Schrödinger's Cormorant

3.3 Time for Grazing

Classical momentum is NOT conserved once we take special relativity into account. A new, relativistic definition of momentum is required to restore the principle.

Nefertiti’s experiment is shown in the animation below: you and she draw a long line in the sand and stand at either end of it. You stand very slightly offset, so that you are just to one side of the line and Nefertiti is off to the other side. Meanwhile, Cormorant and his cousin, Shag, are flying overhead: we’ll look at it all from their points of view later. You and Nefertiti begin by throwing rocks at one another (the rocks are perfectly smooth and spherical, identical in mass and are thrown parallel to the line in the sand, with identical speeds). When the rocks meet the offset in their trajectories means that they strike one another only the slightest of glancing blows. The result is that each is deflected slightly from its original path, as you can see in the animation below.

This is how it looks in your frame of reference (and Nefertiti's):

If we define the line in the sand as the x axis, you can see that before the collision, the stones’ velocities are directed parallel to the axis, so that any perpendicular component of the velocity is zero (vy = 0). Now a key feature of Nefertiti’s experimental design is that the collision is so slight that the difference it makes to the velocities of the rocks along the x-direction is tiny (the deflection is much exaggerated in the picture so you can see clearly what’s happening). Nevertheless, the grazing collision produces a small but non-zero perpendicular component, v+y, and that’s what we are going to focus on (remember that the superscript +, in v+y , means "after the collision").

To measure these y-velocity components following the collision, you are going to measure the time it takes for each rock to move 1 metre (perpendicular distance) from the line in the sand. The symmetry of the setup makes it easy to see what you’re going to find: the y-velocities will be equal and opposite and so the times measured will be the same for each rock.

Question 3-3

Showing that 0 = 0 is always fun, but now for something even more interesting: let’s look at the same collision in an alternative reference frame, moving relative to your own. Time to bring Cormorant back in to the picture. He’s been flying overhead along the x-axis, following the line in the sand, matching the x-component of your rock’s velocity. At the same time, his cousin Shag has been flying in the opposite direction, tracking Nefertiti's rock.

In Cormorant's frame of reference, he - and, therefore, also the rock that he's tracking, - are stationary, leading up to the collision. The animation below shows how the experiment will look in his frame of reference:

The pink rock is always alongside Cormorant, so before the collision it has zero velocity in all directions, in his frame of reference. After the collision the x-component of its now very slightly deflected velocity vector is close enough to zero in Cormorant’s reference frame to be neglected (remember that it was a very slight deflection), so no relativistic worries there.

What about the green rock? Before the collision it has zero y-velocity in Cormorant’s frame of reference, just as it did in yours. No problem there, but it acquires some y-velocity in the collision, and to quantify this, we have to take account of the fact that this rock is flying at a high speed away from Cormorant in the x-direction, which means that we do now have to worry about relativistic effects. You know about two types of effect: time dilation and length contraction. The easy bit is that length contraction happens only in the direction of relative motion: there is no effect in a perpendicular dimension, so Cormorant’s measurement of the perpendicular distance moved by the green rock is not affected by his speedy retreat, which is effectively in the x-direction. However time is another matter.

Before the collision, both rocks are flying parallel to the x-axis, so each has zero velocity in the y-direction. In turn, this means that neither has any momentum in the y-direction, so in Cormorant's frame of reference - as in yours - the total y-momentum is zero before the collision.

In the collision, the rocks are very slightly deflected, so each now acquires a small velocity component in the y-direction. If Cormorant measures the velocity of the rock he's been tracking (pink) and Shag measures the y-velocity of his rock (green), symmetry tells you that the values they get will be equal but opposite.

However the two seabirds are moving at high velocity relative to one another, so we know that, because he is a moving observer, Cormorant will see less time passing for Shag, as the green rock moves its 1m in the y-direction, than passes for himself, as the pink one does the same. So for Cormorant, the green rock will be moving faster in the y-direction than the pink rock. As the velocities of the two rocks are therefore not equal and opposite, it follows that the sum of the y-momenta, which was zero before the collision, is now decidedly not zero.

The horror! The horror! This means that in Cormorant’s frame of reference, the total y-momentum, appears not to be conserved in this collision. This is an incredibly important result, so we had better analyse it quantitatively:

Question 3-4

What you should have uncovered here is a problem of gargantuan proportions. A lesser man, woman or seabird might simply cry or run for the hills but when he got here, Uncle Albert stayed cool and thought – once again – outside the box. OK, he reasoned, conservation of momentum doesn’t work, so let’s redefine momentum. “Sounds a bit like the way politicians try to spin bad news” thinks Cormorant but Uncle Albert is firm about this. His reasoning is that momentum must be conserved, otherwise the Universe makes no sense – so if the classical definition appears to show momentum not being conserved, the classical definition is wrong. His new relativistic definition of momentum is carefully crafted to make its conservation work in any frame of reference:

Notice that, as with all things relativistic, if the velocity v is much less than c, then the factor √(1 – v2/c2) is effectively just equal to 1, and the definition then simplifies to the classical one, p = mv.

We need to take care with the √(1 – v2/c2) term because it looks very similar to the factor √(1 – u2/c2) that we have grown accustomed to in relation to time dilation. There is a crucial difference, however: v is the velocity of an object within a given frame of reference, whereas u is the relative velocity of two different frames of reference. So you can use equation III-1 to find the true momentum of a moving object– as opposed to the classical approximation to it – in whatever frame of reference you find yourself.

Now let's test out this new redefinition and assure ourselves that it really does make the conservation thing happen.

Question 3-5

This exercise should persuade you that Uncle Albert’s reformulation has saved the day: momentum is now conserved, even from the point of view of a moving observer.

How should we interpret this new definition of momentum? At the moment we’ve just accepted it because it works but is there a physical interpretation? One way to look at it would be to suggest that the classical momentum definition still holds but that the mass of a particle increases when it moves. Thus if Cormorant’s mass is m0 when he is at rest, we could say that when he moves at velocity v, his mass increases to m1:

So then, we get:

Of course, Cormorant’s mass increase would only be from your point of view. Cormorant would say that he is stationary in his frame of reference and it is you who is moving and therefore it is you who has gained in mass (so there). And, as with all things relativistic, that pesky factor of √(1 – v2/c2) means that these strange things only become really apparent at velocities that are of a similar magnitude to the speed of light. A quick calculation should make this clear:

Question 3-6

While this idea of “relativistic mass” is quite appealing, it has its problems. There is, for one thing, a feeling that mass should be an intrinsic property of a particle and not subject to variation depending on how fast it’s moving. In addition, while it makes the classical momentum definition work, the “relativistic mass” doesn’t work in the classical equation for kinetic energy: which is where we want to go next. So, many physicists would say that we should continue to treat mass as invariant; it’s momentum that isn’t as simple as mv: it just appears that way, unless your velocity is approaching that of light.

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