# Thread: Relativity and Kinetic Energy

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## Relativity and Kinetic Energy

Alright. According to relativity, there is no such thing as absolute velocity, only relative velocities. So, we should be able to say the same thing about kinetic energies: Ek=(1/2)mv^2, so since velocity is relative, energy must be as well. But that pesky ^2 presents some difficulties.

Imagine a 1-dimensional universe with 3 objects named m1 m2 and m3, each having a mass of 2kg. Their positions are irrelevant, we're just paying attention to their relative velocities and energies.

So, let's look at a few different possible states for these objects.
For notation, Sn refers to state (S1 is state 1, etc) vab refers to the velocity of ma relative to mb (v12 is velocity of 1 relative to 2) and Eab is the same as vAB except for energy. I won't be mentioning 21, 31 or 32, as those will simply be the negative of 12, 13 and 23 respectively

Starting out, let's say all three masses are stationary with respect to eachother. This means that each one has a velocity of 0m/s relative to both of the other two and a kinetic energy of 0J relative to both of the other two. Remember, Ek=(1/2)mv^2

S1:
v12 = v13 = v23 = 0m/s
E12 = E13 = E23 = 0J
Easy, right?

Next, let's change the velocity of m1 to 2m/s

S2:
v12 = v13 = 2m/s
v23 = 0m/s
E12 = E13 = 4J
E23 = 0J
So far so good.

Next, let's change the velocity of m2 to 1m/s

S3:
v12 = 1m/s
v13 = 2m/s
v23 = 1m/s
E12 = 1J
E13 = 4J
E23 = 1J

Now this is where it breaks down. The Energy of 3 relative to 1 should be equal to the sum of the energies of 3 relative to 2 and 2 relative to 1, but it isn't: 1J + 1J != 4J.

Another way to look at it would be to imagine you're in a car which has a total mass of 1000kg with you in it. It starts out stationary relative to the earth. How much energy then does it take to accelerate to 10m/s relative to the earth? If we take the earth as a reference point, then it accelerates from 0m/s to 10m/s, changing its Ek from 0J to 50000J: it takes 50KJ to accelerate the car to 10m/s. What if we take the sun as a reference point, however? The earth itself is travelling at about 30km/s relative to the sun. If you're on the equator and it's the right time of day, you can add almost 0.5km/s to that because of the rotation of the earth, but I'll leave it at 30km/s because that's an easy number to work with. So now, if we're travelling in the same direction as the earth, we're accelerating from 30000m/s to 30010m/s. That's going from 45000000000000J (45TJ) to 450300050000J (45.030005TJ) which is an increase of 30050000J (300MJ.) That is a hell of a lot more than the original 50kJ we calculated, and just a lot of energy period to accelerate a car by 10m/s. But we're not done yet. What happens if we look at the center of the galaxy as our reference point? Our sun travels around the galaxy at a speed of about 235km/s relative to the center. If the earth were at the point in its orbit where it travels in the same direction as the sun, and the car is accelerating in the same direction, that means we're going from 265000m/s to 265010m/s, which translates into an increase in energy of 2650050000J (2.65GJ.) So, how much energy does it take to accelerate the car? 50KJ, 300MJ, 2.65GJ, or something completely different? And why is it that particular one?

The problem intensifies when we consider conservation of energy problems: Any object which is in free fall toward a planet must have a sum of kinetic and gravitational potential energy which remains constant, and for all of these problems, we consider the planet to be stationary even though it clearly is not, and in fact there is no such thing as "stationary."

So, what's going on here? How do we work around this, or fix it?

2. The problem is that you have the wrong understanding of what mass is. You're assuming that kinetic energy is based solely on the rest mass of an object. As far as Newtonian mechanics goes this is correct. However, in SR you need to correct this. As you approach c the mass of an object increases and the proper term to use in your equations is

m = m0 / sqrt (1-v^2/c^2)

Where m0 is the classical rest mass. Note that this tends to infinity as v approaches c.

