# Do forces compose/superpose in special relativity?

1. Feb 17, 2013

### James MC

Hi there,

In Newtonian mechanics, forces "compose". In other words:

$F_{net}=\sum_iF_i$

This states that the net force on a system of particles is the sum of each of the forces on each indiviudal particle. Similarly, the force on particle i due to a system of particles indexed by j is:

$F_i=\sum_jF_{ij}$

The former is sometimes called "the composition of forces", the latter "the superpositon of forces". Similar additivity principles hold for Newtonian gravitational forces.

My question is, do forces add like this in relativity theory? I've found practically no discussion of this online! The only discussions I've found is where one person says that the superposition of forces, in certain cases, is "not allowed in general relativity". And also where one person says that "due to relativity of simultaneity of events we cannot simply sum up the forces applied to the system, for different inertial observers. There are special conditions, when such a summation can be carried out: either the forces are static, or they are applied to the same spatial point."

I'm particularly interested in special relativity, and am not sure I follow what the above author is saying. Does the composition/superposition of forces hold in all cases, in special relativity?

Let me know your thoughts! :)

2. Feb 17, 2013

### Jorriss

Edit. Delete.

3. Feb 17, 2013

### WannabeNewton

In GR you can't add vectors at different points in space - time in the same simple way you would in Newtonian mechanics where you would just parallel translate all the forces to some concurrent point. In GR this no longer works because parallel transport of a vector need not preserve its initial value. You can certainly add vectors that are tangential to space - time at the same point but if they are at different points then it doesn't work unambiguously like it does in Newtonian mechanics.

4. Feb 17, 2013

### James MC

Thanks, that's helpful. So then what about SR, does SR differ from GR in this respect?

My guess is that forces do add in SR, and that this is a consequence of the definition of F in terms of momentum and the conservation of momentum. Thus, when you apply a force F, you are adding F units of momentum to an object per unit time. When you apply a second force F', you are adding F' units of momentum to the object. The two forces add because momentum is a vector conserved quantity in SR: its separate components are separately conserved, and the components of the forces tell you how much of each momentum component is coming in. And I think (think) this reasoning also holds even when the dynamics are nonlinear, so that the argument applies to the F in the expression from wiki :
$F = \gamma(v)^3ma_| + \gamma(v)ma_-$

Does this sound right?

In that case, perhaps what the quote about SR in my first post is saying is that for a given system, different observers can add to calculate the total force, it's just that they will get different results due to the different forces of the component parts as seen in different frames?

5. Feb 17, 2013

### atyy

Off the top of my head, I don't know.

It seems that with 3-force in flat spacetime there shouldn't be a problem with

Ʃ Fi = Ʃ dpi/dt

But then piimivi

And dp/dt=m(vdγ/dt + γdv/dt).

Not sure immediately if there's a good way to continue (I hate forces in SR).

Anyway, just some thoughts for others to comment on.

6. Feb 17, 2013

### Staff: Mentor

That is a correct guess and a good reason. Also, a lot of time you will be working with force densities and momentum densities rather than just forces and momenta.

Basically, forces do add, but it can be a little more complicated. If they are at the same place and time then they add like normal, otherwise you use the usual transformation rules to find out when and where they add.

7. Feb 17, 2013

### bcrowell

Staff Emeritus
SR doesn't have instantaneous action at a distance. If you don't have action at a distance, force becomes a superfluous concept. Instead, you can have (a) collisions of particles (or decays, etc.), or (b) fields.

8. Feb 17, 2013

### Jano L.

The force has a good meaning in relativity, as in Newtonian mechanics - derivaitive of momentum. It does not matter that it is not instantaneous function of position and velocity of other particles. One can take them as functions of past motion of the particles, or functionals of the fields, but that is just mathematical complication, not objection in principle.

The superposition of forces is assumed in relativity, and based on this the theorem of conservation of momentum and energy of fields + matter is derived.

9. Feb 17, 2013

### bcrowell

Staff Emeritus
This only defines the total force acting on a particle. It doesn't define individual forces.

Do you believe that the "functions of past motion of the particles" are uniquely defined? I'm skeptical. And do they take the form of a linear superposition? Again, I'm skeptical. It seems to me that you're going to end up reinvent classical field theory, which doesn't have the kind of addition of forces referred to by the OP. In a field theory like Maxwell's equations, you don't have one object acting on another object at a distance by making a force on it.

