# Confusion over relativistic mass equations

1. Sep 2, 2014

### DiracPool

I'm trying to reconcile the two relativistic mass equations and I 'm getting different results as I push the velocity towards c.

In the first equation, E=mc^2/√(1-v^2/c^2), I'm getting that E approaches infinity as v approaches c.

In the second equation, E=√(m^2c^4+p^2c^2), I'm getting that E approaches √2(mc^2) as v approaches c.

Am I doing something wrong here, or is this an anomaly that has a standard explanation?

2. Sep 2, 2014

### Staff: Mentor

You're doing something wrong here. Show us what you did and someone can probably point you to the error.

The key is that p goes to infinity as v goes to c.

3. Sep 2, 2014

### Staff: Mentor

It looks like you are using $p=mv$ instead of $p=\gamma mv$, but jtbell is right, we cannot know for sure without more details.

4. Sep 2, 2014

### DiracPool

Yeah, I'm using p=mv

E=√(m^2 c^4 + (mv)^2 c^2). Is that not right? Do they really use $p=\gamma mv$ in that equation instead of $p=mv$? I've never seen that.

5. Sep 2, 2014

### Staff: Mentor

$p \ne mv$ in general. Particularly not as v approaches c.

6. Sep 2, 2014

### stevendaryl

Staff Emeritus
The whole point of "relativistic mass" is that you can start with the low-speed equations:

$E = mc^2$
$p = mv$

and then you replace $m$ by $\gamma m$ to get equations that work at any speed:

$E = \gamma mc^2$
$p = \gamma mv$

7. Sep 2, 2014

### DiracPool

So does that mean that both the mass and the momentum terms in that equation are multiplied by gamma?

Yielding E=√[(gamma)^2 m^2 c^4 + (gamma)^2 p^2 c^2]?

Edit: I'm still a bit confused, I always thought that E=mc^2/√(1-v^2/c^2) and E=√(m^2c^4+p^2c^2) were supposed to be equivalent equations, but it seems as if the former is somehow nested inside the second one here if that is the case.

Last edited: Sep 2, 2014
8. Sep 2, 2014

### Staff: Mentor

No.

$E = \sqrt{m^2c^4 + p^2c^2}$ is a relativistic equation, so you need to use the relativistic equation for p, which is

$$p = \gamma mv = \frac {mv} {\sqrt{1 - v^2/c^2}}$$

(m is "rest mass" in these two equations.)

As $p \rightarrow c$ [corrected: As $v \rightarrow c$], the numerator approaches mc. What does the denominator approach to?

Last edited: Sep 3, 2014
9. Sep 2, 2014

### DiracPool

It approaches zero. So, then, in that equation, is it just the momentum term we multiply by gamma, or do we multiply the mass term also? I guess that's all I'm left with.

10. Sep 2, 2014

### Staff: Mentor

No. Generally physicists do not use relativistic mass any more, for a variety of reasons. Unfortunately, pop-sci references to relativistic mass still abound and there are some old or low-quality textbooks that use the term.

Generally, we just use m = rest mass and recognize that momentum is not a linear function of velocity and energy is not a quadratic function of velocity.

11. Sep 2, 2014

### DiracPool

Ok, great. Thanks for the clarification everyone :)

12. Sep 2, 2014

### Staff: Mentor

Oops, I meant $v \rightarrow c$ :yuck:, but I guess you caught that.

13. Sep 2, 2014

### stevendaryl

Staff Emeritus
No, in my equations $m$ means rest mass. If we explicitly say whether we're talking rest mass or relativistic mass, then:

$m_{rel} = \gamma m_{rest}$, then
$E = m_{rel} c^2$
$p = m_{rel} v$

If you haven't already heard, most physicists do NOT use relativistic mass. Nowadays, when people say "mass" they mean "rest mass". So they don't put any subscript like $m_{rest}$, they just write $m$

No. There are a bunch of equivalent definitions for E:

$E = \gamma m_{rest} c^2$
$E = m_{rel} c^2$
$E = \sqrt{m_{rest}^2 c^4 + p^2 c^2}$

(Actually, the first two are only valid for particles with nonzero mass. The last works for massless particles as well.)

There are similarly a bunch of equivalent definitions for p:
$p = \gamma m_{rest} v$
$p = m_{rel} v$

But these are only valid for particles with nonzero rest mass. For a photon, for instance, the momentum is unconnected with speed. For a photon, or any other massless particle,

$E = pc$

They are equivalent, if you use $m_{rest}$ for $m$. (But again, only for particles with nonzero mass.)

14. Sep 3, 2014

### DiracPool

You mean because the velocity condition is replaced by the DeBroglie relationship, right? Instead of P=mv, P=h/λ in the case of a photon?

15. Sep 3, 2014

### stevendaryl

Staff Emeritus
Well, I wouldn't say that. It is true that for a photon, the wave-length and frequency are related to the momentum and energy, but that's quantum mechanics, which goes beyond SR.

Two equations that work for both particles with and without mass are the following:

1. $E = \sqrt{p^2 c^2 + m_{rest} c^4}$
2. $\dfrac{p}{E} = v$

16. Sep 3, 2014

### ChrisVer

The reason is more because the relativistic factor $\gamma$ which appears doesn't work for velocities equal to c [it's singular]. However, that's not a problem for equations 1,2 steven gave at P#15 because they don't contain the factor $\gamma$... in 1 it doesn't exist, and in 2 it's getting cancelled (so you can extend that definition even for $v=c$).

17. Sep 3, 2014

### Staff: Mentor

The second one works only if you use units where c = 1. Showing c explicitly, the way I like to remember it is
$$\frac{pc}{E} = \frac{v}{c}$$
My background is in experimental particle physics, where people usually talk about momentum in energy units (e.g. MeV/c). A particle with E = 1000 MeV and p = 750 MeV/c travels at v/c = 750/1000 = 0.75, i.e. 75% of the speed of light.

Similarly, the way I always remember the first equation is
$$E^2 = (pc)^2 + (mc^2)^2$$
The particle in my example above has (rest) mass $\sqrt{100^2 - 750^2}$ = 661.4 MeV/c2.

Last edited: Sep 3, 2014