Re: Frame Question. Where does the energy go?

On Thu, 12 Feb 2009 16:05:49 +0100, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:

Dr. Henri Wilson wrote:
On Thu, 12 Feb 2009 09:59:46 +0100, "harry" <harald.NOTTHISvanlintel@xxxxxxx>
wrote:

PD wrote:
On Feb 11, 3:27 pm, hw@..(Dr. Henri Wilson) wrote:
According to the PoR and Einstein, all inertial frames are


.....

I suspect you are trolling.
But I will bite anyway.

There is obviously no point in using SR on this problem, since
the speeds are so low. So let's use NM.
The POR is valid in both.

Let the mass of the car be m and the mass of the Earth be M.

Let's first calculate the problem in the rest frame of the Earth.
-----------------------------------------------------------------
Before the braking:
Speed of car v, speed of Earth 0
KE of car: 0.5mv^2 momentum of car: mv
KE of Earth: 0 momentum of Earth: 0
After the braking:
speed of car+Earth V
momentum of car+Earth: (m+M)V
conservation of momentum: (m+M)V = mv
V = (m/(m+M))v
KE of Earth+car: 0.5(m+M)V^2 = 0.5(m^2/(m+M))v^2
Energy dissipated in brakes:
W = 0.5mv^2 - 0.5(m^2/(m+M))v^2 = 0.5(mM/(m+M))v^2

W ~= 0.5mv^2 when m << M

Now let's do the calculation in the inertial frame where
the car is initially at rest.
--------------------------------------------------------
Before the braking:
Speed of car 0, speed of Earth v
KE of car: 0 momentum of car: 0
KE of Earth: 0.5Mv^2 momentum of Earth: Mv
After the braking:
speed of car+Earth V

You love to jump frames, don't you....just as with sagnac..
In the car frame, V = 0
What follows is nonsense.

momentum of car+Earth: (m+M)V
conservation of momentum: (m+M)V = Mv
V = (M/(m+M))v
KE of Earth+car: 0.5(m+M)V^2 = 0.5(M^2/(m+M))v^2
Energy dissipated in brakes:
W = 0.5Mv^2 - 0.5(M^2/(m+M))v^2 = 0.5(mM/(m+M))v^2

W ~= 0.5mv^2 when m << M

The answer to your question is of course that the KE
of the Earth 0.5Mv^2 remains as the KE of the Earth
but for a very tiny part.
And that tiny part is 0.5mv^2 and is dissipated in
the brakes, and it will be the same regardless of
in what inertial frame you choose to calculate it.
That's the POR.

But you knew all this, didn't you?
So you had me, it was a trick question.

I still have you. You tried to give a trick answer and I saw through it.


Or...?



Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm.

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