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

This question already has an answer here:

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?

marked as duplicate by Ben Crowell, Emilio Pisanty, Qmechanic♦Sep 2 '13 at 11:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• I am wondering if you could explain it with work. Work is a change in kinetic energy and $E_{k}=\frac{mv^2}{2}$ so once you start to max out on a velocity due to the limit that light places on us if you did more work you just increase the mass? – user24048 May 6 '13 at 3:02
• possible duplicate of Why does the (relativistic) mass change & why? – Ben Crowell Sep 1 '13 at 14:23
• I'd like to note your teacher is pretty terrible at his job, but your equation is incomplete. – Skyler Nov 10 '14 at 10:45

3 Answers

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. It is good to consider mass as invariant quantity. Defination of mass is m=sqrt(E^2-p^2) taking c=1.. Momentum in special relativity is p=mv/sqrt(1-v^2/c^2). If you plot the graph of this function(momentum is function of velocity). you will see momentum keep increasing and tends to infinity as speed tends to c.. and here 'm' is constant... you can seperate the m/sqrt(1-v^2/c^2) from the expression of momentum then this term is called relativistic mass...this is just another way of saying momentum increasing or energy increasing with speed... what you prove in a cyclotron is momentum dependence on velocity ..you get those experimental results cause graph of momentum vs speed is not a straight line..

• Oh really? Wow, after he said what he said I actually even rebutted his car argument saying that $c$ is very big so it would be unnoticable, but then I just gave up. – user14445 Nov 28 '12 at 16:08
• I find your physics teacher quite entertaining and misleading. Bad for you :P – resgh Nov 28 '12 at 16:10
• I'd be careful with claims that the instructor is wrong. Certainly at everyday velocities the Lorentz factor is vanishingly small, but the theory is clear: all the mindblowing relativistic effects are present in some time amount at any non-zero relative velocity. – dmckee Nov 28 '12 at 18:03
• @dmckee What do you mean, how does this correspond with my teacher's claims? – user14445 Nov 28 '12 at 18:27
• @dmckee Well, doesn't this actually support my view and not his view? – user14445 Nov 28 '12 at 18:32

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

$$E=mc^2$$

• The OP asked a "why" question. This answer simply starts by asserting the result. – Ben Crowell Sep 1 '13 at 14:42
• Then the answer is because relativity holds. – Jorge Lavín Sep 2 '13 at 5:13

Relativistic mass is a relational concept because fermions have a non zero rest mass by way of Higgs field interaction. The increased kinetic energy of a particle accelerated to relativistic speeds is directly proportional to an increase in a particle's inertial mass. A particle's inertial mass can be conveyed as its resistance to change in momentum. So in layman's terms, the heavier an object is the more work you have to put into moving it. Close to c, you reach a point of diminishing return where it will take an infinite amount of energy to accelerate an object. Lorentz transformations dominate the scale at these accelerated energies and you end up with a particle that shrinks to zero. Another way of saying this is that if you accelerate an object that has a non zero rest mass towards the speed of light, what you end up with is black hole traveling awfully darn fast. Time slows down to near zero more so because of general relativity than due to special relativity for this type of singularity because the "ghosts" of entropy via hawking radiation are manifest through quantum gravitational effects.

• Welcome to physics.SE! Sorry to have to greet you with negative comments, but -- fermions have a non zero rest mass by way of Higgs field interaction. This is not really true. Composite particles have mass due to their internal fields. Even for fundamental particles it's not so simple: physics.stackexchange.com/questions/3037/… . Very little of the mass of ordinary matter comes from the Higgs mechanism, and in any case this is not relevant to the question. [...] – Ben Crowell Sep 1 '13 at 14:39
• [...] The middle portion of the answer is correct but is only a description of what happens relativistically. The question was a "why" question. Lorentz transformations dominate the scale at these accelerated energies and you end up with a particle that shrinks to zero. Is this referring to length contraction? Why is that relevant? Another way of saying this is that if you accelerate an object that has a non zero rest mass towards the speed of light, what you end up with is black hole traveling awfully darn fast. No, this is not true. – Ben Crowell Sep 1 '13 at 14:41