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Do forces compose/superpose in special relativity? 
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#1
Feb1713, 08:26 PM

P: 175

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
Feb1713, 08:50 PM

P: 1,072

Edit. Delete.



#3
Feb1713, 09:15 PM

C. Spirit
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Thanks
P: 5,661

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
Feb1713, 10:29 PM

P: 175

Do forces compose/superpose in special relativity?
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
Feb1713, 10:38 PM

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P: 8,792

Off the top of my head, I don't know.
It seems that with 3force in flat spacetime there shouldn't be a problem with Ʃ F_{i} = Ʃ dp_{i}/dt But then p_{i}=γ_{i}m_{i}v_{i} 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
Feb1713, 10:44 PM

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#7
Feb1713, 10:49 PM

Emeritus
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PF Gold
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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
Feb1713, 11:25 PM

PF Gold
P: 1,168

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
Feb1713, 11:52 PM

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


#10
Feb1813, 12:25 AM

PF Gold
P: 1,168

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
Feb1813, 12:42 AM

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#12
Feb1813, 10:01 AM

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PF Gold
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#13
Feb1813, 11:24 AM

PF Gold
P: 1,168

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 noncovariant language. There are many forms of the same theory. We can write Maxwell's equations in the old form, or in a fourtensor 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 noncovariant results of electromagnetic theory, the fourtensor 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. 


#14
Feb1813, 11:34 AM

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#15
Feb1813, 11:48 AM

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#16
Feb1813, 11:53 AM

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Of course, even in nonrelativistic 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
Feb1813, 01:05 PM

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PF Gold
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#18
Feb1813, 05:09 PM

PF Gold
P: 1,168

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 fourvelocity 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? 


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