In newtonian dynamics  are relations between the initial and final speeds $u$ and $v$ of an object which covers a displacement $s$ during time $t$ under constant acceleration $a$, and they are essentially \begin{align} v &amp;= u + at\\ s &amp;= ut + \frac{1}{2}at^2, \end{align} and so on. Is there a corresponding set of equations for relativistic dynamics? I imagine that, for instance, the two equations above would have to factor in length contraction or time dilation, so is there a valid equivalent of 'suvat' equations?
121125
Displacement in an accelerated classical object is: $$s=ut+\frac {at^2}{2}$$ What is the displacement of an accelerated relativistic object?  In Newtonian mechanics there are two types of displacement.   Displacement of an object with velocity constant: $$s=ut$$ Displacement of an accelerated object with acceleration constant: $$s=ut+\frac {1}{2}at^2$$   This is not completely clear, but I think the second displacement should be something different in relativity. Is that true?  What is the result of this integral in relativity? $$s=\int (u+at)dt$$ classicaly $$s=\int (u+at)dt=\int udt +a\int tdt=ut+a\frac {t^2}{2}+x_0$$  What is the relativistic one?  more details: $$v=u+at$$ $$s\to displacement$$ $$a\to acceleration$$ $$u\to initial \,velocity$$ $$v\to final \,velocity$$
29313
I have just started special theory of relativity. The limiting speed known as speed of light fascinated me most. I asked my teacher:      Consider two massless objects moving in the same direction at the speed of light. What would be their relative velocity w.r.t each other?   My teacher tells me that their relative velocity would be zero. But then the speed of light is universal constant regardless of the motion of its frame of reference, so shouldn't their relative velocity be $c$? What is their relative velocity and how? Do objects moving at the speed of light obey law of addition of velocities?
62154
Why is the Lagrangian a function of the position and velocity (possibly also of time) and why are dependences on higher order derivatives (acceleration, jerk,...) excluded?  Is there a good reason for this or is it simply "because it works".
108697
I am very new to relativity, and as I was going through Resnick &amp; Halliday, I saw that momentum was no longer $mv$, rather $\gamma mv$. This was the proof:     $$p =  mv = m \frac{\delta x}{\delta t_0}  $$   where $\delta t_0$ is the proper infinitesimal time interval to cover $\delta x$ distance.     Using the time dilation formula, $\delta t = \gamma \delta t_0$ we can then write    $$ p = m \frac{\delta x}{\delta t_0} = m \frac{\delta x}{\delta t} \frac{\delta t}{\delta t_0} = \gamma m \frac{\delta x}{\delta t} $$   However, since $ \frac{\delta x}{\delta t} $ is just the [observed] particle velocity,    $$ p = \gamma mv $$   On the other hand, when I went on to Feynmann's book I saw that there he talked about using "the new m" (the relativistic mass) to derive all the equations and said      $$p = mv = \frac{m_0 v}{(1 - \frac{v^2}{c^2})} = \gamma m_0 v $$   My problem is that if both these proofs are true, then why doesn't one consider the other's argument, id est, why aren't both of these factors considered together ? If the mass is really changing, why doesn't Resnick &amp; Halliday account for it? That way the momentum would turn out to be $p = \gamma^2 mv$  Am I missing something here?
49433
So I have two particles that have collided in 3 dimensional space. I want the particles to rebound off of each other in an elastic manner. How do I determine the resultant velocity vector if I know: Both particles' initial velocity vector, their masses, and their initial position vectors relative to the origin.  I tried to follow the instructions on  but I am not sure how to make their equations work without polar coordinates / how to translate my data to polar coordinates.  Also, if you answer my question in terms of polar coordinates, can you explain which angle is theta and which is phi? Since there are 2 different conventions, I am not really sure which angle is being referred to in the explanation on the aforementioned website.  EDIT: This isn't for a class. I'm writing a physics engine for a gravity simulation I am making using Unity. I haven't taken kinematics since I was in high school, which is why I am so rusty.
