# Einstein's weak EP?

• B
Mentor
i don't understand why to be honest.
Imagine you jump off a cliff, and let go of a rock just as you do so, so that the rock free falls downward with you. (Assume air resistance is negligible.) The rock's velocity relative to you is constant (assume you push the rock upward slightly as you let it go, so it has a small upward velocity relative to you). Is the rock's velocity relative to the Earth constant?

only1god
Imagine you jump off a cliff, and let go of a rock just as you do so, so that the rock free falls downward with you. (Assume air resistance is negligible.) The rock's velocity relative to you is constant (assume you push the rock upward slightly as you let it go, so it has a small upward velocity relative to you). Is the rock's velocity relative to the Earth constant?
I understand. But imagine that we fabricate here in Earth a mini rocket which only can go at 7 m/s from the start to the end, then we take this little rocket to the space. Wouldn't the velocity be the same?

Mentor
Ok, so here is the math with constant velocity relative to the box. It is a rather silly scenario since no fuel is used for a constant velocity, ##v_0##, relative to the box. But at least we can show that the time is the same:

Box feee floating in space:

Again, standard SUVAT equations give
$$s = ut + \frac{1}{2}at^2$$ $$h = v_0 t +0$$ $$t=\frac{h}{v_0}$$

Box free falling near earth:

The equation for the “rocket” is $$s_{rocket} = ut + \frac{1}{2} a t^2$$ $$s_{rocket}= v_0 t -\frac{1}{2}gt^2$$

The equation for the ceiling of the box is $$s_{ceiling}=h + ut + \frac{1}{2}at^2$$ $$s_{ceiling}=h + 0 -\frac{1}{2}gt^2$$

The rocket reaches the ceiling when $$s_{rocket}=s_{ceiling}$$ $$v_0 t - \frac{1}{2}g t^2 = h -\frac{1}{2}gt^2$$ $$t=\frac{h}{v_0}$$

So it takes the same time in both cases. The equivalence principle holds.

only1god
I understand. But imagine that we fabricate here in Earth a mini rocket which only can go at 7 m/s from the start to the end, then we take this little rocket to the space. Wouldn't the velocity be the same?
Wouldn't the velocity remains also constant?

I will add that your scenario has little to do with general relativity. Newton would have found your claims obviously wrong and ill informed. With the many detailed explanations here, I honestly don't understand what you fail to grasp. I actually think explaining what's wrong with what you claim about your scenario would be considered an easy exercise in a first course in physics. The only reason I don't offer yet another explanation is that others have already explained everything.

Wouldn't the velocity remains also constant?
To repeat ad nauseum, velocity relative to what? Anyway, things you can manufacture into a rocket are e.g. amount of fuel, thrust profile, etc. Even if you clarify velocity relative to what, this is not a thing you can manufacture into a rocket. It is like saying, I have a widget that maintains a height of 5 meters. When I put in orbit around the Earth ? Just like height above something has no meaning without specifying the something, velocity, constant or otherwise has no meaning without a referent. This understanding goes back even centuries before Newton.

Mentor
imagine that we fabricate here in Earth a mini rocket which only can go at 7 m/s from the start to the end
This makes no sense. You can't specify the speed of the rocket since speed is relative. You can specify its thrust, but that doesn't equate to a specific speed.

only1god
This makes no sense. You can't specify the speed of the rocket since speed is relative. You can specify its thrust, but that doesn't equate to a specific speed.
Can something with uniform velocity relative to something, be non-uniform relative to another thing inertial?
In another words, Can a constant velocity relative to something be an acceleration relative to another thing at the same time? (If the thing is inertial)

Can something with uniform velocity relative to something, be non-uniform relative to another thing inertial?
In another words, Can a constant velocity relative to something be an acceleration relative to another thing at the same time? (If the thing is inertial)
Only in one sense - something moving at constant velocity relative to something with proper acceleration can be accelerating relative to something inertial - because this thing must have proper acceleration in order to be at constant relative velocity to something else with proper acceleration. Proper acceleration is something measured with am accelerometer.

Mentor
Can something with uniform velocity relative to something, be non-uniform relative to another thing inertial?
Yes. The "inertial" qualifier is not necessary.

Can a constant velocity relative to something be an acceleration relative to another thing at the same time?
Yes.

(If the thing is inertial)
This qualifier is not necessary.

only1god
Yes. The "inertial" qualifier is not necessary.

Yes.

