How to Prove |\DeltaE/E| = |\Delta\lambda/\lambda|<<1?

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The discussion focuses on proving the relationship |\DeltaE/E| = |\Delta\lambda/\lambda| << 1. Participants explore deriving expressions for Δλ in terms of E and ΔE, ultimately confirming that Δλ/λ = ΔE/E. The conversation emphasizes that if ΔE/E is much less than one, then Δλ/λ will also be much less than one, validating the equality. Clarifications about derivatives and the meaning of notation are also addressed. The conclusion affirms that the proof is satisfactory and all parts of the homework are understood.
ProPatto16
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Homework Statement



show that |\DeltaE/E| = |\Delta\lambda/\lambda|<<1

Homework Equations



\DeltaE>hbar/2pi\Deltat

\lambda=hc/E

The Attempt at a Solution



dunno where to start.
 
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Start by finding an expression for Δλ in terms of E and ΔE.
 
\Delta\lambda=(\DeltaE/E)\lambda

...
 
ProPatto16 said:
\Delta\lambda=(\DeltaE/E)\lambda

...
If λ = hc/E, what is dλ? Can you find the differential?
 
you mean the derivitive of the LHS? so \lambda d\lambda ?? that'd just be 1?
 
Can you find dλ/dE? It is not 1.
 
ah okay just wasnt sure what you were asking.

d\lambda/dE would be \lambdad\lambda = hc/E dE

which would be 1 = -hc/E2 ?
 
no wait. d\lambda wouls be a rate of change so then \Delta\lambda = -hc/E2 ?
 
ProPatto16 said:
ah okay just wasnt sure what you were asking.

d\lambda/dE would be \lambdad\lambda = hc/E dE

which would be 1 = -hc/E2 ?
How do you figure?
One more time. What is

\frac{d}{dE}\left( \frac{hc}{E} \right)
 
  • #10
ProPatto16 said:
no wait. d\lambda wouls be a rate of change so then \Delta\lambda = -hc/E2 ?
Not quite. Δλ = -(hc/E2) ΔE. Now can you find Δλ/λ ?
 
  • #11
oh gee. i feel so dumb -.- this should be so easy.

the notation i use for this differential would be:

hc/E dE
= hcE-1 dE
= (-1)hcE-2
= (-hc)/E2

how else would i derive that with respect to e :/... and i just left h and c as is since theyre constant?
 
  • #12
ProPatto16 said:
oh gee. i feel so dumb -.- this should be so easy.

the notation i use for this differential would be:

hc/E dE
= hcE-1 dE
This is correct except you need E2 in the denominator.
= (-1)hcE-2
= (-hc)/E2
This is not. You just can't drop the dE.
 
  • #13
oh so i did do it right.

well then, \Delta\lambda/\lambda = [(-hc/E2)\DeltaE]/\lambda
 
  • #14
oh i think i got it!... that expands to leave (\DeltaE/E2)(-hc/\lambda) which is \LambdaE/E so then \Delta\lambda/\lambda=\LambdaE/E
 
  • #15
yes i meant to put in deltaE
 
  • #16
whats the relevance of |\Delta\lambda/\lambda|<<1?

does << mean less than? or something else?
 
  • #17
ProPatto16 said:
whats the relevance of |\Delta\lambda/\lambda|<<1?

does << mean less than? or something else?
"<" means "less than"; "<<" means "much less than"; "<<<" means "even less than that".
 
  • #18
well in this particular question deltaE = 4.1neV and E=2.58eV so deltaE/E would be much much much much less than one. which shows that the equality would only hold if deltalambda/lambda was equaly small. would that be sufficient reason?
 
  • #19
You have shown that Δλ/λ = ΔΕ/Ε. This equality holds no matter what. If ΔΕ/Ε << 1, then Δλ/λ << 1 no matter what.
 
  • #20
So it's all good then :)
 
  • #21
ProPatto16 said:
So it's all good then :)
Unless there is more.
 
  • #22
Nah the other parts to the question I could do. I found all the answers just wasn't sure how to actually show that proof. So thanks :)
 

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