Rearranging a Relativity Equation

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Homework Help Overview

The discussion revolves around rearranging a relativity equation to solve for the relative velocity \( u \) between an electromagnetic source and an observer, based on the observed frequency and the source frequency of light. The specific equation under consideration is \( f = \sqrt{\frac{c + u}{c - u}}f_0 \), and the context includes a scenario where a decrease in frequency is examined.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the steps taken to rearrange the equation, with one participant expressing uncertainty about the next steps after isolating terms involving \( u \). Others provide encouragement and suggest moving terms around to isolate \( u \). There is also a question about how to set the values for \( f_0 \) and \( f \) in the context of a frequency decrease.

Discussion Status

The discussion is active, with participants providing feedback on each other's mathematical manipulations. Some participants express confidence in the steps taken, while others seek clarification on the interpretation of the original problem, particularly regarding the expected outcomes related to frequency changes.

Contextual Notes

There is mention of a specific decrease in frequency (3.0%) and a subsequent question about interpreting a poorly worded question related to frequency increase. Participants are navigating through potential formatting errors and clarifying the implications of their calculations.

TFM
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[SOLVED] Rearranging a Relativity Equation

Homework Statement



Rewrite equation:

[tex]f = \sqrt{\frac{c + u}{c - u}}f_0[/tex]

to find the relative velocity u between the electromagnetic source and an observer in terms of the ratio of the observed frequency and the source frequency of light.

What relative velocity will produce a 3.0 decrease in frequency and

Homework Equations



Rearangement of:

[tex]f = \sqrt{\frac{c + u}{c - u}}f_0[/tex]

The Attempt at a Solution



I seem to get stuck when trying to rearrange the equation:

[tex]f = \sqrt{\frac{c + u}{c - u}}f_0[/tex]

[tex]\frac{f}{f_0} = \sqrt{\frac{c + u}{c - u}}[/tex]

[tex](\frac{f}{f_0})^2 = \frac{c + u}{c - u}[/tex]

[tex](c - u)(\frac{f}{f_0})^2 = c + u[/tex]

But I am not sure where to go from here.

Any suggestions?

TFM
 
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TFM said:
[tex](c - u)(\frac{f}{f_0})^2 = c + u[/tex]
You're doing fine; don't stop now.

[tex]c(\frac{f}{f_0})^2 - u(\frac{f}{f_0})^2 = c + u[/tex]

Next: Move all the terms containing u to one side.
 
Would that give:

[tex]c(\frac{f}{f_0})^2 - c = u(\frac{f}{f_0})^2 + u[/tex]

TFM
 
Do uou now factor out the u:

[tex]c(\frac{f}{f_0})^2 = u ((\frac{f}{f_0})^2 + 1)[/tex]

Tnhe divide over to get:

[tex]u = \frac{c(\frac{f}{f_0})^2 - c}{(\frac{f}{f_0})^2 + 1}[/tex]

Does this look right?

TFM
 
Looks good! (I would factor out the c to make it more readable.)
 
That would give:

[tex]u = \frac c((\frac{f}{f_0})^2) - 1}{(\frac{f}{f_0})^2 + 1}[/tex]


What relative velocity u will produce a 3.0 % decrease in frequency

For the actual question, do I put f_0 as 1 and f as 0.97?

TFM
 
TFM said:
That would give:

[tex]u = \frac c((\frac{f}{f_0})^2) - 1}{(\frac{f}{f_0})^2 + 1}[/tex]
:confused: Redo this. (Probably a formatting error.)
 
That definitely isn't right :

[tex]u = \frac{c((\frac{f}{f_0})^2 - 1)}{(\frac{f}{f_0})^2 + 1}[/tex]

TFM
 
Looks good.
 
  • #10
The actual question itself is:

What relative velocity u will produce a 3.0 % decrease in frequency

For this, should I put f_0 as 1 and f as 0.97?

TFM
 
  • #11
TFM said:
For this, should I put f_0 as 1 and f as 0.97?
Makes sense to me.
 
  • #12
It gives, the right answer, Thanks

One last thing, what is the question actually asking to find:

What relative velocity u will produce an increase by a factor of 3 of the observed light?

Does it want the freuqncy to be incresed by three factors?

TFM
 
  • #13
I'm no more a mind reader than you are! :smile: But since the problem seems to be talking about frequency and Doppler shifts, I would assume they mean that the observed frequency is three times the original.
 
  • #14
I calculated it to be 0.8, which is the answer, so I should say it is, as well.

Thats the second poorly worded question I've had this week:rolleyes:

Thanks for all the help,

TFM
 

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