As Wikipedia says:

[...] the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $\frac{1}{2}mv^2$.

Why does this not increase linearly with speed? Why does it take so much more energy to go from $1\ \mathrm{m/s}$ to $2\ \mathrm{m/s}$ than it does to go from $0\ \mathrm{m/s}$ to $1\ \mathrm{m/s}$?

My intuition is wrong here, please help it out!


17 Answers 17


The previous answers all restate the problem as "Work is force dot/times distance". But this is not really satisfying, because you could then ask "Why is work force dot distance?" and the mystery is the same.

The only way to answer questions like this is to rely on symmetry principles, since these are more fundamental than the laws of motion. Using Galilean invariance, the symmetry that says that the laws of physics look the same to you on a moving train, you can explain why energy must be proportional to the mass times the velocity squared.

First, you need to define kinetic energy. I will define it as follows: the kinetic energy $E(m,v)$ of a ball of clay of mass $m$ moving with velocity $v$ is the amount of calories of heat that it makes when it smacks into a wall. This definition does not make reference to any mechanical quantity, and it can be determined using thermometers. I will show that, assuming Galilean invariance, $E(v)$ must be the square of the velocity.

$E(m,v)$, if it is invariant, must be proportional to the mass, because you can smack two clay balls side by side and get twice the heating, so

$$ E(m,v) = m E(v)$$

Further, if you smack two identical clay balls of mass $m$ moving with velocity $v$ head-on into each other, both balls stop, by symmetry. The result is that each acts as a wall for the other, and you must get an amount of heating equal to $2m E(v)$.

But now look at this in a train which is moving along with one of the balls before the collision. In this frame of reference, the first ball starts out stopped, the second ball hits it at $2v$, and the two-ball stuck system ends up moving with velocity $v$.

The kinetic energy of the second ball is $mE(2v)$ at the start, and after the collision, you have $2mE(v)$ kinetic energy stored in the combined ball. But the heating generated by the collision is the same as in the earlier case. So there are now two $2mE(v)$ terms to consider: one representing the heat generated by the collision, which we saw earlier was $2mE(v)$, and the other representing the energy stored in the moving, double-mass ball, which is also $2mE(v)$. Due to conservation of energy, those two terms need to add up to the kinetic energy of the second ball before the collision:

$$ mE(2v) = 2mE(v) + 2mE(v)$$

$$ E(2v) = 4 E(v)$$

which implies that $E$ is quadratic.

Non-circular force-times-distance

Here is the non-circular version of the force-times-distance argument that everyone seems to love so much, but is never done correctly. In order to argue that energy is quadratic in velocity, it is enough to establish two things:

  • Potential energy on the Earth's surface is linear in height
  • Objects falling on the Earth's surface have constant acceleration

The result then follows.

That the energy in a constant gravitational field is proportional to the height is established by statics. If you believe the law of the lever, an object will be in equilibrium with another object on a lever when the distances are inversely proportional to the masses (there are simple geometric demonstrations of this that require nothing more than the fact that equal mass objects balance at equal center-of-mass distances). Then if you tilt the lever a little bit, the mass-times-height gained by 1 is equal to the mass-times-height gained by the other. This allows you to lift objects and lower them with very little effort, so long as the mass-times-height added over all the objects is constant before and after.This is Archimedes' principle.

Another way of saying the same thing uses an elevator, consisting of two platforms connected by a chain through a pulley, so that when one goes up, the other goes down. You can lift an object up, if you lower an equal amount of mass down the same amount. You can lift two objects a certain distance in two steps, if you drop an object twice as far.

This establishes that for all reversible motions of the elevator, the ones that do not require you to do any work (in both the colloquial sense and the physics sense--- the two notions coincide here), the mass-times-height summed over all the objects is conserved. The "energy" can now be defined as that quantity of motion which is conserved when these objects are allowed to move with a non-infinitesimal velocity. This is Feynman's version of Archimedes.

So the mass-times-height is a measure of the effort required to lift something, and it is a conserved quantity in statics. This quantity should be conserved even if there is dynamics in intermediate stages. By this I mean that if you let two weights drop while suspended on a string, let them do an elastic collision, and catch the two objects when they stop moving again, you did no work. The objects should then go up to the same total mass-times-height.

