Why Lagrangian includes 1/2m by Landau?

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• gionole
gionole
I feel like I might be posting many questions, so I hope you're not angry with me on this.

I'm following Landau's book and try to understand why L = K - U but first we need to figure out why it contains K.

Landau discusses two inertial frames where the speed of one frame relative to another is ##\epsilon##.

He mentions:

frame 1: ##L(v^2)##

frame 2: ##L(\hat v^2) = L(v^2) + \frac{\partial L}{\partial v^2} 2v\epsilon##

Then he says ##\frac{\partial L}{\partial v^2}## must be independent of velocity in order to have ##L' = L + \frac{d}{dt}(q,\dot q)## where we can omit ##\frac{d}{dt}(q,\dot q)##.

Everything clear til now, but then he out of nowhere says that L must be proportional to ##\frac{1}{2}mv^2##.. I get why it contains ##v^2##, but why ##\frac{1}{2}m ## ? Note that at this point, he has no idea that L must contain kinetic energy(well, he might have, but whole idea of this derivation is to derive it without knowing things beforehand.
I'm attaching the image of his logic. I'd appreciate if you could explain in his words and not bring complex equations not related to what he says in there. The reason of this is since I'm following his book, I want to exactly know his thought process and don't want to prove it with other methods. Why ##\frac{1}{2}m## ? If I could say ##\frac{1}{4}m##, what would go wrong ? other than the fact that in the end, he would solve it with euler lagrange and get a wrong equation ? he must have had the reason to exactly come up with ##\frac{1}{2}m##

Hope, I'm not too big of a pain. and thanks for all the help. This forum is amazing.

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dextercioby
gionole said:
I feel like I might be posting many questions, so I hope you're not angry with me on this.

I'm following Landau's book and try to understand why L = K - U but first we need to figure out why it contains K.

Landau discusses two inertial frames where the speed of one frame relative to another is ##\epsilon##.

He mentions:

frame 1: ##L(v^2)##

frame 2: ##L(\hat v^2) = L(v^2) + \frac{\partial L}{\partial v^2} 2v\epsilon##

Then he says ##\frac{\partial L}{\partial v^2}## must be independent of velocity in order to have ##L' = L + \frac{d}{dt}(q,\dot q)## where we can omit ##\frac{d}{dt}(q,\dot q)##.

Everything clear til now, but then he out of nowhere says that L must be proportional to ##\frac{1}{2}mv^2##.. I get why it contains ##v^2##, but why ##\frac{1}{2}m ## ? Note that at this point, he has no idea that L must contain kinetic energy(well, he might have, but whole idea of this derivation is to derive it without knowing things beforehand.
I'm attaching the image of his logic. I'd appreciate if you could explain in his words and not bring complex equations not related to what he says in there. The reason of this is since I'm following his book, I want to exactly know his thought process and don't want to prove it with other methods. Why ##\frac{1}{2}m## ? If I could say ##\frac{1}{4}m##, what would go wrong ? other than the fact that in the end, he would solve it with euler lagrange and get a wrong equation ? he must have had the reason to exactly come up with ##\frac{1}{2}m##

Hope, I'm not too big of a pain. and thanks for all the help. This forum is amazing.
Read the sentence just below equation 4.1. Landau isn't deriving Mechanics without some contact with previous work: he's including the (1/2)m because, in the end, that's what's going to appear as a factor if we compare the formulas with Galileo's (et. al.) work. He could simply call it C and then later on derive that C = (1/2)m. You could add that to your derivation, if it makes you feel any better. It's actually a (slightly) better derivation if you do it that way.

-Dan

DrClaude and vanhees71
Interesting. Somehow I have hard time imagining how to do it.

##L(v^2) = cv^2##

##L' = c(v+V)^2 = cv^2 +2cvV + V^2 = cv^2 + \frac{d}{dt}(2crV + V^2)##

How does this yield ##c = \frac{1}{2}m## ?

I am afraid that the English tanslation of the text is not exact. I don't have the original Russian but have two translations of French and Japanese. The both state
$$L=av^2$$
adding "It is accostomed to write constant a as m/2". I don't find it in the English pdf on the web.

