility.

Therefore, $P^{a}S_{a} = 0$ implies $S_{0} = 0$.

Therefore, $S^{a}S_{a} = S^{0}S_{0} + S^{1}S_{1} + S^{2}S_{2} + S^{3}S_{3} = S^{1}S_{1} + S^{2}S_{2} + S^{3}S_{3} = - S_{1}S_{1} - S_{2}S_{2} - S_{3}S_{3}$.

$S_{i}S_{i} >0$, so that $S^{a}S_{a} < 0$, therefore $S^{a}$ is space-like.

 

[spaghetti3451]

b) In the 'mostly minus' convention, $P^{a} = \{-1,-1,0,0 \}$ is one possibility.

Therefore, $P^{a}S_{a} = 0$ implies $S_{0}+S_{1} = 0$, so that $S^{0}=S_{0}=-S_{1}$.

Therefore, $S^{a}S_{a} = S^{0}S_{0} + S^{1}S_{1} + S^{2}S_{2} + S^{3}S_{3} = S_{1}S_{1} - S_{1}S_{1} - S_{2}S_{2} - S_{3}S_{3} = - S_{2}S_{2} - S_{3}S_{3}$.

$S_{i}S_{i} >0$, so that $S^{a}S_{a} < 0$, therefore $S^{a}$ is space-like.

The other possibility is that $P^{a} \propto S^{a}$ for obvious reasons.

 

[spaghetti3451]

Are my solutions correct?

 

[TSny]

Well, in the 'mostly minus' convention, $P^{a} = \{1,0,0,0 \}$ is one possibility.

Yes, it's a possibility. But you want a general proof. You have the right idea, but you can't assume that P⁰ = 1 in the frame where all the Pⁱ = 0. Similarly for part (b) where you assumed that P⁰ and P¹ are both equal to -1.

 

[spaghetti3451]

In that case, does $P^{a} = \{p,0,0,0\}$ for part (a) and $P^{a}=\{p,-p,0,0\}$ for part (b) qualify as four-vectors sufficiently general to give a general proof?

 

[TSny]

In that case, does $P^{a} = \{p,0,0,0\}$ for part (a) and $P^{a}=\{p,-p,0,0\}$ for part (b) qualify as four-vectors sufficiently general to give a general proof?

OK. For part (b), you could rotate the spatial coordinate axes to get $P^{a}=\{p,p,0,0\}$ (without any negative signs). But I think the way you wrote it is OK also.

In post 4 you had

b) In the 'mostly minus' convention, $P^{a} = \{-1,-1,0,0 \}$ is one possibility.

Therefore, $P^{a}S_{a} = 0$ implies $S_{0}+S_{1} = 0$, so that $S^{0}=S_{0}=-S_{1}$.

Therefore, $S^{a}S_{a} = S^{0}S_{0} + S^{1}S_{1} + S^{2}S_{2} + S^{3}S_{3} = S_{1}S_{1} - S_{1}S_{1} - S_{2}S_{2} - S_{3}S_{3} = - S_{2}S_{2} - S_{3}S_{3}$.

$S_{i}S_{i} >0$, so that $S^{a}S_{a} < 0$, therefore $S^{a}$ is space-like.

Logically, can you really claim that $S^{a}S_{a} < 0$ follows from what you have written? Or should you conclude only that $S^{a}S_{a} \leq 0$?

 

[spaghetti3451]

Yes, that proves that $S^{a}$ is space-like or null for part (b)?

But, by the same token, can we also not conclude that $S^{a}$ is space-like or null (and not just space-like) for part (a)?

 

[TSny]

Yes, that proves that $S^{a}$ is space-like or null for part (b)?

Yes. Also, you should go on to show that if S is null, then it must be proportional to P.

But, by the same token, can we also not conclude that $S^{a}$ is space-like or null (and not just space-like) for part (a)?

No, you should be able to prove that in (a), S must be space-like. It can't be null. [Of course, S is assumed to be a nonzero 4-vector in both (a) and (b).]

