Losing the Units: Explaining Why Exponentials are Dimensionless

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Trigonometric functions like sine and cosine are dimensionless because they represent ratios of lengths, meaning their arguments must also be dimensionless. This principle extends to the natural exponential function, where the argument must be dimensionless for the series expansion to hold true. If an argument had dimensions, it would be impossible to combine it with dimensionless quantities in a meaningful way. In physics, this requirement is sometimes overlooked, leading to expressions like e^t being interpreted as e^(t/τ) to ensure dimensional consistency. Understanding these concepts is crucial for accurate mathematical modeling in physics.
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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 explanation as to why we "lose the units" when using the exponential function?
 
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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.
 
Muphrid said:
It is not strictly correct to supply any argument with dimensions to an exponential or trig function. [..]
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.
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
 
Thank you both for your fast insightful and illustrative replies.
 
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