How Do Four-Vectors with Orthogonal Dot Products Determine Each Other's Nature?

[shinobi20]

Homework Statement

Given the basic definition of four-vectors, I want to show some basic properties of two four-vectors contracted and some variation of that.

Relevant Equations

$A \cdot B = A^\mu B_\mu$

Two four-vectors have the property that $A^\mu B_\mu = 0$

(a) Suppose $A^\mu A_\mu > 0$. Show that $B^\mu B_\mu \leq 0$

(b) Suppose $A^\mu A_\mu = 0$. Show that $B^\mu$ is either proportional to $A^\mu$ (that is, $B^\mu = k A^\mu$) or else $B^\mu B_\mu < 0$.

Part (a) is intuitive to me since if the dot product of two four-vectors is zero then either they are perpendicular (if that makes sense in 4-d) or one of the vectors is the zero vector. Since $A^\mu B_\mu = 0$ and $A^\mu A_\mu > 0$, it is trivial that $B^\mu = 0$ so the only scenario left is if one of the four-vectors is timelike ($A^\mu$) and the other is spacelike ($B^\mu$), which is what to be shown. However, I am thinking is there any analytical way of showing part (a)? Can anyone give me a hint on how to do it?

For part (b), I have no idea how to show it analytically too.

 

[PeroK]

For three vectors you have $u \cdot v = |u||v|\cos \theta$.

Is there something analogous for four vectors?

 

[shinobi20]

For three vectors you have $u \cdot v = |u||v|\cos \theta$.

Is there something analogous for four vectors?

I can only think of $\textbf{u} \cdot \textbf{v} = u^0v^0 - |\vec{u}||\vec{v}|\cos \theta$.

 

[PeroK]

I can only think of $\textbf{u} \cdot \textbf{v} = u^0v^0 - |\vec{u}||\vec{v}|\cos \theta$.

Try looking online. Everything is online now!

 

[Ibix]

Try looking online. Everything is online now!

And can be looked up with some rapidity, in fact.

(If you don't get the joke, you haven't understood what @PeroK is talking about)

 

[Gaussian97]

I can only think of $\textbf{u} \cdot \textbf{v} = u^0v^0 - |\vec{u}||\vec{v}|\cos \theta$.

Yes, you can use this relation $0=A\cdot B = A^0B^0-|\vec{A}||\vec{B}|\cos \theta$ with the other condition $A^2=A\cdot A = A_{\mu}A^{\mu}>0$ to get a relation between $B^0$ and $|\vec{B}|$. Then you can prove a) using reduction to absurdity.

For b) is exactly the same, but instead of supposing $A^2>0$ you use $A^2=0$.

 

[shinobi20]

Yes, you can use this relation $0=A\cdot B = A^0B^0-|\vec{A}||\vec{B}|\cos \theta$ with the other condition $A^2=A\cdot A = A_{\mu}A^{\mu}>0$ to get a relation between $B^0$ and $|\vec{B}|$. Then you can prove a) using reduction to absurdity.

For b) is exactly the same, but instead of supposing $A^2>0$ you use $A^2=0$.

$A^0B^0 =|\vec{A}||\vec{B}|\cos \theta \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big) = \Big(\frac{\vec{B}}{B^0}\Big) \cos \theta$

$(A^0)^2 > |\vec{A}|^2 \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big)^2 > 1$

$\Big(\frac{\vec{B}}{B^0}\Big)^2 \cos^2 \theta > 1 \quad \rightarrow \quad \Big(\frac{\vec{B}}{B^0}\Big)^2 > \frac{1}{\cos^2 \theta} \geq 1 \quad \rightarrow \quad (\vec{B})^2 \geq (B^0)^2 \quad \rightarrow \quad 0 \geq (B^0)^2 - (\vec{B})^2 \quad \rightarrow \quad 0 \geq B^\mu B_\mu$

Is this correct? I don't understand how I'm going to prove (a) using contradiction.

And can be looked up with some rapidity, in fact.

(If you don't get the joke, you haven't understood what @PeroK is talking about)

I'm still thinking how to use rapidity to prove (a), more hints?

 

[PeroK]

I can only think of $\textbf{u} \cdot \textbf{v} = u^0v^0 - |\vec{u}||\vec{v}|\cos \theta$.