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DrClaude and dextercioby
gionole said:
Interesting. Somehow I have hard time imagining how to do it.

##L(v^2) = cv^2##

##L' = c(v+V)^2 = cv^2 +2cvV + V^2 = cv^2 + \frac{d}{dt}(2crV + V^2)##

How does this yield ##c = \frac{1}{2}m## ?
It doesn't. Apply it to a problem and compare the answer to the answer given by other methods.

-Dan

topsquark said:
It doesn't. Apply it to a problem and compare the answer to the answer given by other methods.

-Dan
Well, thats true but he did not say it clearly if thsts how he figured it out. Thank you

Thanks @anuttarasammyak . Lots of mistakes in english version.

While I have your attention, I wanted to ask one interesting thing.

I have seen lots of questions on the forums, stackexchange and so on asking why Landau says L must be a function of ##v^2## and not ##v^4##. They think that ##v^4## would work too. but here is the interesting part. If L is ##L(v^4)##, then ##L'(v'^4) = L(v^4) + \frac{\partial L}{\partial v^4} 4v^3 \epsilon##

There's no chance ##\frac{\partial L}{\partial v^4} 4v^3 \epsilon## part can be omitted the same way as it happens for when we had ##L(v^2)##. Even if you assume ##\frac{\partial L}{\partial v^4}## doesn't contain any more ##v##. Let's say ##\frac{\partial L}{\partial v^4} = c##

##L'(v'^4) = L(v^4) + c 4v^3 \epsilon = L(v^4) + c 4 \dot q \dot q \dot q \epsilon =L(v^4) + \frac{d}{dt}(4cq \dot q \dot q)##

If you use euler lagrange for: ##L' = L + \frac{d}{dt}(4cq \dot q \dot q)##, no chance you get the same equation of motion, while when L was a function of only ##v^2##, it yielded the same equations.

I'm not sure where my logic is incorrect, it just can't be a function of ##v^4##, but I have heard people say that it could easily be ##v^4## and logic would work. Am I making a mistake on my logic ?

V^4 is function of v^2. Isn’t it enough?

topsquark
anuttarasammyak said:
V^4 is function of v^2. Isn’t it enough?
yeah, but tried writing it step by step. do you find any mistake in my proof ? what I proof is if L is a function of ##v^4##, in two inertial frames, equation are not the same because for ##v^4## case, ##\frac{d}{dt}f(q,t)## doesn't vanish.

I wonder why people say it could have been ##v^4## and would still work. how

Say Taylor Maclaurin expansion of L(v^2) is
$$L(v^2)=L(0)+a_1v^2+a_2v^4+...$$
The expansion of L(v^4) is
$$L(v^4)=L(0)+b_1v^4+b_2v^8+...$$
The latter is the special case of the former of
$$a_1=0, a_2=b_1,a_3=0, a_4=b_2,...$$
So the investigation of L(v^2) is enough to cover L(v^4). Nature decides whether a_1=0 or not.

dextercioby and topsquark
anuttarasammyak said:
Say Taylor Maclaurin expansion of L(v^2) is
$$L(v^2)=L(0)+a_1v^2+a_2v^4+...$$
The expansion of L(v^4) is
$$L(v^4)=L(0)+b_1v^4+b_2v^8+...$$
The latter is the special case of the former of
$$a_1=0, a_2=b_1,a_3=0, a_4=b_2,...$$
So the investigation of L(v^2) is enough to cover L(v^4). Nature decides whether a_1=0 or not.
1. I have a hard time understanding if your way suggests that it would work for ##v^4## or not.
2. where do you find flaw in my logic in reply #7 ?

1. Say L(v^4),
$$L(v^4) = L((v^2)^2) = F(v^2)$$
You may agree that a function of v^4 is a function of v^2.
Lagrangean as a function of v^4 is Lagrangean as a function of v^2.

2. Why you can disregard the case that
$$\sqrt{v^4}$$
would matter in your Lagrangean L(v^4) ?

The simplest saying that direction of ##\mathbf{v}## does not matter is Lagrangean is a function of v^2 which is ##\mathbf{v}\cdot\mathbf{v}##. It is easier to handle than a function of ##|\mathbf{v}|## which is ##\sqrt{v^2}##.

