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ehrenfest
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Homework Statement
equation 12.28 is only true if [tex]\dot{X'^j} [/tex] is equal to 0, right? Why is that true?
ehrenfest said:Homework Statement
equation 12.28 is only true if [tex]\dot{X'^j} [/tex] is equal to 0, right? Why is that true?
Homework Equations
The Attempt at a Solution
ehrenfest said:but isn't the only way that
[tex] \frac{d}{d\sigma} (X^I \dot{X^J}-\dot{X}^J X^I) = X^I' \dot{X^J}-\dot{X}^J X^I'[/tex]
if \dot{X^J}' equal 0
?
ehrenfest said:I think I see now. It should really be
[tex] \frac{d}{d\sigma} (X^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X^I(\tau, \sigma)) [/tex]
and the sigma derivative of a function of sigma prime is 0.
nrqed said: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'.
ehrenfest said:I think maybe you are missing the dot on top of the x^J. That represents a tau derivative.
There si no d/dsigma on the left sideehrenfest said:[tex]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 ') [/tex]
is equation 12.27 expanded
[tex] \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 ') [/tex]
is equation 12.28 expanded
12.28 follows from 12.27 only if [tex] \frac{d}{d\sigma}(\dot{X}^J(\tau, \sigma')) = 0[/tex] which is true only because the sigma argument has a prime but the sigma in the derivative operator has no prime.
The number 12.28 is considered true because it is the value of a physical quantity that has been measured and verified through experimentation or observation. In this case, it is most likely the value of the velocity, represented by \dot{X'^j}
, which has been found to be equal to 0.
The symbol \dot{X'^j}
represents the derivative of the position vector X'^j
with respect to time. In other words, it represents the rate of change of the position over time, which is often interpreted as the velocity.
The value of 12.28 is most likely obtained through experimental or observational data. Scientists use various methods and instruments to measure physical quantities, such as velocity, and the value of 12.28 is the result of these measurements.
No, the value of 12.28 may not always be true for this equation. It depends on the specific circumstances and conditions under which the measurement was taken. The value of the velocity, represented by \dot{X'^j}
, may vary depending on factors such as the object's mass, environment, and external forces acting on it.
When \dot{X'^j}
equals 0, it means that the rate of change of the position vector, and therefore the velocity, is 0. This can have different interpretations depending on the context, but it generally means that the object is not moving or that its motion is constant and unchanging.