Why not find mass increase in a simple way?

[alba]

The formula usually givento find the relativistic energy of a particle is :## E^2 = p^2c^2 + m^2c^4 ## which is derived from the original Lorentz formula for mass: ##m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} ## which gives the value of the total mass including the rest mass and, subtracting the latter, we get the kinetic energy (net increase of total energy):
##E_k=m_0c^2*\left[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} -1\right]##

If this is correct we know that when v = 0.866c the total energy/mass is twice the rest mass and the Ek is equivalent to one rest mass, in the case of an electron : 0.511 MeV.

My question is:
since E=M, why is it not suggested to find the increased energy (the kinetic energy) in such a simple way? Do you know of any case in which this simple method is not exact, or inconsistent with theory or experiment?
If we accelerate an electron by a kinetic energy of .511MeV we know that it will have energy/mass of 2 electrons, and if we (at LHC) give 7 TeV (7 000 GeV) to a proton of .938 GeV we easily find that the energy/mass has increased by 7000/.938 = 7461 times and the speed is sqrt(1-1/7461^2) =.999999991 c.

Why can't be as easy as that?

 

[Orodruin]

First of all, saying that mass increases with speed is a reliquary from the early days of relativity. You will in general not find physicists talking about relativistic mass. Please see https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

The relation ##E^2 = p^2c^2+ m^2c^4## is more general than the relation you quote as its source. For a massive particle you would find the kinetic energy by taking ##E-mc^2##. If you express it in terms of velocity, it becomes exactly your expression. However, it is not always very interesting to find the kinetic energy in terms of velocity. The necessary information is already contained in the total energy E and the mass m.

 

[alba]

The relation ##E^2 = p^2c^2+ m^2c^4## is more general than the relation you quote as its source. ,
it is not always very interesting to find the kinetic energy in terms of velocity..

In what way that formula is more general?

Why is not interestingto simplify calcs? How do you find , using the "general" formula the speed of a proton with 7 TeV E and the actual mass?Isn't that procedure unnecessarily more complicated?
Also, they say that it is improper to talk about mass increase or relativistic mass, yet , if you want to determine the radius of the circle of a 7TeV proton, you are obliged to consider that its mass has increased by 7461 times.

This method is simple and exact, again, are you aware of any shortcoming?
An instance where it fails?

 

[Orodruin]

In what way that formula is more general?

It is valid for massless particles and derived on more general assumptions.

Why is not interestingto simplify calcs?

Exactly what calculation do you think you are simplifying? If you have E and m, how is your way simpler than ##E-mc^2##?

Also, they say that it is improper to talk about mass increase or relativistic mass, yet , if you want to determine the radius of the circle of a 7TeV proton, you are obliged to consider that its mass has increased by 7461 times.

Reference please. What is the "circle of a proton"? And no, there is nothing that requires you to claim that the mass has increased. What you need to do is to properly account for relativistic kinematics where the inertia is not equal to the mass (and different in different directions! - see the insight).

 

[alba]

Reference please..

At CERN's LHC they run 7 TeV protons P:
can you show how you derive from this datum all other data in a simpler way?, the speed P will reach: .999999991 c, the mass increase : 7461 and the consequent increase of the radius of its circleup to LHC ring's length (27/2pi km) from the expected value of 57 cm?

 

[Orodruin]

the speed P will reach: .999999991 c

The speed is most easily computed using the relation ##v = pc^2/E##. For highly relativistic particles, this can be approximated as
$$
v = \frac{pc^2}{E} = c\sqrt{1 - \frac{m^2c^4}{E^2}} \simeq c \left(1 - \frac{m^2 c^4}{2E^2}\right).
$$

the mass increase

The mass does not increase. Physicists do not use the concept of relativistic mass.

consequent increase of the radius of its circleup to LHC ring's length (27/2pi km) from the expected value of 57 cm?

This depends on the electric field and the change in momentum being equal to the force. The relativistic momentum is not given by the same formula as the classical momentum - that is where the gamma factor and the increase in the radius comes from.

 

[alba]

This depends on the electric field and the change in momentum being equal to the force.
The relativistic momentum is not given by the same formula as the classical momentum - that is where the gamma factor and the increase in the radius comes from.

Can you clarify your sentence in bold? whose EF, the proton's? what force ?
As to momentum,doesn't it change /increase by 7461 times,same as now defunct mass?

