Why does an object with higher speed gain more (relativistic) mass?

Question

Today, in my high school physics class, we had an introductory class on electromagnetism. My teacher explained at some point that an object with a very high speed (he said it started to get somewhat clearly noticable when travelling at 10% of the speed of light) will gain mass, and that that's the reason why you can't go faster than light.

One of my classmates then asked, why is this so? Why does an object with higher speed gain more mass? This of course is a logical question, since it is not very intuitive that a higher speed leads to a higher mass. My teacher (to my surprise (responded saying that it is a meaningless question, we don't know why, in the same way we don't know why the universe was created and those kind of philosophical questions.

I, being interested in physics, couldn't believe this, I was sure that what he said wasn't true. So after a while of thinking I responded saying:

Can't we describe it with Einstein's $E=mc^2?$ If an object gains speed, he gains more (kinetic) energy. With this equality we see that the more energy an object gets, to more massive it becomes.

He then replied saying that this formula is used for different cases, whereupon he gave a vague explanation as to when it is used. He gave me an example to show what I said was incorrect; when a car goes from $10 m/s$ to $40m/s$, according to what I said we would see a big increase in mass, and we don't (this sounded logical to me). So here I am, with the following questions:

• Why does an object with a higher speed have more mass (than the same object with a smaller speed)?

• When is $E=mc^2$ used and why is my argument incorrect in explaining this phenomenon?

Answer 1

In fact you are more or less correct. I assume the increase in mass mentioned is that described in special relativity. The example given by your teacher is incorrect. As the speeds of 10m/s and 40m/s are hardly relativistic, so we can for now assume $E=mc^2$. Increasing the kinetic energy by $\frac{1}{2}mv^2$ thus increases the mass by $$\frac{\frac{1}{2}mv^2}{c^2}=\frac{1}{2}m\frac{v^2}{c^2}$$ This in fact, is INCREDIBLY small, due to the hugeness of $c$. Now back to why the mass of an object increases. According to special relativity, mass and energy are in fact equivalent. Although not related by $E=mc^2$(Actually $E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$), the equivalance means that an increase in the velocity of an object will yes, increase its kinetic energy and thus mass.

Answer 2

If you look at an object at rest, and then you look at the object at some speed $\vec{v}\neq0$ and constant, the special theory of relativity tells you how things change.

There is an invariant (i.e. non changing) mass which we call rest mass $m_0$, and there is a "relativistic" mass $m$ which changes.

You have an static particle near you, do some measurements and the mass that you will have is $m_0$. Now, set the particle into movement in a straight line with constant velocity $\vec{v}$ and measure the mass $m$. You will find that the following is true:

$$m=\gamma m_0 $$

where $\gamma\equiv\gamma(v)$ the Lorentz factor is a function of the speed $v$ of the object

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

and $c$ is the speed of light in the vacuum. You can see that this mass $m$, in the limit $v\to c$ becomes infinite, thus making impossible the object to be moved. However, I think it is more correct think in terms of $m$ as an inertia, in the Newtonian sense (ignoring vectorial character of force and acceleration)

$$F=ma\Rightarrow a =\frac{F}{m}$$

Now fix the force $F$. For a heavy object, $a$ will be smaller than for a light object, thus we can interpret $m$ as the number that tells us how easy is to move that particle. In this sense, we see how $m$ increases with $v$ and thus is harder to move the particle the faster it goes.

The relativistic energy is given by $$E^2=p^2c^2+m^2c^4$$ where $p$ is the momentum of the particle. If you have your particle at rest ($p\propto v =0)$ then is true

$$E=mc^2$$