 

[spaghetti3451]

Yes. Also, you should go on to show that if S is null, then it must be proportional to P.

For the case when both $S^{a}S_{0}=0$ and $P^{a}P_{a}=0$, there are probably many solutions that relate $S^{a}$ and $P^{a}$, but one of them is surely $S^{a}=cP^{a}$. What are the other solutions that could arise?

No, you should be able to prove that in (a), S must be space-like. It can't be null. [Of course, S is assumed to be a nonzero 4-vector in both (a) and (b).]

Ah! If $S^{a}$ is a non-zero four vector in (a), then $S_{i}S_{i}>0$ in (a). Got it!

 

[TSny]

For the case when both $S^{a}S_{0}=0$ and $P^{a}P_{a}=0$, there are probably many solutions that relate $S^{a}$ and $P^{a}$, but one of them is surely $S^{a}=cP^{a}$. What are the other solutions that could arise?

You need to show that there are no other solutions when S is null. That is, if S and P are null and $P^{a}S_{a}=0$, then show that S must be proportional to P.

 

[spaghetti3451]

Well, $S^{a}S_{a}=0$ means that $(S_{0})^{2}=(\vec{S})^{2}$ and $P^{a}P_{a}=0$ means that $(P_{0})^{2}=(\vec{P})^{2}$.

Now, let's divide and obtain $\frac{(S_{0})^{2}}{(P_{0})^{2}} = \frac{(\vec{S})^{2}}{(\vec{P})^{2}}$.

Call the ratio $\frac{(S_{0})^{2}}{(P_{0})^{2}} = c^{2}$.

Therefore, $c^{2} = \frac{(\vec{S})^{2}}{(\vec{P})^{2}} \implies (\vec{S})^{2} = c^{2} (\vec{P})^{2}$.

Therefore, component-wise, the four-vectors are proportional.

Is this correct?

 

[TSny]

Well, $S^{a}S_{a}=0$ means that $(S_{0})^{2}=(\vec{S})^{2}$ and $P^{a}P_{a}=0$ means that $(P_{0})^{2}=(\vec{P})^{2}$.

Now, let's divide and obtain $\frac{(S_{0})^{2}}{(P_{0})^{2}} = \frac{(\vec{S})^{2}}{(\vec{P})^{2}}$.

Call the ratio $\frac{(S_{0})^{2}}{(P_{0})^{2}} = c^{2}$.

Therefore, $c^{2} = \frac{(\vec{S})^{2}}{(\vec{P})^{2}} \implies (\vec{S})^{2} = c^{2} (\vec{P})^{2}$.

OK.

Therefore, component-wise, the four-vectors are proportional.

You have not shown that the individual components of S are proportional to the corresponding individual components of P (with the same proportionality constant). For example, everything you stated above could be satisfied with $(P^0, P^1, P^2, P^3) = (5, 0, 3, 4)$ and $(S^0, S^1, S^2, S^3) = (5, 0, 5, 0)$.

Note that you did not use $P^aS_a = 0$ in your argument above.

 

[spaghetti3451]

Alright, let me start from scratch.

I need to show that if $S^{a}$ and $P^{a}$ are null, and if $P^{a}S_{a} = 0$, then $S^{a} \propto P^{a}$.

Here's my proof.

$S^{a}$ is null $\implies (S_{0})^{2}=(\vec{S})^{2}$.
$P^{a}$ is null $\implies (P_{0})^{2}=(\vec{P})^{2}$.

$P^{a}S_{a}=0 \implies P_{0}S_{0}=(\vec{P}\cdot{\vec{S}}) \implies \sqrt{(\vec{P})^{2}(\vec{S})^{2}}=(\vec{P}\cdot{\vec{S}})$.

Alluding to the rule for dot product in Euclidean geometry, $P^{i} = k S^{i}$ for $i=1,2,3$, where $k$ is a positive real number.

