Question about General Relativity Theory

[swansont]

Relativity, including the principle of equivalence, has been extensively tested, so it's extremely likely that it's correct.

 

So either the answer is no, it is not possible to "change speeds, without having your weight altered," or that if it is, it does not mean the equivalence principle is false.

 

But you have to explain how one might "change speeds, without having your weight altered." Without details it's too vague.

[Johnny5]

Relativity' date=' including the principle of equivalence, has been extensively tested, so it's extremely likely that it's correct.

 

So either the answer is no, it is not possible to "change speeds, without having your weight altered," or that if it is, it does not mean the equivalence principle is false.

 

But you have to explain how one might "change speeds, without having your weight altered." Without details it's too vague.[/quote']

 

Thank you swansont.

 

Let me try to improve things a little...

 

I want to know if the principle of equivalence, and what I am talking about are related, specifically I want to know if the one being true means the other is false.

 

And I know that you are right... I have to explain how one might "change speeds" without having their weight altered... i am trying...

 

Originally, I stumbled upon the idea this way, perhaps this might help...

 

Start with this...

 

[math] \mathbf F = M \mathbf a [/math]

 

Now, consider a reference frame in which an occupant is inside of a rocket, but the engines are off, and the occupant is floating weightless.

 

There are no external forces acting upon the rocket, and there are no external forces acting on the occupant, hence the occupant does not move relative to the interior walls of the ship.

 

So what does it mean to say that the occupant is weightless?

 

Well weight is going to be defined operationally using a scale to measure it.

 

So the reading would be zero right now for weight, that is...

 

[math] \mathbf W = M \mathbf a = M \frac{d \mathbf v}{dt} = 0 [/math]

 

The velocity in the above equation is related to the center of inertia of the occupant, not the ship, and the mass M is the mass of the occupant not the ship.

 

Then, the engines to the ship are turned on, and the ship begins to accelerate relative to the occupant.

 

Now, focus your attention on the reference frame in which the ship, which was initially at rest, now has its nonzero acceleration defined in terms of the coordinates of this reference frame.

 

Let the direction of the acceleration be in the direction of increasing X coordinates of this reference frame. Therefore, the direction of the acceleration is i^.

 

Let the magnitude of the acceleration be denoted by:

 

[math] | \mathbf a | [/math]

 

Now, the acceleration of the ship is equal to the product of its direction in a frame, times the magnitude of the acceleration vector in the frame. So here is a simple formula for the acceleration of the ship in this reference frame:

 

[math] \mathbf a = | \mathbf a | \hat i [/math]

 

Now the acceleration of the ship in this frame is the derivative of the ships velocity in this frame with respect to time... that means this...

 

[math] \mathbf a = \frac{d \mathbf v}{dt} [/math]

 

So we can write this now...

 

[math] | \mathbf a | \hat i = \frac{d \mathbf v}{dt} [/math]

 

Now, the RHS must have the exact same direction as the LHS, or the equality statement would be false. Divide both sides by the direction of the acceleration of the ship to obtain the following statement:

 

[math] | \mathbf a | = \frac{dv}{dt} [/math]

 

In the formula above, the non-bolded letter v denotes the speed of the ship in the frame.

 

Now before the ships engines are turned on, both the ship AND the man are at rest in the frame.

 

After the engines are turned on, the ship begins to accelerate, whilst the man remains at rest.

 

Let the spaceship reach and maintain an acceleration of 9.8 meters per second squared. Denote that value by the letter g.

 

Hence, we can write the following statement which is true in the frame...

 

[math] g = \frac{dv}{dt} [/math]

 

Eventually, the ship's floor reaches the man's feet, and intuitively the man feels pressed to the floor.

 

Let there be a scale between his feet and the floor.

 

Scales measure force, that is what they measure.

 

Let the reading on the scale, whatever it is, be called "the weight of the man in the rocket ship."

 

Now, let inertia be a property of objects, and let the man's inertial mass be represented by the letter M.

 

Now, we know that the following statement is true in the frame...

 

[math] g = \frac{dv}{dt} [/math]

 

Therefore the following statement must also be true in the frame...

 

[math] Mg = M\frac{dv}{dt} [/math]

 

Because the acceleration on the LHS is 9.8 meters per second squared, we should expect the LHS to be equivalent to the man's weight on the surface of the earth. So let us write the following...

 

[math] W = M\frac{dv}{dt} [/math]

 

The LHS IS the man's "weight inside the rocket ship."

