# Relativistic kinetic energy

1. Aug 30, 2006

### Born2Perform

$$K_e = mc^2 -m_0c^2$$

is the kinetic energy of a hight speed particle (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html)

my question is:
according to newton a body to be accelerated requires energy, according to einstein the mass of that body increases with speed, so the body needs more energy, shiouldn't this value
(Ke=gamma1/2mv^2) be added to the $$K_e = mc^2 -m_0c^2$$, giving
$$K_e = mc^2 -m_0c^2 + \gamma \frac {1}{2}mv^2$$
if you understood something, thanks.

Last edited: Aug 30, 2006
2. Aug 30, 2006

### Staff: Mentor

I don't understand why you feel that another term needs to be added to the KE. Perhaps it will be more understandable if you rewrote the relativistic KE equation like this:
$$K_e = m_0 \gamma c^2 -m_0c^2 = (\gamma -1) m_0 c^2$$

3. Aug 30, 2006

### robphy

in the section "Relativistic Kinetic Energy", there is a little button "Show", which will show that your first expression (where $$m$$ is the relativistic mass $$\gamma m_0$$ and $$m_0$$ is the rest mass) is approximately (via a Taylor expansion) Newton's expression for kinetic energy when $$v$$ is very small compared to the speed of light. So, it doesn't need to be added.... it's already in there.

4. Aug 30, 2006

### Zeit

I have already seen that we could say :

$$K_e = \frac{{m_0}{v^2}}{{\gamma^{-2}}+{\gamma^{-1}}}$$

But, I don't have any idea how to prove it, even if it works. We see that when $v = 0$, $\gamma = 1$, so we obtain :

$$K_e = \frac{{m_0}{v^2}}{2}$$

Of course, $K_e = 0$ when $v=0$

Last edited: Aug 30, 2006
5. Aug 30, 2006

### Born2Perform

I think i have misunderstanded, or bad expressed:
an object at a certain speed, must have been "energized" due to his mass increasing, and proportionally to $$E = (\gamma -1)mc^2$$ ok perfect.

but, we also must count the energy spent to accelerate that mass, and, quoting robphy, how can it be included?
of what zeit said i understood nothing, but mine is
$$K_e tot = \gamma \frac {1}{2} mv^2 + (\gamma -1)mc^2$$

where am i wrong?

6. Aug 30, 2006

### rbj

The total energy (from the perspective of some observer)

$$E = m c^2$$

is equal to the sum of the rest energy

$$E_0 = m_0 c^2$$

and the kinetic energy

$$K_e = E - E_0$$ .

$m_0$ is the "rest mass" or "invariant mass" and $m$ is the relativistic mass as seen by the same observer:

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

if you plug all of that in and then take the limit as $|v| << c$, you will get the classical approximation for kinetic energy:

$$K_e \approx \frac{1}{2} m_0 v^2$$

7. Aug 30, 2006

### Staff: Mentor

It's already included in the definition of KE. The energy you spend to accelerate the mass is its KE.

8. Aug 30, 2006

### Born2Perform

Most of the energy is took by the mass increasing, and it is the formula you are showing. but KE is (for me) another thing; Is the energy given to accelerate that mass, which is not the energy spent to make the object mass increase.
And both must be added to have the total KE. i hope somebody understands me...

Last edited: Aug 30, 2006
9. Aug 30, 2006

### Staff: Mentor

I think I understand what you are saying, but what you are saying is not correct. There is no way to separate the energy used to "increase the mass" from the kinetic energy: it is the kinetic energy.

Part of the problem is thinking in terms of "relativistic" mass, which is velocity dependent. Instead, think in terms of rest/invariant mass. That's what you are accelerating and that acceleration is what requires energy to be added. That added energy is energy related to the motion of the particle, which is kinetic energy.

You may also want to read this: Does mass change with velocity?

10. Aug 30, 2006

### WhyIsItSo

Permit me an indulgence if you will. Consider a rocket with unlimited fuel.

In space, the propulsion comes from the inertia of the propellant's mass, yes?

Since from the rocket's perspective, the only velocity that means anything is by comparing its motion to some external reference, and so by definition is a relative velocity. For the rocket as a system, it is effectively a rest mass, or relativistic mass with velocity zero. For the rocket, its acceleration is constant, yet it always remains at rest relativistically...

