# Dimensional Analysis

1. May 2, 2009

### Nusc

$$V(t)=IR(1-e^{t/RC}) \rightarrow ln|IR - V(t)| = ln|IR|-t/RC$$

The first equation below has units of voltage.

the logarithm has units of voltage, but what does ln|IR| and ln|V| physically mean?

The purpose of this question is that if you plot the natural logarithm of the voltage versus time, it should yield a straight line with slope -1/RC and intercept ln|V_c|.

From which you can determine the capacitance given a known resistance.

2. May 2, 2009

### tiny-tim

Hi Nusc!

yes, you're right …

technically the equation is dimensionally wrong …

anything inside a ln() has to be dimensionless, so it should really be ln(|(IR - V)/IR|) = -t/RC,

or ln(|(IR - V)/V0|) = ln(|IR|/V0|) - t/RC, where V0 is some voltage chosen at random just to make the dimensions right.

In practice, we choose V0 = 1 in whatever units we're using, but without bothering to mention it.

3. May 2, 2009

### Nusc

I don't like the idea of introducing an arbitrary voltage.

What if your actually measuring the voltage of the capacitor? The resistance is due to the voltmeter. I need to know what, in your case, V_0 physically is and where does it arise? The voltage of what?

Last edited: May 2, 2009
4. May 3, 2009

### tiny-tim

Hi Nusc!

(try using the X2 tag just above the Reply box )
ok, then use the first formula, ln(|(IR - V)/IR|) = -t/RC.
V0 is arbitrary … it can be anything

in practice, it's 1 volt (so as not to change the maths!)

5. May 3, 2009

### Nusc

You said in practice, can you convince me from an experimental point of view? I just don't like introducing quantities such as V0 or IR without any justification.

I_c(t) + I_R(t) = I

i=dQ/dt implies I_c(t) = dQ/dt = d/dt C*V(t) = C d/dt V(t)

V_R(t) = I_R(t) * R

The above equations imply

C d/dt V(t)+V(t)/R = I

Then we get V(t) = IR(1-e^(-t/RC))

This is understood mathematically but from an experimental point of view I'm not sure whether this is safe. This is important to me.

Thanks

6. May 3, 2009

### tiny-tim

the second equation obviously is a solution of the first, and there's nothing dimensionally suspect about it …

7. May 3, 2009

### Nusc

Sorry, so that second equation you mentioned:
V(t) = IR(1-e^(-t/RC)) is a solution of the first C d/dt V(t)+V(t)/R = I, yes and there is nothing dimensionally suspect about it. What worries me is introducing V0 (or arbitrarily dividing by IR in ln(|(IR - V)/IR|) = -t/RC.)

8. May 3, 2009

### diazona

The division by IR is not arbitrary at all - that's what you get when you solve the differential equation $$C \frac{\mathrm{d}V}{\mathrm{d}t} + \frac{V}{R} = I$$. As you know (I suppose), whenever you do an integral, you wind up with an "arbitrary" constant that is in fact determined by the boundary terms (for example, that $$V(0) = 0$$ or whatever it is). Here IR is that constant.

9. May 4, 2009

### tiny-tim

(and if using, or imagining, V0 worries you , then … don't use it! )​

10. May 4, 2009

### Nusc

If you don't use it then the log of that function doesn't physically make sense, I just needed a justification for V0.

11. May 4, 2009

### tiny-tim

I meant use ln(|(IR - V)/IR|) = -t/RC instead.

The mathematical justification for V0 is that it works.

There's no physical meaning to it … as i said, V0 is arbitrary.

12. May 4, 2009

### Nusc

Let's prove it then:
$$c\frac{d}{dt} V(t)+\frac{V(t)}{R}&=&I,$$
$$I &=& e^{\int^{t}_{t_{0}} \frac{1}{RC} dt} =e^{ \frac{t}{RC}}$$
$$\frac{d}{dt}(V(t)e^{\frac{t}{RC}}) &=& \frac{I}{C}e^{\frac{t}{RC}$$

$$V(t)e^{\frac{t}{RC}}|^{t}_{t0} &=& \frac{I}{c} \int_{t0}^{t} e^{\frac{t'}{RC}} dt'$$
$$V(t)e^{\frac{t}{RC}} - V(t0) &=& \frac{IRC}{C} e^{\frac{t'}{RC}}|^t_{t0}$$
$$V(t)e^{\frac{t}{RC}} - V(t0) &=& IRe^{\frac{t}{RC}}-IRe^{\frac{t0}{RC}}$$

Taking t0 to be 0
$$V(t) - e^{\frac{-t}{RC}}V(0) &=& IR( 1 - e^{\frac{-t}{RC}})$$
$$IR-V(t) &=& e^{\frac{-t}{RC}}(1-V(0))$$
$$ln|IR-V(t)| &=& -\frac{t}{RC}(1-V(0))$$

I don't see where the division of V0 you speak of comes from the differential equation and if you set V0 to be 1 in this case the RHS gives zero. Please prove it to me and don't just say it's true. The RHS has limits so it's not arbitrary.

