# Energy vs Voltage and Sqrt(Voltage): Function of Units

• quietrain
In summary, we discussed the units for plotting a graph of Energy versus Voltage, and how taking a function of a quantity with units can be problematic. We also clarified that certain functions, like x squared, can make sense with units, while others, like cosine, do not. Finally, we explored a potential issue with taking the natural logarithm of quantities with units, and suggested using reference values to address this issue.

#### quietrain

ok, let's 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

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

Hi quietrain! 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.

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

the "x" in those cases must be dimensionless.

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.

Hi Studiot! Studiot said:
… The displacement, y is also measured in metres and is a function of x such that

y = cos(x).

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. Thank you

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

er. so what is the answer to my question in the first post? hehex...

quietrain said:
er. so what is the answer to my question in the first post? hehex...

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.

oh? so if i plot length VS 1/sqrt(VOLTAGE)

my units will be (m) VS (V-1/2)?

Yes.

i see.. thanks a lot!

Say you have cos and expand it around zero

$$\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-...$$

If x is, say, in meters you would have to add meters and square meters which makes no sense.

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.

thanks!

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

Choose a reference pair of values for $B$ and $R$, say $B_0$ and $R_0$, such that $B_0=R_0^x$.

$\begin{displayeqn} \frac{B}{B_0}=\left(\frac{R}{R_0}\right)^x \Rightarrow \log\left(\frac{B}{B_0}\right)=x\log\left(\frac{R}{R_0}\right) \end{displayeqn}$

ah i see thank you very much!