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topsquark
@anuttarasammyak

I get your point, but I'm eager to look at it with different perspective. I want to show that if L contains ##v^4##, plain ##v^4##, without square root for example, e.x - ##L = cv^4##. Then I try to prove this wouldn't be possible as ##\frac{\partial L}{\partial v^4}c4v^3\epsilon## can be written as ##\frac{d}{dt}(c q4 \dot q \dot q \epsilon)## - considering ##\frac{\partial L}{\partial v^4}## does not contain ##v## and I prove in ##7## reply that such ##L' = L + \frac{d}{dt}(c q4 \dot q \dot q \epsilon)## would yield different equation of motion. I'm asking why this proof can not be solid, because if this is correct, this immediatelly proves that it can't contain higher power than ##v^2##. (forget about square root or different representations, we only talk whether L can be ##cv^3##, ##cv^4##...). Maybe I'm wrong about ##c4v^3 == \frac{d}{dt}(c q4 \dot q \dot q \epsilon)##.

gionole said:
then L′(v′4)=L(v4)+∂L∂v44v3ϵ
Say
$$L'(v'^4) = L(v^4) + \frac{\partial L}{\partial v^4} 4v^3 \epsilon$$
and
$$L(v^4)=cv^4$$
, the second term of RHS of the former eqution becomes
$$c4v^3 \epsilon$$
which cannot be a complete time derivative
$$\frac{d}{dt}f(q,t)$$
. So the latter equation form cannot be Lagrangean which is expected to have the same function of velocity in all the IFR.

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topsquark
The idea is to derive the Lagrangian from the demand that the theory you get is in accordance with the structure of Galilei-Newton spacetime, which implies that the theory must be invariant under Galilei transformations. Now asking this for a single particle, assuming a Lagrangian of the form ##L(\vec{x},\dot{\vec{x}},t)## leads from translation invariance in time and space to ##L=L(\dot{\vec{x}})##. Then you have invariance under rotations, which means that ##L=L(\dot{\vec{x}}^2)##.

Now consider the invariance under boosts, i.e.,
$$\vec{x}'=\vec{x}+\vec{v} t, \quad \vec{v}=\text{const}.$$
This implies
$$\dot{\vec{x}}'=\dot{\vec{x}} + \vec{v} t.$$
Now we make ##\vec{v}## infinitesimally small and derive how ##L## changes to linear order in ##\vec{v}##:
$$L(\dot{\vec{x}}^{\prime 2})-L(\dot{\vec{x}}^2)=L'(\dot{\vec{x}}^2) 2 \dot{\vec{x}} \cdot \vec{v} + \mathcal{O}(\vec{v}^2).$$
Now in order to make the Galilei boost a symmetry, this difference must be (up to first order in ##\vec{v}##) a total time derivative, and that's the case if and only if ##L'(\dot{\vec{x}}^2)=C=\text{const}##.

So for a single partice, all you can say is that
$$L=C \dot{\vec{x}}^2,$$
with ##C## an indetermined factor. From Hamilton's principle all that this imply is that according to the Euler-Lagrange equations,
$$C \ddot{\vec{x}}=\text{const}. \; \Rightarrow \; \dot{\vec{x}}=\vec{v}_0=\text{const} \; \Rightarrow \; \vec{x}=\vec{x}_0+\vec{v}_0 t.$$
So from the symmetry under the full Galilei group for one particle you get simply Newton's 1st Law.

