Question about SR and increasing mass

In summary: So the calculation is a lot more messy, but the basic principle stays the same: 4-momentum is conserved. In summary, In summary, when traveling at a speed close to c, an object's mass increases and its time appears to slow down relative to slower moving observers. This also affects its impulse and the conservation of impulse must be considered in a relativistic way. The calculations of momentum and energy must also take into account the mass and speed of the object it collides with.
  • #1
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Let's say you're on a spaceship and you can accelerate to a speed very close to c. As your own time seems to slow down (relative to slow moving observers), your mass also increases.

I know that mass increases in the sense that one requires more energy to give the same acceleration, but does this also mean that your impulse increases ? I assume so. So, if your spaceship would crash into something when you're close to c, it would seem as impulse was violated because (assuming you slow down enough), the mass becomes close to the rest mass again? In that case, would one have to consider a relativistic version of the conservation of impulse?
 
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  • #2
Your own time doesn't seem to slow down. In your own frame you are at rest and therefore you will see the clocks of inertial frames moving with respect to you slow down. In a relativistic collision you will have to use relativistic momentum. Classically you would expect mass to be conserved during a collision, but relativistically this is not true.

Lets do the calculations both classically and relativistically.

A spaceship of mass m going at speed v=4/5c crashes into an object at rest with mass m. (everything is happening in the x-direction)

Classically:
[tex]mv+0*m=M_*u=(m+m)u \Rightarrow m\frac{4}{5}c=2mu \Rightarrow u=\frac{2}{5}c[/tex]

Relativistically:
Momentum as well as total energy is conserved. Total energy is given by [itex]E=\gamma m c^2[/itex] and momentum is given by [itex]p=\gamma mv[/itex]. Combining these into a 4-momentum vector for simplicity yields [itex]p_\mu=((\gamma+1)mc,\gamma m v,0,0)[/itex].

[tex]
p_\mu p^\mu=-(\gamma+1)^2m^2 c^2+\gamma^2 m^2 v^2=-M_*^2 c^2 \Rightarrow M_*=\frac{4}{\sqrt{3}}m[/tex]

And

[tex]
u=\frac{p c^2}{(\gamma+1)mc^2}=\frac{1}{2}c
[/tex]

LennoxLewis said:
assume so. So, if your spaceship would crash into something when you're close to c, it would seem as impulse was violated because (assuming you slow down enough), the mass becomes close to the rest mass again?

Don't forget that this object with which it collides has rest energy as well. Classically its energy would be zero but in relativity this is not so. The total energy as well as momentum is conserved.

Edit: I don't have a lot of time currently so I didn't explain the calculation very well. If you have any questions about it feel free to ask I will try to answer them at a later time.
 
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  • #3
Thanks a lot for the explanation, Cyosis! I do have a question: in the four-vector, where does the (y+1)mc term come from?
 
  • #4
The 4-vector is defined as [itex]p_\mu=(E/c,\vec{p})[/itex]. With E being the total energy. Before the collision we have the energy of the rocket ship [itex]E_r=\gamma mc^2[/itex] and the energy (rest energy) of the object it collides with [itex]E_0=mc^2[/itex]. Therefore the total energy is given by [itex]E_r+E_0=\gamma mc^2+mc^2=(\gamma+1)mc^2[/itex]. Plug it into the definition of the 4-vector and one c will drop out.
 
  • #5
It all makes sense now. Thanks again.
 
  • #6
Note that the [tex]\gamma+1[/tex] is not a general term, but specifically for the case where the rocket collides with an object of equal mass. It is correct for this case, but please don't accidentally over-generalize to other cases.
 
  • #7
Yes DaleSpam raises a good point. I used an equal mass for the the object to simplify the calculation, this however may not have been particular smart for didactic reasons. I emphasize the general 4-momentum is given by [itex]p_\mu=(E/c,\vec{p})[/itex]. Now imagine the object it collides to has mass [itex]\mu[/itex] instead. Then E becomes [itex]\gamma m c^2+\mu c^2=(\gamma m+\mu)c^2[/itex]. Therefore the 4-momentum becomes [itex]p_\mu=((\gamma m+\mu)c,p_x,p_y,p_z)[/itex]. Another thing that is important to realize is that if the object it collides with was originally in motion with a speed other than v then its gamma factor differs from the space ship's gamma factor!
 
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1. How does special relativity (SR) affect the increase in mass?

According to special relativity, as an object approaches the speed of light, its mass increases. This is due to the fact that energy and mass are equivalent, and as the object's velocity increases, so does its energy. This increase in energy results in an increase in mass, as described by Einstein's famous equation E=mc².

2. Is there a limit to how much an object's mass can increase due to SR?

Yes, there is a limit to how much an object's mass can increase due to special relativity. This limit is known as the "infinity mass limit" and it occurs when an object reaches the speed of light. At this point, the object's mass would theoretically become infinite, but this is not physically possible.

3. Does the increase in mass affect the object's size?

No, the increase in mass due to special relativity does not affect the object's physical size. This is because the increase in mass is a result of the object's energy increasing, not its physical dimensions. However, as an object's mass increases, it would require more energy to accelerate it further, making it appear to have a larger size to an outside observer.

4. Can an object's mass decrease due to special relativity?

Yes, an object's mass can decrease due to special relativity. This occurs when an object's velocity decreases, resulting in a decrease in its energy and subsequently, its mass. This phenomenon is known as "mass-energy equivalence" and is a fundamental principle of special relativity.

5. How does special relativity affect the laws of conservation of mass and energy?

In special relativity, the laws of conservation of mass and energy are combined into one law known as the "conservation of mass-energy." This means that mass and energy can be converted into each other, and the total amount of mass-energy in a closed system remains constant. This is demonstrated in nuclear reactions, where a small amount of mass is converted into a large amount of energy, as described by Einstein's equation E=mc².

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