Minimum energy to accelerate a mass?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 1K views
jerromyjon
Messages
1,244
Reaction score
189
I have a similar logical "gap" in my understanding that I still haven't resolved... which seems to be right on point with this thread if I call it "minimum amount of energy required to accelerate a given mass".

From what I know about SR, accelerating mass takes energy which increases exponentially as this mass approaches c. I just don't know at what speed this effect becomes measurable...
 
Physics news on Phys.org
jerromyjon said:
I have a similar logical "gap" in my understanding that I still haven't resolved... which seems to be right on point with this thread if I call it "minimum amount of energy required to accelerate a given mass".

From what I know about SR, accelerating mass takes energy which increases exponentially as this mass approaches c. I just don't know at what speed this effect becomes measurable...
It isn't an exponential growth. The energy of a massive particle moving at velocity v is [itex]E=\gamma mc^2[/itex], where [itex]\gamma=(1-v^2/c^2)^{-1/2}[/itex]. If you Taylor expand the expression for [itex]\gamma[/itex] you get
[tex]E=mc^2<br /> +\frac{1}{2}mv^2<br /> +\frac{3}{4}m<br /> \frac{v^4}{c^2}<br /> +...[/tex]The first term is mass energy. The second is Newtonian kinetic energy. The third and later terms are where relativity disagrees with Newton on energy. So the effect is measurable when that term is measurable.

I don't know if there's an answer to your question in practical terms since I'm not current on the precision of energy measurement devices, and that probably depends on your application anyway. But that's the theoretical basis for an answer.
 
Thank you, I appreciate your response! I'm still trying to learn what is still "greek" to me, but that funky y that you all call gamma... isn't that only zero at zero velocity?
 
Ummm, If I set c=1 then its 1 but I'm not even sure about that c=1 scenario.
 
oh duh nevermind gamma has to be 1 to be at rest energy..
 
Ok. So what I meant was gamma is 1 at rest energy, which would only be at rest velocity. Suppose we do the M&M set-up with "identical" rockets in opposite directions, along with an identical complete system at a known relative velocity parallel to the launch vectors. This is where I'm unsure how to predict how this system evolves, but now I recall something about the relativistic "tether" problem, where the distance changes, severely complicating this visually. I'm going to play with that interactive Minkowski diagram (Thanks Ibix!) and see if I can figure it out...