1.6: Index Notation
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You may be familiar with something called a dot product, which is a way of multiplying two vectors together. By definition, the dot product of two vectors →a and →b is usually written as
→a⋅→b=axbx+ayby+azbz.
If you are familiar with linear algebra, you may know that the previous expression can also be written as
→a⋅→b=(axayaz)(100010001)(bxbybz).
If you aren't familiar with matrix multiplication in linear algebra, you may instead think of it as
→a⋅→b=3∑i,j=1ηijaibj,
where
ηij={1,i=j0,i≠j
is called the metric of the space and where the indices 1-3 represent the spatial components. If you take the dot product of a vector with itself, then you end up with the Pythagorean Theorem, which means that the metric essentially tells you how to find the length of a line. More generally, the metric defines the rules of geometry.
The metric is a function or matrix that can be used to determine the distance between two points. It can be thought of as defining the rules of geometry.
What is the relevance of all this? Recall that the spacetime interval is defined as
Δτ2=Δt2−Δx2−Δy2−Δz2,
which suggests that it can also be written as
Δτ2=(ΔtΔxΔyΔz)(10000−10000−10000−1)(ΔtΔxΔyΔz)=3∑μ,ν=0ημνΔxμΔxν,
where
ημν={1,μ=ν=0−1,μ=ν=1,2,30,μ≠ν
and where x0=t, x1=x, x2=y, and x3=z (note that these are superscripts, not exponents).
Some texts use the opposite sign convention for the metric, where the time component is negative and the spatial components are positive. Both sign conventions work, but some equations involving the metric will look different depending on which sign convention you are using.
For the remainder of this book, we will assume the following conventions:
- If an index appears both "downstairs" and "upstairs," we can drop the summation symbol and assume the summation.
- Roman letters such as i and j are for spatial components only.
- Greek letters such as μ and ν are for all four spacetime components.
With this convention, the spacetime interval is
Δτ2=ημνΔxμΔxν.
The following rules and definitions will also be useful to us.
- Any index that is not summed over is called a free index.
- The free indices on both sides of an equation must be the same.
- Indices can be renamed.
- The same index can't appear downstairs more than once or upstairs more the once.
- Any index can be lowered using the metric (by definition). For example, xμ=ημνxν.
- uμuμ=1, where uμ is the four-velocity. (See Box 1.6.1)
- The inverse metric ημν is defined by ηαμημν=I, where I is the identity matrix (1's on the diagonal and 0's everywhere else). (See Box 1.6.2)
How is a vector with downstairs index different from its upstairs counterpart?
- Answer
-
One of our rules is that xμ=ημνxν. Note that the sum only occurs over ν, since that is the only index that appears both downstairs and upstairs. The index μ is called a free index because it can take on any value 0-3 (i.e. t, x, y, or z).
xμ=ημtxt+ημxxx+ημyxy+ημzxz
The result depends on the value of μ. Let's check each one.
xt=ηttxt+ηtxxx+ηtyxy+ηtzxz=xtxx=ηxtxt+ηxxxx+ηxyxy+ηxzxz=−xxxy=ηytxt+ηyxxx+ηyyxy+ηyzxz=−xyxz=ηztxt+ηzxxx+ηzyxy+ηzzxz=−xz
Therefore xμ=(t−x−y−z).
Note that xμ is exactly the same as xμ except that the signs of the spatial indices have been reversed. There is also nothing special about xμ; we could replace xμ with any four-vector and the result would be that the vector with downstairs index has the same components but with the signs of the spatial components reversed.
For each part below, indicate the indices that are free indices (or say "none").
a) aμνbμcν
b) aμνbνα
c) aαμaβνbμνcσ+aσμaανbμνcβ+aβμaσνbμνcα
d) aμνbμαbνβcαcβ
- Answer
-
a) none
b) μ,α
c) α,β, and σ
d) none
Which of the following violate the index rules?
a) aμ=bαcμβ
b) aμbμνcν+m2=0
c) aμbμν=m2
c) aαβ=aμνbμαbνβ
- Answer
-
a) Violates. Left side has on free index while right side has three.
b) No violation. The first time has no free indices, so it can be added to a scalar.
c) Violates. The left side has a free index while the right side does not.
d) Violates. α and β are free indices on both sides, but β is downstairs on one side and upstairs on the other.
Prove that uμuμ=1, where u is the four-velocity
Show that the inverse metric ημν=(10000−10000−10000−1). (Note that both indices are upstairs.)