# Energy formulas in SR

1. Apr 24, 2015

### PWiz

As I understand it, the formula $E=\gamma m_0 c^2$ gives the total energy of a body in any inertial frame. However, the formula $E=\sqrt{ (m_0 c^2)^2 + (pc)^2}$, which is also supposed to give the total energy of a body (in any inertial frame), does not equal to the first formula. Why is this? I'm guessing that the answer is very simple, but I'm not able to lay my finger on it. Any help is appreciated.

2. Apr 24, 2015

### certainly

$m_0c^2$ is the energy of the particle at rest ($m_0$ is the rest mass). You are no doubt aware that at relativistic speeds mass increases. $E=\sqrt{(m_0c^2)^2+(pc)^2}$ is the energy of the particle with the mass increase taken into consideration. It's just a restatement of the fact that $E=m_vc^2$ where $m_v$ is the mass at velocity $v$.You should have no trouble proving equation 2 if you remember that $m_v=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$. I highly suspect that $\gamma m_0$ in equation 1 is just a convenient way to denote $m_v$

3. Apr 24, 2015

### Orodruin

Staff Emeritus
The answer is very simple indeed. The two formulas are exactly the same if you use the relativistic expression for momentum.

4. Apr 24, 2015

### Orodruin

Staff Emeritus
This is not in line with what most physicists mean when they say "mass". See our FAQ: https://www.physicsforums.com/threads/what-is-relativistic-mass-and-why-it-is-not-used-much.796527/ [Broken]

Last edited by a moderator: May 7, 2017
5. Apr 24, 2015

### certainly

This was exactly the case.
So does this mean that if I accelerate an object of mass $m$ to a high velocity and bring it back at rest the mass will still be $m$?

6. Apr 24, 2015

### PWiz

@certainly Thanks, but I usually avoid using $\gamma m_0$ for $m$ and instead just stick with $\vec p=\gamma m_0 v$, since it causes much less confusion.

@Orodruin Ah, got it! I was thinking how $\sqrt{1+{\gamma}^2 \frac{v^2}{c^2}}$ was equivalent to gamma, but it turns out it was just because I was being lazy and was not explicitly using the gamma function to prove the equivalence of the formulae. But I have one more question though - why do we use the second equation when the first one is so elegant, neat and compact? I can understand that in collisions involving multiple bodies one would use four-momentum vectors trying to find a zero momentum frame and then use scalar products, but for relativistic K.E. calculations, can a simple $m_0c^2(\gamma -1)$ not suffice? I mean first calculating the relativistic momentum in the individual spatial dimensions, then finding the resultant momentum and using the scalar product to calculate the total energy of the body in a frame takes a lot more time than simply plugging in the resultant velocity and mass in the first formula.

7. Apr 24, 2015

### Orodruin

Staff Emeritus
The second form is actually much more elegant in my view and the formulation in terms of four vectors is very powerful. You will find very few particle physicists actually compute gamma factors in terms of velocities.

8. Apr 24, 2015

### stevendaryl

Staff Emeritus
@Orodruin Ah, got it! I was thinking how $\sqrt{1+{\gamma}^2 \frac{v^2}{c^2}}$ was equivalent to gamma, but it turns out it was just because I was being lazy and was not explicitly using the gamma function to prove the equivalence of the formulae. But I have one more question though - why do we use the second equation when the first one is so elegant, neat and compact?[/QUOTE]

The second formula applies even for zero-mass particles such as the photon.

9. Apr 24, 2015

### PWiz

How so? You have to use $E=hf$ regardless of which of the two formulas you choose, since the spatial momentum of particles with 0 rest mass is equal to their energy as they move on null lines.

10. Apr 24, 2015

### stevendaryl

Staff Emeritus
Well, $E=\gamma m_0 c^2$ is undefined when $v=c$ and $m_0 = 0$. But $E = \sqrt{(m_0)^2 c^4 + p^2 c^2}$ is still true.

