# Acceleration to near light speed - where does the energy go?

1. Mar 7, 2013

### RagingPineapple

I've not got much of a background in physics, but this question has bugged me for ages. So I hope you don't mind me asking :)

I know that as an object under acceleration approaches the speed of light, it becomes increasingly more difficult to push it further. I imagine it like pushing against an elastic band: as you approach the limits of the band, each mm of additional stretch takes more effort to achieve than the last one.

My question is - where is the energy going? At human speeds, you can book-keep your energy loss - trace it back and find out how much is lost in friction, for instance.

In this case, if it requires more energy to accelerate the object as it approaches C, then where is this excess energy going? Is the energy being expended on creating the time dilation effect?

Or alternatively, is it all just a mirage created by time dilation? In other words, if we put fuel in a space ship, all our calculations on its use are relative to time (e.g. gallons burned per second). The ship will be enduring time dilation, so what we see as the ship appearing to labour as it approaches C is actually the ship applying less energy per second from our point of view?

My brain hurts. Please ease my pain XD

2. Mar 7, 2013

### A.T.

The object has more energy, than classically at the same speed.

3. Mar 7, 2013

### 1977ub

I think the answer you might be looking for is "momentum." The body which is accelerating has a greater relativistic energy/mass and gains momentum as it is pushed.

4. Mar 7, 2013

### Staff: Mentor

There are two answers (both equally valid, just different ways of interpreting the math):
1) It doesn't go anywhere; the kinetic energy of the object increases just as you'd expect. The relationship between speed and energy is such that the closer you get to c, the more energy it takes to change the speed by a given amount.
2) It goes into an increase on the mass of the object according to Einstein's famous E=mc2.

#1 is the more modern interpretation in which you just consider the total conserved energy of the object, and is far more convenient when you're working with subatomic particles moving close to light-speed. #2 is the explanation you're more likely to find in older texts. Both interpretations are consistent with the general relationship between speed, mass, momentum, and energy: $E^2=(m_0c^2)^2 + (pc)^2$ where $m_0$ is the mass of the object at rest and p is its momentum.

Edit... Should have said something about how the momentum is calculated, as otherwise #1 won't make any sense. The momentum isn't just the mv that you'd expect from classical physics, it's:
$$p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$$
For speeds that are small compared to the speed of light, this reduces to the classical momentum as you'd expect.

Last edited: Mar 7, 2013
5. Mar 7, 2013

### Staff: Mentor

And similarly, the energy is

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

which also increases without limit as v approaches c. So if you continue to do work on an object, at a constant rate, the object's energy continues to increase at a constant rate, but its speed can never exceed c.

6. Mar 18, 2013

### RagingPineapple

So could it be said that, since it requires increasingly large amounts of energy to accelerate an object towards C, it would also require increasingly large amounts of energy to decelerate an object that was travelling close to C?

7. Mar 18, 2013

### ghwellsjr

No, you've got it backwards. In the frame in which the object started out at rest and in which it took increasingly larger amounts of energy to accelerate in the same direction to a speed close to c, it continues to takes a large amount of energy for it to accelerate from that speed in any direction. If it reverses its direction of acceleration (decelerates, as you say) to come back to rest in its original frame, it starts out taking a large amount of energy and then takes decreasing amounts of energy as it comes to rest.

8. Mar 18, 2013

### Staff: Mentor

Almost, but you have the direction of the energy flow backwards. Decelerating an object means reducing its kinetic energy; that energy has to go somewhere as the object slows. So substitute "release" for "require" above and you'll have it.

You might want to take a moment with a calculator and the formulas above to calculate roughly how much kinetic energy a 1 kilogram object traveling at .99c would release if it were suddenly stopped, and then for the same object moving at .999c. To get some sense of what these numbers mean, consider that a large nuclear bomb releases 1016 Joules or so.

[Edit - I think I may have answered a different question than OP was asking. If he was asking about how we would smoothly decelerate the object, go with ghwellsjr's answer in #7]

9. Mar 18, 2013

### RagingPineapple

Yeah, I meant that, but I phrased it wrong ;)

So essentially, all the energy we 'loaded' the object with in accelerating it close to c will come back to bite whatever object gets in its way ;)

An object in a classical universe travelling at .9c would have required less energy to get there, so if it crashed into a wall, less energy would be released than a relativistic object travelling at the same speed. So Einstein gets a bigger bang?