What is derivation of the equation of total energy?

[DimaMazin]

E²=m²c⁴+p²c²

What is derivation of the equation?

[swansont]

Start with [math]E = {\gamma}mc^2[/math]

 

expand gamma, equating it to the right-hand side of the equation. Multiply through by the denominator

 

[math]{\gamma}m^2c^4-{\gamma}m^2v^2c^2 = m^2c^4[/math]

 

 

move the negative term to the other side and note that

 

[math]{\gamma}^2m^2v^2 = p^2[/math]

 

Left-hand side is E² and the right-hand side is m²c⁴+p²c²

Edited September 19, 2014 by swansont
fix typo

[DimaMazin]

Start with [math]E = {\gamma}mc^2[/math]

 

expand gamma, equating it to the right-hand side of the equation. Multiply through by the denominator

 

[math]{\gamma}m^2c^4-{\gamma}m^2v^2c^2 = m^2c^4[/math]

 

 

move the negative term to the other side and note that

 

[math]{\gamma}m^2v^2 = p^2[/math]

 

Left-hand side is E² and the right-hand side is m²c⁴+p²c²

I have understood only the start.

I think [math]{\gamma}^2m^2v^2 = p^2[/math]

[swansont]

I have understood only the start.

I think [math]{\gamma}^2m^2v^2 = p^2[/math]

 

Yes, typo.

 

It's algebra.

 

If you really need to see the two steps I skipped

http://galileo.phys.virginia.edu/classes/252/energy_p_reln.html

 

But note he uses relativistic mass in some of the equations. However, his labeling is clear.

[studiot]

I had a bad latex day recently so I have sketched the derivation out longhand.

You should have enough to fill in the arithmetic, ask if there are any steps you don't follow.

 

[ajb]

The question must now be where does [math]E = \gamma m c^{2}[/math] come from?

 

Another way to get at the equation you seek is to consider the world line action of a free relativistic particle. Depending on how you formulate this, there are several ways, you see that [math]E^{2}-p^{2}= m^{2}[/math] comes out as an equation of motion.

 

That is all classical physical particles must obey this relation, including massless particles. That why we can talk about on-shell and off-shell in quantum field theory. This means if the classical equations of motion are imposed or not. For example, people will talk about virtual particle being off-shell for exactly this reason; they are not constrained to obey the classical equations of motion.

Edited September 19, 2014 by ajb

[DimaMazin]

 

Yes, typo.

 

It's algebra.

 

If you really need to see the two steps I skipped

http://galileo.phys.virginia.edu/classes/252/energy_p_reln.html

 

But note he uses relativistic mass in some of the equations. However, his labeling is clear.

Thank you. Now I understand it all.

[elfmotat]

The easiest way would be to use four-vectors. The four-momentum is:

 

[math]p^\mu = (E/c, \vec{p})[/math]

 

This can be found easily by simply defining four-momentum as mass times four-velocity, where four-velocity is the derivative of position [math]x^\mu =(ct,\vec{x})[/math] with respect to proper time. We know that the magnitude of the four-momentum must be constant for every observer:

 

[math]p_\mu p^\mu = const.[/math]

 

If we consider the frame where a particle is at rest, its momentum becomes:

 

[math]p^\mu = (E/c,0)[/math]

 

But when a particle is at rest its energy is just its rest energy, which is equal to its mass:

 

[math]p^\mu = (mc,0)[/math]

 

Its magnitude squared is:

 

[math]p_\mu p^\mu = m^2 c^2[/math]

 

In some other general reference frame we have:

 

[math]p_\mu p^\mu = E^2/c^2 - |\vec{p}|^2[/math]

 

But as I said these must be equal, so:

 

[math]E^2/c^2 - |\vec{p}|^2= m^2 c^2[/math]

 

You can do a little rearranging to get it into the more familiar form:

 

[math]E^2 = m^2 c^4 + |\vec{p}|^2 c^2[/math]

Edited September 19, 2014 by elfmotat

[DimaMazin]

you see that [math]E^{2}-p^{2}= m^{2}[/math] comes out as an equation of motion.

 

I see different units, how can they work in the equation?

kg²m⁴/s⁴ - kg²m²/s² = kg²

[xyzt]

I see different units, how can they work in the equation?

kg²m⁴/s⁴ - kg²m²/s² = kg²

Try again , with this:

 

[math]E^2-(pc)^2=(mc^2)^2[/math]

 

It is standard practice to set c=1.

[ajb]

Indeed, for notational simplicity I set c = 1. It is very common to do this, but it can be confusing especially if you want to view c as a parameter that sets some scales in the theory.