Question about derivatives

[Johnny5]

I have a question about derivatives.

 

Is there any step in the definition of a differential of time,dt, which mathematically stipulates that the flow of time is unidirectional, or does the mathematics of the "unidirectionality of time" have to come from the thermodynamical concept of entropy?

 

Thank you

[jcarlson]

There isn't any reason mathematically why the differential of time couldnt be negative just like the differential of any other variable in a function. There is no definition in mathematics that I know of that states that the variable "time" is any different from any other variable. However when you apply mathematics like this to physics you also have to take into account the properties of what you are applying the math to, not just the math itself. In the case of time, it is unidirectional, and therefore obtaining a differential of time that was negative in a physics problem would be nonsensical.

[Johnny5]

There isn't any reason mathematically why the differential of time couldnt be negative just like the differential of any other variable in a function. There is no definition in mathematics that I know of that states that the variable "time" is any different from any other variable. However when you apply mathematics like this to physics you also have to take into account the properties of what you are applying the math to, not just the math itself. In the case of time, it is unidirectional, and therefore obtaining a differential of time that was negative in a physics problem would be nonsensical.

 

Oh this answer is excellent, because I see you clearly understand the question. I think in fact there is a reason, and it goes back to the finite discrete difference calculus, which the differential calculus is based upon. Let me explain.

 

Consider the difference operator:

 

[math] \Delta Q = Q2-Q1 [/math]

 

Here we see that the idea of change in the difference calculus is based upon subtraction, where Q denotes some quantity of physics. In order to have delta Q be nonzero, time has to increment or decrement. Now, if we stipulate that Q2 is measured after Q1 is measured, or even better yet stipulate that Q2 is the value of Q after Q1 is the value of Q, then we have inserted the unidirectionality of time into our mathematics haven't we? I mean this is my whole question. The moment Q takes on its value Q2 and the moment Q takes on its value Q1 aren't the same moment in time. Necessarily, one moment must come before the other. By convention Q2 is the value of Q later in time. So doesn't this convention "stipulate" that time flows unidirectionally? Recall that the definition of the derivative is based upon the difference operator.

[jcarlson]

I suppose if we operate on the assumption that time is always the independant variable in a function (where f(t) is dependant on the value of y), which at the moment I can't think of any scenarios where it wouldn't be, it would only be possible for the differential of time dt to be positive, just as its only possible for the differential of x to be positive in a function f(x).

[Johnny5]

I suppose if we operate on the assumption that time is always the independant variable in a function (where f(t) is dependant on the value of y), which at the moment I can't think of any scenarios where it wouldn't be, it would only be possible for the differential of time dt to be positive, just as its only possible for the differential of x to be positive in a function f(x).

 

This didn't make any sense to me.

 

If x denotes the position of something, and its position changes, depending on how you define the differential of position, the answer you get will either be positive or negative.

 

Regards

[jcarlson]

Let me rephrase...

 

In a function f(x), where x is the independant variable, that is, where the value of f(x) changes based on the amount of change in x, the value of x moves at a constant, positive rate along the x-axis, and the value of f(x) varies along the y-axis depending on the current value of x and the function f(x) represents. Therefore it seems impossible to me that the differential of an independant variable in a function could be anything other than positive, and only the differential of the value of the function could be negative.

 

Since time, in every function I can think of, is an independant variable, it makes sense that its differential could never be negative.

 

Distance, on the other hand, is often a function of an independant variable, therefore its differential could be negative, depending on the nature of the function.

 

If for some reason however, the value of time was dependant on another variable, then and only then, could it be negative. However, time, by its nature I would think is not dependant on any other variable, and is therefore unidirectional because its differential cannot be negative.

[Johnny5]

 

If for some reason however' date=' the value of time was dependant on another variable, then and only then, could it be negative. However, time, by its nature I would think is not dependant on any other variable, and is therefore unidirectional because its differential cannot be negative.[/quote']

 

Do you know of any formula relating time to entropy?

