I Square of the difference of four-vectors

Do you know how to take the dot product of two vectors when the metric isn't trivial? (If I'm allowed to phrase it that way.)

 

And do you know how to expand that? In Euclidean 3-space ##\vec a.\vec b=a_xb_x+a_yb_y+a_zb_z##. Do you know the equivalent Minkowski 4-space expression?

 

Right. So when you replace the vector with a difference of two vectors what do you get?

 

First, vector subtraction against an orthonormal basis is always just what you would expect.

Second, you don't need to do this anyway because dot product distributes over vector addition/subtraction.

 

You wrote down an expression for the dot product of two vectors in terms of the components. So if the vectors, instead of ##p_3## and ##p_4##, are both ##p##, and ##p=p_3-p_4## what do you get?

 

Perhaps it would be helpful if you either completed the algebra we've been discussing or said what you think is wrong with 4.29 in the pdf you linked.

 

You mean your expressions in #7? Neither appears to be correct, as PeterDonis said in #8. How did you get them? I haven't tried to replicate 4.29 from your link, but it looks plausible at a quick glance.

 

The issue is that for everyone who has responded on this thread so far, your problem is trivial, yet you seem to have misunderstandings. We feel it is much more useful to help you resolve these than simply give you the answer.

 

I did replicate that equation, just doing the algebra in my head.

 

Thank you. I was confident up to sign errors on the cross terms - it's been a long day...