# How can mass equation rely on 'rest mass'?

1. Apr 11, 2004

### billy_boy_999

how can the equation of relativistic mass, m=m_0/(1-v^2/c^2)^2, determine mass from the variable m_0?

rest mass - 'at rest' relative to what?

2. Apr 11, 2004

### Janus

Staff Emeritus
At rest relative to the frame of reference from which the measurement of mass is made.

3. Apr 11, 2004

### OneEye

I don't think this is right.

Mass measurements are always made at rest relative to the frame of reference of the mass in question.

Further, this would mean that $$m={ {m_0} \over { (1-v^2/c^2)^2 } }$$ calculates the mass relative to some other frame of reference. But this cannot be so, since the mass thus calculated is used to determine the energy needed to accelerate the mass within its own reference-frame.

Worse yet, this would mean that the energy required to accelerate the mass v would vary depending on the frame of reference from which one viewed the mass. Consider two systems, one at v1, the other at v2, which overlook our mass. From the perspective of v1, the energy required to accelerate the mass would be

$$E={ { v m_0} \over { 2(1-{v_1}^2/c^2)^2 } } }$$

while the energy required to accelerate the mass from the perspective of v2 would be

$$E={ { v m_0} \over { 2(1-{v_2}^2/c^2)^2 } } }$$

This doesn't seem right.

But maybe I'm wrong.

Any help?

Last edited by a moderator: Apr 11, 2004
4. Apr 11, 2004

### OneEye

Oops!

Well, duh!

What could make more sense than that different amounts of energy would be required to accelerate a given mass from different inertial frames? The relative velocity is different!!

Open mouth, insert foot.

I withdraw my objection.

5. Apr 12, 2004

### billy_boy_999

thanks Janus, OneEye (btw OneEye - your mistakes are actually helping me understand this better )

okay, so if we can't change the energy required we can change the velocity variable in different reference frames...okay...

but what if two spaceships take off from earth, they both travel in the same direction at 9/10 c...one could say to the other - hey, from here you're at rest and thus at m_0, so it wouldn't take much energy at all to accelerate you past c...how does this translate into different relative velocities?

6. Apr 12, 2004

### Severian596

Hi billy_boy_999,

Remember that the Newtonian method of velocity addition only applies to non-relativistic speeds. Two rockets travelling parallel to each other from the point of view of earth will not necessarily measure their relative velocities as zero. As long as you're talking speeds of v << c then you're safe, otherwise you'll have to use relativistic mechanics to find relative velocities.

The equation for relative velocity w is:

$$w = \frac{u + v}{1+\frac{uv}{c^{2}}}$$

You'll have to look this up for more info, but notice that at "Newtonian" type quantities the term $$\frac{uv}{c^{2}}$$ essentially approaches 0 and the equation is the familiar w = u + v.

NOTE: The relative velocity equation is valid regardless of vectors of direction! Crazy world, isn't it?

7. Apr 12, 2004

### DW

See what a mess of basic physics understanding the "relativistic mass" missnomer makes! Instead of using Planck's relativistic mass paradigm, I think you will have a better grasp of what is happening by using the modern relativistic (also Einstein's) description according to which the mass m does not change with speed. The kinetic energy of the mass in special relativity is then $$KE = (\gamma - 1)mc^2$$. The amount of energy that it has according to the frame for which it is at rest is the mass:
$$m = \frac{E_{0}}{c^2}$$
The the total energy E according to an arbitrary inertial frame is then the sum of these $$E = KE + E_{0}$$ which simplifies to
$$E = \gamma mc^2$$.
This last equation is what you should be using instead of your/Planck's $$m = \gamma m_{0}$$ equation which Einstein dissagreed with using in the beggining and again in his latter years.

Last edited: Apr 12, 2004