# Is this a legitimate derivation of the changing of mass?

• nhmllr
In summary, the conversation discusses the derivation for transformations of length and time in special relativity and the attempt to apply the same methodology to mass. The person proposes that energy can be calculated as force applied for a distance, but the person in the spaceship would calculate it differently due to their velocity. The conversation concludes with the suggestion to analyze a process such as an accelerating rocket in different frames and ensuring the conservation of energy. The person acknowledges their initial mistake and plans to try again.

#### nhmllr

So I had my hand held for the derivations for the transformations of length and time in special relativity, and I wanted to see if I could do the same thing with mass. I came up with something, but I'm not sure if my methodology is correct.

I said that Energy is the force applied on something for a distance.

E = F*d

If you were "standing still" and saw a spaceship accelerating with thrust F for distance d, you would say that the energy exerted was F*d

However, if the person in the spaceship were to say how much energy was exerted, then the would also multiply a distance by a force. However, they would be traveling at a velocity, v, and from their point of view, they were not being accelerated for a distance, d, but rater a distance of

d*sqrt{1-v2/c2}

So they would say that E = F'*d*sqrt{1-v2/c2}

The total sum of energy in a system shouldn't change for people in different reference frames. So

F'*d*sqrt{1-v2/c2} = F*d

m'*a*sqrt{1-v2/c2} = m*a

And because the rate of acceleration can be known to anybody in any reference frame,

m' = m/sqrt{1-v2/c2}

And that's how I did it
I'm a little wary of this little derivation for a few reasons, but I get the correct result. So I'm curious if this works because my methodology is right, or because I made multiple mistakes that "canceled each other out."

Thanks

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The person in the spaceship would say they have had force (which they can feel) applied for zero distance (you don't move relative to yourself) = zero work done. This is consistent with the frame dependence of kinetic energy: the ship has no added kinetic energy in its own frame.

You also state the total energy shouldn't change between frames. This is false. It is conserved in every frame, but the total energy itself is frame dependent.

In other ways you are on the right track. You can try to analyze a process like your accelerating rocket in two different frames (preferably not the rocket's own frame, which is non-inertial) and insist that energy and momentum are conserved in both. But you must be sure your accounting is complete - your rocket can't be subject 'cause less' work. As stated, you have non-conservation of energy in your starting frame. You need to account for where the energy for the work on the rocket comes from. A collision process may be easier.

I will refrain from giving links to common derivations since you indicate you want to try this on your own.

Good luck!

PAllen said:
The person in the spaceship would say they have had force (which they can feel) applied for zero distance (you don't move relative to yourself) = zero work done. This is consistent with the frame dependence of kinetic energy: the ship has no added kinetic energy in its own frame.

You also state the total energy shouldn't change between frames. This is false. It is conserved in every frame, but the total energy itself is frame dependent.

In other ways you are on the right track. You can try to analyze a process like your accelerating rocket in two different frames (preferably not the rocket's own frame, which is non-inertial) and insist that energy and momentum are conserved in both. But you must be sure your accounting is complete - your rocket can't be subject 'cause less' work. As stated, you have non-conservation of energy in your starting frame. You need to account for where the energy for the work on the rocket comes from. A collision process may be easier.

I will refrain from giving links to common derivations since you indicate you want to try this on your own.

Good luck!

Yeah, that's why I was so wary- I wasn't sure that the total of energy was constant for different reference frames. Thinking about it now, it's obvious that isn't true, even in Newtonian mechanics! I was blinded by the correct result, unfortunately. I'll give this another shot. Thanks