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Function of units

  1. Mar 12, 2010 #1
    ok, lets say i am plotting a graph of Energy VS voltage

    so my units will be Joules VS Volt

    so my gradient will be Joules per voltage

    but what if i plot ENergy VS square root of VOltage?

    will it become Joules VS VOLT1/2

    so gradient = Joules per Volt1/2 ?

    my lecturer once said you cannot take a function of anything with units. but what if the equation itself has units?

    like E=mc2. if i plot E against c2, will the units become J VS m2s-2
  2. jcsd
  3. Mar 12, 2010 #2


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    Hi quietrain! :smile:

    I think (s)he was talking about functions like cosx, ex, and logx …

    the "x" in those cases must be dimensionless.

    But (s)he's not talking about powers, like x2 or √x.
  4. Mar 12, 2010 #3

    I don't see why.

    For instance distance, x is measured in metres.

    Now supposing I consider a standing wave.

    The displacement, y is also measured in metres and is a function of x such that

    y = cos(x).

    Why don't you just ask your teacher what was meant? Perhaps you misheard or only heard part of it.
  5. Mar 12, 2010 #4


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    Hi Studiot! :smile:
    No, y is the amplitude times a cosine.

    The amplitude is a length, the cosine is just a dimensionless number.

    And it isn't cosx where x is a length, it's cos(x/L) where L/2π is the wavelength, and x/L is just a dimensionless number. :wink:
  6. Mar 12, 2010 #5
    Thank you
  7. Mar 12, 2010 #6


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    Put another way, a calculation result should not depend on the units you are using.

    "cos(x)" would be different for x measured in meters vs. x measured in feet. But it's okay to have

    cos[x / (1 m)]​

    Then it doesn't matter whether you use (for example) x=0.500 m or equivalently x=1.64 feet, you still get cos(0.500)=0.878
  8. Mar 12, 2010 #7
    er. so what is the answer to my question in the first post? hehex...
  9. Mar 13, 2010 #8
    There are no problems with any of your examples. You just misunderstood what the lecturer said (unless the lecturer was mistaken, but we'll assume that's not the case.)

    Some functions make sense when the argument has units, and some would just be nonsense if you tried to put a dimensionful quantity as the argument. For example, x to any power can make perfect sense if x has units, but taking the cosine of a number with units makes no sense. For example, think about its Taylor series. You would be adding together terms that each have completely different dimensions, which is just total nonsense.

    I think it is an important point, though. "cos(x)" would not just be interpreted differently if you put x in meters or feet---it's just nonsense either way. Sometimes people get lazy with notation (maybe there's really a (1 m^-1) in the argument that someone just didn't bother to write because it's understood) but it's good to keep this in mind.
  10. Mar 13, 2010 #9
    oh? so if i plot length VS 1/sqrt(VOLTAGE)

    my units will be (m) VS (V-1/2)?
  11. Mar 13, 2010 #10


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  12. Mar 13, 2010 #11
    i see.. thanks a lot!
  13. Mar 13, 2010 #12
    Think about it this way:

    Say you have cos and expand it around zero

    [tex]\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-...[/tex]

    If x is, say, in meters you would have to add meters and square meters which makes no sense.
  14. Mar 14, 2010 #13
    ah i see.. so its ok to square a length but not ok to take cosine of a length

    so squaring length gives m2 as units.

  15. Mar 18, 2010 #14
    dreads... another problem,...

    ok, to find the relation of magnetic field B at the center of a loop to radius of that loop, B = Rx

    so in this case , if i want to find x, i have to take ln on both sides, giving me lnB = xlnR , doing this will allow me to plot a graph of lnB vs lnR to get x as the gradient of a straight line graph

    but the problem is taking ln of B and R. B and R has units of T and m... so can i do so? if i can, then what will be my units of lnB and lnR?

    or is there another way that i can manipulate the equation B = Rx so that i can plot a straight line graph? thanks
  16. Mar 18, 2010 #15
    Choose a reference pair of values for [itex]B[/itex] and [itex]R[/itex], say [itex]B_0[/itex] and [itex]R_0[/itex], such that [itex]B_0=R_0^x[/itex].

    \frac{B}{B_0}=\left(\frac{R}{R_0}\right)^x \Rightarrow \log\left(\frac{B}{B_0}\right)=x\log\left(\frac{R}{R_0}\right)
  17. Mar 19, 2010 #16
    ah i see thank you very much!
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