# Help me derive the Doppler effect ratio

1. Aug 31, 2016

### Comscistudent

1. The problem statement, all variables and given/known data
I'm watching this lecture and wanted to attempt to derive the expression for ΔTo/ΔTs

2. Relevant equations
ΔTo = ΔTs + VΔTo / U

3. 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

2. Aug 31, 2016

### Staff: Mentor

That's not what I get, starting from your relevant equation. What happened to ΔTo?
Please show us what you did.

3. Aug 31, 2016

### Comscistudent

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

4. Sep 1, 2016

### Staff: Mentor

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.

5. Sep 1, 2016

### Comscistudent

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

6. Sep 1, 2016

### Staff: Mentor

Since $1 - \frac V U = \frac{U - V}{U}$ you could go from the next to last line above, to this:
ΔTo / ΔTs = $\frac U {U - V}$

7. Sep 1, 2016

### Comscistudent

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!