# What does a curved natural log graph suggest?

Sorry if this is rather simple, but I've only just started learning about using logarithms in experimental physics.

I did an experiment to test the amount of time it would take for an amount of water to leave a burette. I used the starting volume of water in the burette as a control variable, $50cm^3$. I recorded the time it took for the a given volume of water to be left in the burette. For example, $10cm^3$ left took a time of roughly $71\mbox{s}$; $45\mbox{cm}^3$ left took roughly $6\mbox{s}$, and then many values in between.

I would expect this to represent exponential decay, seen as different concentrations and masses of water in the burette would have different effects on the speed of the water leaving the burette. (Correct me if I'm wrong.)

So I plotted a graph of volume against time and it showed exponential decay, but it was only very slightly curved, but curved nonetheless.

So I decided then to plot a graph of $\ln\left(V/\operatorname{cm}^3\right)$ against time/s. However, this did not produce a straight line. If I were to follow the plotted points with a curve, the gradient of the line would have been negative and increased in negative 'magnitude'.

I'm meant to analyse the extent of whether or not my experiment shows exponential decay. I'm quite stuck, because my original graph shows very slight decay, whereas my log graph isn't a straight line. Does the fact that the log graph doesn't produce a straight line show that there isn't exponential decay? Does it not matter? Would it have been straight had there been very few experimental errors/uncertainties (there would have been a lot)?

So I guess, fundamentally, my question is:

What does the curved line on my natural log graph suggest?

-
You should really include the graphs in your question, otherwise it's hard to determine what's going on exactly. –  Wouter Jun 23 '13 at 16:09
Also, just to verify: If the rate at which water flows out of the burette is proportional to the amount of water inside the burette, then exponential decay is what's expected. (but there may be other factors of course) –  NeuroFuzzy Jun 23 '13 at 16:40
@NeuroFuzzy is right. If flow rate is proportional to pressure-at-hole (head) then you should see exponential decay. However, that is only true in the viscous limit. For inviscid flow, flow rate is proportional to square-root of pressure, because of energy balance. –  Mike Dunlavey Jun 23 '13 at 22:13

Some quick scribbling on an envelope suggests that the volume in the burette will vary with time according to:

$$V = V_0 \left(\frac{t_0 - t}{t_0} \right) ^2$$

where $V_0$ is the initial volume (50cc in this case) and $t_0$ is the time the burette takes to empty. So the curve is not an exponential decay. It's actually a section of a parabola, but shifted along the time axis. Some more quick scribbling in Excel and I get a graph that looks like (assuming $t_0$ is 20 seconds):

If you do a graph of ln(V) against time with this data you get a distinctly non-straight line:

It's dangerous to assume that any curve vaguely resembling an exponential is actually an exponential, even though this is a common error amongst budding physicists. You really need to have some mathematical model against which you can evaluate your data.

-
Those two graphs are exactly what I got, would you care to show me how you came to: $V=V_0(\dfrac{t_0-t}{t_0})^2 – Olly Price Jun 23 '13 at 17:30 Exactly what I got in terms of shape anyway. – Olly Price Jun 23 '13 at 17:31 @OllyPrice: that's another question :-) – John Rennie Jun 23 '13 at 17:59 @OllyPrice In that case, try a log-log plot (that is, a plot where both axes are logarithmic). A power-law relation, which is what John suggests, produces a straight line in a log-log plot. – Pulsar Jun 23 '13 at 18:12 However a log-log plot of volume against time will not give you a straight line plot because volume is not proportional to$t^n$for some exponent$n$. You'd have to graph log(volume) against log(($t_0$-$t$)/$t_0\$) to get the straight line, and of course you wouldn't know to do that without your model for how volume should vary. –  John Rennie Jun 24 '13 at 7:17

If you plotted on semi-log paper (or equivalently plotted the log of the dependent variable against the independent variable) and didn't get a straight line (to within uncertainty) then you don't have an exponential relationship.

So ask yourself, how sure are you that you should expect an exponential behavior?

For that matter how sure are you that you can tell a exponential curve from a power law curve? Have you tried plotting on log-log paper (or equivalently plotting the log of the dependent variable against the log of the independent variable)?

If you get a linear relationship except in one region of your graph you might also suspect an unaccounted for systematic effect.

-
A log-log plot won't work in this case –  John Rennie Jun 24 '13 at 7:18

Maybe you could show the graphs. If when you take the logarithm of each point, they don't fall into a straight line, it could be that your data does not really obey an exponential model, but you have to verify that with the propagation of uncertainty. For example, if the data falls something like

and after taking the exponential of the data it goes like,

then it's ok, even if some points get far from the line, because the data, along with it's uncertainty, fits in an acceptable way with the model, which would be a logarithm in this case.

-