# Does SR and GR agree on Newton's Third Law of Motion ?

1. Apr 20, 2013

### SecretOfnumber

Sorry guys not a physicist!

Only wish to know if there is agreement between Newton laws of motion and SR/GR in particular?I know these two don't match in some cases? What are they in simple words please!?

Cheers

Last edited: Apr 20, 2013
2. Apr 20, 2013

### ShreyasR

In an inertial reference frame, Newton's Laws are valid. In a non-IRF, Newton's second law becomes invalid. In SR, you move with a constant velocity. There is no acceleration, no forces. (deals with IRF only)

GR explores what happens when you do add in acceleration effects due to say, gravity. At this point, space becomes "warped", so that instead of motion along a straight line, you have to consider motion along curved paths. (Deals with non-IRFs)

So now you can make the conclusion.?!

3. Apr 20, 2013

### Bill_K

False! Special relativity handles acceleration quite well. General relativity is a theory of gravity. Acceleration alone does not cause space to become "warped".

4. Apr 20, 2013

### SecretOfnumber

Thanks ShreyasR.

Not quite a conclusion!

1)Lets say two spaceships moving on 0.8C towards each other,one releases a tiny rubber ball ,hitting the other ship(on 0.8C+) what happens to the conservation of momentum rule for the observer watching the incident between two ships?

Cheers
SON

5. Apr 20, 2013

### SecretOfnumber

Now I am confused!!)

6. Apr 20, 2013

### PAllen

Let's take them one by one:

1) Newton's first law: Unchanged by SR; For GR, it is true in a local inertial frame to an accuracy limited by spacetime curvature = tidal gravity.

2) Newton's second law: Modified by both SR and GR in the same way. The need for a modification flows directly from the existence of limiting velocity. Apply force to object, add energy to it (force * distance), yet velocity hardly changes when near c. The modified law can be expressed (for similarity to Newton's) as F = m * <proper acceleration>, with the latter defined as derivative of 4-velocity by proper time.

3) Newton's third law: Generally false in SR, and also GR. It cannot be true for objects influencing each other via fields because of finite propagation time for any signal. By the instant action at a distance, Newton's law of gravity could be consistent with the third law; but no field theory consistent with SR can satisfy it. For idealized collisions it remains true in SR and locally in GR.

However, despite the failure of Newton's third law, conservation of momentum and angular momentum is strictly true SR because fields carry momentum and angular momentum. In GR, there are fundamental problems for all conservations laws globally, but locally SR applies. As a result, conservation laws can be readily used at solar system scales in GR. You need to worry about non-conservation when gravitational waves are significant, or on cosmological scales. For gravitational waves, there is no generally valid way to locally describe their energy or momentum.

7. Apr 20, 2013

### Bill_K

To answer the original question, Newton's Laws of Motion are essentially the statement that energy and momentum are conserved, and this holds equally true in special relativity, and general relativity as well, provided we are careful to properly define energy and momentum, and deal only with local quantities. For example for a particle we must include a relativistic factor γ= 1/√(1 - v2/c2). The energy is E = γmc2, while the momentum is p = γmv instead of just mv. Collisions between two particles conserve the sum of these quantities, just as in Newtonian mechanics. As PAllen says, problems involving fields need to take the finite propagation speed into account, but I wouldn't say that renders the Third Law false.

Introductory treatments of special relativity discuss only the simple case of constant velocity. But Newtonian mechanics deals with accelerating particles and accelerating frames, and special relativity being more general it would be contradictory if SR somehow were unable to do likewise. In place of a global reference frame, we often use a local frame tied to one particular observer.

8. Apr 20, 2013

### WannabeNewton

I don't get why there is such a wide spread belief that SR cannot handle accelerations; one has to wonder how this idea was disseminated. Just because SR has a preferred class of observers / frames, doesn't mean it cannot deal with accelerations and forces! By your logic, just because Newtonian mechanics also has a preferred class of frames, it too cannot deal with accelerations and forces.