This is the original demonstration of the laws of elastic collisions by Christian Huygens, who argued that if you drop two masses on pendulums, and let them collide, their center of mass has to go up to the same height, if you catch the balls at their maximum point. From this, Huygens generalized the law of conservation of potential energy implicit in Archimedes to derive the law of conservation of square-velocity in elastic collisions. His principle that the center of mass cannot be raised by dynamic collisions is the first statement of conservation of energy.

For completeness, the fact that an object accelerates in a constant gravitational field with uniform acceleration is a consequence of Galilean invariance, and the assumption that a gravitational field is frame invariant to uniform motions up and down with a steady velocity. Once you know that motion in constant gravity is constant acceleration, you know that

$$ mv^2/2 + mgh = C $$

so that Huygens dynamical quantity which is additively conserved along with Archimedes mass times height is the velocity squared.

  • 4
    $\begingroup$ " The kinetic energy of the second ball is mE(2v)mE(2v) at the start, and after the collision, you have 2mE(v)2mE(v) kinetic energy stored in the combined ball." If you give all the velocity to the second ball then you cannot all of a sudden begin to treat the first one as moving and still possesing kinetic energy. You need to choose your frame of reference and stick to it, otherwise it turns out to be just plain .. cheating ... ;-) $\endgroup$ – bright magus Jul 31 '16 at 21:23
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    $\begingroup$ @brightmagus what do you mean? What Ron does in this answer is a mere Galilean boost of velocities: first of those before collision, then of those after. No cheating here. $\endgroup$ – Ruslan Nov 29 '16 at 10:44
  • $\begingroup$ @Ron Maimon, I have been reading your answers on this subject, but in this answer, I don't get why the sum of the mass times height over all objects will not be constant, if the motion requires me to do work(in both the physical and colloquial sense)? Can you please explain that? $\endgroup$ – codetalker Jun 8 '17 at 13:50
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    $\begingroup$ Two clay balls is a very beautiful argument! Thank you for this illuminating answer! $\endgroup$ – Blex Jan 21 '18 at 13:50
  • $\begingroup$ Reading this reminds me of reading Michael Spivak's Physics for Mathematicians. Great Answer! $\endgroup$ – Joshua Ronis Feb 3 at 22:01

The question is especially relevant from a didactical point of view because one has to learn to distingish between energy (work) and momentum (quantity of motion).

The kinematic property that is proportional to $v$ is nowadays called momentum, it is the "quantity of motion" residing in a moving object, its definition is $p:= mv$.

The change of momentum is proportional to the impulse: impulse is the product of a force $F$ and the timespan $\Delta t$ it is applied. This relation is also known as the second law of Newton: $F \Delta t = \Delta p$ or $F dt = dp$. When one substitutes $mv$ for $p$ one gets its more common form: $F= m \frac{\Delta v}{\Delta t} = ma$.

Now for an intuitive explanation that an object with double velocity has four times as much kinetic energy.
Say A has velocity $v$ and B is an identical object with velocity $2v$.
B has a double quantity of motion (momentum) - that's were your intuition is correct!
Now we apply a constant force $F$ to slow both objects down to standstill. From $F \Delta t = \Delta p$ it follows that the time $\Delta t$ needed for B to slow down is twice as much (we apply the same force to A and B). Therefore the braking distance of B will be a factor of 4 bigger then the braking distance of A (its starting velocity, and therefore also its mean velocity, being twice as much, and its time $\Delta t$ being twice as much, so the distance, $s = \bar{v}\Delta t$, increases 2 x 2 = 4 times).
The work $W$ needed to slow down A and B is calculated as the product of the force and the braking distance $W=Fs$, so this is also four times as much. The kinetic energy is defined as this amount of work, so there we are.


Let me just throw in an intuitive explanation. You could re-phrase your question as:

Why does velocity only increase as the square root of kinetic energy, not linearly?

Well, drop a ball from a height of 1 meter, and it has velocity v when it hits the ground.

Now, drop it from a height of 2 meters. Will it have a velocity of 2v when it hits the ground?

No, because it travels the second meter in a lot less time (because it's already moving), so it has less time to gain speed.


