Interesting Thoughts

1. May 24, 2014

kingofxbox99

I was thinking, and somehow I had this little "thought experiment" type thing relating to objects and the speed of light. (This is just my thoughts, I don't know if it's correct or the validity of my assumptions, etc)

Imagine you're in a (large) room, and that the effects of friction, air resistance, etc. are nonexistant. You walk up to a box at the end of the room, and apply a constant force to it. If the force remains constant, the acceleration will remain constant. Eventually, the box's velocity will get close to the speed of light. Once it hits the speed of light, it's acceleration will instantly drop to 0, and in order to maintain a constant force, it's mass will approach infinity (F = ma). Is this why people always say that your mass would approach infinity as you approach the speed of light? Just a cool little thought I had. :)

2. May 24, 2014

Staff: Mentor

The box will never reach the speed of light. Instead, its acceleration will gradually slow and it will get closer to the speed of light without ever quite getting there.

You cannot use $F=ma$ here; that's a simplified formula that only works for speeds that are small compared with the speed of light. For your problem, you have to make two changes:

First, you have to use the more general $F=\frac{dp}{dt}$ where $p$ is the momentum, instead of $F=ma$.

Second you have to use the relativistic definition of momentum: $p=\gamma{m}{v}$ where $\gamma$ is $\frac{1}{\sqrt{1-v^2}}$ (I'm measuring distances in light-years and times in years so that the speed of light is one in this formula).

Note that if $v$ is small these formulas reduce to the familiar $F=ma$. But in your problem, $v$ is not small.

No. The people who say that are repeating something that they heard from someone else who also doesn't understand relativity. (Or they understand relativity and are really terrible teachers).

Last edited by a moderator: May 25, 2014
3. May 25, 2014

kingofxbox99

Thanks for explaining that! I hate how simplified everything is at the highschool level...

4. May 25, 2014

Staff: Mentor

It's necessary, as it is far easier to learn something like F=MA instead of the relativistic version at that level. You have to walk before you can run, after all. Besides, for everyday physics the classical version you learn in high school works just fine.

5. May 25, 2014

Staff: Mentor

Well, you do have to have classical physics down cold before you're ready to take on relativity, so we teach it first.

Another complication is that many high-school students have not yet had a full year of calculus, so we can't use calculus, just trignometry and algebra. That makes it hard, especially when you consider that when Newton was figuring out classical mechanics, he needed calculus so badly that he had to invent it. $F=\frac{dp}{dt}$ is the most powerful and general expression of Newton's second law; but if you don't have calculus you have to settle for $F=ma$, which falls out of $F=\frac{dp}{dt}$ if you remember that classically $p=mv$ and assume a constant mass.

6. May 25, 2014

DrStupid

That's a very interesting question. When starting from rest the box will be accelerated according to

$$a = \frac{F}{m}\sqrt {1 - \frac{{v^2 }}{{c^2 }}} ^3$$

F and a can only remain constant if m is decreased with velocity (e.g. by emission of radiation) according to

$$m = m_0 \cdot \sqrt {1 - \frac{{v^2 }}{{c^2 }}} ^3$$

That means when the box reaches the speed of light its mass doesn't approaches infinity but zero. Moreover, energy goes zero too:

$$E = m_0 \cdot \left( {c^2 - v^2 } \right)$$

Thus there would be nothing left to accelerate or to apply a force to.