Orthogonal spacelike and timelike vectors and inertial frames

[JD_PM]

TL;DR Summary

1) I wonder if we can simply assert orthonormality for all inertial frames based only on the fact that $S^{\mu}T_{\mu} = 0$ in the original frame and also on the fact that scalars are invariant under Lorentz Transformations.

2) I also wonder if whenever we have two orthogonal vectors $S^{\mu}$ and $T^{\mu}$ there has to be an inertial frame in which $S^{\mu} = (0, \vec s)$ and $T^{\mu} = (t^0, \vec 0)$

I know that any vector $V$ in Minkowski spacetime can be classified in three different categories based on its norm $|V| = \sqrt{V \cdot V} = V^{\mu}V_{\mu}$. These are:

1) If $V^{\mu}V_{\mu} < 0$, $V^{\mu}$ is timelike.

2) If $V^{\mu}V_{\mu} > 0$, $V^{\mu}$ is spacelike.

3) If $V^{\mu}V_{\mu} = 0$, $V^{\mu}$ is lightlike.

I started to play with two vectors in order to check my understanding.

Suppose we have a spacelike vector of the form $S^{\mu} = (0, \vec s)$ and timelike vector of the form $T^{\mu} = (t^0, \vec 0)$, both with respect to a specific inertial frame.

We clearly see that $S^{\mu}$ and $T^{\mu}$ are orthogonal in this frame, as $S^{\mu} T^{\mu} = 0$. Let's show it explicitely (I am using (-,+,+,+) convention):

$$S_{\mu}T^{\mu} = \eta_{\mu \rho} S^{\mu} T^{\rho} = -S^{0} T^{0} + S^{1} T^{1} + S^{2} T^{2} + S^{3} T^{3} = 0$$

OK but here comes the interesting point to me: $S_{\mu}T^{\mu}$ is a scalar (ie it has no free indices) and, as far as I know, scalars are invariant under Lorentz Transformations. Thus we get $S_{\mu}T^{\mu}=0$ again in any other frame. The conclusion I reach is that $S^{\mu}$ and $T^{\mu}$ are orthogonal in any frame. Do you agree with such assertion and the reasoning behind?

And I have another question: does this mean that whenever we have two orthogonal vectors $S^{\mu}$ and $T^{\mu}$ there has to be an inertial frame in which $S^{\mu} = (0, \vec s)$ and $T^{\mu} = (t^0, \vec 0)$?

Thank you.

 

[PeterDonis]

$S_{\mu}T^{\mu}$ is a scalar (ie it has no free indices) and, as far as I know, scalars are invariant under Lorentz Transformations. Thus we get $S_{\mu}T^{\mu}=0$ again in any other frame. The conclusion I reach is that $S^{\mu}$ and $T^{\mu}$ are orthogonal in any frame.

Yes. What you have done is establish that orthogonality is frame-independent, i.e., it is an invariant property of the pair of vectors.

does this mean that whenever we have two orthogonal vectors $S^{\mu}$ and $T^{\mu}$ there has to an inertial frame in which $S^{\mu} = (0, \vec s)$ and $T^{\mu} = (t^0, \vec 0)$?

If $S_\mu$ is spacelike and $T^\mu$ is timelike, yes. Note, however, that it possible for a pair of vectors to be orthogonal without one being spacelike and the other being timelike. You might want to think about what other possibilities there are (i.e., which of the 3 x 2 = 6 unordered pairs of types of vectors allow the pair to be orthogonal).

 

[JD_PM]

If $S_\mu$ is spacelike and $T^\mu$ is timelike, yes.

Alright, I will think about how to prove it and then post what I get.

Note, however, that it possible for a pair of vectors to be orthogonal without one being spacelike and the other being timelike. You might want to think about what other possibilities there are (i.e., which of the 3 x 2 = 6 unordered pairs of types of vectors allow the pair to be orthogonal).

Mmm I will think about this, thank you.

