# Relativistic Kinetic Energy as a solution to a DE

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1. Apr 27, 2015

### AVentura

Has anyone ever seen KE derived this way? The only non-classical assumption is that mass = Energy/c^2 (I know, that's a big one) and that it accumulates as mass as the object is accelerated. (sorry for the plain text formatting, I need to learn how to do it right, I may reply with my first try)

KE(v)=∫ dp/dt dx = ∫ v dp
p = mv , m = (mo + KE(v)/c^2)
so: p = mo*v + KE(v)*v/c^2
and dp = mo*dv + dKE(v)*v/c^2 + KE(v)/c^2*dv

KE(v)=∫ [v*(mo*dv + dKE(v)*v/c^2 + KE(v)/c^2*dv)]

or KE'(v) = v*mo + KE'(v)*v^2/c^2 + KE(v)*v/c^2

The solution to this DE is k/√(1-v^2/c^2)-mo*c^2
If KE(0)=0 then k=mo*c^2
and KE= mo*c^2(gamma-1), which is correct.

I find this interesting because it does not involve using p=m*dx/dτ as momentum, and instead uses p=mv, with the mass growing. I never understood using a moving object's proper time to find momentum. The observer is the only one claiming motion after all.

2. Apr 27, 2015

### Staff: Mentor

No, it isn't. You are also assuming this:

That equation is not implied by the assumption that mass increases as the object is accelerated. It is an additional assumption--and one that is logically equivalent to the "conclusion" you are saying you have "derived". You haven't actually derived anything; you've just shown the implications of assuming that KE(v) is defined as the quoted equation above defines it.

3. Apr 27, 2015

### AVentura

Once you have relationship between energy and mass, is it not classical to say, "You have a mass, and I give you additional mass, therefore you have the sum of those masses"?

4. Apr 27, 2015

### Staff: Mentor

If by "mass" you mean "energy", then if you stick to a single inertial frame, yes, energy is additive. I'm not sure what that is supposed to show.

5. Apr 27, 2015

### DrStupid

As PeterDonis already mentioned, this is an additional assumption, but you do not need it. Your original assumption implies

$dE = v \cdot dp = m \cdot v \cdot dv + v^2 \cdot dm: = dm \cdot c^2$

This results in the differential equation

$\frac{{dm}}{m} = \frac{{v \cdot dv}}{{c^2 - v^2 }}$

and integration gives

$m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$

This equation already shows that your assumption is actually not classic. As it has real solutions for |v|<|c| only, c is a speed that cannot be reached or exceeded by objects with rest mass. Such a speed limit is not consistence with Galilean transformation but with Lorentz transformation.

To get the kinetic energy you just need to include this "mass" into your assumption:

$E_{kin} = E\left( v \right) - E\left( 0 \right) = m_0 \cdot c^2 \cdot \left( {\frac{1}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} - 1} \right)$

6. Apr 27, 2015

### AVentura

^ I like that! I realize that there are lots of ways to show p=gamma*m*v, or KE=mc^2(gamma-1). Just the knowledge (or assumption) that all observers measure the speed of light as c is enough. Thank you for the replies.

I just don't see why p=mdx/dτ is the right way to convey the relativistic form of p, as advanced texts do. It seems perfectly fine to me to conceptualize the phenomenon as an increase of mass.

Last edited: Apr 27, 2015
7. Apr 27, 2015

### Staff: Mentor

The reason this formulation is often used is that it is independent of coordinates, i.e., independent of a choice of inertial frame. That often makes the analysis easier; plus, it generalizes much more easily to curved spacetime. In flat spacetime, given a particular choice of inertial frame, however, your formulation gives the same answers, so it is a valid way of looking at it.

The reason this is often discouraged is that for many people it invites confusion. For example, a common question we get here on PF is whether an object becomes a black hole if it moves fast enough, since it mass keeps increasing. The formulation using proper time $\tau$ makes it much easier for many people to avoid such confusions.

8. Apr 27, 2015

### PAllen

There are a couple of good reasons:

In spacetime, coordinate spatial velocity is not a vector at all. If you ask for a 4-vector that is analogous to velocity you get 4-velocity, which inherently includes scaling of components by gamma.If you then go to an acceleration, you get 4-acceleration as the simplest possible vector.

Then, if you multiply 4-velocity by mass, you get momentum as an 4-vector (or covector, for use in a Lagrangian). You immediately recover the desirable feature that mass only changes due to absorption or emission. All the other 'features' of relativistic mass are 'bugs' from (falsely) pretending spatial velocity is a vector in SR.