Last edited: Apr 24, 2015
11. Apr 24, 2015

### PWiz

Yes, but it's more work again. For objects with non-zero rest mass, one can use $E=\gamma m_0 c^2$ and for massless particles, $E=hf$ can be used. I still don't find a reason to use the length of a four-momentum vector. Can you give an example where it is more convenient to use the second formula?

12. Apr 24, 2015

### Orodruin

Staff Emeritus
Basically any kinematics problem in special relativity.

The "length" of the four-momenutm is $m_0 c$. The power of the four-vector formalism is that inner products between four-vectors are scalars and you can therefore evaluate them in any frame.

I can tell you from experience in correcting exams in special relativity: The people who try to apply conservation of energy and momentum separately rather than learning the four-vector formalism generally make mistakes and get lost in mathematical issues that they really would not need to if they applied four-vector formalism.

Here is an example from the latest exam I put to my students:

13. Apr 24, 2015

### stevendaryl

Staff Emeritus
Vectors are much more convenient to work with than non-vector quantities such as $\gamma$. For one thing, an expression written in terms of 4-vectors is true in any reference frame, so you can pick a reference frame in which the vectors have a particularly simple form to evaluate them.

For example, the expression

$E^2 - p^2 c^2$

is the magnitude of the 4-vector $(E, p^x c, p^y c, p^z c)$. It can be evaluated for a massive particle by looking in its rest frame, where

$E = m_0 c^2$
$p = 0$

So we conclude that $E^2 - p^2 c^2 = m_0^2 c^4$ is true in every frame.

Here's another example of using 4-vectors. Suppose you have a rocket that travels in such a way that, as measured by accelerometers aboard the rocket, the acceleration is constant. How do you compute the rocket's position as a function of time?

That's enormously complicated to do using Lorentz transformations. But in terms of 4-vectors, we can let $U$ be the 4-velocity, and let $A$ be the 4-acceleration, and then the condition of constant acceleration becomes:

$A \cdot A = -g^2$

where g is the magnitude of the acceleration. That equation can readily be integrated to get $U$, which can be integrated to get $(ct, x, y, z)$ as a function of proper time, $\tau$.

14. Apr 24, 2015

### PWiz

@Orodruin Did your students have to include the repulsive force between the electrons in their calculations?
Um shouldn't it be $m_0 c^2$ instead?
@stevendaryl I liked the rocket example. I suppose one could use hyperbolic functions there and avoid vectors, but that would indeed complicate matters.

15. Apr 24, 2015

### Orodruin

Staff Emeritus
No. This type of computations generally assume that the end products are well separated.

No, $m c^2$ has units of energy, not momentum.

16. Apr 24, 2015

### PWiz

Wait, you're defining the four momentum as $p_{\mu}=(\frac E{c},\vec p)$ or as $(E,\vec pc)$? The first one will have a magnitude of $m_0 c$ whereas the second will have $m_0 c^2$.

17. Apr 24, 2015

### Orodruin

Staff Emeritus
As most theoretical physicists, I usually set $c = 1$ and define the four-momentum as $(E,\vec p)$. :)
If I am forced to use units where $c \neq 1$ I would go for the definition $(E/c,\vec p)$. After all, it is called four-momentum and not four-energy.

18. Apr 24, 2015

### PWiz

It's the same thing with spacetime coordinates as well. Some use $(t,x,y,z)$ and some use $(ct,x,y,z)$. The problem is that it becomes painful solving questions which are framed in SI units because you have to remember where to insert the c's and constantly multiply/divide. I usually like to use the $(ict,x,y,z)$ and add $i$ to the 0th component of pretty much every four vector so that I don't have to "untune" my Euclidean metric sense. Is this unwise?

19. Apr 24, 2015

### stevendaryl

Staff Emeritus
The use of $ict$ is discouraged these days. It makes some formulas involving SR seem simpler, but when you go on to General Relativity, using $ict$ doesn't work, anymore.

The preferred way to deal with vectors in General Relativity is to explicitly use the metric tensor. Do you know what the metric tensor is?

