# Kinetic energy at C/2

1. Mar 9, 2014

### bobie

If an electron is travelling at C/2, the Lorenz formula says that mass is about 1.15 me, do we get the value of KE simply multiplying 1/2mv2 by 1.15?
Is it as simple as that?

2. Mar 9, 2014

### dipole

The expression from classical mechanics, $T = \frac{m}{2}v^2$ is only valid for small velocities. The full relativistic expression for the kinetic energy is given by,

$T = (\gamma -1)mc^2$ where $m$ is the rest mass, and $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$.

3. Mar 9, 2014

### bobie

$\gamma = \frac{1}{\sqrt{1-\frac{(c/2)^2}{c^2}}}$
That is the formula I used and I got 1.1547 shall I multiply this or only .15 by .511 MeV?

4. Mar 9, 2014

You have calculated the value for gamma and dipole has given you the equation to calculate the KE. As dipole stated the equation half mv squared "is only valid for small velocities".
Rather than just plugging numbers into the relativistic equation to get the answer try to get a better understanding of the equation and when to use it. A search of the "hyperphysics" site might be a good place to start.

5. Mar 9, 2014

### bobie

Thanks, I suppose 1-γ, 0.15, should be multiplied by .511.

Last edited: Mar 9, 2014
6. Mar 10, 2014

### jartsa

Newtonian:

KE =1/2 * constant mass * v^2

Relativistic:

KE = 1/2 * changing mass * v^2

(Kinetic energy is the energy that we get when we stop an object. Stopping involves a change of mass)

7. Mar 10, 2014

### Staff: Mentor

No, unless you want to invent a new kind of velocity-dependent "relativistic mass" in addition to the one that everybody knows about but almost no physicists use.

8. Mar 10, 2014

### jartsa

Total energy of electron at velocity v = gamma * rest energy

So kinetic energy at velocity v must be .... (gamma * rest energy) - rest energy

9. Mar 10, 2014

### jartsa

Oh yes, that mass would be the "longitudinal mass" that is proportional to gamma^3, not the relativistic mass, also known as transverse mass, that is proportional to gamma.

(I thought it was the relativistic mass. This time I was so cautious that I checked that longitudinal mass seems to work correctly with the newtonian kinetic energy formula.)

10. Mar 10, 2014

### PAllen

NO, that doesn't work either. You would need yet another flavor of 'relativistic mass'. Why do you bother? Physicists just use relativistic formulas, not multiple weird mass definitions to make Newtonian formulas work.

(You would need M = 2(γ-1)mc^2/v^2 as your new 'kinetic relativistic mass'.)

11. Mar 10, 2014

### Staff: Mentor

So you're claiming that $\frac{1}{2}m_Lv^2 = (\gamma - 1)mc^2$ where m is the rest mass and $m_L = \gamma^3 m$? With a bit of algebra, this proposed equation becomes
$$\frac{1}{2}(\gamma^3 m) (\beta c)^2 = (\gamma - 1)mc^2\\ \frac{1}{2} \gamma^3 \beta^2 = \gamma - 1$$
where β = v/c.

Calculating the left and right sides for various values of β between 0 and 1 shows that they are not equal.

(PAllen beat me to it!)

12. Mar 10, 2014

### pervect

Staff Emeritus
A correct formulation of relativistic dynamics for point particles is:

F = dp/dt
p = $\gamma$ m v

where:
F is the force (3-force)
p is the momentum
t is time (coordinate time)
m is the invariant mass

Note that one must use the chain rule to find dp/dt. In the usual case, the invariant mass of the point particle isn't changing, and one would get something like:

$$F = \frac{d \gamma}{dt} m v + \gamma m \frac{dv}{dt} = \frac{d \gamma}{d|v|}\frac{d|v|}{dt} m v + \gamma m \frac{dv}{dt}$$

where v is the 3-velocity vector and |v| is the magnitude of the 3-velocity vector

The chain rule expansion is a bit ugly , if at all possible it's less error prone to stick with F = dp/dt

Last edited: Mar 10, 2014
13. Mar 10, 2014

### PAllen

Yes, for KE in Mev.

14. Mar 10, 2014

### jartsa

Let's consider following example:

A bike cyclists accelerates from 0 to 0.86 c, while taking good care of his energy balance. (Assistants fire food portions for the cyclist to catch)

Longitudinal mass of the cyclist becomes 8 times larger.

Then there's a short service break. From 0.86 c to 0.87 c the longitudinal mass stays constant.

The change of kinetic energy when velocity goes from 0.86 c to 0.87 c is:

Delta KE = 1/2 * longitudinal mass * (0.87 c)^2 - 1/2 * longitudinal mass * (0.86 c)^2

15. Mar 10, 2014

### PAllen

This has more mistakes than lines written. I am not going to bother correcting them all, because you just ignore valid responses and make up new nonsense. Please stop.

16. Mar 10, 2014

### PAllen

To bring Pervect's #12 to work and energy:

With F and P as Pervect defined them (relativistic 3 vectors rather than 4 vectors), some natural results follow:

∫F dot dx is change in energy, and will agree with change of E = mγc^2

17. Mar 10, 2014

### PAllen

irrelevant
irrelevant
impossible
flat out wrong

18. Mar 10, 2014

### jartsa

Excuse me, but it's correct. At speed 0.86 c velocity change is as large as Newton predicts, if we use 8 times more force than what Newton's says to be the correct force for that velocity change. Right?

So eight times more work will be done in this velocity change than what Newton says. E = F*d

19. Mar 11, 2014

### HomogenousCow

Can anyone decipher what he's trying to say

20. Mar 11, 2014

### PAllen

If γ is exactly 2, then, at the moment, 3-force is 8 times Newtonian. But integrating over any finite distance, however small, will not be the same 1/2 (γ^3)m v^2. Note, in particular, that:

∫Fdx $\propto$ v^2/2

requires that dp/dt $\propto$ dv/dt, which is false for relativity. You can't pretend gamma is constant for the integration.

So, I repeat, every part of your persistent argument that you can rescue 1/2 mv^2 using any standard form of relativistic mass is just wrong. See post #12, where Pervect kindly wrote the correct formulas.