# Can continuous acceleration reach speed of light?

1. Nov 20, 2011

### vipinpsharma

Hi,

May be a dumb question; imagine a hypothetical situation of a spaceship in space with no influence of gravity due to earth or nearby moon. Assuming the spaceship has enough fuel, if it injects the fuels outwards, it will accelerate in the opposite direction. Now, the new velocity will be maintained (as there is no resistance to slow it down). Now more fuel (or fuel exhaust if you prefer) is ejected and the ship is accelerate again giving a new increased velocity.

If the ship continues to do this, can it reach speed of light? I know I am missing something here, but given my limited knowledge in this area, I don't know what is wrong in my thinking.

Thanks,
VS

2. Nov 20, 2011

### Pengwuino

No. Each ejection of mass/fuel out the back end of the ship will cause a smaller and smaller increase in velocity. This happens in such a way that you are unable to ever reach the speed of light. You can add energy to something in many different ways, none of which will allow it to reach the speed of light.

3. Nov 20, 2011

### aesir

The simple way to see this is to look at the kinetic energy of the ship, it is $K=(\gamma-1)mc^2$. To reach the speed of light (i.e., $\gamma=\infty$) you would need to inject infinite energy (unless the mass is 0, like for a photon).

4. Nov 20, 2011

### vipinpsharma

Thanks for the reply. This is fascinating as it gives so many more questions:

1) What causes the increase in the velocity smaller and smaller? Does increase in velocity dependent on the instantaneous velocity? I thought once you move the body into motion it goes into a new reference frame and any force/acceleration will measured from this reference frame. Can somebody shed some more light on this?

2) From old school math, I thought the Kinetic Energy is given by 0.5*m*v^2 where 'v' is instantaneous velocity of the ship (not speed of light). And where is this γ coming from in the reply below? Does this basic equation change when we approach relativistic speeds? If it does, why?

3) On another note, when we measure speed of any object object, what is the reference we use? For example, a bullet on the ground has some speed, but with reference to, say our Sun, it has a different speed due to earth's rotation and revolution around the Sun? What is it's speed? May be with reference to some other object in our universe, the bullet has already approached the speed of light?

4) We know we live in a space-time universe, i.e., time is another dimension to three dimensional space. A car moving from one point A to point B with 60mph is also moving through time, albeit very little. However, if a car approaches the speed of light, it is moving less through space but more in time. Applying the same concept to photon of light, does it only move through time and not space? Anything moving at speed of light should progress in the time dimension only and none in space. This should also apply to photons of light?

Thanks,
VS

5. Nov 21, 2011

### aesir

The expression I wrote is the correct relativistic form of kinetic energy, since your speed is comparable to the speed of light (in the reference frame where the ship is at rest at t=0) you can not understand motion with newtonian laws.
$\gamma$ is the Lorentz factor $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
Note that for $v<<c$ the kinetic energy approaches $\frac{1}{2}mv^2$, so for small speeds the newtonian description is correct.

To truly understand all of this you will need to study special relativity.
For an intuitive explanation consider that in SR you can not add speeds linearly, the sum is always lower than the classic counterpart, so as to remain lower than c. speaking sloppy you could say that if you continue adding a constant speed expelling fuel you would gain lower and lower actual speed.

6. Nov 21, 2011

### vipinpsharma

I think I really need to go through Special Relativity. Anybody knows a good introductory reference text?

Thanks,
VS

7. Nov 22, 2011

### harrylin

8. Mar 24, 2013

### Naty1

nope:

easiest way to answer this is to note that everyone, that is all local observers, see light whiz past them at 'c' no matter their speed.....so how could you EVER catch up??

9. Mar 24, 2013

### Staff: Mentor

10. Mar 24, 2013

### arianabedi

I think the text quoted above is as simply as Eisntein would put it for you him self, minus the German accent

11. Mar 25, 2013

### Naty1

There are only two ways to answer this:

from fundamental first principles, nobody knows why.....

OR

because that's what our mathematical models show...as an asymptotic effect.

[v = at is a non relativistic approximation. ]

12. Mar 25, 2013

### Darwin123

Here is a special relativity (SR) answer. I will give an answer which makes sense in special relativity, but maybe not in general relativity (GR).

First, a definition. I like to use the phrase "dynamic acceleration" to signify a force divided by a mass. This differs from kinematic acceleration, which is change in velocity versus time. They are the same only in an inertial frame.

I think what you mean by "acceleration" is "dynamic acceleration". I don't think you mean "kinematic acceleration."

The dynamic acceleration as measured in the space ship can be held constant. The observer at rest with respect to the space ship is in a noninertial frame. He can measure his "dynamic acceleration" by measuring the pseudoforce that keeps him "stationary" with respect to the space ship.

Suppose that he is lying on a couch in the space ship. He can decide to keep the force between him and the couch constant at a comfortable value. The pseudogravity pushes him against the couch so that he is stationary within the space ship. Since if the astronaut knows that the pseudogravity is a "fictitious force", he can deduce that his dynamic acceleration is constant.

He can also keep the pseudogravity constant in his noninertial frame by controlling the rate of fuel consumption. He can adjust the rate of fuel consumption to a rate where the couch has a constant deformation. There are any number of ways that he can keep the pseudogravity constant. Therefore, the effective acceleration in his noninertial frame will be constant.

The catch is this. If he does all these things, his kinematic acceleration in any inertial frame will decrease with time. Any observer belonging to an inertial frame will see his acceleration decrease has he approaches the speed of light in that frame. There is no way that he can keep his acceleration as measured in an inertial frame constant or non-decreasing.

Therefore, the astronaut can never go faster as measured in an inertial frame than the speed of light in a vacuum. Special relativity limits the kinematic acceleration profile of an object as measured in any inertial frame. The acceleration profile can never be such as to push an object faster than the speed of light as measured in an inertial frame.

Note that the astronaut is in a noninertial frame. Therefore, there will be a lot of nonlocal effects that "violate" special relativity. The astronaut lying comfortably on that couch may notice very different objects going faster than light. In fact, he may perceive the speed of light changing with distance from him. However, this doesn't really violate special relativity because he is in a noninertial frame.

General relativity comes in when one asks how the astronaut can determine that the pseudogravity is a "fictitious force". He has to use nonlocal methods of measurement. General relativity starts with the hypothesis that there is no local method of measurement that can distinguish a pseudoforce from the effects of gravity.

He basically has to observe distant objects in order to determine that whether or not he is in an inertial frame. However, general relativity is a more advanced subject than special relativity.

13. Mar 25, 2013

### 1977ub

Trying to figure out / recall the simpler case. Let's say he ejects matter out the back of the ship at a rate which is to him constant. Let's stipulate that his own mass decrease is negligible. (Classically, this would be a constant rate of increase in momentum, and therefore velocity, and therefore a constant acceleration.) For observer at home, his acceleration will lag behind classical expectations as gamma increases. You mention above that he can adjust the force. What I'm wondering is: without making any adjustments, will the constant matter emitter observer experience a constant pseudogravity force? Or to get this constant force will he need to modulate his output?

14. Mar 25, 2013

### 1977ub

Ok never mind. From his perspective, he is in the same situation after each ejection of matter, thus he must experience the same force with each ejection.

15. Mar 25, 2013