# Equivalence Principle question

1. Jun 25, 2010

### Buckethead

According to Einstein's Equvalence Principle inertial mass and gravitational mass are interchangable. If we lived in a universe where these two masses were not equal, how would this translate into everyday experience? For example, if gravitational mass were twice the value of inertial mass would the following be true?:

A 1 kg mass is accelerated at 32 ft/s^2 in space. The mass is weighed on a scale calibrated to this experiment and therefore shows the mass to weigh 1 kg.

The 1 kg mass and scale are placed on the surface of the Earth (where acceleration due to gravity is still 32 ft/s^2) and the mass is shown to weigh 2 kg according to this scale.

Thanks.

2. Jun 25, 2010

### starthaus

If the two masses were not equivalent, then your equation of motion, in Newtonian physics would look like this:

$$m_i\frac{d^2r}{dt^2}=-\frac{GMm_g}{r^2}$$

instead of the well-known form:

$$\frac{d^2r}{dt^2}=-\frac{GM}{r^2}$$

That is, the proportionality constant changes from 1 to $$\frac{m_g}{m_i}$$

In addition to this, the Eotvos experiment and its reenactments would look totally different.

Last edited: Jun 25, 2010
3. Jun 25, 2010

### Buckethead

Thanks for the reply. So in my hypothetical universe the proportionality constant is 2. Does the result of this constant change manifest as indicated in my experiment?

4. Jun 25, 2010

### Ich

That's not how it works. In your experiment, the force is determined by the inertial mass, and as all scales show force, their reading would be the same.
The "equivalence" means that inertial mass is strictly proportional to gravitational mass, not that they are "equal". The proportionality constant is G. If G were different, the surface acceleration of earth would be different, but you still couldn't tell a gravitational field from acceleration.

5. Jun 25, 2010

### Buckethead

Are you sure? For the acceleration part of the experiment I set the force to be such that the acceleration of the weight is 32ft/s^2 or the same as the acceleration on all bodies caused by gravity. I can do this since I have control over the acceleration force. That being the case, in our universe a scale calibrated on Earth would show the weight of the object as 1 kg as would be the case if the weight were put on the same scale in an elevator in space being pulled at 32ft/s^2. The reason is because the inertial mass is equal to the gravitational mass. Now if the inertial mass were suddently to increase to twice it's mass (in our hypothetical universe) then the weight in the elevator (remember it's still going 32ft/s^2) would now show twice the weight it did before or 2 kg. Since the gravitational mass in the hypo universe (by definition of the experiment) remains unchanged, the scale would read 1 kg when the mass was weighed on Earth. Would this not be correct?

6. Jun 26, 2010

### Phrak

What you have done in your thought experiment is induce a local gague transformation on G. G=G(t).

Your scale is now out of calibration. It measures incorrectly by a factor of 2.

7. Jun 27, 2010

### Ich

That's different from your scenario. You had constant -but different- G, which would not violate the equivalence principle. You'd calibrate the elevator to 64 ft/s² (try sensible units next time ), because that's the earth surface acceleration in that universe.
If you change G after calibrating, that's obviously a difference.

8. Jun 28, 2010

### Buckethead

I did switch the experiment around a bit, but the concept of the question remained the same. However, I just realized a flaw in the experiment. I suggested accelerating the elevator at 32ft/s^2 but this would be impossible to determine (I think) if not for a scale calibrated on Earth. You could not use the rocket motors as a ruler as the inertial mass of the exhaust would also be different in the hypothetical universe. I now see how it is impossible to make the inertial mass and gravitational mass equal (as opposed to equivalent) as the concept of acceleration is really related to a measure that originates from gravity. It is not separately calibratable.

I was indeed suggesting the equivalence principle was being violated as this was the heart of the question. I was sure to mention a hypothetical universe where the equivalence principle did not hold and was trying to determine how such a law would then manifest.

9. Jun 29, 2010

### Ich

Of course it's possible to determine acceleration. The point is that, in your original experiment, you wouldn't have 10 m/s² on earth, but 20 m/s², and the force is the same again.

10. Jun 29, 2010

### Chronos

We do not live in such a universe. Your point is irrelevant and unphysical.

11. Jun 29, 2010

### Buckethead

Yes, of course you are right, it's possible to determine acceleration (brain fart on my part)

The key point in my experiment is that the gravitational mass and inertial mass are different. If the gravitational mass of an object is 2x the inertial mass, then an object accelerating at 10m/s^2 in space will experience a certain weight, let's say 1kg. If that same object were then to be placed on a scale on a planet that had a gravity that caused a freefall acceleration of 10m/s^2 then why would the object not weigh 2 kg? If it still weighs a 1kg then I'm totally missing the meaning of gravitational and inertial mass.

Now I understand that in this universe this is not a parameter that can be adjusted, so I'm indeed speaking hypothetically.

12. Jun 29, 2010

### Buckethead

I'm baffled by this. Am I wrong in saying that the Equivalence Principle is a principle based on an observation about 2 entirely different laws. The law of inertia and the law of gravity. They so far (that I am aware of) have not been shown to be one in the same phenomenon, only equivalent for some unknown reason. Or did Einstein determine that they are indeed one and the same phenomenon, unseparable even speculatively?

13. Jun 30, 2010

### Ich

You have two laws with potetially different meanings of mass:
$$F_i=m_i a$$
$$F_g=m_g \, GM_g/r^2$$
Now if we set mg=2*mi
$$F_g=4m_i \, GM_i/r^2$$
but also
$$a_g=4GM_i/r^2$$
and thus
$$F_i=m_i a_g = 4m_i \, GM_i/r^2 = F_g$$
since you accelerate your rocket (by the setup of your experiment) with the acceleration you observed at earth's surface.
You get observable differences only if the ratio mg/mi differs with the composition of the test bodies.

14. Jun 30, 2010

### Buckethead

To get from here
$$F_g=4m_i GM_i/r^2$$
To here:
$$a_g=4GM_i/r^2$$
you are assuming that:
ag=Fg/mi
but you can only say that ag=Fg/mg or ag=Fi/mi as mg and mi are not the same in this hypothethical universe. Correct? (sorry I'm not able to use latex, it keeps putting up old symbols that I erased, go figure).

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