- #1
test1234
- 12
- 2
Hi there, I'm having some difficulty in understanding how the change of variables by considering a retarded time frame can be obtained for this particular eqn I have.
Say I have this original equation,
<br /> \frac{\partial A}{\partial z} + \beta_1 \frac{\partial A}{\partial t}+ \frac{i \beta_2}{2} \frac{\partial^2 A}{\partial t^2}+\frac{\alpha}{2} A = i \gamma |A|^2 A<br />
Considering a frame of reference moving with speed \nu_g (i.e. group velocity), a new variable T can be defined as follows
T=t-\frac{z}{\nu_g} = t-\beta_1 z
where \beta_1=\frac{1}{\nu_g}
Apparently, the differential equation can be reduced to
<br /> \frac{\partial A}{\partial z} -\frac{i \beta_2}{2} \frac{\partial^2 A}{\partial T^2}+\frac{\alpha}{2} A = i \gamma |A|^2 A<br />
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From my understanding, the first order differential term \beta_1 \frac{\partial A}{\partial t} disappears because
<br /> \frac{\partial A}{\partial t} = \frac{\partial A}{\partial T} \times \frac{\partial T}{\partial t}<br />
Since
<br /> T =t-\frac{z}{\nu_g} = t-\beta_1 z \\<br /> \therefore \frac{\partial T}{\partial t} = 1-\frac{\partial}{\partial t} (\beta_1 z) \\<br /> = 1-\beta_1 \nu_g = 0 \phantom{0} (\because \beta_1=\frac{1}{\nu_g})<br />
Hence,
<br /> \frac{\partial A}{\partial t} = 0<br />
However, I can't wrap my head around how to transform the second derivative \frac{\partial^2 A}{\partial t^2}?
<br /> \frac{\partial^2 A}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial A}{\partial t})<br /> = \frac{\partial}{\partial t}(\frac{\partial A}{\partial T} \times \frac{\partial T}{\partial t})<br />
Which seems to give me zero following from the above argument for the first order differential term \frac{\partial A}{\partial t}...
Am I missing something here? I would really appreciate if anyone could help. Thanks in advance!
Say I have this original equation,
<br /> \frac{\partial A}{\partial z} + \beta_1 \frac{\partial A}{\partial t}+ \frac{i \beta_2}{2} \frac{\partial^2 A}{\partial t^2}+\frac{\alpha}{2} A = i \gamma |A|^2 A<br />
Considering a frame of reference moving with speed \nu_g (i.e. group velocity), a new variable T can be defined as follows
T=t-\frac{z}{\nu_g} = t-\beta_1 z
where \beta_1=\frac{1}{\nu_g}
Apparently, the differential equation can be reduced to
<br /> \frac{\partial A}{\partial z} -\frac{i \beta_2}{2} \frac{\partial^2 A}{\partial T^2}+\frac{\alpha}{2} A = i \gamma |A|^2 A<br />
----------------------------------------------------------------------------
From my understanding, the first order differential term \beta_1 \frac{\partial A}{\partial t} disappears because
<br /> \frac{\partial A}{\partial t} = \frac{\partial A}{\partial T} \times \frac{\partial T}{\partial t}<br />
Since
<br /> T =t-\frac{z}{\nu_g} = t-\beta_1 z \\<br /> \therefore \frac{\partial T}{\partial t} = 1-\frac{\partial}{\partial t} (\beta_1 z) \\<br /> = 1-\beta_1 \nu_g = 0 \phantom{0} (\because \beta_1=\frac{1}{\nu_g})<br />
Hence,
<br /> \frac{\partial A}{\partial t} = 0<br />
However, I can't wrap my head around how to transform the second derivative \frac{\partial^2 A}{\partial t^2}?
<br /> \frac{\partial^2 A}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial A}{\partial t})<br /> = \frac{\partial}{\partial t}(\frac{\partial A}{\partial T} \times \frac{\partial T}{\partial t})<br />
Which seems to give me zero following from the above argument for the first order differential term \frac{\partial A}{\partial t}...
Am I missing something here? I would really appreciate if anyone could help. Thanks in advance!
