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"E2 - p2c2 = m2c4" - Meaning of symbols

 
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Apr13-10, 01:40 AM   #1
 

"E2 - p2c2 = m2c4" - Meaning of symbols


What do the symbols mean in the equation-

E2 - p2c2 = m2c4


I know this is so basic, but I am really confused about what all are rest parameters here, and what all involve Lorentz factor...
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Apr13-10, 01:47 AM   #2
 
c = speed of light
m = rest mass of the object
E = object's total energy
p = object's momentum

Eugene.
Apr13-10, 02:07 AM   #3
 
so m is rest mass, while p is momentum with lorentz factor?
Apr13-10, 02:12 AM   #4
 

"E2 - p2c2 = m2c4" - Meaning of symbols


Quote by particlemania View Post
so m is rest mass, while p is momentum with lorentz factor?
That's right. Relativistic momentum is related to the velocity via Lorentz factor

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

Eugene.
Apr13-10, 02:16 AM   #5
 
That means among all variables there (except E) only p has a lorentz factor.


Thanks a lot!
Apr13-10, 07:27 AM   #6
 
No E has the Lorentz factor too:
[tex]E = \frac{mc^2}{\sqrt{1-v^2/c^2}}[/tex]
So you can see that even though E and p depend on the velocity, when you take the difference of their squares, the result is independent of the velocity. The result - [itex]m^2 c^4[/itex] - is said to be invariant (same for all observers).
Apr13-10, 02:11 PM   #7
 
Well the point that E has lorentz factor too was quite obvious if p had and m didnt.
Apr13-10, 02:24 PM   #8
 
Mentor
Note that the formulas in #4 and #6 only hold for massive particles, but the one in #1 holds for massless particles too.
Apr13-10, 05:33 PM   #9
 
@Fredrik : So for massless particles, if my third term becomes 0

then momentum should be

[tex]p = \frac{h}{\lambda}[/tex]

isn't it?
Apr13-10, 05:53 PM   #10
 
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Quote by particlemania View Post
@Fredrik : So for massless particles, if my third term becomes 0

then momentum should be

[tex]p = \frac{h}{\lambda}[/tex]

isn't it?
E=hf
p=E/c
Apr13-10, 07:12 PM   #11
 
Quote by starthaus View Post
E=hf
p=E/c
c=f[tex]\lambda[/tex] completes the thought
Apr17-10, 07:44 PM   #12
 
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As the meaning of symbols has been exhaustively elucidated, I wish to respond to your remark about the importance of this relativist energy equation. You are absolutely right, the equation E2 = m2c4 + p2c4 is one of the most basics in physics. It demonstrates that energy as a whole consists of positive and negative energies.

Everything in nature exists in pairing, the energy is no exception. We may say that this pairing of energy is the prerequisite of the creation (and annihilation). As I stated in my reply to ZirkMan’s “What is spacetime made of?” energy is the only independent reality in nature (the spacetime is merely its structural quality) from which everything else is derived. How?

Allow me to explain a little bit further. Energy as a whole tends to break its symmetry. Eventually it splits into its two opposite elements: the positive and negative energies. As such, the spacetime becomes polarized. When it happens, a hypersurface or more precisely hyper-interface is raising between the two, just like the interface of oil-water system. The 3-space or, in a more technical term, the 3-brane is thus created. I think the study of the brane should be done in this direction.
Apr17-10, 07:53 PM   #13
 
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Quote by aeon.rs View Post
As the meaning of symbols has been exhaustively elucidated, I wish to respond to your remark about the importance of this relativist energy equation. You are absolutely right, the equation E2 = m2c4 + p2c4 is one of the most basics in physics. It demonstrates that energy as a whole consists of positive and negative energies.
How do you figure? Negative energy would only be present with exotic matter, this may be possible in some real-world situations like the Casimir effect but it's an unusual phenomenon, not really implied by that equation.
Apr18-10, 12:13 AM   #14
 
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Quote by JesseM View Post
How do you figure? Negative energy would only be present with exotic matter, this may be possible in some real-world situations like the Casimir effect but it's an unusual phenomenon, not really implied by that equation.
Roger Penrose in his book [“The Road to Reality”, P.614, Vintage Book, London, 2005] elaborates the difficulties with this relativistic energy equation quiet thoroughly. Let me quote and summarize his view. The square root of the expression (m2c4+p2c2)1/2 indeed creates difficulties because it contains an implicit sign ambiguity.

In quantum mechanics the two square roots are complex number and therefore do not tend to separate neatly into positive and negative in a consistent way. In quantum mechanics, however, each of two roots has to be considered as a possibility, so even an unphysical negative energy has to be considered as a physical possibility.

The relativistic expression (m2c4+p2c2)1/2 is, however, more problematic in that we do not normally have a clear-cut procedure for ruling out negative square root.

Paul Dirac found a way to resolve this problem. When he convinced that the negative frequency solutions could not be mathematically eliminated, he put forward an ingenious proposal which got rid of the negative energies, their effect being taken over by introducing the idea of antiparticles and what is called “Dirac sea” of negative energy.

Now, Dirac sea of negative energy represents only the half of the reality. The give the whole picture, as I presented previously, there should be both oceans of positive and negative energies where the 3-interface, our 3-space, is arising in between.
Apr18-10, 12:43 AM   #15
 
"...there should be both oceans of positive and negative energies..."

What's an ocean? Is this a physics term?
Apr18-10, 12:52 AM   #16
 
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Quote by Phrak View Post
"...there should be both oceans of positive and negative energies..."

What's an ocean? Is this a physics term?
aeon.rs referred to the Dirac sea, an idea from quantum field theory.
Apr18-10, 02:35 AM   #17
 
Quote by JesseM View Post
aeon.rs referred to the Dirac sea, an idea from quantum field theory.
hmm.. His oceans seem to be a couple of 3 dimensional hypersurface somewhere else-when.
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