Isn't working with the relativistic Lagrangian AWFUL?

[AxiomOfChoice]

I'm trying to solve a problem for a relativistic electron in an external magnetic field with vector potential $$\vec A$$ using the Lagrangian

$$\mathcal L = -mc^2 / \gamma - e \vec v \cdot \vec A$$

in cylindrical coordinates. But isn't this DREADFULLY TERRIBLE, since when I try to compute $$\dfrac{d}{dt} \dfrac{\partial \mathcal L}{\partial \dot q_i}$$ I'm going to have to take the time derivative of $$\gamma$$, which takes the form

$$\gamma = \left(1 - \frac{1}{c^2} (\dot r ^2 + r^2 \dot \phi^2 + \dot z^2) \right)^{-1/2}$$

Am I making this too hard? Please help!

 

[AEM]

Well, look at it this way. The experience will be good for you. I worked in general relativity before there were such things as algebraic manipulation programs. I have a notebook filled with equations a couple of pages long. It's interesting to find out how your brain can come to see patterns in the symbols that afford simplifications.

Now that I've been gentle, the teacher in me says, Don't Be Lazy!

 

[AxiomOfChoice]

Well, look at it this way. The experience will be good for you. I worked in general relativity before there were such things as algebraic manipulation programs. I have a notebook filled with equations a couple of pages long. It's interesting to find out how your brain can come to see patterns in the symbols that afford simplifications.

Now that I've been gentle, the teacher in me says, Don't Be Lazy!

Hehe, thanks for the words of encouragement. Having run $$\mathcal L = -mc^2 / \gamma - e \vec v \cdot \vec A$$ through the E-L equations, I've come up with a fairly simple EOM:

$$m \gamma^3 \frac{d \vec v}{dt} = e \frac{\partial \vec A}{\partial t} - e \nabla (\vec v \cdot \vec A)$$

Does that look right?

 

[Ben Niehoff]

I should note that your Lagrangian as written is neither Lorentz-invariant, nor gauge-invariant. You need an $$e\Phi$$ term, though I forget whether it should have a minus sign or not.

So, your Lagrangian will work, but only in one reference frame.

 

[AxiomOfChoice]

I should note that your Lagrangian as written is neither Lorentz-invariant, nor gauge-invariant. You need an $$e\Phi$$ term, though I forget whether it should have a minus sign or not.

So, your Lagrangian will work, but only in one reference frame.

I also don't see how I'm going to solve the eventual equations of motion. The $$\gamma$$ term contains first-order time derivatives of all the coordinates, so the equations are going to be mercilessly coupled...aren't they?

 

[AEM]

I also don't see how I'm going to solve the eventual equations of motion. The $$\gamma$$ term contains first-order time derivatives of all the coordinates, so the equations are going to be mercilessly coupled...aren't they?

More than likely. I might ask, how is your vector potential oriented?

 

[Vanadium 50]

It's been ages since I have done this, but my gut tells me Ben is on the right track. Write a Lagrangian that is Lorentz invariant, and at the end, set the time-component of the 4-potential to zero.

 

[AEM]

I will echo Ben and Vanadium 50. You should write your Lagrangian in 4-vector notation and then set the electric potential term equal to zero. Now, to simplify possibly ugly looking equations of motion, don't forget about the role of cyclic coordinates. Refer to Goldstein's Classical Mechanics, 3rd edition, Chapter 7 if you need to refresh your memory.

 

[AxiomOfChoice]

More than likely. I might ask, how is your vector potential oriented?

All I know about $$\vec A$$ is that it needs to be such that the azimuthal component of $$\vec B$$ is zero. So I was thinking $$\vec A = A_\phi \hat \phi$$.

 

[AxiomOfChoice]

I should note that your Lagrangian as written is neither Lorentz-invariant, nor gauge-invariant. You need an $$e\Phi$$ term, though I forget whether it should have a minus sign or not.

So, your Lagrangian will work, but only in one reference frame.

In this case, I know $$\Phi = 0$$.

 

[AEM]

All I know about $$\vec A$$ is that it needs to be such that the azimuthal component of $$\vec B$$ is zero. So I was thinking $$\vec A = A_\phi \hat \phi$$.

I asked my question to make you think about simplifications in the dot product of v and A.

You also know that the azimuthal component of B is zero. That also tells you something about the motion. What you need to do is pick your equations apart and ask about each term: "What does this mean?" "What does this imply?"

 

[Ben Niehoff]

Also, your equations of motion are expressed in terms of lab time, t. Why don't you try expressing them in terms of proper time, $\tau$? You may find something useful.

 

[AxiomOfChoice]

I want to thank you all for your help. I'm sacrificing my Friday night to try to grind through this problem, armed with your hints and suggestions, though I'm sure I'll be back for more help later. Thanks again!

 

[Thefirm]

I look forward to following this...

 

[AxiomOfChoice]

Hmmm...can someone tell me what, in general, you'd get from

$$\frac{\partial}{\partial \vec v} (\vec v \cdot \vec A})?$$

I'm just not sure how to work with the dot product here. Does it obey the product rule, such that

$$\frac{\partial}{\partial \vec v} (\vec v \cdot \vec A}) = \vec A + \vec v \cdot \frac{\partial \vec A}{\partial \vec v} = \vec A?$$

 

[AxiomOfChoice]

Hmmm...can someone tell me what, in general, you'd get from

$$\frac{\partial}{\partial \vec v} (\vec v \cdot \vec A})?$$

I'm just not sure how to work with the dot product here. Does it obey the product rule, such that

$$\frac{\partial}{\partial \vec v} (\vec v \cdot \vec A}) = \vec A + \vec v \cdot \frac{\partial \vec A}{\partial \vec v} = \vec A?$$

I've managed to convince myself (quick, trivial proof...sorry I wasted everyone's time) that my last line is correct. Nvm.

 

[shadi_s10]

hi everyone
I'm reading 'introducing einstein's relativity' of d'inverno
I do not know how to work with the equivalent lagrangian
can anyone help me to prove 11.42 and 11.43?