Relative motion and KE

kojikun · Post by **kojikun** » 2003-02-16 08:32pm

Because motion is relative, isn't KE also relative? an object sittin on the ground is considered to have 0 J of KE because its not moving, but when looked at from outside the earths orbital motion, it has shitloads of KE because its moving at something near 18 miles per second.

Would that then mean that KE is entirely dependent on ones point of view?

If thats true, is it not possible to "accelerate" to any given speed without expending energy by simply changing where a given amount of KE is relative to?

I mean, if your ship has 1 gigajoule of KE relative to mars, but none relative to earth, couldnt you change it so that its 1 gigajoule relative to earth but none relative to mars and in doing so instantly transit from 0 MPH to a larger speed?

also, wouldnt it be possible to instantly reverse direction but keep the same KE, because youre going in the opposite direction at the same speed?

Luke Starkiller · Post by **Luke Starkiller** » 2003-02-16 08:40pm

kojikun wrote:Because motion is relative, isn't KE also relative? an object sittin on the ground is considered to have 0 J of KE because its not moving, but when looked at from outside the earths orbital motion, it has shitloads of KE because its moving at something near 18 miles per second.

Would that then mean that KE is entirely dependent on ones point of view?

If thats true, is it not possible to "accelerate" to any given speed without expending energy by simply changing where a given amount of KE is relative to?

I mean, if your ship has 1 gigajoule of KE relative to mars, but none relative to earth, couldnt you change it so that its 1 gigajoule relative to earth but none relative to mars and in doing so instantly transit from 0 MPH to a larger speed?

also, wouldnt it be possible to instantly reverse direction but keep the same KE, because youre going in the opposite direction at the same speed?

In short: No.

In long: In order to change your KE relative to something you must add energy of some other kind such that your energy in any given frame of refernce is unchanged.

kojikun · Post by **kojikun** » 2003-02-16 08:43pm

and this brings up another interesting question:

when two objects collide, is the impact energy as large as the two amounts of KE relative to a common center (where they impact), or is it relative to one of the two bodies?

if its the first, and lets assume both bodies are of equal speed (say 10m/s) and mass (1g), then its 2 * [(1 * 10^2) / 2] = 100J, but if its relative to either given body its (1 * 20^2)/2 = 200J. So which is it? Each has a (theoretical) KE of 50J, but from either body's perspective the other has quadruple the amount of KE!

Ren · Post by **Ren** » 2003-02-16 10:38pm

kojikun wrote:and this brings up another interesting question:

when two objects collide, is the impact energy as large as the two amounts of KE relative to a common center (where they impact), or is it relative to one of the two bodies?

if its the first, and lets assume both bodies are of equal speed (say 10m/s) and mass (1g), then its 2 * [(1 * 10^2) / 2] = 100J, but if its relative to either given body its (1 * 20^2)/2 = 200J. So which is it? Each has a (theoretical) KE of 50J, but from either body's perspective the other has quadruple the amount of KE!

You deaccelerated your frame of refrence in the second example. It went from 10 m/s to 0 m/s if you keep it constant then both objects are now moving at -10 m/s so Ke is .5*20^2 - .5*2*10^2 = 100J.

Durandal · Post by **Durandal** » 2003-02-17 12:15am

If both the colliding particles are traveling at constant velocity (not accelerating, net force is zero), their reference frames are equivalent, that is, they are both inertial reference frames. In classical mechanics, all inertial reference frames are equivalent, so a frame traveling at a constant velocity of 10 m/s is the same as a stationary frame.

In order to measure the scenario you describe accurately, each observer in each frame must apply coordinate transformations to his system. So, say we have two objects traveling in parallel paths. The first object (O_1) is traveling at 10 m/s. The second object (O_2) is traveling at 5 m/s. Say an observer in O_2's frame wants to measure the velocity of O_1. O_2 must first define a coordinate system. We'll call the x-axis the one that both O_1 and O_2 are moving along, the y-axis as the vector perpendicular to the x-axis, and the z-axis as the orthagonal vector to the x- and y-axes. We'll assume that t is the same for both (which is wrong, but whatever). Therefore, the coordinate system (x,y,z,t) will be used for O_2, and (x',y',z',t') will be the corrected coordinate system.

Therefore, we have the following

x' = x + vt
y' = y
z' = z
t' = t,

where x is the velocity of O_2 relative to itself, v is the measured velocity of O_1 relative to O_2 and t is the time interval on which the velocity of O_1 was measured. Does this help?

Oh, and Ren, Mac Hall fan?

Patrick Degan · Post by **Patrick Degan** » 2003-02-17 12:23am

Energy is an absolute. Like the velocity of light.

Post by **Darth Wong** » 2003-02-17 12:29am

kojikun wrote:and this brings up another interesting question:

when two objects collide, is the impact energy as large as the two amounts of KE relative to a common center (where they impact), or is it relative to one of the two bodies?

