Help me derive the Doppler effect ratio

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SUMMARY

The discussion focuses on deriving the Doppler effect ratio, specifically the expression for ΔTo/ΔTs. The original equation provided is ΔTo = ΔTs + VΔTo / U. The user initially struggled with isolating ΔTo due to its presence on both sides of the equation. After several iterations and corrections, the final derived expression is ΔTo / ΔTs = U / (U - V), which simplifies the relationship between the observed and source time intervals.

PREREQUISITES
  • Understanding of algebraic manipulation
  • Familiarity with the Doppler effect concepts
  • Knowledge of variables representing time intervals (ΔTo, ΔTs)
  • Basic understanding of velocity (V) and its relation to time (U)
NEXT STEPS
  • Study the derivation of the Doppler effect in various contexts (e.g., sound, light)
  • Learn about the implications of the Doppler effect in astrophysics
  • Explore algebraic techniques for solving equations with multiple variables
  • Investigate applications of the Doppler effect in technology, such as radar and medical imaging
USEFUL FOR

Students studying physics, particularly those focusing on wave phenomena, as well as educators and anyone interested in the mathematical derivation of the Doppler effect.

Comscistudent
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Homework Statement


I'm watching this lecture and wanted to attempt to derive the expression for ΔTo/ΔTs

Homework Equations


ΔTo = ΔTs + VΔTo / U

The Attempt at a Solution


I have worked out that ΔTs = 1-V / U, however I'm stuck trying to get ΔTs / ΔTo because of the ΔTo on both sides of the original equation, it just turns into a recursive thing which I know must be wrong.

I'd appreciate any pointers that would help me get there myself rather than a solution. I'm only refreshing myself on algebra currently so I'm doubtless missing something simple
 
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Comscistudent said:

Homework Statement


I'm watching this lecture and wanted to attempt to derive the expression for ΔTo/ΔTs

Homework Equations


ΔTo = ΔTs + VΔTo / U

The Attempt at a Solution


I have worked out that ΔTs = 1-V / U
That's not what I get, starting from your relevant equation. What happened to ΔTo?
Please show us what you did.
Comscistudent said:
, however I'm stuck trying to get ΔTs / ΔTo because of the ΔTo on both sides of the original equation, it just turns into a recursive thing which I know must be wrong.

I'd appreciate any pointers that would help me get there myself rather than a solution. I'm only refreshing myself on algebra currently so I'm doubtless missing something simple
 
Sorry that was a mistake I forgot to divide the ΔTs by ΔTo. Now I've gotten this far but still end up with the variables on both sides

ΔTo = ΔTs + VΔTo / U
ΔTo = ΔTs / ΔTo + V / U
ΔTo / ΔTs = 1 / ΔTo + V/U / ΔTs
ΔTo / ΔTs = 1 / ΔTo + V / UΔTs
 
Comscistudent said:
Sorry that was a mistake I forgot to divide the ΔTs by ΔTo. Now I've gotten this far but still end up with the variables on both sides

ΔTo = ΔTs + VΔTo / U
ΔTo = ΔTs / ΔTo + V / U
The equation above is incorrect. You divided the right side by ΔTo, but not the left side. Any arithmetic operation you apply to one side of an equation, you must apply the same operation to the other side.
Instead, bring the terms with ΔTo to one side, and then factor ΔTo out.
Here's a start.
ΔTo - VΔTo / U = ΔTs
Now factor ΔTo from the two terms on the left side.

Comscistudent said:
ΔTo / ΔTs = 1 / ΔTo + V/U / ΔTs
ΔTo / ΔTs = 1 / ΔTo + V / UΔTs
 
Wow I'm stupid, thanks for putting up with me. I finally got there I think -

ΔTo - VΔTo / U = ΔTs
ΔTo( 1 - V / U ) = ΔTs
ΔTo / ΔTs ( 1 - V / U ) = 1
ΔTo / ΔTs = 1 / ( 1 - V / U )
ΔTo / ΔTs = (1 - V / U )-1
 
Comscistudent said:
Wow I'm stupid, thanks for putting up with me. I finally got there I think -

ΔTo - VΔTo / U = ΔTs
ΔTo( 1 - V / U ) = ΔTs
ΔTo / ΔTs ( 1 - V / U ) = 1
ΔTo / ΔTs = 1 / ( 1 - V / U )
ΔTo / ΔTs = (1 - V / U )-1
Since ##1 - \frac V U = \frac{U - V}{U}## you could go from the next to last line above, to this:
ΔTo / ΔTs = ##\frac U {U - V}##
 
That's even nicer, I wonder why they use the (1 - V / U)-1 in the lecture? Probably works out easier to apply or something, I'm sure I'll find out when I watch the rest of it.

Thanks again!
 

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