Energy=Force

[Guille]

Actually, the title of the thread is a bid exagerated. If I'm wrong please, please tell me because I don't want to go on with my life thinking that I'f demostrate something when I haven't.

ok, I base this in two basic and accepted formulas:

E=mc^2

and

F=ma

from the first one, reareinging:

m=E/c^2

and reareinging the second one...

m=F/a

what means that

F/a=E/c^2

if, and only if,

a=c^2

then we get

F=E



Which means that the force of an object equals to it's energy when it's acceleration is equal th the speed of light squared (90,000,000,000).

tell me if I'm doing something compeltely worng or HAVEN'T UNDERSTOOD something in it. any comment?

[AntonioLao]

a=c^2

this is true only for unitary metric and consequently it is derivable from

[math]\mathbf{a}\cdot\mathbf{r}=c^2[/math]

In quantum mechanics and using the spinor notation, it give rise to a very high frequency oscillation but of very small amplitude superposed on the regular linear motion which we normally see.

[Guille]

this is true only for unitary metric and consequently it is derivable from

[math]\mathbf{a}\cdot\mathbf{r}=c^2[/math]

In quantum mechanics and using the spinor notation, it give rise to a very high frequency oscillation but of very small amplitude superposed on the regular linear motion which we normally see.

how can you derive a=c^2 from ar=c^2. wouldn't that only be if r=0???

By the way, yes, it is not very probable to have a=c^2 because that would meen that a=90,000,000,000meters per second squared, which is a galactic acceleration, has there been any acceleration such as this one? maybe the universe just after the big bang?

[AntonioLao]

has there been any acceleration such as this one? maybe the universe just after the big bang?

the internal oscillation of an electron can be greater than light speed but the observable amplitude is zero.

[Guille]

o, just to add a few things to my demostration,

if,

a=c^2

then, making a and c the subjects in both eqautions would be....

c^2=E/m

a=F/m

which if

c^2=a

then

F/m=E/m

and as the mass of an object (m) is always the same, then

F=E



again!

[AntonioLao]

and as the mass of an object (m) is always the same, then

F=E

this is what puzzled me when i tried to understand the concept of pressure. There seems to be a microscopic definition and also a macroscopic definition but i'm not sure which is which for the following definitions.

pressure p is defined one way by force per unit area as

[math] p = \frac{\mathbf{F}}{\mathbf{A}} [/math]

and another way by energy per unit volume as

[math] p = \frac{E}{V} [/math]

the 1st is the division between two vectors and the other is divsion between two scalar quantities.

And the total pressure of radiation is only a third of p.

[math] \frac{1}{3} p [/math]

[Guille]

pressure p is defined one way by force per unit area as

[math] p = \frac{\mathbf{F}}{\mathbf{A}} [/math]

and another way by energy per unit volume as

[math] p = \frac{E}{V} [/math]

My physics teacher told me that up to grade 12, and in difference with many physics workings, we are going to use little math, basically knowing and understanding formulas and rearenging them. that's easy and I like it to derive things.

from those two formulas, I get that

F/A=E/V

which, if:

A=V

then,

F=E

the problem of this, is that A is surface area and V volume, whihc means that if the object we are talking about is bi-dimensional, then A can be V. or better said, V doesn't exist.

so again,

F=E



These might not be very global demostrations and need the fact that something equals something, but they truly show the high conection that I have always thought about energy and force. because, there is no energy without force, and there is no force without energy. I mean, something needs energy to have the ability of doing work, which is the same as applying a force through a distance (WD=F(times)d), and if there is a force, there is the ability of doing work, or better said, there is energy.

[AntonioLao]

kinetic energy is the integral of an inner product of force with incremental distances.

[math] K.E. = \int \mathbf{F} \cdot d \mathbf{s} [/math]

and the force is the gradient of potential energy

[math] F = \mathbf{\nabla} (P.E.) [/math]

[Guille]

what's the upside-down triangle? what dos it do and is?

[math] F = \mathbf{\nabla} (P.E.) [/math]

[AntonioLao]

what's the upside-down triangle? what dos it do and is?

[math] \nabla [/math] is the gradient operator of vector analysis. It is a derivative operator with respect to space. It transforms a scalar quantity into a vector quantity. It is applied in practice for finding the slopes of quickest descend or ascend on the contours of mountains and hills. For a 3D Cartesian coordinate system, it is given by

[math] \nabla = i \frac{\partial}{\partial x} + j\frac{\partial}{\partial y} + k \frac{\partial}{\partial z} [/math]

where i, j, k are the unit vectors.