Writing the Lagrangians for different frames depending on how "the ball is dropped"

In summary, the article discusses the formulation of Lagrangians for various reference frames in the context of a ball being dropped. It highlights how the choice of frame—whether inertial or non-inertial—affects the expression of the Lagrangian and the resulting equations of motion. By analyzing the system from different perspectives, the importance of frame-dependent variables and the principles of classical mechanics are emphasized, showcasing the versatility of Lagrangian mechanics in solving physical problems.
  • #1
gionole
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I wanna be checking homogeneity of space(only interested in vertical) for simplicity and example we can do is "ball is dropped". To check homogeneity, we use either passive or active transformation and I'm interested in lagrangians.

I heard that we can write lagrangians such as: ##L = \frac{1}{2}m\dot q^2 - mgy## and ##L' = \frac{1}{2} m\dot q'^2 - mg(y'+a)##. This comes from the fact that ##y = y'+a##. (we seem to have y and y' frame).

Question 1: it seems to me that lagrangians that I wrote are an example of passive transformation, because of ##y = y'+a##. It's like the ball is only dropped from single location(one experiment), but we write lagrangians for the ball such as seen from each frame. Is this right ? as in, am I right that this is passive, or can we also call it active ?

Question 2: Active transformation seems such as ball must be dropped from 2 different locations(2 different locations). So we drop a ball from some height, and then we move up and drop it from higher location. How would we go about writing Lagrangians for each experiment ? using the same lagrangians as shown above doesn't seem correct to me, as I think it's passive.
 
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  • #3
@berkeman would love to remove that thread as the question there is not asked correctly. but i can't delete it.
 
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  • #4
gionole said:
@berkeman would love to remove that thread as the question there is not asked correctly. but i can't delete it.
Okay, I closed off the previous thread with a note pointing to this improved version here.
 
  • #5
@berkeman can you close this as well ? Don't want people to spend time on it. I've figured it out. Thanks.
 
  • #6
Sure, thanks for the heads-up. I've closed off this thread now; I'm glad that you figured it out.
 
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FAQ: Writing the Lagrangians for different frames depending on how "the ball is dropped"

What is a Lagrangian and why is it important in physics?

The Lagrangian is a mathematical function that summarizes the dynamics of a system. It is defined as the difference between the kinetic energy and the potential energy of the system. The importance of the Lagrangian lies in its ability to provide the equations of motion for a system through the Euler-Lagrange equations, which are derived from the principle of least action. This approach is particularly useful in mechanics and field theory.

How do you write the Lagrangian for a ball dropped in an inertial frame of reference?

In an inertial frame of reference, where the frame is not accelerating, the Lagrangian for a ball of mass \( m \) dropped from a height can be written as:\[ L = T - V \]where \( T \) is the kinetic energy and \( V \) is the potential energy. For a ball falling under gravity, \( T = \frac{1}{2}mv^2 \) and \( V = mgh \), where \( v \) is the velocity of the ball, \( g \) is the acceleration due to gravity, and \( h \) is the height. Therefore, the Lagrangian is:\[ L = \frac{1}{2}mv^2 - mgh \]

How does the Lagrangian change for a ball dropped in a non-inertial frame of reference?

In a non-inertial frame of reference, such as an accelerating elevator, additional fictitious forces need to be considered. If the frame is accelerating upwards with acceleration \( a \), the effective gravitational acceleration becomes \( g + a \). The Lagrangian in this case would be:\[ L = \frac{1}{2}mv^2 - mg(h - at^2/2) \]where \( t \) is time. This accounts for the additional pseudo-force acting on the ball due to the accelerating frame.

What modifications are needed for the Lagrangian if the ball is dropped with an initial velocity?

If the ball is dropped with an initial velocity \( v_0 \), the kinetic energy term in the Lagrangian must reflect this initial condition. The kinetic energy then becomes \( T = \frac{1}{2}m(v + v_0)^2 \). The potential energy term remains \( V = mgh \). Therefore, the Lagrangian is:\[ L = \frac{1}{2}m(v + v_0)^2 - mgh \]

How can the Lagrangian be used to derive the equations of motion for the ball?

The equations of motion can be derived from the Lagrangian using the Euler-Lagrange equation:\[ \

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