0

(This is a quick one)

Suppose Im standing still (in my perspective) and a hoverboard is flying at $0.8c$ w.r.t me.

I see that the ball (red) is moving at $0.8c$. I measured earlier that the ball is $15$ kg, before it went on the hoverboard. So, its 'rest mass' is $15$ kg for me. But, when its on the hoverboard, its mass will be the 'relativistic mass' $m = γm$ (rest mass), which will be $25$ kg.

So, if I get on the hoverboard, and measure the mass of the ball, will I measure it to be $25$ kg, or $15$ kg?

i.e Is the 'rest mass' (or 'mass measured when object is rest in my frame') same in both cases or not?

enter image description here

Qmechanic
  • 201,751
Rohit Shekhawat
  • 1,305
  • 1
    Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – rob May 11 '23 at 16:21

1 Answers1

2

Rest mass of a body is the mass measured when the body is at rest w.r.t. you. It is always the same. Just make sure that the object's whose rest mass you want to measure is at rest w.r.t. you. The actual mass of the object will be greater than the rest mass if the object is in motion w.r.t. you.

It is impossible to measure the actual mass of a body in motion w.r.t us. We calculate the actual mass of a body in motion using the relativistic mass expression given by theory of relativity.

So, if you board the hoverboard, you will measure the mass of the red sphere to be $15.00$ kg only as the sphere is at rest w.r.t. you in that case (you will measure the rest mass only). At the same time, for an observer on the ground, its actual mass is $25.00$ kg, as that observer measures the speed of the sphere to be $0.8c$.

  • will nobody actually measure that mass to be 25kg then, in any frame? is it just a placeholder value for some other concept? – Rohit Shekhawat May 10 '23 at 15:57
  • 1
    @RohitShekhawat See, this is Relativity. Things are relative. Similarly, mass is also relative based on what frame you are in. –  May 10 '23 at 16:10
  • 1
    @Rohit Shekhawat https://physics.stackexchange.com/questions/760851/is-work-done-relative-according-to-the-theory-of-special-relativity. See this question by me. –  May 10 '23 at 16:11
  • You just destroyed my peace of mind with this comment. I just made my mind that mass measurements will be same for everyone, but you say they are relative. jeeeesus, this is confusing and itriguing. Anyways, will anyone 'measure' 25 kg mass though? yes or no. – Rohit Shekhawat May 10 '23 at 16:32
  • 3
    @Yes, if the sphere is moving at $0.8c$ with respect to you, then the mass will be $25.00$ kg according to you. As you are in the same reference as the sphere, it is moving at $0$ velocity and so it is at rest- thus $15$ kg only. –  May 10 '23 at 16:34
  • Whattt! How can I measure it though, if its moving? Also, will I 'measure' it to be 25 kg, or its mass in E=mc^2 is 25 kg? – Rohit Shekhawat May 10 '23 at 16:36
  • @RohitShekhawat, if you are on the hoverboard, the sphere is at rest according to you. You can simply just measure it to get 15 kg. –  May 10 '23 at 19:16
  • 1
    @RohitShekhawat it is also true that you can't measure the rest mass of a moving body because Relativity theory forbids it. We just calculate it using the Relativistic mass formula –  May 10 '23 at 19:18
  • 2
    @Rohit Seriously, relativistic mass causes so much confusion, and it's not necessary. But if you insist, please see https://en.wikipedia.org/wiki/Mass_in_special_relativity#History_of_the_relativistic_mass_concept Instead, you can work with the momentum. Eg, your ball on the hoverboard has momentum 25 kg × 0.8c in the ground frame. – PM 2Ring May 11 '23 at 01:50
  • 1
    @RohitShekhawat, I have added some stuffs to my original answer. Consider having a look at it. –  May 11 '23 at 07:26
  • @PM2Ring Using "mass" in place of "rest mass" just causes different confusion. – John Doty May 11 '23 at 12:30
  • @JohnD I'm sympathetic to that argument; OTOH, we wouldn't have that issue if the term "relativistic mass" had never been used. ;) Some people discourage the term "rest mass" and insist that we call rest mass "mass", but IMHO that's not so helpful when someone is already confused about this topic. I'm happy to use "rest mass", eg in this answer, but I also use terms like "rest energy", "rest mass-energy", etc. – PM 2Ring May 11 '23 at 14:34
  • @PM2Ring How familiar is the equation $E=mc^2$? How familiar is $E^2-(pc)^2=(mc^2)^2$? Which is closer to laboratory observation (not theoretical rationalization of laboratory observation)? – John Doty May 11 '23 at 14:41
  • @JohnD Yes, most people have heard of $E=mc^2$, but I suspect that the majority of them don't understand it properly. ;) The 2nd equation "merely" requires understanding how Pythagoras' theorem applies in spacetime. Most people have heard of Pythagoras' theorem, but I suspect only a minority would be able to prove it... – PM 2Ring May 11 '23 at 14:57
  • @PM2Ring How Pythagoras' theorem applies in Minkowskian geometry is mathematics, not physics. Its application to physics is more indirect and convoluted than $E=mc^2$ is. So, it's not a good starting point, although it's very useful in some contexts. – John Doty May 11 '23 at 15:29