# Relativistic work/energy

1. Sep 26, 2010

### accountkiller

So, there is a sentence in my book, along with a graph, stating that "relativistic kinetic energy becomes infinite as v approaches c."

But then in a concept problem, it says "A proton is accelerated from rest by a constant force that always points in the direction of the particle's motion. Compared to the amount of kinetic energy that the proton gains during the first meter of its travel, how much kinetic energy does it gain during one meter of travel while it's moving at 99% of the speed of light"

The answer is the same, it says, because there is no change in kinetic energy since there is constant force. But I thought since the first statement above said that kinetic energy --> infinity as v -- c..... means that as v increases, KE increases, right? But that is not so in the problem.

Could someone explain why this is so?
Thanks!

2. Sep 26, 2010

### bcrowell

Staff Emeritus
There are three variables, x, v, and K (kinetic energy). There is a linear relationship between x and K (by the work-kinetic energy theorem). There is a nonlinear relationship between these variables and v.

3. Sep 26, 2010

### Fredrik

Staff Emeritus
The phrase "the answer is the same" doesn't say that the energy never increases, only that it increases by the same amount in both scenarios. (I haven't thought about the actual problem yet, so I'm just commenting on what seems to be a misinterpretation of what you read).

4. Sep 27, 2010

### starthaus

You can find the answer to your problem https://www.physicsforums.com/blog.php?b=1928 [Broken]. The answer is :

$$\Delta W =m_0c^2(\gamma(v)-1)$$ which also happens to be (by definition) the variation of the kinetic energy of the particle.

Last edited by a moderator: May 4, 2017
5. Sep 27, 2010

### Staff: Mentor

Work equals force times distance, so if the force is the same then the energy/distance is the same.