 

[Dale]

The formula usually givento find the relativistic energy of a particle is :## E^2 = p^2c^2 + m^2c^4 ## which is derived from the original Lorentz formula for mass: ##m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} ## which gives the value of the total mass including the rest mass

You are mixing up your formulas and variables here. In the first formula, ##m## is the invariant rest mass. In the second formula ##m## is the now deprecated relativistic mass and ##m_0## is the invariant mass.

Calculating the mass increase is exceptionally easy now. It is 0. Mass now refers to the invariant mass which does not increase.

 

[alba]

Calculating the mass increase is exceptionally easy now. It is 0. Mass now refers to the invariant mass which does not increase.

The invariant mass of the proton is and remains .938 GeV. Yet, I asked, when a P gets 7000 GeV and reaches .99 99 99 99 1 c, bent by a B-field of about 5.5 T, it circles the LHC main ring ( r = 4.24Km in radius), whereas its radius should be about 57 cm (r= p/B= mv 1836*3*10^10/ B 9.7*10^11), if its momentum did not increase by 7461 times and there were no extramass of any kind.

The explanation given by orodruin is not clear, so, why does the radius increase if the total momentum/mass of P stays the same?

 

[Nugatory]

why does the radius increase if the total momentum/mass of P stays the same?

The momentum, which is defined by the relation ##E^2=(m_0c^2)^2+(pc)^2##, does increase.

You are confusing yourself by assuming that the momentum must also be equal to ##mv##, and therefore that you must find a definition of ##m## that makes ##p=mv## true.

 

[Dale]

The invariant mass of the proton is and remains .938 GeV. Yet, I asked, when a P gets 7000 GeV and reaches .99 99 99 99 1 c, bent by a B-field of about 5.5 T, it circles the LHC main ring ( r = 4.24Km in radius), whereas its radius should be about 57 cm (r= p/B= mv 1836*3*10^10/ B 9.7*10^11), if its momentum did not increase by 7461 times and there were no extramass of any kind.

The explanation given by orodruin is not clear, so, why does the radius increase if the total momentum/mass of P stays the same?

I think that your numbers are off, but in any case, if a proton has a KE of 7000 GeV then by ##m^2=E^2-p^2## the momentum is pretty close to 7000 GeV also (in particle physics units). If you use that and correct numbers then you will get the correct radius. As long as you don't incorrectly assume that ##p=mv## it works out, as mentioned by @Nugatory above also.

 

[DrStupid]

Why can't be as easy as that?

Of course it can and it has been done. In http://wikilivres.ca/wiki/Die_magnetische_und_elektrische_Ablenkbarkeit_der_Bequerelstrahlen Kaufmann tried to describe the increase of the mass as an elecromagnetic effect. Later it was clear that he in fact measured the relativistic mass of the electons.

Your problem is not physical but semantic. 100 year ago the sentence "Mass increases with speed." was correct but today it is wrong. The laws of nature are still the same but the meaning of the word "mass" has been changed. In classical mechanics and the early days of relativity it was the factor between momentum and velocity. With this original meaning the sentence is correct. Now it is the factor between four-momentum and four-velocity. With this new meaning the sentence is wrong. Therefore you must not say or write it anymore. Even the still corect wording "Relativistic mass increases with speed." is usually not accepted because many physicists are not willing or able to use this property.

 

[jtbell]

we get the kinetic energy (net increase of total energy):
##E_k=m_0c^2*\left[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} -1\right]##

This is a perfectly valid formula. I've even written it occasionally myself, in the form ##E_k = (\gamma - 1)mc^2## where ##\gamma = 1 / \sqrt{1 - v^2 / c^2}##. (I'm in the camp that always uses ##m## to mean invariant mass.)

It's simply not a very useful formula in practice, because we don't normally accelerate charged particles to a specific velocity ##v##. We accelerate them to a specific kinetic energy, using a specific potential difference. If we accelerate an electron or proton from rest, through a potential difference of 1000 kV, it acquires a kinetic energy of 1000 keV. Simple as that!

