I am reading Susskinds book General Relativity: The theoretical minimum and I got a bit stuck on the transformation rule of contravariant components. The book defines the components of a vector $(V^{’})^m=\frac{\partial Y^m}{\partial X^p}V^p$ where the primed system corresponds to changing from coordinates X to Y. My question is this: a contravariant vector is supposed to transform contrary to be base, but if I let the new basis Y be defined as $Y^m=2 X^m$ that is every basis vector is twice as long, however I do the calculation I find that my components also become twice as large according to the rule above. And from my understanding it is supposed to be the opposite. Where do I go wrong?


1 Answer 1


A tangent vector $\dot\lambda_p =(d\lambda /dt)_p $ to a curve parameterised by $t$ at a point $p$ on it is the map from the set of real functions $f$ defined in neighbourhood of $p$ to $\mathbb{R}$ defined by $$ \dot\lambda_p :f \mapsto \left[ d(f\circ \lambda)/dt\right]_p $$ Given a chart $\phi$ with coordinates $x^i$, the components of $\dot\lambda_p$ w.r.t. chart are $$\dot{ [ x^i \circ \lambda]_p} = \left[\frac{d}{dt}x^i(\lambda(t))\right]_p $$ Denote $\dot\lambda_k(0) =\left(\frac{\partial}{\partial x^k}\right)_p$ and from this any tangent vector $X_p$ (p is denoting point) as $$X_p f=\dot{(f\circ \lambda)}(0)=\dot{[f\circ\phi^{-1} \circ \phi \circ \lambda]} =\sum_k\left(\frac{\partial}{\partial x^k}\right) \left (f\circ\phi^{-1}\right)\dot{\left( x^k \circ \lambda \right)}(0)=\sum_k\left(\frac{\partial}{\partial x^k}\right)_p f \left(X_p x^k\right)$$

So $$X_p = \sum_k \left(X_p x^k\right)\left(\frac{\partial}{\partial x^k}\right)_p$$ The $X_p x^k$ are just the components of $X_p$. Now on using Einstein's summation convention

A vector $X_p=X^k \left(\frac{\partial}{\partial x^k}\right)_p$

Now if we change the coordinates by a transformation $x^i \mapsto y^i(x^m)$

And consider $$X_p f = X^k \left(\frac{\partial f}{\partial x^k}\right)_p = X^k \left(\frac{\partial y^m}{\partial x^k}\right) \left(\frac{\partial f}{\partial y^m}\right)_p = X'^m \left(\frac{\partial f}{\partial y^m}\right)_p $$

Where $X'^m = X^k \left(\frac{\partial y^m}{\partial x^k}\right)$

So what's written in your book is correct.


Now if you consider $y^i =2x^i$ you get $$X'^m = X^k \left(\frac{\partial y^m}{\partial x^k}\right)=2 X^m$$ But basis $$\left(\frac{\partial f}{\partial y^m}\right)_p= \left(\frac{\partial f}{2\partial x^m}\right)_p $$ Which makes $$X_p=X'^m \left(\frac{\partial f}{\partial y^m}\right)_p=2X^k \left(\frac{\partial f}{2\partial x^k}\right)_p=X^k \left(\frac{\partial f}{\partial x^k}\right)_p $$ Which is invariant and your basis actually halves with that coordinate transformation.

  • $\begingroup$ I kind of lied to you in last part, because am lazy. But you sure can get idea. $\endgroup$
    – Pradyuman
    Mar 22 at 10:21
  • $\begingroup$ In the last part before edit I took it as normal differential operator on $\mathbb{R}^n$ which is not the case. It is an operator over some manifold. $\endgroup$
    – Pradyuman
    Mar 22 at 10:47
  • $\begingroup$ Aah I see! So I agree that the formula is correct. But when we write $X^{’m} = 2 X^m$ doesn’t that imply that my components in X’ is now double those in X? But since the basis is twice as long I expected components to be half. $\endgroup$ Mar 22 at 11:20
  • $\begingroup$ You see when components are doubled it's natural for the basis to be halved to keep the vector invariant. $\endgroup$
    – Pradyuman
    Mar 22 at 11:28
  • $\begingroup$ Your choice of coordinate transformation actually halves the basis. $\endgroup$
    – Pradyuman
    Mar 22 at 11:29

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