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Origin of rms speed

  1. Dec 5, 2011 #1
    We've just learned about kinetic theory and gases and my teacher showed us the rms speed. He himself doesn't know why its called the rms speed, or how it originated and I was curious to find out myself. Why go through all the trouble when you could just say the average speed?

    If anyone could explain the story behind the rms speed and the thinking behind it, that would be awesome!
     
  2. jcsd
  3. Dec 5, 2011 #2

    cepheid

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    rms = "root mean square"

    Now, let's parse that phrase to see what it means.. We have a bunch of particles all with different speeds vi (where the subscript 'i' denotes the speed for the "ith" particle i.e. the index i ranges from 1 to N where N is the total number of particles). The squares of the speeds would just be given by vi2. Now let's say you took the average (the mean) of all of these squared velocities, by adding them all up and dividing by the total number of them: (1/N)Ʃi (vi)2. The result of the average is called the "mean square" velocity (it is the mean of all the squared velocities). But this number is now in the wrong units (velocity squared), so to get it into the right units, you take the square root of this result, which is the root mean square velocity.

    Why would you do this instead of just adding up all of the velocities and dividing that sum by the total number? Because velocity is a vector, and if you have a very large number of particles, then the distribution of velocities will be more or less isotropic (which is just a fancy way of saying, "distributed evenly in all directions"). So, on average, velocities in opposing directions will cancel each other out, and you'll be left with 0 mean velocity. This is not that useful if you're trying to get a sense of the typical particle speed in the gas, which is precisely what the rms value tells you.
     
  4. Dec 5, 2011 #3

    berkeman

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    This is an interesting thread to me. In EE, we use RMS because it helps us equate AC (RMS) and DC powers. I wonder if there is something more fundamental going on...
     
  5. Dec 5, 2011 #4
    Sorry for my immature grasp of mathematics, but I don't understand why finding the mean of the velocity squared will not "cancel" the velocities in the first place.
    Do you mind elaborating on this.

    thanks
     
  6. Dec 5, 2011 #5

    cepheid

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    Obviously in part it's just because of the same reasoning I gave above. If a voltmeter told you the average value of an AC signal (which is always 0) that wouldn't be very useful. The RMS value is not only more useful, but also happens to be what the instrument actually measures in the first place. Why is that? Well for any periodic current waveform i(t) (it doesn't necessarily have to be sinusoidal, just periodic), there is some effective value of the current Ieff such that the power dissipated in a resistor by that current is equal to P = (Ieff)2R. In other words, the effective current value is the current value that allows you to write the equation for P in the same form as it appears in the DC case. Now, you can show that the effective current value is just equal to the RMS value of the current waveform, which in this case has the definition [itex] \sqrt{\dfrac{1}{T} \int_T i^2(t)\,dt } [/itex] since this is a continuous-time signal.

    Another magic thing about RMS: the magnitude of the apparent (complex) power S is just equal to the product of the RMS voltage and the RMS current.
     
  7. Dec 5, 2011 #6

    cepheid

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    They won't cancel out because squared numbers are always positive.
     
  8. Dec 5, 2011 #7
    Wait, I thought speed was a scalar quantity and therefore doesn't have any directional value. If this is so, why would the speeds cancel out?

    If we are in fact using velocities, why is it called the rms SPEED?

    Thanks
     
  9. Dec 5, 2011 #8

    cepheid

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    Yeah, the example I gave was more to get across a general sense of why you'd want to calculate the rms value of a discrete set of quantities that could be either positive or negative. For the specific application of speed in a gas, it turns out that rather than summing over a discrete set of quantities, you integrate over a continuous speed distribution. The Wikipedia article has more details on this:

    http://en.wikipedia.org/wiki/Root-mean-square_speed

    So the rms value is a speed (a scalar) and for this application, it is obtained by integrating a continuous distribution function over the range of all possible speeds.
     
  10. Dec 5, 2011 #9
    Thanks, but I haven't learned integration and all that other good stuff yet... so the Wiki page is really confusing to me.

    Thanks anyhow.
     
  11. Dec 6, 2011 #10

    cepheid

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    The intuition for what rms speed is is just the same as what I said in my first post. If you want, you can just think of it as being the square root of the mean of the squares of all the particle speeds. The fancy integration and stuff is just details relating to how you compute the mean. You can simply ignore these details when thinking about the situation intuitively.
     
  12. Dec 6, 2011 #11
    The physical point here is that the translational kinetic energy of a molecule of gas is proportional to the square of the velocity (as in [itex]\frac{1}{2}mv^2[/itex]). So the mean-squared velocity is proportional to the average kinetic energy (which in turn is proportional to the gas temperature). The root-mean-square velocity will be proportional to the square root of temperature.

    In fact, all the "interesting" mean gas speeds are proportional to the square root of temperature:

    Most probable speed: [itex]\widetilde{v}=\sqrt{2RT}[/itex]
    Mean speed: [itex]\overline{v}=\sqrt{\frac{8}{\pi}RT}[/itex]
    RMS speed: [itex]v_{rms}=\sqrt{3RT}[/itex]
    Speed of sound: [itex]a=\sqrt{\gamma RT}[/itex]

    Here R is the mass-based gas constant for the gas in question; note [itex]R=k_B / m[/itex] where m is the molecular mass and [itex]k_B[/itex] is Boltzmann's constant (which is really just the conversion factor from degrees to energy units).

    BBB
     
  13. Dec 6, 2011 #12
    Physicsdreams:
    In the kinetic theory I assume you got an expression for the pressure of a gas. In the equation the speed of 1 molecule, v, appears as v^2.
    When you have to consider all of the molecules you need a representative average of v^2.
    (this is not the same as average speed squared as highlighted already above).
    It means you have to take all the speeds, square each one, add all these squared speeds together then find the average of that number to give the v^2 to use in the equation for pressure. You must not forget that this is a 'speed squared' quantity and to get the equivalent representative speed you must take the square root.
    You can see the effect in a numerical example: suppose you have 3 particles with speeds
    4, 6 and 8. The average speed is 6.
    Now square these speeds to give 16, 36 and 64. the average of these squared speeds is 38.7, now take the square root to get a representative speed = 6.2 (not 6!!!!)
    This is the ROOT MEAN SQUARE speed. What we did was SQUARE the speed, take the MEAN then square ROOT to get back to a speed.
    In electricity when dealing with AC it is meaningless to talk about the average of an AC voltage because one answer is that all AC voltages have an average of zero (a simple average!!!) and this is meaningless. AC goes + and - and to remove this problem the voltage is SQUARED then a MEAN is taken then the squar ROOT is taken to get back to a voltage.....RMS
     
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