# Homework Help: Momentum not consistent with definition in Landau's book?

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1. Dec 14, 2016

### zhouhao

1. The problem statement, all variables and given/known data
I am not sure whether the meaning of the equation $(3)$ which used for deriving momentum is as same as equation $(4)$.I will make a detailed description below.
The lagrangian function for a free particle is $L=-mc^2\sqrt{1-\frac{v^2}{c^2}} \quad (1)$
The action from point $a$ to point $b$ is :$S=\int_a^bLdt=-mc\int_a^bds$, because $ds=cdt\sqrt{1-\frac{v^2}{c^2}}$
$S=-mc\int_a^b\frac{ds^2}{ds}=-mc\int_a^b\frac{dx_idx^i}{ds}=-mc\int_a^bu_idx^i$

In Landau's book:$\delta{S}=-mc\int_a^bu_id\delta{x^i} \quad (2)$

In my calculation,$\delta{S}=-mc[\int_a^b\delta{u_i}dx^i+\int_a^bu_id\delta{x^i}+\int_a^b\delta{u_i}d\delta{x^i}]$
$\int_a^b\delta{u_i}dx^i=0$ is confirmed.But I can not confirm$\int_a^b\delta{u_i}d\delta{x^i}$ equal to zero.I have no idea how Landau does for equation $(2)$.

Let's still proceed in Landau's way:
Subsection integration for equation $(2)$,we get $\delta{S}=-mcu_i\delta{x^i}|_a^b+mc\int_a^b\delta{x^i}\frac{du_i}{ds}ds$

For moving path of the free particle $\delta{S}=0$,$\quad (\delta{x^i})_a=(\delta{x^i})_b=0$, as a result,$\frac{du_i}{ds}=0$
Not fix $\delta{x^i}_b$ to zero,then $\delta{S}=-mcu_i(\delta{x^i})_b \quad (3)$, according to definition of four dimension momentum,$p_i=mcu_i$.

Why equation $(3)$ equal to equation $(4)$,I can not understand?
2. Relevant equations
Definition of four dimension momentum : $p_i=-\frac{\partial{S}}{\partial{x^i}} \quad (4)$

3. The attempt at a solution
The meaning of Equation $(3)$ is just applying a variation to the real moving path from $a$ to $b$ and not fixing $\delta{x^i}_b$ to zero.I can not figure out whether the physical meaning of equation $(3)$ equal to the one of equation $(4)$ or not.

2. Dec 14, 2016

### phyzguy

When doing variation problems like this, the δ quantities are assumed to be infinitesimal variations. So you only keep terms of first order in δ. Terms with two δs, like the one you are worried about, are infinitesimally small compared to terms with one δ, and can be ignored.

3. Dec 14, 2016

### TSny

phyzguy has explained why you should not include a term with two $\delta$'s.

It's worth noting how Landau and Lifshitz (LL) get the result (2) in a nice way.
Since $ds = \sqrt{dx_idx^i}$, $$\delta (ds) = \delta \left( dx_i dx^i \right) ^ {1/2} = \frac{dx_i \delta (dx^i)}{\left( dx_i dx^i \right) ^ {1/2}} = \frac{dx_i \delta (dx^i)}{ds} = u_i d(\delta x^i)$$ Fom this (2) follows.

I'm not sure I understand your question here. Equation (4) is LL's way of defining the four-momentum $p^i$ of the particle. Then using (4) and (3), they get an explicit expression for the components of four-momentum: $p_i = -\frac{\partial S}{\partial x^i} = - \frac{\delta S}{\delta x^i_b}= mcu_i$. From this, the zeroth component of the four-momentum is seen to be the energy of the particle and the spatial components of the four-momentum are seen to be the components of the three-momentum $\vec{p}$ obtained earlier by LL in their equations (9.1) and (9.4).

