QC 6.4 Zwiebach: Deriving the Relativistic Dot Product

[ehrenfest]

Homework Statement

Zwiebach QC 6.4

Why is

$$\frac{\partial}{\partial{X^{\mu}}} \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) ^2 = \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) \cdot \frac{\partial{X^{\mu}}}{\partial{\tau}}$$

The dot is the relativistic dot product.
What I am confused about is how you take the derivative with respect to one component X^mu of the spacetime vectors and get a quantity that still has $$\left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right)$$. Is there an intermediate step someone could show me?

Homework Equations

The Attempt at a Solution

 

[Hurkyl]

Could you explain just what kind of object everything is?

If I just lively think of X as a function of $\tau$ and $\sigma$ with values in a 4-dimensional vector space with the Minkowski inner product, then the partial derivative w.r.t. the i-th component is

$$\frac{\partial}{\partial X_i} \left( \frac{\partial X}{\partial \tau} \cdot \frac{\partial X}{\partial \sigma} \right) = \left( \mathbf{\hat{e}}_i \frac{\partial X_i}{\partial \tau} \right) \cdot \frac{\partial X}{\partial \sigma} + \frac{\partial X}{\partial \tau} \cdot \left( \mathbf{\hat{e}}_i \frac{\partial X_i}{\partial \sigma} \right) = 2\frac{\partial X_i}{\partial \tau} \frac{\partial X_i}{\partial \sigma}$$

($\mathbf{\hat{e}}_i$ is a standard basis vector) Applying the chain rule would give

$$\frac{\partial}{\partial X_i} \left( \frac{\partial X}{\partial \tau} \cdot \frac{\partial X}{\partial \sigma} \right)^2 = 4 \left( \frac{\partial X}{\partial \tau} \cdot \frac{\partial X}{\partial \sigma} \right) \frac{\partial X_i}{\partial \tau} \frac{\partial X_i}{\partial \sigma}$$

Not quite what you have, but maybe you can see how your things are different from my things and fill in the gap.

 

[ehrenfest]

Sorry. My post was rather terse.

$$X^{\mu}(\tau,\sigma)$$ is the mapping function from parameter space (of the parameters tau and sigma) into spacetime. So, in this context what is the basis vector (it is just some pseudo-Euclidean basis vector probably)? Is there a way to do this without going into basis vectors? Why does $$\frac{\partial}{\partial X_i} \left( \frac{\partial X}{\partial \tau} \right) = \left( \mathbf{\hat{e}}_i \frac{\partial X_i}{\partial \tau} \right)$$ ?

 

[Hurkyl]

Because $X = (X_0, X_1, X_2, X_3)$. If you differentiate it with respect to, say, component 2, you get
$$\begin{equation*} \begin{split} \frac{\partial X}{\partial X_2} &= \frac{\partial}{\partial X_2} (X_0, X_1, X_2, X_3) \\&= \left(\frac{\partial X_0}{\partial X_2}, \frac{\partial X_1}{\partial X_2},\frac{\partial X_2}{\partial X_2}, \frac{\partial X_3}{\partial X_2} \right) \\&= (0, 0, 1, 0) = \mathbf{\hat{e}}_2 \end{split} \end{equation*}$$

Hrm, I know I did something wrong last night, but I'm still too sleepy to spot it at the moment...

 

[Jimmy Snyder]

$$\frac{\partial}{\partial{X^{\mu}}} \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) ^2 = \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) \cdot \frac{\partial{X^{\mu}}}{\partial{\tau}}$$

You missed a dot (and/or a prime). The equation is:

$$\frac{\partial}{\partial{\dot{X}^{\mu}}} \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) ^2 = \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) \cdot \frac{\partial{X^{\mu}}}{\partial{\tau}}$$
where [itex]\dot{X}^{\mu} is defined in equation (6.40) on page 100.

 

[Jimmy Snyder]

$$\frac{\partial}{\partial{X^{\mu}}} \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) ^2 = \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) \cdot \frac{\partial{X^{\mu}}}{\partial{\tau}}$$

You missed a dot (and/or a prime). The equation is:

$$\frac{\partial}{\partial{\dot{X}^{\mu}}} \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) ^2 = \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) \cdot \frac{\partial{X^{\mu}}}{\partial{\tau}}$$
where $\dot{X}^{\mu}$ is defined in equation (6.40) on page 100.