Last edited: May 4, 2009
13. May 4, 2009

### tiny-tim

Hi Nusc!

Sorry, I don't understand that at all.

You seem to be introducing V0 for no reason … you don't need it unitl the line in which "ln" first appears.

(are you confusing V0 with V(t0)?)

14. May 4, 2009

### diazona

First, this is the differential equation we want to solve.
$$C\frac{\mathrm{d}}{\mathrm{d}t} V(t)+\frac{V(t)}{R} = I$$
Now, there are many different ways to solve this equation. One is to change variables by $$V(t) = u(t) + IR$$ and rewrite the equation as
$$C\frac{\mathrm{d}}{\mathrm{d}t} u(t)+\frac{u(t) + IR}{R} = I$$
which, by simple algebra, becomes
$$\frac{\mathrm{d}}{\mathrm{d}t} u(t) = -\frac{u(t)}{RC}$$
and then
$$\frac{\mathrm{d}u}{u} = -\frac{\mathrm{d}t}{RC}$$
Now we can integrate on both sides,
$$\int_{t_0}^{t} \frac{\mathrm{d}u}{u} = -\int_{t_0}^{t}\frac{\mathrm{d}t}{RC}$$
When you do the integral on the left, you get
$$\int_{t_0}^{t} \frac{\mathrm{d}u}{u} = \mathrm{ln}\frac{u(t)}{u(t_0)}$$
and the one on the right gives
$$-\int_{t_0}^{t}\frac{\mathrm{d}t}{RC} = -\frac{t}{RC}$$
Equating the two, we get
$$\mathrm{ln}\frac{u(t)}{u(t_0)} = -\frac{t}{RC}$$
and exponentiating:
$$\frac{u(t)}{u(t_0)} = e^{-\frac{t}{RC}}$$
and then multiplying by $$u(t_0)$$:
$$u(t) = u(t_0) e^{-\frac{t}{RC}}$$
Now we can substitute back in $$u(t) = V(t) - IR$$, the inverse of the previous substitution:
$$V(t) - IR = (V(t_0) - IR) e^{-\frac{t}{RC}}$$
and then
$$V(t) = V(t_0)e^{-\frac{t}{RC}} + IR\left(1 - e^{-\frac{t}{RC}}\right)$$
If $$V(t_0) = 0$$, the first term is just zero and we're left with
$$V(t) = IR\left(1 - e^{-\frac{t}{RC}}\right)$$

The need for an arbitrary constant $$V_0$$ comes in when you do the logarithmic integral. In elementary calculus they tell us that
$$\int_{t_0}^{t} \frac{\mathrm{d}V}{V} = \mathrm{ln}V(t) - \mathrm{ln}V(t_0)$$
But then again, in elementary calculus, all variables represent pure numbers - no units. Unfortunately, this doesn't make sense if V has units, because you can't take the logarithm of a unit. For that reason, we introduce a constant $$V_0$$ with the same units as $$V(t)$$ and say that
$$\int_{t_0}^{t} \frac{\mathrm{d}V}{V} = \mathrm{ln}\frac{V(t)}{V_0} - \mathrm{ln}\frac{V(t_0)}{V_0}$$
Why is this valid? Well, as you know, it's a property of logarithms that
$$\mathrm{ln}a - \mathrm{ln}b = \mathrm{ln}\frac{a}{b}$$
So in the first case
$$\mathrm{ln}V(t) - \mathrm{ln}V(t_0) = \mathrm{ln}\frac{V(t)}{V(t_0)}$$
and in the second case
$$\mathrm{ln}\frac{V(t)}{V_0} - \mathrm{ln}\frac{V(t_0)}{V_0} = \mathrm{ln}\frac{V(t)/V_0}{V(t_0)/V_0} = \mathrm{ln}\frac{V(t)}{V(t_0)}$$
They're equal to the same thing and, thus (transitive property of equality) equal to each other.

P.S. Thanks for the words of support tiny-tim ;-)

15. May 4, 2009

### Nusc

The I i defined above is just the integration factor.