 

[PeterDonis]

I will think about how to prove it

You already have. Hint: is every logical step you took in your OP reversible?

 

[JD_PM]

You already have. Hint: is every logical step you took in your OP reversible?

I think yes.

The approach I have in mind is as follows:

Suppose we have the timelike vector $T^{\mu} = (t^0, \vec t)$ with respect to a specific inertial frame. If I am not mistaken we can always apply rotation to go to a frame in which $\vec t$ lies only on one axis; either $x^1$, $x^2$ or $x^3$ would be OK. Let's pick $x^1$. Thus our timelike vector $T^{\mu}$ becomes after rotation (and with respect to another inertial frame)

$$T^{\mu} = (t^0, t^1,0,0)$$

Now the idea would be apply a boost to get a timelike vector with no spatial components.

With the spacelike vector the idea is the same: suppose that we have $S^{\mu} = (s^0, \vec s)$. Then I have to be able to show that it is possible to get, through LTs, a spacelike vector with only spatial components.

In summary: I think that what we have to prove is that there is an inertial frame in which $S^{\mu}$ has only spatial components and $T^{\mu}$ has only the time component $t^0$.

Am I on the right track?

 

[PeterDonis]

Am I on the right track?

Yes, and the things you describe should be easy to prove (as I said, you've basically already done it).

 

[pervect]

Summary:: 1) I wonder if we can simply assert orthonormality for all inertial frames based only on the fact that $S^{\mu}T_{\mu} = 0$ in the original frame and also on the fact that scalars are invariant under Lorentz Transformations.

2) I also wonder if whenever we have two orthogonal vectors $S^{\mu}$ and $T^{\mu}$ there has to be an inertial frame in which $S^{\mu} = (0, \vec s)$ and $T^{\mu} = (t^0, \vec 0)$

You might want to think about how you define an "inertial frame". One definition of an inertial frame is by defining its basis vectors. However, such a definition doesn't encompass the idea that a frame should not rotate, for instance. You'd need some more requirements on your basis vectors to encompass the idea of a non-rotating frame. This would take you fairly far afield into advanced topic, such as Frobenius' theorm, time-like congruences, and Fermi-walker transport. THey're all fascinating topics, but it's not clear if you are ready for them (or interested).

You would also have to clean up a few nits, as you have specified in 1) that the vectors are orthogonal, but you haven't specified any requirement that they are unit length (normal). As others have pointed out, you haven't analyzed the second case very thouroghly, you should consider space-like, null, and time-like vectors.

 

[Orodruin]

If, in (1), S and T are supposed to be basis vectors of an inertial frame, then no. Different inertial frames have different basis vectors and the orthogonality of one of the frames does not mean all frames are just by the invariance of the inner product of the basis vectors in that frame. The inner product of any two vectors is invariant regardless of whether expressed in an inertial frame or not.

 

[pervect]

A space-time diagram might help. We'll do a simple case with one space and one time dimension, so we don't have to worry about issues like spatial rotation.

We're representing vectors on the diagram in the traditional way as little arrows. Hopefully this is self explanatory. Perhaps the details of drawing the space-time diagram need more clarification, but I'm really not sure what to say, I am hoping that they are familiar to the OP, at least enough to make my point.

The vectors x and t represent the orthonormal basis vectors of an observer in some frame S, the vectors x' and t' represent the orthonormal basis vectors in a frame S' that is moving relative to S.

Vector x is orthogonal to the vector t. This is true in all frames. The lengths of these vectors are also frame-independent scalars.

The vector x' is orthogonal to the vector t'. This is also true in all frames.

However x is not x', as you can see on the diagram. And t is not t'. They're different vectors. x and t are the basis vectors of S. x' and t' are the basis vectors of S'.

 

[JD_PM]

You might want to think about what other possibilities there are (i.e., which of the 3 x 2 = 6 unordered pairs of types of vectors allow the pair to be orthogonal).

Let's consider the null vector $N^{\mu} = (0, \vec 0)$.