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I'm currently working through the book Heisenberg's Quantum Mechanics (Razavy, 2010), and am reading the chapter on classical mechanics. I'm interested in part of their derivative of a generalized Lorentz force via a velocity-dependent potential.   I understand the generalized force   $$F_i = -\frac{\partial V}{\partial x_i} + \frac{d}{dt}\left(\frac{\partial V}{\partial v_i}\right)$$  that they derive from a Lagrangian of the form $L = \frac{1}{2}m|\vec v|^2 - V(\vec r,\vec v,t)$.  However, in the next (critical) step of the derivation, the author cites a theorem from Helmholtz saying      ...according to Helmholtz, for the existence of the Lagrangian, such a generalized force can be at most a linear function of acceleration, and it must satisfy the Helmholtz identities.   The three Helmholtz identities are then listed as:  $$\frac{\partial F_i}{\partial \dot{v_j}} = \frac{\partial F_j}{\partial \dot{v_i}},$$  $$\frac{\partial F_i}{\partial v_j} +  \frac{\partial F_j}{\partial v_i} = \frac{d}{dt}\left(\frac{\partial F_i}{\partial \dot{v_j}} +  \frac{\partial F_j}{\partial \dot{v_i}}\right),$$  $$\frac{\partial F_i}{\partial x_j} -  \frac{\partial F_j}{\partial x_i} = \frac{1}{2}\frac{d}{dt}\left(\frac{\partial F_i}{\partial v_j} - \frac{\partial F_j}{\partial v_i}\right).$$  I'm trying to understand where this theorem comes from. Razavy cited a 1887 paper by Helmholtz. I was able to find , but it is in German, so I could not verify whether or not it proved the theorem. Additionally, I could not find it in any recent literature. I searched online and in Goldstein's Classical Mechanics.   The only similar concept that I can find is in the  where we have three equations known as Helmholtz conditions. Are these two concepts one in the same? If so, how should I interpret the function $\Phi$ and the matrix $g_{ij}$ that appear in the Helmholtz conditions I found online?   If the cited theorem from Razavy does not relate from the inverse Lagrangian problem, could I have some help finding the right direction?
85489
I understand how, and why, an accelerating charge emits radiation, and loses energy in the process, as well as the  formula for the power, and its derivation.  However, in classical mechanics, when we want to find the motion of a particle, the process is (theoretically) simple - find all forces on the particle as a function of position and time, plug in Newton's second law and solve the differential equation. It seems to me as if simply because a particle is charged, this process just ceases to work.  When talking about a charged particle, it is simply not enough to find all forces and solve Newton's law - now we need to somehow take into account the energy lost by the particle to radiation. However no one seems to mention how to take that into account, as a term in an equation of motion. It seems like all we know is the total power radiated by the particle - there is no concrete, complete description of how that loss affects its motion.  Of course, in some simple cases we can easily guess how the particle will be affected: for example, for an electron circling in a uniform magnetic field, it is obvious that energy loss due to radiation will cause the electron to spiral inward. You can, at least theoretically, write the Lorentz force acting on the particle, due to the uniform field, plug it into Newton's second law, and also write the Larmor formula, take that loss into account and find a complete description of the particle's motion.   However still, when we write simply Newton's second law here, with the only force being the Lorentz force, we find a circular, not spiral, solution; we have to synthetically add in the Larmor formula to truly recover the particle's motion. Contrary to classical mechanics where Newton's law is a complete description of the motion.   Why is Newton's law no longer enough, or put another way, what term am I missing that will fix the electron's motion, add an additional "force" the will cause it to spiral inward? Is there a general expression for that force? And how does one in general write the equation of motion for a charged particle (or a system of)?   I suspect the answer has something to do with the EM field changing not instantaneously but at the speed of light, since Newton's law of gravitation clearly yields stable, closed orbits and propagates instantaneously. Perhaps I should phrase my question this way:   How does the fact that the EM field propagates at a finite speed affect the way we have to write the equations of motion  for a charged particle (as opposed to the equations of motion for a massive particle, acted upon by an instantaneous, newtonian gravitational field)?   One last note: perhaps this has something to do with Maxwell's equations being Lorentz-invariant and not Galilean? However I find this unlikely to be the case since the electron doesn't have to move in a relativistic speed in order to experience this effect. In any case, this is also why I allowed myself to use Newton's second law and not its (special) relativistic expression - would using that fix the problem?
