I have taught special relativity for many years, but every time I teach it I present the result that mass changes as a function of velocity as a consequence of the modified version of Newton’s 2nd law.

As almost everyone knows, Newton’s 2nd law says that

where is the force applied, is the mass, and is the acceleration felt by the body. In Newtonian mechanics, mass is invariant, but a consequence of special relativity is that nothing can travel faster than the speed of light . This raises the conundrum of why can’t we keep applying a force to a body of mass , causing it to continue accelerating and to ultimately increase its velocity to one greater than the speed of light?

The answer is that Newton’s 2nd law is incomplete. Einstein showed that mass is also a function of velocity, and so we should write

Where is the so-called *Lorentz factor* and is the rest mass (also known as the invariant mass or gravitational mass), the mass an object has when it is at rest relative to the observer. Hence we can argue that, as we approach the speed of light, the applied force goes into changing the mass of the body, rather than accelerating it, leading to a modified version of Newton’s 2nd law

where both velocity and/or mass change as a force is applied. But, because of the fact that until (see Figure 1), very little increase in mass occurs until has reached appreciable values.

However, I have always found this an inadequate explanation of the relativistic mass, as it does not derive it but rather argues for its necessity. So, as I’m teaching special relativity again this year, I decided a few weeks ago to see if I could find a way of deriving it from a simple argument. After several weeks of hunting around I think I have found a derivation which is robust and easy to understand. But, in my searching I came across several “derivations” which were nothing more than circular arguments, and also some derivations which were simply incorrect.

## Two balls colliding

The best explanation that I have found to derive the relativistic mass is to use the scenario of two balls colliding. Although it would be possible, in theory to have the balls moving in any direction, we are going to make things a lot easier by having the balls moving in the y-direction, but with the two reference frames moving relative to each other with a velocity in the x-direction. Also, the balls are going to have the same rest mass, , as measured in their respective frames and (the rest mass of each ball can be measured by each observer in their respective reference frames when they are at rest in their respective frames).

The blue ball moves solely in the y-direction in reference frame , and the red ball moves solely in the y-direction in reference frame . Ball starts by moving in the positive y-direction in reference frame with a velocity , and ball starts moving in the negative y-direction in reference frame with a velocity in frame .

Reference frame is moving relative to frame at a velocity in the positive x-direction. So, as seen in , the motion of ball appears as shown in the left of Figure 2. That is, it appears in to move both in the negative y-direction and the positive x-direction, and so follows the path shown by the red arrow pointing downwards and to the right.

At some moment the two balls collide. After the collision, as seen in , ball will move vertically downwards in the negative y-direction, with a velocity . Ball moves upwards (positive y-direction) and to the right (positive x-direction), as shown by the red arrow in the diagram on the left of Figure 1.

In reference frame the motions of balls and looks like the diagram on the right of Figure 1. In , it is ball which moves vertically, and ball which moves in both the and directions.

## The velocity of ball in

To calculate the velocity of ball as seen in , we have to use the Lorentz transformations for velocity. As we showed in this blog here, if we have an object moving with a velocity in which is moving relative to with a velocity , then the velocity in frame is given by

This equation is true when the velocity is in the x-direction, and the frames are moving relative to each other in the x-direction. So we are going to re-write Equ. (2) as

However, if the velocity of an object is in the y-direction, rather than the x-direction, then we need a different expression. We can derive it from going back to our equations for the Lorentz transformations

This time we write

and

So

Dividing each term in the right-hand side by , we get

Equations (3) and (4) allow us to work out the components of ball ’s velocities in the x-direction and in the y-direction in frame .

After the collision, the velocity of ball becomes . What about ball ?

We can see that will not change, and after the collision will be .

## The momentum before and after the collision

We are now going to look at the momentum of balls and before and after the collision, as seen in frame . We will start off by assuming that the mass is constant for both balls, that is that for both balls, despite the two reference frames moving relative to each other.

If we do this, we can write that the momentum in the x-direction before the collision is given by

The momentum after the collision in the x-direction is given by

So, momentum is conserved in the x-direction. But, what about in the y-direction? Before the collision, the momentum is given by

After the collision, the momentum in the y-direction is given by

.

If we assume that momentum is conserved, we can write

So, if we assume that the mass of both ball and ball in frame is , the momentum in the y-direction is only conserved if . But, is only equal to unity when the relative velocity between the two frames is zero; in other words when the two frames are not moving relative to each other! If and mass is constant, *momentum will not be conserved*.

In physics, the conservation of momentum is considered a law, it is believed to always hold. In order for momentum to be conserved, we can qualitatively see that the mass of ball needs to be greater than the mass of ball as seen in frame , as the speed of ball in the y-direction in frame .

## Allowing the mass to change

We have just shown above that, if we assume both masses are invariant, momentum will only be conserved in the y-direction in the trivial case where the two frames are stationary relative to each other. So, let us now assume that, if , we have to allow the masses to change.