We clearly see that the null vector will be orthogonal to the timelike vector:

$$N_{\mu}T^{\mu} = \eta_{\mu \rho} N^{\mu} T^{\rho} = -N^{0} T^{0} + N^{1} T^{1} + N^{2} T^{2} + N^{3} T^{3} = 0$$

Well, the null vector is orthogonal to any vector, so I guess this is not an interesting case to look at...

We already know that $S^{\mu} T^{\mu} = 0$

I think that playing with the null vector is not what you meant though...

 

[PeterDonis]

Let's consider the null vector $N^{\mu} = (0, \vec 0)$.

By "null vector" I meant a lightlike vector, not the zero vector. That is the usual meaning of "null vector" in relativity.

 

[JD_PM]

I have been thinking on using a general method to see if two vectors can be orthogonal or not.

This is what I have come up with:

Let's just work out the following example: Show whether a lightlike vector can be orthogonal to a timelike vector.

General Method

Step 1: Get an equation for each vector based on its nature.

OK, for the timelike vector we know that

$$-(t^0)^2 + \vec t \cdot \vec t < 0 \rightarrow (t^0)^2 > \vec t \cdot \vec t$$

For the lightlike vector $N^{\mu}$ we know that

$$-(n^0)^2 + \vec n \cdot \vec n = 0 \rightarrow (n^0)^2 = \vec n \cdot \vec n$$

Step 2: Compare both equations

OK, in the end what we want is to analyse the components of $N^{\mu}T_{\mu}$. Thus we should look for a combination that drives us there. Having that in mind and the obtained equations above I think we can claim the following

$$(n^0 t^0)^2 > (\vec n \cdot \vec n) (\vec t \cdot \vec t)$$

Or

$$|n^0 t^0| > \sqrt{(\vec n \cdot \vec n) (\vec t \cdot \vec t)}$$

Step 3: use Schwarz inequality

The Schwarz inequality asserts that

$$|\vec n \cdot \vec t| \leq \sqrt{(\vec n \cdot \vec n) (\vec t \cdot \vec t)}$$

Thus

$$|n^0 t^0| > \sqrt{(\vec n \cdot \vec n) (\vec t \cdot \vec t)} \geq |\vec n \cdot \vec t|$$

Thus we conclude that

$$|n^0 t^0| > |\vec n \cdot \vec t|$$

Step 4: Conclude

OK, we want to show if the following equation holds

$$N^{\mu}T_{\mu} = -(n^0 t^0)^2 + (\vec n \cdot \vec t)^2 = 0$$

We have derived the condition

$$|n^0 t^0| > |\vec n \cdot \vec t|$$

Mmm so it looks like

$$N^{\mu}T_{\mu} \neq 0$$

What do you think of this method?

 

[PeterDonis]

I have been thinking on using a general method to see if two vectors can be orthogonal or not.

You already have a very easy method since you have shown that orthogonality is frame-independent: just pick a convenient inertial frame and look at whether the vectors you are interested in are orthogonal in that frame. If they're orthogonal in that frame, they're orthogonal in every frame.

So the question boils down to: for which of the six possible pairs of vector types, (spacelike, spacelike), (spacelike, timelike), (spacelike, null), (null, null), (null, timelike), (timelike, timelike), can you find vectors of those types that are orthogonal in some convenient inertial frame?

What do you think of this method?

It looks like it has given you a correct answer for that particular case (a null vector and a timelike vector can never be orthogonal), but you could get the same answer much more simply by just looking at what the components of $N^\mu$ are in the frame in which $T^\mu = (t^0, \vec{0})$, since we have already shown that there must always be such a frame, and what that implies about $N_\mu T^\mu$.

 

[vanhees71]

Let's consider the null vector $N^{\mu} = (0, \vec 0)$.

