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Mass Gain in Relativity


Date: 06/06/99 at 09:25:58
From: Robert Holte
Subject: Relativity

I still have a problem with relativity. I understand how it dilates 
space-time, but do not understand the gain in mass at high velocities. 
How does this work? At what speeds do we begin to notice this effect? 
I understand that the gain in mass is on an infinite curve (if 
infinite energy were applied to an object in an attempt to propel an 
object above light speed the mass gain would also be infinite,) but 
what is the math behind this? To be more specific, at what speeds 
would an object experience a 400% gain in mass? ... or 650%?

Thank you.


Date: 06/07/99 at 22:00:40
From: Doctor Rick
Subject: Re: Relativity

Hello, Robert, thanks for writing!

It is better to speak of the "total energy" (or mass-energy) of an 
object and to regard its mass as a fixed property of the object (often 
called its "rest mass"). The total energy of a stationary object is 
equal to its rest mass. In order to accelerate the object, energy must 
be added to it ("kinetic energy" in Newtonian physics). In 
relativistic physics, the rest mass and kinetic energy are not 
separate quantities; due to the equivalence of mass and energy 
according to E = mc^2, the rest mass and kinetic energy go into one 
pot, the "total energy." Due again to the equivalence of mass and 
energy, this total energy can also be expressed as a mass, but let's 
not.

The formula for the total energy of an object with rest mass m and 
velocity v is

  E = mc^2/sqrt(1-(v/c)^2)
    = mc^2(1-(v/c)^2)^(-1/2)

When v/c << 1 (non-relativistic velocities), this is approximately 
equal to

  E ~= mc^2(1 + (v/c)^2/2)
     = mc^2 + (1/2)mv^2

which is just the sum of the rest energy (rest mass converted to 
energy) and the non-relativistic kinetic energy. So the increase in 
total energy is noticed at non-relativistic velocities, but it is 
simply the ordinary kinetic energy. 

The total energy begins to increase noticeably faster than this when 
(v/c)^4 becomes significant. For instance, at 1/10 the speed of light, 
this quantity is 1/10,000 and the effect would only be seen around the 
fourth decimal place. At 1/4 the speed of light, (v/c)^4 is 1/256 or 
about 0.004.

At what velocity will the total energy be increased by a factor "a" 
compared with the rest energy? Let's work it out.

      a = E/(mc^2) = ((1-(v/c)^2)^(-1/2)

   a^-2 = 1 - (v/c)^2

    v/c = sqrt(1 - a^-2)

For an energy-increase factor of 4, we need

  v/c = sqrt(1 - 1/16)
      = 0.968

That is, at a speed of 96.8% of the speed of light, the total energy 
of an object will be 4 times its rest energy. Looked at another way, 
in order to accelerate an object to 96.8% of the speed of light, you 
must impart an additional energy three times the rest energy. For 
comparison, in Newtonian mechanics, the kinetic energy at the speed of 
light would be (1/2)mc^2 - in other words, the energy needed to raise 
an object to the speed of light would be just 0.5 times its 
relativistic rest mass. Under relativity, to get to 96.8% of this 
speed, we needed 6 times as much added energy. 

And it gets worse. As you note, energy expenditures approaching 
infinity will only reduce the 3.2% gap, causing the velocity to 
approach closer to the speed of light. No amount of energy will bring 
the velocity above that of light.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Physics/Chemistry

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