Solving 4-Dimensions Problem: Lagrangnian of Charged Particles

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Homework Help Overview

The discussion revolves around deriving the Lagrangian for a charged particle in a four-dimensional framework, building upon the established three-dimensional case. The original poster presents equations and seeks clarification on the transition from three to four dimensions, particularly regarding the form of the Lagrangian and its implications for motion equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the necessity of including a factor of 1/2 in the four-dimensional Lagrangian, questioning its mathematical justification. They also inquire about applying the four-dimensional Lagrangian to derive the Lorentz force from the motion equations.

Discussion Status

Participants are exploring the differences between three-dimensional and four-dimensional formulations of the Lagrangian. Some express confusion over terminology and notation, while others suggest that the original poster may be conflating concepts. There is a recognition of the need for clarity in the mathematical expressions and their physical interpretations.

Contextual Notes

There is mention of a preference for avoiding tensor notation among participants, which may contribute to misunderstandings. The original poster notes their limited experience with tensors, which could be influencing their approach to the problem.

TheDestroyer
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How do we derive the lagrangnian of a charged particle in 3 dimentions?

Actually this is not the real question but it's a step to it, In three dimentions we have:

[tex]{\cal L} = - m_ \circ c^2 \sqrt {1 - \frac{{v^2 }}{{c^2 }}} + q\overrightarrow v \overrightarrow A - q\phi[/tex]
And As we Know in four dimensional system:
We Have:

[tex] \eqalign{<br /> Speed:U_\mu = \gamma \left( {\overrightarrow v ,ic} \right) \Rightarrow U_\mu ^2 = \gamma ^2 \left( {v^2 - c^2 } \right) = - c^2 \cr <br /> ,,,{\rm{Vector Potential: }}A_\mu = \left( {\overrightarrow A ,i{\phi \over c}}\right)\Rightarrow \cr { \rm{A}}_\mu U_\mu = \overrightarrow A \cdot \overrightarrow v - \phi \cr} [/tex]
Substituting This to the Lagrangian we get, and note that gamma is taken in hamiltons principle ds=L dt in 4-d becomes ds= L dt/Gamma:
[tex] {\cal L} = m_ \circ U_\mu ^2 + qA_\mu U_\mu [/tex]
My Professor says that we should write it like this without doing anything else:
[tex] {\cal L} = {1 \over 2}m_ \circ U_\mu ^2 + qA_\mu U_\mu [/tex]
And that is to make it like the 3d system, 1/2 mv^2, IS THIS CORRECT? THIS IS MY FIRST QUESTION

And my second question, If we used the 4d motion equation on this lagrangian:
[tex] {d \over {dt}}{{\partial {\cal L}} \over {\partial U_\mu }} - {{\partial {\cal L}} \over {\partial x_\mu }} = 0[/tex]
do we get the lorentz force like the 3d system? i didn't get it when i tried, Any one can help?

Can you explain the solution of this motion equation? my professor didn't know it even !

Thanks, Please hurry with the answer
 
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The Lagrangian [tex]L = \frac{1}{2} m \mathbf{v}^2 + q \mathbf{v}\cdot\mathbf{A} - q \phi[/tex] does give the correct physics for low speeds. The more general Lagrangian [tex]L = - m \sqrt{1-\mathbf{v}^2} + q \mathbf{v}\cdot \mathbf{A} - q \phi[/tex] is applicable even to relativistic particles. You can easily check that the more general Lagrangian reduces to the other when v is small (up to an additive constant).

It seems to me that you may be confusing three and four dimensional notation, but I'm not really sure. Let me know if this doesn't help, and I can try again to address your question.
 
No this doesn't help, because i want the lagrangian in four dimentions, i gave the notations for speed and vector potentional above, and mu under them can take values 1,2,3,4, and the brackets after them gives in the first parameter the three position dimension or in another word when mu take the values 1,2,3, and the second one gives the value 4 for mu, and in another word the time component for the vector,

I've written the lagrangian in the four dimensional system, and my question is what my professor insist to put 1/2 beside m while we can't get it mathematically?

and my second question how do we apply the four-dimensional lagrangian on the motion equation specified in the last equation

and note somthing, the result we must get is ofcourse the lorentz force this, I'll write it, and what inside brackets are indexes for tensors or vectors:

m B(mu) = q F(mu,nu) U(nu)

Where B(mu) is the vector of four dimensional acceleration, F(mu,nu) us and U(nu) is the four dimensional speed vector i mentioned above, the tensor of magnetical force in four dimensional system,

Professor says this must be the result, but how, actually I'm getting difficulties using matrices and tensors (he also not only me), plus i asked him if i can use dirac symbols and he refused, i don't know why !

Anyone can show me the way to get the motion equation?

Note: the relativistic lagrangian is not correct, you've forgot c^2,

And thanks for trying to help
 
TheDestroyer said:
How do we derive the lagrangnian of a charged particle in 3 dimentions?

Actually this is not the real question but it's a step to it, In three dimentions we have:

[tex]{\cal L} = - m_ \circ c^2 \sqrt {1 - \frac{{v^2 }}{{c^2 }}} + q\overrightarrow v \overrightarrow A - q\phi[/tex]
And As we Know in four dimensional system:

a first comment: it is a bit confusing because you are using "3-dimensional" and "four-dimensional" in a different way that what is standard! The standard way of saying what you want is that you want a *covariant expression* (that is, in terms of four-vectors). You should not say that you want a "four-dimensional expression", that will confuse everybody!
We Have:

[tex] \eqalign{<br /> Speed:U_\mu = \gamma \left( {\overrightarrow v ,ic} \right) \Rightarrow U_\mu ^2 = \gamma ^2 \left( {v^2 - c^2 } \right) = - c^2 \cr <br /> ,,,{\rm{Vector Potential: }}A_\mu = \left( {\overrightarrow A ,i{\phi \over c}}\right)\Rightarrow \cr { \rm{A}}_\mu U_\mu = \overrightarrow A \cdot \overrightarrow v - \phi \cr} [/tex]
Substituting This to the Lagrangian we get, and note that gamma is taken in hamiltons principle ds=L dt in 4-d becomes ds= L dt/Gamma:
[tex] {\cal L} = m_ \circ U_\mu ^2 + qA_\mu U_\mu [/tex]
My Professor says that we should write it like this without doing anything else:
[tex] {\cal L} = {1 \over 2}m_ \circ U_\mu ^2 + qA_\mu U_\mu [/tex]
And that is to make it like the 3d system, 1/2 mv^2, IS THIS CORRECT? THIS IS MY FIRST QUESTION

Neither is correct. The first term is actually [itex]- mc {\sqrt {U_\mu U^\mu}}[/itex].

Patrick
 
Actually Our professor is trying to simplify everything to avoid using tensors, that's why, he've written it that simple, and I've written the way to get as the professor told us, he's trying to compare everything with 3 dimensional system the way i expressed,

I'm using 4-d vector, for example A is the vector potential has 3 position dimentions and 1 time dimension i*phi/c, and i used the 4-dim speed multiplication like when we multiply any 2 vectors, just multiply x component with the other x component, y with other y, z with other z, and the same about the time dim, time with time to get a scalar ! what can i explain more? if I'm wrong or the simplicity of the professor is for childs just tell me everything you can, and i would be very grateful, thanks

If you believe it I'm now in the third year in physics college and I'm going to finish it and till now i don't know much about tensors ! so please explain everything you can,

Anyone can tell me how to continue with my details? or just tell me how to continue IN ANY OTHER WAY? but detailed way not in words, math + physical expression
 
Lol, No Body Has An Answer?
 

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