What is the derivative of velocity with respect to position?

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What can I expect to get if I take the derivative of velocity with respect to position? Is it zero?
I'm reading a book on Classical Mechanics (No Nonsense Classical Mechanics) and one particular section has me a bit puzzled. The author is using the Euler-Lagrange equation to calculate the equation of motion for a system which has the Lagrangian shown in figure 1. The process can be seen in figure 2. What I don't understand is this:

He seems to consider that the partial derivative of (dx/dt)2 with respect to x is zero. Same thing for the partial derivative of x2 with respect to dx/dt. This doesn't seem obvious to me and I haven't found any explanations elsewhere. Why is it so?

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Looks like derivatives are assumed to commute: d(dx/dt)/dx=d(dx/dx)/dt.
However, if position is a function of time, it does seem meaningful to ask how the velocity is changing from one position to the next. To take it as saying velocity is not changing with position is problematic, since velocity usually does change with position. For any position-dependent force, d^2(x)/dt^2 is a function of x:
d^2(x)/dt^2=f(x) -> dx/dt = ∫f(x)dt = ∫f(x)(dt/dx)dx = v(x)
so to make sense of this you need t(x) to evaluate (dt/dx), which is not always a function
But it is odd that d(dx/dt)/dx=0 implies that velocity is constant with position. Can anybody correct/explain?

edit: differential operators do not, in general, commute, so if it is not a mistake in the text, then there is some special condition that makes it so.

Last edited:
polytheneman
vela
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In Lagrangian mechanics, position and velocity are independent variables.

Last edited:
polytheneman
In the Lagrangian mechanics, position and velocity are independent variables.
Thanks for clearing that up. However, you can have a particle modeled by a Lagrangian which differs in velocity along the positions of the motion, correct?

polytheneman
PeroK
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Summary:: What can I expect to get if I take the derivative of velocity with respect to position? Is it zero?

I'm reading a book on Classical Mechanics (No Nonsense Classical Mechanics) and one particular section has me a bit puzzled. The author is using the Euler-Lagrange equation to calculate the equation of motion for a system which has the Lagrangian shown in figure 1. The process can be seen in figure 2. What I don't understand is this:

He seems to consider that the partial derivative of (dx/dt)2 with respect to x is zero. Same thing for the partial derivative of x2 with respect to dx/dt. This doesn't seem obvious to me and I haven't found any explanations elsewhere. Why is it so?
This is a common misunderstanding. The Lagrangian is initially analysed as an abstract function of the dynamic variables. In which case, ##x## and ##\dot x## are taken to be two abstract independent variables and the functional format of the Lagrangian is analysed.

The Euler-Lagrange equation then specifies how these abstract quantities are related for the specific trajectory of a particle in this system.

This is done by assuming that the action is stationary for an actual trajectory. This is known as the Lagrangian principle and can be shown to be equivalent to Newton's laws. The principle is, however, much more generally and widely applicable than Newton.

Either your text book failed to say any of this or you just skimmed past it!

polytheneman
PeroK
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Thanks for clearing that up. However, you can have a particle modeled by a Lagrangian which differs in velocity along the positions of the motion, correct?
To see what is being done here, let's take the simple example of a free particle. We have a formula for the kinetic energy:

##T = \frac 1 2 m \dot{x}^2##

Now, if you do a normal time differentiation of that equation you get:

##\frac{dT}{dt} = m \dot x \ddot x = 0##

Which follows from Newton's second law.

But, the Lagrangian approach takes ##T = \frac 1 2 m \dot{x}^2## as an equation relating the KE to the variable ##\dot x##. You could draw a graph of this function. You could also ask what is the gradient of this function? To calculate this you would treat ##\dot x## simply as a variable upon which ##T## depends:

##\frac{dT}{d\dot x} = m\dot x##

And, we recognise the RHS as the momentum of a particle. This formal derivative of ##T## with respect to the variable ##\dot x## has thrown up the form of a recognisable quantity.

The Euler-Lagrange equation then describes how the quantities obtained by this process of formal differentiation are related for a particle on a real trajectory - i.e. one that could have been obtained from the Newtonian equations of motion, say. In this simple case, Euler -Lagrange implies that:

##\frac{d}{dt}(\frac{dT}{d\dot x}) = \frac{d}{dt}(m\dot x) = 0##

(This is back to normal time differentiation now.)

And so Euler-Lagrange simply recovers Newton's second law for a free particle: ##m \ddot x = 0##. But, in this case, Newton's second law arose from looking at the way the kinetic energy, ##T##, functionally depended on the dynamic variable ##\dot x##.

jk_er_gamma
If I understand this correctly, you're saying that for a free particle, there is no change in T (and thus no change in dx/dt).
But what about the problem indicates that it was a free particle? Was it that the Lagrangian included (dx/dt)^2?

vela
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However, you can have a particle modeled by a Lagrangian which differs in velocity along the positions of the motion, correct?
I'm not sure what you're asking here.

I'm not sure what you're asking here.
I'm imagining that you start with the assumption that there is an equation of motion (1-dimensional) x(t). Then,
at every point on the graph of that function should be a derivative, the velocity, which you can graph as well. Since x(t) and v(t) have a common parameter of t, you can graph v against x. This graph is not always a horizontal line, so it could have tangents along the curve. The slope of those tangents would indicate nonzero derivative.

vela
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Yes, of course the velocity can change as the particle moves. If you start with the assumption of a specific trajectory ##x(t)##, then clearly ##x## and ##v## aren't independent. They're related through Newton's second law.

In Lagrangian mechanics, however, you do not start with the particle having a trajectory ##x(t)##. You're actually considering all possible combinations of position and velocity and then from that, finding the specific combination that minimizes the action. That specific combination turns out to be the one that satisfies Newton's second law.

jk_er_gamma
In Lagrangian mechanics, however, you do not start with the particle having a trajectory x(t)
Interesting. So that's what is meant by taking position and velocity as independent variables for this process.

fisher garry
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So if it is zero then acceleration is 0?

PeroK
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View attachment 254497

So if it is zero then acceleration is 0?
What is the relevance of this to the rest of the thread?

olgerm
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The $$\frac {d} {dt} \frac {\partial L} {\partial \dot x } - \frac {\partial L} {\partial x} = 0$$ Lagrange equation is written using the following substitution (use the left-hand element to replace the right) $$t \to t$$ $$x(t) \to u$$ $$\dot x(t) \to v.$$ Of course, the partial derivative of the Lagrange function cannot be taken by ## x(t) ## or ## \dot{x} (t) ##, because they are not independent variables. But the partial derivatives of ##u## and ##v## can be calculated for the function ##L (u, v, t)## and then replace it with ## x(t) ## and ## \dot{x} (t) ## as above.
.

vanhees71 and PeroK
What is the relevance of this to the rest of the thread?
It's basically the same as one of the questions I was asking, and I think the main source of confusion.
So if it is zero then acceleration is 0?
I think PeroK was saying that it should be zero for a free particle, since there is no gain or loss of kinetic energy.

At certain steps in the process of examining functions of x and v, they are treated as separate variables and their co-dependence is ignored, since it is a property of functions used to find minimum action, not an analytical moving frame. That is why substituting in dx/dt for v can give confusing answers at certain points in the process.

fisher garry