# Energy-mass equivalence simplification of equations?

1. Aug 23, 2013

### blokpoi

Hi all,

Firstly, I do apologize if this is in the wrong forum or sub-forum, and if the title is a tad misleading, I wasn't quite sure how do phrase it.

Onto the post, however, I am sure most if not all of you have seen the energy mass equivalence, $e=mc^2$

And there is also Einstein's original equation $E^2 = p^2c^2 + m^2c^4$ (I do hope this is the right one, I have a lot of equations lying around, and not to mention I have seen a lot of his "original equation" which are different from this one. Please let me know if this is the wrong one.)

Now my question is, how do you get from $E^2 = p^2c^2 + m^2c^4$ to $e=mc^2$?

If somebody could explain it for me that would be great, thank you.

2. Aug 23, 2013

Staff Emeritus
Set p =0.

3. Aug 23, 2013

### HallsofIvy

$E= e= mc^2$ applies in a coordinate system in which the object is motionless. That means that p= 0 so that $E^2= e^2= p^2c^2+ m^2c^4= 0+ m^2c^4$ and taking the square root of both sides, $E= e= mc^2$.

4. Aug 23, 2013

### vanhees71

This is a typical example for the great confusion introduced by using inadequate outdated ideas about mass. You should only use mass in the sense of "invariant mass" or "rest mass". Then you have the energy-momentum relation
$$E^2=p^2c^2+m^2 c^4.$$
For a particle at rest you have the energy
$$E_0=mc^2.$$
It's important to put this rest energy into the total energy, because then and only then
$$p^{\mu}=\begin{pmatrix} E/c \\ \vec{p} \end{pmatrix}$$
is a four-vector. Then the energy-momentum relation can be written in covariant form as
$$p_{\mu} p^{\mu}=\frac{E^2}{c^2}-\vec{p}^2=m^2 c^2.$$
As it must be this is a Lorentz invariant, i.e., a scalar in Minkowski space.

5. Aug 23, 2013

### astorboy-anmol

Assume p=0. But, actually p doesn't equal to zero. The elaborated equation is the proper equation but we use the simplified equation I.e. E=mc^2 to make our calculations easier.

6. Aug 23, 2013

### ZapperZ

Staff Emeritus
This is false. We set p=0 for the REST mass, because by definition, it has no velocity. It isn't because we want to make it easier. You can't ignore the full equation when the object is moving relativistically, no matter how much simpler you want to make it.

Zz.

7. Aug 23, 2013

### rbj

i'm not gonna kick in my anachronistic understanding of the $E = m c^2$ equations regarding "relativistic mass" as it is not well approved of by the real physicists. (although it was in my first modern physics text by Beyser back in the 70s.) but there is an answer to the OP from that perspective.

8. Aug 23, 2013

### Naty1

9. Aug 23, 2013

### Staff: Mentor

Even Beiser doesn't use it any more. I don't remember when he switched, but my copy of the 6th edition (2003) uses invariant mass, and mentions "relativistic mass" only in a sidebar.

10. Aug 23, 2013

### rbj

i wouldn't doubt that at all. i still think that the $E = m c^2$ makes more simple sense to me when energy is total energy (rest energy + kinetic energy w.r.t. the observer's frame of reference) and mass is the total mass (rest mass + whatever relativistic effect on mass w.r.t. the observer's frame of reference). with that understanding (and an expression for this total or relativistic mass), one can derive

$$E^2 = p^2 c^2 + m_0^2 c^4$$

from

$$E = m c^2$$

$$p = m v$$

and

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

i can remember the bottom three equations but i have more trouble remembering the top one. but i can derive the top one from the bottom three.

11. Aug 23, 2013

### vanhees71

It may make more sense to you, but it is the most troublesome idea for beginners on theory of relativity that there are still so many old-fashioned books around that do not give a clear meaning to the physical quantities in terms of covariant objects. For particle kinematics and dynamics it's manageable in the old-fashioned way, but for more delicate topics (e.g., relativistic statistical thermodynamics) it's a nightmare.

It's a great luck for us that Minkowski has sorted out all these difficulties by identifying the mathematical structure of special relativity and we should teach and learn the subject only in this manifestly covariant way. A lot of trouble in understanding is saved for the beginner.

So the rule of thumb is that intrinsic quantities related with a (massive) particle are best defined as scalars in the rest frame of that particle. E.g., the mass of a particle is defined as a scalar, because it is an intrinsic quantity of the particle. For simplicity we (i.e., theoretical high-energy particle physicists) call this scalar quantity the mass of the particle $m$. Another example is the mean lifetime of an unstable particle. We always define it as the mean lifetime of the decaying particle at rest.

In thermodynamics one defines temperature as a scalar quantity too, and it's measured in the (local) rest frame of the fluid, i.e., in the usual way by a thermometer that is comoving with the medium the temperature measure, and so on.