This is totally wrong. You've written this as a generic statement about all possible ways of developing SR from some set of principles. Not only are there counterexamples, I would be very surprised if you could provide one well constructed example of such a treatment.

[EDIT] Fixed mistakes and clarified in 2nd paragraph.

Last edited: Feb 18, 2013
10. Feb 18, 2013

### Jano L.

Of course, one can think of relativistic theory where forces are not additive. I just wanted to say that the easiest possibility is to assume additivity, as is usually done in electromagnetism.

I think as an example we can take a theory of electric fluid and use similar procedure as in classical electromagnetism, where the Maxwell tensor is derived.

The superposition of forces enters the argument through the integral

$$\int_V \rho\mathbf E + \mathbf j \times \mathbf B\, dV.$$

for the total force acting on the fluid in the region $V$. Together with the Maxwell equations and some equation of state of the fluid, this will lead to conservation theorem from which we can infer the definitions of the momentum of field.

Although such scheme has its deficiencies, it is a relativistic theory and has superposition of forces.

11. Feb 18, 2013

### atyy

Can this be extended to forces on particles at different places at the same time (taking an inertial frame)? I don't immediately see a problem.

12. Feb 18, 2013

### bcrowell

Staff Emeritus
This is not really relativistic, since the volume integral treats space differently from time. You also have your interpretation of the foundational issues backwards. Maxwell's equations are a fundamental, relativistic theory, and they imply conservation of energy-momentum via Poynting's theorem. None of this requires the introduction of the concept of force. Maybe you've been influenced by presentations of freshman mechanics in which force is presented before energy, but even in that nonrelativistic context, that order of presentation is just a matter of tradition. Force essentially has no foundational role anywhere in any modern view of physics. The modern attitude is that symmetries are fundamental, and conservation laws follow from the symmetries.

If you do this for electromagnetic forces, you end up reinventing Maxwell's equations. If you do it for gravity, you have to reinvent GR. Neither of those theories has a concept of object A acting on object B at a distance with a force. In both theories, what acts on an object is a field.

Last edited: Feb 18, 2013
13. Feb 18, 2013

### Jano L.

Ben, this is surprising criterion. Would you say that special theory of relativity is not relativistic, since it treats the spatial coordinates differently than the time coordinate? After all, there are three spatial coordinates and one temporal coordinate, and the metric coefficient has different sign for time coordinate.

I believe that you would not. I also believe that you would not say that Maxwell's equations are not really relativistic just because they are written in terms of volume and surface integrals, which treat space and time differently.

I think what you probably tried to say is that the expression is not covariant/tensor expression. That is true, but that is a different thing. As you certainly know, it is perfectly possible and correct to formulate relativistic theory using non-covariant language. There are many forms of the same theory. We can write Maxwell's equations in the old form, or in a four-tensor form, but the content of the theory is the same.

The same situation occurs with the above expression - and the law of conservation of momentum, and energy (Poynting theorem for fluid). Based on these non-covariant results of electromagnetic theory, the four-tensor of energy- momentum of EM field can be (and historically was) derived for electromagnetic field. I do not claim it is the only correct way.
I was just trying to show an example of a piece of theory where addition of forces is made explicit - the volume integral of density of force.

From your writing I take it that you do not like the concept of force nor the historical considerations in physics. I am sorry to hear that, but that is you choice and you are entitled to it. I can only say that I think you lose much doing so. There are situations in mechanics which do not allow discussion based on energy, but can be analyzed and predicted by discussion of forces. There are also situations in EM which are much easier to analyze using energy than forces. In my view, both force and energy are very important in modern physics and it does not make any sense to throw either of them out as obsolete.

14. Feb 18, 2013

### Jano L.

I am afraid I do not understand your point. Maxwell's equations alone do not form a theory. There are many more things to include. One of them is the Lorentz expression for EM force on a point charge or density of force for electric fluid. Only then one can form some system of equations of motion. If needed, the equations can be stated in terms of four-tensors of momentum of EM field and matter, but the forces can be either retained or defined a posteriori based on those tensors. Both momenta and forces are additive quantities.

