I can not solve this &quot;loop&quot; in my learning: Relativistic kinetic energy is usually defined as (see by example ): $$ T = \int \mathbf{v} d \mathbf{p} = \int \mathbf{v} d(\gamma m \mathbf{v}) = (\gamma-1) m c^2 $$ where $\mathbf{p} = \gamma m \mathbf{v}$ is used as premise. (Four) momentum is defined as: a) from mass multiplied by position derivative respect to proper time (something usually deprecated due to the mix of reference frames for $\mathbf{x}$ and $\tau$): $$\mathbf{p}=m\frac{d\mathbf{x}}{d\tau}=m\frac{d\mathbf{x}}{dt}\frac{dt}{d\tau}=m\mathbf{v}\gamma$$. b) from action related to Lagrangian of the free particle (see ): $$ \mathbf{p} = \frac{\partial S}{\partial \mathbf{q}} = \frac{ \partial L}{\partial \dot{\mathbf{q}}} $$ where $ S=\int L dt $ and the Lagrangian $ L = - \frac{ 1 }{\gamma} m c^2 $ is obtained from the total energy $E=\gamma m c^2$ and: b1)  total energy from kinetic (loop!) $T=(\gamma-1)mc^2$ plus rest energy $mc^2$ or, b2) from energy-momentum relation $E^2 = p^2c^2+m^2c^4 $ and energy-momentum relation (see ) from four-momentum norm (loop!). I can not find a linear (non-circular) sequence of definitions that allows me to define and evaluate (four) momentum and kinetic energy in relativistic context.
82475
Motivating the non-relativistic momentum $$\mathbf{p} = m\mathbf{v}$$ is quite easy: it is meant to represent the quantity of motion of the particle, and since the mass is one measure of quantity of matter it should be proportional to mass (how much thing is moving) and should be proportional to velocity (how fast and to where it is moving). Now, in Special Relativity the momentum changes. The new quantity of motion becomes $$\mathbf{p} = \dfrac{m\mathbf{v}}{\sqrt{1-\dfrac{v^2}{c^2}}}$$ Or, using $\gamma$ the Lorentz factor $$\mathbf{p} = \gamma(v) m\mathbf{v}$$ where I write $\gamma(v)$ to indicate that the velocity is that of the particle relative to the frame in which the movement is being observed. The need for this new momentum is because the old one fails to be conserved and because using the old one in Newton's second law leads to a law which is not invariant under Lorentz transformations. So the need for a new momentum is perfectly well motivated. What I would like to know is how can one motivate that the correct choice for $\mathbf{p}$ is the $\gamma(v)m\mathbf{v}$. There are some arguments using the mass: considering a colision, requiring momentum to be conserved, transform the velocity and then find how mass should transform. Although this work, it doesn't seem natural, and it is derived in one particular example. On my book there's even something that Einstein wote saying that he didn't think it was a good idea to try transforming the mass from $m$ to $M = \gamma(v)m$, that it was better to simply keep $\gamma$ on the new momentum without trying to combine it with the mass. So I would like to know: without resorting to arguments based on transformation of the mass, how can one motivate the new form of momentum that works for special relativity?
52767
In classical mechanics with 3 space dimensions the orbital angular momentum is defined as  $$\mathbf{L} = \mathbf{r} \times \mathbf{p}.$$  In relativistic mechanics we have the 4-vectors $x^{\mu}$ and $p^{\mu}$, but the cross product in only defined for 3 dimensions. So how to define orbital angular momentum e.g. in special relativity in terms of 4-vectors? Or more generally in $d$ dimensions?
106728
I'm trying to figure out how to derive the transformation matrix for the Lorentz boost. Consider two inertial frames A and B, and let B move at a constant velocity V with respect to A. All the derivations I've seen assume that this implies that A moves at a velocity -V with respect to B. This does not seem obvious to me, can someone explain why this is true?  Edit : To clarify, I understand "inuitively" why this is true. What I'm interested in doing is deriving the Lorentz boost formula axiomatically - i.e., using the invariance of the speed of light as my starting point and showing that the Lorentz boost is the only linear transformation that satisfies this. What I don't get is how this reversal of velocity follows from this.
70630
Basically, which is the correct formula for thermal energy, and is this the same as kinetic energy? My notes are pretty conflicting on this topic, and I'm getting pretty confused.