4. Dec 15, 2016

### zhouhao

Thank you very much!
I am confused with $\frac{\partial S}{\partial x^i} =\frac{\delta S}{\delta x^i_b}$.
Definition:
Let $x^i(t),t \in [t_a,t_b]$ represent the path $\widetilde{ab}$ of a free particle moving from $a$ to $b$.
$\delta{x^i(t)}$ means a variation path $\delta{\widetilde{ab}}$ to $\widetilde{ab}$ and $\delta{x^i(t_a)}=0$

$\delta{S}=-mc\{\int\limits_{\widetilde{ab}+\delta{\widetilde{ab}}}u_idx^i-\int\limits_{\widetilde{ab}}u_idx^i\}=-mcu_i\delta{x^i(t_b)}$

$dS=-mc\{\int_{(x^0,x^1,x^2,x^3)(t_b)}^{(x^0,x^1,x^2,x^3)(t_b)+dx^i}ds\}=-mc\{\int_{(x^0,x^1,x^2,x^3)(t_b)}^{(x^0,x^1,x^2,x^3)(t_b)+dx^i}u_idx^i\}$

$dS$ depend on the path $\widetilde{dx^i}$ from point $b$ to $(x^0(t_b)+dx^0,x^1(t_b)+dx^1,x^2(t_b)+dx^2,x^3(t_b)+dx^3)$.

For example, calculating $\frac{\partial{S}}{\partial{x^3}}$:

$\frac{\partial{S}}{\partial{x^3}}=-mc\frac{1}{dx^3}\{\int_{(x^0,x^1,x^2,x^3)(t_b)}^{(x^0,x^1,x^2,x^3+dx^3)}u_idx^i\}$

If $\widetilde{dx^i}$ is not derivable at continuous point $b$ of $\widetilde{ab}$ ,then $\frac{\partial{S}}{\partial{x^3}}\neq-mcu_3(t_b)$

It seems that $\frac{\partial S}{\partial x^i} =\frac{\delta S}{\delta x^i_b}$ not absolutely right?

Last edited: Dec 15, 2016
5. Dec 15, 2016

### TSny

I'm sometimes slow at grasping things. I don't understand your phrase "If $\widetilde{dx^i}$ is not derivable at continuous point $b$ of $\widetilde{ab}$". Can you clarify this? Maybe give a specific example of "$\widetilde{dx^i}$ is not derivable".

6. Dec 15, 2016

### zhouhao

Thank you for your help and sorry for the ambiguous.I try to give a detailed clarification.It is so glad to get someone's response from somewhere of Earth.

$\widetilde{dx^i}$ means a infinitesimal path from point $b$,it just like a small additive curve to $\widetilde{ab}$.
Exactly speaking,$\widetilde{dx^i}$ is $(dx^0,dx^1,dx^2,dx^3)$,the different ratio $dx^0:dx^1:dx^2:dx^3$ represent different direction of the infinitesimal path.

For different direction,we can get corresponding $u_i=\frac{dx^i}{ds}$ which may equal to $u_i(t_b)$ or not equal to $u_i(t_b)$(equal means derivable) -----$u_i(t_b)$ is the speed at point $b$ of $\widetilde{ab}$.
If choosing a correct ratio $dx^0:dx^1:dx^2:dx^3$ to let $u_i$ equal to $u_i(t_b)$,then $dS=-mcu_i(t_b)dx^i$.

But choosing a correct ratio does not mean $\frac{\partial{S}}{\partial{x^i}}=-mcu_i(t_b)$,because according math definition $\frac{\partial{S}}{\partial{x^i}}=-mc\frac{1}{dx^i}(\int_{x^i,x^j,\cdots}^{x^i+dx^i,x^j,\cdots}ds)$ requaires :$dx^i\neq0 \quad dx^{j{\neq}i}=0$ and the chosen ratio may not satisfy the requirement.

Since $dS=u_i(t_b)dx^i$ require specific ratio of $dx^0:dx^1:dx^2:dx^3$, $\frac{\partial{S}}{\partial{x^i}}$ not absolutely equal to $u_i(t_b)$;
$\frac{\partial{S}}{\partial{x^i}}{\neq}\frac{\delta{S}}{\delta{x^i}}$,I do not know how to understand the four momentum now.

Last edited: Dec 15, 2016
7. Dec 15, 2016

### TSny

There is no need to require the ratio $dx^0:dx^1:dx^2:dx^3$ to correspond to the direction of $u^i_b$. Equation (9.11) in LL will still be valid.

The black line $ab$ in the figure below represents the actual trajectory of the free particle from event $a$ to event $b$. For a free particle, the actual path is a straight line.

