# Relative velocity and energy

1. May 10, 2013

### DiracPool

Is the energy gained by a particle moving at a high velocity a product of acceleration or can it arise in an object moving at constant velocity? I mean....OK, thought experiment. I haved a 100 GeV spherical mass moving through space relative to the earth at 10 m/s. I now accelerate that mass to a velocity, v', relative to the earth whereby now the traveling mass has an energy of 200 GeV. I now stop accelerating the mass. I'm assuming now that the mass will continue on indefinitely traveling at v' with an energy of 200 GeV relative to the Earth. Is that correct?

I guess the larger question I'm asking is, is it only the acceleration of the mass that imparts energy to that object? Such that the actually velocity per se has nothing to do whatsoever with imparting energy to an object? In other words, accleration gives energy/mass to an object and deceleration removes energy/mass? While the actual velocity per se of the object has nothing to do with it?

2. May 10, 2013

### ash64449

i don't know whether i am fully right or wrong,but if other posters can see anything wrong,they can point out. Well,making mistakes is not mistakes!!!

By looking at the velocity we can understand how much energy it has.

And we provide energy with the help of force.

And we know that force causes acceleration.

So energy is provided by acceleration.

To achieve higher velocity,we need to put more force,so more energy.

Earlier kinetic energy was given by the equation $\frac{1}{2}mv^2$

Now because of the effects of relativity,it has been replaced by the equation:

$\frac{mc^2}{\sqrt {1-\frac{v^2}{c^2}}}$

3. May 10, 2013

### ash64449

if v=0, you can see that $E=mc^2$

So we find from the equation that mass in rest is same as energy and energy increases when something moves,so we can also say that mass increases when something moves(not sure!)
edit: from the above total energy calculation equation,you can find that when an object is made to move at slower velocity,you will get rest mass plus kinetic energy of the object. For slower velocity that kinetic energy part is same.for bigger velocity,kinetic energy will not be same as kinetic energy found out from special relativity. Infact,it is a bigger number. We see from the above conclusion that objects resistance to acceleration has increased(inertia). So mass is increased. But you should not confuse with the mass that calculates the total amount of matter. Total amout of matter is still same but inertia has increased.. This one is called relativistic mass.

Last edited: May 10, 2013
4. May 10, 2013

### ash64449

and i have also heard that there are two types of masses.
For example,light bends under the influece of gravity(general theory of relativity). Source of gravity is energy. Light is energy.so light bends. Since energy is mass, light has this type of mass.. But don't consider the mass that you think.light has no mass,the way you think,because there are two types of masses...
One of them is relativistic mass and the other is rest mass..

5. May 10, 2013

### ash64449

6. May 10, 2013

### pervect

Staff Emeritus
I'm not sure exactly what you mean.

We can say that energy is a function of velocity - relativistically

E = m/sqrt(1-v^2/c^2). (m here is rest mass.)

Acceleration is the rate of change of velocity with time, if the acceleration is zero the energy is constant.

To impart velocity to an object and make it accelerate requires some force. This force does work in accelerating an object to a higher velocity / higher energy. Work = force * distance, and power = rate of work = force * velocity.

7. May 10, 2013

### Staff: Mentor

How does an object come to be moving at constant velocity if it doesn't accelerate to that velocity?
Yes.
You cannot slice apart two sides of the same coin.

8. May 10, 2013

### DiracPool

I'm not sure what you mean by that last statement?

Ok, here's a related question that may "root out" a larger picture here. Take the exact same scenario I used in the original post, only replace the earth as the "stationary" frame relative to the traveling sphere, with me in a rocket ship. The first phase of the thought experiment is identical to the above, the sphere is accelerated until its energy is 200 GeV, and then it starts coasting.

Ok, now in the second phase, I accelerate myself in my rocket ship in the opposite direction to the moving sphere until the moving sphere is traveling at a velocity, say v'', relative to me whereby its energy is now 400 GeV.

In this instance the sphere's velocity relative to me and its energy have increased again, but the sphere has not been given any energy. It has not been accelerated. So here is a case that an object that is moving at a constant velocity gains energy/mass. I know I painted myself into a corner on this one and the response is going to be "it's all relative", i.e., in principle the sphere was actually accelerated "relative to me" when I took off in the opposite direction.

That answer is unsettling, though. Perhaps someone can unsettle me? Or is my unsettledness unfounded?

9. May 10, 2013

### Staff: Mentor

No, this is exactly backwards. The speed is the only thing that matters, the energy is a function of speed only, not acceleration and not velocity (except insofar as they affect speed in a given frame). If an object accelerated in one frame then the object's speed and energy increased in that frame. But there exists another frame where the object decelerated and its speed and energy decreased, and there exists another frame where the object changed direction and its speed and energy remained constant.

10. May 10, 2013

### ash64449

just curious. Can you tell me an example?

11. May 10, 2013

### Staff: Mentor

Every example. Let's use low velocities so that relativistic effects can be ignored. Let an object start at rest in frame A and accelerate to a velocity of 10 m/s to the right. Let frame B be the frame moving 10 m/s to the right wrt frame A. Let frame C be the frame moving 5 m/s to the right wrt frame A.

In frame A the speed increases from 0 m/s to 10 m/s. In frame B the speed decreases from 10 m/s to 0 m/s. In frame C the speed starts at 5 m/s and ends at 5 m/s.

12. May 10, 2013

### ash64449

so isn't same amount of force applied in different frame and ended up with different velocity?

13. May 10, 2013

### Staff: Mentor

Yes, the same force, the same acceleration, the same change in velocity, but different changes in speed and therefore different changes in KE.

14. May 11, 2013

### BruceW

yeah, it is all relative. But this is general relativity you are talking about now. If you want to use the reference frame of the spaceship, where the spaceship is accelerating with respect to the earth, then these two reference frames are not 'inertial reference frames' with respect to each other. Therefore, you're going to get gravitational fields appear in one reference frame where there was none in the other reference frame.

15. May 11, 2013

### Bill_K

No, no, no! Special relativity alone is perfectly adequate to handle acceleration, and acceleration is not the same as gravity.

16. May 11, 2013

### BruceW

special relativity is not enough to handle using reference frames with relative acceleration. It is only able to handle inertial reference frames. And in DiracPool's post, he is talking about what happens according to the reference frame of an accelerating rocket. This is possible in GR, but the metric will be weird (which can be interpreted as a kind of 'gravity' where there was none before).

17. May 11, 2013

### Staff: Mentor

This is not correct. See: http://www.edu-observatory.org/physics-faq/Relativity/SR/acceleration.html

"It is a common misconception that Special Relativity cannot handle accelerating objects or accelerating reference frames. It is claimed that general relativity is required because special relativity only applies to inertial frames. This is not true."

What defines GR is the Einstein field equations, i.e. curved spacetime or tidal gravity. As long as your spacetime is flat then you can use SR, otherwise you need to use GR and the EFE.

18. May 11, 2013

### BruceW

I see. so if we have a situation where we have:
- reference frame [1]
- reference frame [2] (which is non-inertial with respect to reference frame [1] )
- the coordinates of an object according to reference frame [1]
- a mapping from coordinates in reference frame [1] to reference frame [2]

Then we can calculate the coordinates of the object with respect to reference frame [2]. Note, this doesn't require flat spacetime at all. (If I have interpreted it right). This is pretty interesting. I will keep it in mind in the future, thanks.

19. May 11, 2013

### Staff: Mentor

Yes, that is all correct with one very minor exception.