This qualifier is not necessary.
Doesn't that imply one of the two things are accelerating? This doesn't happen in the space-box because it's at rest, so something that have constant velocity relative to another thing would also have constant velocity relative to the space-box because it's at rest.

@only1god , I notice you have made no response to either of @Dale ’s posts with simple calculations. I claim there is no point in any further discussion till you do so. If there is something you don’t understand in those, please ask. These posts are really elementary, so there is no point in discussing anything till you understand them

Doesn't that imply one of the two things are accelerating? This doesn't happen in the space-box because it's at rest, so something that have constant velocity relative to another thing would also have constant velocity relative to the space-box because it's at rest.
In your original scenario, the rocket would be accelerating relative to the box by the same amount, whether it was in free fall near Earth or in free fall far away from any large mass. The fuel consumption would be identical.

Mentor
Doesn't that imply one of the two things are accelerating?
"Accelerating" is relative--at least, it is as you are using the term. For example, if you jump off a cliff and let a rock go just as you do so, you and the rock are accelerating relative to the Earth, but you are not accelerating relative to the rock and the rock is not accelerating relative to you.

There is a non-relative concept of acceleration called "proper acceleration", which basically means "feeling weight". This sense of "acceleration" is not relative to anything; it's something you can measure locally. For example, when you stand on Earth, you feel weight, and hence have nonzero proper acceleration; but if you jump off a cliff, you don't feel weight, and hence have zero proper acceleration. In your scenarios, the person at rest in the "box" feels no weight (they are in free fall), but the rocket does when its engine is on.

If we re-frame your scenarios using proper acceleration, it becomes even easier to see that the equivalence principle works. In each scenario, as just noted above, you have zero proper acceleration (you are weightless), and so does the box (since it is at rest relative to you), and the rocket has the same nonzero proper acceleration when its engine is on. Its proper acceleration and your lack of it (and the box's lack of it) in turn completely determines the rocket's motion relative to you (and the box) in both scenarios.

this doesn't happen in the space-box
Yes, it does. When the rocket's engine fires in the space-box, it accelerates relative to you and the box (in the first sense of "acceleration" above).

You appear to lack an extremely basic understanding of how rockets work.

because it's at rest
The box is at rest relative to you, but the rocket in the "space box" scenario is not when its engine fires.

Again, you appear to lack an extremely basic understanding of how rockets work.

Mentor
something that have constant velocity relative to another thing would also have constant velocity relative to the space-box because it's at rest.
If the rocket has constant velocity relative to the box, it is in free fall (weightless, engine not firing). This is true in both scenarios ("space box" and "box free-falling above the Earth"), as I have already said repeatedly.

In the "space box" scenario for this case (rocket moving at constant velocity relative to box), nothing at all has any acceleration anywhere.

In the "box free-falling above the Earth" scenario for this case (rocket moving at constant velocity relative to box), the Earth's surface has nonzero proper acceleration. This means the Earth is accelerating relative to the box and the rocket. You can phrase this either way: you can adopt the box frame and say that the Earth's surface is accelerating, or you can adopt the Earth's surface frame and say that the box and the rocket are both accelerating (and as @Bandersnatch pointed out a while back, both of their accelerations relative to Earth are the same, so they cancel out and the speed of the rocket relative to the box remains constant).

only1god
you appear to lack an extremely basic understanding of how rockets work.
Even if you could be right, you shouldn't say this, or at least you shouldn't say "extremely". I could also say you have an extremely lack of understanding others.

• weirdoguy
Even if you could be right, you shouldn't say this, or at least you shouldn't say "extremely". I could also say you have an extremely lack of understanding others.
So here is where thing stand. You described a proposed experiment and predicted result. This is fine. You have been told that your predictions are flat out wrong, and the reasons for this have been provided logically by @Nugatory amd @Bandersnatch and @PeterDonis, and with high school math by @Dale. You have not engaged with this with any attempt to understand, and I suspect without any intellectual honesty. This is not ok at all. Why are you here?

FYI, we have had some discussions here of really subtle issues with the various equivalence principles. I was looking forward to something along these lines. In contrast yours is really so obviously wrong that any good student graduating high school physics would easily understand what’s wrong with it.

Last edited:
• vanhees71, berkeman, Doc Al and 3 others
Mentor
With that, I think it is time to close this thread. As @PAllen mentioned, the idea has been clearly explained multiple times by multiple people. So there is not much more to say that is constructive.

• vanhees71