Its (total) energy is then $$E = E_k + mc^2 = 1000~\rm{keV} + 511~\rm{keV} = 1511~\rm{keV}.$$ Its momentum (in energy units) is $$pc = \sqrt{E^2 - (mc^2)^2} = \sqrt{(1511~\rm{keV})^2 - (511~\rm{keV})^2} = 1422~\rm{keV}.$$ And if for some reason we want its speed, we use Orodruin's formula: $$v = \left( \frac {1422~\rm{keV}} {1511~\rm{keV}} \right) c = 0.941c = 2.82 \times 10^8~\rm{m/s}$$

 

[alba]

...Its momentum (in energy units) is $$pc = \sqrt{E^2 - (mc^2)^2} = \sqrt{(1511~\rm{keV})^2 - (511~\rm{keV})^2} = 1422~\rm{keV}...$$

if, as I suppose, then you divide it by the unit( 511k) in order to find the radius, can someone explain what is the difference? it seems just as an unnecessarily complicated way to say the same thing and get the same result.
What have you done is multiply the trig functions by 1/cos (which is M or E, the total mass energy) so that the hypotenuse becomes the momentum of c, the cos (1/M) becomes the unity, and the sine (now the tangent 2.78) becomes the momentum of the actual speed .94108 c.

Can you explain what is the point/sense/use of this more complicated procedure?,
especially how:

"It is valid for massless particles and derived on more general assumptions."

I can't see how this can be valid for photons or massless particles, since for them momentum is E/c whereas here you are still using, in a covert manner, the energy of the rest mass of the electron, so yo , at the end of the day, are still considering the momentum mv of massive particles.

What am I missing?

 

[Nugatory]

Can you explain what is the point/sense/use of this more complicated procedure?,

In particle physics we almost never know the velocity up front (see jtbell's post #23 above) and we almost never need the velocity as part of whatever solution we're looking for, so formulas that relate energy and momentum directly without involving velocity are easier to use.

especially how:
"It is valid for massless particles and derived on more general assumptions."

I can't see how this can be valid for photons or massless particles, since for them momentum is E/c whereas here you are still using, in a covert manner, the energy of the rest mass of the electron, so yo , at the end of the day, are still considering the momentum mv of massive particles.

A massless particle is one with ##m=0##, so ##p=E/c## is just the ##m=0## special case of ##pc=\sqrt{E^2-(mc^2)^2}##.

 

[Dale]

at the end of the day, are still considering the momentum mv of massive particles.

You have been corrected on this point multiple times by multiple posters. It is not acceptable to persist in this. Momentum is given by ##m^2=E^2-p^2## in natural units.

it seems just as an unnecessarily complicated way to say the same thing and get the same result

Relativistic mass is just another name for total energy. So any formula that is less complicated with relativistic mass can be written in that simplified manner using energy instead.

The invariant mass is a separate concept from energy, and there are many formulas (particularly in particle physics) that are simpler with the invariant mass.

So use E (total energy) whenever it simplifies the formula and use m (invariant mass) whenever that simplifies the formula.

 

[vela]

Can you explain what is the point/sense/use of this more complicated procedure?

I have to admit, after going over this thread several times, I still have no idea what you mean when you claim a procedure is "more complicated" or that your method is "simplifying" a calculation. Jtbell's calculations are about as straightforward and simple as you can get. How would you solve the problem?

 

[alba]

I have to admit, after going over this thread several times, I still have no idea what you mean when you claim a procedure is "more complicated" or that your method is "simplifying" a calculation. Jtbell's calculations are about as straightforward and simple as you can get. How would you solve the problem?

At LHC they run protons at 7TeV = 7000 GeV ( rest mass is .93827 GeV),so they are adding the equivalent of 7000/.938 = 7461 proton rest masses (= 1/cosθ (=0.000134)) both if you please to consider it mass or energy : it is not a name or a procedure that shapes or changes reality.

Its speed will be sinθ= sqrt(1-1/7461^2) * c : v = .99 99 99 99 101*2.99*10^10 , its momentum (tanθ) will increase by 7461 times and so will the radius of its circle: in a 5.5-T MF the orbiting radius will increase from 57 cm to r = 57*7461= 4.243 Km, the actual radius of LHC ring (26.6 km).

What you are doing with your formula is setting cos = cos*1/cos (unit), sin=sin/cos = tan (momentum of particle at that KE/speed), 1(=c) /cos (momentum of particle at c) and then working back the info.