Therefore, $(\vec{P})^{2} = k^{2} (\vec{S})^{2} \implies (P_{0})^{2} = k^{2} (S_{0})^{2} \implies P_{0}=kS_{0}$.

What do you think?

 

[TSny]

$S^{a}$ is null $\implies (S_{0})^{2}=(\vec{S})^{2}$.
$P^{a}$ is null $\implies (P_{0})^{2}=(\vec{P})^{2}$.

$P^{a}S_{a}=0 \implies P_{0}S_{0}=(\vec{P}\cdot{\vec{S}}) \implies \sqrt{(\vec{P})^{2}(\vec{S})^{2}}=(\vec{P}\cdot{\vec{S}})$.

In getting to the last equation, can you assume that $P_{0}S_{0} = \sqrt{(\vec{P})^{2}(\vec{S})^{2}}$ or only that $P_{0}S_{0} = \pm \sqrt{(\vec{P})^{2}(\vec{S})^{2}}$?

Alluding to the rule for dot product in Euclidean geometry, $P^{i} = k S^{i}$ for $i=1,2,3$, where $k$ is a positive real number.

Does $k$ necessarily have to be positive?

Therefore, $(\vec{P})^{2} = k^{2} (\vec{S})^{2} \implies (P_{0})^{2} = k^{2} (S_{0})^{2} \implies P_{0}=kS_{0}$.

From this argument, you can only conclude that $P_{0}= \pm kS_{0}$.

What do you think?

I like this approach. It does not require going to a specific reference frame. You just need to take care of the sign issue.

You can use this approach to also prove part (a) without using a specific frame.

 

[spaghetti3451]

In getting to the last equation, can you assume that $P_{0}S_{0} = \sqrt{(\vec{P})^{2}(\vec{S})^{2}}$ or only that $P_{0}S_{0} = \pm \sqrt{(\vec{P})^{2}(\vec{S})^{2}}$?

Well, $(S_{0})^{2}=(\vec{S})^{2} \implies S_{0} = \pm \sqrt{\vec{S}^{2}}$ and $(P_{0})^{2}=(\vec{P})^{2} \implies P_{0} = \pm \sqrt{\vec{P}^{2}}$.

Therefore, $P_{0}S_{0} = \sqrt{(\vec{P})^{2}(\vec{S})^{2}}$, don't you think?

 

[TSny]

Well, $(S_{0})^{2}=(\vec{S})^{2} \implies S_{0} = \pm \sqrt{\vec{S}^{2}}$ and $(P_{0})^{2}=(\vec{P})^{2} \implies P_{0} = \pm \sqrt{\vec{P}^{2}}$.

Therefore, $P_{0}S_{0} = \sqrt{(\vec{P})^{2}(\vec{S})^{2}}$, don't you think?

The $\pm$ sign in $S_{0} = \pm \sqrt{\vec{S}^{2}}$ is not necessarily "correlated" with the $\pm$ sign in $P_{0} = \pm \sqrt{\vec{P}^{2}}$ .

There is no reason why $P_{0}S_{0}$ must be positive.

 

[spaghetti3451]

Let $P^{i}=cS^{i}$, where $c$ is a non-zero real number.

Therefore, $P^{a}S_{a}=0 \implies P^{0}S_{0}+P^{i}S_{i} =0 \implies P^{0}S_{0}+cS^{i}S_{i} = 0 \implies P^{0}S_{0}-cS_{i}S_{i} = 0 \implies P^{0}S_{0}-c(S_{0})^{2}=0 \implies S_{0}(P_{0}-cS_{0})=0 \implies S_{0}=0\ \text{or}\ P_{0}=cS_{0} \implies S_{0}=0\ \text{or}\ P^{0}=cS^{0}$.

Now, $S_{0}=0 \implies (\vec{S})^{2} = (S_{0})^{2}=0$, but $S^{a}$ cannot be a zero vector.

Therefore, $P^{a} \propto S^{a}$.

Is this correct?

 

[TSny]

That looks good to me.