Effective mass from the Lagrangian

In summary, the Lagrangian given is solved using Euler-Lagrange equations to get a system with an effective mass and a v-dependent force. This effective mass can be interpreted as the special relativistic v-dependent mass and the v-dependent force may be the magnetic force.
  • #1
Malamala
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Hello! I have the following Lagrangian:

$$L = \frac{1}{2}mv^2+fv$$

where ##v = \dot{x}##, where x is my coordinate and f is a function of v only (no explicit dependence on t or x). What I get by solving the Euler-Lagrange equations is:

$$\frac{d}{dt}(mv+f+\frac{\partial f}{\partial v} v) = 0$$
$$m\ddot{x} + \frac{\partial f}{\partial v}\ddot{x} + \frac{\partial f}{\partial v}\ddot{x} + \frac{\partial^2 f}{\partial v^2}\ddot{x} = 0$$
$$(m+2\frac{\partial f}{\partial v}+ \frac{\partial^2 f}{\partial v^2})\ddot{x} = 0$$

Is this correct? Can I think of this system as a particle of effective mass ##M = m+2\frac{\partial f}{\partial v}+ \frac{\partial^2 f}{\partial v^2}## moving without any force acting on it? Thank you!
 
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  • #2
You made a trivial error, ##\partial^2f/\partial v^2## is multiplied with ##\dot{x}##, not ##\ddot{x}##. Hence, in addition to an effective mass multiplying ##\ddot{x}##, there is also a ##v##-dependent force. Otherwise, the idea of effective ##v##-dependent mass seems OK to me. After all, an effective ##v##-dependent mass appears also in old formulations of special relativity. In fact, with a right choice of ##f##, you can reproduce the special relativistic ##v##-dependent mass exactly. What remains to be seen is whether the ##v##-dependent force could be interpreted as the magnetic force, I leave it as a research/exercise problem for the others.
 
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  • #3
Demystifier said:
You made a trivial error, ##\partial^2f/\partial v^2## is multiplied with ##\dot{x}##, not ##\ddot{x}##. Hence, in addition to an effective mass multiplying ##\ddot{x}##, there is also a ##v##-dependent force. Otherwise, the idea of effective ##v##-dependent mass seems OK to me. After all, an effective ##v##-dependent mass appears also in old formulations of special relativity. In fact, with a right choice of ##f##, you can reproduce the special relativistic ##v##-dependent mass exactly. What remains to be seen is whether the ##v##-dependent force could be interpreted as the magnetic force, I leave it as a research/exercise problem for the others.
Thank you! For the ##\partial^2f/\partial v^2## term, don't we have ##\frac{d}{dt}(\partial f/\partial v)v = \partial^2f/\partial v^2 \frac{dv}{dt}v = \partial^2f/\partial v^2 \ddot{x}\dot{x}##? So indeed I did a mistake, but that term would still contribute as an effective mass by ##\partial^2f/\partial v^2 \dot{x}## (I missed the ##\dot{x}## term before), no? Or am I doing my derivatives wrong?
 
  • #4
Malamala said:
Thank you! For the ##\partial^2f/\partial v^2## term, don't we have ##\frac{d}{dt}(\partial f/\partial v)v = \partial^2f/\partial v^2 \frac{dv}{dt}v = \partial^2f/\partial v^2 \ddot{x}\dot{x}##? So indeed I did a mistake, but that term would still contribute as an effective mass by ##\partial^2f/\partial v^2 \dot{x}## (I missed the ##\dot{x}## term before), no? Or am I doing my derivatives wrong?
You are right. There is no "force", everything can be put into the effective mass. I made an error too.
 
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What is "Effective mass from the Lagrangian"?

"Effective mass from the Lagrangian" refers to a concept in physics that is used to describe the behavior of particles in a material or system. It is a measure of how a particle moves in response to external forces, and is derived from the Lagrangian, a mathematical function that describes the dynamics of a system.

How is effective mass calculated from the Lagrangian?

The effective mass is calculated by taking the second derivative of the Lagrangian with respect to the position of the particle. This value represents the inertia of the particle and can vary depending on the properties of the material or system.

What is the significance of effective mass in physics?

Effective mass is an important concept in physics as it helps us understand the behavior of particles in different materials and systems. It can affect the speed, acceleration, and other properties of particles, and is often used in the study of semiconductors, superconductors, and other condensed matter systems.

How does effective mass differ from mass?

Effective mass is a concept that is used to describe the behavior of particles in a specific material or system, while mass is a fundamental property of matter that remains constant regardless of the environment. Effective mass takes into account the interactions and forces present in a material, whereas mass is an intrinsic property of a particle.

Can effective mass be negative?

Yes, effective mass can be negative in certain cases. This can occur when the particle is moving in a material with a complex energy landscape, such as a crystal lattice, and experiences a force in the opposite direction of its motion. However, in most cases, effective mass is positive and represents the inertia of the particle in the given system.

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