2538
I'm looking for a rather intuitive explanation (or some references) of the difference between the metric of a curved space-time and the metric of non-inertial frames.  Consider an inertial reference frame (RF) with coordinates $\bar x^\mu$, in flat spacetime $\eta_{\mu \nu}$ (Minkowski metric).   If I have well understood, on one hand, I can go to an accelerated RF by change of coordinates $x^\mu(\bar x)$. The metric is given by: $$\tag{1}g_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}$$ On the other hand, I know that a curved space-time with metric $q_{\mu \nu}$ cannot be transformed to Minkowski $\eta_{\mu \nu}$ by coordinate transformation. In other words there does NOT exist any coordinate $x^\mu(\bar x)$ such that (in the whole coordinate patch): $$\tag{2}q_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}\qquad \leftarrow \text{(does not exists in curved space)}$$   So far, everything is more or less ok... But my question is:    What is the difference between $q_{\mu \nu}$ and $g_{\mu \nu}$? I mean, in both cases a particle would "feel" some fictitious forces (in which I include the weight force due to the equivalence principle).  What physical situation can $q_{\mu \nu}$ describe and $g_{\mu \nu}$ cannot?   I additionally know that by change of coordinates $q_{\mu \nu}$ is locally Minkowski. But still, I can't see clearly the difference.
63686
Frequently I see the expression  $$F = \frac{dp}{dt} = \frac{d}{dt}(mv) = \frac{dm}{dt}v + ma,$$ which can be applied to variable mass systems.  But I'm wondering if this derivation is correct, because the second law of Newton is formulated for point masses.   Furthermore if i change the inertial frame of reference, only $v$ on the right side of the formula $F = \frac{dm}{dt}v+ma$ will change, meaning that $F$ would be dependent of the frame of reference, which (according to me) can't be true.   I realize there exists a formula for varying mass systems, that looks quite familiar to this one, but isn't exactly the same, because the $v$ on the right side is there the relative velocity of mass expulsed/accreted. The derivation of that formula is also rather different from this one.   So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake.
68485
In the Wikipedia article , it says     After Newtonian gravitation was found to be inconsistent with special relativity, . . .   I don't see how Newtonian gravitation itself is inconsistent with special relativity.  After all, Newton's Universal Law of Gravitation is completley analogous to Coulomb's Law, so it would seem that, if there were an analogous "gravitational magnetic field", one could formulate a theory of gravitation in exact analogy with Maxwell's Theory of Electromagnetism, and of course, this would automatically be consistent with Special Relativity.  What about this approach to gravitation does not work?  The only problem I could see with this is the lack of evidence for a "gravitational magnetic field".  That being said, gravity is "weak" as it is, and my guess is that it would be extremely difficult to set up an experiment in which a massive body were moving fast enough for this "magnetic field effect" to be obsevable.  EDIT:  As has been pointed out to me in the answers, Newton itself is inconsistent with SR.  Similarly, however, so is Coulomb's Law, yet, electromagnetism is still consistent with SR.  