Collision physics are dominated by momentum, not KE.

if its the first, and lets assume both bodies are of equal speed (say 10m/s) and mass (1g), then its 2 * [(1 * 10^2) / 2] = 100J, but if its relative to either given body its (1 * 20^2)/2 = 200J. So which is it? Each object has a (theoretical) KE of 50J, but from either body's perspective the other has quadruple the amount of KE!

OK, let's take scenario 1: observer situated so that both 1kg objects appear to be moving at 10 m/s towards each other. The net momentum of the two objects is zero, since their velocity vectors cancel out. The KE is 50 J per object, for a total of 100 J.

Now, let's take scenario 2: observer situated on one of the two objects, so that the other one appears to be approaching at 20 m/s. The net momentum of the two objects is -20 kg*m/s, so that when they collide, the average velocity of the resulting agglomeration should be 10 m/s backwards (from the observer's perspective). The first object has 0 KE while the second object has 200 J of KE.

On the surface, it seems as if we have doubled the KE of the situation by simply changing the position of our observer! However, this does NOT take into account the situation after the impact.

You see, in the first scenario, both objects lose 100% of their velocity in the impact, so 100% of their KE must be converted into other forms of energy during the collision. Delta-KE for the system is 100 J. But in the second scenario, the first object gains 10 m/s going backwards while the second object still retains half of its original velocity. So Delta-KE for the system is more complicated. Object 1 gains 50 J of KE while object 2 goes from 20 m/s to 10 m/s, so it loses 150 J of KE. Therefore, Delta-KE for scenario 2 is +50J-150J=-100J. Exactly the same as scenario 1.

Make sense?

kojikun · Post by **kojikun** » 2003-02-17 12:36am

I've said it before but it stands to be said again: I love you mike.

What about the alteration of POV to change speed/direction?

Ren · Post by **Ren** » 2003-02-17 12:39am

Darth Wong wrote: Make sense?

A lot more sense than my explanation, which is basically the same thing just more confused

Durandal · Post by **Durandal** » 2003-02-17 12:57am

kojikun wrote:I've said it before but it stands to be said again: I love you mike.

What about the alteration of POV to change speed/direction?

Then you're no longer in an inertial reference frame. Then you go into integral calculus.

Morat · Post by **Morat** » 2003-02-17 02:06am

What about the alteration of POV to change speed/direction?

That doesn't work.

Imagine a fat guy. The fat guy can't just say he's thin relative to the moon, and therefore become thin. In order to be thin he has to actually lose weight.

kojikun · Post by **kojikun** » 2003-02-17 05:21pm

yes but morat, we're not talking about absolutes. relative POV is just that: there is no absolute reference frame (except maybe light which may explain why there needs to be a speed limit like c).

Durandal · Post by **Durandal** » 2003-02-17 05:35pm

There's no absolute reference frame, but that doesn't mean that there aren't absolute quantities that are the same in every reference frame, like the speed of light and energy. You simply have to apply coordinate transformations to your own frame of reference in order to correct for them if you're not stationary.

kojikun · Post by **kojikun** » 2003-02-17 05:48pm

but energy ISNT absolute in every reference frame. If youre in the space shuttle, to you a pen floating along side you has nearly no kinetic energy, but from earths POV the thing has more energy then a car at 60mph. There is no absolute with energy, its all relative.

Mad · Post by **Mad** » 2003-02-18 03:32pm

kojikun wrote:What about the alteration of POV to change speed/direction?

Doesn't work. Both objects will be moving relative to each other, but their direction is not going to change unless an accelerative force is applied to one of them.

I mean, if your ship has 1 gigajoule of KE relative to mars, but none relative to earth, couldnt you change it so that its 1 gigajoule relative to earth but none relative to mars and in doing so instantly transit from 0 MPH to a larger speed?

No. If it has 0 KE relative to earth, it will remain having 0 KE until a force is applied to it.

also, wouldnt it be possible to instantly reverse direction but keep the same KE, because youre going in the opposite direction at the same speed?

No, because momentum will have to go from +x to -x, which would require acceleration to change.

Post by **Darth Wong** » 2003-02-18 03:38pm

I think that it can be helpful in these cases to use the term "work" instead of "energy". The term "energy" tends to make people think in terms of absolutes, whereas the term "work" naturally lends itself to an understanding of the relativistic nature of the situation.

It takes a certain amount of work to accelerate an object from velocity A to velocity B. You can arbitrarily say that any object has 0 KE by simply setting your inertial frame of reference to match that object, and this can confuse people, but the notion of work makes more intuitive sense to the average person: if you are to accelerate that object in any way, you must perform work on it. Leaving it alone does not require any work.

Durandal · Post by **Durandal** » 2003-02-18 04:05pm

kojikun wrote:but energy ISNT absolute in every reference frame. If youre in the space shuttle, to you a pen floating along side you has nearly no kinetic energy, but from earths POV the thing has more energy then a car at 60mph. There is no absolute with energy, its all relative.

