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I'm trying to understand something that's coming from my Marion & Thornton (4th edition 1995 on p. 264 in a section titled "Conservation Theorems Revisited"). The topic is conservation of energy and introduction of the Hamiltonian from Lagrange's equations.

We're told that the Lagrangian ([tex]L[/tex]) cannot depend explicitly on time. I think my problem is that I don't really understand what this means. Let me explain why.

I understand that if [tex]L[/tex] cannot depend

[tex]\frac{\partial}{\partial t}L=0 [/tex]

Then it seems that there is still one hope left for [tex]L[/tex] to be dependent on time, namely if it depends

[tex]\frac{d}{d t}L=\sum_j\frac{\partial L}{\partial {q_j}} \dot{q_j}+\sum_j\frac{\partial L}{\partial \dot{q_j}} \ddot{q_j} + \frac{\partial}{\partial t}L = \sum_j\frac{\partial L}{\partial {q_j}} \dot{q_j}+\sum_j\frac{\partial L}{\partial \dot{q_j}} \ddot{q_j}[/tex]

[tex]\dot{q}[/tex] is clearly the total derivative of [tex]q[/tex] with respect to time. But it seems to me that we can arrive at an equation for [tex]q[/tex] that can depend explicitly on time.

In other words, what I'm saying is that the particle's position will have some dependence on time, and we can express that dependence in an explicit manner, not to me a very astonishing idea since that's what much of freshman physics is about.

But [tex]L[/tex] depends explicitly on [tex]q[/tex] since the first term in the second equation above is nonzero.

So if [tex]L[/tex] depends explicitly on [tex]q[/tex] and [tex]q[/tex] can be expressed explicitly in terms of time, then why can we not just plug the equation for [tex]q[/tex] in terms of [tex]t[/tex] into the expression for [tex]L[/tex] and then take a nonzero partial time derivative of [tex]L[/tex]?

Clearly I'm missing something and I would appreciate someone filling me in.

Many thanks for reading this!

We're told that the Lagrangian ([tex]L[/tex]) cannot depend explicitly on time. I think my problem is that I don't really understand what this means. Let me explain why.

I understand that if [tex]L[/tex] cannot depend

*explicitly*on time then its*partial*derivative with respect to time is necessarily zero:[tex]\frac{\partial}{\partial t}L=0 [/tex]

Then it seems that there is still one hope left for [tex]L[/tex] to be dependent on time, namely if it depends

*implicitly*rather than*explicitly*on time. In such a case it is not true that the total derivative is zero:[tex]\frac{d}{d t}L=\sum_j\frac{\partial L}{\partial {q_j}} \dot{q_j}+\sum_j\frac{\partial L}{\partial \dot{q_j}} \ddot{q_j} + \frac{\partial}{\partial t}L = \sum_j\frac{\partial L}{\partial {q_j}} \dot{q_j}+\sum_j\frac{\partial L}{\partial \dot{q_j}} \ddot{q_j}[/tex]

[tex]\dot{q}[/tex] is clearly the total derivative of [tex]q[/tex] with respect to time. But it seems to me that we can arrive at an equation for [tex]q[/tex] that can depend explicitly on time.

In other words, what I'm saying is that the particle's position will have some dependence on time, and we can express that dependence in an explicit manner, not to me a very astonishing idea since that's what much of freshman physics is about.

But [tex]L[/tex] depends explicitly on [tex]q[/tex] since the first term in the second equation above is nonzero.

So if [tex]L[/tex] depends explicitly on [tex]q[/tex] and [tex]q[/tex] can be expressed explicitly in terms of time, then why can we not just plug the equation for [tex]q[/tex] in terms of [tex]t[/tex] into the expression for [tex]L[/tex] and then take a nonzero partial time derivative of [tex]L[/tex]?

Clearly I'm missing something and I would appreciate someone filling me in.

Many thanks for reading this!

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