6543
The action  $$S=\int L \;\mathrm{d}t$$  is an important physical quantity. But can it be understood more intuitively? The Hamiltonian corresponds to the energy, whereas the action has dimension of energy × time, the same as angular momentum.  I've heard the action being described as a measure of change, although I don't know how this description can be justified.
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?
74745
Let $(M,g)$ be a  Riemannian manifold, $\gamma:I\rightarrow M$ a geodesic. Let $c:(-\varepsilon,\varepsilon)$ is defined to be another geodesic with  $$c(0)= \gamma(0),\ \ \ c^\prime(0)=X.$$ Let $V(s)$,$W(s)$ be parallel vector fields along $c$.  The derivative that I want to compute is  $$\frac{\partial}{\partial s}\exp_{c(s)} \big(t\cdot(V(s)+sW(s))\big)$$ at $s=0$.  I looked at examples, but there is no such thing as the $c(t)$ there so I can not figure it out.  Thank you very much, I hope it is not a too simple calculation question for this forum.
79344
Intuitively, I would expect the change in kinetic energy of a particle to be frame independent. It just doesn't "feel" right that between two points in time-space, one frame should measure a change in kinetic energy of a particle different to another frame.  Is my intuition right? Is the change in kinetic energy of a particle frame independent?
74686
In a theoretical physics homework problem, I have to show the following: $$\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$$  Where $T$ is the Energy-Momentum-Tensor, $j$ the generalized current and $F$ the Field-Tensor. We use the $g$ for the metric tensor, I think in English the $\eta$ is more common.  I know the following relationships:   Current and magnetic potential with Lorenz gauge condition: $$\mathop\Box A^\mu = \mu_0 j^\mu$$ Energy-Momentum-Tensor: $$T^{\mu\nu} = \frac1{\mu_0} g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu} + \frac1{4\mu_0} g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$$ Field-Tensor: $$F^{\mu\nu} = 2 \partial^{[\mu} A^{\nu]} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$$ d'Alembert operator: $$\mathop\Box = \partial_\mu \partial^\mu$$ Bianchi identity: $$\partial^{[\mu} F^{\nu\alpha]} = 0$$   So far I have set all the definitions into the formula I have to show, but I only end up a lot of terms from antisymmetrisation and product rule. I also drew all what I have in , but I still cannot see how to tackle this problem.  Could somebody please give me a hint into the right direction?
46448
It is well known from Noether's Theorem how from continuous symmetries in the Lagrangian one gets a conserved charge which corresponds to linear momentum, angular momentum for translational and rotational symmetries and others.  Is there any elementary argument for why linear or angular momentum specifically (and not other conserved quantities) are conserved which does not require knowledge of Lagrangians? By elementary I mean, "if this is not so, then this unreasonable thing occurs".   Of course, we can say "if we want our laws to be the same at a different point in space then linear conservation must be conserved", but can we derive mathematically the expression for the conserved quantity without using the Lagrangian?  I want to explain to a friend why they are conserved but he doesn't have the background to understand the Lagrangian formalism.
54863
In classical mechanics, massless particles don't exist because for $m=0$, $p=0$.  The relativistic relation between energy, mass and spatial momentum is: $E^2= (pc)^2 + (mc^2)^2$ . So it is said that setting $m=0$ in the first equation you get $E=pc$.  How could setting $m=0$ in that equation give you $E=pc$ whilst $p$ appears in the equation and we know $p=γmu$? If you set $m=0$ you will have indeterminacy due to "$γm$". It seems to me like we are doing a "trick" in order to get the $E=pc$. Perhaps there is another proof for this relation?
