In this video, we’re going to talk
about relativistic momentum. And we’ll introduce this topic
through an experiment.
Imagine we have a box sitting on a
table top. At the outset, the box is
stationary. We’ll say that it has a mass 𝑚 sub
zero. Now, if we were to weigh this box
on a scale, we could use the weight that the scale read out along with the
acceleration due to gravity 𝑔 to solve for the value of 𝑚 zero.
Having done that, let’s imagine
further that we now put the box in motion. The box is moving at a speed 𝑣
across the table top. And say it was possible to, again,
weigh the box while it’s in motion. If we did that, once again using
the result on the scale and the acceleration due to gravity to solve for the box’s
mass 𝑚, then we might be curious to know, are the two masses equal? That is, does the mass of the box
change by putting it into motion?
Now, in the classical world, the
answer would be yes, the mass of the box is the same. It doesn’t change by putting it
into motion. But when we account for relativity,
we find that, strangely enough, that if we were able to very very precisely
calculate the mass of this box, we would find that 𝑚 does not equal 𝑚 sub
zero. And actually, if we accelerated the
box even more to a speed of two 𝑣 and again measured its mass, what we could call
𝑚 two, then, once again, we would find unequal masses. 𝑚 sub two is not equal to 𝑚 or 𝑚
This effect that we’re seeing can
be explained by a principle that an object’s mass changes with its speed. If we let 𝑚 sub zero be the
object’s mass when it’s not in motion, then to get the mass when it is in motion, we
multiply by 𝛾, the Lorentz factor, equal to one over the square root of one minus
𝑣 squared over 𝑐 squared. So, 𝛾 connects two different
masses, the rest mass 𝑚 sub zero and the relativistic mass we’re calling 𝑚. To better understand this
relationship, we can make a plot of the mass ratio with velocity.
If we were to plot relativistic
mass as a ratio of relativistic mass to rest mass against speed 𝑣, we could see
that when the speed 𝑣 was equal to zero, in that case, 𝛾 is one and 𝑚 equals 𝑚
sub zero. So, the ratio of 𝑚 to 𝑚 zero
would be one. At the other end, when 𝑣
approaches the speed of light 𝑐, then we see that 𝛾 goes to infinity, which would
mean that 𝑚 over 𝑚 sub zero would go to infinity as well.
The overall shape of this ratio
looks roughly like this. Where when the velocity of our
object is less than about half the speed of light, we don’t see much change in the
mass of our object due to the relativistic effects. This helps explain why our
principal may seem very counterintuitive. Our daily life involves massive
objects whose speed is much closer to zero than it is to 𝑐.
So, we now have an equation for
relativistic mass 𝑚. What if we multiplied both sides of
this equation by the speed 𝑣? In that case, looking at the
left-hand side of this equation in particular, you may recognize that equation. Mass times velocity is equal to
momentum 𝑝. And we’ve now arrived at the
formula for relativistic momentum. Relativistic momentum 𝑝 is equal
to 𝛾 times the rest mass of an object times its speed 𝑣.
Note that the object’s speed
appears in two places in this equation. First, it appears in the velocity
𝑣. And it also appears under the
square root sign in 𝛾. So, relativistic momentum 𝑝 is
equal to classical momentum 𝑚 sub zero times 𝑣 multiplied by 𝛾. 𝛾 is what makes this momentum
relativistic. Let’s try two exercises to put this
new information into practice.
Find the momentum of a helium
nucleus having a mass of 6.68 times 10 to the negative 27th kilograms that is moving
We can call the given mass, 6.68
times 10 to the negative 27th kilograms, 𝑚. And we can call the nucleus’ speed,
0.200𝑐, 𝑣. We want to solve for the nucleus’s
momentum. We’ll call that 𝑝. Now, starting off, we may not know
at first whether to use the classical or the relativistic formulation for
momentum. The clue in the problem statement
that tells us to use relativistic is the fact that the speed of this nucleus is
comparable to the speed of light. That means relativistic effects
will be important to include.
