Another way to look at e = mc squared

  • 11 Replies
  • 1516 Views

0 Members and 1 Guest are viewing this topic.

*

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • 4189
  • The graviton sucks
    • View Profile
Another way to look at e = mc squared
« on: 18/06/2015 01:46:38 »
Momentum is derived by the equation p = mv, where p is the momentum, m is the mass and v its velocity. Since the maximum velocity of anything in the universe is c then the potential maximum momentum is p = mc. If we then integrate this we find that

m[tex]\int \frac{1}{2} c^2\, dc[/tex] + C

equates to kinetic energy. In this case equaling (1/2)mc^2 + C. This is understandable since the potential maximum kinetic energy for the mass has been derived from the potential maximum momentum. If we disregard C then the kinetic energy is half the total mass energy.

Kinetic energy is in its proper form as ke = (1/2)mv^2. If we then say e = mv^2 what would this mean?

EDIT: corrected to "If we disregard C then the kinetic energy is half the total mass energy."
« Last Edit: 19/06/2015 13:34:18 by jeffreyH »
Fixation on the Einstein papers is a good definition of OCD.

*

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • 4189
  • The graviton sucks
    • View Profile
Re: Another way to look at e = mc squared
« Reply #1 on: 18/06/2015 16:24:08 »
Taking m0 as rest mass and mrel as relativistic mass then solving [tex]m_0c^2 = m_{rel}v^2[/tex] may provide some interesting insights. As the relativistic mass increases would we be able to find a solution before reaching v = c?
Fixation on the Einstein papers is a good definition of OCD.

*

Offline PmbPhy

  • Neilep Level Member
  • ******
  • 2804
    • View Profile
Re: Another way to look at e = mc squared
« Reply #2 on: 18/06/2015 17:44:34 »
Momentum is derived by the equation p = mv, where p is the momentum, m is the mass and v its velocity. Since the maximum velocity of anything in the universe is c then the potential maximum momentum is p = mc. If we then integrate this we find that

m[tex]\int \frac{1}{2} c^2\, dc[/tex] + C

equates to kinetic energy. In this case equaling (1/2)mc^2 + C. This is understandable since the potential maximum kinetic energy for the mass has been derived from the potential maximum momentum. If we disregard C then the kinetic energy is half the systems total energy.

Kinetic energy is in its proper form as ke = (1/2)mv^2. If we then say e = mv^2 what would this mean?
It doesn't mean anything. Just because you can do something mathematically it doesn't mean that there's any physical connection to it.

*

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • 4189
  • The graviton sucks
    • View Profile
Re: Another way to look at e = mc squared
« Reply #3 on: 19/06/2015 13:30:33 »
Momentum is derived by the equation p = mv, where p is the momentum, m is the mass and v its velocity. Since the maximum velocity of anything in the universe is c then the potential maximum momentum is p = mc. If we then integrate this we find that

m[tex]\int \frac{1}{2} c^2\, dc[/tex] + C

equates to kinetic energy. In this case equaling (1/2)mc^2 + C. This is understandable since the potential maximum kinetic energy for the mass has been derived from the potential maximum momentum. If we disregard C then the kinetic energy is half the systems total energy.

Kinetic energy is in its proper form as ke = (1/2)mv^2. If we then say e = mv^2 what would this mean?
It doesn't mean anything. Just because you can do something mathematically it doesn't mean that there's any physical connection to it.

It is a meaningless quantity and the question was rhetorical. Sorry for the confusion.
Fixation on the Einstein papers is a good definition of OCD.

*

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • 4189
  • The graviton sucks
    • View Profile
Re: Another way to look at e = mc squared
« Reply #4 on: 20/06/2015 15:07:33 »
We can take the equation

[tex]m_0 c^2 = \frac {1} {\sqrt{1 - \frac{v^2}{c^2}}} m_0 v^2[/tex]

and derive

[tex]c^2 \sqrt{1 - \frac{v^2}{c^2}} = v^2[/tex].

From this it can be seen that the velocity on the left side of the equation cannot be the same as the result. So the result then must be a coordinate value for velocity. This also shows that if v = c then the coordinate value of v must be invalid because of the square root term. So coordinate v tends towards zero as proper v approaches c.
Fixation on the Einstein papers is a good definition of OCD.