We clearly see that the null vector will be orthogonal to the timelike vector:

$$N_{\mu}T^{\mu} = \eta_{\mu \rho} N^{\mu} T^{\rho} = -N^{0} T^{0} + N^{1} T^{1} + N^{2} T^{2} + N^{3} T^{3} = 0$$

Well, the null vector is orthogonal to any vector, so I guess this is not an interesting case to look at...

We already know that $S^{\mu} T^{\mu} = 0$

I think that playing with the null vector is not what you meant though...

By definition the zero-vector is not perpendicular to anything. Perhaps you mean a "null vector", i.e., a light-like vector where $N_{\mu} N^{\mu}=0$ but $N^{\mu} \neq (0,\vec{0})$?

 

[SiennaTheGr8]

By definition the zero-vector is not perpendicular to anything. Perhaps you mean a "null vector", i.e., a light-like vector where $N_{\mu} N^{\mu}=0$ but $N^{\mu} \neq (0,\vec{0})$?

This might just be a question of definition, but I thought that the zero-vector $(0, \vec 0)$ is orthogonal to all four-vectors:

$(0, \vec 0) \cdot (a^t, \vec a) = (0)(a^t) - \vec 0 \cdot \vec a = 0$

 

[JD_PM]

You already have a very easy method since you have shown that orthogonality is frame-independent: just pick a convenient inertial frame and look at whether the vectors you are interested in are orthogonal in that frame. If they're orthogonal in that frame, they're orthogonal in every frame.

So the question boils down to: for which of the six possible pairs of vector types, (spacelike, spacelike), (spacelike, timelike), (spacelike, null), (null, null), (null, timelike), (timelike, timelike), can you find vectors of those types that are orthogonal in some convenient inertial frame?

Absolutely, it is much shorter (I am afraid I always attempt exercises at first using lengthy methods ).

This is what I get for each case:

1) For $S^{\mu} \hat S_{\mu}$

Let $S^{\mu} = (0, \vec s_1)$ and $\hat S^{\mu} = (0, \vec s_2)$ (note that I have used the hat to emphasize that the two spacelike vectors are different).

Thus

$$S_{\mu} \hat S^{\mu} = \eta_{\mu \rho} S^{\mu} \hat S^{\rho} = -S^{0} \hat S^{0} + S^{1} \hat S^{1} + S^{2} \hat S^{2} + S^{3} \hat S^{3} = \vec s_1 \cdot \vec s_2 = 0$$

Note we can always construct the spatial components of the spacelike vectors such that they are orthogonal in the original frame. Thus the inner product $\vec s_1 \cdot \vec s_2$ vanishes.

Thus

$$S^{\mu} \hat S_{\mu} = 0$$2) For $S^{\mu}T_{\mu}$ I have already shown that $S^{\mu}T_{\mu} = 0$3) For $S^{\mu} N_{\mu}$

Let $S^{\mu} = (0, \vec s)$ and $N^{\mu} = (n^0, \vec n)$.

Thus

$$S_{\mu} \hat N^{\mu} = \eta_{\mu \rho} S^{\mu} N^{\rho} = -S^{0} N^{0} + S^{1} N^{1} + S^{2} N^{2} + S^{3} N^{3} = \vec s \cdot \vec n = 0$$

Here I am not entirely sure but I think we should also be able to construct the spatial components of the spacelike vector and the lightlike vector such that they are orthogonal in the original frame. Thus the inner product $\vec s \cdot \vec n$ vanishes.

Thus

$$S^{\mu} N_{\mu} = 0$$

4) For $N^{\mu} \hat N_{\mu}$

Let $N^{\mu} = (n_1^0, \vec n_1)$ and $\hat N^{\mu} = (n_2^0, \vec n_2)$.

Thus

$$N_{\mu} \hat N^{\mu} = \eta_{\mu \rho} N^{\mu} \hat N^{\rho} = -N^{0} \hat N^{0} + N^{1} \hat N^{1} + N^{2} \hat N^{2} + N^{3} \hat N^{3} = -n_1^0 n_2^0 + \vec n_1 \cdot \vec n_2 \neq 0$$

Thus

$$N^{\mu} \hat N_{\mu} \neq 0$$
5) For $T^{\mu} N_{\mu}$

Let $N^{\mu} = (n^0, \vec n)$ and $T^{\mu} = (t^0, \vec 0)$.