Yes it is. That's one of the principles of the special theory of relativity. The speed of light is the same to every observer, regardless of the frame of reference. That means that energy is constant, as well. If there's no such thing as absolute energy, why are astronomers looking for the missing energy in the universe?

Furthermore, relative to the Earth, the pen has gravitational potential energy, not kinetic energy. And it has an utterly minuscule amount, at that. The equation PE = mgh only applies to objects on the surface of the Earth. Otherwise, potential energy is inversely proportional to the distance between two objects.

This is why I keep going on about coordinate transformations. Haven't you read anything I've written?

EDIT: Galilyean transformations will not work on relativistic scales. Energy is treated somewhat differently in special relativity, as it includes both the object's rest energy and kinetic energy. Also, special relativity unifies momentum and energy conservation, whereas Newtonian mechanics has them as two separate laws. So, things get a bit complex when you get to relativity with significant fractions of c. However, on Newtonian scales, you must apply coordinate transformations to get accurate results.

kojikun · Post by **kojikun** » 2003-02-18 06:32pm

Mike, what if the transition was instant? Im talking pure theory here, i know there is no instant, but for the sake of argument as long as there isnt a "change in energy"..

durandal, c != energy.

Durandal · Post by **Durandal** » 2003-02-18 06:38pm

kojikun wrote:Mike, what if the transition was instant? Im talking pure theory here, i know there is no instant, but for the sake of argument as long as there isnt a "change in energy"..

durandal, c != energy.

Yes, but the two are intrinsically linked. If energy was always depended upon your frame of reference, we couldn't very well conserve energy, could we?

EDIT: Oh, and by the way, the proper coordinate transformations for relativistic scales are the Lorentz transformations. I should've mentioned that earlier.

Durandal · Post by **Durandal** » 2003-02-18 06:59pm

All right, time for me to put my foot in my mouth.

Physics for Scientists and Engineers, Volume II, Chapter 39.7, p.1273 wrote:Finally, note that because the mass m of a particle is independent of its motion, m must have the same value in all reference frames. For this reason, m is often called the invariant mass. On the other hand, because the total energy and linear momentum of a particle both depend on velocity, these quantities depend on the reference frame in which they are measured.

And after looking through my notes again, special relativity does not require that mass or kinetic energy be conserved, but it does require that total energy (KE + mc^2). Invariant mass is given by

M = sqrt( [(E1 + E2)^2]/c^4 + (|p1 + p2|^2)/c^2 ).

You use this equation to calculate the invariant mass of a nucleon that has split into two pieces. Invariant energy is, of course, E = mc^2.

The conclusion I'd draw from this is that, an object going really fast according to observer 1 and really slow according to observer 2 will appear much less massive to observer 1.

Morat · Post by **Morat** » 2003-02-18 07:47pm

Mike, what if the transition was instant? Im talking pure theory here, i know there is no instant, but for the sake of argument as long as there isnt a "change in energy"...

The equations don't work if you switch reference frames. You can't just say, "I have 50 kJ of KE relative to A and 0 relative to B, so I'll just swap them and get 0 kJ relative to A and 50 kJ relative to B." Your change of energy is 50 kJ in both reference frames, thus you have to do 50 kJ of work to perform that acceleration.

Durandal · Post by **Durandal** » 2003-02-18 07:55pm

Here's another relevant example. According to the first law of thermodynamics, there's absolutely nothing wrong with my monitor spontaneously changing some of its heat energy into kinetic energy and flying around in my room. The reason you never see such a phenomenon is because the second law doesn't allow it.

kojikun · Post by **kojikun** » 2003-02-18 11:04pm

Morat, but theres not ACTUAL CHANGE. I'm talking instances. Theres no actual change in amount of energy, so no work need be done.

Durandal, how does that apply to this??

Mad · Post by **Mad** » 2003-02-18 11:49pm

kojikun wrote:Morat, but theres not ACTUAL CHANGE. I'm talking instances. Theres no actual change in amount of energy, so no work need be done.

Depends on the reference, doesn't it? It may go from 100 m/s (+x) to 100 m/s (-x) in one reference frame, but in another, it just went from 150 m/s (+x) to 50 m/s (+x). Guess what? Its KE just changed...

You might say that the math works out in the first reference frame, and that's all that matters. But the math only works for KE in that one reference frame. It doesn't work for velocity or momentum; those things have to be taken into account, as well. Further, since all reference frames are valid, it follows that if something doesn't work out in one reference frame, then it won't work in any other reference frames, either.

Durandal, how does that apply to this??

His point is that there are other rules of physics that apply besides just the KE equations.

Durandal · Post by **Durandal** » 2003-02-19 12:00am

kojikun wrote:Durandal, how does that apply to this??

Just because something doesn't violate conservation of energy doesn't mean that it is permissible by the Physics Gods.

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Relative motion and KE

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