- #1

Rick16

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- TL;DR Summary
- In situations, where a function is not given explicitly, but just in a general form like f(ax+by), I am often lost when such functions are being manipulated.

Here are two examples:

The first one is from Taylor's Mechanics, section 6.2:

Taylor takes the function ##f(y+\alpha\eta, y' + \alpha\eta', x)## and differentiates it with respect to ##\alpha##. In the expression for the function, ##y## and ##\eta## depend on ##x##, ##\alpha## is an independent parameter. Here is what he gets: $$\frac {\partial f} {\partial \alpha}= \eta \frac {\partial f} {\partial y} +\eta' \frac {\partial f} {\partial y'}$$

In order to simplify the situation, I am just considering the function ##f(y+\alpha\eta)##. If I understand correctly, this function could be something like ##f=(y+\alpha\eta)^2## or ##f=e^{y+\alpha\eta}## or ##f=\frac {\pi} {\sqrt{y+\alpha\eta}}##, but it could not be something like ##f=y^2+\alpha\eta##, i.e. the term ##y+\alpha\eta## should appear unchanged in the explicit expression for the function. Is this correct?

The simplest such function would then be ##f=y+\alpha\eta##. If I take the derivative of this function with respect to ##\alpha##, I get: ##\frac {\partial f} {\partial \alpha}=\eta##. If I take the derivative of ##f=(y+\alpha\eta)^2##, I get ##\frac {\partial f} {\partial \alpha}=2(y+\alpha\eta)\eta##. Taking the derivative of ##f=e^{y+\alpha\eta}## gives ##\frac {\partial f} {\partial \alpha}=\eta f##. So I actually do get ##\frac {\partial f} {\partial \alpha}=\eta \frac {\partial f} {\partial y}## in all of these cases. But I cannot see how I can arrive at this expression without looking at specific cases. Also, looking at only a few specific cases does not give me the guarantee that the result would be the same in all cases.

The second example is from Landau's Mechanics, §4:

Landau writes the Lagrangian of a free particle as ##L(v^2)## in a reference frame ##K##. He then considers a reference frame ##K'## that moves with a velocity ##\vec{\epsilon}## relative to ##K##, so that the velocity of the particle in the frame ##K'## is ##\vec{v'}=\vec{v}+\vec{\epsilon}##.

He then writes the Lagrangian ##L'=L(v')^2=L(\vec{v}+\vec{\epsilon})^2=L(v^2+2\vec{v}\vec{\epsilon}+\epsilon^2)## and in the next step ##L(v')^2=L(v^2)+\frac {\partial L} {\partial v^2} 2\vec{v}\vec{\epsilon}##. Where does this come from? It looks like a Taylor expansion, but I cannot make sense of it. Why does he take the derivative with respect to ##v^2##? ##L'## does not depend on ##v^2##, it depends on ##(v')^2##. And I don't see how the other terms fit in either.

Apart from these two examples, could someone recommend a text to me where this kind of manipulations is systematically explained?

The first one is from Taylor's Mechanics, section 6.2:

Taylor takes the function ##f(y+\alpha\eta, y' + \alpha\eta', x)## and differentiates it with respect to ##\alpha##. In the expression for the function, ##y## and ##\eta## depend on ##x##, ##\alpha## is an independent parameter. Here is what he gets: $$\frac {\partial f} {\partial \alpha}= \eta \frac {\partial f} {\partial y} +\eta' \frac {\partial f} {\partial y'}$$

In order to simplify the situation, I am just considering the function ##f(y+\alpha\eta)##. If I understand correctly, this function could be something like ##f=(y+\alpha\eta)^2## or ##f=e^{y+\alpha\eta}## or ##f=\frac {\pi} {\sqrt{y+\alpha\eta}}##, but it could not be something like ##f=y^2+\alpha\eta##, i.e. the term ##y+\alpha\eta## should appear unchanged in the explicit expression for the function. Is this correct?

The simplest such function would then be ##f=y+\alpha\eta##. If I take the derivative of this function with respect to ##\alpha##, I get: ##\frac {\partial f} {\partial \alpha}=\eta##. If I take the derivative of ##f=(y+\alpha\eta)^2##, I get ##\frac {\partial f} {\partial \alpha}=2(y+\alpha\eta)\eta##. Taking the derivative of ##f=e^{y+\alpha\eta}## gives ##\frac {\partial f} {\partial \alpha}=\eta f##. So I actually do get ##\frac {\partial f} {\partial \alpha}=\eta \frac {\partial f} {\partial y}## in all of these cases. But I cannot see how I can arrive at this expression without looking at specific cases. Also, looking at only a few specific cases does not give me the guarantee that the result would be the same in all cases.

The second example is from Landau's Mechanics, §4:

Landau writes the Lagrangian of a free particle as ##L(v^2)## in a reference frame ##K##. He then considers a reference frame ##K'## that moves with a velocity ##\vec{\epsilon}## relative to ##K##, so that the velocity of the particle in the frame ##K'## is ##\vec{v'}=\vec{v}+\vec{\epsilon}##.

He then writes the Lagrangian ##L'=L(v')^2=L(\vec{v}+\vec{\epsilon})^2=L(v^2+2\vec{v}\vec{\epsilon}+\epsilon^2)## and in the next step ##L(v')^2=L(v^2)+\frac {\partial L} {\partial v^2} 2\vec{v}\vec{\epsilon}##. Where does this come from? It looks like a Taylor expansion, but I cannot make sense of it. Why does he take the derivative with respect to ##v^2##? ##L'## does not depend on ##v^2##, it depends on ##(v')^2##. And I don't see how the other terms fit in either.

Apart from these two examples, could someone recommend a text to me where this kind of manipulations is systematically explained?