Thus

$$T_{\mu} N^{\mu} = \eta_{\mu \rho} T^{\mu} N^{\rho} = -T^{0} N^{0} + T^{1} N^{1} + T^{2} N^{2} + T^{3} N^{3} = -t^0 n^0 \neq 0$$

Thus

$$T^{\mu} N_{\mu} \neq 0$$
6) For $T^{\mu} \hat T_{\mu}$

Let $T^{\mu} = (t_1^0, \vec 0)$ and $\hat T^{\mu} = (t_2^0, \vec 0)$.

Thus

$$T_{\mu} \hat T^{\mu} = \eta_{\mu \rho} T^{\mu} \hat T^{\rho} = -T^{0} \hat T^{0} + T^{1} \hat T^{1} + T^{2} \hat T^{2} + T^{3} \hat T^{3} = -t_1^0 t_2^0 \neq 0$$

Thus

$$T_{\mu} \hat T^{\mu} \neq 0$$

How do you see it PeterDonis? :)

PS: Thank you for suggesting doing these extra cases! It made me get a deeper understanding and even made me come up with an extra method!

 

[JD_PM]

By definition the zero-vector is not perpendicular to anything. Perhaps you mean a "null vector", i.e., a light-like vector where $N_{\mu} N^{\mu}=0$ but $N^{\mu} \neq (0,\vec{0})$?

Hi vanhees71, I misunderstood its meaning, as PeterDonis suggested here:

By "null vector" I meant a lightlike vector, not the zero vector. That is the usual meaning of "null vector" in relativity.

 

[PeterDonis]

This is what I get for each case

You need to be clear about what you are proving. You are not proving that, for example, all pairs of spacelike vectors are orthogonal. You are only proving that there exist pairs of spacelike vectors that are orthogonal.

we can always construct the spatial components of the spacelike vectors such that they are orthogonal in the original frame.

No, you can't "always" do that You can only do it if $\vec{s}_1 \cdot \vec{s}_2 = 0$. That is what your proof shows. And whether or not that is true for any given pair of spacelike vectors is frame-invariant. (Can you see why?)

Here I am not entirely sure but I think we should also be able to construct the spatial components of the spacelike vector and the lightlike vector such that they are orthogonal in the original frame.

Again, you can't always do this. You can only do this if $\vec{s} \cdot \vec{n} = 0$, and whether or not that is true is frame invariant. (Again, can you see why?)

For $S^{\mu}T_{\mu}$ I have already shown that $S^{\mu}T_{\mu} = 0$

More precisely, you have shown that there exist pairs of vectors $S^\mu$, $T^\mu$, with $S$ spacelike and $T$ timelike, for which $S_\mu T^\mu = 0$. But you have not shown that this is true for all such pairs of vectors. To make sure you see this, you might want to see if you can find a spacelike vector that is not orthogonal to some chosen timelike vector.

Thus
$$
N^{\mu} \hat N_{\mu} \neq 0
$$

This is not correct for all pairs of null vectors; there are pairs of null vectors which are orthogonal. Here is such a pair: the vectors with components in some chosen inertial frame of $A^\mu = (a, a, 0, 0)$ and $B^\mu = (b, b, 0, 0)$, with $a \neq b$. It is easily calculated that $A^\mu B_\mu = 0$.

Your conclusions 5) and 6) are correct, because you are fortunate that if one of the vectors is timelike, you can always find a frame in which only one component is nonzero, which makes it much easier to draw correct general conclusions.

 

[JD_PM]

You need to be clear about what you are proving. You are not proving that, for example, all pairs of spacelike vectors are orthogonal. You are only proving that there exist pairs of spacelike vectors that are orthogonal.

