Dot product of four vectors

  • #1
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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?

Thanks in advance.
 

Answers and Replies

  • #2
BvU
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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 :smile:.

And you are unhappy with co/contravariant notation too ?
 
  • #3
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Not sure I understand why you would like to keep signs positive ?

And you are unhappy with co/contravariant notation too ?
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
PeterDonis
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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.
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
DrGreg
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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.
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
Matterwave
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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.
A negative is missing there...
 
  • #7
nrqed
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A negative is missing there...
No, DrGreg's expression is correct.
 
  • #8
DrGreg
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A negative is missing there...
The one or three negatives (depending on your metric sign convention) are hiding inside the subscript notation.
 
  • #9
Matterwave
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Yeah, you're right. :)
 
  • #10
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If you really think you need to calculate the expression you gave, you've probably made a mistake in your calculation.
You're right - I did ;)

Thanks for all the replies nonetheless.
 

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