Relativistic calculation of speed when the momentum is known

AI Thread Summary
The discussion focuses on calculating the speed of an electron given its momentum, which is 1.48 x 10^-21. The initial approach using classical mechanics results in a speed exceeding the speed of light, necessitating a relativistic calculation. The correct formula for relativistic speed is derived as v = (p*c) / √(m^2*c^2 + p^2). Participants explore the transformation of the momentum equation and the steps to isolate v, emphasizing the importance of proper unit handling. The conversation highlights the need for clarity in deriving the relativistic speed formula from momentum.
Karagoz
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Homework Statement


We know the momentum of an electron, which is: 1,48*10^-21.

Momentum is m*v (mass*speed)

If we divide the momentum by the mass of the electron to find electron's speed, it'll give a value where v> 3*10^8 m/s.

Since speed can't be above speed of light, we have to calculate it relativistic to find the speed of light of the electron.

Homework Equations



upload_2018-5-2_17-39-35.png

gives:
upload_2018-5-2_17-39-50.png


The Attempt at a Solution



With the formula above, the problem is easy to solve.

But I don't get how that formula is transformed.

When I try it, this is what I get:

p = mv/√(1 - v^2/c^2)

divide by m, and multiply by √(1 - v^2/c^2)

v = p/m * √(1 - v^2/c^2)

^2 both sides

v^2 = p^2/m^2 * (1 - v^2/c^2) = p^2/m^2 - (p^2*v^2)/(c^2*m^2)

making some changes so both have same divsor:

v^2 = (p^2*c^2)/(m^2*c^2) - (p^2*v^2)/(m^2*c^2)

How do they get the equation where:

v = (p*c) / √(p^2 + m^2*c^2)

??
 

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Karagoz said:
v^2 = (p^2*c^2)/(m^2*c^2) - (p^2*v^2)/(m^2*c^2)
Collect together the terms that contain v2 and factor out the v2.
(The toolbar has a superscript tool.)
 
Start with your first Relevant equation and solve for v! Collect all the terms with v on one side, square everything to get rid of the square root, and solve for v2
 
Karagoz said:
We know the momentum of an electron, which is: 1,48*10^-21.
Also, remember that numbers are useless without appropriate units.
 
p = mv/√(1 - v^2/c^2)

p = mv/√(1 - v^2/c^2)

divide by m, and multiply by √(1 - v^2/c^2)

v = p/m * √(1 - v^2/c^2)

^2 both sides

v^2 = p^2/m^2 * (1 - v^2/c^2) = p^2/m^2 - (p^2*v^2)/(c^2*m^2)

making some changes on the right side so both have same divsor:

v^2 = (p^2*c^2)/(m^2*c^2) - (p^2*v^2)/(m^2*c^2)

This was where I left:

v^2 = (p^2*c^2)/(m^2*c^2) - (p^2*v^2)/(m^2*c^2)

Add + (p^2*v^2)/(m^2*c^2) on both sides.

v^2 + (p^2*v^2)/(m^2*c^2) = (p^2*c^2)/(m^2*c^2)

Get a common divisor on left side.

(v^2*m^2*c^2)/(m^2*c^2) + (p^2*v^2)/(m^2*c^2)

Simplify the left side

v^2(m^2*c^2 + p^2) / (m^2*c^2) = (p^2*c^2) / (m^2*c^2)

Multiply both sides by: (m^2*c^2)

v^2(m^2*c^2 + p^2) = (p^2*c^2)

divide by (m^2*c^2 + p^2)

v^2 = (p^2*c^2)/(m^2*c^2 + p^2)

SQROOT both sides:

v = (p*c) / √(m^2*c^2 + p^2)

v = (p*c) / √(p^2 + m^2*c^2)Do you know a more simplified way?
 
Last edited:
p/m = v / √(1 - v2 / c2)
p2 / m2 = v2 / (1 - v2 / c2) = 1 / (1/v2 - 1/c2)
(1/v2 - 1/c2) = m2 / p2
1/v2 = (m2 / p2) + (1/c2))
That separates the v to the left. Now take the lcm on the right hand side, invert, and take the square root.
 
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