# Relativistic mass,momentum and energy.

1. Jul 30, 2009

### Urmi Roy

Hi,
I found a derivation of the forumla for relativistic momentum in Resnick and Halliday,which says that --

Therefore, p= mv= m(dx/dt) (classiscal mechanics) -----eqn(1)

(actually its delta x by delta t but I couldn't find the symbol for delta)

Where dx is the distance travelled a moving particle as viewed by an observer watching the particle.

However, to be in accordance to relativity,it should be--

p=mv' =m(dx/dt') (in relativity) ----eqn (2)

Where dt' is time required to travel a particular distance measured by an observer moving with the particle.

therefore,
p= m(dx dt) /(dt dt') = m[dx*(gamma)]/dt

(gamma is the time dilation factor).

p=mv(gamma)

My question related to this is-

Why are the time and space coordinates taken with respect to different observers?

Last edited: Jul 30, 2009
2. Jul 30, 2009

### vin300

Because distance measured by stationary observer on the time elapsed in the moving coordinates is the proper velocity, proper velocity times inertial mass gives proper momentum which is a conserved four vector as momentum in 3D
Or you could write it as relativistic mass time velocity but there is an unresolved disagreement on the use of relativistic mass

3. Jul 30, 2009

### Staff: Mentor

Hi Urmi Roy,

IMO, a better way to look at it is in terms of 4-vectors. You want a definition of four-momentum (energy, momentum, and mass) which is covariant under Lorentz transforms. To get that you can use only invariant or covariant quantities, like the four-position and the proper time.

4. Jul 30, 2009

### Urmi Roy

Right,I understand this--so what it implies is that the correct value (the one we're trying to calculate through the formula) is the one,where both the coordinates are measured from their 'proper' frames.

However,the fact lies that,when we're making the measurements from one particular frame of reference,we would calculate the momentum as what we see ourselves--say that I,in the stationary frame mentioned earlier, want to measure the momentum of the particle--I could look up and just observe the distance it travelled in a particular span of time w.r.t. to me and divide the dist. by my time---wouldn't that be the momentum w.r.t to me?

The formula for relativistic momentum seems to calculate a value of momentum which is common to all reference frames viewing the event--at all relative speeds possible,inspite of the fact that the term 'proper' is relative,(since the observer who is moving in my reference frame,might say that it's me who's moving).

According to what I said earlier,different people may calculate different momenta from their different reference frames--how do I explain the discrepancy between my thought and the result of the relativistic momentum formula?

5. Jul 30, 2009

### Urmi Roy

I'm ashamed to say this,DaleSpam but I don't think I know enough yet to talk in terms of these terms!! I only know the very basics of relativity,in purely physical terms.

6. Jul 30, 2009

### ZikZak

No, the momentum with respect to you is $p= m (\gamma v) = \gamma m v$. Only in the Newtonian limit is it $mv$.

I don't understand your confusion. The relativistic momentum is $\gamma m v,$ which depends on the observed coordinate velocity v. It is certainly not common to all reference frames, since the body's velocity is different in each one.

As far as considering 4-vectors and so forth, I think that the important point to take away from that is that we get to define momentum in whatever way happens to be useful. It happens that the quantity $\gamma m {\bf v}$ is conserved in Relativity, and it converges to the Newtonian momentum in the nonrelativistic limit, so that is what we use and call the momentum.

7. Jul 30, 2009

### Urmi Roy

I'll go back to my question...
What I mean is that in the calculation that I put in my opening post,the relativistic momentum is calculated using the space coordinates and time coordinates for different observers,which is explained by what vin300 said,but the thing is that if I have the required instruments and in my reference frame and I calculate the momentum of the particle by using the space coordinate and time coordinate according to my frame,then that is the momentum according to me--which is different from that calculated by the formula.Thus,the result of the relativistic formula is not that momentum that I observe--this is the discrepancy I'm talking about.

8. Jul 30, 2009

### ZikZak

When you make these measurements you're talking about, why are you using anything other than the relativistic momentum formula to calculate relativistic momentum? When you measure mass and velocity, you're not measuring momentum; you're measuring mass and velocity, from which you can derive the momentum.... if you use the correct formula.

If you want to measure momentum directly, you would measure the total impulse required to bring the object to rest, an experiment that would yield the value $\gamma m v.$

9. Jul 30, 2009

### Staff: Mentor

10. Jul 30, 2009

### Urmi Roy

Thanks,ZikZak for paying attention to my problem.Actually,I'm so used to momentum being defined as mv that its posing quite a problem at this initial stage of my study of relativity.

So basically,just like when Newton put down the expression for momentum in classical mechanics,he did so after making many experimental observations and rounding them up to give us the mathematical expression for it----if we did the same sort of experiments at relativistic speeds,we would get $\gamma m v.$.Right?

Then,the person in that particular reference frame would look back retrospectively and say that the experimental values make sense if we define momentum by making a product of gamma,the rest mass of the object and the ratio of the distance travelled by the time elapsed(both in their respective proper frames of reference).Right?

