- #1
deuteron
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- TL;DR Summary
- What physical meaning does covariance and contravariance have?
Hi,
I am very confused about the mathematics related to special relativity.
I have understood, that a four-vector with an upper index has the form:
$$A^\mu = (A^0 , A^1, A^2, A^3)$$
where lowering the index would make the indices other than the ##0##th negative:
$$A_\mu = (A_0, -A^1, -A^2, -A^3)$$
In order to lower the index, we would need to multiply the four-vector with the Minkowski metric.
I understand, that the four vector ##x^\mu = (ct, x^1,x^2,x^3)## gives the temporal and spatial positions of an event.
What I don't understand is what "multiplying by the Minkowski metric" physically does to the four vector, and what the physical difference between ##x_0## and ##x^0## is. What physical difference does it make when the four-vector has a lower index or an upper index?
I am in general very confused about this, so any recommendation of a source on the mathematics of special relativity is very welcome
I am very confused about the mathematics related to special relativity.
I have understood, that a four-vector with an upper index has the form:
$$A^\mu = (A^0 , A^1, A^2, A^3)$$
where lowering the index would make the indices other than the ##0##th negative:
$$A_\mu = (A_0, -A^1, -A^2, -A^3)$$
In order to lower the index, we would need to multiply the four-vector with the Minkowski metric.
I understand, that the four vector ##x^\mu = (ct, x^1,x^2,x^3)## gives the temporal and spatial positions of an event.
What I don't understand is what "multiplying by the Minkowski metric" physically does to the four vector, and what the physical difference between ##x_0## and ##x^0## is. What physical difference does it make when the four-vector has a lower index or an upper index?
I am in general very confused about this, so any recommendation of a source on the mathematics of special relativity is very welcome