Can Total Energy Be Less Than Rest Energy in Special Relativity?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
tkav1980
Messages
47
Reaction score
1
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.
 
Physics news on Phys.org
tkav1980 said:
In special relativity, At what speed does an object need to travel for its Total energy to be less than its resting energy

Impossible.
[tex]E^2=m^2c^4 +p^2 c^2=E_0 ^2 + (pc)^2 \implies E \geq E_0[/tex]
 
Nabeshin said:
Impossible.
[tex][/tex]

But i don't see a solution for velocity? Even if that number was a negative number wouldn't there be at least a value that could be assigned. Sorry if I am annoying I've 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. wouldn't 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:
tkav1980 said:
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.

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

Find [tex]v[/tex] such that [tex]TE<RE[/tex] :-)
 
tkav1980 said:
But i don't see a solution for velocity? Even if that number was a negative number wouldn't there be at least a value that could be assigned. Sorry if I am annoying I've 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. wouldn't there be a solution in terms of velocity. at least a way to mathmatically solve for it weather the answer is real or imaginary?
In units such that c=1, we have [itex]E=\gamma m -m[/itex], where E is the kinetic energy, [itex]\gamma m[/itex] is the total energy, and m is the rest energy. You're looking for a velocity v such that [itex]\gamma m< m[/itex]. This is equivalent to [itex]\gamma<1[/itex], but

[tex]\gamma=\frac{1}{\sqrt{1-v^2}}>1[/tex]

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 [itex]-E^2+p^2=A[/itex]. If [itex]A\leq 0[/itex], we write [itex]A=-m^2[/itex], where m is defined to be >0, and is called the "mass" of the particle. If we insist on writing [itex]A=-m^2[/itex] 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 [itex]A=n^2[/itex], where n=im>0. The equation that we would write as [itex]E^2=p^2+m^2[/itex] for massive particles would be written as [itex]E^2=p^2-n^2[/itex]. 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.