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Square Root in Special Relativity

  1. Feb 19, 2007 #1
    Can anyone explain what the square root does? I know 3 squared equals the square root of 9. I want to learn what is actually happening behind the math, if that makes sense. Consider the following: I’m calculating the wavelength of light, and traveling towards the source.

    Speed of light = 300 Units per second
    Wavelength = .75 Units
    My Speed = 25 Units per second

    The square root of 275/325 equals .919866211
    .919866211 multiplied by .75 equals the measured wavelength .689899658256

    Why do we square root and what is actually happening?
    I am pretty sure that the number 325 has the effect (or whatever) added to it, and the number 275 has the effect subtracted from it.

    Does the square root mean the effect of 325 is applied to the wavelength, and then the effect of 275, then 325, then 275, then 325, and so on? I did this 24 times and came up with a wavelength of .689832748504272 which is close, but still wrong. It didn’t matter which effect was applied first either.

    Any ideas?
  2. jcsd
  3. Feb 19, 2007 #2


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    [itex]3^2=9, \hspace{1cm} \sqrt{9}=3[/itex]. This is not what you wrote!

    What are you doing here? You appear to be just throwing numbers together, for example where does 275 come from? Why do you want to take sqrt(275/325)-- it just appears from nowhere!
  4. Feb 19, 2007 #3
    Sorry... The 275 is my speed minus the speed of light. The 325 is my speed plus the speed of light. And I'm using the formula from the special theory of relativity.
  5. Feb 19, 2007 #4


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    "The formula" for what? There are a fair few equations from special relativity. It may seem like I'm being picky, but, since I don't know how much you know, and you are not expressing your ideas in a clear way, it is nigh on impossible to help you!

    Please write down the equation you are talking about, and explain the terms in the equation, then I will try and help.
  6. Feb 19, 2007 #5

    Doc Al

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    To understand the meaning of the relativistic Doppler formula (and why it has a square root), study how it is derived: Relativistic Doppler Shift
  7. Feb 19, 2007 #6
    It’s hard to explain what I’m looking for… but I am aware of the theory.

    I will try to explain this in a different way, lets say we have two separate percentages.

    A = .916666666 %

    B = 1.083333333 %

    These two percentages affect the wavelength at the same time.
    The measured wavelength will equal the old wavelength multiplied by .91986621108 %
    We calculate it using the formula below.

    The square root of A / B = .91986621108 %

    How does A & B merge to equal .91986621108

    This is more of a math question than a theory question, but I figured someone would know.
  8. Feb 19, 2007 #7


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    [grrr... I took a too-long break while writing this post. :grumpy: The two posts that now precede it weren't there when I started to write it!]

    It looks to me like grounded is trying to calculate the relativistic Doppler effect. One way to write the formula for it is:

    [tex]\lambda_{observed} = \lambda_{emitted} \sqrt {\frac{c + v}{c - v}}[/tex]

    if the source is moving away from the observer. If the source is moving towards the observer (which seems the be case for grounded's example), exchange the + and - signs, or equivalently, make v a negative number.

    This calculation looks OK, for the source moving towards the observer.

    To understand why the formula is the way it is, you need to look at how it is derived. It's basically the same derivation as for the non-relativistic Doppler effect, but also takes into account time dilation of the moving source. The simplest version I can find on the Web at the moment is this one, which assumes that you already know the non-relativistic Dopper effect formula:


    I'm sorry, I don't understand what you're saying here. The square root simply means "calculate the square root of whatever is inside" (once).

    [tex]\sqrt {\frac{c - v}{c + v}} = \sqrt{\frac{300-25}{300+25}} = \sqrt{\frac{275}{325}} = \sqrt{0.8461...} = 0.9198...[/tex]

    (I've omitted most of the decimal places here, but kept them in my calculator.)
    Last edited: Feb 19, 2007
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