Thus, how do we know that it is not the case that Newton's Law is the special "static" case of a more general gravitomagnetic theory exactly analogous to Maxwell's Theory:  Let $\mathbf{G}$ be the gravitation field, let $\rho$ be the mass density, and $\mathbf{J}$ be the mass current density, let $\gamma _0$ be defined so that $\frac{1}{4\pi \gamma _0}=G$, the Universal Gravitation Constant, let $\nu _0$ be defined so that $\frac{1}{\sqrt{\gamma _0\nu _0}}=c$, and suppose there exists a field $\mathbf{M}$, the gravitomagnetic field, so that the following equations hold: $$ \vec{\nabla}\cdot \mathbf{G}=-\frac{\rho}{\gamma _0} $$ $$ \vec{\nabla}\cdot \mathbf{M}=0 $$ $$ \vec{\nabla}\times \mathbf{G}=-\frac{\partial \mathbf{M}}{\partial t} $$ $$ \vec{\nabla}\times \mathbf{M}=\nu _0\mathbf{J}+\gamma _0\nu _0\frac{\partial \mathbf{G}}{\partial t} $$ where theses fields would produce on a mass $m$ moving with velocity $\mathbf{v}$ in our inertial frame the Lorentz force $\mathbf{F}=m\left( \mathbf{G}+\mathbf{v}\times \mathbf{M}\right)$.  This theory would automatically be consistent with SR and would reduce to Newton's Law of Gravitation for the case of gravitostatics:  $\frac{\partial \mathbf{G}}{\partial t}=\frac{\partial \mathbf{M}}{\partial t}=\mathbf{J}=0$.  (To be clear, you can't just set the time derivatives equal to $\mathbf{0}$ as it seems I have done.  Upon doing so, you obtain the corresponding static theory, which is technically incorrect (as you can easily see because it won't be relativistically invariant), but is nevertheless often a useful approximation.)  My question can thus be phrased as:  without appealing directly to GR, what is wrong with this theory?
65341
I would like to recover the (timelike) geodesics equations via the variational principle of the following action:  $$ \mathcal{S}[x] = -m \int d\tau = -m \int \sqrt{-g_{\mu\nu}\,dx^{\mu}\,dx^{\nu}} $$  Using an arbitrary auxiliary parameter $\lambda$, then one is able to rewrite the action and obtains: $$ d\tau = \sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$ $$ -m \int d\tau = -m \int (d\tau / d\lambda) d\lambda =  -m\int\sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$  Now we can vary the paths $x^{\mu}(\tau) \rightarrow x^{\mu}(\tau) + \delta x^{\mu}(\tau)$.  I need the following equation to be true, so i get the geodesics equations, but I don't know how one can say the following: $$ \int d\tau \,\delta g_{\mu\nu}\frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} = \int d\tau\, g_{\mu\nu, \rho} \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} \delta x^{\rho} $$ I know there is a simpler action, which leads to the same equation of motion, but there one has to do the same manipulation, which I sadly don't understand.  Thanks for helping me!
22513
I am studying quantum physics and there is something I don't understand:   I know that for any particle $E=hf$ (Einstein relation) and  $v=\lambda f$ ($v$ is the speed of the particle). I also know that the kinetic energy is $E_k=\frac{mv^2}{2}$.    Solving those 3 equations for $\lambda$:  $$h\frac{v}{\lambda}=\frac{mv^2}{2},$$ I finally find $$\lambda=\frac{2h}{mv}=\frac{2h}{p},$$ which is not consistent with the De Broglie relation $$\lambda=\frac{h}{p}.$$ Where am I wrong in my development?
118793
Consider a particle that is described by $x(t)=(2.0-0.1t) \cos(0,5t)$ and $y(t)=(2.0-0.1t) \sin(0,5t)$ t in seconds and x,y in meters.    In previous subquestions we were asked to determine an expression for the module of the velocity vector and the tangent acceleration. The expressions obtained are:  $v=\sqrt{k^2 + w^2(r_0-kt)^2}$  $a_t=-\frac{kw^2(r_0-kt)}{\sqrt{k^2+w^2(r_0-kt)^2}}$  with $k=0.1$ and $w=0.5$ and $r_0=2.0$.  Now my question is in the next question, to find an expression with the normal acceleration.   My attempt was to take   $a_c = \frac{v^2}{R}$  $R = \sqrt{x^2+y^2}= 2.0-0.1t$  $a_c=\frac{k^2}{r_0-kt} + w^2$  The answer given by my textbook is however:  $a_c= 4k^2w^2 + w^4(r_0-kt)^2-\frac{k^2w^4(r_0-kt)^2}{{k^2+w^2(r_0-kt)^2}}$  I don't understand how this formula was derived. It seems like the third term of the sum is $-a_t^2$ but I can't understand the other two terms and what variables ares involved. And also I want to know why my answer is wrong or if it is equivalent to the one given, I can't understand that. Thanks!
