# Losing the Units

1. Oct 2, 2012

### starzero

Hi All and sorry if this is too easy a question but here goes....

Sines, Cosines and the rest of the trig functions are the ratio of two lengths and thus are dimensionless quantities.

That is if I plug in a value for t in sin(ωt) there are no units.

For example the solution of

x'' + ω^2 x = 0 with x(0) = x0 and x'(0) = v0 is given by

x(t) = x0 cos(ωt) + v0/ω sin(ωt)

The units come from the initial conditions not the sine or cosine.

So here is the question...

Same is true ( I believe ) when using the natural exponential function exp(t).

How does one simply explain this.

I tried to reason it out using eulers formula exp(iω) = cos(ω) + i sin(ω) figuring that again we get ratios of lengths,
however in the case where the real part is non-zero we get another exponential (which is not the ratio of lengths)

exp(a +ib) = exp(a)(cos(b) + i sin(b))

Is there a simple explaination as to why we "lose the units" when using the exponential function?

2. Oct 2, 2012

### Muphrid

It is not strictly correct to supply any argument with dimensions to an exponential or trig function. Example: $\omega t$ has no dimensions (frequency is 1/time, and time/time is dimensionless).

Why is it incorrect? Consider a series expansion of the exponential function.

$$e^x = 1 + x + \frac{1}{2} x^2 + \ldots{}$$

$x$ must be a dimensionless parameter. If it had, say, dimensions of length, how would we add length to 1 and to length squared?

Nevertheless, in physics this strict need is sometimes ignored. As an example, you should take as implicit that $e^t$ is in fact $e^{t/\tau}$ where $\tau$ is 1 in whatever units of time you're working with.

3. Oct 2, 2012

### harrylin

Yes indeed, and it may be useful to give an example (of not ignoring this):
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html

4. Oct 2, 2012

### starzero

Thank you both for your fast insightful and illustrative replies.