The only real physical reason (which is not really a fully satisfying answer) is that $E \sim v^2$ is what experiments tell us. For example, gravitational potential energy on the Earth's surface is proportional to height, and if you drop an object, you can measure that the height it falls is proportional to the square of its speed. Thus, if energy is to be conserved, the kinetic energy has to be proportional to $v^2$.

Of course, you could question why gravitational potential energy is proportional to height, and once that was resolved, question why some other kind of energy is proportional to something else, and so on. At some point it becomes a philosophical question. The bottom line is, defining kinetic energy to be proportional to the square of the speed has turned out to make a useful theory. That's why we do it.

On the other hand, you could always say that if it were linear in velocity, it would be called momentum ;-)

P.S. It may be worth mentioning that kinetic energy is not exactly proportional to $v^2$. Special relativity gives us the following formula:

$K = mc^2\left(1/\sqrt{1 - v^2/c^2} - 1\right)$

For low speeds, this is essentially equal to $mv^2/2$.

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    $\begingroup$ I believe this doesn't really address the question; kinetic energy is defined to be $v^2$ as per definition of work done by Newton's law (same holds for the relativistic expression). Then that this matches with conservation of the energy is another matter (because you would need the correct definitions of the potentials, which in turn come looking at the work done by the conservative bit of the force). $\endgroup$ – gented Nov 11 '16 at 22:47
  • $\begingroup$ The question is really asking why kinetic energy depends on $v^2$ rather than $v$? $\endgroup$ – jim Jul 11 '18 at 15:24

Just to post another, more mathematical, version of this that is not dependent on thermodynamics, but rather just vector calculus and Newton's laws, let's consider Newton's second law:

$$\sum {\vec F} = m{\vec a}$$

Now, apply the definition of work, $W = \int d{\vec s} \cdot{\vec F}$

We have, assuming that $s$ is the actual path traveled by the particle, and using some clever changes of variables:

$$\begin{align} \sum W &= m\int d{\vec s(t)}\cdot {\vec a}\\ &=m\int dt\frac{d{\vec s}}{dt}\cdot {\vec a}\\ &= m\int dt \,{\vec v} \cdot {\vec a}\\ &= m\int dt\,{\vec v}\cdot \frac{d{\vec v}}{dt}\\ &= m\int {\vec v} \cdot d{\vec v}\\ &= \frac{1}{2}m\left(v_{f}^{2} - v_{i}^{2}\right)\\ &= \Delta {\rm KE} \end{align}$$

So, we see that the definition of work is synonymous with quadratic dependence on the velocity. Who cares? Well, now, we fix some requirements on the force. Namely, we assume our forces are conservative. What does this mean? Well, it means that our force is curl free $\rightarrow {\vec \nabla} \times {\vec F}=0$. This is mathematically equivalent to many things, but the most important two are that $\int d{\vec s}\cdot {\vec F}$ doesn't depend on the path you integrate over, but only the endpoints of the curve, and second, that ${\vec F} = -{\vec \nabla}\phi$ for some function $\phi(x,y,z,t)$. Once you know this, it's relatively easy to show that $\int {\vec ds}\cdot {\vec F} = \phi_{0} - \phi_{f}$

Then, you have:

$$0 = \Delta {\rm KE} + \sum \Delta {\rm PE}_{i}$$

where the sum is over the potentials for the various forces (and I deviously substituted PE for $\phi$, since we are obviously talking about potential energy now.) We have now proved that the total energy does not change. Therefore, the standard definition of work gives us a conserved quantity, which we can call the energy (so long as we assume the absence of nonconservative forces, but in the presence of these, energy is not conserved, and we start having to worry about losses to heat and radiation).

  • $\begingroup$ This doesn't answer the question. Why is K.E. $1/2mv^2$? So basically Ron's answer seems like the only one that actually tries to answer this. Even though it needs to appeal to another definition of K.E. which is also not very intuitive. $\endgroup$ – philmcole Jun 10 '18 at 18:10
  • $\begingroup$ @philmcole: the thrust of this answer is "If you believe in newtonian mechanics, you get a conserved quantity that is equal to $\frac{1}{2}mv^{2}$. How is that not a why? The answer ultimately has to come from Newtonian mechanics somewhere. $\endgroup$ – Jerry Schirmer Jun 18 '18 at 15:23

As Piotr suggested, accepting the definition of work $W=\mathbf{F}\cdot d\mathbf{x}$, it follows that the kinetic energy increases quadratically. Why? Because the force and the infinitesimal interval depend linearly on the velocity. Therefore, it is natural to think that if you multiply both quantities, you need to end up with something like $K v^{2}$, where $K$ is an 'arbitrary' constant.

