Re: Frame Question. Where does the energy go?

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
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?

Henri Wilson. ASTC,BSc,DSc(T)

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

.....
Oh dear, here we go. Looks like it's time to teach Hank some basics of
freshman physics.

go on....

But this is very good - if he manages to understand Newtonian mechanics, he may be ready to explore SRT. :-)

hahahahahwahhawhawhahhohohoho!

Not one of you can explain this, can you.....
I haven't had so much fun in years.

Hahahahahahahohohohohoo!

Where does all the energy go?

Harald




Henri Wilson. ASTC,BSc,DSc(T)

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

.....

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
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.

Or...?

--
Paul

http://home.c2i.net/pb_andersen/
.

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