Otherwise,as I said before,since the term 'proper' is relative,is there no clear reason that we have to take the distance from the stationary observer's view and the time as per the moving observer's view?

Last edited: Jul 31, 2009
11. Jul 31, 2009

### Fredrik

Staff Emeritus
This post might be helpful, even if you don't understand all of it. It defines four-velocity. Four-momentum is the rest mass times the four-velocity.

12. Jul 31, 2009

### Urmi Roy

Thanks Dalespam and Fredrik, for giving me those links,I'll take some time to go through and uderstand them.
In the mean time,I'd like to clear up my last problems in regard to this topic.

1. E=mc2 is defined for even an object in its rest frame--where m is the rest mass. Now, since Einstein said that the kinetic energy provided to a body emerges as inertia,how can we explain what this energy,(that exists in the body when it is at rest),actually is?After all, the body in this state has no kinetic energy.

2.When we define kinetic energy of a body in relativity,we say that it is the difference of the total energy and the energy of the rest mass.This definition could also apply to potential energy of a body,if it didn't have any motional energy, couldn't it?

Also,please go through the last part of my last post and confirm if my understanding is correct.

Thanks for your cooperation,everyone.

13. Jul 31, 2009

### vin300

The coefficient of restitution of your post on my head is equal to unity
This urges me to learn

14. Jul 31, 2009

### ZikZak

That is indeed what I was getting at. "Momentum" is a useful concept for various reasons, one of which is that it is conserved. So if you were to do these series of experiments with sufficient accuracy, you would find that it was $\gamma mv,$ not $mv$, that was conserved for massive particles in collisions. The only reason that Newton did not notice this was that in his experiments, $\gamma$ would have always been indistinguishable from unity.

15. Aug 1, 2009

### Urmi Roy

16. Aug 1, 2009

### Fredrik

Staff Emeritus
I think the reason why he says "to be in accordance to relativity,it should be" is that if x and t are both four-vector components, then dx/dt isn't a four-vector component, but if you replace dt with the coordinate independent dt', what you get is a four-vector component.

He's just trying to guess how to modify the non-relativistic definition to get something that looks relativistic. It's not really possible to justify the definition this way. Only experiments can do that. He should have been more clear about the fact that his line of reasoning only leads to a definition that might be appropriate for SR. (In his defense, he did say "should", not "must").

17. Aug 1, 2009

### Urmi Roy

Isn't there any better way we can justify this formula,if the derivation given in "Resnick and Halliday" is ambiguous?
I would rather have an 'logical explanation' of the phenomenon than just a plain mathematical derivation,if that is possible.The reason I stuck to the above given derivation is that it has more,physical logic than mathematics,but if it's ambiguous,I guess I need a better derivation.

Also,my last two questions remain unanswered (post 12)!

18. Aug 2, 2009

### Fredrik

Staff Emeritus
It's not a derivation. It's a definition. Definitions can't be derived. The additional stuff in there is just an attempt to make you see that we have good reasons to think that this particular definition might work.

This is why I dislike the traditional presentation of SR so much. Everyone misunderstands it exactly the same way you did. (I did too). Many professors who teach SR as a part of a course in classical mechanics or electrodynamics still don't get it. Those calculations in the early parts of a traditional presentation of SR are not derivations. They are only supposed to help us guess the "correct" theory.

1. Physics doesn't tell us what anything "actually is", but feel free to rephrase the question if you'd like.
2. I actually don't know how potential energy works in SR, but if you apply "this definition", which in units such that c=1 says that $E=\gamma m-m$, to a system at rest ($\gamma=1$), we get E=0.

19. Aug 3, 2009

### Urmi Roy

Just to do some extra reading on this topic,I spent some time googling,and I came accross a website,where the author said--"I think of momentum as being inversely proportional to the wavelength observed. I disconnect it, conceptually, from the objects velocity entirely."

What does this actually mean?

20. Aug 4, 2009

### Urmi Roy

In another physicsforums thread,I found that "What I get from dextercioby's and pervect's posts is that the reason the 4-position is differentiated wrt the proper time, is because the 4-velocity will then be the same for all observers, where if instead, each observer used their respective times, they would get different answers."

This was on https://www.physicsforums.com/showthread.php?t=68455

Perhaps I can now rephrase my question and put it like this--
If 4-velocity is defined as $$\vec{V} = \frac{d\vec{X}}{d\tau} \ \ \ \ \ \ \ \ \ where \ \tau \ \ is \ the \ proper \ time$$

then momentum is M$$\vec{V}$$$$\ gamma$$----this agrees with the experimental findings,as Zik Zak suggested(post #8)---however, the 4-vectors are merely mathematical tools used to aid our calculations,so it is the gamma factor that always modifies the 'mv' product to yield the correct value (which agrees with experiment)---but again,is this not just a coincidence----the same gamma factor that occurs in all relativistic quantities, modifies a product(mv),{where v is the 4-velocity,a mathematical aid to assist calculations}, and again,helps us to get the correct answer!

Am I getting it wrong somewhere?

Last edited: Aug 4, 2009