# Two equations of generalized forces

• A
• Kashmir
In summary, the conversation discusses the definition of generalized forces and how it relates to Lagrangian mechanics. It is mentioned that in some special cases, the generalized force can be written in the form of a velocity dependent potential, which can then be used to derive Lagrange's equation. However, there is a contradiction as to whether this holds true for monogenic systems. The conversation concludes with an example using the Lorentz force to show how the generalised force can be derived from the velocity dependent potential.
Kashmir
Wikipedia article under generalized forces says

Also we know that the generalized forces are defined as
How can I derive the first equation from the second for a monogenic system ?

What is the definition of Fi in the second equation?

dextercioby said:
What is the definition of Fi in the second equation?
##F_i## is the force on the ith particle

You have it the wrong way around! There are some mechanical systems for which ##Q_j## can be written in the form ##Q_j = \dfrac{d}{dt} \dfrac{\partial U}{\partial \dot{q}^j} - \dfrac{\partial U}{\partial q^j}##, where ##U## is called a velocity dependent potential.

[for example, the Lorentz force is 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)##.

vanhees71
ergospherical said:
You have it the wrong way around! There are some mechanical systems for which ##Q_j## can be written in the form ##Q_j = \dfrac{d}{dt} \dfrac{\partial U}{\partial \dot{q}^j} - \dfrac{\partial U}{\partial q^j}##, where ##U## is called a velocity dependent potential.

[for example, the Lorentz force is 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)##.
ergospherical said:
You have it the wrong way around! There are some mechanical systems for which ##Q_j## can be written in the form ##Q_j = \dfrac{d}{dt} \dfrac{\partial U}{\partial \dot{q}^j} - \dfrac{\partial U}{\partial q^j}##, where ##U## is called a velocity dependent potential.

[for example, the Lorentz force is 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 ?
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

that’s correct. in some (special) cases the generalised force takes that second form. there’s nothing to prove!

Kashmir
ergospherical said:
that’s correct. in some (special) cases the generalised force takes that second form. there’s nothing to prove!
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 :)

ergospherical said:
that’s correct. in some (special) cases the generalised force takes that second form. there’s nothing to prove!
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)##

e.g. the Lorentz force
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##.

Kashmir, dextercioby and vanhees71
Let us recall a formula for a rigid body, Assume that the rigid body experiences a net force ##\boldsymbol F## and a torque ##\boldsymbol M_A## about a point ##A##. The point ##A## belongs to the rigid body. Then
$$Q_i=(\boldsymbol F,\frac{\partial \boldsymbol v_A}{\partial \dot q_i})+(\boldsymbol M_A,\frac{\partial \boldsymbol \omega}{\partial \dot q_i})$$

## 1. What are two equations of generalized forces?

The two equations of generalized forces are the Lagrange's equations and Hamilton's equations. These equations are used to describe the motion of a system by considering the forces acting on it.

## 2. How is Lagrange's equation derived?

Lagrange's equation is derived from the principle of least action, which states that the path taken by a system between two points is the one that minimizes the action, a quantity related to the system's energy.

## 3. What are the variables used in Hamilton's equations?

The variables used in Hamilton's equations are the generalized coordinates and momenta of a system. These variables are related to the position and velocity of the system, respectively.

## 4. What is the significance of generalized forces?

Generalized forces are important in describing the motion of a system because they take into account all the forces acting on the system, including non-conservative forces such as friction. This allows for a more accurate description of the system's behavior.

## 5. Can generalized forces be used to solve any type of system?

Yes, generalized forces can be used to solve any type of system, as long as the equations of motion can be written in terms of the system's generalized coordinates and momenta. This includes both classical and quantum mechanical systems.

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