In this defintion, the total energy of a body is E = mc^2 where m is defined as above. As v goes to zero this becomes the rest energy defined by the object's rest mass.

So the proper statement for the momentum of an object is

P = m * v = m0 * v /sqrt(1-v^2/c^2) (understand this is a vector quantity.)

A little substitution and you get an equivalent expression for the total energy in terms of the momentum and the rest energy.

E^2 = p^2 * c ^2 + m0^2 * c^4

Any good intro text on special relativity should explain all of this in more detail and hopefully clear up any misunderstanding you, or other readers might have.

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Well, in none of those cases does the mass change significantly enough to really affect the kinetic energy by that much, and in fact, it only makes it (slightly) worse. Your new equation for kinetic energy is just Ek=m0v^2/(2Sqrt(1-v^2/c^2)). You can plug the numbers into that, and you'll get similar results. I'm not talking about the total energy here, just the change in kinetic energy: the amount of energy required to change an object's velocity.

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Kinetic energy itself isn't invariant between frames of reference (neither, for that matter, is the velocity) so there's no reason to expect the terms to add up.

You have to use four-vectors.

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Why shouldn't they? If I have an object falling toward the earth from rest, it will gain a certain amount of potential energy (mgΔh) which will be the same no matter which frame of reference you use. That change in potential energy should be equal to the change in kinetic energy, but if the amount of energy it takes to change your velocity by a certain amount changes depending on your frame of reference, then how can you possibly calculate what your final velocity will be? I'm aware that in these problems, we always use the earth (or whatever the object is falling toward) as our frame of reference, but shouldn't it work no matter what frame of reference we use?
You have the same problem with a rocket ship, for example. If you have a certain amount of usable energy in the form of fuel, what frame of reference do you use to determine what speed that will get you to? It will be different whether you use the earth, the sun, the center of the galaxy, or something else. And these are at such low speeds that the relativistic effects on time and mass are essentially negligible.

6. I don't quite understand your problem. At the start of your post you quite correctly point out that energy is not absolute - it's relative to whatever frame of reference we happen to be using. Then you complain because your sums prove this!

As swansnt says, there is no reason why the figures from different frames should tally.

Take a simple example: if a spaceship is travelling very close to the speed of light relative to the Earth, it will require an enormous amount of energy (relative to the Earth) to accelerate it by just 1 m/s. But if you are in another spaceship travelling alongside the first one at the same speed, it will only take a small amount of energy to accelerate the other ship by 1 m/s. You and the Earth are measuring the same energy and coming up with very different results. That's relativity.

You have the same problem with a rocket ship, for example. If you have a certain amount of usable energy in the form of fuel, what frame of reference do you use to determine what speed that will get you to? It will be different whether you use the earth, the sun, the center of the galaxy, or something else. And these are at such low speeds that the relativistic effects on time and mass are essentially negligible.
You've answered your own question. You can use any frame of reference you like, but the answer will not necessarily be the same each time. Why should it be?

By the way, I'm not at all sure that the relativistic effects Eta C speaks of are irrelevant. In your earlier example, you found that the increase in kinetic energy relative to the Sun was 250 MJ greater than that relative to the Earth. By E=mc˛, that corresponds to a relativistic increase in mass of about 0.0000000028 kg. A speed of about 700 m/s would be enough to add that to the mass of a 1000 kg car!

7. Originally Posted by uniqueuponhim
Well, in none of those cases does the mass change significantly enough to really affect the kinetic energy by that much, and in fact, it only makes it (slightly) worse. Your new equation for kinetic energy is just Ek=m0v^2/(2Sqrt(1-v^2/c^2)). You can plug the numbers into that, and you'll get similar results. I'm not talking about the total energy here, just the change in kinetic energy: the amount of energy required to change an object's velocity.
But that quantitiy is not kinetic energy. It's the total energy of the object. In relativity kinetic energy, as defined in Newtonian mechanics, is not a conserved quantity. Total energy, including the mass increase, is. As you approach c any additional work (in the physics sense) done on an object increases its mass, not its velocity. This is not the place to teach you physics 108 (to use the old UI syllabus). Do some reading and clear up your misconceptions.

