Skip to article frontmatterSkip to article content
Site not loading correctly?

This may be due to an incorrect BASE_URL configuration. See the MyST Documentation for reference.

Appendix

Notation

Some useful math symbols

Derivatives and partial derivatives

I will often write

ixi\del_i \equiv \frac{\partial}{\partial x^i}

Some useful tensors

  1. The Kronecker delta function. If I,J=i,,NI,J = i,\ldots,N are some indices then

δJI={1I=J0IJ\delta^I_J = \begin{cases} 1 & I = J \\ 0 & I \neq J \end{cases}
  1. The totally antisymmetric tensor

In dd dimensions, the tensor ϵi1,,id\epsilon_{i_1,\ldots, i_d} is totally antisymmetric in the exchange of all indices. Thus the only nonvanishing components are ones for which i1,,idi_1,\ldots,i_d are a permutation of 1,2,,d1,2,\ldots,d. We fix the tensor completely by edemanding that ϵ1,2,,d=1\epsilon_{1,2,\ldots,d} = 1.

Einstein summation convention

The Einstein summation convention is that all repeated indices are summed over unless otherwise stated explicitly. Thus, for some dd-dimensional vectors ViV^i, WiW^i

ViWij=1dVjWjV^i W^i \equiv \sum_{j = 1}^d V^j W^j

Technically speaking I am being a little careless. In curved spaces (for example, on the sphere), there are geometrically distinct objects ViV^i, ViV_i, and I should only be summing over pairs of indices in which one is raised and one is lowered. In Euclidean space, however, there is a standard map which states that numerically, Vi=ViV^i = V_i, so I will ignore this issue for the time being.

(We can think of Vi,WiV^i, W^i as elements of a vector space, and Vi,WiV_i, W_i their dual vectors; we will discuss this language when we get to linear algebra).

Derivatives of expressions involving the Einstein summation convention often confuse people. For a general vector field VI(x)V^I(x),

xJVIWI=WIJVI+VIJWI \frac{\del}{\del x^J} V^I W_I = W^I \del_J V^I + V^I \del_J W^I

Now if VI=WI=xIV^I = W^I = x^I, then JxI=δJI\del_J x^I = \delta^I_J. You can convince yourselves that

δJIVI=VJ\delta^I_J V^I = V^J

and therefore,

J(xIxI)=2xJ\del_J (x^I x^I) = 2 x^J

More on the totally antisymmetric tensor

  1. Contraction identities in dd dimensions A. ϵi1,,idϵi1,,id=d!\epsilon_{i_1,\ldots,i_d}\epsilon^{i_1,\ldots,i_d} = d!. B. In d2d - 2, ϵijϵik=δjk\epsilon_{ij} \epsilon^{ik} = \delta^k_j. C. In d=3d = 3, ϵkmnϵkij=δmiδnjδniδmj\epsilon_{kmn}\epsilon^{k i j} = \delta^i_m \delta^j_n - \delta^i_n \delta^j_m. D. In d=3d = 3, ϵijkϵijl=2δkl\epsilon_{ijk}\epsilon^{ijl} = 2\delta^l_k. E. There are similar identities in higher dimensions, which we will leave aside for now.

  2. Definition of cross product.

For general dd, the cross product of two vectors UiViU^i V^i is a d2d-2-rank tensor:

(U×V)i1,,id2=ϵi1,,id2,id1,idUid1Vid({\vec U} \times {\vec V})_{i_1,\ldots,i_{d-2}} = \epsilon_{i_1,\ldots,i_{d-2}, i_{d--1}, i_d} U^{i_{d-1}} V^{i_d}

For d=2d = 2, this is a scalar (actually a “pseudo-scalar”):

(U×V)=ϵijUiVj=A1V2U2V1({\vec U}\times {\vec V}) = \epsilon_{ij} U^i V^j = A^1 V^2 - U^2 V^1

For d=3d = 3,

(U×V)i=ϵijkUjVk({\vec U}\times {\vec V})^i = \epsilon_{ijk} U^j V^k

Or alternatively,

UiVjUjVi=ϵijk(U×B)kU^i V^j - U^j V^i = \epsilon_{ijk} ({\vec U}\times {\vec B})^k

Some useful functions

  1. The Heaviside step function θ(x)\theta(x):

θ(x)={0,if x<01,ifx>0\theta(x) = \begin{cases} 0, & \text{if}\ x < 0 \\ 1, & \text{if} x > 0 \end{cases}