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Relativistic Mass and Taylor Series: A Calculus Problem with Physics Applications, Exams of Relativity Theory

The formula for relativistic mass and its Taylor series. It explains the practical significance of the interval of convergence of the series. a math problem solution from the University of Michigan Department of Mathematics.

Typology: Exams

Pre 2010

Uploaded on 05/11/2023

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9. (11 points) The theory of relativity predicts that when an object moves at speeds close to the
speed of light, the object appears heavier. The apparent, or relativistic, mass m, of the ob ject
when it is moving at speed vis given by the formula
m=m01v2
c21/2
where cis the speed of light and m0is the mass of the object when it is at rest.
(a) (8 points) Write the first four nonzero terms of the Taylor series for min terms of v. (Hint: You
may want to use the binomial series.)
Since mis given by
m=m01v2
c21/2
,
and 0 v < c, we may use the binomial series
(1 + x)p= 1 + px +p(p1)
2! x2+p(p1)(p2)
3! x3+· · · ,
with x=v2/c2and p=1/2, to obtain:
m=m01 + 1
2c2v2+3
8c4v4+5
16c6v6+· · · .
(b) (3 points) The series you derived in part (a) converges for vin the interval [0, c). Interpret the
practical significance of this interval of convergence in the context of this problem (that is, as far
as the relativistic mass of an object is concerned.)
When an object’s speed (v) is less than the speed of light (c) the relativistic mass (m) of the
object is finite. In this case, the Taylor series for the relativistic mass (derived above) converges
to the actual relativistic mass.
University of Michigan Department of Mathematics Fall, 2005 Math 116 Exam 2 Problem 9 (relativity) Solution

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  1. (11 points) The theory of relativity predicts that when an object moves at speeds close to the speed of light, the object appears heavier. The apparent, or relativistic, mass m, of the object when it is moving at speed v is given by the formula

m = m 0

v^2 c^2

where c is the speed of light and m 0 is the mass of the object when it is at rest.

(a) (8 points) Write the first four nonzero terms of the Taylor series for m in terms of v. (Hint: You may want to use the binomial series.)

Since m is given by

m = m 0

v^2 c^2

and 0 ≤ v < c, we may use the binomial series

(1 + x)p^ = 1 + px + p(p − 1) 2! x^2 + p(p − 1)(p − 2) 3! x^3 + · · · ,

with x = −v^2 /c^2 and p = − 1 /2, to obtain:

m = m 0

2 c^2 v^2 +

8 c^4 v^4 +

16 c^6 v^6 + · · ·

(b) (3 points) The series you derived in part (a) converges for v in the interval [0, c). Interpret the practical significance of this interval of convergence in the context of this problem (that is, as far as the relativistic mass of an object is concerned.)

When an object’s speed (v) is less than the speed of light (c) the relativistic mass (m) of the object is finite. In this case, the Taylor series for the relativistic mass (derived above) converges to the actual relativistic mass.

University of Michigan Department of Mathematics Fall, 2005 Math 116 Exam 2 Problem 9 (relativity) Solution