Timelike and lightlike/null vectors

  • Thread starter purakanui
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  • #1
purakanui
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I have another problem that I am stuck on.

Show that a timelike vector cannot be orthogonal to a null vector.

Timelike:
X[tex]^{2}[/tex] = g(sub a b)X[tex]^{a}[/tex]X[tex]^{b}[/tex] > 0

Null:
X[tex]^{2}[/tex] = g(sub a b)X[tex]^{a}[/tex]X[tex]^{b}[/tex] = 0

In order for them to be orthogonal...

g(sub a b)X[tex]^{a}[/tex]Y[tex]^{a}[/tex] = 0

I know from the line element of Minkowski space-time you have

ds[tex]^{2}[/tex] = dt[tex]^{2}[/tex] - dx[tex]^{2}[/tex] - dy[tex]^{2}[/tex] - dz[tex]^{2}[/tex]

I have played around with substituting the line element into the equation that specifies if it is orthogonal, but to no success. Any help would be great!

Sorry for the bad writting, I couldn't get latex to do subscripts, it only did them in superscripts....

Thanks,

Chris
 

Answers and Replies

  • #2
PAllen
Science Advisor
8,874
2,085
Here's a possible argument. Treat each 4 vector as e.g. (x0, r) where r is a 3 vector, x0 the timelike component. Let Y be a null vector, X be timelike vector. Then we have:

(1) Y0^2 - ry dot ry = 0, where dot is euclidean 3 dot product.

(2) X0 Y0 - rx dot ry = 0

(3) X0^2 - rx dot rx > 0

From which:

x0 > norm(rx) , 3 norm, euclidean, by (3)

Y0 = norm (ry) by (1)

X0 Y0 > norm(rx) norm(ry) >= rx dot ry

which contradicts (2).
 
  • #3
purakanui
10
0
thanks =)
 

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