What does this symbol represent?

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SUMMARY

The discussion centers on the interpretation of the symbol "mc" in the kinetic energy equation for a particle moving at relativistic speeds. The correct interpretation is that "mc" is a typographical error and should be "m0," representing the rest mass of the particle. The equation discussed is T = (p²c² + m₀²c⁴)¹/² - m₀c², where T denotes kinetic energy, p is momentum, c is the speed of light, and m₀ is the rest mass. The confusion arises from the misrepresentation of the rest mass in the equation.

PREREQUISITES
  • Understanding of relativistic physics concepts
  • Familiarity with the equation for kinetic energy in relativistic contexts
  • Knowledge of momentum and its relation to energy
  • Basic grasp of the speed of light as a constant in physics
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  • Study the derivation of the relativistic kinetic energy equation
  • Learn about the significance of rest mass (m₀) in relativistic physics
  • Explore the implications of momentum in relativistic contexts
  • Review common typographical errors in scientific equations and their impact on understanding
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This discussion is beneficial for students in nuclear engineering, physics enthusiasts, and anyone studying the principles of relativity and kinetic energy equations.

Ian Baughman
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In my nuclear engineering class we are looking at relativity right now. For one of our homework problems we have to derive an equation for a particle moving at relativistic speed showing the kinetic energy in terms of the particles momentum. In the answer they have a term mc and I have no idea what this represents. I can't find it in the book or online. I was wondering if somebody maybe knew what it meant?
 
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Without proper context it is not possible to answer your question. Are we supposed to guess what the answer looks like and where this symbol appears?
 
The answer is:
T = (p2c2+mo2c4)1/2-mcc2
I know:
p = momentum
T = Kinetic Energy
c = speed of light
mo = Rest mass​

I'm just confused on what mc represents.
 
I think that's just a typographical error, and ##m_c## should really be ##m_0## (the "rest mass", which also appears inside the parentheses). Or even a mechanical printing flaw that turned an "0" into a "c".
 
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Based on the expression, it has to be m0. The expression subtracts the rest energy from the total energy.
 
Ian Baughman said:
The answer is:
T = (p2c2+mo2c4)1/2-mcc2I know:
p = momentum
T = Kinetic Energy
c = speed of light
mo = Rest mass
I'm just confused on what mc represents.

As everyone mentioned above, this is a typo and was meant to be m0 (rest mass).
On the LHS of the equation you have T which is the "kinetic energy"... the kinetic energy would be equal to the total energy of your particle (E) minus the rest mass energy it has even when it's at rest (the m0).
 

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