# E = mc2 . behaviour of m?

1. Nov 11, 2005

### Lakshminarayanan.V

e = mc2..... behaviour of m?????

hi all

scenario: in space.

when an article is being accelerated to the speed of light it is said that it can achieve a max of v=c and its mass starts to increase ( it adds energy to mass rather than velocity.). fine. now without using e = mc2 or any other einstein derived eqn pls explain, how a body (that is accelerated) adds the energy to mass, rather why does it become heavier and heavier as it reaches speeds close to that of light? if u cud explain it considering the constituents of the article (atoms) that wud go very well with the sort of explanation i need. thank you.

reg.

vln

2. Nov 11, 2005

### daniel_i_l

just like time and space, mass is relative to. If you agree that you can mesure mass with a ruler and a clock, then since those mesurments are relative, the mass that you mesure will be relative to. It is important to note that the rest-mass dosn't change.
you might want to look here:

Last edited by a moderator: Apr 21, 2017
3. Nov 11, 2005

### sruthisupriya

i don't know if I am right, but we know that energy acquired by a moving body increases its KE(velocity) and viceversa.So, when the velocity approaches c,the energy goes on increasing.but, the velocity cannot go beyond c and hence the energy has to manifest itself in another form.And at that point, the only way to do so is to add up to the mass.sorry, but e=mc^2 cannot just stop being there.

4. Nov 11, 2005

### mitchellmckain

This idea of increased mass is very misleading. If you get on a ship and accelerate to 99.5% of the speed of light, nothing on the ship changes in any way. However if a particle of mass m is moving toward you with this velocity its total energy is 10 mc^2, which is the rest mass energy mc^2 plus a kinetic energy of 9 mc^2. There is a similar increase in momentum to 10mv. Some people like to call this kinetic energy and momentum increased or relativistic mass. This would mean that it is the mass that increases from m to M = 10 m and then momentum would be given by the usual formula Mv and instead of kinetic energy we have increased mass energy given by Mc^2.
I think this is a little absurd and apparently the opinion of many of educators in relativity has been shifting in this direction. See my thread entitled "Idea of increased mass at relativistic speeds" lower in the list of threads, for more details.

5. Nov 11, 2005

### MichaelW24

I would say the mass does not increase, as the value of "m" in equations does not increase. For example the poster above said "if a particle of mass m is moving toward you with this velocity its total energy is 10 mc^2", which is true but clearly m still stands for the rest mass, which does not change.

Instead, I'd say that the momentum changes, and the momentum in relativity is given by p = gamma*m*v, so again the mass m does not change. However by applying a constant force on a particle of mass m, it will accelerate from rest, but usually you assume that a constant force means a constant acceleration, only the case when mass is constant. If you say that the force is equal to d(p)/dt, (the more general defenition), and then use the modified formula for momentum, then the momentum can continually increase, increasing the tital energy of the particle. The only thing, is that at any instant when the particle is travelling at v, the increase in velocity is overtaken by the increase in the value gamma*m as v gets close to c..

Thats how I understand it

6. Nov 11, 2005

### Jimmy Snyder

An article cannot be accelerated to the speed of light. No matter how much you accelerate it, it will be going at less than that speed. Even at slow speeds, as you add energy to the article, some of that energy goes to increased speed, and some to increased mass, but the increased mass is not noticable. As the speed increases beyond some value, you begin to notice that not all of the energy you supply is showing up as increased velocity and the measured mass is increasing. The speed at which this happens depends upon the precision of your measuring instruments. As you continue to add energy, you will find it increasingly difficult to measure any difference in the speed of the article, but it will be increasing. Again, the point at which this happens depends upon your measuring instrument. When it does happen, it will seem as though all of the energy you add shows up as mass. In modern particle accelerators electrons reach a state pretty close to this.

7. Nov 11, 2005

### MichaelW24

But surely the mass isn't increasing, gamma is increasing which allows the momentum to steadily rise with a constant force; the mass "m" which is used in all equations always refers to the rest mass. Unless you define another kind of mass as "gamma*m" it may get quite confusing when you say the mass, which newcomers like me would take to mean "m", increases...

8. Nov 11, 2005

### Jimmy Snyder

I'm a newcomer myself. Certainly there is a frame invariant kind of mass called rest mass and this doesn't change with velocity. But there also is inertial mass that resists change in velocity. The energy that you add to the article in order to make it go faster, fails to have the effect you would expect if you were pushing rest mass. There is more inertia than rest mass can account for. This is what I am calling increased mass.

9. Nov 11, 2005

### MichaelW24

Ok, this extra inertial mass should be (gamma - 1)m I think... do you know if this contributes to gravity? Its quite confusing as I was given a hint that the m in F=ma isnt nessecarily the same as the m in, for example, W = mg describing a body's weight.

10. Nov 11, 2005

### Jimmy Snyder

Unfortunately I don't know. But I have strong reason to believe it does. The energy that you put into the particle was contributing to gravity. Now that it has been converted to mass, it seems that it should continue to do so.

11. Nov 11, 2005

### Pippo

I do have a strange idea.
Mass creates the space, if a mass moves there is a sort of compression of the space in the direction of the mass movements, this compression make the distance to be covered by the movement multiplied by a factor depending by the speed (here the other strange idea is that the space, being create by a mass, moves at C speed !)....I can continue if someone is interested, but don't kill me !! Tks

Last edited: Nov 11, 2005
12. Nov 11, 2005

### pmb_phy

Sure - The weight of a body has to do with acceleration. Acceleration is dependant of time and time is relative. These things together yield an increase in weight.