Relativistic momentum 𝑝 is equal
to 𝛾 times rest mass, 𝑚 sub zero, times 𝑣, where 𝛾 is one over the square root
of one minus 𝑣 squared over 𝑐 squared. So, 𝑝 equals 𝑚𝑣 over the square
root of one minus 𝑣 squared over 𝑐 squared, where 𝑣 is given to us in the problem
statement as two-tenths the speed of light 𝑐. And 𝑚 also is given to us in the
statement. So, we’re now ready to plug in to
solve for momentum.
When we enter our values to this
equation, the factor of the speed of light 𝑐 under the square root sign cancels
out. But in the numerator, it
remains. In this exercise, we’ll treat 𝑐 as
exactly 3.00 times 10 to the eighth meters per second. When we make that substitution and
then enter these values into our calculator, we find that 𝑝 is 4.09 times 10 to the
negative 19th kilograms meters per second. That’s the relativistic momentum of
this helium nucleus.
Now, let’s do one more example
involving relativistic momentum.
Calculate the speed of a
1.00-microgram-mass dust particle that has the same momentum as a proton moving at
0.999𝑐. The rest mass of a proton is 1.67
times 10 to the negative 27th kilograms.
We can call the 1.00-microgram mass
of the dust particle 𝑚 sub 𝑑. And we can call the proton’s mass
of 1.67 times 10 to the negative 27th kilograms 𝑚 sub 𝑝. The proton’s speed, 0.999𝑐, we’ll
call 𝑣 sub 𝑝. We want to solve for the speed of
the dust particle, which we’ll call 𝑣 sub 𝑑.
We can start off our solution by
calculating the relativistic momentum of the proton. We recall the relationship for
relativistic momentum, that 𝑝 is equal to rest mass 𝑚 sub zero times 𝑣 over the
square root of one minus 𝑣 squared over 𝑐 squared. If we call the relativistic
momentum of our proton 𝑝 sub 𝑝, then that’s equal to 𝑚 sub 𝑝 𝑣 sub 𝑝 over the
square root of one minus 𝑣 sub 𝑝 squared over 𝑐 squared. In the problem statement, we’re
told 𝑚 sub 𝑝 and 𝑣 sub 𝑝, so we can plug those into this equation now.
When we do, when we look under the
square root sign, we see that the factors of 𝑐 cancel one another out. But the factor of 𝑐 in the
numerator remains. We’ll treat the speed of light 𝑐
as exactly 3.00 times 10 to the eighth meters per second. Using that value and entering these
terms on our calculator to solve for 𝑝 sub 𝑝, result is 1.119 times 10 to the
negative 17th kilograms meters per second. We’ve now solved for 𝑝 sub 𝑝, the
momentum of the proton, but what we’re after is 𝑣 sub 𝑑, the speed of the dust
If we call 𝑝 sub 𝑑 the momentum
of the dust particle, we know that that’s equal to 𝑝 sub 𝑝. But we don’t yet know whether we’ll
need to include relativistic effects to solve for it. If we rewrite the mass of the dust
particle into units of kilograms from units of micrograms, then it’s equal to
1.10 [1.00] times 10 to the negative ninth kilograms. If we took the ratio of the mass of
the dust particle to the mass of the proton, the dust particle is roughly 10 to the
18th times more massive than the proton. Which means that the speed of the
dust particle, 𝑣 sub 𝑑, related to the speed of the proton is the inverse of that,
roughly 10 to the negative 18th times as fast.
This tells us that 𝑣 sub 𝑑 is
slow enough that we may reasonably neglect relativistic effects when we calculate
it. That means we can replace 𝑝 sub 𝑑
with 𝑚 sub 𝑑 times 𝑣 sub 𝑑. Solving for 𝑣 sub 𝑑, it’s equal
to the proton’s momentum divided by the dust particle’s mass. These values we know and can plug
in. When we calculate this fraction, we
find that 𝑣 sub 𝑑, the speed of the dust particle, is 1.12 times 10 to the
negative eighth meters per second. That’s how fast the dust particle
In summary of relativistic
momentum, we want to be careful to distinguish between rest mass, which we’ve marked
as 𝑚 sub zero, and relativistic mass, which we’ve called simply 𝑚. We also want to keep in mind the
contrast between classical momentum and relativistic momentum. Notice that the only difference
between the two equations is 𝛾. And finally, when in doubt whether
to use classical or relativistic momentum, look at the speed of the objects
involved. Higher speeds are a sign that
relativistic effects must be included.