A much more interesting question is why the Lagrangian depends on the velocity squared. Given the homogeneity of space, it can not contain explicitly $\mathbf{r}$ and given the homogeneity of time, it can not depend on the time. Also, since space is isotropic, the Lagrangian can not contain the velocity $\mathbf{v}$. Therefore, the next simplest choice should be that the Lagrangian must contain the velocity squared. I do think that the Lagrangian is more fundamental in nature than the other quantities, however, its derivation involves the definition of work or equivalently, energy. So probably you won't buy the idea that this last explanation is the true cause of having the kinetic energy increasing quadratically, although, I think it is much more satisfactory than the first explanation.


In comes down to definitions.

Momentum is defined as $p = mv$. Momentum grows linearly with velocity making momentum a quantity that is intuitive to understand (the more momentum the harder an object is to stop). Kinetic energy is a less intuitive quantity associated with an object in motion. KE is assigned such that the instantaneous change in the KE yields the momentum of that object at any given time:

$\frac{dKE}{dv} = p$

A separate question one might ask is why do we care about this quantity? The answer is that in a system with no friction, the sum of the kinetic and potential energies of an object is conserved:

$\frac{d(KE + PE)}{dt} = 0 $


For every relatively equal (in percents) increase of the speed, the applied force must be present over increasingly (quadratically) long travel distance. F=m*a. At the same time force*distance=work, where work=energy.


The general form of the kinetic energy includes higher order corrections due to relativity. The quadratic term is only a Newtonian approximation valid when velocities are low in comparison to the speed of light c.

There is another fundamental reason for which kinetic energy cannot depend linearly with the velocity. Kinetic energy is a scalar, velocity is a vector. Moreover, if the dependency was linear this would mean that the kinetic energy would vary by substituting $\mathbf{v}$ by $-\mathbf{v}$. I.e. the kinetic energy would depend of the orientation, which again makes no sense. The Newtonian quadratic dependence and the relativistic corrections $v^4$, $v^6$... satisfy both requirements: kinetic energy is a scalar and invariant to substituting $\mathbf{v}$ by $-\mathbf{v}$.


I think it follows from the first law of Thermodynamics. It turns your definition of work into a conserved property called energy. If you define work in the $Fdx$ style (as James Joule did) then the quadratic expression for kinetic energy will follow with the symmetry arguments.

In his excellent answer, Ron Maimon cleverly suggests using heat to avoid a reference to work. To determine the number of calories he uses a thermometer. A perfect thermometer will measure $\partial{E}/\partial{S}$ so when he's done defining entropy, he still needs a non-mechanical definition of work. (In fact, I believe it is Joule's contribution to show that the calorie is a superfluous measure of energy.) The weakness in Ron's answer is that he also needs the second law of thermodynamics to answer the question.

To see this explicitly, write the first law in terms of the Gibbs equation: $$ dE = TdS + vdp + Fdx $$ This equation defines $v = \partial{E}/\partial{p}$. For a conservative system set $dE=0$ and to follow Huygens, set $dS=0$ to get $vdp = - Fdx$ and to follow Maimon we set $dx=0$ to get $vdp = -TdS$. These are two ways of measuring kinetic energy.

Now to integrate. Huygens assumes $p$ is only a function of $v$. For small changes in $v$ we make the linear approximation $p = mv$, where $m \equiv dp/dv$. Plug that in, integrate, and you get the quadratic dependence. In fact, it's not too hard to see that if you use gravity for the force that $F = mg$ which leads to $$ \frac{1}{2} m v^2 + mgh = C . $$ Raimon also has to assume the independence of $p$ on $S$. To integrate he will have to evaluate $T$ as a function of $S$ (and possibly $p$) or use the heat capacity.

