Re: Frame Question. Where does the energy go?

On Fri, 13 Feb 2009 20:39:17 +0100, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxx> wrote:

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

Henri's original post:
| According to the PoR and Einstein, all inertial frames are
| effectively equal.
|
| So if I am driving my car at 100km/hr along a road, the Earth is
| moving at 100km/hr in my frame. I calculate that its KE is, say,
| 10^30 joules.
|
| When I stop my car, the earth no longer appears to be moving.....and
| its KE is now zero.
|
| Where has all that energy gone?
|
|....or was Einstein completely wrong?


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.

Ah! It was no trick question! :-)

Of course not.

I suspected you had made a trap, but you hadn't.
You didn't know better! :-0

Ah! You cannot find an answer. You jumped frames to hide your ignorance...and
made a fool of yourself in the process.

So it's still a mystery to you where the Earth's KE
in the car's rest frame goes _while you are braking_!!! :-)

That's right. When you brake, the Earth's speed goes from 100km/hr to zero, in
your frame. When does its KE go?

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...?

So it was the "Or..?" after all. :-)

Sorry for underestimating your ignorance and stupidity.
I should have learned by now that you don't know any physics,
not even good ol' Newtonian mechanics.

According to Henri Wilson, Newtonian mechanics is nonsense!

Here it is again:
You are taveling inertially in your car at 100km/hr along a flat road.

According to Newton, when you brake, your speed goes from 100km/hr to zero in
the EARTH FRAME. Your KE of 50000J is transfered to the earth+you or dissipated
in heat.

According to Newton, when you brake, the earth's speed goes from 100km/hr to
zero in YOUR FRAME. Its K.E. of 5000000000000000000000000000000J is hardly
transfered to you or dissipated in heat. So where does it go?

Don't worry, I'm sure Diaper or Dopey Dono will come up with a likely answer.


Henri Wilson. ASTC,BSc,DSc(T)

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

......
.

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