Understanding the Units in Relativistic Momentum Calculations

But if you meant you divided by 1,000,000 and then multiplied by c/0.8 to get the correct answer in MeV/c, then it makes sense. This method essentially converts the mass of the electron into MeV/c^2, which is the unit of mass used in special relativity, and then multiplies it by c/0.8 to get the momentum in MeV/c. So both methods should give the same answer, as long as the units are consistent. Does that make sense?In summary, the momentum of an electron whose speed is 0.600c can be found using the equation p = \frac{m * v}{\sqrt{1 - \frac{v^2}{
  • #1
jk4
I was working a problem in a Modern Physics book:
Find the momentum (in MeV/c) of an electron whose speed is 0.600c.

My first approach was:
mass of electron = 9.1E-31 kg

[tex]\sqrt{1 - \frac{(0.600c)^{2}}{c^{2}}} = 0.800[/tex]

[tex]p = \frac{9.1E-31 * 0.600c}{0.800} = 2.04E-22[/tex] (ignoring units)

then I needed to convert to MeV/c so with some messing around I ended up dividing by 1,000,000 and then multiplying by c to get the exact answer in the book. But this bothered me because I thought I had MeV then multiplying by c to get the answer in the book in MeV/c which doesn't make sense...

I then realized if I first convert the mass into [tex]\frac{MeV}{c^{2}}[/tex] then the units work out perfectly. But I'm still curious why I get the same answer doing it the first way, could someone please help me understand why it works out?
 
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  • #2
jk4 said:
I was working a problem in a Modern Physics book:
Find the momentum (in MeV/c) of an electron whose speed is 0.600c.

My first approach was:
mass of electron = 9.1E-31 kg

[tex]\sqrt{1 - \frac{(0.600c)^{2}}{c^{2}}} = 0.800[/tex]

[tex]p = \frac{9.1E-31 * 0.600c}{0.800} = 2.04E-22[/tex] (ignoring units)

then I needed to convert to MeV/c so with some messing around I ended up dividing by 1,000,000 and then multiplying by c to get the exact answer in the book. But this bothered me because I thought I had MeV then multiplying by c to get the answer in the book in MeV/c which doesn't make sense...

I then realized if I first convert the mass into [tex]\frac{MeV}{c^{2}}[/tex] then the units work out perfectly. But I'm still curious why I get the same answer doing it the first way, could someone please help me understand why it works out?
According to this, 1 MeV = 1.60217653 * 10^-13 J (and 1 J = 1 kg * m^2 / s^2). So, 1 MeV/c = (1.60217653 * 10^-13 / 299792458) kg * m / s = 5.34428565 * 10^-22 kg * m / s. And if the mass of an electron is 9.11 * 10^-31 kg, then its momentum at 0.6c is (9.11 * 10^-31 * 0.6 * 299792458 / 0.8) kg * m / s = 2.05 * 10^-22 kg * m / s. This is 0.384 the size of 5.34428565 * 10^-22 kg * m / s which I found above for the value of 1 MeV/c, so the momentum of the electron at 0.6c should be 0.384 MeV/c...is this different from what you got?
 
Last edited:
  • #3
I got 0.383MeV/c so you that's what I got. I was just confused about how the units worked themselves out in my first method of computing the answer, but thanks to your post I see it.
 
  • #4
jk4 said:
I got 0.383MeV/c so you that's what I got. I was just confused about how the units worked themselves out in my first method of computing the answer, but thanks to your post I see it.
So is your question from the original post resolved? I was confused about what you meant when you said you divided by 1,000,000 and multiplied by c to get the correct answer, since 2.04*10^-22 * 299792458 / 1,000,000 = 6.12 * 10^-20, which isn't even close to 0.384.
 

Related to Understanding the Units in Relativistic Momentum Calculations

What is a unit of momentum?

A unit of momentum is a measurement of the quantity of motion of an object. It is calculated by multiplying the mass of an object by its velocity.

What is relativistic momentum?

Relativistic momentum is a concept in physics that takes into account the effects of special relativity on the momentum of an object moving at high speeds. It is calculated using the equation p = mv/√(1-v^2/c^2), where p is the relativistic momentum, m is the mass of the object, v is its velocity, and c is the speed of light.

What are the units of relativistic momentum?

The units of relativistic momentum are kilogram meters per second (kg*m/s) in the SI system of units. In other systems of units, such as the CGS system, the units may be expressed as grams centimeters per second (g*cm/s).

What is the difference between relativistic momentum and classical momentum?

Relativistic momentum takes into account the effects of special relativity, such as time dilation and length contraction, on the momentum of an object moving at high speeds. This is in contrast to classical momentum, which assumes that time and space are absolute and do not change with an object's velocity.

How is relativistic momentum used in practical applications?

Relativistic momentum is used in many practical applications, including particle accelerators, nuclear reactors, and space travel. It is also an important concept in the field of astrophysics, where the high speeds and energies of celestial objects require the use of relativistic equations.

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