Amr Morsi

It uses the definition of a Bounded Sequence.

But, you shouldn't go the first step directly. You must use Left and Right limits (for both x and y).

Check the following: http://en.wikipedia.org/wiki/Uniform%5Fcontinuity

Go to the last point in "Properties" section.

O.K. ajb. Do you know any other deduction for the zero "mass" of the photon?

I thought the previous was the reason.

area=0.5*base*height=0.5*base*sqrt(S^2-(9.5)^2), where S is the length of the side.

Then, d(area)/dt=d(area)/dS*dS/dt= 0.5*19*S/sqrt(S^2-(9.5)^2) * 2 ft^2/min

At height=8 ft, S=sqrt(8^2+(9.5)^2) ft.

Substitute with S in the rate of change of area , and you will get the result.

Since the sequence is bounded from above, therefore it reaches a certain number (Ao), as n tends to infinity. Then, lim A(n), as n tends infinity, equals Ao. Therefore, it has a limit.

1. You will have to check the Left Limit for the variable x. It will be lim f(x,y)=xo*y

2. Check the Right Limit for the variable x. It will be also lim f(x,y)=xo*y.

3. Therefore, the Left Limit of f(x,y), as x tends to xo, is equal to the Right Limit.

4. Repeat the previous, as y tends yo. And, you will get the same results.

5. Then lim f(x,y), as (x,y) tends to (xo,yo), is equal to f(xo,yo).

Note: There are some simple functions that can be considered continuous, unless it is required to prove so....... such as f(x)=x, f(x)=x^2.

Sorry, another factor is remaining which is lim f(x), as x tends to infinity, exists and is equal to zero.

And, we have the theorm "If a real-valued function f(x) is continuous on (0,infinity) and its limit, as x tends to infinity, exists (and is finite), then f(x) is uniformly continuous."

You are right, timo. A is not invetible and so A^2. Sorry for that error.

The reason that photon does have a zero rest mass is that its gamma is infinity. Therefore, mo=m/infinity=zero.

But, as you said, Einstein Tensor depends upon the Metric Tensor, therefore the metric (which is sufficient to describe the spacetime curvature) is defined in The Einstein Field Equation. Then, the basic background of gravity is defined in " The Einstein Field Equation".

It is, then, what is remaining, is the way that spacetime curvature is acting upon the mass situated at certain position.

That a function to be continuous on an interval, it has to be continuous at every point in the interval.

O.K...... I was talking like that because f(x,y) is obviously continuous,

and lim f(x,y), as (x,y) tends to (xo,yo), is f(xo,yo).

If you need an exact proof, then find the left limit which will be f(xo - epsilon , yo - epsilon). And, find the right limit which will be f(xo + epsilon , yo + epsilon). Then, since both are equal then the limit is found.

You misunderstood me ajb. I agree with you that it's indirectly implied. But, we can't say that "this energy is not included in the Einstein Field Equations".

How is gravitational energy not involved in Einstein Field Equation?!

It is totally built upon the effect of Mass on the spacetime curviture, which in turns define gravity. Gravitational energy is implied in the Einstein Tensor in the Equation.

But, don't forget. There is another part that is remaining which is "An object in a gravitational field follows the shortest distance between two points". It is completing the description of the gravitaional effect on masses.

For sure, many of the properties you stated about energy waves are correct. But, there is a difference between Probability Wave and Energy Wave.

PW is describing the probability of existence of a particle under certain potential. Of course, the particle has energy, but, it is the wave nature of the particle not the energy.

Energy waves, however, can be described by the motion of mediators, which are particles with special specifications. And, I agree with you, these mediators can be described quantatively by Quantum Field Nature according to the properties of the source of the field.

When gravitons interact with a mass, this doesn't imply a change at the source. But, as this mass is affected, its own field of cource changes (at a velocity of c).

You can check it with a small lamp..... Intensity of light is proportional to the voltage applied.

The RMS of a sine wave is A/sqrt(2). The average of half a cycle is 2A/pi.

Where A is the amplitude of the sine wave.

I think the energy in the strings (energy of all harmonics) is related to the energy of the particle involved. This includes the kinetic energy and the rest energy. Potential energy is included in the factors affecting the strings.

The integration is in the positive direction of theta. That's why it is the green trace.

I think that Fnet is not zero, it is equal to -dp/dt; where p is the momentum of gases.

Vdm/dt is an additional term due to the change of the total mass of the rocket. But, it is not the only effect on the acceleration.

Perhaps because the angle phi in your problem is defined upon -pi to pi. So, when crossing pi it is transferred to a negative value. This may be the reason.

You are right Bignose. But, isn't the power here is most significant: (n!)^n. Plus, at large n the sequence turns to zero.

It will get to the result, alejandrito. Where is the difficulty? Is it the right hand side?

Substitute with 8*pi*Tmn instead of Gmn. And, then raise the index n. I think this will come very close to eq 2 and eq 3.

Multiplying with gmn will make the equation loose its nature tensor...... It will be scalar.