# Special relativity question.

1. May 7, 2010

### tkav1980

Ok here's the short story. A friend of mine is in grad school for physics. I have a BS in physics, however i graduated in 2002 and haven't used anything i learned as i don't work in the field. I always try to think up questions to stump him. This time he got me. Here's what he asked me. In special relativity, At what speed does an object need to travel for its Total energy to be less than its resting energy. This one is way out of my league at this point.

If i posted this in the wrong place i apologige.

p.s. i asked him about the object having zero mass and didnt get a response.

2. May 7, 2010

### Nabeshin

Impossible.
$$E^2=m^2c^4 +p^2 c^2=E_0 ^2 + (pc)^2 \implies E \geq E_0$$

3. May 7, 2010

### tkav1980

But i dont see a solution for velocity? Even if that number was a negative number wouldnt there be at least a value that could be assigned. Sorry if im annoying ive been at work a long time at this point in my day. The brain shut off, but from my understanding of what he asked the total energy of the particle ( the rest energy plus the kenetic energy) must be less than it's rest energy. wouldnt there be a solution in terms of velocity. at least a way to mathmatically solve for it weather the answer is real or imaginary?

Last edited: May 7, 2010
4. May 7, 2010

### starthaus

$$TE=\frac{m_0c^2}{\sqrt{1-(v/c)^2}}$$
$$RE=m_0c^2$$

Find $$v$$ such that $$TE<RE$$ :-)

5. May 8, 2010

### Fredrik

Staff Emeritus
In units such that c=1, we have $E=\gamma m -m$, where E is the kinetic energy, $\gamma m$ is the total energy, and m is the rest energy. You're looking for a velocity v such that $\gamma m< m$. This is equivalent to $\gamma<1$, but

$$\gamma=\frac{1}{\sqrt{1-v^2}}>1$$

However, all of the above is for massive particles...and by that I mean particles with mass m>0.

Every particle satisfies an equation of the form $-E^2+p^2=A$. If $A\leq 0$, we write $A=-m^2$, where m is defined to be >0, and is called the "mass" of the particle. If we insist on writing $A=-m^2$ even when A>0 (which would be appropriate if we intend to call m the "mass"), m must be imaginary. These particles are called tachyons.

It's a bit more convenient to write $A=n^2$, where n=im>0. The equation that we would write as $E^2=p^2+m^2$ for massive particles would be written as $E^2=p^2-n^2$. I haven't really thought about how to define momentum, rest energy or total energy for tachyons, but it's clear that these are the things you need to work out if you're going to really answer the question.

The really short answer is of course v>c (if it's possible at all...I haven't verified that it is), but the answer isn't complete without the appropriate definitions.