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sciencegem
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Hi,
This is a worked example in the text I'm independently studying. I hope this isn't too much to ask, but I am stupidly having trouble understanding how one step leads to the other, so was hoping someone could give me a little more of an in-depth idea of the derivation. Thanks.
Consider waves on a string of mass m and length l. Let us define the mass density p=m/l, tension T and displacement from the equilibrium ψ(x,t). The kinetic energy T can then be written as T=(1/2)∫[itex]^{l}_{0}[/itex]dxp([itex]\partial[/itex]ψ/[itex]\partial[/itex]t)^2 and the potential energy V=(1/2)∫[itex]^{l}_{0}[/itex]dxT([itex]\partial[/itex]ψ/[itex]\partial[/itex]x)^2. The action is then
S|ψ(x,t)|=∫dt(T-V)=∫dtdxL(ψ,[itex]\partial[/itex]ψ/[itex]\partial[/itex]t,[itex]\partial[/itex]ψ/[itex]\partial[/itex]x)
where
L(ψ,[itex]\partial[/itex]ψ/[itex]\partial[/itex]t,[itex]\partial[/itex]ψ/[itex]\partial[/itex]x)=p/2([itex]\partial[/itex]ψ/[itex]\partial[/itex]t)^2 - T/2([itex]\partial[/itex]ψ/[itex]\partial[/itex]x)^2
is the Lagrangian density. We then have immediately
0=δS/δψ=[itex]\partial[/itex]L/[itex]\partial[/itex]ψ - (d/dx)[itex]\partial[/itex]L/[itex]\partial[/itex]([itex]\partial[/itex]ψ/[itex]\partial[/itex]x) - (d/dt)[itex]\partial[/itex]L/[itex]\partial[/itex]([itex]\partial[/itex]ψ/[itex]\partial[/itex]t)
=0 + T([itex]\partial[/itex]^2ψ/[itex]\partial[/itex]x^2) - p([itex]\partial[/itex]^2ψ/[itex]\partial[/itex]t^2)
Of which the wave equation falls out effortlessly.
This is a worked example in the text I'm independently studying. I hope this isn't too much to ask, but I am stupidly having trouble understanding how one step leads to the other, so was hoping someone could give me a little more of an in-depth idea of the derivation. Thanks.
Homework Statement
Consider waves on a string of mass m and length l. Let us define the mass density p=m/l, tension T and displacement from the equilibrium ψ(x,t). The kinetic energy T can then be written as T=(1/2)∫[itex]^{l}_{0}[/itex]dxp([itex]\partial[/itex]ψ/[itex]\partial[/itex]t)^2 and the potential energy V=(1/2)∫[itex]^{l}_{0}[/itex]dxT([itex]\partial[/itex]ψ/[itex]\partial[/itex]x)^2. The action is then
S|ψ(x,t)|=∫dt(T-V)=∫dtdxL(ψ,[itex]\partial[/itex]ψ/[itex]\partial[/itex]t,[itex]\partial[/itex]ψ/[itex]\partial[/itex]x)
where
L(ψ,[itex]\partial[/itex]ψ/[itex]\partial[/itex]t,[itex]\partial[/itex]ψ/[itex]\partial[/itex]x)=p/2([itex]\partial[/itex]ψ/[itex]\partial[/itex]t)^2 - T/2([itex]\partial[/itex]ψ/[itex]\partial[/itex]x)^2
is the Lagrangian density. We then have immediately
0=δS/δψ=[itex]\partial[/itex]L/[itex]\partial[/itex]ψ - (d/dx)[itex]\partial[/itex]L/[itex]\partial[/itex]([itex]\partial[/itex]ψ/[itex]\partial[/itex]x) - (d/dt)[itex]\partial[/itex]L/[itex]\partial[/itex]([itex]\partial[/itex]ψ/[itex]\partial[/itex]t)
=0 + T([itex]\partial[/itex]^2ψ/[itex]\partial[/itex]x^2) - p([itex]\partial[/itex]^2ψ/[itex]\partial[/itex]t^2)
Of which the wave equation falls out effortlessly.