Relativistic mass,momentum and energy.

[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 traveled 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?

 

[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

 

[Dale]

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.

 

[Urmi Roy]

Because distance measured by stationary observer on the time elapsed in the moving coordinates is the proper velocity,

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 traveled 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?

 

[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.

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.

 

[ZikZak]

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 traveled 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?

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

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).

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.

 

[Urmi Roy]

I'll go back to my question...

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 traveled 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?

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?

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.

 

[ZikZak]

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.

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 $\gamma m v.$

 

[Dale]

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.

No need to be ashamed. Try the following Wikipedia and Hyperphysics pages:
http://en.wikipedia.org/wiki/Four-vector
http://en.wikipedia.org/wiki/Four-momentum
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html

If those don't clear things up please let me know.

 

[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.

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 $\gamma m v.$

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 traveled 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?

 

[Fredrik]

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.

 

[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.

 

[vin300]

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.

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

 

[ZikZak]

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 traveled by the time elapsed(both in their respective proper frames of reference).Right?

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.

 

[Urmi Roy]

Thanks,ZikZak,but what about this part of it?

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?

 

[Fredrik]

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").

 

[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)!

 

[Fredrik]

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)!

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. 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?

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.

 

[Urmi Roy]

Just to do some extra reading on this topic,I spent some time googling,and I came across 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?

 

[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?

 

[Fredrik]

I think the point of the first comment is just that the author prefers to think of momentum as something unrelated to velocity. I don't know why he does that.

The 4-velocity isn't the same for all observers. It's not Lorentz invariant, it's Lorentz covariant, which means that its components change from one inertial frame to another as described by the Lorentz transformation. Proper time on the other hand, is invariant. It's a coordinate independent property of a curve. That's why differentiation of a 4-vector with respect to proper time yields another 4-vector.

What is a "coincidence" in mathematics? Strictly speaking, nothing is. When people describe a mathematical result as a "coincidence", they usuallly mean (I think) that it's hard to see any connection between the calculations that led to this particular result, and the mathematics that you would usually associate with the result. If this is what we mean by a coincidence, then the presence of gamma in a definition in SR is hardly a coincidence. If you mean something else, then maybe it is.

 

[Urmi Roy]

What is a "coincidence" in mathematics? Strictly speaking, nothing is. When people describe a mathematical result as a "coincidence", they usuallly mean (I think) that it's hard to see any connection between the calculations that led to this particular result, and the mathematics that you would usually associate with the result.

Yes, indeed,this is what I meant, and you're right,the last post I made was a rather naive one--sent by me,who doesn't have too much of knowledge in this topic!

I think I'll be sorted if you do me a last favour of perhaps suggesting a website where I could perhaps look through the entire derivation of relativistic momentum and understand which expression came from where (except wikipedia,where the derivation was a bit too complicated).

 

[Fredrik]

What exactly is it that you'd like to see derived? As I said before, p=mu, where u is the 4-velocity, is a definition, so it can't be derived. u="the normalized tangent vector of the world line" is another definition.

 

[Dale]

If you would like the derivation of the Newtonian limit using this definintion then that is fairly straight-forward.

 

[vin300]

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?

.

My pleasure, Urmi, that you aroused the thinker in me to answer this really out of the box question.
Mass and energy are properties of a body(say it any way). A mass, in itself and its vicinity(the meaning of vicinity depends on mass) has energy which combine to give a total of mc^2
When you speak of potantial energy, it implies overlapping of two fields to hype these properties.
It thus does not constitute the "" of the body originally, it is an external additive property.
In relativity, this mass is included in the rest mass.
Note that, similarly, thermal energy is included in rest energy and angular energy is included in kinetic energy

Relativistic momentum is mv, the same as in Newtonian, with an exception that m does not remain m0
To understand momentum better, you must know Noether's theorem

 

[Urmi Roy]

Thanks,vin300,perhaps,since you said that the potential energy of a body may be included and accounted for in the expression of rest energy,aswell as the thermal energy,we could say that the inertia posessed by the rest mass is due to the 'potential and thermal energies' posessed by it.This seems to explain the existence of rest mass (inertia) as a manifestation of the invariant potential energy --just as kinetic enrgy is manifested in the relativistic mass.

But,this also implies that the rest mass may change if the conservative field force changes,and also if the temperature of the body changes.How far is that true?

 

[Urmi Roy]

Sorry Fredrik and DaleSpam for abruptly leaving this thread for the last few days,especially since you were so regular in providing your help.

In the mean time,however,I've been trying to get things straight in my head.

I found a website, http://savefile.com/files/2159145
(which you'll have to download to view) in which the approach is such that ignorant people like me can understand the origin of the covariant formulae in relativistic physics and also how we came to include the 'proper time', 'proper velocity' in these formulae.

If you happen to read through it one day, please confirm if what I say in the next paragraph(atleast the lines in bold letters) are okay.


