Does kinetic energy vary by frame of reference?

[Bigman]

Kinetic energy is 1/2mv^2, and velocity varies depending on velocity of the observer, so does kinetic energy also vary depending on the velocity of the observer? for example, someone sitting in a bus who slides a 2 kg brick down the center isle of the bus at 5 m/s could say the has a kinetic energy of 25 joules, while someone on the side of the road who observes the brick's velocity as 55m/s (the bus is doing 50m/s) would say the bricks kinetic energy is 3025 joules. would they both be right?

 

[Chris Hillman]

Yes, KE is an observer-dependent quantity!

Kinetic energy is 1/2mv^2, and velocity varies depending on velocity of the observer, so does kinetic energy also vary depending on the velocity of the observer?

The relativistic expression for the kinetic energy of a particle with mass m moving at velocity v wrt some inertial observer (in flat spacetime) is
$${\rm KE} = m \, \left( -1 + \cosh(\operatorname{arctanh}(v)) \right) = m \left( -1 + \frac{1}{\sqrt{1-v^2}} \right) = \frac{1}{2} \, m v^2 + \frac{3}{8} \, m v^4 + O(v^6)$$
where the first term in the last expression is the Newtonian expression for the kinetic energy (of a particle with mass m moving at velocity v wrt some inertial observer). All the "relativistic correction terms" here are positive, so the relativistic expression is always greater than the Newtonian expression, but they agree pretty nearly when v is much smaller than unity. (I am using "geometric units" in which c=1.) It was from examining the expression $m \cosh{\operatorname{arctanh}{v}}$ that Einstein deduced the famous equivalence of mass and energy, incidently!

Now, holding m fixed and varying v it is easy to see that the KE varies in both Newtonian and relativistic physics, so KE is an observer-dependent quantity.

 

[bernhard.rothenstein]

kinetic energy transformation

The relativistic expression for the kinetic energy of a particle with mass m moving at velocity v wrt some inertial observer (in flat spacetime) is
$${\rm KE} = m \, \left( -1 + \cosh(\operatorname{arctanh}(v)) \right) = m \left( -1 + \frac{1}{\sqrt{1-v^2}} \right) = \frac{1}{2} \, m v^2 + \frac{3}{8} \, m v^4 + O(v^6)$$
where the first term in the last expression is the Newtonian expression for the kinetic energy (of a particle with mass m moving at velocity v wrt some inertial observer). All the "relativistic correction terms" here are positive, so the relativistic expression is always greater than the Newtonian expression, but they agree pretty nearly when v is much smaller than unity. (I am using "geometric units" in which c=1.) It was from examining the expression $m \cosh{\operatorname{arctanh}{v}}$ that Einstein deduced the famous equivalence of mass and energy, incidently!

Now, holding m fixed and varying v it is easy to see that the KE varies in both Newtonian and relativistic physics, so KE is an observer-dependent quantity.

Please tell me if the kinetic energy of a particle is K in I and K' in I' then how are they related. Do you know a place where the problem is discussed. Thanks in advance.

 

[Doc Al]

Kinetic energy is 1/2mv^2, and velocity varies depending on velocity of the observer, so does kinetic energy also vary depending on the velocity of the observer? for example, someone sitting in a bus who slides a 2 kg brick down the center isle of the bus at 5 m/s could say the has a kinetic energy of 25 joules, while someone on the side of the road who observes the brick's velocity as 55m/s (the bus is doing 50m/s) would say the bricks kinetic energy is 3025 joules. would they both be right?

The short answer is yes. Kinetic energy, like velocity, is different for observers moving at different speeds.

 

[jtbell]

Please tell me if the kinetic energy of a particle is K in I and K' in I' then how are they related.

The total energy $E = K + m_0 c^2$ and momentum $p$ in two different inertial reference frames are related by the Lorentz transformation:

$$p^{\prime}c = \gamma (pc - \beta E)$$

$$E^{\prime} = \gamma (E - \beta pc)$$

because $E$ and $pc$ form a four-vector just like $ct$ and $x$.

$$x^{\prime} = \gamma (x - \beta ct)$$

$$ct^{\prime} = \gamma (ct - \beta x)$$

 

[bernhard.rothenstein]

kinetic energy transformation

The total energy $E = K + m_0 c^2$ and momentum $p$ in two different inertial reference frames are related by the Lorentz transformation:

$$p^{\prime}c = \gamma (pc - \beta E)$$

$$E^{\prime} = \gamma (E - \beta pc)$$

because $E$ and $pc$ form a four-vector just like $ct$ and $x$.

$$x^{\prime} = \gamma (x - \beta ct)$$

$$ct^{\prime} = \gamma (ct - \beta x)$$

Thanks for your answer. My question is how does transform kinetic energy? It would be correct to approach as
dW(k)=F(x)dx
dW'(k)=F'(x)dx'
and to take into account F(x)=F'(x)
(W(k), W'(k) kinetic energy in I and I', F(x) and F'(x) the OX components of the force acting on the particle.

 

[Bigman]

man, i see a lot of symbols i don't recognize (like the upper case beta and the trig functions with the 'h' on the end), but i think i get the jist of it. at the very least, i know that the answer to my question is yes :D thanks for the responses guys

 

[pervect]

man, i see a lot of symbols i don't recognize (like the upper case beta and the trig functions with the 'h' on the end), but i think i get the jist of it. at the very least, i know that the answer to my question is yes :D thanks for the responses guys

$\beta = v/c$

The hyperbolic trig functions referred to are described for instance at http://en.wikipedia.org/w/index.php?title=Hyperbolic_function&oldid=167748987

 

[Dale]

If you want to transform KE between reference frames I would recommend using the four-momentum. Energy is the timelike part of the four-momentum and momentum is the spacelike part. Together they transform like any other four-vector.