Relativistic mass

ebvalaim 2022-03-21 0 Comments

When Special Relativity is being introduced at school (if it is at all - the curriculum might depend on your country and it can change in time), one of the notions being discussed is so called "relativistic mass".

One of the consequences of relativity is that faster moving objects are harder to accelerate, which means that their inertia increases. And since it is being said from the beginning of the physics lessons that mass is the measure of inertia, it is tempting to try to explain this effect with an increase in mass. So, the notion of mass is being split into "rest mass" - the mass an object has at rest - and a "relativistic mass" - the mass of the object in motion, larger than the rest mass. The equations also become prettier right away, since if we denote the relativistic mass by $m$, we can always write $E = mc^2$, and momentum can be expressed using the formula known from classical physics $p = mv$ (versions with the rest mass also have an ugly square root in the denominator - we'll see it later). This is the life!

If you are following articles or discussions about relativity on the internet, you probably noticed relativistic mass being mentioned in multiple contexts. It is often used to explain the impossibility of reaching the speed of light ("because the mass would grow to infinity"), or sometimes someone will ask whether an object can become a black hole by going fast enough (it can't). The relativistic increase in mass is being treated as fact in such situations, as something certain.

Well, I'd like to disturb this state of affairs slightly with this article ;) Because, as it turns out, the notion of relativistic mass loses a lot of its appeal upon closer scrutiny. As a result, relativistic mass is rarely being used in academia and you can encounter it pretty much only at school, in discussions on the internet and in popular science publications. Let's take a closer look at the reasons behind that.

The increase in inertia

Wait a minute - I just said that inertia of objects in motion increases, and mass is the measure of inertia. And now I'm saying that the mass doesn't increase? What is all this about?

In pre-relativistic physics, it was simple. If you applied a force to an object, it would accelerate - and the ratio of the force to the acceleration was precisely mass. So the harder an object is to accelerate - which means, the smaller the acceleration caused by the same force is - the larger its mass.This can be expressed with an equation as $a = \frac{F}{m}$.

What's more, an object would always accelerate in the same direction in which the force was acting (which might seem obvious... but let's not get ahead of ourselves). This means that the relationship above can be written using vectors and it will still hold: $\vec{a} = \frac{\vec{F}}{m}$. This way we took into account that forces can act in various directions and we can still calculate the acceleration.

So what does relativity change in this picture?

Let's keep the distinction between the rest mass and relativistic mass for now. An object at rest has mass $m_0$. According to the idea of relativistic mass, it increases in motion to:

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

This ugly square root in the denominator is usually denoted by $\frac{1}{\gamma}$, which lets us express the relativistic mass as:

$$m = m_0 \gamma$$

When the velocity is 0, $\gamma = 1$ and relativistic mass is simply equal to the rest mass.

That's cool. We introduced the relativistic mass because we wanted to account for the increase in inertia by the increase in mass. So, can we still write $\vec{a} = \frac{\vec{F}}{m} = \frac{\vec{F}}{m_0 \gamma}$?

Well... almost. As it turns out, we can - but only if the force is perpendicular to the direction of motion! If it is parallel, the result is different - then we need to calculate the acceleration like this: $\vec{a} = \frac{\vec{F}}{m_0 \gamma^3}$ (I'll derive this below for the curious).

Wait... what does it even mean? Well, it means that the object has larger inertia in the direction of motion than perpendicular to it. So if we want to find the acceleration of the object when a force is applied, we can't just divide it by the object's mass - we have to decompose the force into parallel and perpendicular components, divide them by different numbers and then add the results to get a single acceleration. In effect it is possible that when the force acts at an angle to the velocity, the acceleration might not even be in the direction of the force!

The conclusion is that the "relativistic mass" doesn't solve the problem of inertia. There isn't even a single number that would reasonably measure inertia, because now it depends on the direction! (It is still possible to introduce a quantity measuring inertia - only now it has to be a tensor of rank 2, which can be written as a matrix.)

That's the first point against the relativistic mass.

Collisions

One of the ways of comparing masses of different objects is making them collide. If a more massive object hits a less massive one, then the former will only slow down slightly, and the latter will get bounced at a significant speed. On the contrary, if a less massive object hits a more massive one, the former bounces away, and the latter picks up only a small amount of speed.

In a special case in which an object hits another one that is at rest and has the same mass - the former will stop, and the latter will fly away with the same speed the former had initially. Newton's cradle is one of the better known illustrations of this fact.

The exact formulae for the velocities after the collision depending on the velocities before the collision and the masses of objects can be derived using the conservation laws for energy and momentum. We won't be doing this here, we'll just use the intuitive understanding described above.

The question now is this: what happens if the two objects colliding have the same rest masses, but vastly different relativistic masses? For example, what if one ball with a mass of 1 gram is at rest, and a second identical one going fast enough to have a relativistic mass of 100 grams (so $\gamma = 100$) hits it? Will they behave like objects with different masses (the moving one will slow down, and the resting one will start moving with some speed), or like equals (the moving one will stop, and the resting one will start moving with the same speed the moving one had before)?

