Libby Owens
2022-07-16
Answered

The total energy of an object comes from the time part of the four-momentum, and so isn't a Lorentz invariant. On the other hand, is the potential energy of a compressed spring a Lorentz invariant?

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Cheyanne Charles

Answered 2022-07-17
Author has **13** answers

The potential energy is not a Lorentz invariant, it is just a field energy for the spring parts, including the most significant contribution, the increased electronic energy from the compression of the electronic wavefunction, and it transforms along with the field momentum as a four-vector. In general, it isn't useful to separate energy into potential energy and kinetic energy in special relativity, the energy must have local flow, and is described by field energy, a stress-energy tensor, and it is due to a combination of local fields.

Karsyn Beltran

Answered 2022-07-18
Author has **5** answers

It depends of what potential energy you are using. If you refer to a potential energy that depends only on position variables, $V=V(x)$, this is not Lorentz invariant. If you refer to potential energy that depends both on position and time variables, this can be Lorentz invariant.

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According to my book the spring constant is given by $k={V}^{\u2033}({X}_{o})$. Where $V(X)$ is the potential energy function. if I use the function $V(X)={X}^{6}+{X}^{4}$ the spring constant is zero at ${X}_{o}=0$. However, the plot looks like a stable equilibrium that would allow harmonic oscillation near ${X}_{o}$. So why is k zero in this case?

asked 2022-09-27

Kinetic energy in a spring

I am building a spring launcher for an engineering project and need to calculate spring compression to achieve a projectile velocity of 4 m/s. The mass ${m}_{p}$ of the projectile is 5 g, the mass of the spring is 7 g and the spring constant is 1000 N/m.

Is it right to assume that initial potential energy in the spring = kinetic energy of the marble + kinetic energy of the spring at the point of release?

I can calculate the marble's kinetic energy easily enough using ${E}_{k}=\frac{1}{2}{m}_{p}{v}^{2}$. However, how would I calculate the spring's kinetic energy? One end is secured, while the other end would be traveling at the same speed as the marble. I think I would have to use integration somehow but not completely sure how to.

I am building a spring launcher for an engineering project and need to calculate spring compression to achieve a projectile velocity of 4 m/s. The mass ${m}_{p}$ of the projectile is 5 g, the mass of the spring is 7 g and the spring constant is 1000 N/m.

Is it right to assume that initial potential energy in the spring = kinetic energy of the marble + kinetic energy of the spring at the point of release?

I can calculate the marble's kinetic energy easily enough using ${E}_{k}=\frac{1}{2}{m}_{p}{v}^{2}$. However, how would I calculate the spring's kinetic energy? One end is secured, while the other end would be traveling at the same speed as the marble. I think I would have to use integration somehow but not completely sure how to.

asked 2021-03-02

Two point charges ${q}_{1}=+2.40$ nC and ${q}_{2}=-6.50$ nC are 0.100 m apart. Point A is midway between them and point B is 0.080 m from $q1$ and 0.060 m from $q}_{2$. Take the electric potential to be zero at infinity.

(a) Find the potential at point A.

(b) Find the potential at point B.

(c) Find the work done by the electric field on a charge of 2.50 nC that travels from point B to point A.