Now notice that we required the changes in $v$ to be small. In fact, kinetic energy is not always proportional to $v^2$. If you go close to the speed of light the whole thing breaks down and for light itself there is no mass, but photons do have kinetic energy equal to $c p$ where $c$ is the speed of light. Therefore, it's better to think of kinetic energy as $$ E_{kin} = \int v dp $$ and just carry out the integration to find the true dependence on $v$.

So, in summary, I suggest the "why" of the question is the same as the "why" of the first law.


Basically, momentum is related to force times time, and KE is related to force times distance. It is all a metter of frame of reference, either time or distance. The relationship between time and distance for a starting velocity of zero is $d = \frac{at^2}{2}= \frac{tV}{2}$. Plug this into the the equations you get the KE$ = \frac{pV}{2} = \frac{p^2}{2m}$

Woolah - magic!


Imagine a cannon that fires a bed spring at 10 m/s (22 mph); the momentum of the fired bed spring carries it 25 meters. A second cannon fires the bed spring at 20 m/s (44 mph); the momentum of the spring carries it 50 meters. Doubling the speed, doubles the momentum.

In a second set of tests the first cannon fires the spring into an immovable brick wall from a distance of one meter. Upon impact the spring which has a length of 20 cm compresses to a length of 16 cm (20% compressed). When the second cannon fires the spring into the wall, traveling at twice the speed, the 24 cm spring compresses to a length of 4 cm (80% compressed); this of course requires 4 times more energy than the first collision with the wall. Double the speed, quadruple the kinetic energy!

Please correct me if I'm off base. thanks


I have a quantitative answer which is a thought experiment avoiding all but the simplest equations.

An object going from velocity v=0 to v=1 needs to be pushed or pulled in some way. In my explanation I will use the same method to push the object from v=0 to v=1 then from v=1 to v=2, then v=2 to v=3, etc. I will show how the energy of movement embodied in the object goes up from 0 to 1 to 4 to 9, etc.

Start with two identical balls, m1 and m2. Between the two balls is a spring, s1, which is held in compression. Assume the mass of the spring is very small. The potential energy in the spring is PE=2 and all 3 actors have velocity v=0.

A. v=0. All objects have 0 velocity so kinetic energy KE=0.

B. v=1. Release the spring and m1 shoots off to the left with velocity v=1. m2 goes in the opposite direction with v=-1. The kinetic energy of both balls is the same and is KE=1 because all of the potential energy of the spring has been transferred symmetrically to the balls.

C. v=2. Now place another identical ball, m3, just to the right of m1 and also travelling at v=1 and with a compressed spring, s2, between them. Nothing has changed about m1, it is still happily travelling at v=1. So what's the total energy of the m1, s2 and m3 system? It's 1+2+1=4 being m1's KE, s2's PE and m3's KE.

Now release the spring and m1 shoots off to the left with v=2 and m3's velocity goes from v=1 to v=0 making its KE=0. Because we've said that the mass of the spring is very small so its KE is almost zero then all of the energy which was in the system before the spring was released is now in m1. So the KE of m1 is KE=4. Phew, KE is proportional to v squared!

D. v=3. Simply repeat the process to make m1 go from v=2 to v=3 by pushing off another identical ball, m4. First, work out the total energy of the two ball and spring system before the spring is released. It's 4+2+4=10. After the spring is released m4 has v=1 which we've established is equivalent to KE=1. So m1 has the remaining energy of the system which is KE=9.

E. v=4. Repeat the process. Energy of system before the spring is released, 9+2+9=20. KE of m1 after spring is released, KE=20-4=16.

I'm not happy with assuming away the mass of the spring so a neater explanation has a spring attached to each ball and the balls interact via their springs which are in contact.


Kinetic energy is defined as $\frac{1}{2}mv^2$ (in classical mechanics at least).

When the motion of an object is subjected to a physical law that is constant through time (for instance $\ddot{r}=-\frac{GM}{r^2}$ where GM is a constant), then when you integrate both sides with respect to distance and multiply by the mass $m$ of the object you get:

$$\frac{1}{2}mv_2^2 - \frac{GMm}{r_2} = \frac{1}{2}mv_1^2 - \frac{GMm}{r_1}$$

Assuming that the law is constant through time, then between the initial and final states the object's quantity $\frac{1}{2}mv^2 - \frac{GMm}{r}$ is conserved through time as well.