9. Apr 27, 2015

### AVentura

Is it possible that these are all just conveniences that become available once we start including the work we do accelerating a mass, into the mass as we push? And that additional mass is Work/c^2?

Or I should say, nature requires we account for that work into mass, and then everything else follows?

Last edited: Apr 27, 2015
10. Apr 27, 2015

### PAllen

Where is the 'work' in 4-velocity?

Consider also that the most useful definition of total energy in SR (c=1) is:

E2 = m2 + p2

Here, m is invariant. Besides m, it is momentum that contributes. Momentum grows non-linearly with velocity because of the invariant limiting speed.

11. Apr 27, 2015

### AVentura

I just know that to accelerate an object something must apply a force over a distance to that object. The work done in the process adds an amount of mass (or apparent mass) equal to that work/c^2.

12. Apr 27, 2015

### PAllen

No, work is energy. The idea that energy = mass is due the false simplification of:

E2 = m2c4 + p2c2

into:

E = mc2

13. Apr 27, 2015

### AVentura

The KE in an object (rotational, oscillatory, heat) contribute to the mass of the object, do they not?

Perform work on a spring, you increase its mass (as in rest mass) too.

14. Apr 27, 2015

### PAllen

That's because p doesn't change for the body as a whole. The complete equation explains so much more than the incomplete equation.

Focus again on a point someone made earlier: why doesn't accelerating an object eventually lead to a BH? Because the invariant quantity mass (and the corresponding invariant - total curvature measured from infinity) does not change at all. Meanwhile, adding energy to a body from 'all sides' so its p remains constant, does eventually lead to a BH.

15. Apr 27, 2015

### AVentura

I get that. But i still wonder if increasing your momentum increases your overall mass (I know I'm not supposed to say that) and it's this that makes a velocity of c impossible for objects with non zero rest mass. Drstupid's use of the infinitesimal dm (however discouraged) is quite elegant to me.

16. Apr 27, 2015

### PAllen

This all depends on mv, with v spatial as the definition of momentum. I (and many others) claim this is not correct for SR. Momentum and velocity are vector quantities in Newtonian mechanics, and one expects them to be vector quantities in SR. However, spatial coordinate velocity is not a vector quantity in Minkowski spacetime. As soon as you replace it with the closest analog that is a vector, you see that the 'speed limit' is a feature of the geometry not of 'growing mass'.

17. Apr 27, 2015

### PAllen

To add to my prior observation, consider the algebra of velocities. Velocities add as:

(u+v)/(1+ uv/c2)

There is no mass, work, etc. here, just the algebraic structure of SR. This algebra is even true for u or v = c. Again, 'growing mass' has nothing to do with the invariant speed limit.

18. Apr 27, 2015

### AVentura

I can see that being correct.

But isn't that just why a traveler's claimed speed added to a frame I see moving does not exceed c to me? I think I can still claim the reason why he can't beat c is because he is getting so heavy during acceleration. And claim he is simply wrong about his speed due to his time dialation (and length contraction).

Last edited: Apr 27, 2015
19. Apr 27, 2015

### PAllen

And he can claim the same about you. Then you get into all the foolish false paradoxes of how can both ratios (his mass to mine, mine to his) approach infinity. I am not claiming there is no way to reconcile all this in some consistent framework involving relativistic mass; what I am suggesting is that all of this becomes irrelevant, even inexpressible, once you accept that nature is telling you the algebra and geometry of the universe are not what you thought from Newton.

Note that there are formal derivations showing that being unable to detect absolute (non-accelerated) motion, combined with isotropy and homogeneity, lead uniquely to either Minkowski algebra/geomertry or Euclidean 3-space X time fiber bundle. Nothing about mass, work or energy are needed. Then, the fact that we observe the algebra of SR velocity addition settles which applies to our universe, without a single question about mass, force, energy, or even light.

20. Apr 27, 2015

### AVentura

Thanks for information. I'm going to take a look at minkowski. I've always liked the Fitzgerald prospective (relativistic effects are real, be it length or weight), and it helped me predict that the CERN neutrino speed experiment was an error (they would have gotten a different result later in the day). I realize that there may be more fundamental reasons at play, that I simply don't grasp yet. But my viewpoint also helped me independently discover the bell spaceship paradox as well. I won't give up this view easily, but I realize I must try for the sake of my own knowledge. Thanks again, everyone.