From Earth, if we could observe this rocket indefinitely, as it increases speed relative to us, we would note that its mass increases, correct? Hence we determine that the rocket can never reach c because that would mean infinite mass hence infinite energy. But wait! The propellant the rocket is ejecting would appear to be increasing mass towards infinity also. This seems to suggest that most of the discussion about the energy required to accelerate a body to c means it's impossible is completely sidestepped. It seems to come from the perspective that the observer is providing the energy, not the observed object itself.

Question 1
I am guessing their is going to be a time dilation issue here; that to us (the observer), the rocket's acceleration will appear to decrease. The net effect being that our observation is that the rocket will take inifinite time to reach c.

Is that a valid treatment of the issue?

Question 2
Since we know intellectually that the rocket is in fact still accelerating, therefore is consistently increasing its speed relative to us, yet our perceptions tell us its rate of acceleration is decreasing (asymptoticly?). Time dilation is the only logical explanation. All other factors of mass and force are in fact constant. This seems to mean that time dilation is, in simple terms, the foundation of all relativity.

It seems to follow that our perception of increased mass is misleading. It is somehow the time dilation that is responsible for this perception. The mass hasn't in fact changed at all, has it?

Have I just grasped the fundamental concept of relativity, or did I miss something badly?

11. Aug 30, 2006

### pervect

Staff Emeritus
This objection, and related objections (see for instance) http://arxiv.org/abs/physics/0504110

are why I prefer to explain the impossiblity of reaching 'c' in terms of the velocity addition formula and avoid "relativistic mass" in the explanation altogether.

As the above reference points out, 'relativistic mass' does not explain why when a rocket is moving at nearly 'c', and you take off and move in the opposite direction at a high velocity that the your combined velocity is less than 'c'.

But the velocity addition formula explains this quite well.

i.e v = (v1 + v2) / (1 + v1*v2/c^2).

If you study the velocity addition formula, you will see that If v1<c and v2<c, then (v1+v2)/(1+v1*v2/c^2) < c.

The velocity addition formula not only explains this, it explains why a rocket will never reach the speed of light, no matter how long and how hard it accelerates.

For instance, lets suppose we have a rocket accelerating at 10 m/s^2, roughly one gravity.

Every second, it adds 10 m/s to its velocity. But this addition is done using the relativistic velocity addition formula, not by direct addition.

Because the sum of any two velocities less than 'c' is also less than c, we can see that the rocket will never reach the speed of light, in spite of the fact that it is constantly adding to its velocity.

The velocity of the rocket starts out less than c.

We add some velocity, dv, less than c using the relativistic velocity addition formula to the velocity of the rocket.

We know that the resulting velocity will be less than the "direct sum", and because the velocities being added together are less than 'c', the result will also be less than 'c'.

12. Aug 30, 2006

### WhyIsItSo

pervect,

Thanks for explanation. Thing is, I follow you and can accept the formula and just use it, or I can understand why that formula is true. Not in a strict mathematical sense, but the concept behind it.

"Give me a fish, I eat for a day. Teach me to fish, I eat for life".

13. Aug 30, 2006

### pervect

Staff Emeritus
The velocity addition formula can be derived from the Lorentz transform. One of the easiest ways to understand SR in my opinion is the Bondi K calculus, which people tend to either like or hate. (I have no which way you'll react,you might even be one of the rare people who winds up in the middle). Mathematically the K-calculus approach is very undemanding, one needs high school algebra, and to accept the basic assumption that there is a constant doppler shift that is a function only of relative velocity, and that the speed of light is constant.

"Relativity and common sense" by Bondi takes this approach, and is available (or was availabe) in a cheap dover edition. You can also find this approach online as well, for instance.

http://www.geocities.com/ResearchTriangle/System/8956/Bondi/intro.htm

I'm not sure how complete the online treatment is offhand.

Note that at some point you'll have to accept that reality is the way it is, rather than ask why it is that way. You can ask "why is the speed of light constant" or "why is the doppler shift constant" all you like, and you won't get a much better answer than "That's the way the universe works".

What the K calculus approach will do is show you how coordinates transform (the Lorentz transform) can be derived, giving a few basic assumptions.

