I'm searching for a specific book with the derivation of ##E\propto v^2##?

[exponent137]

TL;DR Summary

I ask for the reference for the following derivation of $E\propto v^2$ or a similar one.
https://www.physicsforums.com/threads/the-final-explanation-to-why-kinetic-energy-is-proportional-to-velocity-squared.78484/#post-609992

Twice I found the following derivation of $E\propto v^2$ in a little distinct forms.
https://www.physicsforums.com/threa...tional-to-velocity-squared.78484/#post-609992
The derivation is in post #9, if it is not shown properly.

This means, this derivation is without $E=Fdx$ formula, without calculus of variations, without special relativity, etc.

I am interested, what is a reference for this or any other similar derivation. What I heard additionally that this is not in the book
"Landau and Lifgarbagez Classical Mechanics ", that this was written by a russian author, somewhere after 1950, translated in English, in a red cover.

The most probably, this is neither this book of Arnold:
https://loshijosdelagrange.files.wo...tical-methods-of-classical-mechanics-1989.pdfas I searched in it.

Similar but not so specific questions were given twice already. But they were not answered.

NOTE:
Can I ask that you leave this question unanswered if you do not know a precise answer. In such a case it will stay in "UNANSWERED THREADS", this means that someone will notice this thread some day and will answer to this. In the opposite case, this thread will be hidden somewhere in PF, and no one will answer to it.

 

[mitochan]

Hi.
SR says energy of a particle of mass m is
$$E=\frac{mc^2}{\sqrt{1-v^2/c^2}}=mc^2\{1+1/2(v/c)^2+3/8(v/c)^4+...\}$$
$$=mc^2 + 1/2 \ mv^2 + 3mc^2/8 \ (v/c)^4 + ...$$
With the constant term $mc^2$ and higher $(v/c)^{2n}$ ,where n>1, terms we cannot say E is proportional to $v^2$ any more.

 

[exponent137]

Hi.
SR says energy of a particle of mass m is
$$E=\frac{mc^2}{\sqrt{1-v^2/c^2}}=mc^2\{1+1/2(v/c)^2+3/8(v/c)^4+...\}$$
$$=mc^2 + 1/2 \ mv^2 + 3mc^2/8 \ (v/c)^4 + ...$$
With the constant term $mc^2$ and higher $(v/c)^{2n}$ ,where n>1, terms we cannot say E is proportional to $v^2$ any more.

Yes, but SR can also be used for the derivation of $E\propto v^2$ in Newtonian mechanics.

But here I wish a reference for the derivation, which I linked. Now a theory behind it is not so important for me, as the reference for this book, where the derivation for https://www.physicsforums.com/threa...tional-to-velocity-squared.78484/#post-609992 is written.

If no one knows this reference, maybe someone knows some hint for it. I also searched in https://en.wikipedia.org/wiki/List_of_Russian_physicists, but even Vladimir Igorevich Arnold is not written in the link, https://en.wikipedia.org/wiki/Vladimir_Arnold. But here https://en.wikipedia.org/wiki/List_of_Russian_mathematicians he is.

 

[Ibix]

What I heard additionally that this is not in the book
"Landau and Lifgarbagez Classical Mechanics ", that this was written by a russian author, somewhere after 1950, translated in English, in a red cover.

Incidentally, since you sound uncertain about it, this book is by Landau and Lifshitz. I believe Greg ran an overzealous profanity filter over PF at some point, and it edited the latter's name in quite a few posts.

 

[exponent137]

Incidentally, since you sound uncertain about it, this book is by Landau and Lifshitz. I believe Greg ran an overzealous profanity filter over PF at some point, and it edited the latter's name in quite a few posts.

Now I see that "Lifgarbagez" is wrong. I thought that both options are correct.
But, I copied this name from PF.

 

[mitochan]

$E\propto v^2$ in Newtonian mechanics.

L-L texts in section 4 says for L, Lagrangean of a free particle,
$$\frac{\partial L}{\partial v^2}$$ does not depend on $v$.
This is the most direct derivation I know.

 

[exponent137]

L-L texts in section 4 says for L, Lagrangean of a free particle,
$$\frac{\partial L}{\partial v^2}$$ does not depend on $v$.
This is the most direct derivation I know.

Maybe the most direct, but too abstract.

Derivation https://www.physicsforums.com/threa...tional-to-velocity-squared.78484/#post-609992
is also not ideal. But, I wish only reference for this.

 

[vanhees71]

I've no idea what you are after. The most simple derivation of why $E_{\text{kin}}=m \vec{v}^2/2$ is a useful quantity I know of is the "work-energy theorem". It follows from Newtons equation,
$$m \ddot{\vec{x}}=\vec{F},$$
valid along the trajectory of the particle. Multiplying with $\dot{\vec{x}}$ gives
$$\mathrm{d}_t \left (\frac{m}{2} \dot{\vec{x}}^2 \right)=\dot{\vec{x}} \cdot \vec{F}.$$
If $\vec{F}$ is conservative, i.e.,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}),$$
also the right-hand-side is a total time derivative and thus you get the energy conservation law
$$\mathrm{d}_t \left (\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x}) \right)=0.$$
That's, why it is useful to define $E_{\text{kin}}$ in its standard form. I guess, it's hard to find a mechanics textbook, where this is not derived.

 

[exponent137]

I've no idea what you are after. The most simple derivation of why $E_{\text{kin}}=m \vec{v}^2/2$ is a useful quantity I know of is the "work-energy theorem". It follows from Newtons equation,
$$m \ddot{\vec{x}}=\vec{F},$$
valid along the trajectory of the particle. Multiplying with $\dot{\vec{x}}$ gives
$$\mathrm{d}_t \left (\frac{m}{2} \dot{\vec{x}}^2 \right)=\dot{\vec{x}} \cdot \vec{F}.$$
If $\vec{F}$ is conservative, i.e.,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}),$$
also the right-hand-side is a total time derivative and thus you get the energy conservation law
$$\mathrm{d}_t \left (\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x}) \right)=0.$$
That's, why it is useful to define $E_{\text{kin}}$ in its standard form. I guess, it's hard to find a mechanics textbook, where this is not derived.

I please you still for reference this derivation. I think that precisely such derivation does not exist in "Landau, Lifshitz"?

But still ever I please for derivation given in the first post.

 

[vanhees71]

I don't know a specific source, but where is the need? The calculation is simple enough, isn't it? I'm pretty sure it's in almost any introductory textbook. Landau&Lifshitz vol. I is of course among the best classical-mechanics texts ever, particularly because it doesn't bother its readers with mechanics without using Hamilton's action principle ;-)).