- #1
harjyot
- 42
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I was trying to Go from the uncertainty principle to its energy-time counter part. i know the maths is a bit off,but the idea is correct?
dx=position
p=momentum
e=energy
[itex]\upsilon[/itex]=frequency
[itex]\lambda[/itex]=wavelength
c=velocity of electromagnetic radiations
dt=time
now ,
[itex]\lambda[/itex]=h/p....(i)
c=[itex]\upsilon[/itex].[itex]\lambda[/itex]....(ii)
e=h.[itex]\upsilon[/itex]
e=(h.c)/[itex]\lambda[/itex]
replacing [itex]\lambda[/itex]'s value here from (i)
e=(h.c)/(h/p)
e=c.p
now c = velocity of light , it can be written as dx/dt
e= (dx/dt).p
multiplying by dt on both sides
e.dt=(dx/dt).dt.p
e.dt=dx.p
Therefore frome this relation if we straight away incorporate this in place of the
σx.σp≥h/4π
cannot we get
σe.σt≥4π
dx=position
p=momentum
e=energy
[itex]\upsilon[/itex]=frequency
[itex]\lambda[/itex]=wavelength
c=velocity of electromagnetic radiations
dt=time
now ,
[itex]\lambda[/itex]=h/p....(i)
c=[itex]\upsilon[/itex].[itex]\lambda[/itex]....(ii)
e=h.[itex]\upsilon[/itex]
e=(h.c)/[itex]\lambda[/itex]
replacing [itex]\lambda[/itex]'s value here from (i)
e=(h.c)/(h/p)
e=c.p
now c = velocity of light , it can be written as dx/dt
e= (dx/dt).p
multiplying by dt on both sides
e.dt=(dx/dt).dt.p
e.dt=dx.p
Therefore frome this relation if we straight away incorporate this in place of the
σx.σp≥h/4π
cannot we get
σe.σt≥4π