# Dimensions in logarithms after integration

1. Jun 21, 2010

### quantum13

1. The problem statement, all variables and given/known data
Let v = 1 / kt

v = m/s
k = 1/m
t = s

v = dx/dt so dx = dt / kt

integrating,

x = ln (kt)/k + C

However the argument of a logarithm is dimensionless. But an integration is a perfectly normal thing to do. So how come this integration results in a dimensional logarithm argument?

2. Jun 21, 2010

### kuruman

What is constant C? Doing it right gives you

$$x=\int^{t_1}_{t_2}\frac{dt}{kt}=\frac{1}{k}(ln(t_2)-ln(t_1))=\frac{1}{k}ln\left(\frac{t_2}{t_1}\right)$$

The argument of the natural log is dimensionless as it should be.

3. Jun 21, 2010

### quantum13

d'oh the summer makes me forget even how to do basic integrals

but what if you do the integral in a non-definite way and use the constant C turning it into an initial value problem

dx = dt / kt
x = ln(t) / k + C
and use t=0, x=x0

but with this way i get ln(0), so how do i figure out C?

i know im making some mistake but i cant seem to figure it out..

4. Jun 21, 2010

### mo_0820

I want to know $$v_{t=0}=?$$

5. Jun 22, 2010

### RoyalCat

Your equation for velocity is undefined for t=0, so it's no surprise that the position is equally undefined there.

You can't solve it as an initial value problem because of the specific function. You end up with dimensions in the argument. You can't talk about a function $$f(t)=x$$ of how much distance the particle has traveled since time $$t=0$$ because that specific integral diverges, so you must take the definite integral for $$f(t)=x$$ and only speak of differences in this function for $$t>0$$

And since you can't take the definite integral of one side of the equation, and the indefinite integral of the other side, you must therefore take the definite integral of both.

For instance, if we define $$x(t=1) = x_0$$ as an initial condition, we avoid the whole thing about the function being undefined at our chosen initial value, but the function $$f(t)=x$$ as specified by the indefinite integral is still poorly defined.

$$kx=\ln{t}+kC$$

(Note how in the next step, if we were to measure time in any other units, we would not receive $$\ln{1}$$ but some other value!)

$$kx_0=\ln{1}+kC$$
$$C=x_0$$

$$x=\frac{kx_0+\ln{t}}{k}$$

Bringing both terms into the logarithm:
$$x=\frac{\ln{(e^{kx_0}\cdot t)}}{k}$$

And so our argument for the logarithm still has dimensions of time, which proves we made a serious mistake somewhere along the way.

6. Jun 22, 2010

### kuruman

In physics there are no indefinite integrals, if you think about it for a moment, because mathematical expressions model physical reality. To take a simpler case, suppose you want to solve v = dx/dt = constant, the simplest differential equation there is. You can immediately write the mathematical expression x = vt + C if you are doing "math". What does this mean physically?

Physically you would say, I know that the rate of change of position with respect to time is constant. To find the overall change of position I need to write its incremental change as

dx = v dt

and add all such increments. But how exactly are you going to add them? You need a starting and ending point for both position and time, so you say

$$\int^{x}_{x_0}dx=\int^{t}_{t_0}vdt$$

(x-x0)=v(t-t0)

The symbols represent the starting space-time point (x0, t0) and the ending space-time point (x, t) of the summation. When you write x = vt, the assumption is that (a) the clock that measures time starts when motion starts (t0=0) and (b) that the object is at the origin when the clock starts (x0=0). Just because they are swept under the rug does not mean that they are not or should not be there. To put it in a nutshell, any summation requires a starting and an ending point and if you really don't know what these are, then you cannot do the summation.

Last edited: Jun 22, 2010