It's a bit (not too difficult) math to learn in the beginning, namely some vector algebra in pseudo-Euclidean (Minkowskian) spacetime, but it's worth the effort, at least much more worth the efford than to get confused by unsharply defined quantities in the beginning.

The equations of motion of a particle are best described in the manifestly covariant either. It starts with the idea to define a scalar measure of time. As you know from elementary kinematics, time intervals of events change (e.g., in the above example an instable particle moving with velocity $v<c$ has a lifetime of $t=\tau/\sqrt{1-v^2/c^2}$, where $\tau$ is the lifetime of the particle in a frame where it is at rest). For a moving particle, an invariant time measure is defined in a natural way as that time which is measured by an observer who is comoving with the particle, the socalled proper time of the particle. Fortunately you don't have to perform the Lorentz transformations from the laboratory frame (where you like to describe the motion) to the momentary restframe of the particle to measure a little proper-time interval $\mathrm{d} \tau$ and then transform back to the lab-frame time $t$, but you can use Minkowski's vector algebra. For a particle you define the space-time vector $x^{\mu}=(c t,\vec{x})$. During a little time interval $\mathrm{d} t$ it changes by $\mathrm{d} x^{\mu} = \mathrm{d} t (c,\vec{v})$. Then the expression
$$\mathrm{d} x^{\mu} \mathrm{d} x_{\mu} = \mathrm{d} t (c^2-\vec{v}^2)$$
is invariant (i.e., a scalar) under Lorentz transformations. Since $|\vec{v}|<c$ for all possible motions of massive particles, this is always a positive expression, and thus we can define the proper time as
$$\mathrm{d} \tau=\frac{\mathrm{d} t}{c} \sqrt{c^2-v^2}=\mathrm{d} t \sqrt{1-v^2/c^2}.$$
Now instead of working with cumbersome non-covariant quantities you better define covariant ones. One is four-velocity
$$u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} {\mathrm{d} t}{\mathrm{d} \tau}=(c,\vec{v})/\sqrt{1-v^2/c^2}=\gamma_v(c,\vec{v}).$$
Now we can define the four-momentum as
$$p^{\mu}=m u^{\mu},$$
where $m$ is the scalar (invariant) mass of the particle. Obviously we have
$$p_{\mu} p^{\mu}=m^2 c^2.$$
For small velocities $|\vec{v}| \ll c$ we have
$$p^{\mu} \approx m (c +v^2/(2c),\vec{v}),$$
which shows that it makes sense to identify the temporal component with the energy of the particle (up to a factor of $c$), i.e.,
$$E=c p^0=\frac{m c^2}{\sqrt{1-v^2/c^2}}=m c^2 \gamma_v$$
and the three spatial components with the momentum of the particle.

The equations of motion are then given in the covariant form
$$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=K^{\mu},$$
where $K^{\mu}$ is obviously a four-vector, called the Minkowski force.

Now it looks as if there are four equations instead of 3, but that's not true, because there is a constraint, because of $p_{\mu} p^{\mu}=m^2 c^2=\text{const}$. Taking the derivative with respect to $\tau$ gives
$$p_{\mu} \frac{\mathrm{d}p^{\mu}}{\mathrm{d} \tau}=p_{\mu} K^{\mu}=0.$$

An important example is the motion of a particle in an external electromagnetic field, neglecting the backreaction of the radiation emitted due to the acceleration of the charge,
$$m \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=\frac{q}{c} F^{\mu \nu}(x) u_{\nu},$$
where $F^{\mu \nu}=-F^{\nu \mu}(x)$ is the antisymmetric field-strength tensor (containing 6 field components, which in the conventional three-dimensional form are $\vec{E}$ and $\vec{B}$. The spatial comonents of this equation of motion indeed read
$$m \frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau}=q \left ( u^0 \vec{E}/c+\frac{\vec{u}}{c} \times \vec{B} \right)=q \gamma_v \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right).$$
Multiplying this equation by $\mathrm{d} \tau/\mathrm{d} t=1/\gamma_v$ you can derive the non-covariant equation of motion
$$\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=m \frac{\mathrm{d}}{\mathrm{d} t} \frac{\vec{v}}{\sqrt{1-v^2/c^2}}=q \left (\vec{E}+\frac{\vec{v}}{c}\times \vec{B} \right).$$
The temporal component of the equation of motion, by the way, reads in this case
$$\frac{\mathrm{d} p^0}{\mathrm{d} \tau}=\frac{q}{c} \vec{E} \cdot \vec{u}$$
or written in the non-covariant way
$$\frac{\mathrm{d} E_{\text{kin}}}{\mathrm{d} t}=q \vec{E} \cdot \vec{v},$$
which tells you that magnetic forces do not contribute to the work on the particle.

As you see, it's always easy to translate from the manifestly covariant form of the equations to the more conventional non-covariant forms. Finding the fundamental laws in relativistic physics is, however, much easier using the manifestly covariant form, particularly when it comes to the question, how to change from one inertial frame to another.