We will assume that mass is a function of speed. For ball , the momentum in the x-direction is still zero, both before and after the collision. For ball , we will now write the momentum in the x-direction, both before and after the collision, as

What about in the y-direction? For ball , before the collision we can write

Where is the mass of ball in frame which is affected by its velocity in frame , which is .

For ball as seen in frame we can write that the momentum in the y-direction before the collision is given by

Where , the Lorentz factor due to the relative velocity between and .

After the collision, we can write the momentum for ball in the y-direction as being

And, for ball we can write

Equating the momentum in the y-direction before and after the collision, we have

For ball , we will write

where

(that is, depends on the speed of ball in frame , and that speed is ).

So, the momentum of ball in the y-direction is given by

For ball , we will write

Where depends on the speed of ball as seen in frame . (**Note:** the mass does not depend on just the y-component of ball ‘s speed (as is often incorrectly stated), *it depends on its total speed*).

To calculate the value of we note that it is made up of the x-component and the y-component . But, , and we showed above that , where this .

Using Pythagoras to calculate , we have

so

Using this value of we can write

But, the terms can be factorised as

And so we can write

But, , so we can write

This means that we can write the momentum for ball in the y-direction as

Comparing this to Equ. (7), the equation for , we can see that they are equal, as required.

So, we have proved that, to conserve momentum, we need mass to be a function of speed, and specifically that

Where is the speed of the ball in a particular direction in frame .

on 22/09/2017 at 15:29 |Di EnwBlimey. This seems very complicated, da iawn! But which birth sign is it for? Can you do one for me, Aries, with a prediction or two. Diolch

on 22/09/2017 at 17:35 |RhEvansIt’s not complicated at all.

on 27/08/2018 at 17:34 |Emmett HumeI believe that the arrows for the direction of the blue ball as perceived in frame S’ need to be reversed – that is point to the left.

on 27/08/2018 at 18:37RhEvansYou’re correct. Well spotted, you’re the first person to point that (deliberate?) error out 😜

on 31/10/2017 at 20:08 |enricouvaThanks. Definitely the most straightforward and yet sound-enough derivation I’ve seen.

on 01/11/2017 at 11:53 |RhEvansThank you.

on 08/11/2017 at 15:10 |What Do X-Rays Bring to Mind? | Sciences In the Mural Of Life[…] mass : p = γmv. A non-circular and sound-enough derivation of relativistic mass, γm, is found here. As for the term γ itself, although it was originally conceived as a “fudging factor” […]

on 10/11/2017 at 17:28 |Serious PhysicistAmazing amazing work! Now I feel like there is someone who exactly understands the pain of learning things superficially. You are seriously a great physics instructor. Please supplement your explanation with more diagrams.

on 10/11/2017 at 21:04 |RhEvansThank you. Such positive comments make writing my blog worthwhile 🙂

on 29/11/2017 at 12:49 |enricouvaIn your graphic of the Lorentz transformations, in the expression for time, there is a typo—the exponent of 2 should not be there. Luckily the typo had a short half- life 🙂 —- it did not persist when you differentiated through the expression in the subsequent derivation.

on 29/11/2017 at 13:36 |RhEvansI’ll check. Thanks.

on 30/11/2017 at 15:03 |RhEvansOh yes, I am not sure how that squared term got in there. I will correct it.

on 18/12/2017 at 18:34 |Daniel Hoppe HansenIn the parts where you derived equation 7 and 8, how did you determine the starting equation, where the relativistic mass is equal to a ball-specific gamma times the rest mass? I feel as if I’m missing an obvious connection, but I’m not seeing it.

on 18/12/2017 at 19:39 |RhEvansI’ll get back to you soon. I currently have very limited (and expensive) internet access. I’ll answer your question over the Xmas break. Thanks.

on 04/04/2018 at 14:33 |TonyHi. I have hunting for clear pro/con explanations in the current debate on is it necessary or a red herring to introduce any mass other than m°. I didn’t know there was a debate till I started reviewing spec relativity after 50yrs. I ended uo going back to my Univ of Toronto text from 1962!! This Goble and Baker has the very same essence (but not followed through quite as far) as your post. These are still the only two I have seen which seem to achieve a logical deduction of non-rest mass. Good post and glad to see it. It pleases me to see that my 1960’s BSc course and text is still relevant. Now I have to get back to the current debate sometime.

on 03/02/2019 at 04:06 |Don LawnAs a lapsed physicist I was considering trying to re-derive this myself. I would have failed.

on 01/11/2019 at 08:14 |shivani groverThis was tremendously helpful..thank you!

on 07/12/2019 at 13:14 |RandyI was doing some calculations on the momentum of a photon and obtained this result where a 2 appeared in the formula.

gamma = 1/(2-v^2/c^2)^(1/2)

This expression allows for a relativistic particle like a photon to have mass.

What would have to change in your proof to make this possible?