I am not introducing V0 for no reason, V(t0) is just V(0) for t0 = 0.

16. May 4, 2009

### Nusc

I got the same result in a different way. So I'll just quote what I don't understand.
You mention logarithmic integral, but when I talk about the logs in first few posts, I just took the log of the solution for V(t). There's no reason to integrate again.

Recall,

students are measuring the voltage of a charging capacitor. The resistance comes from the voltmeter 10M ohms.
So the idea is that using the equation for a log, students get a linear fit and they can determine the capacitance. The log solution I wrote at the begining works, but the physical meaning of logarithm with units inside does not make sense physically.

Because what your saying when you choose V0 = 1 V means that the capacitor is initially charged at 1 volt. This is not true. The capacitors are discharged when you short the circuit.

And please show me explicitly how you get from:
$$V(t) = V(t_0)e^{-\frac{t}{RC}} + IR\left(1 - e^{-\frac{t}{RC}}\right)$$
to
$$ln(|(IR - V)/V(t_0|) = -t/RC$$

Last edited: May 4, 2009
17. May 5, 2009

### Nusc

Any ideas?

18. May 6, 2009

### tiny-tim

No, V0 is not V(0), V0 is arbitrary.

In V(t) = IR(1 - e-t/RC), you have already chosen the constant of integration so that V(0) = 0.

You then rewrite that as IR - V(t) = IRe-t/RC,

and in order to take logs, you then have to divide by an arbitrary V0 (which we always take to be 1 volt, because dividing by 1 is so easy ),

to get ln((IR - V(t))/V0) = ln(IR/V0) - t/RC

(or, of course, you can just write ln((IR - V(t))/IR) = -t/RC)

19. May 6, 2009

### Nusc

Thanks!

20. May 6, 2009

### diazona

um... does that mean all is clear now?

21. May 27, 2009

### Nusc

No. It does not physically make sense to me from an experimental view to divide out the voltage.

And introducing this arbitrary voltage complicates the error analysis. Does anyone have any suggestions?

22. May 27, 2009

### diazona

I don't think I even understand what you mean by making sense from an experimental view. Do you have some reason to doubt that dividing by a constant voltage is a valid thing to do?

The arbitrary voltage is an exact value so it doesn't change the error analysis at all.

23. May 28, 2009

### Nusc

It's valid mathematically of course. But what does it mean experimentally to divide by an arbitrary voltage? Forget this for now.

24. May 28, 2009

### diazona

It doesn't mean anything experimentally. It's only necessary because the natural logarithm is defined only for unitless quantities, and that's a purely mathematical reason.

Note that if the natural logarithm were defined for quantities with units (and if that definition were consistent with everything else in mathematics) then you wouldn't need to divide by the arbitrary voltage. You can try to come up with such a definition, if you want. But I don't think it's possible.

EDIT: actually, now that I think about it... you could say that the arbitrary voltage is related to our choice of a unit system. After all, the volt itself is a fairly arbitrary choice. You could just as well use, say, the elementary charge divided by Planck energy as your unit of voltage. So I guess, in some sense, dividing by an arbitrary voltage is equivalent to choosing a unit to measure the potential difference in. If your voltmeter displays in volts, that's your arbitrary voltage. Of course, it makes sense that the equations must be independent of the arbitrary voltage because a physical system will behave the same way no matter which units you choose to measure it in.

Last edited: May 28, 2009
25. May 28, 2009

### Staff: Mentor

This actually is a fairly subtle point that usually doesn't matter, but seems to be a source of confusion here. If my answer adds more confusion feel free to ignore it.

Division by an arbitrary voltage is the same thing as selecting your units for measuring the potential difference. For example, suppose we are measuring the height of a man. We might use a meter stick and find that the man is 2 m tall. What this actually means is that the dimensionless ratio between the height of the man and the length of the meter stick is 2. I.e. L(man)/L(meter) = 2 -> L(man) = 2 m. In this sense all dimensionful measurements are actually dimensionless ratios to some standard with the same dimensions.

So if you divide by one Volt you are simply stating that you are using base SI units. On the other hand, if you divide by one statvolt then you are using base CGS units. If you divide by some other voltage then you are using some other units.

In this case the equation works out such that the choice of units does not matter, but that is not always the case. In other situations, where the choice of units matters there will be an explicit division by a non-arbitrary voltage and the argument to the transcendental function will be explicitly non-dimensional. The voltage on the bottom of such an equation defines a kind of "natural" unit of potential difference for the system in question.