The answer lies again within the conservation laws, only this time the relativistic ones have to be used. I'll write the full equations below, and here I'll just tell you the result: as it turns out, the balls will behave like equals. So, relativistic mass is irrelevant in collisions - the only mass that counts is the rest mass.

That's another point against the relativistic mass.

Duplication of notions

This isn't a physical argument, strictly speaking, more like a technical one, but it still carries some weight.

Physicists like to simplify their lives. One of the simplifications they like to make is eliminating the constants of nature from the equations. Let me explain.

Let's take the speed of light as an example. We can check in some books or on the internet that it is equal to 299 792 458 m/s. The number is rather ugly, but we have no say in what the nature chose as the speed of electromagnetic waves... or do we?

This particular number only comes from the units we chose. If we wanted to express the speed of light in feet per second, the number would be different. If we chose furlongs per fortnight, we'd get another different number. Hmm... What if we made our lives simpler and chose units such that the number was somewhat simpler? For example, if it was... just 1?

We can do that and that's exactly what physicists do. Example units like that could be a second and a light-second. Or a year and a light-year. Or any time unit and the distance light travels during that time.If we choose such a system of units, we'll have $c = 1$. And just like that, $c$ disappears from all the equations, because whether we multiply or divide by 1, it doesn't ever change anything.

(Physicists like to take it a step further and eliminate more constants. The units in which $c = G = \hbar = 1$ - so ones in which the speed of light, the gravitational constant and the reduced Planck constant are all 1 - are a popular choice. These units are called "natural units" or... "Planck units". The basic units in this system are Planck length, Planck time and Planck mass.)

Alright, but why am I mentioning all this? Well, let's see what happens to the famous Einstein formula for energy in such a system:

$$E = mc^2 = m_0 \gamma c^2$$

When we introduce units in which $c=1$, we get:

$$E = m = m_0 \gamma$$

The relativistic mass is always equal to energy in such units! So it is a de facto duplicate of the notion of energy. Everywhere where we would use the relativistic mass before, we can just substitute in energy (in other units: $\frac{E}{c^2}$) and nothing will change. Why would we need such an additional notion, then?

And that's yet another point against the relativistic mass.

Summary

As you can see, introducing the notion of relativistic mass doesn't really get us much. It isn't great for measuring inertia, it's useless in collisions, and is de facto a duplicate of energy. For these reasons, physicists pretty much stipped using this notion - now, when mass is being mentioned, it almost always means rest mass.

And because of that, I'm asking you, dear Readers - let's stop saying that mass increases for objects in motion. Let's stop saying that it becomes infinite at the speed of light. We can go ahead and substitute "inertia" for "mass" in these contexts, or - since it is a notion almost equivalent to relativistic mass - just use "energy". Both inertia and energy tend to infinity as speed tends to $c$. Let mass remain a property that is constant for a given object.

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The equations

Two key equations we will need are the relativistic expressions for energy and momentum:

$$E = m \gamma c^2$$

$$\vec{p} = m \gamma \vec{v}$$

$m$ is the rest mass in both of these equations - we forget that relativistic mass is even a thing now.

And, as always:

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

Inertia

It will be our goal to express the acceleration $\vec{a} = \frac{d\vec{v}}{dt}$ using the force, mass and velocity.

We'll assume the following equation (which is correct in non-relativistic physics, too) as the definition of the force:

$$\vec{F} = \frac{d\vec{p}}{dt}$$

We have the formula for momentum, so we just need to start differentiating ;) We get:

$$\vec{F} = m\frac{d\gamma}{dt}\vec{v} + m \gamma \frac{d\vec{v}}{dt} = m\frac{d\gamma}{dt}\vec{v} + m\gamma \vec{a}$$

The second term strongly resembles the force known from Newtonian physics (if we used the relativistic mass), but there is still the first term. Let's focus just on the derivative of $\gamma$ for now:

$$\frac{d\gamma}{dt} = \frac{d}{dt} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = -\frac{1}{2} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}^3} \left(-\frac{1}{c^2} \right) \frac{d(v^2)}{dt} = \frac{\gamma^3}{2c^2} \frac{d(v^2)}{dt}$$

We'll notice now that $v^2$ is simply $\vec{v} \cdot \vec{v}$, which is the dot product of the velocity vector with itself. The dot product behaves just like a regular product with respect to differentials, so we get:

$$\frac{d(v^2)}{dt} = 2\vec{v}\cdot\frac{d\vec{v}}{dt} = 2\vec{v}\cdot\vec{a}$$

Hence:

$$\vec{F} = m \frac{\gamma^3 \vec{v}\cdot\vec{a}}{c^2} \vec{v} + m\gamma \vec{a}$$

(Let's remember this form of the equation, we'll come back to it in a moment.)

Now, let us take a note of a property of the dot product: namely that $\vec{a} \cdot \vec{b} = ab\cos \alpha$, where $a$ and $b$ are the magnitudes of the respective vectors, and $\alpha$ is the angle between them. In particular, if the vectors are perpendicular, the dot product is zero, and if they are parallel, it is equal to the product of the magnitudes (up to the sign).

