# Hartle Gravity - Simple basis vector question

1. Jan 16, 2010

### CFDFEAGURU

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$$^t{}$$e(sub t) + a$$^x{}$$e(sub x) + a$$^y{}$$e(sub y) + a$$^z{}$$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$$^\alpha{}$$ are (-2,0,0,1)

Is a timelike, spacelike, or null?

The value of a$$\alpha$$ = -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
2. Jan 16, 2010

### Fredrik

Staff Emeritus
It's not. If this is special relativity, the basis vectors are just the standard basis vectors of $\mathbb R^4$. $e_0=(1,0,0,0),\ e_1=(0,1,0,0)$ 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 $u^2=g_{\alpha\beta}u^\alpha u^\beta$ is <0, =0, or >0 respectively. If we're talking about the Minkowski metric with the -+++ convention, you have $u^2=-(u^0)^2+(u^1)^2+(u^2)^2+(u^3)^2$.

3. Jan 16, 2010

### CFDFEAGURU

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

4. Jan 16, 2010

### Fredrik

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

$$a^2=a^T\eta a=-(a^0)^2+(a^1)^2+(a^2)^2+(a^3)^2=(-2)^2+0^2+0^2+1^2=-3<0$$

5. Jan 16, 2010

### CFDFEAGURU

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