Summing it up:
the simple and logic way of finding the radius of the particle is to multiply the regular radius of any proton (57) by 7 T eV / .938 GeV, that is 57*7461 cm.,you haven't yet shown how you figure that out.

If you want to consider the 7461 times increase as mass or energy is surely not solved by changing the order of calculations or a formula, it is a conceptual theoretical problem that has to be solved theoretically, and you should explain why pure energy and not mass responds to a B-Field increasing its inertia and consequently the radius of the orbit: usually Energy does not respond to magnetic fields

I hope my question is at last clear. Thanks you all for your efforts and a Happy New Year!

 

[Dale]

the simple and logic way of finding the radius of the particle is to multiply the regular radius of any proton (57) by 7 T eV / .938 GeV, that is 57*7461 cm.,you haven't yet shown how you figure that out.

The exact same way, except we call the 7 TeV quantity energy, and the .938 GeV quantity mass. As opposed to calling them both mass. None of the physics is changed, just the terminology.

If you want to consider the 7461 times increase as mass or energy is surely not solved by changing the order of calculations or a formula

We are not even changing the order of calculations.

it is a conceptual theoretical problem that has to be solved theoretically

What is the theoretical problem? There is none. We have a conserved Minkowski four-vector, the four-momentum, the timelike component is called energy, the spacelike components are called momentum, and the norm is called mass.

you should explain why pure energy and not mass responds to a B-Field increasing its inertia and consequently the radius of the orbit: usually Energy does not respond to magnetic fields

No, you should explain why you think this incorrect statement is true. For a particle to "respond to a B field" it must have more energy than its rest mass. Otherwise it is at rest and the magnetic force is zero. So energy does respond to magnetic fields.

Your expectation that it doesn't is simply wrong. You need to examine your underlying assumptions here. You are assuming some incorrect things about energy and magnetic fields, and probably inertia too.

 

[alba]

The exact same way, except we call the 7 TeV quantity energy,
We are not even changing the order of calculations.
.

I will not discuss here the theoretical side.
You will greatly oblige me if at last you will show your procedure, if you do, you will see that your claim is not true. You have to do the square root Lorentz calcs in your peculiar way.

You have repeatedly berated me for not listening to what you say, I did nothing of the kind, it is you who have surely missed the fact that I am skipping the Lorentz formula altogether. As you rightfully said, we give kinetic energy and that is the starting point, the method I suggested uses only the mass-energy equivalence, so I get the final result simply by dividing the supplied energy by the rest energy/mass: one and only one step 7000/.938. You can't possibly devise a simpler way or any other way as simple as that. Show me wrong, if you please!

If , as you keep saying it is only an issue of linguistics or semantics, you change the terminology and not the procedure.

 

[DrStupid]

the method I suggested uses only the mass-energy equivalence

No, it doesn't. The mass-energy equivalence is the equivalence between rest mass and rest energy. (Read Einstein's original paper if you don't believe it.) You are referring to the equivalence between relativistic mass and total energy. That's something different. I already told you that you are physically correct but semantic wrong. You are confusing the terms "relativistic mass" and "mass". Physicists can't understand you if you are using the wrong terms and if you use the correct terms most of them don't want to understand you in this particular case because the concept of relativistic mass is too old fashioned for their taste.

And I need to agree with vela. I do not see any advantage in your way. It is not even easy to see the way at all.

 

[alba]

I do not see any advantage in your way. It is not even easy to see the way at all.

The terminology changes too often,but I made my procedure clear.
You do not see anyadvantage, but alas after so many posts, you or nobody has showed in detail how is your better way of doing it. I hope yu are kind enough to show me how many ops/stes do you need to find the radius of a proton with 7 TeV kinetic energy.

 

[Dale]

I get the final result simply by dividing the supplied energy by the rest energy/mass: one and only one step 7000/.938. You can't possibly devise a simpler way

I don't know what you mean by "final result" but ##\text{final result}=E_k/m## is just as simple for me as for you.

If , as you keep saying it is only an issue of linguistics or semantics, you change the terminology and not the procedure.

Yes, exactly. The procedure is the same, but "mass" always refers to "invariant mass", and "relativistic mass" is not used but instead we use "total energy".

 

[alba]

No, it doesn't. The mass-energy equivalence is the equivalence between rest mass and rest energy. You are referring to the equivalence between relativistic mass and total energy.