91613
I'm not familiar with any complicated physics equation, however I do understand some basics. Suppose there is two objects, both of them are moving away from each other in a 3-dimensional space, which they both have the speed of half the speed of light ($c/2$). Relative to the reference frame of object A, object B would be moving away from itself equal to the speed of light.     A          B &lt;-----      -----&gt;   -c/2       c/2   to a 3rd observer    0          c    to A   -c          0    to B   This seems to break the law of "nothing can move faster than the speed of light". I believe that there is some other explanation for this case. I have done some research, and there are quite a few questions on this topic. Take  and  for example. However, they are about adding velocities together, which isn't quite the same as in the case I was describing.   is also similar to my case, however the two objects described are both moving toward the same direction, and I'm not shooting any beam or laser toward the other object.  Most probably in reality there are some extremely complex laws and equations which makes this question more complicated. However, the two objects are not interacting with each other and I'm not trying to "detect" the speed of the other object in the reference frame of A or B. Also none of the objects are moving at a significant fraction of the speed of light relative to the third observer, so special relativity doesn't apply(?) I just could not think of any reason why wouldn't B be moving away from A in the speed faster than light.  A simple explanation would be great.
67625
I got badly lost in derivation of relativistic formulas for energy and momentum.  I stumbled upon relativistic action as follows (which should explain relativistic motion of a classical particle):  $$ S = \int Cds=C\int_{t_i}^{t_f}\sqrt{c^2-(x')^2}dt $$  Where $C$ is some constant (depends on what kind of physics we put in equation) and $s$ is relativistic interval. Later on Lagrangian $$ L(x')\equiv C\sqrt{c^2-(x')^2} $$ is used in derving relativistic energy and momentum.  I am familiar with Lagrangians and symmetry rules which connect energy and momentum to Lagrangian formalism. What I do not understand is this action - weren't action sum over time? Why all of a sudden it is sum over relativistic interval?
32545
I know this may seem a weird question, but it always bothers me. My physics book (Resnick,Halliday,Walker), and also various sites never say anything beyond acceleration.   But when a moving body is being acted by a variable force , its acceleration will definitely change: it will either increase or decrease. Then there will be rate of change of acceleration with respect to time. So, why don't books mention this? What is the cause for not measuring $\frac{d\vec{a}}{dt}$ ? If it exists, what is the use of it?
3958
Take two different reference frames. The first one - say $\lbrace{W}\rbrace$ - is inertial and it's fixed to the Earth surface. The second one $\lbrace{B}\rbrace$ is moving and in general is rotating and accelerated, so $\lbrace{B}\rbrace$ it is not inertial.  The equations of motion for the origin of the frame $\lbrace{B}\rbrace$ can be formulated on both frames as:  $$ \sum_{i}\,  ^wF_{ext} = m\, ^w\dot{v}$$ in the inertial frame and $$ \sum_{i}\,  ^bF_{ext} = m\, ^b\dot{v} - ^b\omega \times m ^bv$$ on the non-inertial one, where $^f x$ represents the vector $x$ expressed on the frame $f$.  Let an accelerometer to be installed over $\lbrace{B}\rbrace$ which by design subtracts gravity from its output.  I'm not sure if it is measuring the term $^b\omega \times m ^bv$ or not so, what is the accelerometer exactly giving as output?