20. Apr 24, 2015

### PAllen

It is old fashioned, but if it helps you that is all that matters. I think it has fewer potential problems that relativistic mass. In fact, the only major limitation is the generalization to arbitrary coordinates SR or to GR - it only remains simple in SR with standard coordinates. For example, Dirac and Bondi developed nice approaches in SR that use coordinates that have two lightlike coordinates and two spatial coordinates, with no time coordinate at all. That works just fine with metric approaches but I don't see how you deal with such formulations using ict.

21. Apr 24, 2015

### PWiz

Isn't that because in SR spacetime is represented by flat Minkowski space, whereas in GR spacetime is a curved pseudo-Riemannian manifold (which is locally flat)?
Yes, but only the Euclidean metric tensor $$g= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$
and the Lorentzian metric tensor (with c=1)
$$η_{μν}= \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$
used for taking the norm of two vectors in their respective manifolds.
I don't know the mathematical details of GR, but I'm pretty sure a different metric tensor is used with polar coordinates (I think the Schwarzschild metric), although $ds^2=η_{μν} dx^μ dx^ν$ still holds true (I guess).
@PAllen Which alternate coordinate systems in SR are you talking about?

22. Apr 24, 2015

### stevendaryl

Staff Emeritus
Well, for example, you can do SR in polar coordinates, in which case the metric tensor no longer takes on the form of $\eta_{\mu \nu}$.

The point is that using $ict$ is a way to get the minus sign into expressions such as $-(ct)^2 + x^2 + y^2 + z^2$, so you can pretend that you are using Euclidean vectors. But the more general approach is to let the $-1$ be part of the metric tensor, and keep all components as real numbers.

23. Apr 24, 2015

### PWiz

The Lorentzian metric changes to a Euclidean one if you use $ict$. Isn't it better that way? You then have only have one metric tensor to deal with.

24. Apr 24, 2015

### PAllen

Consider u = x-ct, v=x+ct. A line of u constant describe a light ray in the +x direction, a line of constant v describes a light ray in the -x direction. Using these coordinates, (ignoring YZ plane), you end up with the amazingly simple metric:

ds2 = du dv

and I am not setting c=1, it just drops out if transform to these coordinates.

So where do you put your ict here?

25. Apr 24, 2015

### stevendaryl

Staff Emeritus
No, it's really not the better approach. The approach that you're talking about ignores the distinction between covariant vectors (ones with components $A^\mu$) and contravariant vectors (ones with components $A_\mu$). That distinction is extremely important for curved spacetime, but it's also important for flat spacetime, if you're using curvilinear coordinates.

For example, let's just do 3-dimensional vectors in spherical coordinates,nonrelativistically. Then a point in space is described by three numbers, $r, \theta, \phi$. The path of a particle would similarly be described by three numbers: $\frac{dr}{dt}, \frac{d\theta}{dt}, \frac{d\phi}{dt}$, which you can think of as components of the velocity vector. But if you're computing the speed $v$, it's not simply $v^2 = \frac{dr}{dt}^2 + \frac{d\theta}{dt}^2 + \frac{d\phi}{dt}^2$. Instead, it's more complicated:
$v^2 = \frac{dr}{dt}^2 + r^2 \frac{d\theta}{dt}^2 + r^2 sin^2(\theta) \frac{d\phi}{dt}^2$. That can be understood as

$\sum_{i,j}\ g_{ij} v^i v^j$

where $g_{rr} = 1$, $g_{\theta \theta} = r^2$, $g_{\phi \phi} = r^2 sin^2(\theta)$.

Of course, it is possible to get a notion of a velocity vector that uses a trivial metric, by letting the components of the velocity vector be:

$\frac{dr}{dt}, r \frac{d\theta}{dt}, r sin(\theta) \frac{d\phi}{dt}$

That allows the metric to be simple, but in exchange for making the velocity vector more complicated. In a sense, what you'd be doing is incorporating $\sqrt{g}$ into the definition of the velocity vector (which is the same thing that is going on in using $ict$). That's a convoluted thing to do, and only works in the cases where the metric tensor is diagonal, which isn't always the case.