 

Regards

[Johnny5]

Let me rephrase...

 

In a function f(x)' date=' where x is the independant variable, that is, where the [b']value of f(x) changes based on the amount of change in x[/b], the value of x moves at a constant, positive rate along the x-axis, and the value of f(x) varies along the y-axis depending on the current value of x and the function f(x) represents.

 

Would you mind explaining the bold part a little more?

[jcarlson]

Do you know of any formula relating time to entropy?

 

Regards

 

I dont know of any specific formulae relating time to entropy, although I would venture to say that there is, seeing as how time is a measurement of the change in the entropy of the universe. However, entropy can theoretically only increase, not increase, so therefore time would only be able to move foreward and not backwards, so the differential would still be positive.

[jcarlson]

Would you mind explaining the bold part a little more?

 

 

What I meant was, that in a function of a variable, for example [math] f(x) = x^{2} + 3x - 2 [/math], the value of the function is dependant upon the value of the variable (in this example, x), which continuously moves down the x-axis, so that the change in the variable (again, in this example, x), and thus the differential, is always positive.

[Johnny5]

What I meant was, that in a function of a variable, for example [math] f(x) = x^{2} + 3x - 2 [/math'], the value of the function is dependant upon the value of the variable (in this example, x), which continuously moves down the x-axis, so that the change in the variable (again, in this example, x), and thus the differential, is always positive.

 

You say that the differential is always positive, do you mean df, or do you mean dx?

 

The following is basic calculus:

 

[math] y = f(x) = x^{2} + 3x - 2 [/math]

 

The picture is of a parabola. And we can discover its vertex.

 

Taking the first derivative we have:

 

[math] dy/dx = \frac{df}{dx} = 2x + 3 [/math]

 

Setting that equal to zero gives:

 

[math] 0 = 2x + 3 [/math]

 

So if dy/dx=0 then x=-3/2

 

dy/dx is the slope of the tangent line as a function of x.

 

Setting dy/dx=0 is the criterion for the tangent to the curve to be a horizontal line. This is the place where the function assumes a local max, or local min.

 

The second derivative is given by:

[math] d^2y/dx^2 = 2 [/math]

 

Positive so concave up.

 

So when x=-3/2, the function has a local minimum.

 

 

[math] y = (-3/2)^{2} + 3(-3/2) - 2 = -4 [/math]

 

So the vertex of the parabola is at (x,y)=(-3/2,-4).

 

Picking up from here, what do you mean that dx is necessarily positive?

 

Regards

[jcarlson]

You say that the differential is always positive' date=' do you mean df, or do you mean dx?

 

...

 

Picking up from here, what do you mean that dx is necessarily positive?

 

Regards[/quote']

 

I mean dx is always positive, and that the derivative df/dx will only be negative due to the differential df, never because of the differential dx.

[Dave]

I have to say that from a strictly mathematical point of view, we don't really care all that much about what the variables in the equation actually represent. There's nothing in any definition of the derivative of a function to say something completely different for time-based functions.

[Johnny5]

I have to say that from a strictly mathematical point of view, we don't really care all that much about what the variables in the equation actually represent. There's nothing in any definition of the derivative of a function to say something completely different for time-based functions.

 

My question was, does the difference calculus take this into account.

 

A - B

 

is totally different from

 

B - A

 

In order to define the difference operator [math] \Delta [/math] the values of the physical quantity which changed are necessarily values at different moments in time. So a convention must be used, in order to specify which is the latter and which is the earlier value. So I am thinking that there is already a convention in place, for the unidirectionality of the "flow of time".

 

To make this totally clear, let Q denote an arbitary physical quantity. The difference in Q is defined as:

 

[math] \Delta Q = Q2 - Q1 [/math]

 

So if I don't choose a convention right now, a student will come along and ask, which came first the Q2 or the Q1. Am I right in saying this?

 

Regards