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
    Commented Mar 22, 2023 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
    Commented Mar 22, 2023 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. $\endgroup$ Commented Mar 22, 2023 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
    Commented Mar 22, 2023 at 11:28
  • $\begingroup$ Your choice of coordinate transformation actually halves the basis. $\endgroup$
    – Pradyuman
    Commented Mar 22, 2023 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.