Relativistic effect of a free particle

[Lambda96]

Homework Statement

Show that $L[\gamma]$ takes the following form for small $\frac{v}{c}$ $L[\gamma] = \alpha_1 + \alpha_2 \int_{t_i}^{t_f} \mathrm{d}t \, \frac{1}{2} m v(t)^2 + \mathcal{O}\left(\left(\frac{v}{c}\right)^4\right)$

Relevant Equations

none

Hi,

I am stuck with the following task

I have developed a Taylor expansion for $L[\gamma]=\sqrt{c^2-v^2}$ up to the third order for the position $a=0$, for this I have rewritten $L[\gamma]$ as follows:

$$L[\gamma]=\sqrt{c^2-v^2}=c \sqrt{1-\frac{v^2}{c^2}}$$

Then I did the following $x=\frac{v^2}{c^2}$ and got the following expression for Taylor:

$$L[\gamma] \approx c - \frac{v^2}{2c}+ \frac{v^4}{8 c^3}-\frac{v^6}{16 c^5}+\mathcal{O}\left(\left(\frac{v^2}{c^2}\right)^4\right)$$

Then I formed the integral

$$L[\gamma]=\int_{t_i}^{t_f} dt \quad c - \frac{v^2}{2c}+ \frac{v^4}{8 c^3}-\frac{v^6}{16 c^5}+\mathcal{O}\left(\left(\frac{v^2}{c^2}\right)^4\right)$$

Then integral can now be divided

$$L[\gamma]=\int_{t_i}^{t_f} dt \quad c - \int_{t_i}^{t_f} dt \frac{v^2}{2c}+ \int_{t_i}^{t_f} dt \frac{v^4}{8 c^3}- \int_{t_i}^{t_f} dt \frac{v^6}{16 c^5}+ \int_{t_i}^{t_f} dt \mathcal{O}\left(\left(\frac{v^2}{c^2}\right)^4\right)$$

For the first part I would now assume the following $\int_{t_i}^{t_f} dt \quad c = c(t_f -t_i)=\alpha_1$ but unfortunately I can't get any further with the rest.

 

[PeroK]

Hmm? The book looks a bit wonky to me.

Where did the $m$ come from? And, if the constant $\alpha_2$ has a factor of $1/c$, then the $(v/c)^4$ doesn't look right.

 

[PeroK]

PS the dimensions of the quantities in the final integral are inconsistent. The first term, in $v^2$, has dimensions of velocity squared. The subsequent terms are dimensionless.

 

[Orodruin]

The problem requires $\mathcal O((v/c)^4)$, not $\mathcal O((v^2/c^2)^4)$ …
Just discard terms of order higher than $v^2$ and you have the solution after identifying $\alpha_2 = 1/mc$.

Hmm? The book looks a bit wonky to me.

Not to me. Looks perfectly fine.

Where did the $m$ come from?

From setting $\alpha_2 = 1/mc$. Not to say I agree with the approach. The Lagrangian needs a factor of $m$ to have the right units. It cannot have units of time or length.

And, if the constant $\alpha_2$ has a factor of $1/c$, then the $(v/c)^4$ doesn't look right.

It is $\mathcal O((v/c)^4)$.

PS the dimensions of the quantities in the final integral are inconsistent. The first term, in $v^2$, has dimensions of velocity squared. The subsequent terms are dimensionless.

Looks perfectly consistent to me once you factor in the $\mathcal O$ being multiplied by some dimensional constant (ie, c).

 

[PeroK]

You don't see a problem with there being no $m$ in equation (4) but an $m$ in equation (5)?

I still don't get how an expression like $mv^2 + \mathcal O((\frac v c)^4)$ can be dimensionally consistent.

Ultimately, to study successfully from a book you have to be on the same wavelength as the author. I note also that the author has $v(t)$ and $v$ in the same expression.

That's all too sloppy, wild and woolly for my taste.

 

[Orodruin]

You don't see a problem with there being no $m$ in equation (4) but an $m$ in equation (5)?

No. There are also $\alpha$s in (5) which are not in (4). These have to be adjusted appropriately.

I still don't get how an expression like $mv^2 + \mathcal O((\frac v c)^4)$ can be dimensionally consistent.

$\mathcal O((\frac v c)^4)$ means $\alpha (v/c)^4 + \mathcal (v/c)^5$. Again $\alpha$ is dimensional. Not the best of notations, but still.

Ultimately, to study successfully from a book you have to be on the same wavelength as the author. I note also that the author has $v(t)$ and $v$ in the same expression.

They mean different things though. The $v$ is outside the integral and so there is no $t$ to consider. The author may be using $\mathcal O$ notation somewhat sloppily, but I think it is quite clear what is intended.

That's all too sloppy, wild and woolly for my taste.

 

[PeroK]

They mean different things though. The $v$ is outside the integral and so there is no $t$ to consider. The author may be using $\mathcal O$ notation somewhat sloppily, but I think it is quite clear what is intended.