--------"As far as I understood,the article begins by stating the need to define the physical quantities of momentum and force in their covariant forms,so we can always refer to the same expressions each time we want to make a calculation in our study of relativity.
The expression for momentum is already given in its relativistic (and hence covariant) form--which we can arrive at by many ways,(like from the consevation of momentum fromula in wikipedia).

Anyway,the funny thing is that when we take the (coordinate time) derivative of the covariant momentum,we get an expression,which is found to be equally valid in all reference frames,since as we see later,the (gamma cubed multiplied by the coordinate acceleration) quantity is equal for all the frames,and hence,the force felt by the body is the same in all reference frames(since dp/dt is the force).
However,this invariance in the force is not mathematically expressed when we use Newton's formula F=ma (a=coordinate acceleration)--so we have to redefine the formula and put it as m*(gamma cubed)*(coordinate acceleration),where m is the rest mass in order to incorporate the concept.Now,if we rearrange the terms,we find that we can also incorporate the expression for proper time,and hence we obtain the formula for covariant force in terms of proper time.Ofcourse,from this expresion,we can obtain the expression for covariant momentum also, in terms of proper velocity,by dividing by rest mass and integrating."

Thanking you all,
Urmi.

 

[Urmi Roy]

By the way,is Noether's theorem in no way comprehensible by a person who ha just passed high school? DH told me in one of his posts that usually one reads it at the late graduation or postgraduation level.

 

[vin300]

The potential energy in relativity does not contribute to the rest mass.Mass represents inertia which is resistance to motion which is scalar, while in a gravitationally potent field, there is resitance in a dir and assistance in the other, thus not letting pot en to equate to the scalar

 

[bcrowell]

The potential energy in relativity does not contribute to the rest mass.Mass represents inertia which is resistance to motion which is scalar, while in a gravitationally potent field, there is resitance in a dir and assistance in the other, thus not letting pot en to equate to the scalar

No, this is incorrect. Potential energy, like all forms of energy, is a scalar, not a vector. Potential energy does contribute to inertial mass. If some forms of energy contributed to inertial mass and others didn't, then you could violate conservation of momentum (as well as Lorentz invariance) by converting energy from one form into another.

I have the Halliday and Resnick treatment in front of me now. It strikes me as a complete muddle. They start out by saying that the classical p=mv won't be conserved relativistically, which is correct, although they never bother to say why. (One could either offer experimental evidence or a theoretical proof. They do neither.) Then they say you can either abandon conservation of momentum or try to redefine momentum; this is reasonable. But then they simply posit $p=m\Delta x/\Delta t_o$, where $\Delta x$ is measured in one frame and $\Delta t_o$ in another (the particle's rest frame). No motivation. Totally ugly.

Here is my own way of explaining it: http://www.lightandmatter.com/html_books/6mr/ch01/ch01.html#Section1.3 Fredrik is correct that $p=m\gamma v$ is only a definition. However, it's a definition that's subject to two very tight constraints: (1) it has to result in conservation of momentum, and (2) it has to obey the correspondence principle by being consistent with the classical definition of momentum in the appropriate limit.

 

[vin300]

I didn't imply pot.en. is not a scalar.
To make it simple, if you have a mass in a gravitational field, you cannot have a mass more than you actually have.If it's otherwise, there's a limit after which the mass won't stay in the field(and it's too small to cause a considerable change of mass), like when a satellite revolving the Earth gains a velocity more than than the critical valocity and actually has more mass, it gets out of the field. So now you know how things go.
It is definitely not the other way, for that would mean that where there's more potential energy(and hence more mass and more energy as you said), it needs even more energy to exit.What you should note is that in Newtonian physics the potential energy is negative and you made it positive.

 

[Dale]

if you have a mass in a gravitational field, you cannot have a mass more than you actually have.If it's otherwise, there's a limit after which the mass won't stay in the field(and it's too small to cause a considerable change of mass), like when a satellite revolving the Earth gains a velocity more than than the critical valocity and actually has more mass, it gets out of the field. So now you know how things go.
It is definitely not the other way, for that would mean that where there's more potential energy(and hence more mass and more energy as you said), it needs even more energy to exit.What you should note is that in Newtonian physics the potential energy is negative and you made it positive.

I don't really follow this. Since gravity is always attractive the binding energy is always negative resulting in a mass deficit, not "a mass more than you actually have". bcrowell is correct, all forms of energy contribute to the invariant mass of a system. The wikipedia article on binding energy and mass deficit is pretty decent. It particular, the 3rd paragraph of the mass deficit section here specifically talks about the case under discussion of two gravitationally bound objects:
http://en.wikipedia.org/wiki/Binding_energy#Mass_deficit

 

[vin300]

It's a language problem.Thought contribution to mass can only mean more mass.
As a correction to post #29,if in an experimental setup a mass could move to higher up in potential, it gains more mass and slows down due to inertia.