Understanding kinetic energy under Galilean transformations

[hungry_r2d2]

Hello,

Consider the following two situations. There is a train of mass M, going at
V=10 m/s with respect to the train station. There is a mass m passenger
on that train, who starts walking at v=1m/s parallel to the direction of the
train motion. The kinetic energy of this system (train+passenger) w.r.t.
the train station is

K_1 = (1/2) \left( M V^2 + m (v+V)^2 \right).

However, if the passenger is not moving, the kinetic energy of the system is
lower by

DK = (m/2) (v^2 + 2Vv) .

On the other hand, the passenger measures his kinetic energy to be

K_p = mv^2/2 .

I think I understand the algebra (or are there mistakes?), but I don't get
the physical content, i.e. where does the energy come from? Also, if DK were
measurable, one would be able to determine V, which (imho) contradicts
the Galilean relativity principle.

Thanks a lot!

 

[Cleonis]

To compare energies you have to stick with a single reference for velocity.

The natural choice of velocity reference is the train station.

Let's change the scenario somewhat. Let's say that the train is a freight train, with an empty weight of 10 tonnes, and it's is carrying a huge truck, also weighing 10 tonnes. The train has a velocity of 10 m/s. Let's say the gaps between the train wagons are bridged in such a way that the truck can drive along the length of the train. What happens if start up that truck, and you bring it to a velocity with respect to the train of 1 m/s?

Then clearly the velocity of the truck with respect to the train station will not top out at 11 m/s. In pushing itself forward the truck pushes the train back, so the train loses some of its velocity.


When you take that effect into account the finding is that Kinetic Energy is consistent with galilean relativity.

 

[hungry_r2d2]

Thanks for the answer, however it doesn't answer my question (probably because I formulated it unclearly).

I realize it is possible to calculate the velocity change of the train due to the truck by imposing conservation of impulse. However, it is possible to overcome this effect by giving the passenger a jetpack and imagining it does not interact with the train.

I still would like to get more input about how K_2 and DK are related.

 

[Bill_K]

hungry_r2d2, It's easy to see that kinetic energy does not transform under a Galilean transformation the simple way you might think it does. Consider a two-particle system:

M = m₁ + m₂
P = m₁v₁ + m₂v₂
E = 1/2 (m₁v₁² + m₂v₂²)

In a frame that moves with velocity V:

E' = 1/2 (m₁(V + v₁)² + m₂(V + v₂)²)
= E + P.V + 1/2 M V²

The extra term you see in DK is clearly the P.V cross term. The "expected" result that E' = E + 1/2 M V² only holds when P = 0, namely when the original frame was the center of mass frame. Only in that case can you write the energy as a sum of two terms - as the books say it, "the energy in the center of mass plus the energy of the center of mass."

 

[hungry_r2d2]

Again, thanks for help. Sorry for bumping, but I still don't get it.

Imagine that the passenger powers up his jet pack and gains E worth of kinetic energy in the train coordinate system. (We may assume 100% efficiency and negligible loss of mass in the jet pack). Thus the passenger gains the velocity

v = \sqrt{2 m E}

with respect to the train. But what is his velocity with respect to the train station?

1) Is it

u = \sqrt{V^2+v^2}

due to energy conservation?

2) Or is it simply V+v? If this case is correct, then the final energy of the system with respect to the station is K₁, but then it means that energy is not conserved and that mvV of energy came in from somewhere. In that case, where does this energy come from? Also, if the passenger observes that he needs DK energy to gain v, then it can calculate the velocity of the train, and then the Galilean relativity does not work anymore.

Most probably, I am confused and wrong at multiple places. I would be very thankful, if somebody pointed out all of them.

 

[hungry_r2d2]

All right, I think I now understand the problem. The main point is that inside the train momentum is always conserved, even if a jet pack is involved. I attach an a more complete solution in the PDF. I acknowledge major input by R.K.