Losing the Units: Explaining Why Exponentials are Dimensionless

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Discussion Overview

The discussion centers around the dimensionality of exponential functions and trigonometric functions, specifically addressing why these functions are considered dimensionless when used in equations. The scope includes conceptual clarification and technical reasoning related to mathematics and physics.

Discussion Character

  • Conceptual clarification
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that trigonometric functions like sine and cosine are ratios of lengths, making them dimensionless when evaluated, as seen in the equation x(t) = x0 cos(ωt) + v0/ω sin(ωt).
  • Another participant argues that it is incorrect to assign dimensions to the arguments of exponential or trigonometric functions, emphasizing that expressions like ωt must be dimensionless.
  • A further explanation is provided regarding the series expansion of the exponential function, stating that if the argument had dimensions, it would be problematic to add terms of different dimensions.
  • One participant suggests that in practical physics, the strict requirement for dimensionless arguments is sometimes overlooked, using the example of e^t being interpreted as e^{t/τ} where τ is a unit of time.

Areas of Agreement / Disagreement

Participants generally agree on the need for dimensionless arguments in mathematical functions, but there is a recognition that this requirement is sometimes ignored in practical applications, leading to a nuanced discussion without a clear consensus.

Contextual Notes

Some limitations are noted regarding the strictness of dimensional analysis in mathematical expressions, particularly in physics, where practical usage may diverge from theoretical rigor.

starzero
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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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