$A^0B^0 =|\vec{A}||\vec{B}|\cos \theta \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big) = \Big(\frac{\vec{B}}{B^0}\Big) \cos \theta$

$(A^0)^2 > |\vec{A}|^2 \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big)^2 > 1$

$\Big(\frac{\vec{B}}{B^0}\Big)^2 \cos^2 \theta > 1 \quad \rightarrow \quad \Big(\frac{\vec{B}}{B^0}\Big)^2 > \frac{1}{\cos^2 \theta} \geq 1 \quad \rightarrow \quad (\vec{B})^2 \geq (B^0)^2 \quad \rightarrow \quad 0 \geq (B^0)^2 - (\vec{B})^2 \quad \rightarrow \quad 0 \geq B^\mu B_\mu$

Is this correct? I don't understand how I'm going to prove (a) using contradiction.

Your proof for b) looks all right.

The idea behind proof by contradiction is:

(*) We know $A \cdot B = 0$ and $A \cdot A > 0$.

If we assume that $B \cdot B > 0$ and reach a contradiction, then this proves that $B \cdot B \le 0$ (given the premise (*) above).

Another way to think of this is that if we assume all three things are true and reach a contradiction, then we know that at most two of them can be true.

 

[Gaussian97]

$A^0B^0 =|\vec{A}||\vec{B}|\cos \theta \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big) = \Big(\frac{\vec{B}}{B^0}\Big) \cos \theta$

$(A^0)^2 > |\vec{A}|^2 \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big)^2 > 1$

$\Big(\frac{\vec{B}}{B^0}\Big)^2 \cos^2 \theta > 1 \quad \rightarrow \quad \Big(\frac{\vec{B}}{B^0}\Big)^2 > \frac{1}{\cos^2 \theta} \geq 1 \quad \rightarrow \quad (\vec{B})^2 \geq (B^0)^2 \quad \rightarrow \quad 0 \geq (B^0)^2 - (\vec{B})^2 \quad \rightarrow \quad 0 \geq B^\mu B_\mu$

Is this correct? I don't understand how I'm going to prove (a) using contradiction.

Yes, that's a correct way to prove a). Now you can try with b) which is essentially the same. Also, note that you have proved not only that $B^2 \leq 0$ but the more restrictive property that $B^2<0$. Which also is a direct consequence of b).

 

[shinobi20]

Yes, that's a correct way to prove a). Now you can try with b) which is essentially the same. Also, note that you have proved not only that $B^2 \leq 0$ but the more restrictive property that $B^2<0$. Which also is a direct consequence of b).

For part (b)

$A^0B^0 =|\vec{A}||\vec{B}|\cos \theta \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big) = \Big(\frac{\vec{B}}{B^0}\Big) \cos \theta$

$(A^0)^2 = |\vec{A}|^2 \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big)^2 = 1$

$\Big(\frac{\vec{B}}{B^0}\Big)^2 \cos^2 \theta = 1 \quad \rightarrow \quad \Big(\frac{\vec{B}}{B^0}\Big)^2 = \frac{1}{\cos^2 \theta}$

If $~\frac{1}{\cos^2 \theta} > 1~$ then

$(\vec{B})^2 > (B^0)^2 \quad \rightarrow \quad 0 > (B^0)^2 - (\vec{B})^2 \quad \rightarrow \quad 0 > B^\mu B_\mu$

If $~\frac{1}{\cos^2 \theta} = 1~$ then

$(\vec{B})^2 = (B^0)^2 \quad \rightarrow \quad 0 = (B^0)^2 - (\vec{B})^2 \quad \rightarrow \quad 0 = B^\mu B_\mu$

such that if $~B^\mu~$ is a null vector, it must satisfy the general equality $B^\mu = k A^\mu$.

I think what I've done is correct.

Your proof for b) looks all right.

The idea behind proof by contradiction is:

(*) We know $A \cdot B = 0$ and $A \cdot A > 0$.

If we assume that $B \cdot B > 0$ and reach a contradiction, then this proves that $B \cdot B \le 0$ (given the premise (*) above).

Another way to think of this is that if we assume all three things are true and reach a contradiction, then we know that at most two of them can be true.

I understand what you mean now on how to prove using contradiction, but I'm curious on what @Ibix mean by using rapidity to prove this, I still can't find a way to use rapidity to show this.