Pete

13. Nov 11, 2005

### pervect

Staff Emeritus
There area some technicalities that make computing the gravitational field of a moving object difficult, in fact I have a long-standing argument with another poster here on the topic. (Of course I think I'm right :-) - and the other poster might argue that it's easy, but in spite of his arguments I think he's still wrong :-)).

Without going into too much detail, though, like arguing, for instance "what the heck is a gravitational field in GR anyway?", you can think of the gravitational field of a moving mass as being definitely non-uniform. If the particle is moving to the left and right across the page, the field in the direction of motion will be "weak"

(weak field)----->(weak field)

Given the same direction of motion, the field transverse to the direction will be "strong"

(strong field)
----->
(strong field)

(I've drawn two separate diagrams due to the vagaries of ascii art on the system.)

This is reasonably close to the truth by anybodys definition of "gravitational field" that matches the Newtonian definition in the Newtonian limit, and it matches the behavior of the electric field (which exhibits similar but not numerically identical behavior), and should convey the essentials without getting into a great deal of highly technical arguments.

You may recall that there is an intergal of the electric field which gives a conserved charge for the electrostatic case. In the gravitational case, there is an intergal of a pseudo-tensor (which is not actually thought of as a field as far as I know) that gives the "gravitational mass" of a moving particle in an asymptotically flat space-time. This pseudo-tensor does indeed give an ADM mass (as it is called) of gamma*m, where m is the rest mass.

I believe the Bondi mass will also be gamma*m, but I haven't gone through the calculation. (The textbook treatment I have for the Bondi mass is too complex to get a result, and I'd have to order the original paper from Bondi from the library to get a simpler coordinate-based treatement).

The Bondi and the ADM mass are two different definitions of mass in GR. (There is a third sort of mass, the Komar mass, that doesn't apply because it requires a static space-time and the metric of a moving mass is not static).

The fact that GR has the need for three different sorts of mass should serve as a warning that the subject is highly technical, and that the above discussion has only "scratched the surface".

14. Nov 11, 2005

### mitchellmckain

The problem is that this idea of an increased resistance to velocity is hidebound to a single reference frame. Put a small ship inside a big ship and accelerate it to 99.5% of the speed of light. Now if the small ship comes out, does it have any great difficulty accelerating away from the big ship due to increased mass? NO! There is no increased mass and the little ship can accelerate just fine until it is traveling away from the big ship at 99.5% of the speed of light. It is only if you stick to the view point of observers back where the big ship started, because relative to them the small ship has accelerated to only 99.9987% of the speed of light and if they want they can invent this idea of relativistic mass to explain where the energy went or they can understand that their inertial frame is not the only frame in which things can be measured and realize that nothing has happened to the mass of the ship at all.

Yes it contributes to gravity because all energy contributes to gravity so you are still free to call it an increase of mass or call it kinetic energy. The real question you have to ask is if you really want to use a different mass in F=ma depending on which direction the force is in. For the relativistic mass M = gamma m can only be used in F = M a if the force is in the same direction as the relativistic velocity. For a force in a perpendicular direction you must use the rest mass in F = m a. But it makes more sense to me, just change the formula to F = gamma m a, where the mass does not change and it is only gamma which depends on the velocity in the direction of the force. It makes more sense to me, to call the additional energy added by motion, kinetic energy rather than an addition to the mass.

15. Nov 12, 2005

### pmb_phy

This problem is a lot easier than you all think. All one needs to do is use Einstein's 1905 SR paper and the principle of equivalence. Note that if m = relativistic mass then if a particle is moving on a scale then its moving perpendicular to the particle's acceleration. If F = transverse acceleration and a = acceleration of scale = local acceleration due to gravity then F = ma. This relation is derived based on two principles - Lorentz contraction and time dilation. Therefore the answer to the question posed has everything to do with the properties of spacetime.

Pete

16. Nov 12, 2005

### Pippo

Dear all,
is it possible to explain in few words why the speed of the light is in the middle of the relativistic theory? How it can be related with space/time, does it exist a comphrensible explanation ? Thank you for any answer

Last edited: Nov 12, 2005
17. Nov 12, 2005

### Jimmy Snyder

This is true. I had mentioned that rest mass is frame invariant, but forgot to add that this increased inertia is not.

18. Nov 12, 2005

### Harmeet Singh

OK,Ill explain it to you in a different way.We know that the force between 2 charges depend on epsilennot and the force between two current carrying conductors depends upon munot.The relation between c,epsilen,and,munot
is c=1/(munot*epsilennot)^1/2.
Now since all inertial frames are supposed to be equivalent the forces between 2 charges and two current carrying conductors should be the same in all inertial frames.Now if there exists a frame where the value of c is different then the forces between two charges and two current carrying conductors will be different in them because as c changes epsilon and mu also changes.But this cannot be true because the laws of nature are the same in every inertial frame.Hence the speed of light has to be a universal constant.To make this true all the meaurements related to mass length and time vary in different inertial frames accordingly only to keep the speed of light constant for different observers.

19. Nov 12, 2005

### Pippo

If C is costant for two observers it means that it has nothing to do with them but only with the space time around them, so C being the only absolute value it is considered as a reference point for the relativistic theory, is just a reference point. Is this correct?

Last edited: Nov 12, 2005
20. Nov 12, 2005

### mitchellmckain

No. c represents a measurable quantity, the speed of light in a vacuum. Is it only a coincidence that electrodynamic depends on a constant that happens to be the speed of light in our inertial frame? Is the speed of light in a vacuum different in different directions? Does this measurement of the speed of light in a vacuum depend on how fast the earth is moving? If so, relative to what? The postulate of special relativity answers all these questions in the negative. To answer any in the affirmative would mean that a planet moving at a different velocity would derive different laws or constants of physics.