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Do we ever talk about rate of change of momentum with respect to movement in spacetime? That should be a good quantity, given that spacetime is more fundamental than time.

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    $\begingroup$ I don’t understand your question considering you can also define a force in terms of the opposite derivative of potential energy with respect to position (for conservative forces at least). Anyways, $F=dp/dt$ the momentum part of the equation incorporates space, and thus to define a force we use a change in space, a change in time, and mass(a form of energy). $\endgroup$
    – Bandoo
    Mar 11, 2020 at 11:25
  • $\begingroup$ @Bandoo How do we define potential energy without Force? $\endgroup$
    – Ryder Rude
    Mar 11, 2020 at 14:00
  • $\begingroup$ The below answer by Dale does a better job explaining how energy is related to force. But the point of my comment was to show that $F=dp/dt$ incorporates all elements of space, time , and energy. and thus the definition is not truly biased towards time. $\endgroup$
    – Bandoo
    Mar 11, 2020 at 14:42
  • $\begingroup$ You might think what your headline says, only because you have a direct representation for time with the acceleration a in F=ma => F=m*dv/dt but here you omit that v is a vector in space .. so you have that component as well $\endgroup$
    – eagle275
    Mar 12, 2020 at 9:08
  • $\begingroup$ @eagle275 it's still very asymmetric in space and time $\endgroup$
    – Ryder Rude
    Mar 12, 2020 at 9:27

3 Answers 3

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Do we ever talk about rate of change of momentum with respect to movement in spacetime?

Yes. In spacetime just like time and space are unified into one spacetime so also energy and momentum are unified into one concept called the four-momentum. The four-momentum is a four dimensional vector where the first element is energy and the remaining three elements are the momentum with energy having the same relationship to momentum as time has to space.

Equipped with the concept of the four-momentum we can define the rate of change of the four-momentum. This quantity is called the four-force and has exactly the properties that you would expect in a relativistic generalization of force. The timelike component of the four-force is power. So in relativity the four-force unites power and force in the same way as spacetime unites space and time.

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    $\begingroup$ Really cool to know that power and force are unified. They always seemed similar to me (as the quantities which drive the instantaneous motion of the body) $\endgroup$
    – Ryder Rude
    Mar 11, 2020 at 13:57
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    $\begingroup$ Four-momentum and four-force need better names. The latter sounds like a lame comic-book crime-fighting team. $\endgroup$
    – jez
    Mar 14, 2020 at 1:55
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    $\begingroup$ @jez But lame comic-book crime-fighting teams are awesome, so I love it :) $\endgroup$ Mar 14, 2020 at 17:00
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Force was defined a long time before special and general relativity, as models of observations and data came on the scene.

Force is defined in one of the laws of Newtonian mechanics.

Second law

In an inertial frame of reference, the vector sum of the forces $F$ on an object is equal to the mass $m$ of that object multiplied by the acceleration $a$ of the object: $F = ma$. (It is assumed here that the mass $m$ is constant.)

In physics theories, laws are extra axioms imposed on the solutions of the differential equations so as to pick those solutions that fit observations and data.

Then why is the definition of force biased towards time?

Because that is what the observations and data need for the theory of mechanics to fit experiments and observations and to predict future behavior. Classical mechanics is validated over and over again in the phase space region of small (with respect to velocity of light) velocities and small masses.

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So if you have a path through spacetime, there is a natural length to that path—except you have to choose whether you want a timelike length (in which case the path never goes faster than light) or a spacelike length (in which case it always does.) If you try to mix these you typically get complex numbers involved as you take the square root of negative intervals.

Assuming you like causality for everyone, timelike paths make more sense and their length is known as proper time, measured by taking a clock on the spaceship as it makes that precise journey.

This idea that particles stay within the light cone and therefore have time like paths is why you has more interest in time derivatives rather than space derivatives in relativity. Space derivatives make some sense for fields that are spread out over spacetime, see also (relativistic) classical field theory textbooks for more on this: but for particles, spatial derivatives make no sense.

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  • $\begingroup$ Spatial derivatives have some use for particles too. If you define the derivative of momentum with respect to space as space-force (SF) and you have the SF on a particle as a function of time, then you'll find that the quantity $m\ln{|v(t)|}-\int_{t_0}^t SF(t)dt$ remains conserved, where $t_0$ is an arbitrarily chosen time point. $\endgroup$
    – Ryder Rude
    Mar 11, 2020 at 14:16
  • $\begingroup$ On second thoughts, I think the expression works only for one dimension because in higher dimensions, SF would be a vector. But still, this seems like a nice conserved quantity. This doesn't even put restrictions on SF (like it doesn't have to be conservative or anything). $\endgroup$
    – Ryder Rude
    Mar 13, 2020 at 6:17

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