# Homework Help: Are a and b spacelike, timelike, or null?

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1. Sep 12, 2015

### ope211

Hello! I am working on homework for my general relativity class, and I am very confused about how to tell the difference between spacelike, timelike, and null vectors, and the book is very unhelpful.

Relevant equations
Consider two four vectors a and b whose components are given by:
a^α=(-2, 0, 0, 1)
b^α=(5, 0, 3, 4)

The attempt at a solution

The book says: The length of a four vector is the absolute value of the space-time difference between its tail and tip. Four vectors that are spacelike have a tail and tip separation that is spacelike, timelike vectors have a separation that is timelike, and null vectors have null separation (length zero.)

I think I can rule out that these vectors are null, since they don't have length zero, but how do I tell the difference between timelike and spacelike vectors? As you can see, the book is extremely unhelpful and I've looked all over for examples but I can't find any. The only thing I found that makes a little sense is that one has a positive scalar product and the other a negative one, but how can I use this to understand both vectors?

I feel like this is a dumb question, but I hope you guys can help!

2. Sep 12, 2015

### TSny

Hello and welcome to PF!

Which of the four components refers to the time component, the first or the last?

Can you show us explicitly how you would calculate the length of each of your vectors?

3. Sep 12, 2015

### ope211

Hello,
Unfortunately the problem itself doesn't say, but the book's convention seems to be (t, x, y, z)

I calculate the length as |a|=sqrt((-2)^2+0+0+1^2)=sqrt(5)
and |b|=sqrt(5^2+0+3^2+4^2)=sqrt(50)

Another thing I was confused about was what it meant by "The length of a four vector is the absolute value of the space-time difference between its tail and tip," so I calculated it the way I would for any vector. Is it possible this is my problem?

4. Sep 12, 2015

### TSny

Lengths of four vectors in (flat) spacetime are calculated differently than the length of a vector in Euclidean space. When calculating the length (or square of the length) of a four vector, the time component is treated differently than the spatial components. However, different texts use different sign conventions when defining the square of the length of a four vector. Check your textbook or notes for the sign convention that you are using. Often the term "spacetime interval" is used for the square of the "distance" between two points in spacetime.

5. Sep 13, 2015

### ope211

I thought this might be true! I may just have to wait and see if my professor covers it in the lecture, because for the life of me I can't find anything else on the topic in the book and no examples anywhere else. I appreciate the help!