Thus, our $\vec{v} \cdot \vec{a}$ can be written as $va_{\|}$, where $a_{\|}$ is the component of the acceleration that is parallel to the velocity, or, to be more precise - the projection of acceleration onto the direction of velocity (which will be negative if the angle between acceleration and velocity exceeded 90 degrees). The perpendicular component has no effect here.

On the other hand, $\vec{v}$ can be written as $v\vec{e}_v$ - where $\vec{e}_{v}$ is a unit vector (a vector of magnitude 1) with the direction and sense the same as the velocity.

Thanks to these "tricks", we can write the first term as:

$$m \frac{\gamma^3 \vec{v}\cdot\vec{a}}{c^2} \vec{v} = m\gamma^3\frac{v^2 a_{\|}}{c^2} \vec{e}_v$$

Now let's note that $a_{\|}\vec{e}_v$ is a vector that has the same direction as the velocity, a sense dependent on the sign of the projection of acceleration on the direction of the velocity, and a magnitude equal to that projection - so it's simply the parallel component of the acceleration $\vec{a}_{\|}$! Let's also decompose $\vec{a}$ in the second term in the force into $\vec{a}_{\|} + \vec{a}_{\bot}$ and we'll get:

$$\vec{F} = m\gamma^3\frac{v^2}{c^2}\vec{a}_{\|} + m\gamma(\vec{a}_{\|} + \vec{a}_{\bot}) = m\gamma\left(1 + \gamma^2\frac{v^2}{c^2}\right)\vec{a}_{\|} + m\gamma\vec{a}_{\bot}$$

Now the only thing that's left is to simplify $1 + \gamma^2\frac{v^2}{c^2}$:

$$1 + \gamma^2\frac{v^2}{c^2} = 1 + \frac{\frac{v^2}{c^2}}{1 - \frac{v^2}{c^2}} = \frac{1 - \frac{v^2}{c^2}}{1 - \frac{v^2}{c^2}} + \frac{\frac{v^2}{c^2}}{1 - \frac{v^2}{c^2}} = \frac{1}{1 - \frac{v^2}{c^2}} = \gamma^2$$

Hence the final equation:

$$\vec{F} = m\gamma^3\vec{a}_{\|} + m\gamma\vec{a}_{\bot}$$

Thus:

$$\vec{a}_{\|} = \frac{\vec{F}_{\|}}{m \gamma^3}$$

$$\vec{a}_{\bot} = \frac{\vec{F}_{\bot}}{m \gamma}$$

Coming back for a moment to the equation I told you to remember - I mean the form which still contained $(\vec{v}\cdot\vec{a})\vec{v}$. As it turns out, such a term can be written as:

$$(\vec{v}\cdot\vec{a})\vec{v} = (\vec{v}\otimes\vec{v})\vec{a}$$

where $\vec{v}\otimes\vec{v}$ is a 3x3 matrix with components given by:

$$a_{ij} = v_i v_j$$

The whole equation can then be written as:

$$\vec{F} = m\gamma\left(\mathbf{1} + \frac{\gamma^2}{c^2} \vec{v} \otimes \vec{v} \right)\vec{a}$$

where $\mathbf{1}$ is the identity (unit) matrix. This form expresses the force as the acceleration multiplied by a 3x3 matrix - it is this matrix I meant as a possible measure of the object's inertia.

Collisions

Let us consider a collision of two objects with rest masses equal to $m$, of which one is moving with velocity $v$, and the other one is at rest. The object at rest has energy and momentum given by:

$$E_1 = mc^2$$

$$p_1 = 0$$

and the moving object:

$$E_2 = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$p_2 = \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$

The total energy and momentum of the system are then:

$$E = E_1 + E_2$$

$$p = p_1 + p_2$$

Let's assume that the objects will be moving with velocities $v_1$ and $v_2$ after the collision. The total energy and momentum will then be:

$$E = \frac{mc^2}{\sqrt{1-\frac{v_1^2}{c^2}}} + \frac{mc^2}{\sqrt{1-\frac{v_2^2}{c^2}}}$$

$$p = \frac{mv_1}{\sqrt{1-\frac{v_1^2}{c^2}}} + \frac{mv_2}{\sqrt{1-\frac{v_2^2}{c^2}}}$$

Due to the conservation of energy and momentum, these have to be exactly equal to the values from before the collision. This gives us two equations for two unknowns $v_1$, $v_2$.

There is one obvious solution, and that is $v_1 = v$, $v_2 = 0$ - this is simply the situation from before the collision. But thanks to the symmetry of the problem, there is a different, equally obvious solution: $v_1 = 0$, $v_2 = v$. This second solution then has to correspond to the situation after the collision (it can be proven that there are no other solutions) - so the objects will behave such that the moving one will stop, and the other one will start moving with the same speed the first one had before the collision.

Which means they will behave exactly like objects with equal masses.