 

M is the man's inertial mass, which we have assumed is a property of the man.

 

And dv/dt is the change in the spaceships speed in the reference frame, with respect to change in time (as measured by a clock at rest in the frame).

 

Let us ignore the refernece to dt for now, and focus on this...

 

[math] W = M dv [/math]

 

Now in the case where the inertial mass of the man is unchangeable we have this:

 

[math] W = M(V2-V1) [/math]

 

Then I began to wonder whether or not the inertial mass of an object could be varied. Manipulated.

 

Then I worked on the following associated numerical problem...

 

Suppose that the inertial mass of a man, as computed using a scale on earth yields a value of 100 kilograms of inertia. Let us write this as follows:

 

[math] M = 100 Kg [/math]

 

Now, from numerous simple experiments we know that the magnitude of the acceleration due to earth's gravity near the surface of the earth is 9.8 meters per second squared roughly. This value has already been denoted by the letter g.

 

Let W denote the weight of the man on earth, using the same exact scale he will later use in the spaceship. So we have this...

 

[math] W = Mg = (100 Kg) g [/math]

 

Suppose that the maximum force that the man can stand is

 

[math] F_{maximum} [/math]

 

 

Any force greater than this will cause him to blackout.

 

Whatever number it is, it is pretty much the same for all human beings.

 

But its a number, a constant, not a variable.

 

Let

 

[math] F_{comfortable} [/math]

 

denote a force which most of us would feel comfortable at.

 

This too, is pretty much a constant for most human beings.

 

For any specific man, it has an exact value.

 

I don't want to keep on writing out the word "comfortable" so let me shorten it to...

 

[math] F_c [/math]

 

Ok, so we have a 100 kilogram man, inside a spaceship which is uniformly accelerating at 9.8 meters per second in some reference frame, and his body is being subjected to an external force F_c. That is...

 

[math] F_c = (100) g [/math]

 

Now, suppose that the center of mass of our solar system is at rest in the reference frame I have been talking about.

 

At an acceleration rate of g, it would take the man a very very long time to reach a star which was a distance 50 light years away from our sun.

 

So in order to improve the situation, we desire to have his ship accelerate at the following rate in the rest frame of our sun...

 

[math] 1000 g [/math]

 

That is one thousand times the acceleration of earth's gravity. Keep in mind that highly trained jet fighter pilots black out between 6-8 g's.

 

So now just plug this acceleration value into the comfortable formula.

 

So we have this...

 

[math] F_c = (100) (1000 g) [/math]

 

Now here is the problem. For any specific man, the LHS is an exact number, it cannot vary.

 

But we just increased the acceleration rate of the ship 1000 fold. Therefore, the only way for the statement to remain true, is for the man to have his inertial mass decreased by a factor of 1000, in order to maintain equality.

 

That is:

 

[math] F_c = \frac{100}{1000} (1000 g) [/math]

 

So if there is any process in nature, through which the inertial mass of a man could be lowered, then we could superaccelerate the man.

 

So now, my question is, if it's possible to vary the inertial mass of an object, what does that mean for the truth value of the principle of equivalence?

 

In other words

 

If inertial mass cannot be varied does this go against the equivalence principle?

 

and

 

If inertial mass can be varied, does this go against the equivalence principle.

 

I do not see that the truth value of the equivalence principle can be independent of the truth value of the possibility of inertial mass variance.

 

Maybe it will help if I say this...

 

The general theory of relativity was supposed to equate gravitational frames with uniformly accelerating ones.

 

And it is intuitive and normal to assume that if the spaceship in the above example is accelerating at g, and your mass is M, then your weight is what it is here on earth. And this was the whole basis of the General theory.

 

So my question comes in here...

 

Suppose that superacceleration is possible.

 

Then a gravity frame, and a uniformly accelerating frame cannot be equivalent. <--- see that's what I want to hear about... whether or not this is correct.

 

Because if you could be accelerating at say 1,000,000 g's and your weight appeared to be what it is on earth, that logically goes against the principle, does it not?

 

I want to know if this is right or wrong. So that's why I need to hear from an expert on the general theory of relativity.

[swansont]

Because if you could be accelerating at say 1' date='000,000 g's and your weight appeared to be what it is on earth, that logically goes against the principle, does it not?

[/quote']

 

Yes, it would seem to be contradictory.

[Johnny5]

Yes, it would seem to be contradictory.

 

Thank you

 

Do you think you could help me say what I am trying to say better, using mathematics?