# Dot product of four vectors

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1. Dec 16, 2015

### tomdodd4598

Hi there,

I understand that taking the dot product of two four vectors automatically applies the metric tensor to the second vector. Is there a way to take write the dot product, using vector notation, in a way which keeps the signs of all of the components positive?

2. Dec 16, 2015

### BvU

Not sure I understand why you would like to keep signs positive ?

For simple metrics like flat minkowsky you can use (x, y, z, ict) as four vectors and use the standard dot product. But now you have an imaginary component, which many will consider even worse than negative .

And you are unhappy with co/contravariant notation too ?

3. Dec 16, 2015

### tomdodd4598

I am ok with the co/contra notation, it's just that I have a situation where I have two four vectors, A and B (in a Lagrangian), and would like a nice way to write A⁰B⁰+A¹B¹+A²B²+A³B³.

4. Dec 16, 2015

### Staff: Mentor

Why do you want a nice way to write that quantity? It doesn't contain any physics; the physical quantity is the dot product using the standard metric.

5. Dec 16, 2015

### DrGreg

There's no nice way to write this because, relativistically speaking, that quantity has no useful interpretation, as it is coordinate-dependent and different observers would disagree what its value was. On the other hand$$g(\textbf{A}, \textbf{B}) = A_\mu B^\mu = A_0 B^0 + A_1 B^1 + A_2 B^2 + A_3 B^3$$makes sense. If you really think you need to calculate the expression you gave, you've probably made a mistake in your calculation.

6. Dec 16, 2015

### Matterwave

A negative is missing there...

7. Dec 16, 2015

### nrqed

No, DrGreg's expression is correct.

8. Dec 16, 2015

### DrGreg

The one or three negatives (depending on your metric sign convention) are hiding inside the subscript notation.

9. Dec 16, 2015

### Matterwave

Yeah, you're right. :)

10. Dec 16, 2015

### tomdodd4598

You're right - I did ;)

Thanks for all the replies nonetheless.