Now you can think about 2 particles. At the end you get that the Lagrangian must be of the form
$$L=C_1 \dot{\vec{x}}_1^2 + C_2 \dot{\vec{x}}_2^2 -V(|\vec{x}_1-\vec{x}_2|),$$
i.e., on the fundamental level you have a central conservative force, and the equations of motion read
$$2 C_1 \ddot{\vec{x}}_1=-\frac{\partial}{\partial \vec{x}_1} V(|\vec{x}_1-\vec{x}_2|)=-\frac{\vec{x}_1-\vec{x}_2}{|\vec{x}_1-\vec{x}_2|} V'(|\vec{x}_1-\vec{x}_2|),$$
$$2 C_2 \ddot{\vec{x}}_2=-\frac{\partial}{\partial \vec{x}_2} V(|\vec{x}_1-\vec{x}_2|)=-\frac{\vec{x}_2-\vec{x}_1}{|\vec{x}_1-\vec{x}_2|} V'(|\vec{x}_1-\vec{x}_2|).$$
So you see that you get the 2nd and 3rd Newtonian postualtes. Since Galilei and Newton the convention has been that the force on a particle is proportional to its acceleration and that the proportionality constant is called the "mass of the particle". Thus we have to set ##C_1=m_1/2## and ##C_2=m_2/2##.

How you call the parameters of the model is a matter of convention, and the choice of the mass in this way is simply pretty convenient concerning all kinds of numerical factors.

topsquark
anuttarasammyak said:
Say
$$L'(v'^4) = L(v^4) + \frac{\partial L}{\partial v^4} 4v^3 \epsilon$$
and
$$L(v^4)=cv^4$$
, the second term of RHS of the former eqution becomes
$$c4v^3 \epsilon$$
which cannot be a complete time derivative
$$\frac{d}{dt}f(q,t)$$
. So the latter equation form cannot be Lagrangean which is expected to have the same function of velocity in all the IFR.
Is not this what I said ? The only difference is I represented ##c4v^3## to be ##\frac{d}{dt}(c4q\dot q \dot q)## and if you do euler-lagrange, you fail to get same EOM. I think we're on the same page. Don't you ?

Edit

Ah, I can't represent ##c4v^3## to be ##\frac{d}{dt}(c4q\dot q \dot q)##. Now, clear.

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anuttarasammyak
The point is that two Lagrangians are equivalent if they only differ by a total time derivate of a function ##f(q,t)##, i.e., ##f## must not depend on ##\dot{q}##.

gionole
vanhees71 said:
The point is that two Lagrangians are equivalent if they only differ by a total time derivate of a function ##f(q,t)##, i.e., ##f## must not depend on ##\dot{q}##.
Got it. Thanks so much.

I think I only got one question remaining on this topic if you guys got time and don't mind.

For freely moving particle, It's said ##L## can't depend on velocity vector, but magnitude.

I'm looking for the counter-argument. Let's say it depends on velocity vector. Then, how would Lagrangian be written in terms of velocity vector and why would it yield the wrong solution for freely moving particle? (no need for bringing potential in this).

Looking for the math proof. I know Euler-lagrange. If it depends on velocity vector, ##L = \frac{1}{2}m(v_i + v_j)## and this yields ##\frac{d}{dt}(\frac{1}{2}m) = 0##. Would this be correct approach to prove what I'm asking ? I'm trying to get the idea why Landau makes it depend on ##v^2##. (I'm not asking why it doesn't depend ##v^4##). My main question is to rigorously show why it can't depend on vector, and if it did, what would it break for freely moving particle ? It mentions isotrophy of space, but want to see in math proof how it's wrong. Note that no need to bring potential in this. We only focus on freely moving particle.

That's rotational invariance. An infinitesimal rotation is paramerized as
$$\delta \vec{x}=\delta \vec{\omega} \times \vec{x},$$
and then, assuming ##L=L(\dot{\vec{x}})## (which follows from time and space translation invariance), this implies
$$\delta L=(\delta \vec{\omega} \times \dot{\vec{x}}) \cdot \frac{\partial L}{\partial \dot{\vec{x}}} = \delta \vec{\omega} \cdot \left (\dot{\vec{x}} \times \frac{\partial L}{\partial \dot{\vec{x}}} \right).$$
This is 0 for all ##\delta \vec{\omega}## if
$$\dot{\vec{x}} \times \frac{\partial L}{\partial \dot{\vec{x}}}=0.$$
This implies that
$$\frac{\partial L}{\partial \dot{\vec{x}}}=C \dot{\vec{x}} \; \Rightarrow \; L=\frac{C}{2} \dot{\vec{x}}^2$$
with ##C=\text{const}##.