52873
thanks for any help.  I'm trying to show that in a 2body problem, angular momentum is conserved given that $\dfrac{dp}{dt}=\dfrac{-GMm(rv)}{r³}$, where p is momentum, t time, G gravitational constant, M mass of 1 object, m mass of the other, (rv) is the vector between them and r is the magnitude of the vector between them.  I've had a lot of attempts but I don't seem to get very far.  Where L is angular momentum, $r_1$ is position vector of mass M and $r_2$ is position vector of mass m, $p_1$ is momentum of mass M and $p_2$ is momentum of mass m.  L=$r_1Xp_1+r_2Xp_2$  If $\dfrac{dL}{dt}=0$ then dL will be conserved over time, applying product rule             $L=\dfrac{dr_1}{dt}Xp_1+r_1X\dfrac{dp_1}{dt}+\dfrac{dr_2}{dt}Xp_2+r_2X\dfrac{dp_2}{dt}$  noticing that $p_1=m\dfrac{dr_1}{dt}$ and $p_2=m\dfrac{dr_2}{dt}$ clearly $p_1$ is parallel to$ \dfrac{dr_1}{dt}$ and $p_2$ is parallel to $\dfrac{dr_2}{dt}$ and hence when cross producted with them, the result is 0.  Giving $L=r_1X\dfrac{dp_1}{dt}+r_2X\dfrac{dp_2}{dt}$  I gather I must now show that either these two terms are 0, or they are equal and     opposite. I can't see any reason why they would be 0, and (slightly guessing) it seems to    make sense to me that they are terms related to how the angular momentum of one body changes    with the other body, and I assume they must therefore be equal and opposite. However when I    do  $r_1X\dfrac{dp_1}{dt}=-r_2X\dfrac{dp_2}{dt}$      I notice $\dfrac{dp_1}{dt}=\dfrac{-GMm(rv)}{r³}$ and $\dfrac{dp_2}{dt}=\dfrac{GMm(rv)}{r³}$ (since (rv)'s direction is flipped)     Therefore $r_1X\dfrac{-GMm(rv)}{r³}=-\dfrac{r_2XGMm(rv)}{r³}$     Defining $w=\dfrac{GMm}{r³}$:  $r_1X(rv)(-w)=r_2X(rv)(-w)$     Scalar products are commutative so the (-w)'s easily cancel giving  $r_1X(rv)=r_2X(rv)$  Which then the only solutions are if all components of $r_1$ are equal to all components of      $r_2$, however angular momentum should clearly be conserved in all positions, not just when the masses are ontop of each other.  Thanks again for any help.
65274
I'm trying to understand the energy-momentum tensor $T^{\mu\nu}$ but I'm confused about the units. My textbook says the components of $T^{\mu\nu}$ are $\mathrm{Jm^{-3}}$. Four-momentum is is given by$$P^{\mu}=\left(E/c,\mathbf{p}\right)=\left(E/c,p_{x},p_{y},p_{z}\right)$$  The $E/c$ component of $P^{\mu}$ has units $\mathrm{Jsm^{-1}}$. The definition of $T^{\mu\nu}$ is “the rate of flow of the $\mu$ component of four-momentum across a surface of constant $\nu$.” Using this definition, how do you get the rate of flow of the $E/c$ component of four-momentum across a surface of constant time (ie the $T^{00}$ component) to have the correct units of $\mathrm{Jm^{-3}}$? Surely you need to multiply $E/c$ by something with $\mathrm{s^{-1}m^{-2}}$ units, but what exactly?  Thank you
43513
,I got one confusion when reading Goldstein's Classical Mechanics (page 20, third edition). After getting the equation $$ \sum \left\{\left[\frac{\mathrm{d}}{\mathrm{d}t}{\left(\frac{\partial T}{\partial \dot{q_{j}}} \right)} - \frac{\partial T}{\partial q_{j} } \right] - Q_{j} \right\} \ \delta q_{j} = 0 $$ then it says that     Note that in a system of Cartesian coordinates the partial derivative of $T$ with respect to $q_{j}$ vanishes. Thus, speaking in the language of differential geometry, this term arises from the curvature of the coordinates $q_{j}$. In polar coordinates, e.g., it is in the partial derivative of $T$ with respect to an angle coordinate that the centripetal acceleration term appears.   My question is:  Is the above statement general, i.e., that the kinetic energy $T$ does not depend on the position. I wonder why velocity can't depend on the particle's position vector. I mean, why couldn't we have cases where $\vec{v} = \vec{v}(\vec r ,t)$, so that the kinetic energy depends on $q_{j}$ or $\vec{r}$?    I will use the example of the second answer in this post   the second answer,   suppose we have a velocity distribution of gas whose function is $\vec {v}=\vec {v}(x,y,z,t)$  then the partial derivative with respect to $x$ is  $$\frac{\partial T} {\partial x}=mv\frac{\partial v} {\partial x}$$  since $v$ depends on $x$, I think in this case. $\frac{\partial T}{\partial q_{j} }$ does not vanish.
