How to Find the Derivative of F = f(x)/f(x+dx)?

[sha1000]

TL;DR Summary

Derivative of F = f(x)/f(x+dx)

Hello,

I'm struggling with this for some time.

So I have the function: f(x) = sqrt(1 - 1/x)
The derivative of this function can be easily calculated.

Now we define the function:
F(x) = f(x)/f(x + dx) = sqrt(1 - 1/x)/sqrt(1 - 1/(x+dx))

I have a hard time to find F'(x) due to the presence of dx in the equation. (Lack of math knowledge)

I would really appreciate if someone can help me with this; some hints on how to proceed.

Thank you in advance.

 

[Office_Shredder]

Can you give a little more context on how you got to this problem? My guess is the dx is basically just a constant, but it's hard to know.

 

[sha1000]

Can you give a little more context on how you got to this problem? My guess is the dx is basically just a constant, but it's hard to know.

Thank you for the reply.

The context comes from the gravitational time dilation: t' = t • sqrt( 1 - 2GM/R) with c =1

This gives the time dilation as a function of R compared to the flow of time at R -> infinity.

From the point of view of local observer (A) standing on the top of the tower; the time dilation on the first floor (B) can be calculated using the equation:

t'(B) = t(A)• sqrt(1 - 2GM/R)/ sqrt( 1 - 2GM/(R + h))
with h = distance between A and B.
And t(A) = t(r-> infinity) since the observer at point A does not feel any time dilation.

So I would like to get the derivative of this function. Local derivative of the time dilation equation when h -> 0.

 

[PeroK]

Now we define the function:
F(x) = f(x)/f(x + dx) = sqrt(1 - 1/x)/sqrt(1 - 1/(x+dx))

In your time dilation equation, $x$ is fixed, so should be replaced by a constant, and $dx$ is just your variable. You should have:
$$F(h) = \frac{f(R)}{f(R+h)}$$
Then may calculate $\frac{dF}{dh}$.

 

[sha1000]

In your time dilation equation, $x$ is fixed, so should be replaced by a constant, and $dx$ is just your variable. You should have:
$$F(h) = \frac{f(R)}{f(R+h)}$$
Then may calculate $\frac{dF}{dh}$.

It's a good start but I'm looking for something different.

In this equation x is fixed. Thus, it permits to calculate the time dilation as a function of distance in the reference frame of the observer at some fixed distance R.

From here it is possible to calculate the local derivative F'(h) when h->R. Afterwards I want to express this local derivative as a function of R.

Sometimes I have a hard time to express my questions, hope it is not too confusing.

 

[PeroK]

It's a good start but I'm looking for something different.

In this equation x is fixed. Thus, it permits to calculate the time dilation as a function of distance in the reference frame of the observer at some fixed distance R.

From here it is possible to calculate the local derivative F'(h) when h->R. Afterwards I want to express this local derivative as a function of R.

Sometimes I have a hard time to express my questions, hope it is not too confusing.

That happens automatically. For example, when you do a derivative:
$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$Then, we imagine $h$ varying about a fixed $x$ for the purpose of calculating the derivative at $x$. Note, however, that $x$ may take any value. What we have done, therefore, is mathematically calculated the derivative for all points $x$. That's why the derivative is now a function of $x$, with the result we got valid generally for all $x$.

The important thing is that $x$ does not vary while we are taking the limit $h \rightarrow 0$.

 

[Office_Shredder]

I'm pretty confused about what you actually want to compute. Can you describe it in words with no equations? Like, I want to estimate how much time dilates if I move a small distance.

 

[Mark44]

Summary:: Derivative of F = f(x)/f(x+dx)

Now we define the function:
F(x) = f(x)/f(x + dx) = sqrt(1 - 1/x)/sqrt(1 - 1/(x+dx))

In your time dilation equation, x is fixed, so should be replaced by a constant, and dx is just your variable. You should have:
$$F(h) = \frac{f(R)}{f(R+h)}$$
Then may calculate $\frac{dF}{dh}$.

I agree with @PeroK here. In the equation in the first quote, F is a function of x alone, and dx should be considered to be a constant. The equation in the second quote treats x as a constant, but allows h (= dx of the first equation) to vary.

 

[sha1000]

That happens automatically. For example, when you do a derivative:
$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$Then, we imagine $h$ varying about a fixed $x$ for the purpose of calculating the derivative at $x$. Note, however, that $x$ may take any value. What we have done, therefore, is mathematically calculated the derivative for all points $x$. That's why the derivative is now a function of $x$, with the result we got valid generally for all $x$.

The important thing is that $x$ does not vary while we are taking the limit $h \rightarrow 0$.

Thank you again.

Si basically I need to derive the equation and apply the limit h->0.

1) F(h) = sqrt (1-2GM/R)/sqrt(1-2GM/(R+h)))

F'(h) = - sqrt(1-2GM/R) * GM/(R+h)²\sqrt(1-2GM/(R+h))^3/2

2) If h->0

F'(R) = - GM/R²(1-2GM/R)

Do I have it right?

 

[PeroK]

Thank you again.

Si basically I need to derive the equation and apply the limit h->0.

