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

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

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

<br /> \frac{\partial}{\partial X_i} \left(<br /> \frac{\partial X}{\partial \tau}<br /> \cdot<br /> \frac{\partial X}{\partial \sigma}<br /> \right)^2 =<br /> 4<br /> \left(<br /> \frac{\partial X}{\partial \tau}<br /> \cdot<br /> \frac{\partial X}{\partial \sigma}<br /> \right)<br /> \frac{\partial X_i}{\partial \tau} \frac{\partial X_i}{\partial \sigma}<br />

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 <br /> \frac{\partial}{\partial X_i} \left(<br /> \frac{\partial X}{\partial \tau}<br /> \right)<br /> =<br /> \left( \mathbf{\hat{e}}_i \frac{\partial X_i}{\partial \tau} \right)<br /> ?

 

[Hurkyl]

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

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 \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.