I assumed that the $\mathcal O$ was under the integral sign. It's certainly ambiguous, with the $dt$ not at the end of the integrand. The integral term needs a bracket round it to make things clear for me.

It seems the OP suffered from the same difficulty as I did to know what the author intended. If a textbook is only clear to those who know the subject inside out, then it may be of little use to those trying to learn the subject for the first time.

The OP and I have to go through all the permutations of what the author might have meant.

 

[Orodruin]

I assumed that the $\mathcal O$ was under the integral sign. It's certainly ambiguous, with the $dt$ not at the end of the integrand. The integral term needs a bracket round it to make things clear for me.

It seems the OP suffered from the same difficulty as I did to know what the author intended. If a textbook is only clear to those who know the subject inside out, then it may be of little use to those trying to learn the subject for the first time.

The OP and I have to go through all the permutations of what the author might have meant.

That is one thing, it is quite another thing to claim expressions are dimensionally inconsistent and certainly wrong when it is not the case.

I do not think the OP suffers from the same problems as you do either. The main issue of the OP seems to be misreading the $\mathcal O((v/c)^4)$ as $\mathcal O((v^2/c^2)^4)$, which is a very different thing. Apart from that, OP seems to be doing well.

 

[Lambda96]

Thank you PeroK and Orodruin for your help

Do I then have to proceed as follows?

$$\begin{align*}
L(\gamma)&=c- \frac{c}{2} x + \mathcal{O}^2\left(\left(x\right)\right) \\
L(\gamma)&=c- \frac{v^2(t)}{2c} + \mathcal{O}^2\left(\left( \frac{v^2}{c^2}\right)\right)\\
L(\gamma)&=c- \frac{v^2(t)}{2c} + \mathcal{O}\left(\left( \frac{v}{c}\right)^4\right)
\end{align*}$$

I would then integrate the expression above

$$\begin{align*}
L(\gamma)&=\int_{t_i}^{t_f} dt \quad c- \frac{v^2(t)}{2c} + \mathcal{O}\left(\left( \frac{v}{c}\right)^4\right) \\
L(\gamma)&=\int_{t_i}^{t_f} dt \quad c+ \int_{t_i}^{t_f} dt \quad- \frac{v^2(t)}{2c} + \mathcal{O}\left(\left( \frac{v}{c}\right)^4\right)\\
L(\gamma)&=c(t_f+t_i) -\frac{1}{mc} \int_{t_i}^{t_f} dt \quad \frac{1}{2}m v^2(t) + \mathcal{O}\left(\left( \frac{v}{c}\right)^4\right)\\
L(\gamma)&=\alpha_1 + \alpha_2 \int_{t_i}^{t_f} dt \quad \frac{1}{2}m v^2(t) + \mathcal{O}\left(\left( \frac{v}{c}\right)^4\right)
\end{align*}$$

Then $\alpha_1=c(t_f - t_i)$ and $\alpha_2=-\frac{1}{mc}$, but shouldn't I write $-{mc}\mathcal{O}\left(\left( \frac{v}{c}\right)^4\right)$ in the integral?

 

[PeroK]

That is one thing, it is quite another thing to claim expressions are dimensionally inconsistent and certainly wrong when it is not the case.

That was because it wasn't clear that he had taken the second term outside the integral. As I said before, if someone is going to write $\int dt$, they need to be clear about where the integrand ends. I can see now that $v$ is the average speed - but that is undefined.

These are the little ways to sow confusion. You can't ask questions of a textbook, so the obligation is on the author to be unambiguous.

 

[PeroK]

That is one thing, it is quite another thing to claim expressions are dimensionally inconsistent and certainly wrong when it is not the case.

You're misquoting me. The expressions I wrote down were dimensionally inconsistent. They were my understanding of what the author was doing.

Also, I never used either the words "certainly" or "wrong" in any of my posts on this thread.

 

[PeroK]

... but shouldn't I write $-{mc}\mathcal{O}\left(\left( \frac{v}{c}\right)^4\right)$ in the integral?

I get an extra factor of $c$ as well. $c\Delta t$ in fact. I'm tempted to say the original $L$ has dimensions of velocity-time, but $\mathcal{O}\left(\left( \frac{v}{c}\right)^4\right)$ is dimensionless. Maybe that's wrong?

 

[Orodruin]

The $\mathcal O$ notation just tells you how an expression depends on a small parameter. As I said earlier:

$\mathcal O((\frac v c)^4)$ means $\alpha (v/c)^4 + \mathcal O(v/c)^5$. Again $\alpha$ is dimensional. Not the best of notations, but still.

 

[ergospherical]

It would be helpful to motivate why the constants take the form that they do. It would be better to start from $S[\gamma] := -mc^2 L[\gamma] = -mc^2 \int_{\gamma} d\tau$, since now $S[\gamma]$ has explicit dimensions of action (with $mc^2$ being the only simple choice for something with units of energy for a single particle).

 

[ergospherical]

$-mc L[\gamma]$ rather, I hadn't noticed that they've used $ds$ rather than $d\tau$.

 

[Lambda96]

Thank you PeroK, Orodruin and ergospherical for your help