Actually, on reflection, I've mis-stated a little. Total energy is conserved within any single reference frame. However, under a Lorentz transormation, energy and momentum will change. The invariant of the energy-momentum four vector is the rest mass of an object. That remains the same in all frames.

And on reading the rest of the responses, Swansont got to the heart of the matter better than I. Even under Newtonian mechanics kinetic energy is not invariant under Galilean transformations.

8. You might want to look here, at the formula for a Lorentz transformation of the energy-momentum four vector of an object when measured from a different reference frame.

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Ok... let me try this one more time:

2 identical spaceships take off from the earth. Starting out, they are both moving very slowly: the first one moves at 1m/s and the second one moves at 2m/s. From the point of view of each of the spaceships, they are moving at 1m/s relative to each-other, and from the point of view of the earth, they are moving at 1m/s relative to each-other. Next, they speed up to 10m/s and 11m/s, respectively. Again, from the point of view of the spaceships, they are moving 1m/s relative to each-other, and relative to the earth, they are moving 1m/s relative to each-other. The ships could continue to accelerate, and until they start approaching the speed of light, their speed relative to each-other is the same from either frame of reference, and would continue to be the same if we used the sun or the center of the galaxy as a frame of reference. No problems so far: relativity seems to work fine with velocities. But when you plug in energies into those numbers, you get something different for every single one of those situations. My point isn't that the kinetic energy relative to the reference point is changing, I'm fine with that: It is that the kinetic energies of the two ships relative to EACH-OTHER changes every time you change your reference point, and by massive amounts, and without even coming close to C.
Let's go back to the car I talked about in the first post, the one with a mass of 1000kg. Relative to the earth, it takes 50kJ to accelerate that car from 0m/s to 10m/s. So, let's say we burn 50kJ worth of fuel and accelerate it to 10m/s. But wait: what if we look at it with the sun as a reference point. Now, the car is starting out at 30000m/s and expends 50kJ of energy to accelerate. This is enough to change it's velocity to... 30000.00167m/s relative to the sun, or 1.67mm/s relative to the earth. If we add in the relativistic effects of travelling at 30000m/s, that just makes the car even more massive, meaning it goes even less than 1.67mm/s relative to the earth, and if you want to talk about total energy rather than kinetic, then again, it would be travelling even less than 1.67mm/s relative to the earth from the sun's frame of reference. 50kJ in one frame of reference accelerates the car 6000 times more than it does in another, very similar frame of reference which wasn't even close to approaching the speed of light relative to the first. Unlike kinetic energy, the amount of energy I can get out of burning a certain amount of fuel is absolute. That means that if a certain amount of fuel contains 50kJ of usable energy, it will contain 50kJ of usable energy no matter what your frame of reference is. So if I burn 50kJ worth of fuel, do I accelerate to 10m/s or do I accelerate to 1.67mm/s? It cannot be both; an object travelling at 10m/s relative to the earth will be travelling at 10m/s relative to the earth whether you use the earth as your reference frame or the sun. And whatever velocity I do accelerate to, why is it that particular one? What makes that frame of reference so special?

10. I think that your problem is that you're imagining a ship accelerating by a pure transformation of chemical energy to kinetic energy, but of course we don't know how to do that. Instead, we use rockets, that send part of the mass backward so that the rest goes forward. Since you're talking about relatively slow speeds, the whole thing is actually a question of Galilean relativity, just using Newton's laws. I did this all in Excel, and I'd be happy to send you the spreadsheet if anyone wants to see the details.