You can very well have certain, permissible classical fields (e.g. electromagnetic field) propagating on minkowski space-time and have test particles interacting with these fields, within the framework of SR. We can for example solve Maxwell's equations in flat space-time $\partial^{a}F_{ab} = -4\pi j_{b}, \partial_{[a}F_{bc]} = 0$ for the electromagnetic field strength tensor $F_{ab}$ and get the associated electromagnetic field. A test particle interacting with the electromagnetic field will feel a 4-acceleration $u^{a}\partial_{a}u^{b} = \frac{q}{m}F^{b}_{}{}_{c}u^{c}$. Electromagnetism is well incorporated into SR. SR doesn't lack some kind of ability to talk about forces and accelerations.

9. Apr 20, 2013

### Agerhell

1. The first law of Newton holds in special and general relativity. The first law states that if there is no force there is no acceleration. Now gravity in general relativity is sometimes looked upon not as a force but as an effect of space-time curvature. This does not matter as the first law only applies where there is no gravity (which means no gravitational force in Newtonian terminology and no space-time curvature in general relativity).

2. According to the second law of Newton you have:

$\frac{d}{dt}(m\bar{v})=\bar{F}$

In plain language this means that the force (which is a quantity with a direction, which the bar symbolizes) is equal to how the quantity mass times velocity (which nowadays is called momentum but Newton called "motion") varies with time. As the velocity also has a direction, it has a bar over it in the formula.

This can also be written:

$$\frac{dm}{dt}\bar{v}+m\frac{d\bar{v}}{dt}=\bar{F}$$

Classically one assumes the mass to be constant (dm/dt=0) in which case:

$$m\frac{d\bar{v}}{dt}=\bar{F}$$

As how the velocity varies with time, ($d\bar{v}/dt$), is the same thing as the acceleration this can be written:

$$m\bar{a}=\bar{F}$$

which is probably what you recognize as the second law of Newton? The second law of Newton written in this way is not true in relativity. If you consider only electromagnetic forces and not gravity you can assume that the mass is not constant but varies with velocity as:

$$m=m_0\gamma=m_0\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

This expression can be stuffed into the first or second expression above. After solving something called a "differential equation" you get:

$$\frac{{\rm d}\bar{v}}{{\rm d}t}=\frac{1}{\gamma^3}(\bar{F}\cdot\hat{v})\hat{v}-\frac{1}{\gamma}(\bar{F}\times\hat{v}) \times\hat{v}$$

This version of Newtons second law works if the force is of electromagnetic origin, but not if the force is gravity. Perhaps it is possible to let the mass of the orbiting body vary not only with the velocity, but also with the position within the gravitational field and somehow get the second law of Newton to hold in the presence of gravity, but used in the usual way, the second law of Newton do not hold if there is gravity involved.

3. The third law of Newton says that if there is change in motion in one direction there is an equal and opposite change in motion somewhere ells. Formulated in this way, the third law of Newton holds true also in relativity, I guess.

10. Apr 20, 2013

### WannabeNewton

Would it be restricted only to fields though? What about the case of a stationary, asymptotically flat space-time in which an observer at infinity holds a particle stationary using a really long massless string (stationary in the sense that the particle follows an orbit of the time-like killing vector field). In such a case, the particle locally feels some force $F^{b} = m\nabla^{b}\ln V$ from the end of the string it hangs from, where $V$ is the red-shift factor, but the force exerted by the observer at infinity on the other end of the string turns out to have magnitude $F_{\infty} = VF$.

If we were using Newtonian mechanics, on the other hand, we would instead expect the magnitude of the force felt by the particle on one end of the string to be equal to that of the force exerted by the observer on the other end, if the string is to be massless. The way it is usually explained in mechanics texts (e.g. Kleppner) is that the information about the exerted force is carried from one end of the string to the other via tension, as per Newton's 3rd law, and if the string is massless then the forces at the two ends would have to agree.

11. Apr 20, 2013

### PAllen

Good point. Correct is that it is only true in relativity for for collisions. For anything else, it is not true as stated by Newton. I meant fields as the the common exception, but it isn't the only one, as you note.

12. Apr 20, 2013

### WannabeNewton

Of course the main crux of your post, that there is a finite propagation time for the information, is at the heart of all this so that itself is an all encompassing statement. Fields are probably the most physically intuitive counter examples though, as you noted.