1
$\begingroup$

Is the concept of energy increasing as it approaches the speed of light based on the fact that this is only true relative to the observer?

Lets say, there's a scenario where a person in a rocket ship is going past the Earth approaching the speed of light relative to somebody standing on the Earth. From the perspective of the man on the Earth, the energy of the person in the rocket ship & the rocket ship itself have an increased energy compared to their energy at rest, but from the perspective of the person on board the ship, he and the rocket ship have the energy at rest and the Earth has an increased energy compared to it being at rest.

To add on to this, lets say the man on Earth got on his own second spaceship and eventually caught up close to the first spaceship and is approaching the same speed as the first spaceship. Relative to the second spaceship, would the increased energy of the first spaceship gradually lower down to it's energy at rest up until they become the exact speed?

Is this all true?

(I want everyone to understand that I am focusing on the increase of energy due to an object/particle at speeds approaching the speed of light, I already understand how energy increases in Newtonian physics due to the equation of KE.)

$\endgroup$
  • $\begingroup$ The concept of the mass increasing as a massive object approaches the speed of light is no longer in vogue with the physics community. Rather, think of it as momentum or energy increasing rather than mass. $\endgroup$ – K7PEH Aug 18 '15 at 17:18
  • $\begingroup$ So this would mean that energy increases as speed increases; does that apply to what I was saying, in that relative to the observer on the space ship, he and his ship are at their 'rest energy' and relative to the observer on Earth, he and the Earth are at their 'rest energy'? $\endgroup$ – Neil Graham Aug 18 '15 at 17:21
  • $\begingroup$ Energy is not conserved when viewed by observers in different frames. Neither observer would see anything unusual in their computations. Note that part of your question hints at the desire to be both observers, that is being simultaneously in different frames. Since this is not possible, maybe the question is not valid. But, I hold back on saying it is not valid because someone smarter than me would probably give a good answer -- which is why I am only commenting. $\endgroup$ – K7PEH Aug 18 '15 at 17:40
  • 1
    $\begingroup$ If you removed all occurrences of the word "invariant" and if you changed the word "mass" to "momentum" everywhere, this would be perfectly correct $\endgroup$ – Jim Aug 18 '15 at 17:49
1
$\begingroup$

Yes, the energy and the increase in energy all depend on your reference frame, but this is NOT special to relativity! The same thing happens in classical mechanics.

I wrote a similar answer to the question, "Can you tell your absolute speed in space?"

Consider the regular Newtonian mechanics equation, $\mathrm{Ke}=\frac{1}{2}mv^2$. If you weigh 50kg, are moving at 0 meters per second and want to accelerate by 1 meter per second, you need $0.5*50*1=25$ joules of energy. If you're moving at $1000$ meters per second (roughly three times the speed of sound in our atmosphere) and you want to accelerate by one meter per second, you need $0.5\times 50\times(1001^2-1000^2)=50025$ joules of energy.

This fact does not mean there's a special reference frame, nor that you can tell your velocity through space, nor anything else weird! There's no special relativity here, just everyday classical mechanics.

You should better understand energy in classical mechanics before you look at energy in special relativity!

$\endgroup$
  • $\begingroup$ I understand the concepts of Newtonian physics, but what I am getting at is the fact that the energy is not linear to 1/2mv <sup>2</sup> as it approaches the speed of light. Originally this question was geared towards questioning why mass was increasing closer to the speed of light, but too many people explained that the energy was increasing at an increasing rate rather than the mass. $\endgroup$ – Neil Graham Aug 18 '15 at 18:42
  • $\begingroup$ @NeilGraham I think you should be more careful and I do still think this answers your question. The train of thought, "if energy goes to infinity as you approach the speed of light, let's just measure our energy to find our absolute speed in the universe" is a fallacy, but it's just as tempting in classical mechanics due to the square in classical kinetic energy. The title (specifically the phrase "if there is no definite speed in the universe") is what made me address that misconception. There's nothing to say to the bulk of the OP but "yes, that's right, energy is relative". $\endgroup$ – user12029 Aug 18 '15 at 19:20
  • $\begingroup$ I see what your saying. $\endgroup$ – Neil Graham Aug 18 '15 at 19:22
0
$\begingroup$

Is the concept of mass increasing as it approaches the speed of light based on the fact that this is only true relative to the observer?

When you say mass increases what you mean is energy increases. You don't want to have a different mass for forces in different directions so the idea of mass changing with speed is now abandoned. You have energy instead of mass increase when speed increases and you have to learn when to use energy versus mass. For instance energy is the source of gravity, not mass.

Mass becomes something that tells you how to balance energy and momentum, if mass is zero equal amounts of both. If mass if not zero then energy exceeds momentum by an amount that depends on mass.

And then there is the point that energy does depend on your frame. Energy depends on your observer, as does momentum.

From the perspective of the man on the Earth, the mass of the person in the rocket ship & the rocket ship itself have an invariant mass that is greater than their rest mass

Absolutely not. You can't say an invariant mass relative to an observer. An invariant thing doesn't depend on the observer. For instance the balance of energy and momentum of a system doesn't depend on the observer, and that is the mass of the system. And it is not the sum of the masses of the parts.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.