Let $S\left(t_b, x_b \right)$ represent the action for this path. This action also depends on $\left( t_a, x_a \right)$, but since this point is considered fixed, we don't bother to indicate the functional dependence of $S$ on this point.

Let the end point $b$ be moved to $b'$, while $a$ is kept fixed. The brown line shows the new actual trajectory between $a$ and $b'$. The action for this path is $S\left(t_{b \,'}, x_{b \,'} \right)$. The variation in the end point from $b$ to $b'$ is indicated by the blue arrow $\delta$. The components of $\delta$ are $\delta t$ and $\delta x$. Or more generally, $\delta$ would have space time components that LL represent by $\delta x^i = \left(\delta x^0, \delta x^1, \delta x^2, \delta x^3 \right)$. Again, there is no need for $\delta x^0:\delta x^1:\delta x^2:\delta x^3$ to equal $u^0_b:u^1_b:u^2_b:u^3_b$. Equation (9.11) of LL will be valid for arbitrary $\delta x^i$. That is, you have $\delta S = -m c u_{i_b} \delta x^i$, where $u_{i_b}$ is the four-velocity at $b$ for path $ab$. So, this equation would be valid if you choose, say, $\delta x^3 \neq 0$ and all other $\delta x^i = 0$.

Last edited: Dec 15, 2016
8. Dec 16, 2016

### zhouhao

You are absolutely right and give a very good picture.
As I understand, $dS$ is different with $\delta{S}$.
I would use your picture to give a clarification.

Straight line $ab$ is actual path of a free particle and $ab'$ is a infinitesimal variation path to $ab$.
$u_i^b$ is the speed at point $b$.
$\delta{S}=-mc(\int\limits_{ab'}u_idx^i-\int\limits_{ab}u_idx^i)\approx-mcu_i^b\delta{x_b^i}$.

However,I think $dS=-mc\int\limits_{bb'}u_idx^i=-mc(\sqrt{\delta{x^i_b}\delta{x_i^b}})$
and $\frac{\partial{S}}{\partial{x^i}}=\frac{dS}{\delta{x^i}}=mci\neq\frac{\delta{S}}{\delta{x^i}}$ when only $\delta{x^i}\neq0$

This nonconsistence confuse me.

9. Dec 16, 2016

### TSny

I think we're narrowing it down to the specific difficulty. As I said, it can take me a while.
I don't think that what you are calling $dS$ is relevant.
It is important to keep in mind how $S(t, x)$ is defined. [$x$ can be extended to 3 dimensional space $(x^1, x^2, x^3)$.] We first pick a reference spacetime point $a$: $\left(t_a, x_a \right)$ The value of $S(t_b, x_b)$ at some spacetime point $(t_b, x_b)$ is the action $\int_{t_a}^{t_b} L dt = -mc\int\limits_{ab}u_idx^i$ along the actual path of a free particle that travels between events $(t_a, x_a)$ and $(t_b, x_b)$.

If we are interested in the partial derivative $\frac{\partial{S}}{\partial{x}}$ of the function $S(t, x)$ at some point $(t_b, x_b)$, it would be defined in the standard way as limit as $\Delta x \rightarrow 0$ of $\frac{S(t_b, x_b + \Delta x) - S(t_b, x_b) }{\Delta x} = \frac{\delta S}{\delta x}$. Here the $\delta S$ and $\delta x$ notations are in agreement with how you are using this notation. Of course, for the partial derivative we chooses $\delta t = 0$ while $\delta x \neq 0$. In three dimensions of space, if we wanted $\frac{\partial{S}}{\partial{x^3}}$ we would choose $\delta x^3 \neq 0$ and $\delta t = \delta x^1 = \delta x^2 = 0$.

You use the notation $dS$ to represent $-mc\int\limits_{bb'}u_idx^i$. But I don't see how this quantity comes into play in the discussion in LL. If we make an arbitrarty variation of $b$ to $b'$, the change in the action would be given by $S(t_b', x_b') - S(t_b, x_b) = \delta S \neq dS = -mc\int\limits_{bb'}u_idx^i$.

When we vary $b$ to $b'$ the change in action is the difference in action of the brown and black paths in the diagram. The change in action is not given by the action along the blue displacement.