- #1

Kashmir

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Also we know that the generalized forces are defined as

How can I derive the first equation from the second for a monogenic system ?

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- Thread starter Kashmir
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- #1

Kashmir

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Also we know that the generalized forces are defined as

How can I derive the first equation from the second for a monogenic system ?

- #2

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What is the definition of **F**_{i} in the second equation?

- #3

Kashmir

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##F_i## is the force on the ith particleWhat is the definition ofF_{i}in the second equation?

- #4

ergospherical

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[for example, the

If this is the case, then it follows that Lagrange's equation ## \dfrac{d}{dt} \dfrac{\partial T}{\partial \dot{q}^j} - \dfrac{\partial T}{\partial q^j} = Q_j## can be written in Lagrangian form by taking ##L(q,\dot{q},t) = T(q,\dot{q},t) - U(q,\dot{q},t)##.

- #5

Kashmir

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velocity dependent potential.

[for example, theLorentz forceis derivable from the velocity dependent potential ##U = q\phi - q \dot{\mathbf{r}}\cdot \mathbf{A}##]

If this is the case, then it follows that Lagrange's equation ## \dfrac{d}{dt} \dfrac{\partial T}{\partial \dot{q}^j} - \dfrac{\partial T}{\partial q^j} = Q_j## can be written in Lagrangian form by taking ##L(q,\dot{q},t) = T(q,\dot{q},t) - U(q,\dot{q},t)##.

Thank you for your reply but Why have I it the wrong way around ?velocity dependent potential.

[for example, theLorentz forceis derivable from the velocity dependent potential ##U = q\phi - q \dot{\mathbf{r}}\cdot \mathbf{A}##]

If this is the case, then it follows that Lagrange's equation ## \dfrac{d}{dt} \dfrac{\partial T}{\partial \dot{q}^j} - \dfrac{\partial T}{\partial q^j} = Q_j## can be written in Lagrangian form by taking ##L(q,\dot{q},t) = T(q,\dot{q},t) - U(q,\dot{q},t)##.

You also seem to say the same thing which I'm saying.

In all cases isn't the generalized force ##Q_{j}=\sum_{i=1}^{n} \mathbf{F}_{i} \cdot \frac{\partial \mathbf{r}_{i}}{\partial q_{j}}## and for some situations (monogenic) it reduces to ##\mathcal{F}_{i}=-\frac{\partial \mathcal{V}}{\partial q_{i}}+\frac{d}{d t}\left(\frac{\partial \mathcal{V}}{\partial \dot{q}_{i}}\right)## which I can't prove how.

Please tell me what I'm doing wrong. I don't have a teacher.

God bless you

- #6

ergospherical

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- #7

Kashmir

- 439

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Maybe I'm not clear.

Do you agree that we have defined generalized forces In all cases as ##Q_{j}=\sum_{i=1}^{n} \mathbf{F}_{i} \cdot \frac{\partial \mathbf{r}_{i}}{\partial q_{j}}## .

If that's true then the definition of generalized forces for monogenic systems is the same.

But it's said that the generalized force in the monogenic system is different given as ##\mathcal{F}_{i}=-\frac{\partial \mathcal{V}}{\partial q_{i}}+\frac{d}{d t}\left(\frac{\partial \mathcal{V}}{\partial \dot{q}_{i}}\right)##.

So there is a contradiction. And the only way to resolve it is to prove that they are equivalent for the monogenic case,which I cant.

I hope I'm clear. Thank you again :)

- #8

Kashmir

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What are those special cases? Under those special cases we should be able to move from ##Q_{j}=\sum_{i=1}^{n} \mathbf{F}_{i} \cdot \frac{\partial \mathbf{r}_{i}}{\partial q_{j}}## to ##\mathcal{F}_{i}=-\frac{\partial \mathcal{V}}{\partial q_{i}}+\frac{d}{d t}\left(\frac{\partial \mathcal{V}}{\partial \dot{q}_{i}}\right)##

- #9

ergospherical

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for example take the generalised coordinates to be the cartesian coordinates ##\mathbf{r} = (x,y,z)##

and let ##U(\mathbf{r}, \dot{\mathbf{r}}) = q\phi(\mathbf{r}) - q\dot{\mathbf{r}} \cdot \mathbf{A}(\mathbf{r})## be a velocity dependent potential

then ##\dfrac{\partial U}{\partial x_j} = q\dfrac{\partial \phi}{\partial x_j} - q\dot{x}_i \dfrac{\partial A_i}{\partial x_j} ##

and ##\dfrac{d}{dt} \dfrac{\partial U}{\partial \dot{x}_j} = -q \dfrac{dA_j}{dt}##

therefore\begin{align*}

Q_j = \dfrac{d}{dt} \dfrac{\partial U}{\partial \dot{x}_j} - \dfrac{\partial U}{\partial x_j} &= -q \dfrac{dA_j}{dt} - q\dfrac{\partial \phi}{\partial x_j} + q\dot{x}_i \dfrac{\partial A_i}{\partial x_j} \\

&= q \dot{x}_i \left( \dfrac{\partial A_i}{\partial x_j} - \dfrac{\partial A_j}{\partial x_i}\right) - q\dfrac{\partial \phi}{\partial x_j}

\end{align*}the first term can be rewritten as\begin{align*}

[\dot{\mathbf{r}} \times (\nabla \times \mathbf{A})]_j &= \epsilon_{jkl} \dot{x}_k (\nabla \times \mathbf{A})_l \\

&= \epsilon_{jkl} \dot{x}_k \epsilon_{lmn} \dfrac{\partial A_n}{\partial x_m} \\

&= (\delta_{jm} \delta_{kn} - \delta_{km} \delta_{jn})\dot{x}_k \dfrac{\partial A_n}{\partial x_m} \\

&= \dot{x}_k \left( \dfrac{\partial A_k}{\partial x_j} -\dfrac{\partial A_j}{\partial x_k} \right)

\end{align*}(just replace the dummy index ##k \rightarrow i##), and putting ##\nabla \times \mathbf{A} \equiv \mathbf{B}## as well as ##-\dfrac{\partial \phi}{\partial \mathbf{r}} \equiv \mathbf{E}## gives\begin{align*}

Q_j = q (\dot{\mathbf{r}} \times \mathbf{B})_j + qE_j

\end{align*}i.e. the Lorentz force

it's easy for this example, in the case of cartesian coordinates, to see that the generalised force is nothing but the actual force ##Q_j = \mathbf{F} \cdot \dfrac{\partial \mathbf{r}}{\partial x_j} = \mathbf{F} \cdot \mathbf{e}_j = F_j##.

- #10

wrobel

Science Advisor

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$$Q_i=(\boldsymbol F,\frac{\partial \boldsymbol v_A}{\partial \dot q_i})+(\boldsymbol M_A,\frac{\partial \boldsymbol \omega}{\partial \dot q_i})$$

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