# Properties of Four-Vectors

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.

Homework Helper
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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##.

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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!

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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)

• PeroK
Homework Helper
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##.

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?

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

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.

Homework Helper

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)

##\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

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

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.

Homework Helper
For part (b)

##\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

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

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?

Homework Helper
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##.

Last edited:
Homework Helper
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.

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