Limit as the Speed of Light becomes Infinite

[Crosson]

Many of the equations in Relativity reduce to their classical versions in the limit as the speed of light becomes infinite (Example: Hamiltons equations). So is it possible that the equation:

E = mc^2

is telling us that rest mass (particles) can't exist in a universe with an infinite upper speed limit? Does it tell us that there would be no physics in such a world? (You either have infinite energy or zero)?

 

[dextercioby]

Just one question from me:what were the assumptions made to reach \mbox{E}=\mbox{m}\mbox{c}^{2}...?

Daniel.

 

[Jameson]

I'm not an expert on this kind of physics, but I'm not sure I understand the question. The equation E = mc^2 was integrated from the equation:

m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}

if I'm not mistaken. How are you saying that a particle can only have infinite or zero energy? I just need a little clarification so maybe I can follow along.

The only way I see an inifinite mass if when the speed of the particle exceeds the speed of light, and if the particle is at rest then m = m_0. Where are you getting something undefined from E= mc^2?

Thanks,
Jameson

 

[dextercioby]

It can be shown both formulas (Einstein's and mass formula) can be derived from general principles...

Daniel.

 

[Crosson]

The equation E = mc^2 is a valid equation for massive particles at rest. It comes from the more general equation for relativistic energy:

E = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}

The question remains, why does this equation break down as c goes to infinity?

 

[dextercioby]

Because it was derived in the case c finite ...?

Daniel.

 

[pervect]

Many of the equations in Relativity reduce to their classical versions in the limit as the speed of light becomes infinite (Example: Hamiltons equations). So is it possible that the equation:

E = mc^2

is telling us that rest mass (particles) can't exist in a universe with an infinite upper speed limit? Does it tell us that there would be no physics in such a world? (You either have infinite energy or zero)?

It's tellingus that that's E=mc^2 is not the right formula for energy when the speed of light is infinite.

If you take the limit of the Lorentz transform as c->infinity

x' = gamma(x-vt)
t' = gamma(t-vx/c^2)

you get

gamma = 1/sqrt(1-v^2/c^2) ->1

thus

x' = x-vt
t' = t

which is the familiar galilean transform from Newtonian mechanics. Note that in Newtonian mechanics E=.5*m*v^2, as opposed to the relativistic mechanics

E^2 - (pc)^2 = m^2c^4

which is what gives E=mc^2 when p=0.

 

[Crosson]

My point is that many derived equations of relativity (such as Hamiltons equations, relativistic momentum, schwarschild radius) reduce to classical (and more importnat, sensible) expressions in the limit c goes to infinity. It is standard to see this sort of thing done in textbooks.

Why ignore a very important case in which the limit does not lead to an expected result? I think special relativity implies that 'rest mass' (whatever that is) could not exist in a universe with infinite speed limit.

Note that in Newtonian mechanics E=.5*m*v^2, as opposed to the relativistic mechanics

E^2 - (pc)^2 = m^2c^4

which is what gives E=mc^2 when p=0.

Note that I am familiar with elementary relativity, and that the equation you are showing me is exactly the equation that begs the question I am asking.

 

[Dragongod]

Crosson you are right. Pervect was agreeing with you. He said E=mc^2 is not the right formula. According to Einstein a photon can never be at rest mass because c=v. A photon doesn't exist if it has no mass. Also I was wondering if the speed of light is really infinite. If the speed of light is infinite then how come we say how fast it is such as it goes around the world 7 times in one second. If it was really infinite wouldn't we not be able to give it a numberical value?

 

[pervect]

My point is that many derived equations of relativity (such as Hamiltons equations, relativistic momentum, schwarschild radius) reduce to classical (and more importnat, sensible) expressions in the limit c goes to infinity. It is standard to see this sort of thing done in textbooks.

OK, I agree with this, and I pointed out that it's related to the way that the Lorentz transforms reduce to the Gallilean transforms when c->infinity, so there is a formal basis for saying that relativity reduces to classical mechanics when c-> infinity.

Why ignore a very important case in which the limit does not lead to an expected result? I think special relativity implies that 'rest mass' (whatever that is) could not exist in a universe with infinite speed limit.

I guess I don' follow your question. To put it very simplistically, my view is that the limit of 1/c as c-> infinty exists, but the limit of c as c-> infinity doesn't exist. (One might say informally that it's infinite, but mathematically there is no limit). So one does not expect everything to have a well defined limit as c-> infinity.

To expand on this further, the Lorentz interval x^2+y^2+z^2 - c t^2 only makes sense mathematically if c is finite. When c is infinte, there isn't any corresponding interpretation of this quantity. Thus, classical mechanics doesn't have any corresponding concept of the Lorentz interval. Space and time are totally separate entities in classical mechanics, they never mix.

The length of the Lorentz interval gives proper time when the 4 vector is space-time (x,y,z,t), and it gives the invariant mass when the 4-vector is the energy-momentum 4-vector (dx/dtau, dy/dtau, dz/dtau, dt/dtau). This yields the very basic equation that's the root of your question.

Since the former concept doesn't exist in classical mechanics (there is no Lorentz interval in classical mechanics), the later concept doesn't exist either (there is nothing equivalent to the invariant mass of the energy-momentum 4-vector).

Sorry if this sounds too basic.

 

[dextercioby]

E^{2}=m^{2}c^{4}+\vec{p}^{2}c^{2} is the essence of SR.Anything else is just secundary.Incidentally,it can be shown that a spinless relativistic free particle has the Hamiltonian

H(\vec{p})=\sqrt{\vec{p}^{2}c^{2}+m^{2}c^{4}}

Daniel.

 

[Crosson]

Thank both of you, Pervect and dextercolby. It seems that the bottom line is that the 4-momentum enters into the theory in a more fundamental than all of those derived quantities which come from the lorentz transform in one way or another. (indeed, we would obtain the LT by looking for a transform which left the 4-momentum invariant).