wrobel
vanhees71 said:
That's rotational invariance. An infinitesimal rotation is paramerized as
$$\delta \vec{x}=\delta \vec{\omega} \times \vec{x},$$
and then, assuming ##L=L(\dot{\vec{x}})## (which follows from time and space translation invariance), this implies
$$\delta L=(\delta \vec{\omega} \times \dot{\vec{x}}) \cdot \frac{\partial L}{\partial \dot{\vec{x}}} = \delta \vec{\omega} \cdot \left (\dot{\vec{x}} \times \frac{\partial L}{\partial \dot{\vec{x}}} \right).$$
This is 0 for all ##\delta \vec{\omega}## if
$$\dot{\vec{x}} \times \frac{\partial L}{\partial \dot{\vec{x}}}=0.$$
This implies that
$$\frac{\partial L}{\partial \dot{\vec{x}}}=C \dot{\vec{x}} \; \Rightarrow \; L=\frac{C}{2} \dot{\vec{x}}^2$$
with ##C=\text{const}##.
I haven't fully moved to invariance topic and ##w## symbol is unclear to me. At this point, I prefer different proof.

My proof is that If ##L## depends on velocity vector, i.e ##L = \frac{1}{2}m(v_x\vec i+ v_y\vec j)## , this yields ##\frac{d}{dt}(\frac{1}{2}m) = 0## which yields ##0=0##. Since it's meaningless, we move to higher order of velocity vector = ##|\vec v|^2##. What would be wrong with this ? I didn't even need isotrophic property

gionole said:
Looking for the math proof. I know Euler-lagrange. If it depends on velocity vector, L=12m(vi+vj) and
What are vi and vj ? One of them is constant ?

anuttarasammyak said:
What are vi and vj ? One of them is constant ?
use ##v_{x\vec i} + v_{y\vec j}##

Thanks.
gionole said:
, i.e L=12m(vxi→+vyj→) , t

Your Lagrangean is not scalar but 2d vector which I am not accustomed to deal with. Action which is integral of Lagrangean is also vector in your theory ?

anuttarasammyak said:
Thanks.Your Lagrangean is not scalar but 2d vector which I am not accustomed to deal with. Action which is integral of Lagrangean is also vector in your theory ?
@anuttarasammyak

Yes, Action would end up the vector as well. Do you think if action were a vector, by varional calculus, we wouldn't derive euler-lagrange in the first place ? I don't know why, but in my opinion, it still should work for ##\vec L## the same way. Thoughts ?

---

I also wanted to ask. Landau says that if we got 2 inertial frames, equations of motion have the same form. Imagine a train 1 moving with constant speed of 50km/h and a train 2 moving with 55km/h(relative to train 1, it moves by 5km/h). Imagine a table in both trains and we roll the same mass of ball with the same force in both trains.

Train 1: equation of motion of the ball: ##x_0 + v_0t + at^2## where ##a = \frac{F}{m}## (##F## is force I roll it with). In train 2, the equation of motion would end up having the same form.

If now, train 1 is inertial and train 2 non-inertial, then by Landau's logic, we should end up having different forms of equation, but look, in train 2's frame, ball's acceleration would be: ##a = \frac{F}{m} - \frac{F_{fictitious}}{m}##, but we still would have the same form of equations in both trains.(the only difference would be acceleration value). What is meant by different form in terms of my exact example ?

gionole said:
I haven't fully moved to invariance topic and ##w## symbol is unclear to me. At this point, I prefer different proof.

My proof is that If ##L## depends on velocity vector, i.e ##L = \frac{1}{2}m(v_x\vec i+ v_y\vec j)## , this yields ##\frac{d}{dt}(\frac{1}{2}m) = 0## which yields ##0=0##. Since it's meaningless, we move to higher order of velocity vector = ##|\vec v|^2##. What would be wrong with this ? I didn't even need isotrophic property
This cannot be a right ansatz in the first place, because ##L## is not a vector quantity to begin with.

What do you mean by "##w## symbol"?

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