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Hartle Gravity - Simple basis vector question

  1. Jan 16, 2010 #1
    Helo all,

    I have a very simple question about basis Four-Vectors and Components. In Hartle's book, Gravity, he uses the following equation to show the components of the 4-vector, a

    a =a[tex]^t{}[/tex]e(sub t) + a[tex]^x{}[/tex]e(sub x) + a[tex]^y{}[/tex]e(sub y) + a[tex]^z{}[/tex]e(sub z)

    Sorry for the half LaTex half something else but I couln't get the subscript LaTex command to work right. All it did was create another superscript.

    Here is my question:

    The e(subs) are a unit vector so there value should be -1 for the t component and 1 for the x, y, and z components correct?

    For example, (this is problem 5.1)

    The components of the 4-vector a[tex]^\alpha{}[/tex] are (-2,0,0,1)

    Is a timelike, spacelike, or null?

    The value of a[tex]\alpha[/tex] = -2(-1) + 0(1) + 0(1) + 1(1) = -1

    Since -1 < 0, a is timelike.

    Is the above correct ?

    Thanks
    Matt
     
    Last edited: Jan 16, 2010
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  3. Jan 16, 2010 #2

    Fredrik

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    It's not. If this is special relativity, the basis vectors are just the standard basis vectors of [itex]\mathbb R^4[/itex]. [itex]e_0=(1,0,0,0),\ e_1=(0,1,0,0)[/itex] and so on. If it's general relativity, we're talking about basis vectors for the tangent space, and every coordinate system defines a basis as described here.

    To determine if a four-vector u is timelike, null or spacelike, you must check if [itex]u^2=g_{\alpha\beta}u^\alpha u^\beta[/itex] is <0, =0, or >0 respectively. If we're talking about the Minkowski metric with the -+++ convention, you have [itex]u^2=-(u^0)^2+(u^1)^2+(u^2)^2+(u^3)^2[/itex].
     
  4. Jan 16, 2010 #3
    Fredrik,

    Thanks.

    The "unit" vector was causing me the confusion.

    So the correct calculation for the value of a is

    The value of a = -(-2(-2))+ 0(0) + 0(0) + 1(1) = -3

    and a is timelike.

    glamotte was also helping me out with this but the "unit" vector was causing me some confusion.

    Thanks
    Matt
     
  5. Jan 16, 2010 #4

    Fredrik

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    It's the correct calculation, but what you're calculating isn't a, it's a2:

    [tex]a^2=a^T\eta a=-(a^0)^2+(a^1)^2+(a^2)^2+(a^3)^2=(-2)^2+0^2+0^2+1^2=-3<0[/tex]
     
  6. Jan 16, 2010 #5
    Yes, sorry for that. I understood it to be a^2, I just didn't show it in my post.

    Thanks for the correction.

    Matt
     
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