So, we're going to start by having a lab frame, which we'll consider at rest, and two small test ships, each of which has a mass of 1000 kg (900 kg for the ship itself, and 100 kg of fuel that's burned). Ship A has an initial velocity of 5 m/s, relative to the lab frame, and ship B has an initial velocity of 10 m/s. We'll also define frames A and B as moving along with each ship at its initial velocity. Both of them then execute a burn, accelerating by 5 m/s (note that this means that ship A is no longer in frame A, which hasn't changed; we're not going to try to deal with accelerating reference frames here). Now, how do they do that? They can't just convert fuel to kinetic energy. Instead, they'll send the fuel flying backward at some velocity. By conservation of momentum, we can see that each ship will have to send its 100 kg of fuel flying backward at 45 m/s relative to itself in order to make the remaining 900 kg of ship move 5 m/s faster. Since momentum is linear in velocity, it's hopefully obvious to you that we'd get the same results in any frame, but you're free to check the numbers. So, how does the energy look for these two ships in all three frames?

In the lab frame, ship A together with its fuel are moving at 5 m/s, for a total of 12.5 kJ of energy. After the burn, the ship is moving at 10 m/s, giving it 45 kJ, while the fuel is moving 40 m/s backward (45 m/s relative to the ship, of course) for 80 kJ, or a grand total of 125 kJ. Obviously, this was an inelastic collision (in reverse) and we had to put in 112.5 kJ of energy, which came from burning the fuel.

Ship B (and its fuel), on the other hand, started at a velocity of 10 m/s, giving it an initial kinetic energy of 50 kJ. After the burn, the ship is moving at 15 m/s, so it now has a kinetic energy of 101.25 kJ, while it's 100 kg of fuel is moving backward at 35 m/s (again, this is 45 m/s relative to the ship), giving it 61.25 kJ, for a total of 162.5 kJ of energy. Again, we see that the "collision" was inelastic, and we had to add 112.5 kJ of energy. Hey, look, it's exactly the same amount!

Let's try this same thing in frame A. In this frame, ship A starts with a velocity of 0 m/s and ends with a velocity of 5 m/s. So, it starts with no kinetic energy. After accelerating, the ship has 11.25 kJ, while its fuel, moving backward at 45 m/s has 101.25 kJ, so we've had to add (you guessed it) 112.5 kJ of energy. Ship B measured from frame A would be exactly the same as ship A measured from the lab frame, so I'll skip that. I'll also leave the measurements from frame B (where ship A starts out moving backward and then comes to rest) as an exercise for the reader.

If velocities were really high, the math would all get more complex, since we'd have to account for special relativistic effects, but the results would be the same. Similarly, I've simplified things by assuming that all the fuel is sent out the back at the same final velocity, which obviously wouldn't be the case, but the overall result wouldn't change.

Does this help things make sense? Note, by the way, that if there were such thing as an inertial drive, that could just magically change the kinetic energy without throwing stuff out the back to maintain conservation of momentum, we'd run into exactly the problem that you suggest. That's one reason to suspect that such a drive is not possible, sadly.

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Thank you for answering the question, that makes a lot more sense now. It just seemed kind of counter-intuitive to me that the relative velocities of two objects to each-other would remain constant no matter what the refernce frame was but that the relative kinetic energies wouldn't.
Also, I wasn't suggesting some magical inertial drive that converted chemical energy directly into kinetic energy, I just figured people would know what was going on and I wouldn't have to explain it: A reaction happens, releasing a certain amount of energy, which causes the gases present to heat up, thereby increasing their kinetic energy, which they would then transfer to the ship by being blown out the back. My problem was that you can only release a certain amount of energy in a given reaction, and after taking into account inefficiencies, etc, only a certain amount of that energy is "usable," that's why I was saying things like "a certain amount of fuel contains 50kJ of usable energy." But you've explained it very well, thank you.

12. Originally Posted by uniqueuponhim
Also, I wasn't suggesting some magical inertial drive that converted chemical energy directly into kinetic energy...
Just to clarify, I didn't think you really were doing that intentionally, but it happens that the solution to the problem is mostly remembering to pay attention to the kinetic energy of the reaction mass as well, as you can see above, so you were sort of doing that implicitly. Plus, that gave an opportunity to mention one of the reasons physicists think that such a drive is unlikely.

Originally Posted by uniqueuponhim
But you've explained it very well, thank you.
Very glad I could be of help.

