# Question about a general way to prove E^2 = (cp)^2 + (mc^2)^2

1. Aug 24, 2013

### lightarrow

For particles with non zero mass m we know that:

E = mc2γ (1)
P = mvγ (2)

with the usual meaning of the symbols.
From those two equations is easy to prove the equation:

E2 = (cp)2 + (mc2)2 (3)

which doesn't contain the factor γ any longer and so it can even be used for particles of speed = c as photons. But is this really correct? We started from equations (1) and (2) which *are not* applicable to v = c particles.

Is there a more general way to prove (3)?

--
lightarrow

2. Aug 24, 2013

### Staff: Mentor

You have to know something about massless particles in order to derive an equation for massless particles.
E=pc, for example, is sufficient - formula (3) follows as a direct consequence for m=0.
E=hf and $p=\frac{hf}{c}=\frac{h}{\lambda}$ can be used, too.

3. Aug 24, 2013

### Staff: Mentor

I think that the most general way to prove it is to show that P=(E,p) forms a four-vector. Then define m=|P| and (3) follows automatically.

4. Aug 24, 2013

Staff Emeritus
I would say the best thing to do is to start with (3) and use that to prove (1) and (2).

5. Aug 24, 2013

### lightarrow

Right, but then would it be correct to prove that a photon's mass is zero substituting E = cp in E2 = c2p2 + m2c4?
Since the last equation, strictly speaking, was derived only for m ≠ 0 particles. What do you think?

6. Aug 24, 2013

### lightarrow

Good. How I prove that (E,p) is a four-vector?

7. Aug 24, 2013

### Bill_K

I think both equations are true, and it's an idle exercise to worry about which one is derived from what. Similarly, in electromagnetism does ∇·B = 0 imply the existence of A, or is it the other way around??

8. Aug 24, 2013

### lightarrow

I know but mine it's not a strictly mathematical problem: I'm trying to follow a "physical path", starting from experimental results and SR.

9. Aug 24, 2013

### Staff: Mentor

SR, formulated with 4-vectors in a Minkowski space, is what DaleSpam suggested. You get all equations out of this formalism.

10. Aug 24, 2013

### WannabeNewton

$p\cdot p = 0$ for a photon so $-E^2 + |\vec{p}|^2 = 0$. For massive particles, $p\cdot p = -m^2$ so $E^2 = |\vec{p}|^2 + m^2$.

11. Aug 24, 2013

### pervect

Staff Emeritus
As I recall, there's a technique to get the relativistic momentum relationships by considering the kinematics of the collision of billiard balls - i.e. their velocities must transform according to the usual relativistic formulae, and this constraint is sufficient to give an expression for the momentum as a function of velocity. I think Goldstein has this.

Photons aren't billiard balls, but I suspect that by considering a pair of photons producicing a particle pair, you could derive a similar expression. I haven't tried to work out the detials though.

A more direct route than pair production might involve two photons bouncing off a small lightweight mirror, and insisting that the mirror's trajectory isn't modified, as an analogy to the "billiard balls".

Last edited: Aug 24, 2013
12. Aug 24, 2013

### Staff: Mentor

Simply show that it transforms correctly. I.e. show that $E'=\gamma (E-p_x v)$ and $p_x'=\gamma (p_x -vE)$

13. Aug 24, 2013

### Bill_K

Can I use (1) and (2)? :tongue2:

14. Aug 24, 2013

### robphy

In physics, sometimes when studying particular phenomena (e.g., interactions of massive particles) we get a glimpse (e.g., your (1) and (2) and your (3)-applied-only-to-massive-particles) of the full story. Then, one may be led to consider theoretical-extensions ( (3) to the zero-mass case)... and checking it with experiment.

Now... thinking about it a little more... ( seeking a "physical path" to (3) )...
Here could be another alternative view incorporating (1)-and-(2) and (3).
At face value,
(1) is a plot of E as a function of m and v
(2) is a plot of p as a function of m and v
and you notice that (3) algebraically results from (1) and (2) and the definition [and elimination] of gamma.
Apparently, (1) and (2) don't apply to the v=c case because gamma will be infinite,
which forces m=0 in order to somehow make E and p finite for a typical photon. ( The applicability of (3) as a constraint isn't so obvious.)

So, instead of (1) and (2) consider
pc/E=v/c ( by taking (2)/(1) )
.... in rapidity terms, this says (mc^2 sinhQ)/(mc^2 cosh Q)=tanh Q.
The Euclidean analogue is that the slope is the ratio of the components of a vector.
In particular, if the slope is 1 or -1, the magnitudes of the components are equal.

Practically speaking, consider experiments for various choices of m and v and plot their data on a pc vs E graph... [assuming you know how to get p and E from the experimental data]*.
You'll notice that variable-m constant-v experiments will trace out [a certain subset of] radial lines on the pE-plane.
You'll notice that constant-m variable-v experiments will trace out [a certain subset of] hyperbolas on the pE-plane (3)... including the asymptotic hyperbola (lines with slopes 1 and -1) corresponding to the zero-mass case.

*I think the tricky point for motivation is getting the expression for the relativistic momentum.

Last edited: Aug 24, 2013
15. Aug 24, 2013

### Staff: Mentor

Also note that starting from Maxwell's equations, one can show that the energy and (magnitude of) momentum densities in an electromagnetic wave are related by E = pc.

16. Aug 25, 2013

### lightarrow

Thanks to everyone for the answers.

--
lightarrow

17. Aug 25, 2013

### rbj

it's how i learned that $m = \gamma m_0$ (the now frowned-upon anachronism). but it required keeping the $p=mv$ relation.

18. Aug 26, 2013

### kg4pae

The proof depends on the fact that
$$\gamma^2=\frac{1}{1-\frac{v^2}{c^2}}$$
which in this context is true. Implicitly we are assuming all the caveats of SR and (for this proof)
$$m\ne 0$$

It is possible to relax the requirement that
$$m\ne 0$$
by proceeding from a topological argument using the invariance of distances in space-time. You would still have the caveats of SR and so the same $$\gamma$$.

Try to substitute the first expression for $$\gamma$$ into (1) and (2) and then formulate the right-hand side of (3). Simplify and see what you get.

For the more general way, using the topological invariance of distances in space-time we start with
$$p_\mu p^\mu=p_0 p^0+\vec{p}\cdot\vec{p}=-m^2 c^2$$
and get
$$-(p^0)^2+|\vec{p}|^2=-m^2 c^2$$
or
$$p^2+m^2 c^2=(p^0)^2$$
This only leaves the interpretation of $$p^0$$. Based on units and consistency with situations where
$$m\ne 0$$
we take
$$p^0=E/c$$.

Last edited: Aug 26, 2013
19. Aug 26, 2013

### kg4pae

This follows from a math theorem that says for any vector A the following is true:
$$\mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{A})=0$$
So if $$\mathbf{\nabla} \cdot \mathbf{B}=0$$ then there must exist an A such that
$$\mathbf{B}=\mathbf{\nabla} \times \mathbf{A}$$

Last edited: Aug 26, 2013