If instead of $-\frac{GM}{r^2}$ the physical law is some other function $f(r)$ constant through time, then the object's quantity $\frac{1}{2}mv^2 - F(r)$ where F is a primitive of f is conserved through time as well.

That quantity is called energy. Then we give a name to the two terms: the term that depends on velocity ($\frac{1}{2}mv^2$) is referred to as kinetic energy, and the term that depends on distance ($-F(r)$) is referred to as potential energy.

It is useful to define these quantities, because if we assume that the acceleration of an object is a function of distance constant through time (as is the case with the law of gravitation, Coulomb's law, Hooke's law, ...), and if we know the value of $F(r)$ and the value of the velocity at a given distance $r_1$ (which are both derived from measurements), then we can deduce directly the velocity of the object at any other distance without having to calculate the integral of $f(r)$ every time.

Since kinetic energy is a defined quantity it is meaningless to ask why it increases quadratically with velocity, it does because it is defined that way. The above argument gives a reason as to why it is defined that way.

Why does it take so much more energy to go from 1 m/s to 2 m/s than it does to go from 0 m/s to 1 m/s?

It isn't harder to accelerate something from 1 m/s to 2 m/s than from 0 m/s to 1 m/s, at a constant acceleration it takes the same time, however it takes 3 times more distance (so it takes 4 times more distance to accelerate from 0 m/s to 2 m/s than from 0 m/s to 1 m/s).

Let's say you accelerate your object at some constant rate so that it takes a time $\tau$ to go from 0 m/s to 1 m/s. Then it will take the same time $\tau$ to go from 1 m/s to 2 m/s.

Its velocity as a function of time will be $v(t) = \frac{1}{\tau}t$. In particular, $v(\tau) = 1$ and $v(2\tau) = 2$. Its distance traveled as a function of time will be $d(t) = \frac{1}{2\tau}t^2$

It takes a distance $d(\tau) = \frac{\tau}{2}$ to accelerate it from 0 m/s to 1 m/s, while it takes a distance $d(2\tau) = 2\tau$ to accelerate it from 0 m/s to 2 m/s.

As you can see, $d(2\tau) = 4d(\tau)$. At no point do you need to invoke kinetic energy to explain this observation, it takes 4 times more distance because the object is moving faster between $\tau$ and $2\tau$ than between $0$ and $\tau$. Similarly, at a constant deceleration rate it takes 4 times more distance to brake to a stop at velocity $2v$ than at velocity $v$, not because kinetic energy makes it somehow harder to brake when we are going faster, but simply because it takes twice longer to brake (the time to go from $2v$ to $v$ is the same as the time to go from $v$ to $0$), and because we are moving faster than $v$ (hence covering more distance) during half the braking time.

  • $\begingroup$ Only this really answered the question. $\endgroup$ – Milind R Oct 19 '15 at 6:00
  • $\begingroup$ No, KE is not defined as $\frac{1}{2}mv^2$. It's a consequence of Newton's 2nd law, which is a consequence of the fact that $x$ and $\dot{x}$ are enough to specify the state of a system uniquely. $\endgroup$ – Arturo don Juan Sep 27 '17 at 18:32

One way to look at this question of yours is as follows:

$$ E(v) = \frac{m v^2}{2} \; . $$

So, if we multiply the velocity by a certain quantity, i.e., if we scale the velocity, we get the following,

$$ E(\lambda v) = \frac{m (\lambda v)^2}{2} = \lambda^2 \frac{m v^2}{2} = \lambda^2 E(v)\; . $$

That is, if you scale your velocity by a factor of $\lambda$, your Energy is scaled by a factor of $\lambda^2$ — this should answer your question (just plug in the numbers).

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    $\begingroup$ That like saying x=y because y=x $\endgroup$ – The Imp Sep 6 '15 at 19:25

Kinetic energy's quadratic makes perfect sense if our reality is not actually first order in space, and is instead simply a measurement of the relative rate that an object is passing through time. The space of our existence then becomes the space of simultaneous time, at any given point in time, as it progresses.