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Would it be possible for you to give an example of a ship accelerating to a speed at which there would be relativistic effects on time and mass? For example, if you had a 10000kg ship with 90000kg of fuel and you wanted to accelerate to 0.9C relative to the earth - both from the earth reference frame and from the 0.9C reference frame? I'd like to see how conservation of momentum works at those speeds, and how the added mass affects the calculations.

14. Originally Posted by uniqueuponhim
Would it be possible for you to give an example of a ship accelerating to a speed at which there would be relativistic effects on time and mass? For example, if you had a 10000kg ship with 90000kg of fuel and you wanted to accelerate to 0.9C relative to the earth - both from the earth reference frame and from the 0.9C reference frame? I'd like to see how conservation of momentum works at those speeds, and how the added mass affects the calculations.
Arg! Yes, I can do that, but it isn't going to be pretty. I think to get something reasonable, I'll have to abandon my treatment of the fuel all being expended instantaneously and sent backward at a single velocity, although I think I'll still probably do it in several discrete bits rather than add integration into the mess. It will take a little time to work through, though, so don't look for it before the weekend.

Oh, and as a nitpick, the speed of light is always represented by a little "c", never a big "C". Physicists are always running out of letters to use for things (that's why they've had to branch out into Greek), so they always distinguish case.

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## kinetic energy IS invariant- you've missed the point

Hi,

This is a conundrum I considered when I was at school about thiry years ago, so it's good to find it on the internet now, and not surprising that people are still missing the real answer.

The intertial drive explanation is wrong, but close. A car or ship doesn't conserve moment, when accelerating, by jettisoning fuel with an equivalent momentum. It's no different to walking. When we start to walk our momentum increases, even though we don't jettison any fuel at all. Momentum is conserved becase the act of walking causes a miniscule recoil of the earth.

The best example I can give to illustrate this is as follows. Imagine you conduct an experiment in which stand on the top of a ramp and release a trolley to roll down the ramp. You can calculate the speed of the trolley at the bottom of the ramp by refereing to conservation of energy- the potential energy (mgh) lost by dropping down the height of the ramp has to equal the kinetic energy gained by the trolly (0.5mvsquared). To make the maths easy, lets suppose g is 10, the height of the ramp is .45metres and the trolley weighs a kilogram. The speed is the square root of 2gh. In our example 2gh is 9 so the speed is 3 metres per second.

Now imagine this is witnessed by someone in a plane travelling at 1000 metres per second in the direction of the upward slop of the ramp. They initially see the trolley moving at 1000 metres per second. When the trolley gets to the bottom of the ramp they see it moving at 1003 metres per second. They calculate the gain in kinetic energy as 0.5m *(1003 squared -1000squared) Which is 0.5m*9 plus 0.5m*6000, whereas the observer on the ramp only sees a gain of 0.5m*9- a huge discrepancy. Notice the discrepancy always has the value muV, where m is the mass of the trolly, u is its final speed as see from the ramp and V is the speed of the observer in the plane. The discrepancy arises because we have missed out the KE of the ramp, which also changes when viewed from the plane. Suppose the ramp and observer weighed, say, 200kg. Then to conserve momentum, the recoil speed of the ramp (assuming the ramp is not fixed to the ground) needs to be 1/200th of the speed of the 1kg trolley- say 0.015 meters per second. This recoil seems to be zero from the view of the observer on the ramp, who is recoiling too. But for the person in the plane, they see the speed of the ramp reducing from 1000 metres per second to 999.985 metres per second, and when you calculate the corresponding loss of KE lo and behold it exactly balances out the seeming discrepany of extr energ gained by the trolley. You can see that this works whatever the mass of the ramp, since its additional change in KE is also MUV, where M is the mass of the ramp and U is its recoil speed, and we know that to concerve momentum MU must equal mu. This even works then if the trolley is fixed to the earth, in which case the miniscule recoil speed of the whole earth still equates to the right amount of KE.

Hope this helps

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