B Why is squaring used in physics equations?

[Pjeter Schornstein]

Hey all!
I't my first question here I hope I'm posting it in the correct thread.

I know that both Newton Gravity equation and Einstein Energy equation both use square. How did they come to use it in the first place?

Thank you.

 

[anuttarasammyak]

You know the area of sphere is $4\pi r^2$ where r is its radius. When some amounts emerging from center source arrives and distributed on the sphere, it is diluted divided by $4\pi r^2$.

 

[PeroK]

Hey all!
I't my first question here I hope I'm posting it in the correct thread.

I know that both Newton Gravity equation and Einstein Energy equation both use square. How did they come to use it in the first place?

Thank you.

Newtonian gravity obeys an inverse square law, where the force reduces in proportion to the square of the distance. Newton had no choice about that - that was mother nature's decision. There is some logic in it as the surface of a sphere increases as the radius squared. If you imagine a gravitational field thinning out as it gets further from the source, then an inverse square law makes sense.

The $c^2$ in $E = mc^2$ is essentially a conversion factor from units of mass to units of energy. It is related to the Lorentz "gamma" factor $\gamma = \frac 1 {\sqrt{1- v^2/c^2}}$, which arises when you analayse spacetime under the assumption that the speed of light is invariant.

 

[Ibix]

I know that both Newton Gravity equation and Einstein Energy equation both use square. How did they come to use it in the first place?

Because the equations don't make accurate predictions if you don't. And the units don't make sense. You'd have to look at the specific derivations to know how each term arises for a specific case.

 

[anuttarasammyak]

As for energy formula in both
E=\frac{1}{2}mv^2
in Newtonian mechanics and
E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}
in relativity theory, $v^2$ appears. That is because not direction but magnitude of velocity matters. $v^2$ has no information of direction. Same for $c^2$. Replacing $v\rightarrow -v$ and/or $c\rightarrow -c$ do not change energy formula.

 

[A.T.]

I know that both Newton Gravity equation and Einstein Energy equation both use square. How did they come to use it in the first place?

Are these two the only formulas you know, that use squaring? Or how come you picked these two?

 

[Leo Liu]

The distance in the equation of gravity is squared because it matches our observations -- the planets orbit around the sun in ellipses because of the inverse square; this phenomenon is stated as the Kepler's 1st law.

Despite of it's being empirically derived, I can provide a bit of intuition here using Gauss's law for gravity. Let us assume that the objects we are dealing with are perfect spheres or point masses. The left side of the equation is gravity flux passing through an imaginary Gaussian sphere.

$$\oint \vec g \cdot d\vec A=-4\pi GM_{enclosed}$$
The surface area A of the Gaussian sphere is 4π(R+d)^2, where R is the radius of the sphere and d is the distance from the surface of the sphere to the imaginary Gaussian sphere. Since the gravitational field of a spherical object is uniform, and is perpendicular to each tiny region of the sphere, we can write:
$$g\,4\pi (R+d)^2=-4\pi GM_{enclosed}$$
$$\text{We place a point mass or another sphere m on the surface of the Gaussian sphere.} $$
$$\text{Notice that the radius of the Gaussian sphere is the same as the distance between the centre of mass of the two objects (M and m).}$$
$$mg=-\frac{GM_{enclosed}m}{(R+d)^2}$$
$$\text{Let R+d=r:}$$
$$F=-\frac{GMm}{r^2}$$
It is noteworthy that the equation derived merely states the relationship between the magnitude of each variable. The direction of the force can be determined by conducting experiments.
In short, the square comes from the surface area of the Gaussian surface on which m is placed.

 

[etotheipi]

$$\oint \vec g \cdot d\vec A=-4GM_{enclosed}$$

I think the RHS should be $-4\pi G M$, but it's just a small change.

$$mg=-\frac{GM_{enclosed}m}{(R+d)^2}$$

It's standard to treat $g = |\vec{g}| > 0$, so the LHS probably ought to have a negative sign in there too, but this is an even smaller quibble than the last one

 

[Herman Trivilino]

That is because not direction but magnitude of velocity matters. v² has no information of direction

And neither does $v$. In those formulas $v$ is the speed, a quantity that is never negative.

 

[Herman Trivilino]

I know that both Newton Gravity equation and Einstein Energy equation both use square. How did they come to use it in the first place?

Lots of formulas involve multiplication, such as the formula for the area of a rectangle, which is length times width. If the length equals the width then the area is the square of either quantity simply because the square of a quantity equals the quantity multiplied by itself.

 

[jack action]

I know that both Newton Gravity equation and Einstein Energy equation both use square. How did they come to use it in the first place?

It is more or a less an happy coincidence that is better explained with calculus.

For example, the amount of energy $dE$ due to work is, by definition, proportional to a force $F$ displaced by distance $dx$:
$$dE = F dx$$
If the only force considered is due to the acceleration $a$ of a mass $m$, then $F=ma$ or:
$$dE = madx$$
But we can relate the distance traveled $dx$ and the acceleration $a$ to the velocity $v$, the elapsed time $dt$ and the velocity change $dv$:
$$v=\frac{dx}{dt}$$
$$a=\frac{dv}{dt}$$
Rearranging both equations to eliminate $dt$, we get:
$$adx=vdv$$
Therefore:
$$dE = mvdv$$
As you can see, there is no square in this equation. The amount of energy is proportional to 3 things: The mass, the velocity and the change in velocity. When you use calculus to solve this equation, you get:
$$\int_{E = E_0}^{E_f} dE = \int_{v = v_0}^{v_f}mvdv$$
$$E_f - E_0 = \frac{1}{2}m \left(v_f^2 - v_0^2\right)$$
That is the true equation. This means: "When you change the velocity from $v_0$ to $v_f$, then you will change the energy level from $E_0$ to $E_f$ ."

Now, if one assumes $v_0 = 0$ and $E_0 = 0$, then it simplifies to:
$$E_f = \frac{1}{2}mv_f^2$$
Or simply:
$$E = \frac{1}{2}mv^2$$
So even though it looks like there is a variable that is squared, the fundamental equation is actually considering 2 different measurements, i.e. the velocity itself and the change in velocity. It is actually a sum of small changes that accumulates to a difference between the squared final value and the squared initial value.

Repeating this exercise with an equation considering a squared distance - for example, $r^2$ - you may find that it is actually considering a distance $r$ and a change in distance $dr$. Or, in the case of an area coming from a sphere or circle, it is still due to 2 different measurements in two different directions, but because of the symmetry, calculating an area will lead to a square value of the radius since it is the same basic value in all directions (like $\int dA = 4\pi r^2$ in post #15).

 

[davenn]

As for energy formula in both
E=\frac{1}{2}mv^2

that should be KE = ...

or sometimes used E_K = ...

Kinetic Energy

 

[DaveE]

While not directly an answer to your original question, the 1/r² laws are common in physics. Anytime you have a fixed amount of "stuff", like light from a star for example, that is then spread out evenly over a sphere (i.e. it goes equally everywhere in 3-D space), then the stuff density (the amount of stuff you have at anyone point or area) will decrease as 1/r²; like the amount of starlight that hits your eye.

For example, how does the thickness of a spherical balloon skin vary as you inflate the balloon? 1/r². Same amount of material spread over a sphere with increasing radius.

 

[Dale]

A very long digression was relocated to the general math forum

https://www.physicsforums.com/threads/digression-on-squaring.993884/#post-6394524