I am not, but are you saying that mass=energy applies only to rest mass? Is it not univerisally valid? I am referring to KE added to the rest mass, is the equivalence valid for it?

 

[alba]

I don't know what you mean by "final result" but ##\text{final result}=E_k/m## is just as simple for me as for you.

Yes, exactly. The procedure is the same, but "mass" always refers to "invariant mass", and "relativistic mass" is not used but instead we use "total energy".

I make no bones of whatever you wish me to call it, since may be very soon it will change. Since your objection is only to words, any terms suits me fine. Now, after so many posts, can you please at last show me what is the 'correct' way in which you determine the radius , step by step, and show me why my way is not simpler and better than yours?

 

[DrStupid]

are you saying that mass=energy applies only to rest mass?

No, I'm saying that the term "mass-energy equivalence" refers to the equivalence of rest mass and rest energy.
There is also an equivalence between relativistic mass and total energy (that's what you are referring to) but it is not called mass-energy equivalence because relativistic mass is not mass.

I am referring to KE added to the rest mass, is the equivalence valid for it?

The equivalence is correct but the terms "mass" for the sum of KE and rest mass as well as "mass-energy equivalence" for this equivalence are wrong.

Again: You can't expect that somebody agrees with you if you are using wrong terms.

 

[alba]

The equivalence is correct but the terms "mass" for the sum of KE and rest mass as well as "mass-energy equivalence" for this equivalence are wrong..

All right, let's make something clear :
rest mass is equivalent to energy
toenergy of any kind bound to massive particles is equivalent to mass, it applies r thermal energy, KE, gluons etc...is that right?
I was referring to that principle if it is valid. The proton has rest mass equivalent to .938 GeV plus KE that is equivalent to 7461 rest masses of the p.The increased energy of the p is different from its rest mass, it is not relativistic mass, yet is equivalent to masslike heat in a hit bar of gold

besides those being basic notions, if I were wrong I could never get the exact result

 

[russ_watters]

I have to admit, after going over this thread several times, I still have no idea what you mean when you claim a procedure is "more complicated" or that your method is "simplifying" a calculation. Jtbell's calculations are about as straightforward and simple as you can get. How would you solve the problem?

It looks to me like the OP's method was a "cheat" where you are provided with specific input data that skips most of the work in the problem (being "given" kinetic energy instead of it being part of the calculation). It is similar to how in engineering, we roll-up several constants and typical variables into simplified equations that work for a specific set of conditions but aren't generally/broadly applicable. That's all the OP did. It isn't wrong, per se, it just isn't as useful as the OP thinks it is.

 

[DrStupid]

rest mass is equivalent to energy

No, it's not. Rest mass is equivalent to rest energy.

 

[alba]

No, it's not. Rest mass is equivalent to rest energy.

OK, (rest) energy,...

 

[DrStupid]

OK, (rest) energy,...

Did you understand that kinetic energy and mass are not equivalent?

 

[Vanadium 50]

Can we close this? Alba's question "why" has been answered, and he still feels everybody else is doing it wrong. More discussion is unlikely to change anything.

 

[jartsa]

So we have a proton that is 1000 times more energetic than a proton at rest. And we want to know how much it resists when we try to change its velocity?

Let's see this equation:
p=vE/c²

The larger the E, the larger the change of momentum associated with some small delta v.

The energetic proton needs to be given 1000 times larger impulse in order to change it velocity a small amount, compared to the impulse needed to change the velocity of the still standing proton the same small amount.

This only works with transverse forces.

 

[jbriggs444]

Let's see this equation:
p=vE/c²
[...]
The energetic proton needs to be given 1000 times larger impulse in order to change it velocity a small amount, compared to the impulse needed to change the velocity of the still standing proton the same small amount.

That factor of 1000 holds good for a transverse acceleration where the energy is held constant, but the direction of the velocity is changed by the impulse. For an applied impulse with a non-zero component in the direction of the velocity, the relationship is not so simple. ##\frac{dE}{dv}## enters in.

Edit: Sorry, as Orodruin pointed out, you already said as much. *blush*

 

[Orodruin]

That factor of 1000 holds good for a transverse acceleration where the energy is held constant, but the direction of the velocity is changed by the impulse. For an applied impulse with a non-zero component in the direction of the velocity, the relationship is not so simple.

To be fair, he said as much:

This only works with transverse forces.