Oh I see.

No, you can't "always" do that You can only do it if $\vec{s}_1 \cdot \vec{s}_2 = 0$. That is what your proof shows. And whether or not that is true for any given pair of spacelike vectors is frame-invariant. (Can you see why?)

Ups sorry, I expressed myself incorrectly. What I meant is exactly what you said: only when $\vec{s}_1 \cdot \vec{s}_2 = 0$ we can do it. I think it is because the inner product is invariant under Lorentz transformations.

Again, you can't always do this. You can only do this if $\vec{s} \cdot \vec{n} = 0$, and whether or not that is true is frame invariant. (Again, can you see why?)

I expressed myself incorrectly again. The same above reasoning applies here.

More precisely, you have shown that there exist pairs of vectors $S^\mu$, $T^\mu$, with $S$ spacelike and $T$ timelike, for which $S_\mu T^\mu = 0$. But you have not shown that this is true for all such pairs of vectors. To make sure you see this, you might want to see if you can find a spacelike vector that is not orthogonal to some chosen timelike vector.

But we were aimed at just showing that there exists a pair, weren't we? That is what I showed at #1.

This is not correct for all pairs of null vectors; there are pairs of null vectors which are orthogonal. Here is such a pair: the vectors with components in some chosen inertial frame of $A^\mu = (a, a, 0, 0)$ and $B^\mu = (b, b, 0, 0)$, with $a \neq b$. It is easily calculated that $A^\mu B_\mu = 0$.

Your conclusions 5) and 6) are correct, because you are fortunate that if one of the vectors is timelike, you can always find a frame in which only one component is nonzero, which makes it much easier to draw correct general conclusions.

But at 5) and 6) I showed it for just a particular case. weren't you asking for that? Or are 5) and 6) wrong for a particular case?

 

[PeterDonis]

I think it is because the inner product is invariant under Lorentz transformations.

The 4-inner product, $(S_1)^\mu (S_2)_\mu$, is certainly invariant under Lorentz transformations. But we are talking about the 3-inner product, $\vec{s}_1 \cdot \vec{s}_2$. This "inner product" is not invariant under Lorentz transformations. However, there is a more restricted statement that is invariant under Lorentz transformations. Can you see what it is?

we were aimed at just showing that there exists a pair, weren't we?

Yes; but it's also a good exercise to find a pair that is not orthogonal; it might help to see the difference between the two cases (orthogonal vs. not orthogonal).

at 5) and 6) I showed it for just a particular case

Your proof in both 5) and 6) is general: it shows that it is impossible to find a null vector orthogonal to any timelike vector, or to find any pair of timelike vectors that are orthogonal.

My comment was about your overall method, which was not considering all possibilities for each case. But for the cases involving timelike vectors, you can reduce the possibilities to just one by choosing the frame where the timelike vector has only one component. You can't do that for the (spacelike, spacelike), (spacelike, null), or (null, null) cases, so you have to consider multiple possibilities for those cases if you want to find a general proof.

are 5) and 6) wrong

No. See above.

 

[vanhees71]

Maybe, I just restate what was already said, but maybe a summary helps to sort the somewhat confusing thread ;-).

In the following I'm using the west-coast convention and $c=1$, i.e., the metric components are $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$.

A four-vector $V$ is called

-timelike if $V \cdot V>0$

-lightlike if $V \cdot V =0$ (mathematicians also talk about a "null vector", which however leads to confusion sometimes as can be observed in the thread above)

-spacelike if $V \cdot V<0$.

Next we note that a pseudocartesian basis always consists of four vectors $e_{\mu}$ with
$$e_{\mu} \cdot e_{\nu}=\eta_{\mu \nu}.$$

For the following we note that the usual Schmidt decomposition from Euclidean vector spaces also works for Minkowski space. We need it in the following form:

For a given time-like vector $u$ with there exists a pseudo-Cartesian basis $e_{\mu}$ such that $u^{\mu}=(u^0,0,0,0)$ wrt. to this basis.

