Orthogonality of timelike and null vector

In summary, we can show orthogonality of timelike and null vectors by considering a space-time with a point ##p##, a time-like vector ##X## and a non-zero vector ##Y## orthogonal to ##X##. By using an orthonormal basis and applying the Cauchy-Schwarz inequality, we can prove that ##Y## is space-like.
  • #1
nomather1471
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Can we show orthogonality of timelike and null vector?
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  • #2
If X is timelike, and Y is orthogonal to X, then Y is spacelike.
 
  • #3
I just want to expand upon what George said. Let ##(M,g)## be a space-time and ##p\in M##. Furthermore, let ##X## be a time-like vector in ##T_p M## and ##Y## a non-zero vector in ##T_p M## such that ##g_p(X,Y) = 0##. Finally, let ##\{e_i \}## be an orthonormal basis for ##T_p M## so that ##X = X^i e_i## and ##Y = Y^i e_i##.

Then ##(X^2)^2 + (X^3)^2 + (X^4)^2 < (X^1)^2## and ##X^1 Y^1 = X^2 Y^2 + X^3Y^3 + X^4Y^4##.
Note this immediately implies that ##(X^1)^2 > 0## and that ##(Y^2)^2 + (Y^3)^2 + (Y^4)^2 > 0##.

Therefore [tex](X^1 Y^1)^2 \leq ((X^2)^2 + (X^3)^2 + (X^4)^2)((Y^2)^2 + (Y^3)^2 + (Y^4)^2)< (X^1)^2((Y^2)^2 + (Y^3)^2 + (Y^4)^2)[/tex] thus ##(Y^1)^2 < (Y^2)^2 + (Y^3)^2 + (Y^4)^2## i.e. ##Y## is space-like.
 
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  • #4
WannabeNewton said:
Finally, let ##\{e_i \}## be an orthonormal basis

WLOG, choose ##X = \{e_0 \}##.
 
  • #5
A time-like vector can be Lorentz transformed to be purely time-like:[itex]x^\mu=(t,0)[/itex].
A null vector always has x=t, so they can't be orthogonal.
 
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  • #6
WannabeNewton said:
I just want to expand upon what George said. Let ##(M,g)## be a space-time and ##p\in M##. Furthermore, let ##X## be a time-like vector in ##T_p M## and ##Y## a non-zero vector in ##T_p M## such that ##g_p(X,Y) = 0##. Finally, let ##\{e_i \}## be an orthonormal basis for ##T_p M## so that ##X = X^i e_i## and ##Y = Y^i e_i##.

Then ##(X^2)^2 + (X^3)^2 + (X^4)^2 < (X^1)^2## and ##X^1 Y^1 = X^2 Y^2 + X^3Y^3 + X^4Y^4##.
Note this immediately implies that ##(X^1)^2 > 0## and that ##(Y^2)^2 + (Y^3)^2 + (Y^4)^2 > 0##.

Therefore [tex](X^1 Y^1)^2 \leq ((X^2)^2 + (X^3)^2 + (X^4)^2)((Y^2)^2 + (Y^3)^2 + (Y^4)^2)< (X^1)^2((Y^2)^2 + (Y^3)^2 + (Y^4)^2)[/tex] thus ##(Y^1)^2 < (Y^2)^2 + (Y^3)^2 + (Y^4)^2## i.e. ##Y## is space-like.
I think you have to show that:
[tex](X^1 Y^1)^2 \leq ((X^2)^2 + (X^3)^2 + (X^4)^2)((Y^2)^2 + (Y^3)^2 + (Y^4)^2)[/tex]
 
  • #7
nomather1471 said:
I think you have to show that:
[tex](X^1 Y^1)^2 \leq ((X^2)^2 + (X^3)^2 + (X^4)^2)((Y^2)^2 + (Y^3)^2 + (Y^4)^2)[/tex]

I had forgotten completely that one has the freedom to Lorentz boost to a frame wherein ##X = e_0## as stated by George and Clem above, which makes the calculation much simpler. Nevertheless, the above inequality follows directly from the Cauchy-Schwarz inequality.
 
  • #8
I am student at geometry so your notation is quite different for me but thanks for everyone for all replies...
 

FAQ: Orthogonality of timelike and null vector

What is the concept of orthogonality in the context of timelike and null vectors?

The concept of orthogonality refers to the perpendicularity or independence of two vectors in a vector space. In the context of timelike and null vectors, it means that these two types of vectors are perpendicular or independent of each other in a four-dimensional spacetime.

What is a timelike vector and how does it relate to orthogonality?

A timelike vector is a vector with a positive squared length in a four-dimensional spacetime. It represents the direction of movement through time. In terms of orthogonality, timelike vectors are always orthogonal to null vectors, meaning they are perpendicular and do not share any components or directions.

What is a null vector and how is it related to orthogonality?

A null vector has a squared length of zero in a four-dimensional spacetime and represents a direction that is neither purely spatial nor purely temporal. In the context of orthogonality, null vectors are always orthogonal to timelike vectors, meaning they are perpendicular and do not share any components or directions.

Why is the concept of orthogonality important in relativity and spacetime?

The concept of orthogonality is important in relativity and spacetime because it helps to define the relationship between timelike and null vectors, which are fundamental in describing the dynamics of particles and fields in a four-dimensional spacetime. It also plays a crucial role in understanding the geometry of spacetime and how it is affected by the presence of matter and energy.

How does the concept of orthogonality apply to the principle of causality in relativity?

In relativity, the principle of causality states that the cause of an event must precede the effect in time. This is directly related to the orthogonality of timelike and null vectors, as it ensures that events cannot occur simultaneously or in reverse order in a four-dimensional spacetime. The orthogonality of these vectors also helps to define the light cone structure, which is essential in understanding causality in relativity.

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