15. Feb 18, 2013

### Staff: Mentor

Yes, but of course in other inertial frames the forces may not be at the same time.

16. Feb 18, 2013

### Staff: Mentor

I think this is perhaps going a little too far. The fields still exert forces on the particle, and a collision can certainly be considered to exert a force. I agree that there is no action at a distance, but I don't think that obviates force in general.

Of course, even in non-relativistic mechanics you can use Lagrangians and Hamiltonians to do physics without forces. So clearly they are not necessary, but that doesn't mean that they can't be used when it is convenient to do so.

17. Feb 18, 2013

### bcrowell

Staff Emeritus
As always in relativity, the relevant symmetry is the symmetry of the Poincare group.

Maxwell's equations can be written in manifestly covariant form. Your equation for the total force can't.

If you can write something in manifestly covariant form, that proves that it can be expressed properly for relativity. Since you haven't written your equation for the total force in manifestly covariant form, the burden of proof is on you to show that it's meaningful relativistically. You're not going to be able to do that, because it's not.

I assume this means that you've changed your mind about your original erroneous claim in #8 that force was necessary for the development of dynamics in SR.

18. Feb 18, 2013

### Jano L.

The proof is bit long so I will not write all the steps in their full extent here. You can find a good start in Heitler's book, 2nd ed., Chap. I, sec. I. He derives the Poynting theorem and the Maxwell stress tensor theorem in this way: he integrates the density of force $\rho\mathbf E + \mathbf j\times \mathbf B$ and density of power $\mathbf j\cdot \mathbf E$ to obtain total force and power that the body receives.

Assume part of the fluid inside a volume region $V$. Heitler's equations 22a, 22b are, in changed notation

$$\int_V \partial_t \bigg( g^k_{matter} + g^k_{field} \bigg) d^3x = \int_V \partial_s M^{ks} d^3 x,$$

$$\int_V \partial_t \bigg( \epsilon_{matter} + \epsilon_{field} \bigg) d^3x = -\int_V \partial_s S^{s} d^3 x,$$

where $g^k$-s are densities of momentum, $\epsilon$-s densities of energy, $M^{ks}$ is the Maxwell stress tensor and $S^s$ the Poynting vector. If the boundary terms in RHS are negligible, these equations express conservation of momentum and energy within the volume $V$.

To see that these equations are relativistic, take the RHS to the left side and express everything as one volume integral:

$$\int_V \partial_t \bigg( g^k_{matter} + g^k_{field}\bigg) -\partial_s M^{ks} d^3x = 0,$$
$$\int_V \partial_t \bigg( \epsilon_{matter} + \epsilon_{field} \bigg) + \partial_s S^{s} d^3x = 0.$$

These equations can be written as

$$\int_V \partial_\mu \bigg( T_{\mathrm{field}}^{\rho\mu} + T_{\mathrm{matter}}^{\rho\mu}\bigg) d^3x = 0$$

using the tensors

$$T_{\mathrm{field}}^{\rho\mu} = {{F^{\rho\nu}} F^{\mu}_{~~\,\nu}} - \frac{1}{4} \delta^{\rho\mu} F^{\sigma\nu} F_{\sigma\nu} = \left( \begin{matrix} \epsilon_{field}& ~\mathbf S \\ \mathbf S & - \mathbf M\\ \end{matrix} \right),$$

$$T_{\mathrm{matter}}^{\rho\mu} = \rho_0 u^\rho u^\mu,$$

and the continuity of mass flow $\partial_\mu (\rho_0 u^\mu) = 0$. Here $\rho_0$ is the rest mass density and $u^\mu$ is the local four-velocity of the fluid $dx^\mu/d\tau$.

Finally, since the volume $V$ is arbitrary, we obtain

$$\partial_\mu \bigg( T_{\mathrm{field}}^{\rho\mu} + T_{\mathrm{matter}}^{\rho\mu}\bigg) = 0,$$

which is exact, tensor equation.

To summarize, model of fluid whose motion is acted upon by EM forces, described by the above force and power density, leads to conservation law that is the same in every inertial frame. This is an evidence that the concept of force works well in relativity.

Can you please explain why did you claim that the above expression for force density is not meaningful in relativistic theory?