This is easy to prove given the general form of a rotation-free Lorentz boost. Say $u^{\mu}=(u^{0},\vec{u})$ is timelike, then you just perform a Lorentz boost with boost velocity $\vec{v}=\vec{u}/u^{0}$, which reads (with the usual Lorentz factor $\gamma=(1-\vec{v}^2)^{-1/2}$ and $\hat{v}=\vec{v}/|\vec{v}|$).
$$\hat{\Lambda} = \begin{pmatrix} \gamma & -\gamma \vec{v}^{\text{T}} \\ -\gamma \vec{v} & (\hat{1}-\hat{v} \otimes \hat{v})+\gamma \hat{v} \otimes \hat{v} \end{pmatrix}.$$
Note that, because $u^{\prime \mu}$ is timelike, you have $|\vec{v}|<1$, so that all this makes sense. Then, indeed,
$$\hat{\Lambda} \begin{pmatrix} u^0 \\ \vec{u} \end{pmatrix} = \begin{pmatrix} \gamma - \gamma \vec{v} \cdot \vec{u} \\ -\gamma u^0 \vec{v} + \vec{u}+(\gamma-1) (\hat{v} \cdot \vec{u}) \hat{v} \end{pmatrix}.$$
To see that the spatial part indeed vanish, we simply need to use the definition of $\vec{v}$ and $\hat{v}$, from which $u^0 \vec{v}=\vec{u}$ and thus $\hat{v} \cdot \vec{u}=\hat{v} \cdot u^0 \vec{v}=u^0 |\vec{v}|$, leading to
$$-\gamma u^0 \vec{v} + \vec{u} + (\gamma-1) (\hat{v} \cdot \vec{u}) \hat{v}=-\gamma \vec{u} + \vec{u} + (\gamma-1) u^0 \vec{v} = -(\gamma-1) \vec{u} + (\gamma-1) \vec{u}=0.$$
Now it's easy to show that

If $u$ is time like and $V$ and arbitrary four-vector with $u \cdot V=0$, then $V$ is spacelike or 0.

To see this, as we have just shown explicitly above, we can find a pseudo-Cartesian basis such that $u=(u^0,0,0,0)$. In this basis $u \cdot V=u^0 V^0=0$ since $u$ is time-like $u^0 \neq 0$ and thus $V^0=0$, i.e., $V=(0,\vec{V})$, which is spacelike if $\vec{V} \neq 0$ or the zero-vector for $\vec{V}=0$, which proves our claim.

Another important theorem is that

If two vectors $a$ and $b$ are lightlike vectors (with $a \neq 0$) and $a \cdot b=0$, then there's some $\lambda \in \mathbb{R}$ with $b=\lambda a$.

Proof: From $a \cdot b=0$ we have $a^0 b^0=\vec{a} \cdot \vec{b}=\pm |\vec{a}| |\vec{b}|$ since from $a \cdot a$ we have $(a^0)^2=\vec{a}^2$ and thus $|\vec{a}|=\pm a^0$ and by the same argument $|\vec{b}|=\pm b^0$. Now from the Cauchy-Schwarz inequality we know that from $|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}|$ it follows that $\vec{b}=\lambda \vec{a}$. With the ansatz $b^0=\lambda' a^0$ this implies
$$b=\begin{pmatrix} \lambda' a^0 \\ \lambda \vec{a} \end{pmatrix},$$
and again from $a \cdot b=0$
$$\lambda' (a^0)^2=\lambda \vec{a}^2 \; \Rightarrow\; \lambda'=\lambda \; \Rightarrow \; b=\lambda a.$$

 

[JD_PM]

However, there is a more restricted statement that is invariant under Lorentz transformations. Can you see what it is?

Honestly I do not see what you mean here.

 

[PeterDonis]

I do not see what you mean here.

Actually, on thinking it over, the statement I had in mind is not true anyway. So disregard.