 

[Gaussian97]

For part (b)

$A^0B^0 =|\vec{A}||\vec{B}|\cos \theta \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big) = \Big(\frac{\vec{B}}{B^0}\Big) \cos \theta$

$(A^0)^2 = |\vec{A}|^2 \quad \rightarrow \quad \Big(\frac{A^0}{\vec{A}}\Big)^2 = 1$

$\Big(\frac{\vec{B}}{B^0}\Big)^2 \cos^2 \theta = 1 \quad \rightarrow \quad \Big(\frac{\vec{B}}{B^0}\Big)^2 = \frac{1}{\cos^2 \theta}$

If $~\frac{1}{\cos^2 \theta} > 1~$ then

$(\vec{B})^2 > (B^0)^2 \quad \rightarrow \quad 0 > (B^0)^2 - (\vec{B})^2 \quad \rightarrow \quad 0 > B^\mu B_\mu$

If $~\frac{1}{\cos^2 \theta} = 1~$ then

$(\vec{B})^2 = (B^0)^2 \quad \rightarrow \quad 0 = (B^0)^2 - (\vec{B})^2 \quad \rightarrow \quad 0 = B^\mu B_\mu$

such that if $~B^\mu~$ is a null vector, it must satisfy the general equality $B^\mu = k A^\mu$.

That's almost correct, in fact, you have proved that $B^2 \leq 0$. But you should prove that if $B^2=0$ then $B\propto A$.

 

[shinobi20]

That's almost correct, in fact, you have proved that $B^2 \leq 0$. But you should prove that if $B^2=0$ then $B\propto A$.

Isn't it because $A^\mu$ is a null vector and since it was showed that $B^\mu$ is also a null vector, so they must be proportional to each other? If there is any analytical way of showing the proportionality, can you please give me a hint?

 

[Gaussian97]

Isn't it because $A^\mu$ is a null vector and since it was showed that $B^\mu$ is also a null vector, so they must be proportional to each other? If there is any analytical way of showing the proportionality, can you please give me a hint?

Well, if what you say it's true you must prove it.
As a hint, I will prove to you that this is not enough, consider for example $A^{\mu} = (1, 0, 0, 1)$ and $B^{\mu}=(1, 1, 0, 0)$. It's trivial to prove that $A^2=B^2=0$. But is also trivial to see that they are not proportional, so there must be something more going on.

 

[shinobi20]

Well, if what you say it's true you must prove it.
As a hint, I will prove to you that this is not enough, consider for example Aμ=(1,0,0,1)Aμ=(1,0,0,1) and Bμ=(1,1,0,0)Bμ=(1,1,0,0). It's trivial to prove that A2=B2=0A2=B2=0. But is also trivial to see that they are not proportional, so there must be something more going on.

I think my proof of part (b) is lacking for the latter part, I'll add some more information.

If $\frac{1}{\cos^2 \theta} = 1$, then $\cos \theta =\pm 1$ and $B^\mu B_\mu = 0$ which means that the magnitude of the temporal part is equal to the spatial part.

Having $\cos \theta =\pm 1$ means that $\vec{B} = k\vec{A}$ for some constant $k$, i.e., they are either parallel or antiparallel (proportional to each other).

So, $A^0 B^0 = |\vec{A}||\vec{B}| \cos \theta = k |\vec{A}|^2$.

Since $(A^0)^2=|\vec{A}|^2$, we have $A^0 B^0 = k (A^0)^2 \quad \rightarrow \quad B^0 = k A^0$.

OR more simply, $B^0 = |\vec{B}| = k |\vec{A}| = k A^0$.

Thus, $B^0 − \vec{B} = k A^0 - k \vec{A} = k (A^ 0 − \vec{A}) \quad \rightarrow \quad B^\mu = k A^\mu$.

 

[Gaussian97]

Ok perfect, now I think that's all the exercise.
Maybe for this last part is easier to do that, since $A^0=|\vec{A}|$ and $B^0=|\vec{B}|$ then $\vec{A}=k\vec{B}\Longrightarrow A^0=kB^0$.

Congrats

 

[shinobi20]

Ok perfect, now I think that's all the exercise.
Maybe for this last part is easier to do that, since $A^0=|\vec{A}|$ and $B^0=|\vec{B}|$ then $\vec{A}=k\vec{B}\Longrightarrow A^0=kB^0$.

Congrats

Whoa, I just thought of that while looking at my solution again so I edited it. However, I'm still curious about what @Ibix meant for proving this using rapidity.

 

[Ibix]

Apologies - I don't think I read the question carefully enough. If the vectors are both timelike or both spacelike then you can note that their inner product is $|A||B|\cosh\phi$ - but this doesn't help in this case.

A way that does work is to note that if $A^\mu A_\mu>0$ then there exists a coordinate system in which $A^0$ is the only non-zero component of $A^\mu$ and the metric is diagonal. The first result is trivial then. Similarly if $A^\mu A_\mu=0$ then there exists a coordinate system in which $A^0=A^1$ are the only non-zero components of $A^\mu$. The second result is then straightforward.