*

Offline PmbPhy

  • Neilep Level Member
  • ******
  • 2804
    • View Profile
Re: Another way to look at e = mc squared
« Reply #5 on: 20/06/2015 17:42:05 »
Quote from: jeffreyH
We can take the equation

[tex]m_0 c^2 = \frac {1} {\sqrt{1 - \frac{v^2}{c^2}}} m_0 v^2[/tex]
You can't just write down equations Jeff. I explained that there has to be a physical meaning to them. In this case your equation is wrong. The left side of your equation is a constant whereas the right side isn't, it depends on v. Therefore that equation is wrong to begin with.

I'm surprised at you Jeff. I'm not used to seeing you make serious mistakes like this.

*

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • 4189
  • The graviton sucks
    • View Profile
Re: Another way to look at e = mc squared
« Reply #6 on: 20/06/2015 17:51:06 »
You haven't seen where I am going yet. This is totally a unbalanced and invalid equation. There is a reason for it.
Fixation on the Einstein papers is a good definition of OCD.

*

Offline PmbPhy

  • Neilep Level Member
  • ******
  • 2804
    • View Profile
Re: Another way to look at e = mc squared
« Reply #7 on: 20/06/2015 18:04:33 »
Quote from: jeffreyH
You haven't seen where I am going yet. This is totally a unbalanced and invalid equation. There is a reason for it.
It's just plain bad juju to not explain it all in one post and not give the reason why you're posting nonsense.

*

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • 4189
  • The graviton sucks
    • View Profile
Re: Another way to look at e = mc squared
« Reply #8 on: 20/06/2015 19:23:53 »
Well in the equation

[tex]c^2 \sqrt{1 - \frac{v^2}{c^2}} = v^2[/tex]

The mass terms have been removed. The velocities cannot be equivalent. If we set the gamma velocity to be escape velocity and the result to be a coordinate velocity

[tex]c^2 \sqrt{1 - \frac{v_e^2}{c^2}} = v_c^2[/tex]

then taking the square root of the result gives the attached plot. The plot assumes a Schwarzschild solution. For a photon emitted exactly at the horizon and moving perpendicular to the horizon the physics breaks down. For the Kerr solution frame dragging will impart an initial velocity which will result in a geodesic path for the photon. At the poles of the Kerr solution we run into the same problem as with the Schawzscild solution. It is at the poles that we see relativistic jets. Could these result from photon collisions at these special points in the Kerr solution? A photon that would be stationary at that point would get a kick from a photons following a geodesic that crosses the pole.
Fixation on the Einstein papers is a good definition of OCD.

*

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • 4189
  • The graviton sucks
    • View Profile
Re: Another way to look at e = mc squared
« Reply #9 on: 25/06/2015 02:52:49 »
The graph below can be compared to that of light against escape velocity. The question to ask is what can this tell us about the environment when approaching an event horizon?

« Last Edit: 25/06/2015 02:55:32 by jeffreyH »
Fixation on the Einstein papers is a good definition of OCD.

*

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • 4189
  • The graviton sucks
    • View Profile
Re: Another way to look at e = mc squared
« Reply #10 on: 27/06/2015 08:41:59 »
An analysis of velocity is all well and good but what about kinetic energy. The first attached plot kinetic energy over time for a vertical projectile under the influence of earth gravity. The second is a comparison of non-relativistic kinetic energy against relativistic. Since the gravitational field does not act upon a body in the same way as any other external force, that is it acts upon all parts of a body at once, then it may be possible to consider only the proton and its kinetic energy as a stand in for mass. If we calculate kinetic energy at light speed for the proton what would we find?
Fixation on the Einstein papers is a good definition of OCD.

*

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • 4189
  • The graviton sucks
    • View Profile
Re: Another way to look at e = mc squared
« Reply #11 on: 27/06/2015 09:11:29 »
Attached is a plot of proton kinetic energy with both relativistic and non-relativistic comparison. Since this analysis is considering the proton as a stand in for any mass then it would appear that gravitation should find it much easier to accelerate objects to relativistic speed than other external forces. Since the per unit energy is spread out and no compression of the mass happens. In fact tidal forces should aid the process by increasing the separation of atoms within the mass. At which point we are considering individual particles. The question is, what difference is there inertially between the force of gravity and other forces?
Fixation on the Einstein papers is a good definition of OCD.