15509
Recently, I have been looking at the Lagrange equations that describe particles of different spin. I know lagrange takes the form $L=$kinetic-potential, and it seems these equations do take this form. But I dont see how they correspond to the kinetic or potential. How are these equations derived? The Klein-Gordon equation spin 0 $$L=c^2\partial _\lambda \phi \partial^\lambda \phi ^* -\left(\frac{mc}{\hbar}\right)^2\phi \phi^*$$ Dirac Lagrangian spin 1/2 $$L=i\hbar c \tilde\psi \gamma ^\mu \partial _\mu \psi-mc^2\tilde\psi  \psi $$ The Proca equation spin 1 $$L=-\frac{1}{16 \pi}F^{\mu \nu} F_{\mu \nu} +\frac{1}{8 \pi} \left(\frac{mc}{\hbar}\right)^2A^\mu A_\mu$$ I suppose I mean to say, is there one equation that you can get all three of these equations from?
6496
I am stuck with an exercise in Sean Carroll's Spacetime and Geometry (Chapter 4, Exercise 3). The goal is to show that the continuity of the energy-momentum tensor, i.e. \begin{equation} \nabla_\mu T^{\mu\nu}=0\tag{1} \end{equation} is equivalent to the geodesic equation in the case of a free particle. The energy-momentum tensor of a free particle with mass $m$ moving along its worldline $x^\mu (\tau )$ is \begin{equation} T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.\tag{2} \end{equation} Taking the covariant derivative of this tensor gives $$\begin{align} \nabla_\mu T^{\mu\nu}=&amp;m\int d \tau \nabla_\mu\left[ \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right]\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}\cr &amp;+m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\nabla_\mu\left[\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}\right].\tag{3} \end{align}$$ The first covariant derivative of the right-hand side of the above equation reduces to an ordinary partial derivative, as the argument is a scalar. This allows us to apply partial integration to this term. The second covariant derivative has an argument that is not explicitly dependent on $y^\sigma$, so the covariant derivative can be written as a multiplication of this tensor with the appropriate Christoffel symbols. This finally leads us to $$\begin{align} &amp;-m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{d^2x^\nu}{d\tau^2} \cr &amp;+  m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\left[ \Gamma^\mu_{\mu\sigma}\frac{dx^\sigma}{d\tau}\frac{dx^\nu}{d\tau} + \Gamma^\nu_{\mu\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\sigma}{d\tau} \right].\tag{4} \end{align}$$ The continuity equation requires \begin{equation} -\frac{d^2x^\nu}{d\tau^2} + \Gamma^\mu_{\mu\sigma}\frac{dx^\sigma}{d\tau}\frac{dx^\nu}{d\tau} + \Gamma^\nu_{\mu\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\sigma}{d\tau}=0.\tag{5} \end{equation} This is the geodesic equation with an extra term, i.e. the term in the middle and with an incorrect sign for the first term. Can I get rid of this term in the middle by changing the parameter $\tau$ of the worldline? What about the incorrect sign? What did I do wrong?
16693
If there is a system, described by an Lagrangian $\mathcal{L}$ of the form   $$\mathcal{L} = T-V = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) + \frac{k}{r},\tag{1}$$   where $T$ is the kinetic energy and $V$ the potential energy, it is also possible to define the total energy $E$ of the system   $$ E = T + V =\frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) - \frac{k}{r}.\tag{2}$$  If the angular momentum $M$ is defined by $$ M = m r^{2} \dot{\phi},\tag{3}$$ then $E$ can be written as  $$ E = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}-\frac{k}{r}}_{V_\textrm{eff}\left(r\right)},\tag{4}$$  where the last two terms are written as new, "effective" potential $V_\textrm{eff}\left(r\right)$.    In addition, using the definition of $M$, the Lagrangian $\mathcal{L}$ can be written as  $$ \mathcal{L} = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}+\frac{k}{r}}_{-V_\textrm{eff}\left(r\right)},\tag{5}$$  where the sign of $V_\textrm{eff}\left(r\right)$ has been changed, since $\mathcal{L} = T-V$. But from this argument, it appears that there are two different possible ways to construct the same effective potential. This seems to me contradictory. Where is my mistake?