14. Aug 30, 2006

### clj4

Hi,

I use a similar proof with a slightly different mathematical approach:

The speed formula shows the speed as viewed from frame S' if we know the speed in frame S.
Let the rocket have an intial speed v_0 wrt to the observer (frame S')
If we apply a boost dv wrt to the rocket frame (S), the speed as viewed by S' is:

v_1=(v_0+dv)/(1+v_0*dv/c^2)

If we apply another boost equal to dv wrt to the rocket (now frame S is moving at speed v_1 wrt S') the resultant speed is:

v_2=(v_1+dv)/(1+v_1*dv/c^2)

After n such boosts, the rocket speed becomes:

v_n=(v_n-1+dv)/(1+v_n-1*dv/c^2)

Let v=lim(v_n) . Taking the limit of the above expression we get:

v=(v+dv)/(1+v*dv/c^2)

Solving for v we get v=c.

In other words, v can only approach c after an INFINITE number of boosts dv (no matter what value dv). The approach towards the limit is from below since at each iteration v_n<c.

This is my preferred "proof". Like pervect's , it is purely kinematic and it sidesteps the messy "relativistic mass" notion. Thank you for your attention.

Last edited: Aug 30, 2006
15. Aug 31, 2006

### WhyIsItSo

pervect,

Thanks once again. I have the page and will (attempt) to digest it tomorrow.

In the mean time, what of my thought that all of these phenomena really come down to time distortions?

I mean, it would appear that mass is constant; Ah..., for a given body, any velocity-derived change in mass (which is only seen by some other observer) is not in fact occuring. It seems to be a given that c is a constant, and only red/blue-shifting occurs.

It looks to me as though the only workable answer to these apparent paradoxes lies in very strange things happening to time. That seems to be the only true "variable" in these equations I've seen. Equations showing relationships for changes in mass, for example, do have units of time involved.

Mathematically, I understand the value of the flexibility of seeing equations in terms of, for example, mass. But is it a valid thought that TIME is the only thing that really changes?

16. Sep 2, 2006

### Born2Perform

17. Sep 2, 2006

### masudr

(Proper) time just happens to be a useful variable for parametrizing worldlines of massive particles/systems. So most quantities corresponding to systems can be given as functions of proper time.

It is not correct to say that time is the only thing that really changes: it is merely a useful variable. Everything changes with respect to time (or some other suitable parameter); it's not just time that changes.

18. Sep 2, 2006

### Born2Perform

Can you give me a proof that those two things (the energy spent to accelerate mass, and the energy spent to make the mass increase) are not two separate things?
I read from all that $$K_e = mc^2 - m_0c^2$$ includes alredy
$$K_e = \gamma\frac {1}{2}mv^2$$, but nobody gave me this proof.
a body needs to be accelerated anyway, the energy spent to increase its mass should be another thing and not included.

Last edited: Sep 2, 2006
19. Sep 2, 2006

### pervect

Staff Emeritus
Energy is conserved. The total energy of a system E is given by the formula you mentioned, E_tot = gamma * m

See for instance the translated original of Einstein's 1905 paper
http://www.fourmilab.ch/etexts/einstein/specrel/www/

or any relativity textbook. There are a few online, as I recall. Do some reading, don't just make stuff up, please.

Given that energy is conserved, if a system originally has an energy E1, and winds up with an energy E2, the amount of energy needed to change its state is just E2-E1.

20. Sep 2, 2006

### Andrew Mason

That is because is it not correct.

$$E_K = (\gamma - 1) m_0c^2$$ or

$$E_K = \sqrt{p^2c^2 + m_0^2c^4} - m_0c^2$$

since:

$$E^2 = p^2c^2 + m_0^2c^4$$

and rest energy is:

$$E_0 = m_0c^2$$

$$E^2 - E_0^2 = p^2c^2$$

I would suggest that you read carefully what has been posted here, particularly Pervect and Doc Al. The increase in energy does not increase the real mass of the moving body. It just appears to do this to the observer. What we observe is an increase in its momentum.

A force F being applied to the body for a time t increases the body's momentum by Ft. If the body is already travelling at close to c, the increase in speed, v, is very small so it appears to have an incease the mass. This increase in momentum increases the kinetic energy according to the above energy equation:

$$E^2 - E_0^2 = p^2c^2$$

AM

Last edited: Sep 2, 2006