Imagine we’re in a rocket.

We have an imaginary engine, that provides a constant amount of thrust. How much power does such an engine provide? In essence the power depends linearly on the velocity:

And the velocity depends (by definition) on the frame of reference.

So imagine that we started from a planet a while back. Our thrust is constant (remember?) so our acceleration is constant, and so velocity grows linearly.

Does that make sense?

If this was a classical rocket engine (and we can somehow neglect that our rocket looses mass), then the Power would continue to grow. At some point, the power will exceed the power present in the fuel (power = energy/time). That doesn’t make sense to me.

Also, lets say we hook up with another rocket that has the same velocity as us. After having drinks and dinner together, we start up our engine and head off. Depending on our frame of reference, we either have enormous power (if we keep our original frame of reference – the planet we started from) or very little power (if we use the other rocket as a frame of reference). That also makes no sense.

Our assumption was that we have an imaginary engine that provides a constant amount of thrust. When faced with conclusions that make no sense, it is customary to question the assumptions. Our imaginary engine.

So, have I then proved that it is impossible to create such an imaginary engine? Really?

Related to the above, we have the definition of kinetic energy:

Here again, we have that the energy depends on the frame of reference. Energy seems such a crucial cornerstone of physics, that I don’t understand how it can depend on the frame of reference. When e.g. the energy content of fuel or batteries or whatever are absolutes values.

I’ve been toying with this for years, never taking it seriously, but I am perplexed. I was just trying to explore the characteristics that such an imaginary engine would have and I don’t know what to make of it. Do you?

> so our acceleration is constant, and so velocity grows linearly.

Velocity doesn’t grow lineary, because the weight of the rocket will increase

according to Einsteins law of general relativity. Ofcourse this change is not

noticable at low speeds, but when you are reaching speed of light, the

weight of the rocket will grow infinitely. Accelaration will slow and now matter

how much power you use, you are not able to reach speed of the light.

Sure, that will be true as we approach the speed of light. But as I said in my title, it is about “Classical Mechanics” – disregarding relativity.

Even when we operate at speeds where we can disregard relativity there are problems in my little example that I can’t fathom. Any relativity problems only come in addition.

Pieter,

I think your analysis is spot on. But I don’t think that classical mechanics was constructed to make sense power and energy across shifting points of reference. Yes, power varies with increasing speed, so the speed of the other frame of reference would be added (OR SUBTRACTED) to arrive at power, and energy too. An Eveready battery powering a toy bunny walking within another frame of reference (in the same direction as it) would really be filled with a lot of energy. But we quickly learn: don’t compare energies across frames of reference.

“Energy seems such a crucial cornerstone of physics, that I don’t understand how it can depend on the frame of reference”

Energy is indeed not frame indifferent (think of potential energy: where do you put the origin? kinetic energy: what is the velocity? the batteries that you mention: what is the “origin” in your chemical scale?)

However, in classical mechanics, differences in energy are frame indiferent! The energy that you battery loses will increase height or velocity by a given amount, no matter the origins.

This creates many problems in learning. Most of the equations that we see involving energy would lead to less confusions if they were written as differences (with deltas), so arbitrariness due to origins disappear. Starting by Einstein’s: E=c^2m is better understood if written as dE=c^2dm