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Abstract

     The value of the kinetic energy of a moving body of momentum  P  is not the same value in Special Relativity (SRT) as it is in Newtonian Mechanics (NM).  This paper examines, explains, and corrects this dilemma using two methods.

Kinetic Energy in SRT and NM

     Kinetic energy of a freely moving body, which will be denoted by  KS in SRT and  KN  in NM, is the product of the force  F applied times the distance  x´ over which it is applied, or

K = F x´         1         

Net applied force results in change in momentum  P .

F = dP/dt         2

So

KS = (dP/dt)x´         3

in SRT.  Also, NM kinetic energy must equal

KN = (dP/dt)x´         4

     In SRT

I2 = (ct)2 – x2 = (ct´)2 – x´2 = …         5

where  I is the Relativistic Interval Equation which is the same in all reference frames for any two events,  c is the speed of light, and  t in these examples is the time during which the light originating in the non-moving reference frame at  x=0  has traveled the distance  I  at speed  c in that frame.   The  x´  is the distance an object or reference frame has moved perpendicularly while the light has moved the distance  I .   The  x=0  when considering the light rays moving perpendicular to the direction of motion of  x´ .   Then the formula looks like

I2 = (ct)2 = (ct´)2 – x´2 = (ct´´)2 – x´´2 …         6

The  x (not  x´ )  in these examples will not change because it is the non-moving reference frame,  x=0 .

     Speed is distance per unit time.  If you know the distance an object traveled,  x´ ,  the distance a timing signal (light) traveled,  I ,  and the speed,  c , of that timing signal, then NM velocity,  V ,  would be

V = (x´/I)c = (x´/ct)c = x´/t         7

     Momentum  P in NM is

P = mV = m(x´/t) = m(x´/I)c         8

According to SRT

x´= γ(x+vt)         9

t´ = γ(t + (v/c2)x)         10

so for  x=0 ,  in which case  I=ct ,

x´= γvt = γ(v/c)ct = γ(v/c)I         11

t´ =γt         12

where  x´  is the distance the moving frame traveled while the light traveled the distance  I  at speed  c ,  γ is the gamma function,  γ=1/(1 – (v/c)2)1/2 , t´  is calculated from  (I/c)  according to the SRT Eq. 12, and  v is the SRT velocity (not the same as NM velocity  V ), which is  x´/t´  by dividing Eq. 11 by Eq. 12 or

x´/t´ = γvt/γt = v         13

By Eq. 11

x´/I = γ(v/c)         14

and momentum  P in SRT has the same value as in NM, namely

P = m(x´/I)c = mγ(v/c)c = mγv         15

using Eqs. 8 and 14.  Since in NM  P=mV ,  then

V = γv         16

x´/I = (γv)/c = γ(v/c) = (P/m)/c         17

meaning the ratio of the distance the object traveled to the distance the light traveled is  (P/m)/c .  Therefore, distance traveled  x´ would be

x´ = (P/m)I/c = (1/m)Pct/c = (1/m)Pt    .         18

However,  P is varying at the rate  (dP/dt) = F = a constant.   P at any given time is

P = (dP/dt)t         19

Using Eqs. 18 and 19

dx´ = (1/m)(dP/dt)t dt         20

x´ = (1/m)(dP/dt) ∫t dt         21

x´ = (1/2m)(dP/dt) t2        22

and since  K=Fx´ ,  from Eqs. 3 and 22 one gets

KS = (dP/dt) x´ = (1/2m) (dP/dt)2 t2         23       

which is

KS = (1/2m) (m d(γv)/dt)2 t2 = ½ m(γv)2         24

in SRT.  In NM it is

KN = (1/2m) (m dV/dt)2 t2 = ½ mV2         25

Another Method

     The usual formula given for  KS in SRT is

KS = mc2 (γ–1)    .         26

This is derived from

dK = F dx´ = (dP/dt) dx´ = (dx´/dt) dP    .         27

The correct  P=mγv was used for the  P of  dP .   However, the  (dx´/dt)  which equals  P/m , the correct form and the same as NM velocity, is not what was used for the integration.  Instead,  P/γm was used for  (dx´/dt)  which is  v or  (dx´/dt´ ) ,  where the denominator is the dilated time, was used.  Thus the integration result of  mc2(γ–1) for KS .

     If calculated correctly by this  “(dx´/dt) dP”  method of Eq. 27, it looks like this.

dKS = F dx´ = (dP/dt) dx´ = (dx´/dt) dP = (1/m) P dP         28

KS = (1/m) ∫PdP         29

Therefore, by the method used to calculate the  “(γ-1)”  version (Eq. 27) for  KS ,  when done properly using  (1/m)P instead of  (1/m)(P/γ)=v ,  the result is

KS = (1/2m) P2 = ½ m(γv)2 = ½ mc2(γ(v/c))2         30

just as found above in Eq 24.  Which, by the way, since P=mV in NM, is

KN = (1/2m) P2 = ½ mV2 = ½ mc2(V/c)2 = ½ mc2(γ(v/c))2         31

Conclusion

     The correct equation for kinetic energy in Special Relativity (SRT) is

KS = ½ m(γv)2 = ½ mc2(γ(v/c))2 = ½ mc2(V/c)2

equal to the Newtonian Mechanics (NM) value

KN =½ mV2 = ½ mc2(V/c)2

The difference between the old KS=mc2(γ–1)  and the correct  KS= ½ m(γv)2 = ½ mc2(γ(v/c))2  is < 1%  for

(v/c) < 0.2 .