Why is 12.28 True if \dot{X'^j} = 0?

[ehrenfest]

Homework Statement

equation 12.28 is only true if $$\dot{X'^j}$$ is equal to 0, right? Why is that true?

Homework Equations

The Attempt at a Solution

 

[nrqed]

Homework Statement

equation 12.28 is only true if $$\dot{X'^j}$$ is equal to 0, right? Why is that true?

Homework Equations

The Attempt at a Solution

No. It follows from 12.27 which does not require anything like that

 

[ehrenfest]

but isn't the only way that

$$\frac{d}{d\sigma} (X^I \dot{X^J}-\dot{X}^J X^I) = X^I' \dot{X^J}-\dot{X}^J X^I'$$

if \dot{X^J}' equal 0

?

 

[nrqed]

but isn't the only way that

$$\frac{d}{d\sigma} (X^I \dot{X^J}-\dot{X}^J X^I) = X^I' \dot{X^J}-\dot{X}^J X^I'$$

if \dot{X^J}' equal 0

?

No. I am not sure why you conclude this. Notice that for I not equal to J, the expression in parenthesis is zero even before taking a derivative.

 

[ehrenfest]

I think I see now. It should really be

$$\frac{d}{d\sigma} (X^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X^I(\tau, \sigma))$$

and the sigma derivative of a function of sigma prime is 0.

 

[nrqed]

I think I see now. It should really be

$$\frac{d}{d\sigma} (X^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X^I(\tau, \sigma))$$

and the sigma derivative of a function of sigma prime is 0.

I am sorry, maybe I am too slow tonight (after 8 hours of marking) but I don't quite see what you mean. The expression in parenthesis is zero whenever I is not equal to J. Those X's are operators. And they commute when I is not equal to J. when I = J, they don't commute but give a delta function of sigma - sigma'.

 

[Jimmy Snyder]

Note that $\sigma$ and $\sigma'$ are independent variables in this equation.

 

[ehrenfest]

I am sorry, maybe I am too slow tonight (after 8 hours of marking) but I don't quite see what you mean. The expression in parenthesis is zero whenever I is not equal to J. Those X's are operators. And they commute when I is not equal to J. when I = J, they don't commute but give a delta function of sigma - sigma'.

I think maybe you are missing the dot on top of the x^J. That represents a tau derivative.

 

[nrqed]

I think maybe you are missing the dot on top of the x^J. That represents a tau derivative.

You are right, I forgot to say that it's the conmmutator of X with dot X. But my point is the same: X^I and (dot X)^J commute whenever I is not equal to J. So the derivative is zero trivially because the expression in parenthesis is zero before even taking thee derivative and not because there is no sigma dependence (well, there is sigma dependence trivially because it's zero!)

 

[ehrenfest]

$$X^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X^I(\tau, \sigma) = 2 \pi \alpha' \eta^{IJ} \delta(\sigma - \sigma ')$$
is equation 12.27 expanded

$$\frac{d}{d\sigma} (X'^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X'^I(\tau, \sigma)) = 2 \pi \alpha' \eta^{IJ} \frac{d}{d \sigma}\delta(\sigma - \sigma ')$$
is equation 12.28 expanded

12.28 follows from 12.27 only if $$\frac{d}{d\sigma}(\dot{X}^J(\tau, \sigma')) = 0$$ which is true only because the sigma argument has a prime but the sigma in the derivative operator has no prime.

 

[nrqed]

$$X^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X^I(\tau, \sigma) = 2 \pi \alpha' \eta^{IJ} \delta(\sigma - \sigma ')$$
is equation 12.27 expanded

$$\frac{d}{d\sigma} (X'^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X'^I(\tau, \sigma)) = 2 \pi \alpha' \eta^{IJ} \frac{d}{d \sigma}\delta(\sigma - \sigma ')$$
is equation 12.28 expanded

There si no d/dsigma on the left side

12.28 follows from 12.27 only if $$\frac{d}{d\sigma}(\dot{X}^J(\tau, \sigma')) = 0$$ which is true only because the sigma argument has a prime but the sigma in the derivative operator has no prime.

Ok, I see what your question was. yes, of course, $\frac{d}{d \sigma} f(\sigma') = 0$. Sorry, I did not understand your question because this was implicit for me.