In this scenario, changing the kinetic energy of an object creates a push (think "force") to a new space of simultaneity as time moves faster or slower for an object.

Why does Newton's first law work? Is it because time passes at a constant rate, when it is not changing? And when it changes, does it do so as a square / square root because time is experienced as passing in all directions?

Time would then be thought of as passing omni-directionally, while motion obviously exists linearly. Combine these and you have the square root of spacetime! Just like the expanding or contracting radius of a sphere! (I always seem to have the Higg's Mexican hat flash before my eyes when I think of force going a particular direction for some reason, even though it doesn't directly relate here as far as I can tell lol).

As time passes more slowly, space correspondingly shrinks for every object as to align with light-speed invariance for every object at all times. Lorentz contraction results and becomes predominant at the rate that time passes approaches zero for one object relative to the other. The space of simultaneous time at a point then rounds out the picture.

The symmetries of SR and energy conservation are fun things to think through in this scenario.

Reality actually has to exist in this way, as best I can tell. There is no leeway for it to be wrong since light-speed invariance also applies to the infinitesimal of acceleration. The infinitesimal can only be a reflection of a new space of simultaneity being created if the accelerating object is to maintain light-speed constancy in all instances.

Kinetic energy is not actually and truly a scalar that exists in space. And time is actually a real thing that really does pass. The passing of time is no illusion.

Kinetic energy requires the passing of time because it is the quantitative relative dilation of that very rate that time passes that defines what kinetic energy actually and truly is!

The result is that the square in kinetic energy aligns perfectly with Lorentz invariance to perfectly explain our reality as we measure it to be, if we just accept that time passing is a real thing, and give it the primacy in our thinking that it deserves to have!

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    $\begingroup$ This appears to be mostly nonsense. $\endgroup$ – Kyle Kanos Jan 27 '16 at 17:46
  • $\begingroup$ Cool, thanks for the comment Kyle. If you want to talk about things, I'm here. If not, that's cool too. One thing at a time would be nice if you do. I'm willing to go until you change my mind. $\endgroup$ – Craig Heile Jan 27 '16 at 17:51
  • $\begingroup$ Seriously, the invite is open to everyone and anyone. I've thought about this extensively for years, so I am not defensive. $\endgroup$ – Craig Heile Jan 27 '16 at 17:52
  • $\begingroup$ If you want an example, then your first two paragraphs contain meaningless statements. $\endgroup$ – Kyle Kanos Jan 27 '16 at 17:53
  • $\begingroup$ To be honest, I don't sense you think I am worthy of your time. Which, again, is fine, I just don't want to push it if you would rather be doing something else. Answer this or not, your choice, is the passing of time an illusion? $\endgroup$ – Craig Heile Jan 27 '16 at 17:55

Your intuition is NOT wrong and I will show you why. Say you have a machine that can accelerate a ball to 5 meters per second consistently. Put that machine on a car that is traveling at 5 meters per second and have it accelerate a ball. How fast is that ball traveling? The answer is 10 meters per second (5+5=10) as any physicist will tell you. If the machine is not moving, it changes the ball's velocity from rest to 5 m/s. When it is moving, it changes the ball's velocity from 5 to 10 m/s. The machine does not magically know that it is moving and so it always uses the same amount of energy, say electrical, whenever it accelerates the ball. If the kinetic energy formula were actually valid, that machine would not be able to cause the ball to accelerate to 10 m/s in this situation.

Those that tell you differently are only parroting the usual physics lessons. They can "prove" the kinetic energy formula is valid but you have to assume that work (the product of force and distance) is a proven fact; it is not. They might show an experiment that equates potential energy with kinetic that seems valid; this also depends on the assumption that work is a valid scientific concept. The example I gave in the first paragraph can be "explained" away by a clever physicist, maybe, but it is a direct test proving your intution is correct.

  • $\begingroup$ The error in your argument is that as every action has a reaction, the car will be pushed backwards by the recoil of your machine slightly. The faster the car is going, the more energy it loses by the recoil because of the square law, balancing the increase in energy of the ball by the same law. $\endgroup$ – rghome Nov 10 '16 at 22:10

protected by Qmechanic Feb 21 '14 at 22:16

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