1) F(h) = sqrt (1-2GM/R)/sqrt(1-2GM/(R+h)))

F'(h) = - sqrt(1-2GM/R) * GM/(R+h)²\sqrt(1-2GM/(R+h))^3/2

2) If h->0

F'(R) = - GM/R²(1-2GM/R)

Do I have it right?

You are tying yourself in knots here. You either differentiate a function according to the rules of differentiation; or, you differentiate a function from first principles using the limit-based definition. You're mixing up the two.

In general if $F(h) = f(R+h)$, then $F'(h) = f'(R+h)$, by the chain rule. Taking $h \rightarrow 0$ to get the derivative $f'(R)$ is unnecessary. You can simply differentiate $f(R)$ directly.

 

[sha1000]

You are tying yourself in knots here. You either differentiate a function according to the rules of differentiation; or, you differentiate a function from first principles using the limit-based definition. You're mixing up the two.

In general if $F(h) = f(R+h)$, then $F'(h) = f'(R+h)$, by the chain rule. Taking $h \rightarrow 0$ to get the derivative $f'(R)$ is unnecessary. You can simply differentiate $f(R)$ directly.

Sorry for my delayed answers. Too much work lately.

F(h)= sqrt(1 - 2GM/R)/ sqrt( 1 - 2GM/(R + h))

The simple derivative of F(h) depicts the rate of change of the time dilation of the observer B (R = h) compared to the fixed observer A (R fixed).

I want to quantify this derivative when h --> 0. And then express this quantity as a function of the fixed R.

I'm looking for the local derivative between A (R fixed) and B (h-->0).

2) I'm pretty sure the simple derivative of F(R) does not give me that answer. Since it shows the rate of change of time dilation between the observer at infinity and the observer A (R).

 

[sha1000]

I'm pretty confused about what you actually want to compute. Can you describe it in words with no equations? Like, I want to estimate how much time dilates if I move a small distance.

Well, I'll try to do my best.

1) So there is a time dilation effect in GR as a function of R.

The equation T(R) = T(inf)sqrt(1- 2GM/R) gives the dilated time compared to the time of the observer placed at some distance R --> infinity

The derivative of this function gives the rate of change of dilated time compared to the time rate at infinity.

2) Now take the case of two observers:
- observer A --> placed on the top of the tower
- observer B --> placed at the base of the tower

By using the equation T(R) we can calculate the time dilation at A and B as compared to the observer placed at infinity.

We can also calculate the time dilation observed by A (top of the tower) at point B (tower base).
For that we can use the following equation:

t'(B) = t(A)• sqrt(1 - 2GM/R)/ sqrt( 1 - 2GM/(R + h))
h = distance between A and B.

Now I would like to quantify the derivative (the rate of change) of this equation at point A (fixed) in its immediate surrounding (h --> 0). Let's call it "local derivative".

And finally I would like to find how this local derivative varries as a function of distance of point A from the center of Earth.I hope it is not too confusing.

 

[PeroK]

I hope it is not too confusing.

This looks physically meaningless to me. You could fix the height of the tower, $h$ and calculate the time dilation between the top and bottom as a function of $R$, the radial coordinate of the base. Or, you could fix $R$ and vary $h$. But, taking $h = 0$ and varying $R$ doesn't give a meaningful quantity that is of any relevance to any measurement that I can see.

That said, there should be no problem differentiating your function wrt $h$ and then setting $h = 0$. But, I don't see what possible use or physical meaning that derivative could have.

 

[sha1000]

This looks physically meaningless to me. You could fix the height of the tower, $h$ and calculate the time dilation between the top and bottom as a function of $R$, the radial coordinate of the base. Or, you could fix $R$ and vary $h$. But, taking $h = 0$ and varying $R$ doesn't give a meaningful quantity that is of any relevance to any measurement that I can see.

That said, there should be no problem differentiating your function wrt $h$ and then setting $h = 0$. But, I don't see what possible use or physical meaning that derivative could have.

Probably there is no real physical observable behind this.

But I just want to be sure that my mathematical development stands true for the described problem.

Does is seem correct for you?

1) F(h) = sqrt (1-2GM/R)/sqrt(1-2GM/(R+h)))

F'(h) = - sqrt(1-2GM/R) * GM/(R+h)2\sqrt(1-2GM/(R+h))3/2

2) If h->0

F'(R) = - GM/R2(1-2GM/R)

 

[PeroK]

Probably there is no real physical observable behind this.

But I just want to be sure that my mathematical development stands true for the described problem.

Does is seem correct for you?

1) F(h) = sqrt (1-2GM/R)/sqrt(1-2GM/(R+h)))

F'(h) = - sqrt(1-2GM/R) * GM/(R+h)2\sqrt(1-2GM/(R+h))3/2

2) If h->0

F'(R) = - GM/R2(1-2GM/R)

That's not mathematically valid. $F$ is a function of $h$. It doesn't become a function of $R$ when you set $h = 0$. The derivative you have calculated is $F'(0) \equiv F'(h)\big{|}_{h = 0}$.

Alternatively, you can define $F(R, h)$ as a function of two variables and what you have calculated is the partial derivative wrt $h$, then set $h = 0$:
$$\frac{\partial F}{\partial h}(R, 0) \equiv \frac{\partial F(R, h)}{\partial h} \big{|}_{h = 0}$$

 

[sha1000]

I was blind to see that this was the partial derivative.

Thank you very much for your help.