Video: Relativistic Momentum

In this video we learn how to calculate the relativistic momentum of an object, and see how the mass of an object varies with its velocity, especially as its velocity approaches a significant proportion of the speed of light.

10:11

Video Transcript

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 π‘š 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 π‘š 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 π‘š, 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 π‘š does not equal π‘š 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 π‘š two, then, once again, we would find unequal masses. π‘š sub two is not equal to π‘š or π‘š sub zero.

This effect that we’re seeing can be explained by a principle that an object’s mass changes with its speed. If we let π‘š 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 π‘š sub zero and the relativistic mass we’re calling π‘š. 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 π‘š equals π‘š sub zero. So, the ratio of π‘š to π‘š 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 π‘š over π‘š 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 π‘š. 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 π‘š 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 at 0.200𝑐.

We can call the given mass, 6.68 times 10 to the negative 27th kilograms, π‘š. 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, π‘š sub zero, times 𝑣, where 𝛾 is one over the square root of one minus 𝑣 squared over 𝑐 squared. So, 𝑝 equals π‘šπ‘£ 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 π‘š 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 π‘š sub 𝑑. And we can call the proton’s mass of 1.67 times 10 to the negative 27th kilograms π‘š 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 π‘š 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 π‘š 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 π‘š 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 particle.

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 π‘š 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 moves.

In summary of relativistic momentum, we want to be careful to distinguish between rest mass, which we’ve marked as π‘š sub zero, and relativistic mass, which we’ve called simply π‘š. 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.

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